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 |
|---|---|---|---|---|---|---|---|---|---|
2. Given that the altitude to the hypotenuse of right $\triangle A B C$ is 4. Then the minimum value of the area of $\triangle A B C$ is $\qquad$ . | 2. 16 .
Let the lengths of the two legs of the right triangle $\triangle ABC$ be $a$ and $b$, and the length of the hypotenuse be $c$. From the area relationship, we have $ab = 4c$.
By the Pythagorean theorem, we know $c^2 = a^2 + b^2 \geq 2ab = 8c$.
Thus, $c \geq 8$. Therefore, $S_{\triangle ABC} = 2c \geq 16$.
When ... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,072 |
3. In rectangle $A B C D$, $A B=12, A D=3, E$ and $F$ are points on $A B$ and $D C$ respectively. Then the minimum length of the broken line $A F E C$ is $\qquad$ . | 3. 15 .
As shown in Figure 5, construct the symmetric points $A_{1}$ and $C_{1}$ of $A$ and $C$ with respect to $DC$ and $AB$, respectively. Connect $A_{1}C_{1}$ to intersect $AB$ and $DC$ at points $E_{1}$ and $F_{1}$, respectively, and connect $A_{1}F$ and $C_{1}E$. Draw a perpendicular from $A_{1}$ to the extension... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,073 |
4. If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$ are six different positive integers, taking values from $1, 2, 3, 4, 5, 6$. Let
$$
\begin{aligned}
S= & \left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\left|x_{3}-x_{4}\right|+ \\
& \left|x_{4}-x_{5}\right|+\left|x_{5}-x_{6}\right|+\left|x_{6}-x_{1}\right| .
\end{alig... | 4. 10 .
Since equation (1) is a cyclic expression about $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$, we can assume $x_{1}=6, x_{j}=1(j \neq 1)$. Then
$$
\begin{array}{l}
S \geqslant\left|\left(6-x_{2}\right)+\left(x_{2}-x_{3}\right)+\cdots+\left(x_{j-1}-1\right)\right|+ \\
\quad\left|\left(x_{j+1}-1\right)+\left(x_{j+... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,074 |
Three. (20 points) As shown in Figure 1, quadrilateral $ABCD$ is a cyclic quadrilateral, diagonals $AC$ and $BD$ intersect at point $E$, and the extensions of $DA$ and $CB$ meet at point $F$. Given that $\angle CAD=60^{\circ}$ and $DC=DE$, prove that:
$A$ is the circumcenter of $\triangle BEF$.
保留源文本的换行和格式，直接输出翻译结果。 | Three, because $D C=D E$, so,
$$
\angle A E B=\angle D E C=\angle D C E \text {. }
$$
Also, $A, B, C, D$ are concyclic
$$
\begin{array}{l}
\Rightarrow \angle E C D=\angle A C D=\angle A B D=\angle A B E \\
\Rightarrow \angle A E B=\angle A B E \Rightarrow A B=A E .
\end{array}
$$
Let $\angle A B E=\alpha$. Then
$$
\b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,075 |
Four. (25 points) Do there exist two positive integers $a, b$, satisfying $a \leqslant b$, such that the equation in $x$
$$
x^{2}-a b x+a+b=0
$$
has two integer solutions? If they exist, find all such positive integers $a, b$; if not, provide an explanation. | Therefore, $x_{1}, x_{2}$ are both positive integers, and
$$
x_{1} x_{2}-x_{1}-x_{2}=a+b-a b \text {, }
$$
i.e., $\square$
$$
\begin{array}{l}
\left(x_{1}-1\right)\left(x_{2}-1\right) \\
=-(a-1)(b-1)+2 .
\end{array}
$$
By $\left(x_{1}-1\right)\left(x_{2}-1\right) \geqslant 0$, we know
$$
0 \leqslant M \leqslant 2(M=(... | a=2, b=3 ; a=b=2 ; a=1, b=5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,076 |
Five. (25 points) Prove that the distances from any two different integer points to point $P(\sqrt{2}, \sqrt{3})$ are not equal (an integer point is a point where both the x-coordinate and y-coordinate are integers).
| Five, assume the conclusion does not hold.
Then there exist two distinct integer points $A(a, b)$, $B(c, d)$ ($a, b, c, d$ are integers) such that $A P = B P$.
$$
\begin{array}{l}
\text { Hence } A P^{2} = B P^{2} \\
\Rightarrow (a - \sqrt{2})^{2} + (b - \sqrt{3})^{2} \\
= (c - \sqrt{2})^{2} + (d - \sqrt{3})^{2} \\
\Ri... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,077 |
Example 6 Find the value of $\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}+15^{1998}}{7^{1998}+35^{1998}}}$. | $\begin{array}{l}\text { Solve the original expression }=\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}\left(1+5^{1998}\right)}{7^{1998}\left(1+5^{1998}\right)}} \\ =\left(\frac{7}{3}\right)^{999} \times\left(\frac{3}{7}\right)^{999}=\left(\frac{7}{3} \times \frac{3}{7}\right)^{999}=1 .\end{array}$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,078 |
1. Let $a, b, c$ be real numbers, and satisfy
$$
a+b+c=15, a^{2}+b^{2}+c^{2}=100 \text {. }
$$
Then the product of the maximum and minimum values of $a$ is $\qquad$ | $-1 . \frac{25}{3}$.
Since $100-a^{2}=b^{2}+c^{2}$
$$
\geqslant \frac{1}{2}(b+c)^{2}=\frac{1}{2}(15-a)^{2},
$$
Therefore, $3 a^{2}-30 a+25 \leqslant 0$.
Since the maximum and minimum values of $a$ are the two roots of the equation
$$
3 a^{2}-30 a+25=0
$$
and the maximum and minimum values of $a$ can be achieved, by V... | \frac{25}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,079 |
2. Given that $\triangle A B C$ is an equilateral triangle, $M$ and $N$ are the midpoints of sides $B C$ and $B A$ respectively, $O$ is the circumcenter of $\triangle B M N$, and there is a point $D$ on side $A C$ such that the area of $\triangle A O D$ is $\frac{1}{n}$ of the area of $\triangle A B C$ $\left(n>1, n \i... | 2. $\frac{3}{2 n-3}$.
Since the distance from point $O$ to $AC$ is equal to $\frac{2}{3}$ of the distance from point $B$ to $AC$, we have:
$$
\begin{array}{l}
\frac{S_{\triangle A O D}}{S_{\triangle A B C}}=\frac{\frac{1}{2} A D \cdot \frac{2}{3}}{\frac{1}{2} A C \cdot 1}=\frac{1}{n} \Rightarrow \frac{A D}{A C}=\frac{... | \frac{3}{2 n-3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,080 |
3. Person A writes down the positive integers $1, 2, \cdots$, 2009 on the blackboard, then turns away from the blackboard, and asks Person B to erase some of these numbers and then add the remainder of the sum of the erased numbers when divided by 7. After several such operations, only two numbers remain on the blackbo... | 3.5.
Since $1+2+\cdots+2009 \equiv 0(\bmod 7)$, therefore, the single digit $a$ satisfies $100+a \equiv 0(\bmod 7)$. Hence $a \equiv 5(\bmod 7)$. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,081 |
4. The number of solutions to the equation $\pi^{x-1} x^{2}+\pi^{x^{2}} x-\pi^{x^{2}}=x^{2}+x-1$ is $\qquad$ ( $\pi$ is the ratio of a circle's circumference to its diameter). | 4. 2 .
The original equation is transformed into
$$
x^{2}\left(\pi^{x-1}-1\right)+(x-1)\left(\pi^{x^{2}}-1\right)=0 \text {. }
$$
When $x \neq 0,1$, $x^{2}\left(\pi^{x-1}-1\right)$ and $(x-1)\left(\pi^{x^{2}}-1\right)$ have the same sign.
Therefore, $x=0,1$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,082 |
6. Given the quadratic function
$$
y=x^{2}-\frac{2 n+1}{n(n+2)} x+\frac{n+1}{n(n+2)^{2}}
$$
the length of the segment intercepted on the $x$-axis is $d_{n}$. Then
$$
\sum_{n=1}^{100} d_{n}=
$$
$\qquad$ | 6. $\frac{7625}{10302}$.
Given $d_{n}=\sqrt{\left[\frac{2 n+1}{n(n+2)}\right]^{2}-\frac{4(n+1)}{n(n+2)^{2}}}=\frac{1}{n(n+2)}$,
then $\sum_{n=1}^{100} d_{n}=\frac{1}{2} \sum_{n=1}^{100}\left(\frac{1}{n}-\frac{1}{n+2}\right)$ $=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{101}-\frac{1}{102}\right)=\frac{7625}{10302}$. | \frac{7625}{10302} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,084 |
7. Given the ellipse $\Gamma_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the hyperbola $\Gamma_{2}: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>b>0)$. The left and right foci of the ellipse $\Gamma_{1}$ are $F_{1}$ and $F_{2}$, respectively, and the left and right foci of the hyperbola $\Gamma_{2}$ are $F_{3}$ ... | 7. $\frac{3 \sqrt{2}}{4}$.
By Ceva's Theorem, we have $\frac{F_{3} F_{1}}{F_{1} F_{2}} \cdot \frac{F_{2} B}{B C} \cdot \frac{C A}{A F_{3}}=1$. By Menelaus' Theorem, we have $\frac{F_{3} F_{4}}{F_{4} F_{2}} \cdot \frac{F_{2} B}{B C} \cdot \frac{C A}{A F_{3}}=1$.
