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8. Let $P$ be any point inside an $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, and let the lines $A_{i} P (i=1,2, \cdots, n)$ intersect the boundary of the $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ at another point $B_{i}$. Prove:
$\sum_{i=1}^{n} P A_{i} \geqslant \sum_{i=1}^{n} P B_{i}$. (Feng Zhigang) | 8. Let $t=\left[\frac{n}{2}\right]+1$, and set
$$
A_{n+j}=A_{j}(j=1,2, \cdots, n) \text {. }
$$
Notice that, the distance between any vertex of a regular $n$-sided polygon and any point on its boundary does not exceed the length of its longest diagonal $d$. Therefore, for any $i(1 \leqslant i \leqslant n)$, we have
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,325 |
1. For integers $a_{1}, a_{2}, \cdots, a_{k}$, let
$$
n=\sum_{i=1}^{k} a_{i},\binom{n}{a_{1}, \cdots, a_{k}}=\frac{n!}{\prod_{i=1}^{k}\left(a_{i}!\right)} .
$$
Let $d=\operatorname{gcd}\left(a_{1}, a_{2}, \cdots, a_{k}\right)$ denote the greatest common divisor of $a_{1}, a_{2}$,
$\cdots, a_{k}$. Prove: $\frac{d}{n}\b... | 1. Let $a_{1}=d x_{1}, a_{2}=d x_{2}, \cdots, a_{k}=d x_{k}$. Then $\left(x_{1}, x_{2}, \cdots, x_{k}\right)=1$.
By Bezout's theorem, there exist integers $u_{1}, u_{2}, \cdots, u_{k}$, such that $\sum_{i=1}^{k} u_{i} x_{i}=1$. Therefore, $\sum_{i=1}^{k} u_{i} a_{i}=d$. Let
$$
\begin{array}{l}
S_{i}=\frac{a_{i}}{n}\bin... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,326 |
2. A set $S$ of points in space satisfies the property: the distances between any two points in $S$ are all different. Assume the coordinates $(x, y, z)$ of the points in $S$ are all integers, and $1 \leqslant x, y, z \leqslant n$. Prove: the number of elements in the set $S$ is less than
$$
\min \left\{(n+2) \sqrt{\fr... | 2. Let $|S|=t$. Then for any $\left(x_{1}, y_{1}, z_{1}\right)$, $\left(x_{2}, y_{2}, z_{2}\right) \in S$, we have
$$
\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(z_{1}-z_{2}\right)^{2} \leqslant 3(n-1)^{2}
$$
(because the distance between any integer points satisfying $1 \leqslant x, y, z \leqslant ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,327 |
3. Given any four points $A_{1}, A_{2}, A_{3}, A_{4}$ not all collinear in a plane, such that
$$
A_{1} A_{2} \cdot A_{3} A_{4}=A_{1} A_{3} \cdot A_{2} A_{4}=A_{1} A_{4} \cdot A_{2} A_{3} .
$$
Let $O_{i}$ be the circumcenter of $\triangle A_{k} A_{j} A_{l}$ $(\{i, j, k, l\}=$ $\{1,2,3,4\})$. Assume that for each index ... | 3. If four points $A_{1}, A_{2}, A_{3}, A_{4}$ form a concave quadrilateral, let's assume $A_{4}$ is inside $\triangle A_{1} A_{2} A_{3}$, as shown in Figure 1.
Construct $\triangle A_{1} A_{3} P \backsim \triangle A_{1} A_{2} A_{4}$. Then $\angle A_{3} A_{1} P=\angle A_{4} A_{1} A_{2}$.
Thus, $\angle A_{4} A_{1} P=\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,328 |
4. For a finite set $X$ of positive integers, define
$$
\sum(X)=\sum_{x \in X} \arctan \frac{1}{x} .
$$
Let $S$ be a finite set of positive integers such that $\sum(S)<\frac{\pi}{2}$. Prove: There exists at least one finite set $T$ of positive integers such that $S \subset T$, and $\sum(T)=\frac{\pi}{2}$. | 4. Notice that when $\tan \alpha, \tan \beta$ are both rational numbers, $\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}$ is also a rational number. Therefore, $\tan \left(\sum(S)\right)$ is a rational number.
It is well-known that $\sum_{k=1}^{n} \frac{1}{k}$ diverges as $n \rightar... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,329 |
As shown in Figure 3, $AB$ is the diameter of $\odot O$, $C$ is the midpoint of $\overparen{AB}$, $M$ is the midpoint of $AC$, and $CH \perp BM$ at $H$. Prove:
$$
OH=\frac{1}{2} AH .
$$ | Proof: It is known that $A C = B C, A C \perp B C$, $\angle C A B = \angle C B A = 45^{\circ}$.
Also, $C H \perp B M$, then $\triangle C M H \backsim \triangle B M C$.
Therefore, $C M^{2} = M H \cdot M B$.
Noting that $A M^{2} = C M^{2}$, then $A M^{2} = M H \cdot M B$.
Thus, $\triangle A M H \backsim \triangle B M A$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,330 |
Example 1 In a right triangle $ABC$, the three sides are $a$, $b$, and $c$, where $c$ is the hypotenuse. Prove:
$$
\tan \frac{A}{2}+\tan \frac{B}{2}=\frac{2 c}{a+b+c} .
$$ | Explanation: From the half-angle of a triangle, we think of the incenter of the triangle. Let's assume the incenter of $\triangle A B C$ is $I$.