Thus, $\frac{F_{3} F_{1}}{F_{1} F_{2}}=\frac{F_{3} F_{4}}... | \frac{3 \sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,085 |
8. Given a regular 200-gon $A_{1} A_{2} \cdots A_{200}$, connect the diagonals $A_{i} A_{i+9}(i=1,2, \cdots, 200)$, where $A_{i+200}=$ $A_{i}(i=1,2, \cdots, 9)$. Then these 200 diagonals have $\qquad$ different intersection points inside the regular 200-gon. | 8. 1600.
For each diagonal $A_{i} A_{i+9}(i=1,2, \cdots, 200)$, $A_{i+1}, A_{i+2}, \cdots, A_{i+8}$ each draw two diagonals intersecting with $A_{i} A_{i+9}$. And each such intersection point is counted twice, so the number of different intersection points of these 200 diagonals inside the regular 200-gon does not exc... | 1600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,086 |
9. (14 points) Given that $C$ and $D$ are two points on the semicircle $\odot O$ with diameter $AB$, the chords $AD$ and $BC$ intersect at point $E$. Points $F$ and $G$ are on the extensions of $AC$ and $BD$ respectively, and satisfy $AF \cdot BG = AE \cdot BE$. If the orthocenters of $\triangle AEF$ and $\triangle BEG... | 9. (1) As shown in Figure 1, given $\angle C A D = \angle C B D$,
$$
\frac{A F}{A E} = \frac{B E}{B G} \text{, }
$$
we get $\triangle A E F \sim \triangle B G E$.
Since
$$
B C \perp A F, A D \perp B G,
$$
we have $\angle C E F + \angle B E G = \angle C E F + \angle C F E = 90^{\circ}$.
Thus, $\angle F E G = 90^{\circ}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,087 |
10. (15 points) Given a sequence of positive numbers $\left\{a_{n}\right\}(n \geqslant 0)$ that satisfies $a_{n}=\frac{a_{n-1}}{m a_{n-2}}(n=2,3, \cdots, m$ is a real parameter $)$. If $a_{2009}=\frac{a_{0}}{a_{1}}$, find the value of $m$. | 10. Since $a_{2}=\frac{a_{1}}{m a_{0}}, a_{3}=\frac{\frac{a_{1}}{m a_{0}}}{m a_{1}}=\frac{1}{m^{2} a_{0}}$,
$$
\begin{array}{l}
a_{4}=\frac{\frac{1}{m^{2} a_{0}}}{m \frac{a_{1}}{m a_{0}}}=\frac{1}{m^{2} a_{1}}, a_{5}=\frac{\frac{1}{m^{2} a_{1}}}{m \frac{1}{m^{2} a_{0}}}=\frac{a_{0}}{m a_{1}}, \\
a_{6}=\frac{\frac{a_{0}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,088 |
Example 7 Let $x=2 \frac{1}{5}, y=-\frac{5}{11}, z=-2 \frac{1}{5}$. Then $x^{2}+x z+2 y z+3 x+3 z+4 x y+5=$ | Solve: From the given, we have
$$
x y=-1, x+z=0, y z=1 \text {. }
$$
Therefore, the expression to be found is
$$
\begin{array}{l}
=x(x+z)+2 y z+3(x+z)+4 x y+5 \\
=2 \frac{1}{5} \times 0+2 \times 1+3 \times 0+4 \times(-1)+5=3 .
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,089 |
11. (15 points) Divide the sides of an equilateral $\triangle ABC$ with side length 3 into three equal parts, and draw lines parallel to the other two sides through each division point. The 10 points where the sides of $\triangle ABC$ and these parallel lines intersect are called grid points. If $n$ grid points are cho... | 11. $n_{\min }=5$.
Let the two equal division points on side $AB$ from point $A$ to $B$ be $D$ and $E$, the two equal division points on side $BC$ from point $B$ to $C$ be $F$ and $G$, and the two equal division points on side $CA$ from point $C$ to $A$ be $H$ and $I$, with the central grid point being $K$.
If the mi... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,090 |
1. Given the set $A=\{x \in \mathbf{R}|| x-2 \mid \leqslant 1\}$, set $B=\left\{x \in \mathbf{R} \left\lvert\, \frac{x-5}{2-x}>0\right.\right\}$. Then $A \cap B=$ | - 1. $\{x \mid 2<x \leqslant 3\}$.
It is easy to see that $A=\{x \mid 2 \leqslant x \leqslant 3\}, B=\{x \mid 2<x<5\}$. | \{x \mid 2<x \leqslant 3\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,091 |
3. Let a tangent line of the circle $x^{2}+y^{2}=1$ intersect the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Then the minimum value of $|AB|$ is $\qquad$ . | 3. 2 .
By symmetry, without loss of generality, assume the point of tangency is
$$
P(\cos \theta, \sin \theta)\left(0<\theta<\frac{\pi}{2}\right) \text {. }
$$
Then $|P A|=\tan \theta,|P B|=\cot \theta$.
Thus, $|A B|=|P A|+|P B|=\tan \theta+\cot \theta \geqslant 2$.
Equality holds if and only if $\tan \theta=\cot \th... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,093 |
4. Given the function
$$
f(x)=\left\{\begin{array}{ll}
2^{2-x}, & x<2 ; \\
\log _{3}(x+1), & x \geqslant 2 .
\end{array}\right.
$$
If the equation $f(x)=m$ has two distinct real roots, then the range of the real number $m$ is $\qquad$ (express in interval form). | 4. $(1,+\infty)$.
As shown in Figure 6, the graphs of the functions $y=f(x)$ and $y=m$ are plotted in the same Cartesian coordinate system. It is easy to see that when $m>1$, the equation $f(x)=m$ has two distinct real roots. | (1,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,094 |
5. Let $f(x)$ be an odd function defined on $\mathbf{R}$, $f(1)=2$, when $x>0$, $f(x)$ is an increasing function, and for any $x, y \in \mathbf{R}$, we have $f(x+y)=f(x)+f(y)$. Then the maximum value of the function $f(x)$ on the interval $[-3,-2]$ is | 5. -4 .
Since $f(x)$ is an odd function and is increasing on $(0,+\infty)$, therefore, $f(x)$ is also increasing on $(-\infty, 0)$. Thus, $f(-3) \leqslant f(x) \leqslant f(-2)$.
$$
\begin{array}{l}
\text { Also, } f(2)=f(1)+f(1)=4, \text { then } \\
f(-2)=-f(2)=-4 .
\end{array}
$$
Therefore, the maximum value of the ... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,095 |
6. For $n \in \mathbf{N}_{+}$, if $2^{n} n+1$ is a multiple of 3, then the set of remainders when $n$ is divided by 6 is | 6. $\{1,2\}$.
$$
\begin{array}{l}
2^{n} n+1=(3-1)^{n} n+1 \\
=3 M+(-1)^{n} n+1(M \in \mathbf{N}) .
\end{array}
$$
If $3 \mid\left(2^{n} n+1\right)$, then $3 \mid \left[(-1)^{n} n+1\right]$.
When $n=6 k\left(k \in \mathbf{N}_{+}\right)$,
$$
(-1)^{n} n+1=6 k+1 \equiv 1(\bmod 3) \text {; }
$$
When $n=6 k+1(k \in \mathbf... | \{1,2\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,096 |
7. As shown in Figure $2, A B$ is the diameter of the semicircle $\odot O$, and $C, D$ are two moving points on the semicircle, with $C D / / A B$. If the radius of the semicircle $\odot O$ is 1, then the maximum value of the perimeter of trapezoid $A B C D$ is $\qquad$ . | 7.5.
In Figure 2, connect $A C$, and draw $C H \perp A B$ at point $H$. Let $\angle A B C=\theta\left(0<\theta<\frac{\pi}{2}\right)$. Then $A D=B C=A B \cos \theta=2 \cos \theta$.
Thus, $B H=B C \cos \theta=2 \cos ^{2} \theta$.
Therefore, $C D=A B-2 B H=2-4 \cos ^{2} \theta$.
Hence, the perimeter of trapezoid $A B C D... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,097 |
8. As shown in Figure 3, in $\triangle A B C$, $A B=3, A C=5$. If $O$ is the circumcenter of $\triangle A B C$, then the value of $\overrightarrow{A O} \cdot \overrightarrow{B C}$ is | 8. 8 .
Let $D$ be the midpoint of side $B C$, and connect $O D, A D$. Then $O D \perp B C$. Therefore,
$$
\begin{array}{l}
\overrightarrow{A O} \cdot \overrightarrow{B C}=(\overrightarrow{A D}+\overrightarrow{D O}) \cdot \overrightarrow{B C} \\
=\overrightarrow{A D} \cdot \overrightarrow{B C}+\overrightarrow{D O} \cdo... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,098 |
9. A hemispherical container with a base contains three small balls that are pairwise externally tangent. If the radii of these three small balls are all 1, and each small ball is tangent to the base and the spherical surface of the hemisphere, then the radius of the hemisphere $R=$
| 9. $\frac{3+\sqrt{21}}{3}$.
The centers of three small spheres $O_{1}, O_{2}, O_{3}$ form an equilateral triangle with a side length of 2, then the radius of the circumcircle is $\frac{2 \sqrt{3}}{3}$.
Let the center of the hemisphere be $O$, and the small sphere $O_{1}$ touches the bottom of the hemisphere at point ... | \frac{3+\sqrt{21}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,099 |
Example 8 Find the sum:
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{60}\right)+\left(\frac{2}{3}+\frac{2}{4}+\cdots+\frac{2}{60}\right)+ \\
\left(\frac{3}{4}+\frac{3}{5}+\cdots+\frac{3}{60}\right)+\cdots+\left(\frac{58}{59}+\frac{59}{60}\right)
\end{array}
$$ | $$
\begin{aligned}
= & \frac{1}{2}+\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)+\cdots+ \\
& \left(\frac{1}{60}+\frac{2}{60}+\cdots+\frac{59}{60}\right) \\
= & \frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\cdots+\frac{59}{2} \\
= & \frac{1}{2}(1+2+\cdots+59) \\
= & \frac{1}{2} \times \fr... | 885 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,100 |
10. Divide a line segment of length $a$ into three segments; the probability that these three segments can form a triangle is $\qquad$ | 10. $\frac{1}{4}$.
Let the lengths of the three segments be $x, y, a-(x+y)$. Then
$$
\left\{\begin{array}{l}
0 < x < a, \\
0 < y < a, \\
0 < a - (x + y), \\
x + a - (x + y) > y, \\
y + a - (x + y) > x
\end{array} \Rightarrow \left\{\begin{array}{l}
x + y > \frac{a}{2}, \\
y < \frac{a}{2}, \\
x < \frac{a}{2}.
\end{arra... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,101 |
11. Let $0<\alpha<\pi, \pi<\beta<2 \pi$. If for any $x \in \mathbf{R}$, we have
$$
\cos (x+\alpha)+\sin (x+\beta)+\sqrt{2} \cos x=0
$$
always holds, find the value of $\alpha \beta$. | Given the equation can be transformed into
$$
(\cos \alpha+\sin \beta+\sqrt{2}) \cos x+(\cos \beta \cdots \sin \alpha) \sin x=0 \text {. }
$$
The necessary and sufficient condition for the above equation to hold for any $x \in \mathbf{R}$ is
$$
\begin{array}{l}
\left\{\begin{array}{l}
\cos \alpha+\sin \beta+\sqrt{2}=0... | \frac{21\pi^2}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,102 |
12. As shown in Figure 4, let points $A(-\sqrt{5}, 0)$ and $B(\sqrt{5}, 0)$, and the incenter of $\triangle ABC$ moves on the line $x=2$.
(1) Find the equation of the locus of point $C$;
(2) If two rays are drawn from point
$M(2,0)$, intersecting the locus found in (1)
at points
$P$ and $Q$, and $\overrightarrow{M P} \... | 12. (1) Let the incircle of $\triangle ABC$ touch side $AB$ at point $D$. Then
$$
\begin{array}{l}
|CA| - |CB| = |AD| - |BD| \\
= (\sqrt{5} + 2) - (\sqrt{5} - 2) = 4
\end{array}
$$
(2) Let $l_{PQ}: x = my + a (a > 2)$. Substituting into $\frac{x^2}{4} - y^2 = 1$, we get
$$
\left(m^2 - 4\right) y^2 + 2am y + a^2 - 4 = 0... | \left(\frac{10}{3}, 0\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,103 |
13. Given the function $f(x)=\frac{16 x+7}{4 x+4}$, the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}>0, b_{1}>0, \\
a_{n}=f\left(a_{n-1}\right), b_{n}=f\left(b_{n-1}\right)(n=2,3, \cdots) .
\end{array}
$$
(1) Find the range of $a_{1}$ such that for any positive integer $... | 13. (1) Notice that
$$
f(x)=\frac{16(x+1)-9}{4(x+1)}=4-\frac{9}{4} \cdot \frac{1}{x+1} \text {. }
$$
Then $a_{n+1}-a_{n}$
$$
\begin{array}{l}
=\left(4-\frac{9}{4} \cdot \frac{1}{a_{n}+1}\right)-\left(4-\frac{9}{4} \cdot \frac{1}{a_{n-1}+1}\right) \\
=\frac{9}{4} \cdot \frac{a_{n}-a_{n-1}}{\left(a_{n}+1\right)\left(a_{... | 0<b_{n}-a_{n} \leqslant \frac{1}{8^{n-1}}(n=1,2, \cdots) | Algebra | proof | Yes | Yes | cn_contest | false | 722,104 |
One. (40 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=4, a_{n+1} a_{n}+6 a_{n+1}-4 a_{n}-8=0$. Let $b_{n}=\frac{6}{a_{n}-2}\left(n \in \mathbf{N}_{+}\right)$. Find:
(1) The general term formula of the sequence $\left\{b_{n}\right\}$;
(2) The sum of the first $n$ terms $S_{n}$ of the sequence $\lef... | (1) From $b_{n}=\frac{6}{a_{n}-2} \Rightarrow a_{n}=\frac{6}{b_{n}}+2$.
Substituting into $a_{n+1} a_{n}+6 a_{n+1}-4 a_{n}-8=0$, we get $b_{n+1}=4 b_{n}+3$.
Then, $b_{n+1}+1=4\left(b_{n}+1\right)$, where $b_{1}+1=4$.
Thus, $b_{n}+1=4 \times 4^{n-1}=4^{n} \Rightarrow b_{n}=4^{n}-1$.
(2) From $b_{n}=\frac{6}{a_{n}-2} \Ri... | \frac{2}{3} \times 4^{n+1}+4 n-\frac{8}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,105 |
II. (40 points)
As shown in Figure $5, P A$ and $P B$
are two tangents to
$\odot O$, touching at
points $A$ and $B$, respectively. A line through point $P$
intersects $\odot O$ at points $C$ and
$D$, and intersects chord $A B$ at point
$Q$. Prove:
$$
P Q^{2}=P C \cdot P D - Q C \cdot Q D .
$$ | II. As shown in Figure 9, connect $O A$, $O B$, and $O P$. Let $O P$ intersect $A B$ at point $H$, and let the midpoint of $C D$ be $M$. Connect $O M$. Then $O M \perp C D$, $O H \perp A B$.
Therefore, $O$, $H$, $Q$, and $M$ are concyclic.
Thus, $P Q \cdot P M = P H \cdot P O$.
Also, in the right triangles $\triangle P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,106 |
$$
\begin{array}{l}
\text { Three. (50 points) Let } \\
(x+1)^{p}(x-3)^{q} \\
=x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\cdots+a_{n-1} x+a_{n} \\
\left(p, q \in \mathbf{N}_{+}\right) \text {. } \\
\end{array}
$$
(1) If $a_{1}=a_{2}$, prove that $3 n$ is a perfect square;
(2) Prove that there exist infinitely many pairs of pos... | Three, (1) It is known that $p+q=n$.
$$
\begin{array}{l}
\text { Since }(x+1)^{p}(x-3)^{q} \\
= {\left[x^{n}+p x^{p-1}+\frac{p(p-1)}{2} x^{p-2}+\cdots\right], } \\
{\left[x^{q}-3 q x^{q-1}+\frac{9 q(q-1)}{2} x^{q-2}+\cdots\right] } \\
= x^{n}+(p-3 q) x^{n-1}+\left[\frac{p(p-1)}{2}+\right. \\
\left.\frac{9 q(q-1)}{2}-3... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,107 |
(1) For any $x>0, y>0$, we have
$$
\frac{1}{1+x} \geqslant \frac{1}{1+y}-\frac{1}{(1+y)^{2}}(x-y) \text {; }
$$
(2) $\sum_{k=0}^{n} \mathrm{C}_{n}^{k} \frac{3^{k}}{3^{k}+1} \geqslant \frac{3^{n} \times 2^{n}}{3^{n}+2^{n}}$. | (1) From
$$
\begin{array}{l}
\frac{1}{1+y} \times \frac{x-y}{(1+y)^{2}} \\
=\frac{1}{1+y}-\frac{(1+x)-(1+y)}{(1+y)^{2}} \\
=-\frac{1+x}{(1+y)^{2}}+\frac{2}{1+y} \\
=-(1+x)\left(\frac{1}{1+y}-\frac{1}{1+x}\right)^{2}+\frac{1}{1+x} \\
\leqslant \frac{1}{1+x}
\end{array}
$$
Equality holds if and only if $x=y$. Therefore,... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,108 |
1. Given the set
$$
M=\left\{x|5-| 2 x-3 \mid \in \mathbf{N}_{+}\right\} \text {. }
$$
Then the number of all non-empty proper subsets of $M$ is ( ).
(A) 254
(B) 255
(C) 510
(D) 511 | - 1. C.
Given $5-|2 x-3| \in \mathbf{N}_{+}$, i.e., $|2 x-3|=0,1, 2,3,4$, we have
$$
M=\left\{-\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3, \frac{7}{2}\right\} .
$$
Since $M$ contains 9 elements, the number of all non-empty proper subsets of $M$ is $2^{9}-2=510$. | 510 | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,109 |
2. The points $(x, y)$ in the plane that satisfy the constraints $\left\{\begin{array}{l}x \geqslant 2, \\ x+y \leqslant 0, \\ x-y-10 \leqslant 0\end{array}\right.$ form a region $D$, and the region $D$ symmetric to the line $y=2 x$ is region $E$. Then the distance between the closest two points in regions $D$ and $E$ ... | 2. B.
The intersection points of three lines are $A(2,-2)$, $B(2,-8)$, and $C(5,-5)$. The region $D$ is $\triangle ABC$. Point $A$ is closest to the line $y=2x$, with a distance of $\frac{|12 \times 2 - 1 \times (-2)|}{\sqrt{5}} = \frac{6 \sqrt{5}}{5}$. Therefore, the distance between the closest points in regions $D$... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,110 |
3. Given that the three sides $a, b, c$ of $\triangle A B C$ form a geometric sequence, and the angles opposite to $a, b, c$ are $\angle A, \angle B, \angle C$ respectively. Then the range of $\sin B+\cos B$ is ( ).
(A) $\left(1,1+\frac{\sqrt{3}}{2}\right]$
(B) $\left[\frac{1}{2}, 1+\frac{\sqrt{3}}{2}\right]$
(C) $(1, ... | 3. C.
$$
\begin{array}{l}
\text { Given } a c=b^{2}=a^{2}+c^{2}-2 a c \cos B \\
\quad \geqslant 2 a c-2 a c \cos B \\
\Rightarrow \cos B \geqslant \frac{1}{2} \Rightarrow 0<\angle B \leqslant \frac{\pi}{3} .