As shown in Figure 1, connect $A I$,
$B I$, and draw $I D \perp A B$ at point
D.
Obviously, $\angle I A D=\frac{\angle A}{2}$,
$$
\angle I B D=\frac{\angle B}{2},
$$
and the ... | \tan \frac{A}{2}+\tan \frac{B}{2}=\frac{2 c}{a+b+c} | Geometry | proof | Yes | Yes | cn_contest | false | 721,331 |
Example 2 As shown in Figure 3, in $\triangle A B C$, $\angle C=90^{\circ}, B C=a, C A=b$, $A B=c, A E$ and $B F$ are two angle bisectors. Let the area of the shaded quadrilateral $A B E F$ be $S$. Prove:
$$
\frac{1}{S}=\frac{1}{b c}+\frac{1}{c a}+\frac{1}{a b} .
$$ | Explanation: Clearly, equation (1) $\Leftrightarrow S=\frac{a b c}{a+b+c}$.
By the angle bisector property of a triangle, we easily get
$$
\begin{array}{l}
C E=\frac{a b}{b+c}, C F=\frac{a b}{a+c} . \\
\text { Hence } S=S_{\triangle A B C}-S_{\triangle C E F} \\
=\frac{1}{2} a b-\frac{1}{2} \cdot \frac{a b}{b+c} \cdot ... | \frac{1}{S}=\frac{1}{b c}+\frac{1}{c a}+\frac{1}{a b} | Geometry | proof | Yes | Yes | cn_contest | false | 721,332 |
Example 1 Non-negative real numbers $a, b, c$ satisfy $a+b+c=1$, and let $S=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$. Prove:
$$
\frac{9}{4} \leqslant S \leqslant \frac{5}{2} \text {. }
$$ | Explanation: When $a=b=c=\frac{1}{3}$, $S=\frac{9}{4}$. It is easy to see that the left side of equation (1) $S \geqslant \frac{9}{4}$ is a conventional extremum.
By the corollary of the mean inequality, we have
$$
\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right)(1+a+1+b+1+c) \geqslant 3^{2},
$$
which means $S \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,333 |
Example 2 Non-negative real numbers $a, b, c$ satisfy $a+b+c=1$, let
$$
S=\sqrt{a+\frac{1}{4}(b-c)^{2}}+\sqrt{b+\frac{1}{4}(c-a)^{2}}+\sqrt{c+\frac{1}{4}(a-b)^{2}} \text {. }
$$
Prove: $S \leqslant 2$. | Explanation: This problem is adapted from the 2007 Girls' Mathematical Olympiad. The original problem was to prove:
$$
S=\sqrt{a+\frac{1}{4}(b-c)^{2}}+\sqrt{b}+\sqrt{c} \leqslant \sqrt{3}.
$$
Here, $S$ is a non-cyclic symmetric expression. In the adapted version, when $a=b=c=\frac{1}{3}$, $S=\sqrt{3}$ is not the maxim... | S \leqslant 2 | Inequalities | proof | Yes | Yes | cn_contest | false | 721,334 |
Example 3 Given that $a, b, c$ are non-negative real numbers, and $a+b+c=1$. Prove:
$$
S=\sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1} \geqslant 2+\sqrt{5} \text{. }
$$ | Explanation: When $a=b=c=\frac{1}{3}$, $S=\sqrt{21}>2+\sqrt{5}$ is not the minimum value.
Consider the three inequalities:
$$
\begin{array}{l}
\sqrt{4 a+1} \geqslant \lambda a+1, \\
\sqrt{4 b+1} \geqslant \lambda b+1, \\
\sqrt{4 c+1} \geqslant \lambda c+1 .
\end{array}
$$
What value should the parameter $\lambda$ take... | S \geqslant 2+\sqrt{5} | Inequalities | proof | Yes | Yes | cn_contest | false | 721,335 |
Example 4 Let non-negative real numbers $a, b, c$ satisfy $a+b+c=$
1. Prove:
$$
S=10 a^{3}-9 a^{5}+10 b^{3}-9 b^{5}+10 c^{3}-9 c^{5} \leqslant \frac{9}{4} \text {. }
$$ | Let $f(x)=10 x^{3}-9 x^{5}$. Then
$$
S=f(a)+f(b)+f(c) \text {. }
$$
Given $a+b+c=1$, there must be two numbers whose sum does not exceed $\frac{2}{3}$ (let's assume $b+c \leqslant \frac{2}{3}$). We will prove:
$$
f(b)+f(c) \leqslant f(b+c),
$$
which is
$$
\begin{array}{l}
10 b^{3}-9 b^{5}+10 c^{3}-9 c^{5} \leqslant 1... | \frac{9}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 721,336 |
Example 5 Let non-negative real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$
\begin{array}{l}
S=\sqrt{1-3 a^{2}}+\sqrt{1-3 b^{2}}+\sqrt{1-3 c^{2}} \\
\geqslant 1+\sqrt{2 \sqrt{3}-3} .