\end{array}
$$
Therefore, $\sin B+\cos B=\sqrt{2} \sin \left(B+\frac{\pi}{4}\right) \in(1, \sqrt{2}]$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,112 |
4. Let $M$ be the set of midpoints of all edges of a cube. Then the number of planes that pass through and only through three points in $M$ is ( ).
(A) 56
(B) 81
(C) 136
(D) 145 | 4. A.
Through three points in $M$, there are $\mathrm{C}_{12}^{3}=$ 220 planes (including duplicates), among which, planes passing through and only through four points in $M$ total $C_{4}^{3} \times(9+12)=84$; planes passing through six points total $\mathrm{C}_{6}^{3} \times 4=80$. Therefore, the number of planes pas... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,113 |
5. Let $a_{n}=2^{n}, b_{n}=n \ (n=1,2, \cdots), A_{n} 、 B_{n}$ be the sums of the first $n$ terms of the sequences $\left\{a_{n}\right\} 、\left\{b_{n}\right\}$, respectively. Define
$$
c_{n}=a_{n} B_{n}+b_{n} A_{n}-a_{n} b_{n} \text {. }
$$
Then the sum of the first 10 terms of the sequence $\left\{c_{n}\right\}$ is (... | 5. D.
When $n \geqslant 2$,
$$
\begin{aligned}
c_{n}= & \left(A_{n}-A_{n-1}\right) B_{n}+\left(B_{n}-B_{n-1}\right) A_{n}- \\
& \left(A_{n}-A_{n-1}\right)\left(B_{n}-B_{n-1}\right) \\
\Rightarrow & c_{n}=A_{n} B_{n}-A_{n-1} B_{n-1} .
\end{aligned}
$$
Also, $c_{1}=a_{1} b_{1}=A_{1} B_{1}$, then the sum of the first $n... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,114 |
7. For a regular tetrahedron \(ABCD\), the center of the circumscribed sphere is \(O\), and \(E\) is the midpoint of edge \(BC\). The dihedral angle \(A-BO-E\) is \(\qquad\). | II. $7 . \frac{2 \pi}{3}$.
Since point $E$ is on side $B C$, the dihedral angle $A-B O-E$ is equal to the dihedral angle $A-B O-C$. Also, $O$ is the center of the circumscribed sphere of the regular tetrahedron $A B C D$, so $B O \perp$ plane $A C D$. Let the projection of point $B$ on plane $A C D$ be $O^{\prime}$. Th... | \frac{2 \pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,116 |
8. The product of the maximum and minimum values of the function $f(x)=\frac{x-x^{3}}{1+2 x^{2}+x^{4}}$ is . $\qquad$ | 8. $-\frac{1}{16}$.
Let $x=\tan \theta\left(\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right)$. Then
$$
\begin{array}{c}
f(x)=\frac{\tan \theta\left(1-\tan ^{2} \theta\right)}{\left(1+\tan ^{2} \theta\right)^{2}} \\
=\sin \theta \cdot \cos \theta \cdot \cos 2 \theta \\
=\frac{1}{4} \sin 4 \theta \in\left[-\f... | -\frac{1}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,117 |
9. A wire of length 1 is cut into three pieces. Then the probability that these three pieces can exactly form a triangle is | 9. $\frac{1}{4}$.
Let the three segments be of lengths $x, y, 1-x-y$. Then the total sample space is $x>0, y>0, x+y<1$. The area of the region where the triangle can be formed is $x+y>\frac{1}{2}$, with an area of $\frac{1}{8}$.
Thus, the required probability is $\frac{1}{4}$. | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,118 |
10. Let $a(00$, then the range of values for $t$ is . $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 10. $t>\frac{1}{\sqrt{a}}$ or $\sqrt{a}<t<1$; when $x>\frac{1}{2}$, $f(x)>0$.
So, $f(x)>0 \Leftrightarrow 0<x<\frac{1}{2}$ or $x>1$
$\Leftrightarrow \log _{a} t>\log _{a} \frac{1}{\sqrt{a}}$ or $\log _{a} \sqrt{a}<\log _{a} t<\log _{a} 1$.
$\Leftrightarrow t>\frac{1}{\sqrt{a}}$ or $\sqrt{a}<t<1$. | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,119 |
11. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}=\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}}-1 \text {. }
$$
Then $a_{1}+a_{2}+\cdots+a_{2000}=$ | 11. $\frac{2009}{2010}$.
Notice that
$$
\begin{array}{l}
1+\frac{1}{k^{2}}+\frac{1}{(k+1)^{2}}=\frac{k^{4}+2 k^{3}+3 k^{2}+2 k+1}{k^{2}(k+1)^{2}} \\
=\left[\frac{k^{2}+k+1}{k(k+1)}\right]^{2} .
\end{array}
$$
$$
\begin{array}{l}
\text { Then } a_{k}=\sqrt{1+\frac{1}{k^{2}}+\frac{1}{(k+1)^{2}}}-1 \\
=\frac{k^{2}+k+1}{k... | \frac{2009}{2010} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,120 |
12. Given that $P$ is a moving point on the hyperbola $\frac{x^{2}}{8}-\frac{y^{2}}{4}=1$, $F_{1}$ and $F_{2}$ are its left and right foci, and $O$ is the origin. Then the range of $\frac{\left|P F_{1}\right|+\left|P F_{2}\right|}{|O P|}$ is $\qquad$. | 12. $(2, \sqrt{6}]$.
By the symmetry of the hyperbola, we can assume $P(x, y)$ $(x \geqslant 2 \sqrt{2})$.
By the focal radius formula, we get
$$
\begin{array}{l}
\left|P F_{1}\right|=\frac{\sqrt{6}}{2} x+2 \sqrt{2}, \\
\left|P F_{2}\right|=\frac{\sqrt{6}}{2} x-2 \sqrt{2}, \\
|O P|=\sqrt{x^{2}+y^{2}}=\sqrt{x^{2}+\frac... | (2, \sqrt{6}] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,121 |
Example 10 Calculate:
$$
\frac{2009^{2}-2008^{2}}{19492009^{2}-19492008 \times 19492010+2 \times 2008} .
$$ | Let $19492009=a, 2008=b$. Then the original expression $=\frac{(b+1)^{2}-b^{2}}{a^{2}-(a-1)(a+1)+2 b}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,122 |
14. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with eccentricity $e=\frac{\sqrt{6}}{3}$, and the distance from the origin to the line passing through points $A(0,-b)$ and $B(a, 0)$ is $\frac{\sqrt{3}}{2}$.
(1) Find the equation of the ellipse;
(2) Given a fixed point $E(-1,0)$, the line $y=k x... | 14. (1) Let $l_{A B}: b x-a y-a b=0$.
According to the problem, we have
$$
\frac{c}{a}=\frac{\sqrt{6}}{3}, \frac{a b}{\sqrt{a^{2}+b^{2}}}=\frac{\sqrt{3}}{2} .
$$
Solving these, we get $a=\sqrt{3}, b=1$.
Thus, the equation of the ellipse is $\frac{x^{2}}{3}+y^{2}=1$.
(2) Substituting $y=k x+t$ into the ellipse equatio... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,124 |
15. As shown in Figure 1, in $\triangle ABC$, $AB > AC$, and $AE$ is a tangent to the circumcircle of $\triangle ABC$,
$D$ is a point on $AB$, and
$AD = AC = AE$. Prove:
that line $DE$ passes through the incenter of $\triangle ABC$. | 15. As shown in Figure 3, the angle bisector $CI$ of $\angle ACB$ intersects $DE$ at point $I$, and $AI$, $CE$ are connected.
Since $AE$ is the tangent to the circumcircle of $\triangle ABC$, we have
$$
\begin{array}{l}
\angle ACB = 180^{\circ} - \angle DAE. \\
\text{Also, } AD = AE, \text{ then} \\
180^{\circ} - \ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,125 |
16. Given the sequence $\left\{a_{n}\right\}$, the adjacent terms $a_{2 k-1}$ and $a_{2 k}$ are the roots of the equation
$$
x^{2}-\left(3 k+2^{k}\right) x+3 k \cdot 2^{k}=0
$$
(1) Find the sum of the first $2 n$ terms of the sequence $\left\{a_{n}\right\}$, $S_{2 n}$;
$$
\begin{array}{l}
\text { (2) Let } f(n)=\frac{... | 16. (1) Notice the equation
$$
x^{2}-\left(3 k+2^{k}\right) x+3 k \cdot 2^{k}=0
$$
has two roots $x_{1}=3 k, x_{2}=2^{k}$. Therefore,
$$
\begin{array}{l}
S_{2 n}=\sum_{k=1}^{n}\left(a_{2 k-1}+a_{2 k}\right)=\sum_{k=1}^{n}\left(3 k+2^{k}\right) \\
=\frac{3 n^{2}+3 n}{2}+2^{n+1}-2 .
\end{array}
$$
(2) $T_{1}=\frac{1}{a_... | \frac{1}{6} \leqslant T_{n} \leqslant \frac{5}{24} | Algebra | proof | Yes | Yes | cn_contest | false | 722,126 |
1. Let $a, b, c, p$ be real numbers, not all equal, and
$$
a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}=p \text {. }
$$
Then $a^{2} b^{2} c^{2}=(\quad)$.
(A) $-p$
(B) $p$
(C) $p^{2}$
(D) $\sqrt{p}$ | - 1. C.
Given $p=c+\frac{p-\frac{1}{c}}{\left(p-\frac{1}{c}\right) p-1}$. Simplifying, we get $\left(p^{2}-1\right)\left(p-c-\frac{1}{c}\right)=0$. When $p=c+\frac{1}{c}$, we have $a=b=c$, which is a contradiction.
Therefore, $p^{2}=1$.
From the given information,
$$
\begin{array}{l}
a+\frac{1}{b}=b+\frac{1}{c} \Right... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,127 |
2. As shown in Figure 1, in the right triangle $\triangle ABC (\angle C=90^{\circ})$, there are three squares with side lengths $a$, $b$, and $c$ respectively. Then the relationship satisfied by $a$, $b$, and $c$ is ( ).
(A) $a+c=b$
(B) $a c=b^{2}$
(C) $a^{2}+c^{2}=b^{2}$
(D) $\frac{1}{b}+\frac{1}{c}=\frac{1}{a}$ | 2. A.
Obviously, Rt $\triangle D E F \backsim \mathrm{Rt} \triangle H G M$. Then
$$
\begin{array}{l}
\frac{D E}{H G}=\frac{E F}{G M} \Rightarrow \frac{a}{b-c}=\frac{b-a}{c} \\
\Rightarrow a c=(b-a)(b-c) \\
\Rightarrow b(a+c)=b^{2} .
\end{array}
$$
Since $b \neq 0$, thus, $a+c=b$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,128 |
3. Given that $\triangle A B C$ is an equilateral triangle, its height is
4. If the distances from point $P$ to sides $A B, A C$ are $1, 2$ respectively, then the distance from point $P$ to side $B C$ is ( ).
(A) 1
(B) 1 or 3 or 5
(C) 1 or 3
(D) 1 or 3 or 5 or 7 | 3. D.
As shown in Figure 5, let point $P$ be inside $\triangle ABC$, and $AH$ be the altitude of $\triangle ABC$. The perpendiculars from $P$ to the three sides of $\triangle ABC$ are $PD$, $PE$, and $PF$, respectively. Let the side length of $\triangle ABC$ be $a$.
From the area relationship, we have
$$
\begin{array}... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,129 |
4. Given an acute triangle $\triangle A B C$ with the three interior angles $\angle A$, $\angle B$, $\angle C$ and the opposite sides of lengths $a$, $b$, $c$ respectively, and a point $O$ on the plane of $\triangle A B C$, and the distances from $O$ to the sides $B C$, $A C$, $A B$ are $x$, $y$, $z$ respectively. The ... | 4. D.
(1) As shown in Figure 7 (a), when $O$ is the centroid of $\triangle ABC$, from $ax = by = cz$, we get
$$
x: y: z = \frac{1}{a}: \frac{1}{b}: \frac{1}{c} = bc: ca: ab.
$$
(2) As shown in Figure 7 (b), when $O$ is the circumcenter of $\triangle ABC$, let the circumradius of $\triangle ABC$ be $R$. Then
$$
\angle B... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,130 |
5. Let $a$, $b$, $c$ ($a<5$) be positive real numbers, and satisfy $a^{2}-a-2 b-2 c=0$ and $a+2 b-2 c+3=0$. Then the size relationship between $a$, $b$, and $c$ is ( ).
(A) $a<b<c$
(B) $b<c<a$
(C) $b<a<c$
(D) $c<a<b$ | 5. C.
Given $\left\{\begin{array}{l}a^{2}-a=2 b+2 c, \\ a+3=2 c-2 b .\end{array}\right.$
then $4 c=a^{2}+3$,
$$
4 b=a^{2}-2 a-3=(a-1)^{2}-4>0 \Rightarrow a>3 \text {. }
$$
Also, $a>0 \\
\Rightarrow c>a, \\
4(b-a)=a^{2}-6 a-3=(a-3)^{2}-12b .
\end{array}
$$
Therefore, $c>a>b$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,131 |
6. As shown in Figure 2, modules (1)-(5) are each composed of 4 small cubes with an edge length of 1, and module (6) is composed of 15 small cubes with an edge length of 1. Now, three modules are selected from modules (1)-(5) to be placed on module (6) to form a large cube with an edge length of 3. The number of ways t... | 6. B.
From modules (1)-(5) in Figure 2, selecting three modules to complete the task is not unique.
In fact, from the position of the gap in module (6), one can first choose module (5) and embed it in the gap of module (6), at which point the accumulated shape is as shown in Figure 8. Next, select two modules from mo... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 722,132 |
Example 1 Given $a^{2}+b^{2}=6 a b$, and $a>b>0$. Then $\frac{a+b}{a-b}=$ $\qquad$
$(2001$, Beijing Middle School Mathematics Competition (Grade 8)) | Let $a=x+y, b=x-y$. Substituting into the given equation and simplifying, we get $4 x^{2}=8 y^{2}$.
Since $a>b>0$, it follows that $x>y>0$.
Thus, $\frac{x^{2}}{y^{2}}=2 \Rightarrow \frac{x}{y}=\sqrt{2}$.
Notice that $a+b=2 x, a-b=2 y$.
Therefore, $\frac{a+b}{a-b}=\frac{2 x}{2 y}=\frac{x}{y}=\sqrt{2}$. | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,133 |
Example 2 Calculate the value of $\sqrt{14+6 \sqrt{5}}-\sqrt{14-6 \sqrt{5}}$ is ( ).
(A) 1
(B) $\sqrt{5}$
(C) $2 \sqrt{5}$
(D) 5
(2000, National Junior High School Mathematics League) | Let $\sqrt{14+6 \sqrt{5}}=a+b$,
$$
\sqrt{14-6 \sqrt{5}}=a-b .
$$
(1) $\times$ (2), (1) $)^{2}+(2)^{2}$ respectively yield
$$
\begin{array}{l}
a^{2}-b^{2}=4 \text {, } \\
a^{2}+b^{2}=14 \text {. } \\
\end{array}
$$
(4) - (3) gives $b^{2}=5$.
Since $b>0$, thus, $b=\sqrt{5}$.
Therefore, the desired expression $=(a+b)-(a-... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,134 |
1. Let $a<b<0$, and $a^{2}+b^{2}=4 a b$. Then the value of $\frac{a+b}{a-b}$ is ( ).
(A) $\sqrt{3}$
(B) $\sqrt{6}$
(C) 2
(D) 3
(2002, National Junior High School Mathematics Competition) | Prompt: Example 1. Answer: (A).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,135 |
9. (16 points) Let the ellipse $C: \frac{(x-1)^{2}}{16}+\frac{(y-1)^{2}}{9}=1$, and the line $l: y=a x+b$. If two tangents are drawn from any two points $M, N$ on the line $l$ to the ellipse $C$, and the lines connecting the points of tangency are $m, n$, then $m \parallel n$. Try to find the locus of the point $(a, b)... | 9. Let \( M\left(x_{0}, y_{0}\right) \) be any point on the line \( l: y=a x+b \).
From the ellipse \( C: 9 x^{2}+16 y^{2}-18 x-32 y-119=0 \), the chord of contact corresponding to \( M \) is
\[
9 x_{0} x+16 y_{0} y-9\left(x+x_{0}\right)-16\left(y+y_{0}\right)-119=0 \text {. }
\]
Substituting \( y_{0}=a x_{0}+b \) in... | a+b=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,136 |
10. (20 points) As shown in Figure 1, rectangle $O A B C$ is a billiard table, with $O A=12, O C=6$. A ball is hit from point $O$ and can bounce off the four edges of the table infinitely. It is known that the ball passes through the center of rectangle $O A B C$, $D(6,3)$.
Figure 1
(1) Find the number of all integer ... | 10. Translate the rectangle $O A B C$ and points $D, E$ as a whole to the right and upward, to get a series of rectangles.
Let the images of $D, E$ in the $i(i=0,1, \cdots)$-th column and $j(j=0,1, \cdots)$-th row of rectangles be $D_{i j}, E_{i j}$. Then
$$
\begin{array}{l}
D_{i j}=(12 i+6,6 j+3) ; \\
E_{2 t, 2 t}=(2... | 11 \text{ integer points, } f(E)_{\max }=2 \sqrt{37} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,137 |
11. (20 points) Given $k>\sqrt{2009}+\sqrt{2010}$. Prove: for any positive integers $m, n$,
$$
|n \sqrt{2009}-m|>\frac{1}{k n} \text {. }
$$ | 11. First, prove:
$$
n \sqrt{2009}-[n \sqrt{2009}]>\frac{1}{2 n \sqrt{2009}} \text {. }
$$
In fact, equation (1)
$$
\begin{array}{l}
\Leftrightarrow\left(n \sqrt{2009}-\frac{1}{2 n \sqrt{2009}}\right)^{2}>([n \sqrt{2009}])^{2} \\
\Leftrightarrow 2009 n^{2}-1+\frac{1}{4 \times 2009 n^{2}}>([n \sqrt{2009}])^{2} \text {.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,138 |
One. (40 points) Given $0<\alpha<\beta \leqslant \frac{\pi}{2}$. On the unit circle $\odot O$, there are two fixed points $A$ and $B$, with $\angle A O B=2 \alpha$. $M$ is a moving point on $\odot O$. On the line $M A$, there exists a point $P$ such that $\angle A P K=180^{\circ}-\beta$ (where $K$ is the midpoint of si... | As shown in Figure 3, on the major arc $\overparen{A B}$, take point $C$ such that $\angle 4 O C=2 \beta$. Take the midpoint $D$ of $B C$, connect $A D$ and extend it to intersect $\odot O$ at point $E$. Draw $C F / / D A$ to intersect $\odot O$ at another point $F$.
By the midline theorem, we know $C M / / D K$.