\end{array}
$$ | Let $f(x)=\sqrt{1-3 x^{2}}\left(0 \leqslant x \leqslant \frac{\sqrt{3}}{3}\right)$. Then $S=f(a)+f(b)+f(c)$.
As shown in Figure 1, draw the graph of $f(x)$
within $\left[0, \frac{\sqrt{3}}{3}\right]$
(a part of an ellipse).
For any $k \in \left[0, \frac{\sqrt{3}}{3}\right]$, and a small positive number $\varepsilon>0$... | 1+\sqrt{2 \sqrt{3}-3} | Inequalities | proof | Yes | Yes | cn_contest | false | 721,337 |
Example 6 Let $a, b, c \geqslant 0$, and $a b+b c+c a=\frac{1}{3}$. Prove:
$$
S=\frac{1}{a^{2}-b c+1}+\frac{1}{b^{2}-a c+1}+\frac{1}{c^{2}-a b+1} \leqslant 3 .
$$ | Explanation: When $a=b=c=\frac{1}{3}$, $S=3$.
Also, when $a=0, b=\frac{\sqrt{3}}{3}, c=\frac{\sqrt{3}}{3}$, $S=3$ as well.
Therefore, the maximum value 3 is both conventional and unconventional, which is quite rare. Moreover, $S$ has no other extremum (the lower bound of $S$ is $\frac{5}{2}$, $S>\frac{5}{2}$ but cannot... | 3 | Inequalities | proof | Yes | Yes | cn_contest | false | 721,338 |
Example 7 Let real numbers $x, y, z$ all be not equal to $\frac{b}{a}$, and satisfy $x y z=\left(\frac{b}{a}\right)^{3}$. Prove:
$$
\left(\frac{b x}{a x-b}\right)^{2}+\left(\frac{b y}{a y-b}\right)^{2}+\left(\frac{b z}{a z-b}\right)^{2} \geqslant\left(\frac{b}{a}\right)^{2} .
$$ | Explanation: Let $\frac{b x}{a x-b}=m, \frac{b y}{a y-b}=n, \frac{b z}{a z-b}=p$.
Then $x=\frac{b m}{a m-b}, y=\frac{b n}{a n-b}, z=\frac{b p}{a p-b}$.
$$
\begin{array}{l}
\text { By } x y z=\left(\frac{b}{a}\right)^{3} \\
\Rightarrow \frac{b m \cdot b n \cdot b p}{(a m-b)(a n-b)(a p-b)}=\left(\frac{b}{a}\right)^{3} \\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,339 |
Example 1 Let $f$ be a positive-valued function defined on the set of positive integers, with $f(1)=1$ and
$$
f(n+1) \geqslant(n+1) f(n)-2 n \quad (n \geqslant 1) .
$$
Try to find $f(n)$.
Analysis: Dividing both sides of equation (1) by $(n+1)!$ yields
$$
\begin{array}{l}
\frac{f(n+1)}{(n+1)!} \geqslant \frac{f(n)}{n!... | Solution: Let $f(n)=n!g(n)-2$. Substituting into equation (1) yields
$$
\begin{array}{l}
(n+1)!g(n+1)-2 \\
\geqslant(n+1)(n!g(n)-2)-2 n .
\end{array}
$$
Simplifying, we get $g(n+1) \geqslant g(n)$.
Thus, the analytical expression for the given recursive inequality is
$$
f(n)=n!g(n)-2 \text {, }
$$
where any function ... | f(n)=n!g(n)-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,340 |
Example 2 Let $f(n)$ be a positive function defined on the set of natural numbers, $f(1)=1$ and satisfies
$$
f(n-1)-f(n) \geqslant n f(n-1) f(n)(n \geqslant 2) \text {. }
$$
Find $f(n)$. | Analysis: Dividing both sides of equation (1) by $f(n-1) f(n)$ yields
$$
\begin{array}{l}
\frac{1}{f(n)}-\frac{1}{f(n-1)} \geqslant n(n \geqslant 2) . \\
\text { Then } \frac{1}{f(n)}-[n+(n-1)+\cdots+1] \\
\geqslant \frac{1}{f(n-1)}-[(n-1)+\cdots+1] . \\
\text { Let } g(n)=\frac{1}{f(n)}-[n+(n-1)+\cdots+1] .
\end{array... | f(n)=\frac{2}{n(n+1)+2 g(n)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,341 |
Example 3 Let $f$ be a positive-valued function defined on all positive integers, $f(1)=\frac{1}{2}$ and
$$
f(n+1) \leqslant \frac{3 f(n)}{2 f(n)+1}(n \geqslant 1) \text {. }
$$
Find $f(n)$. | Analysis: It is easy to know that $f(n)>0$. Taking the reciprocal of both sides of equation (1) yields
$$
2+\frac{1}{f(n)} \leqslant \frac{3}{f(n+1)}(n \geqslant 1) \text {. }
$$
Subtracting 3 from both sides of equation (2) gives
$$
\frac{1}{f(n)}-1 \leqslant 3\left(\frac{1}{f(n+1)}-1\right) \text {. }
$$
Multiplyin... | f(n)=\frac{3^{n}}{3^{n}+g(n)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,342 |
Example 3 In the right triangle $ABC$, $\angle C=90^{\circ}$, $BC=a$, $CA=b$, $AB=c$. If $\frac{b}{c+a}+\frac{a}{c+b}=\frac{13}{15}$, find: $a: b: c$. | Explanation: Using the Pythagorean theorem to transform the two ratios on the left side of the given equation, we have
$$
\begin{array}{l}
\frac{b}{c+a}=\frac{c-a}{b}, \frac{a}{c+b}=\frac{c-b}{a}. \\
\text { Therefore, } \frac{13}{15}=\frac{b}{c+a}+\frac{a}{c+b}=\frac{c-a}{b}+\frac{c-b}{a} \\
=\frac{a c-a^{2}+b c-b^{2}... | a: b: c=5: 12: 13 \text{ or } 12: 5: 13 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,343 |
Example 4 Let $f(1)=t(t>2), f(n)$ satisfies the fractional linear recurrence inequality
$$
f(n+1) \geqslant \frac{1}{2}\left(f(n)+\frac{4}{f(n)}\right) .
$$
Try to find $f(n)$. | The characteristic equation of the given recursive inequality corresponding to the recursive sequence is $x=\frac{1}{2}\left(x+\frac{4}{x}\right)$, which has two roots $\pm 2$. It is easy to see that
$$
\begin{array}{l}
f(n+1) \geqslant \frac{1}{2}\left(f(n)+\frac{4}{f(n)}\right) \\
\geqslant \sqrt{f(n) \frac{4}{f(n)}}... | not found | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,344 |
Example 1 Let the quadratic function $f(x)$ satisfy
$$
f(x+2)=f(-x+2)
$$
and its graph intersects the $y$-axis at the point $(0,1)$, and the segment it intercepts on the $x$-axis is $2 \sqrt{2}$. Find the analytical expression of $f(x)$.
(2002, Anhui Province High School Mathematics Competition) | Explanation: Utilizing the symmetry of the graph of a quadratic function, and combining it with the fact that the length of the segment it intercepts on the $x$-axis is $2 \sqrt{2}$, we know that the intersection points of $f(x)$ with the $x$-axis are $(2+\sqrt{2}, 0)$ and $(2-\sqrt{2}, 0)$. Therefore, we can use the z... | f(x)=\frac{1}{2} x^{2}-2 x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,345 |
Example 2 Given the quadratic function
$$
f(x)=4 x^{2}-4 a x+\left(a^{2}-2 a+2\right)
$$
the minimum value on $0 \leqslant x \leqslant 1$ is 2. Find the value of $a$.
(2001, Hunan Province High School Mathematics Olympiad Selection Contest) | Notice that
$$
f(x)=4\left(x-\frac{a}{2}\right)^{2}-2 a+2 \text {. }
$$
It is easy to see that the graph opens upwards, and the axis of symmetry is $x=\frac{a}{2}$. Therefore, we can solve the problem by classifying it according to the three possible positions of the axis of symmetry $x=\frac{a}{2}$ relative to the cl... | a=0 \text{ or } 3+\sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,346 |
Example 4 Given that $a, b, c$ are positive integers, and the quadratic equation $a x^{2}+b x+c=0$ has two real roots whose absolute values are both less than $\frac{1}{3}$. Find the minimum value of $a+b+c$.
(2005, National High School Mathematics League, Fujian Province Preliminary Contest) | Let $x_{1}$ and $x_{2}$ be the roots of the equation $a x^{2} + b x + c = 0$. By Vieta's formulas, we have
$$
x_{1} + x_{2} = -\frac{b}{a}, \quad x_{1} x_{2} = \frac{c}{a}.
$$
Thus, $x_{1}9$.
Therefore, the roots $x_{1}$ and $x_{2}$ of $a x^{2} + b x + c = 0$ are in the interval $\left(-\frac{1}{3}, 0\right)$. Hence, w... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,348 |
Example 5 Let $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$, and the equation
$$
\cos 2 x-4 a \cos x-a+2=0
$$
has two distinct solutions. Try to find the range of values for $a$.
(2007, National High School Mathematics Competition Gansu Province Preliminary) | Explanation: From the problem, we have
$$
2 \cos ^{2} x-1-4 a \cos x-a+2=0,
$$
which simplifies to $2 \cos ^{2} x-4 a \cos x-a+1=0$.
Let $t=\cos x$. Given $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$, we know $0 \leqslant t \leqslant 1$.