Below... | \frac{1}{\sin \beta} \sqrt{2\left(\sin ^{2} \alpha+\sin ^{2} \beta\right)-\sin ^{2}(\alpha+\beta)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,139 |
II. (40 points) Let the number of ways to choose $k$ pairwise coprime numbers from the set $\{1,2, \cdots, 28\}$ be $T(k)$. Find $T(2)+T(3)+\cdots+T(12)$. | Obviously, the prime numbers in $\{1,2, \cdots, 28\}$ are 2, 3, $5,7,11,13,17,19,23$, a total of 9.
When $k \geqslant 11$, the $k$ numbers taken out, except for 1, must include at least 10 numbers. By the pigeonhole principle, there must be two numbers divisible by the same prime number, which means these two numbers ... | 15459 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,140 |
Three. (50 points) A and B play a game. A randomly selects a positive integer pair $(k, n)(2 \leqslant k \leqslant 36,1 \leqslant n \leqslant$ $216)$, and B performs the following operation: divides $[0,36]$ into $k$ segments, $[0,36]=\bigcup_{i=1}^{k}\left[a_{i-1}, a_{i}\right]\left(a_{0}=0<a_{1}<\cdots<a_{k}\right.$ ... | Three, first find the minimum value of $F(k)$.
For any $a \in [a_{i-1}, a_{i}](i=1,2, \cdots, k)$, take $f(a)=\min \left\{2(a-a_{i-1}), a_{i}-a\right\}$.
Let $a_{i}-a_{i-1}=b_{i}(i=1,2, \cdots, k), a-a_{i-1}=b$.
Then $f(a)=\min \left\{2 b, b_{i}-b\right\}=g(b)\left(0 \leqslant b \leqslant b_{i}\right)$.
Obviously, $g(b... | \frac{241}{1512} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,141 |
Four, (50 points) Distribute 2010 red cards and 2010 white cards arbitrarily to 2010 players participating in the game, with each person receiving two cards. All players sit in a circle facing down. The game rule is that each operation requires each player to simultaneously follow the following principle: if a player h... | In 2009.
Let $n=2$ 010. A wheel (divided into $n$ sections) can be placed in a circle of $n$ players. Each player, according to the game rules, when the section they face contains a white card and they hold a red card, they place the red card in the section and take back the white card; in all other cases, it is assume... | 2009 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,142 |
The initial 267 four-digit number $w_{1}$ and the sum of its four digits is the four-digit number $w_{2}, w_{2}$ and the sum of its four digits is the four-digit number $w_{3}, w_{3}$ and the sum of its four digits is the four-digit number $w_{4}, w_{4}$ and the sum of its four digits is the four-digit number $w_{5}, w... | Let $w_{5-i}=a_{i} b_{i} c_{i} d_{i}(i=0,1, \cdots, 4)$. Then $1001 a_{0}+101 b_{0}+11 c_{0}+2 d_{0}=2009$.
Obviously, $a_{0}=1$ (otherwise, if $a_{0}=2$, then $101 b_{0}+11 c_{0}+2 d_{0}=7$, which only has the solution $b_{0}=c_{0}=0, d_{0}=\frac{7}{2}$, which is not an integer).
Thus, $101 b_{0}+11 c_{0}+2 d_{0}=1008... | 1921 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,143 |
As shown in Figure 1, given a circle $\odot O$ with radius $r$ that is tangent to two sides $AB$ and $BC$ of a square $ABCD$ with side length $4r$. A tangent line is drawn from point $A$ to $\odot O$, intersecting $BC$ at point $E$. Prove that $AE$ is a tangent to the circle with diameter $CD$. | Proof As shown in Figure 1, let the points of tangency of $\odot O$ with $\triangle ABE$ be $F$, $G$, and $S$.
$$
\begin{array}{l}
\text { From } S_{\triangle A B E}=\frac{1}{2} \times 4 r \cdot B E \\
=\frac{1}{2} r(A B+B E+A E),
\end{array}
$$
we get $2 B E=r+A E$.
Then $B E=r+A E-B E=r+A E-(E G+r)$ $=A E-E G=A E-E ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,144 |
Given Rt $\triangle ABC, \angle C=90^{\circ}$, take any point $M$ on $AC$ and any point $N$ on $BC$. Let the line through $M$ perpendicular to $AC$ intersect the line through $N$ perpendicular to $BC$ at point $P$. Connect $AN$ and $BM$ intersecting at point $L$. Let $H_{1}$ and $H_{2}$ be the orthocenters of $\triangl... | Proof: As shown in Figure 2, let the line passing through point $A$ and perpendicular to $AC$ intersect the line passing through point $B$ and perpendicular to $BC$ at point $Q$. Then $Q$ is a fixed point.
Let $AN$ intersect $PM$ at point $X$, and $BM$ intersect $PN$ at point $Y$. Connect $H_{1}A$, $H_{1}M$, $PL$, and ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,145 |
2. Solve the equation $\frac{19 x-x^{2}}{x+1}\left(x+\frac{19-x}{x+1}\right)=84$. (2006, Ningbo City, Zhejiang Province Junior High School Mathematics Competition) | Prompt: Example 6. Answer: $x_{1}=3, x_{2}=4, x_{3}=6+$ $\sqrt{29}, x_{4}=6-\sqrt{29}$. | x_{1}=3, x_{2}=4, x_{3}=6+\sqrt{29}, x_{4}=6-\sqrt{29} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,146 |
Let $p, q (p>q)$ be two prime numbers, satisfying $p \equiv 3(\bmod 4)$, and $k$ is a given positive integer greater than 3. Prove: the equation $x^{4}+k y^{4}=p q$ has at most one positive integer solution. | Prove a lemma first.
Lemma $p \equiv 3(\bmod 4)$, if two positive integers $m, n$ satisfy $p \mid\left(m^{2}+n^{2}\right)$, then $p \mid m, p \mid n$.
Proof of the lemma If one of $p \mid m$ or $p \mid n$ holds, the conclusion is easily seen to be true.
Otherwise, $(p, m)=1, (p, n)=1$.
Let $p=4 t+3\left(t \in \mathbf{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,147 |
3. Let $x$ and $y$ be real numbers, and satisfy $x^{2} + xy + y^{2} = 3$. Then the maximum and minimum values of $x^{2} - xy + y^{2}$ are
| Prompt: Copy Example 4. Answer: 9 and 1.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 9 \text{ and } 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,148 |
4. If $\left\{\begin{array}{l}x^{2}+x y+y=14, \\ y^{2}+x y+x=28,\end{array}\right.$ find $x+y=$ $\qquad$
(2001, TI Cup National Junior High School Mathematics Competition) | Let $x=a+b, y=a-b$.
Transform the original system of equations to
$$
\begin{array}{l}
\left\{\begin{array}{l}
2 a^{2}+2 a b+a-b=14, \\
2 a^{2}-2 a b+a+b=28
\end{array}\right. \\
\Rightarrow a=3 \text { or }-\frac{7}{2} .
\end{array}
$$
From $x+y=2 a$, we get $x+y=6$ or -7. | 6 \text{ or } -7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,149 |
Example 2 Given that $\triangle ABC$ is a determined triangle, $A_{1}$, $B_{1}$, and $C_{1}$ are the midpoints of sides $BC$, $CA$, and $AB$ respectively, and $P$ is a moving point on the circumcircle of $\triangle ABC$. $PA_{1}$, $PB_{1}$, and $PC_{1}$ intersect the circumcircle of $\triangle ABC$ at another point $A'... | Proof: As shown in Figure 5, let $A_{0}, B_{0}, C_{0}$ be the three vertices of the triangle formed by the intersections of the lines $A A^{\prime}$, $B B^{\prime}$, and $C C^{\prime}$.
We need to prove:
$S_{\triangle A_{0} B_{0} C_{0}}$
$$
=\frac{1}{2} S_{\triangle A B C} \text {. }
$$
From this, we can conclude that... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,152 |
Example 3 Let the circle $C_{a}$ be internally tangent to the circumcircle of $\triangle A B C$ and tangent to sides $A B$ and $A C$, and let $r_{a}$ be the radius of circle $C_{a}$. Similarly define $r_{b}$ and $r_{c}$, and let $r$ be the inradius of $\triangle A B C$. Prove that $r_{a}+r_{b}+r_{c} \geqslant 4 r$. ${ ... | Proof As shown in Figure 6, let circle $C_{a}$ be tangent to $AB$, $AC$, and the circumcircle of $\triangle ABC$ at points $D$, $E$, and $F$, respectively. Let $M$ and $N$ be the midpoints of arcs $\overparen{AB}$ and $\overparen{AC}$, respectively, and let $I$ be the incenter of $\triangle ABC$.
At this point, point ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,153 |
Example 4 Convex quadrilateral $ABCD$ is inscribed in circle $\Gamma$, and a circle that intersects side $BC$ is internally tangent to circle $\Gamma$, and is tangent to $BD$, $AC$ at points $P$, $Q$ respectively. Prove that the incenter of $\triangle ABC$ and the incenter of $\triangle DBC$ both lie on line $PQ$.
(200... | Proof As shown in Figure 7, let the center of circle $\Gamma$ be $O$, and the center of the circle that intersects $BC$ and is internally tangent to circle $\Gamma$ be $O_{1}$, with the tangency point being $T$.
Obviously, points $O$, $O_{1}$, and $T$ are collinear.
Let $DB$ and $CA$ intersect at point $H$, and line $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,154 |
Example 5 As shown in Figure 8, point $P$ is inside $\triangle A B C$, and the projections of $P$ onto sides $B C$, $C A$, and $A B$ are $D$, $E$, and $F$, respectively. Through point $A$, perpendiculars are drawn to lines $B P$ and $C P$, with the feet of the perpendiculars being $M$ and $N$, respectively. Prove that ... | Proof From the given conditions, we have
$$
\begin{array}{l}
\angle A E P=\angle A F P=\angle A M P \\
=\angle A N P=90^{\circ} .
\end{array}
$$
Therefore, points $A, N, F, P, E, M$ all lie on a circle with $A P$ as its diameter.
Thus, for the cyclic hexagon $A F N P M E$, the intersections of its three pairs of oppo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,155 |
Example 3 Given real numbers $x, y, z$ satisfy $x-y=8$, $xy+z^{2}=-16$. Then $x+y+z=$ $\qquad$ . | Let $x=a+b, y=a-b$. Then $x-y=2b \Rightarrow b=4$.
Since $xy+z^{2}=-16$, we have $(a+b)(a-b)+z^{2}+16=0$. Also, $-b^{2}=-16$, then $a^{2}+z^{2}=0$. By the property of non-negative numbers, we know $a=0, z=0$. Therefore, $x=4, y=-4$. Hence, $x+y+z=4-4+0=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,156 |
Example 6 Given that $\triangle ABC$ is an acute triangle, the circle $\odot K$ with diameter $AB$ intersects $AC$ and $BC$ at points $P$ and $Q$ respectively. The two tangents to $\odot K$ from points $A$ and $Q$ intersect at point $R$, and the two tangents to $\odot K$ from points $B$ and $P$ intersect at point $S$. ... | Proof As shown in Figure 9, let $R Q$ intersect $P S$ and $A C$ intersect $R K$, and $B C$ intersect $K S$ at points $W$, $Y$, and $N$ respectively. Connect $P K$, $W K$, $Q K$, $W N$, $W Y$, and $B P$. Then
$$
\begin{array}{l}
\angle Y K W \\
=\angle Y K Q-\angle W K Q \\
=\frac{1}{2}(\angle A K Q-\angle P K Q) \\
=\f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,157 |
Example 7 In the right triangle $\triangle ABC$, $\angle A=90^{\circ}, \angle B > \angle C, O$ is the circumcenter of $\triangle ABC$, $l_{A}$ and $l_{B}$ are two tangent lines of $\odot O$, touching at points $A$ and $B$ respectively. Let $BC$ intersect $l_{A}$ at point $S$, $AC$ intersect $l_{B}$ at point $D$, $AB$ i... | Prove as shown in Figure 10, let the extension of $BA$ intersect the tangent to $\odot O$ at point $C$ at point $E'$.
Applying Pascal's theorem to $\triangle ABC$, we know that points $S$, $D$, and $E'$ are collinear. Therefore, point $E'$ coincides with point $E$.
Thus, the positions of points $R$ and $Q$ are as show... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,158 |
1. In $\triangle ABC$, there is a circle that is internally tangent to the circumcircle of $\triangle ABC$ and is tangent to $AB$ and $AC$ at points $P$ and $Q$, respectively. Prove that the midpoint of $PQ$ is the incenter of $\triangle ABC$.
(Adapted from the 20th IMO problem) | Let $D$ be the common tangent point of the two circles, and the lines $D P, D Q$ intersect the circumcircle at points $K, H$ respectively. Applying Pascal's theorem to the cyclic hexagon $A B H D K C$, we know that the three pairs of opposite sides $A B$ and $D K, B H$ and $K C, D H$ and $A C$ intersect at points $P, I... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,159 |
3. The circumcenter of the convex quadrilateral $ABCD$ is $O$. It is known that $AC \neq BD$, and $AC$ intersects $BD$ at point $E$. If $P$ is a point inside the quadrilateral $ABCD$ such that
$$
\angle PAB + \angle PCB = \angle PBC + \angle PDC = 90^{\circ} \text{.}
$$
Prove that $O$, $P$, and $E$ are collinear. ${ }... | ```
\begin{array}{l}
\text { By } 90^{\circ}=\angle P A B+\angle P C B \\
=\frac{1}{2} \angle A^{\prime} O B+\frac{1}{2} \angle C^{\prime} O B=\frac{1}{2} \angle A^{\prime} O C
\end{array}
```
we know that $A^{\prime} C^{\prime}$ is the diameter of the circumcircle.
Similarly, $B^{\prime} D^{\prime}$ is also its diame... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,161 |
4. In an isosceles right $\triangle ABC$, $\angle A=90^{\circ}, AB=$ $1, D$ is the midpoint of $BC$, $E$ and $F$ are two other points on $BC$, $M$ is the other intersection point of the circumcircle of $\triangle ADE$ and the circumcircle of $\triangle ABF$, $N$ is the other intersection point of line $AF$ and the circ... | Given: With point $A$ as the inversion center and $r=1$ as the radius of the inversion circle. Denote the image of point $X$ as $X^{*}$. Then
points $B, F, D, E, C$ are collinear
$\Leftrightarrow$ points $A, B^{*}, F^{*}, D^{*}, E^{*}, C^{*}$ lie on the same circle, $M$ is one of the intersection points of the circumci... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,162 |
Question 1 Let $E$ and $F$ be points on the sides $AD$ and $BC$ of a convex quadrilateral $ABCD$, respectively, such that $AE: ED = BF: FC$. The ray $FE$ intersects the rays $BA$ and $CD$ at points $S$ and $T$, respectively. Prove that the circumcircles of $\triangle SAE$, $\triangle SBF$, $\triangle TCF$, and $\triang... | Proof: As shown in Figure 1, let the circumcircles of $\triangle TCF$ and $\triangle TDE$ intersect at another point $P$ (different from point $T$).
We need to prove: $P, A, E, S$ and $P, B, F, S$ are concyclic.
Notice that $P, E, D, T$ are concyclic, then
$$
\begin{array}{l}
\angle PDE = \angle PTE, \\
\angle PED = 18... | null | Geometry | proof | Yes | Yes | cn_contest | false | 722,163 |
Intersection point $P$ is a fixed point
Question 2 Given in a convex quadrilateral $ABCD$, moving points $E, F$ are on the rays $AD, BC$ respectively, satisfying $DE: CF = AD: BC$. If $CD$ intersects $FE$ at point $G$, prove: the circumcircles of $\triangle DEG$ and $\triangle CFG$ intersect at another point different ... | Proof As shown in Figure 2, let $BA$ and $CD$ intersect at point $M$.
From Question 1, we know that $\triangle MAD \sim \triangle MBC$ and $\triangle GDE \sim \triangle GCF$ have a common point (denoted as $P$).
Since $A, B, C, D, M$ are all fixed points, the intersection point $P$ of the circumcircles of $\triangle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,164 |
Question 3 As shown in Figure 3, in the convex quadrilateral $ABCD$, $AD$ is parallel to $BC$, and $E$, $F$ are points on sides $AD$, $BC$ respectively, such that $AE: ED = BF: FC$. Ray $FE$ intersects rays $BA$, $CD$ at points $S$, $T$, and $BA$ intersects $CD$ at point $Q$. Prove that the circumcircles of $\triangle ... | Prove: As shown in Figure 3, let the circumcircles of $\triangle SAE$, $\triangle SBF$, $\triangle TDE$, $\triangle TCF$, $\triangle QAD$, and $\triangle QBC$ be $\odot O_{1}$, $\odot O_{2}$, $\odot O_{3}$, $\odot O_{4}$, $\odot O_{5}$, and $\odot O_{6}$, respectively. Let the intersection of ray $DA$ and $CB$ be point... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,165 |
Question 4 As shown in Figure 4, in quadrilateral $ABCD$, ray $BA$ intersects $CD$ at point $Q$, and ray $DA$ intersects $CB$ at point $J$. $E$ and $F$ are arbitrary points on the sides $AD$ and $BC$ of the convex quadrilateral $ABCD$, respectively. Ray $FE$ intersects rays $BA$ and $CD$ at points $S$ and $T$, respecti... | Proof Consider quadrilateral TSAD.
By Miquel's theorem for quadrilaterals, the circumcircle of $\triangle T S Q$ passes through point $M$.
Now consider quadrilateral TSBC.
Thus, the circumcircle of $\triangle T S Q$ also passes through point $M^{\prime}$.
Therefore, points $T 、 M 、 M^{\prime} 、 S 、 Q$ are concyclic.
Si... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,166 |
Example 4 Given real numbers $x, y$ satisfy $4 x^{2}-5 x y+4 y^{2}=5$, let $S=x^{2}+y^{2}$. Then $\frac{1}{S_{\text {max }}}+\frac{1}{S_{\text {min }}}=$ $\qquad$ ( $\max$ denotes the maximum value, $\min$ denotes the minimum value). | Let $x=a+b, y=a-b$. Substituting into the given equation and simplifying, we get $3 a^{2}+13 b^{2}=5$.
Therefore, $b^{2}=\frac{5}{13}-\frac{3}{13} a^{2}\left(0 \leqslant a^{2} \leqslant \frac{5}{3}\right)$.
Thus, $S=x^{2}+y^{2}=(a+b)^{2}+(a-b)^{2}$
$$
=2 a^{2}+2 b^{2}=\frac{20}{13} a^{2}+\frac{10}{13} \text {. }
$$
Th... | \frac{8}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,167 |
Question 5 As shown in Figure 5, in quadrilateral $ABCD$, ray $BA$ intersects $CD$ at point $Q$, and ray $DA$ intersects $CB$ at point $J$. $E$ and $F$ are arbitrary points on the sides $AD$ and $BC$ of the convex quadrilateral $ABCD$, respectively. Ray $FE$ intersects rays $BA$ and $CD$ at points $S$ and $T$, respecti... | Prove that, as shown in Figure 5, from Question 4 we know that points $T, M, M^{\prime}, S, Q$ are concyclic.