Thus, the original equation having two distinct solutions is equivalent t... | \frac{3}{5} \leqslant a < 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,349 |
Proposition $P$ is any point inside the tetrahedron $A_{1} A_{2} A_{3} A_{4}$. The line $A_{1} P$ intersects the sphere determined by $P, A_{2}, A_{3}, A_{4}$ at point $B_{1}$. The line $A_{2} P$ intersects the sphere determined by $P, A_{1}, A_{3}, A_{4}$ at point $B_{2}$. The line $A_{3} P$ intersects the sphere dete... | Proof: It is easy to see that
$$
v_{1} \overrightarrow{P A_{1}}+v_{2} \overrightarrow{P A_{2}}+v_{3} \overrightarrow{P A_{3}}+v_{4} \overrightarrow{P A_{4}}=\mathbf{0} \text {. }
$$
Let $\lambda=\frac{A_{1} P}{P B_{1}}$. Then $\overrightarrow{P A_{1}}=-\lambda \overrightarrow{P B_{1}}$, substituting into equation (3) ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,350 |
Let the four-digit number $w_{1}=\overline{a b c d}$ be a perfect square. By splitting it in the middle, we get two two-digit numbers $x_{1}=\overline{a b}$ and $y_{1}=\overline{c d} ; w_{2}=3 x_{1} y_{1}+1$ is a perfect square. By splitting $w_{2}$ in the middle, we get two two-digit numbers $x_{2}$ and $y_{2} ; w_{3}... | Solution: From the fact that 9 times $w_{4}$ is the four-digit number $w_{5}$, and $w_{5}$ is also a perfect square, we have
$$
\begin{array}{l}
w_{4}=33^{2}=1089, \\
w_{5}=9 \times 33^{2}=99^{2}=9801 .
\end{array}
$$
From $w_{4}=x_{3} y_{3}+1$ being the perfect square 1089, we have
$$
\begin{array}{l}
1089-1=33^{2}-1... | w_{1}=1156 \text{ or } 1444, w_{2}=1849, w_{3}=1764, w_{4}=1089, w_{5}=9801 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,351 |
Given $\triangle A B C$ with its incircle touching side $B C$ at point $D, A D$ intersects the incircle again at point $E$, and the tangent line through $E$ intersects $A B$ and $A C$ at points $F$ and $G$ respectively. Prove:
$$
\frac{1}{A B}+\frac{1}{A F}=\frac{1}{A C}+\frac{1}{A G} .
$$ | Proof: As shown in Figure 3, let the incircle touch side $AB$ at point $T$, and connect $TD$ and $TE$.
By the area relationship, we have
$$
\begin{array}{c}
\frac{AF}{FT}=\frac{S_{\triangle AFE}}{S_{\triangle FTE}} \\
=\frac{AE \sin \theta}{TE \sin \alpha}, \\
\frac{BT}{AB}=\frac{S_{\triangle BTD}}{S_{\triangle BAD}}=\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,352 |
As shown in Figure 4, the convex quadrilateral $ABCD$ is circumscribed around $\odot O$. The lines containing the pairs of opposite sides intersect at points $E$ and $F$, and the diagonals intersect at point $G$. Prove that $OG \perp EF$.
Translate the above text into English, preserving the original text's line break... | Proof: As shown in Figure 4, let the points of tangency of circle $\odot O$ with sides $AB$, $BC$, $CD$, and $DA$ be $M$, $N$, $P$, and $Q$ respectively, and let $OE$ intersect $PM$ at point $H$. The radius of $\odot O$ is $r$.
By Newton's theorem, $AC$, $BD$, $PM$, and $QN$ intersect at point $G$.
It is easy to prove... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,353 |
Example 4 As shown in Figure 4, on the hypotenuse $AB$ of the right triangle $\triangle ABC$, take $BP = BC, AQ = AC$.
Construct $PM \perp BC$ at point $M$, $QN \perp AC$ at point $N$, and $PM$ intersects $QN$ at point $G$. Let the perimeter of rectangle $GMNC$ be $p_{1}$, and the perimeter of $\triangle PQG$ be $p_{2}... | Let $BC = a$, $CA = b$, $AB = c$. Then
$$
PQ = a + b - c.
$$
From $\triangle PQG \sim \triangle ABC$, we get
$$
\begin{array}{l}
QG = \frac{a}{c} \cdot PQ = \frac{a}{c}(a + b - c), \\
PG = \frac{b}{c} \cdot PQ = \frac{b}{c}(a + b - c).
\end{array}
$$
Thus, $p_2 = PQ + QG + PG$
$$
= (a + b - c)\left(1 + \frac{a}{c} + ... | \frac{4}{5}<\frac{p_{1}}{p_{2}}<1 | Geometry | proof | Yes | Yes | cn_contest | false | 721,354 |
252 Given that $a, b, c$ are positive real numbers satisfying $abc=1$. Prove:
$$
\frac{1}{a(a+b)}+\frac{1}{b(b+c)}+\frac{1}{c(c+a)} \geqslant \frac{3}{2} .
$$ | Prove: Let $a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z}(x, y, z$ be positive real numbers). Then the original inequality is equivalent to
$$
\frac{x^{2}}{y^{2}+z x}+\frac{y^{2}}{z^{2}+x y}+\frac{z^{2}}{x^{2}+y z} \geqslant \frac{3}{2} \text {. }
$$
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{\left[\left... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,355 |
Example 5 Two right trapezoids have their oblique sides and upper bases both equal to 1, and their acute angles are complementary. Their areas are $S_{1}$ and $S_{2}$, respectively. Prove: $1<S_{1}+S_{2}<2$.
---
The above text has been translated into English, maintaining the original text's line breaks and format. | Explanation: By combining two right trapezoids into the form shown in Figure 5, extend $AD$ and $EF$ to intersect at point $K$. It is easy to see that quadrilateral $ABEK$ is a rectangle. Extend $DC$ to intersect $BE$ at point $H$, then $CH \perp BE$. Let $BH = x$ and $CH = y$. Then $x^2 + y^2 = 1$. It is easy to see t... | 1 < S_1 + S_2 < 2 | Geometry | proof | Yes | Yes | cn_contest | false | 721,356 |
Example 6 In the right triangle $\triangle ABC$, $\angle C=90^{\circ}$, the incircle $\odot O$ touches $AC$, $BC$ at points $E$, $F$, respectively. The rays $BO$, $AO$ intersect the line $EF$ at points $M$, $N$. Prove: $\frac{1}{5}<\frac{S_{\triangle OMN}}{S_{\triangle ABC}}<\frac{1}{4}$. | As shown in Figure 6, let $BC = a$, $CA = b$, $AB = c$, and connect $OE$ and $AM$. Then
$$
OE \perp AC.
$$
From the conditions shown in the figure, it is easy to see that
$$
\begin{array}{l}
\angle AEM = \angle CEF = 45^{\circ}. \\
\text{Also, } \angle AOM = \frac{1}{2}(\angle CAB + \angle CBA) = 45^{\circ},
\end{arra... | \frac{1}{5} < \frac{S_{\triangle OMN}}{S_{\triangle ABC}} < \frac{1}{4} | Geometry | proof | Yes | Yes | cn_contest | false | 721,357 |
1. In Rt $\triangle ABC$, $\angle C=90^{\circ}$, $O$ is the incenter, $OE \perp BC$ at $E$, $OF \perp AC$ at $F$. Extend $AC$ to $P$, extend $BC$ to $Q$, such that $AP=BQ=AB$. Try to compare the area of $\triangle CPQ$ with the area of quadrilateral $OECF$. | (提示: Let $B C=a, C A=b, A B=c$. Then
$$
\begin{array}{l}
S_{\triangle C P Q}=\frac{1}{2}(c-a)(c-b) \\
=\frac{1}{2} a b-\frac{1}{2} c(a+b-c) .
\end{array}
$$
Using the equivalent relationship to substitute $a b$ in the formula, we get $\left[\frac{1}{2}(a+b-c)\right]^{2}$. Therefore, $\left.S_{\triangle C P Q}=S_{\text... | S_{\triangle C P Q}=S_{\text {quadrilateral OECF }} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,358 |
2. On the hypotenuse $AB$ of the right triangle $\triangle ABC$, take $BP = BC$, $AQ = AC$. Construct $PM \perp BC$ at point $M$, $QN \perp AC$ at point $N$, and let $PM$ intersect $QN$ at point $G$. Try to compare the area of rectangle $GMCN$ with the area of $\triangle PQG$. | From Fig. 4, we know
$$
\begin{array}{l}
C M \cdot C N=\left(a-\frac{a^{2}}{c}\right)\left(b-\frac{b^{2}}{c}\right) \\
=\frac{a b}{c^{2}}(c-a)(c-b) .
\end{array}
$$
And $S_{\triangle P Q G}=\frac{1}{2} Q G \cdot P G$
$$
\begin{array}{l}
=\frac{1}{2}(a+b-c) \frac{a}{c}(a+b-c) \frac{b}{c} \\
=\frac{a b}{c^{2}} \cdot \fr... | S_{\text {quadrilateral } C M C N}=S_{\triangle P Q G} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,359 |
3. In the acute triangle $\triangle C E F$, $\angle E C F=45^{\circ}, C D$ is the altitude. Prove that $1<\frac{C D}{E F}<\frac{5}{4}$. | (Prompt: Extend $F E$ to point $A$, $E F$ to point $B$, such that $\angle E C A = \angle E C D$, $\angle F C B = \angle F C D$, then $\triangle A B C$ is a right triangle. Let the three sides of $\triangle A B C$ be $a$, $b$, $c$. It is easy to see that
$$
\begin{array}{l}
E F = a + b - c, \\
C D = \frac{a b}{c} = \fra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,360 |
4. Convex quadrilateral $A B C D$ is inscribed in $\odot O$, its perimeter is $l$, $A B \parallel D C$ and $\overparen{A B}+\overparen{D C}=\overparen{B C}+\overparen{A D}$. Given that the radius of $\odot O$ is 10. Prove: $48<l<57$. | (It is easy to get $\overparen{B C}=\overparen{A D}=90^{\circ}$. As shown in Figure 7, draw $M N \perp A B$ through point $O$, with the feet of the perpendicular $M$ and $N$ being the midpoints of $A B$ and $D C$ respectively. It is easy to prove that Rt $\triangle O B M \cong$ Rt $\triangle C O N$. From $B M+O M > O B... | 48<l<57 | Geometry | proof | Yes | Yes | cn_contest | false | 721,361 |
Example 6 In the right triangle $\triangle ABC$, it is known that $CD$ is the altitude on the hypotenuse $AB$, and $O$, $O_{1}$, $O_{2}$ are the points of intersection of the angle bisectors of $\triangle ABC$, $\triangle ACD$, and $\triangle BCD$ respectively. Prove:
(1) $O_{1}O \perp CO_{2}$;
(2) $OC=O_{1}O_{2}$.
(19... | Proof: (1) Method 1.
As shown in Figure 10, given that $O_{1}$ and $O$ are both on the angle bisector of $\angle A$, let this bisector intersect $\mathrm{CO}_{2}$ at point $E$.