Let this circle be $\odot O_{1}$. Let the circumcircles of $\triangle T D E$ and $\triangle S B F$ be $\odot O_{2}$ and $\odot O_{3}$, respectively, and let $\odot O_{2}$ and $\odot O_{3}$ intersect at points ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,168 |
Example 1 Let $a, b, c > 0$, and $a+b+c=1$. Prove:
$$
\sqrt{a^{1-a} b^{1-b} c^{1-c}} \leqslant \frac{1}{3} \text {. }
$$
(2008, Australian Mathematical Olympiad) | $$
\begin{array}{l}
\frac{1-a}{2}+\frac{1-b}{2}+\frac{1-c}{2}=1 . \\
\text { Then } \sqrt{a^{1-a} b^{1-b} c^{1-c}}=a^{\frac{1-a}{2}} b^{\frac{1-b}{2}} c^{\frac{1-c}{2}} \\
\leqslant \frac{1-a}{2} \cdot a+\frac{1-b}{2} \cdot b+\frac{1-c}{2} \cdot c \\
=\frac{1}{2}\left[(a+b+c)-\left(a^{2}+b^{2}+c^{2}\right)\right] \\
\l... | \frac{1}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 722,170 |
Example 2 Let $a, b, c > 0$, and $abc = 1$. Prove: $a^{b+c} b^{c+a} c^{a+b} \leqslant 1$. | Prove that
$$
\frac{c}{a+b+c}+\frac{a}{a+b+c}+\frac{b}{a+b+c}=1 .
$$
$$
\begin{array}{l}
\text { Then }\left(a^{b+c} b^{c+a} c^{a+b}\right)^{\frac{1}{a+b+c}} \\
=(a b)^{\frac{c}{a+b+c}}(b c)^{\frac{a}{a+b+c}}(c a)^{\frac{b}{a+b+c}} \\
\leqslant \frac{c}{a+b+c} \cdot a b+\frac{a}{a+b+c} \cdot b c+\frac{b}{a+b+c} \cdot c... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,171 |
Example 3 Let $a, b, c > 0$. Prove:
$$
a^{a} b^{b} c^{c} \geqslant(a b c)^{\frac{a+b+c}{3}} \text {. }
$$
(1974, USA Mathematical Olympiad) | $$
\begin{array}{l}
\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1 . \\
\text { Then }\left(\frac{1}{a}\right)^{\frac{a}{a+b+c}}\left(\frac{1}{b}\right)^{\frac{b}{a+b+c}}\left(\frac{1}{c}\right)^{\frac{c}{a+b+c}} \\
\leqslant \frac{a}{a+b+c} \cdot \frac{1}{a}+\frac{b}{a+b+c} \cdot \frac{1}{b}+\frac{c}{a+b+c} \cdot \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,172 |
Example 4 Let $a$, $b$, $c$ be the lengths of the sides of a triangle. Prove:
$$
(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c} \leqslant a^{a} b^{b} c^{c} .
$$ | $$
\begin{array}{l}
\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1 . \\
\text { Then }\left(\frac{a+b-c}{a}\right)^{\frac{a}{a b+c}}\left(\frac{b+c-a}{b}\right)^{\frac{b}{a+b+c}}\left(\frac{c+a-b}{c}\right)^{\frac{c}{a+b+c}} \\
\leqslant \frac{a}{a+b+c} \cdot \frac{a+b-c}{a}+\frac{b}{a+b+c} \cdot \frac{b+c-a}{b}+ \\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,173 |
Example 5 Let $a, b, c > 0$. Prove: $a^{a} b^{b} c^{c} \geqslant a^{c} b^{a} c^{b}$.
Analysis: On the left side of this inequality, the exponents are $a, b, c$, while on the right side, the exponents are $c, a, b$. Since $a^{c} b^{a} c^{b}=b^{a} c^{b} a^{c}$, the exponents on both sides are actually the same. | $$
\begin{array}{l}
\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1 . \\
\text { Then }\left(\frac{b}{a}\right)^{\frac{a}{a+b+c}}\left(\frac{c}{b}\right)^{\frac{b}{a+b+c}}\left(\frac{a}{c}\right)^{\frac{c}{a c b+c}} \\
\leqslant \frac{a}{a+b+c} \cdot \frac{b}{a}+\frac{b}{a+b+c} \cdot \frac{c}{b}+\frac{c}{a+b+c} \cdot... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,174 |
Example 6 Let $a, b, c$ be the lengths of the sides of a triangle. Prove:
$$
\begin{array}{l}
(a+b-c)^{a+b-c}(b+c-a)^{b+c-a}(c+a-b)^{c+a-b} \\
\geqslant a^{b} b^{c} c^{a} .
\end{array}
$$ | $$
\begin{array}{l}
\frac{a+b-c}{a+b+c}+\frac{b+c-a}{a+b+c}+\frac{c+a-b}{a+b+c}=1 . \\
\text { Then }\left(\frac{a}{a+b-c}\right)^{\frac{c+b-c}{a+b+c}}\left(\frac{b}{b+c-a}\right)^{\frac{b+c-a}{a+b+c}}\left(\frac{c}{c+a-b}\right)^{\frac{c+a-b}{a+b+c}} \\
\leqslant \frac{a+b-c}{a+b+c} \cdot \frac{a}{a+b-c}+\frac{b+c-a}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,175 |
1. As shown in Figure 1, two circles $\Gamma_{1}$ and $\Gamma_{2}$ intersect at points $A$ and $B$. A line through point $B$ intersects circles $\Gamma_{1}$ and $\Gamma_{2}$ at points $C$ and $D$, respectively. Another line through point $B$ intersects circles $\Gamma_{1}$ and $\Gamma_{2}$ at points $E$ and $F$, respec... | 1. Connect $A C, A D, A E, A F, D F$.
From $\angle A D B=\angle A F B, \angle A C B=\angle A E F$ and $C D=E F$
$\Rightarrow \triangle A C D \cong \triangle A E F$
$\Rightarrow A D=A F$
$\Rightarrow \angle A D F=\angle A F D$
$\Rightarrow \angle A B C=\angle A F D=\angle A D F=\angle A B F$
$\Rightarrow A B$ is the an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,176 |
2. Let integer $k \geqslant 3$, and the sequence $\left\{a_{n}\right\}$ satisfies $a_{k}=2 k$, and for all $n>k$, we have
$$
a_{n}=\left\{\begin{array}{ll}
a_{n-1}+1, & \text { if } a_{n-1} \text { and } n \text { are coprime, } \\
2 n, & \text { if } a_{n-1} \text { and } n \text { are not coprime. }
\end{array}\right... | 2. Suppose $a_{l}=2 l(l \geqslant k)$. Let $p$ be the smallest prime factor of $l-1$. Then
$$
(l-1, i)=\left\{\begin{array}{ll}
1, & 1 \leqslant i<p ; \\
p, & i=p .
\end{array}\right.
$$
Therefore, $(2 l+i-2, l+i-1)=\left\{\begin{array}{ll}1, & 1 \leqslant i<p ; \\ p, & i=p .\end{array}\right.$
By the given condition,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,177 |
Example 5 Given real numbers $a, b$ satisfy $a^{2}+a b+b^{2}=1$, and $t=a b-a^{2}-b^{2}$. Then the range of $t$ is $\qquad$ $t_{\max }=$ $\qquad$ ,$t_{\text {min }}=$ $\qquad$ .
(2001, TI Cup National Junior High School Mathematics Competition) | Let $a=x+y, b=x-y$.
Substitute into the given equation:
$$
(x+y)^{2}+(x+y)(x-y)+(x-y)^{2}=1 \text {. }
$$
Simplify to get $y^{2}=1-3 x^{2}$.
Since $y^{2} \geqslant 0$, we have $0 \leqslant x^{2} \leqslant \frac{1}{3}$.
Thus, $t=a b-a^{2}-b^{2}$
$$
\begin{array}{l}
=(x+y)(x-y)-(x+y)^{2}-(x-y)^{2} \\
=-\left(x^{2}+3 y^{... | t_{\max }=-\frac{1}{3}, t_{\min }=-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,178 |
3. Let complex numbers $a, b, c$ satisfy: for any complex number $z$ with modulus not exceeding 1, we have $\left|a z^{2}+b z+c\right| \leqslant 1$. Find the maximum value of $|b c|$.
(Li Weiguo, problem contributor) | 3. Let $f(z)=a z^{2}+b z+c$,
$$
\begin{array}{l}
g(z)=z^{-2} f(z)=a+b z^{-1}+c z^{-2}, \\
h(z)=\mathrm{e}^{\mathrm{i} \alpha} g\left(\mathrm{e}^{\mathrm{i} \beta} z\right)=c^{\prime} z^{-2}+b^{\prime} z^{-1}+a^{\prime} .
\end{array}
$$
Choose appropriate real numbers $\alpha, \beta$ such that $c^{\prime}, b^{\prime} \... | \frac{3 \sqrt{3}}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,179 |
4. Given that $m, n$ are integers greater than 1, and $a_{1}<a_{2}<\cdots<a_{m}$ are all integers. Prove: There exists a subset $T$ of the set of integers, with the number of elements
$$
|T| \leqslant 1+\frac{a_{m}-a_{1}}{2 n+1},
$$
such that for each $i \in\{1,2, \cdots, m\}$, there is $t \in T$ and $s \in[-n, n]$, s... | 4. Let $a_{1}=a, a_{m}=b$, perform the division with remainder $b-a=(2 n+1) q+r(q, r \in \mathbf{Z}, 0 \leqslant r \leqslant 2 n)$. Take $T=\{a+n+(2 n+1) k \mid k=0,1, \cdots, q\}$. Then $|T|=q+1 \leqslant 1+\frac{b-a}{2 n+1}$, and the set
$$
\begin{array}{l}
B=\{t+s \mid t \in T, s=-n,-n+1, \cdots, n\} \\
=\{a, a+1, \... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,180 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.