Since $\angle A = \angle D C B$, then $\angle E A C = \angle O_{2} C B$.
Thus,
$$
\begin{aligned}
& \angle E A C + \\
& \angle A C E \\
= & \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,364 |
II. (50 points) If all the coefficients of a polynomial are natural numbers, it is called a "natural polynomial". For a positive integer $n$, let $A(n)$ denote the number of different natural polynomials $P(x)$ satisfying $P(2)=n$. Prove: $\lim _{n \rightarrow \infty} \frac{\log _{2} A(n)}{\left(\log _{2} n\right)^{2}}... | First, we prove that for any positive integer $m$, we have
$A(2 m+1)=A(2 m)$
$$
=A(2 m-1)+A(m) \text {. }
$$
In fact, for any natural polynomial $P(x)$ satisfying $P(2)=2 m+1$, since $P(2)$ is odd, the constant term of $P(x)$ is odd. Let $Q(x)=P(x)-1$. Then $Q(x)$ is a natural polynomial, and
$$
Q(2)=P(2)-1=2 m \text ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,365 |
Three. (50 points) Each square of an $m \times n(m, n \geqslant 5)$ chessboard is randomly colored either black or white. Each operation consists of changing the color of five consecutive squares in the same row, column, or diagonal to the opposite color. Can all the squares be changed to the opposite color after a fin... | When $5 \mid m n$, the goal can be achieved; when $5 \nmid m n$, the goal cannot be achieved.
(1) If $5 \mid m n$, by the fact that 5 is a prime number, we can assume without loss of generality that $5 \mid n$. Then the $m \times n$ chessboard can be divided into several $1 \times 5$ rectangles, and operating on each $... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,366 |
(1) Find the minimum value of the smallest side length of a peculiar triangle;
(2) Prove that there are infinitely many isosceles peculiar triangles;
(3) How many non-isosceles peculiar triangles are there? | (1) Let $a, b, c (a \leqslant b \leqslant c)$ be the side lengths of a peculiar triangle. Then, by Heron's formula, we have
$$
\begin{aligned}
16 \Delta^{2}= & (a+b+c)(a+b-c) . \\
& (a-b+c)(-a+b+c) .
\end{aligned}
$$
Since $(a, b, c)=1$, at least one of $a, b, c$ must be odd. If there is an odd number of odd numbers a... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,367 |
In $\triangle A B C$, points $E$ and $F$ are both on the large side $B C$, with $B E = B A$ and $C F = C A$. Also, $D E \parallel A B$ and $D F \parallel A C$. Prove that the circumcircles of $\triangle A B F$, $\triangle A C E$, and $\triangle D E F$ intersect at one point. | Proof: Since the base angles of an isosceles triangle are acute, we have:
$$
\begin{array}{l}
\angle B E A = \angle B A E < 90^{\circ}, \\
\angle C F A < 90^{\circ}, \\
\angle E A F < \angle B A E < 90^{\circ}.
\end{array}
$$
Therefore, $\triangle A E F$ is an acute triangle. Let its circumcenter be $O$, then $O$ must... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,368 |
As shown in Figure 4, let quadrilateral $ABCD$ be a square, $E$ be any point on side $AD$ (not coinciding with $A$ or $D$), and $EC$ intersect $BD$ at point $F$. Let $x$, $y$, and $z$ be the areas of the shaded regions.
(1) Try to find the equation that $x$, $y$, and $z$ satisfy;
(2) If the values of $x$ and $y$ are gi... | (1) There must be $x+y=z$. The reason is as follows.
From $z=S_{\triangle F B C}=S_{\triangle E B C}-S_{\triangle E B F}$
$$
=S_{\triangle A B D}-S_{\triangle E B F}=S_{\triangle D E F}+S_{\triangle A B E}=x+y,
$$
Therefore, the three numbers $x, y, z$ should satisfy the equation $x+y=z$.
(2) From $B C / / E D$, we ge... | \frac{D E}{A E}=\frac{1+\sqrt{k+1}}{k} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,369 |
Let $O$ be any point inside $\triangle ABC$, and extend $AO$, $BO$, $CO$ to intersect the circumcircles of $\triangle BOC$, $\triangle COA$, $\triangle AOB$ at points $A'$, $B'$, $C'$ respectively. Prove:
$$
\frac{OA}{OA'} + \frac{OB}{OB'} + \frac{OC}{OC'} \geqslant \frac{3}{2}.
$$ | Proof: As shown in Figure 5, let
$$
\begin{array}{l}
\angle B O A^{\prime}=\alpha, \\
\angle C O A^{\prime}=\beta, \\
\angle C O B^{\prime}=\gamma .
\end{array}
$$
By Ptolemy's theorem, we have
$$
\begin{array}{l}
O A^{\prime} \cdot B C \\
=O B \cdot A^{\prime} C+O C \cdot A^{\prime} B .
\end{array}
$$
In $\triangle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,370 |
Let $\left\{a_{n}\right\}$ be a positive arithmetic sequence, $n$ and $p$ be positive integers, and $p>1$. Prove:
$$
\prod_{i=1}^{n}\left(1+\sqrt[p]{a_{i}}\right) \geqslant\left(1+\sqrt[2p]{a_{1} a_{n}}\right)^{n} .
$$ | $$
\begin{array}{l}
a_{i} a_{n+1-i}=\left[a_{1}+(i-1) d\right]\left[a_{1}+(n-i) d\right] \\
=a_{1}^{2}+(n-1) a_{1} d+(i-1)(n-i) d^{2} \\
\geqslant a_{1}^{2}+(n-1) a_{1} d \\
=a_{1}\left[a_{1}+(n-1) d\right]=a_{1} a_{n}, \\
\quad\left(1+\sqrt[p]{a_{i}}\right)\left(1+\sqrt[p]{a_{n+1-i}}\right) \\
\quad=1+\left(\sqrt[p]{a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,371 |
Example 7 As shown in Figure 11, let $P$ and $Q$ be two points on the hypotenuse $AB$ of the right triangle $\triangle ABC$, satisfying
$$
\begin{array}{l}
AQ = AC, \\
BP = BC.
\end{array}
$$
Take points $M$ and $N$ on $AC$ and $BC$ respectively, such that $CM = CN$ and equal to the altitude from $C$ to $AB$. Let $G$ ... | Proof: As shown in Figure 11, construct $CD \perp AB$ at point $D$. Then $\angle ACD = \angle B$, and $CD = CN$.
Given $AQ = AC$, we have
$\angle B + \angle DCQ = \angle ACD + \angle DCQ = \angle ACQ = \angle AQC = \angle B + \angle QCB$.
Thus, $\angle DCQ = \angle QCB$, which means $QC$ bisects $\angle DCB$.
Similarly... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,372 |
1. In trapezoid $A B C D$, it is known that $A D / / B C(B C>$ $A D), \angle D=90^{\circ}, B C=C D=12, \angle A B E=45^{\circ}$. If $A E=10$, then the length of $C E$ is $\qquad$
(2004, "Xinli Cup" National Junior High School Mathematics Competition) | (提示: Extend $D A$ to point $F$, complete the trapezoid into a square $C B F D$, then rotate $\triangle A B F$ $90^{\circ}$ around point $B$ to the position of $\triangle B C G$. It is easy to see that $\triangle A B E \cong \triangle G B E$. Therefore, $A E$ $=E G=E C+A F=10$. Let $E C=x$, then $A F=10-$ $x, D E=12-x, ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,373 |
2. In trapezoid $A B C D$, it is known that $A D / / B C, A D \perp$ $C D, B C=C D=2 A D, E$ is a point on side $C D$, $\angle A B E=45^{\circ}$. Then $\tan \angle A E B=$ $\qquad$
(2007, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | (提示: Extend $D A$ to point $F$, such that $A F=A D$, to get the square $B C D F$. Then extend $A F$ to point $G$, such that $F G=C E$. Then $\triangle B F G \cong \triangle B C E$. It is also easy to prove that $\triangle A B G \cong \triangle A B E$. Let $C E=x, B C=2 a$. Then $F G=x, A F=A D=a, A E=A G=x+a, D E=2 a-x... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,374 |
3. In $\triangle A B C$, it is known that $A B=c, A C=b, B C$ $=a$, the heights on sides $B C$ and $A C$ are $h_{a}$ and $h_{b}$, respectively, and $a \leqslant$ $h_{a}, b \leqslant h_{b}$. Find the degree measures of the three interior angles of $\triangle A B C$.
(1999, Tianjin City Junior High School Mathematics Com... | (In any triangle, there are $h_{a} \leqslant b$, $h_{b} \leqslant a$. It is also easy to know that $a \leqslant h_{a} \leqslant b \leqslant h_{b} \leqslant a$, which means $a=b=h_{a}$ $=h_{b}$. Therefore, $\triangle A B C$ is an isosceles right triangle. Hence $\angle A=$ $\left.\angle B=45^{\circ}, \angle C=90^{\circ}... | \angle A = 45^{\circ}, \angle B = 45^{\circ}, \angle C = 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,375 |
4. From vertex $A$ of square $A B C D$, draw two rays such that the angle between them is $45^{\circ}$, intersecting $B C$ and $C D$ at points $E$ and $F$, and intersecting $B D$ at points $P$ and $Q$. Prove that:
$S_{\triangle A E F}=2 S_{\triangle A P Q}$.
(1990, Sichuan Province Junior High School Mathematics Compet... | (提示: Connect $Q E$. It is easy to know that $A, B, E, Q$ are concyclic. Therefore, $\angle A Q E=90^{\circ}$, which means $E Q \perp A F$. Similarly, $P F \perp A E$. By property 5, $P Q$ bisects the area of $\triangle A E F$, so $\left.S_{\triangle A K F}=2 S_{\triangle A P(}.\right)$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,376 |
5. Let quadrilateral $A B C D$ be a square, points $M, N$ lie on sides $B C, C D$ respectively, and $\angle M A N=45^{\circ}$. Prove:
$$
\frac{S_{\text {square } A B C D}}{S_{\triangle M A N}}=\frac{2 A B}{M N} \text {. }
$$
(1997, Shanghai Junior High School Mathematics Competition) | (Draw $A H \perp M N$ at point $H$. It is easy to know that $A H = A B$. Also, $S_{\triangle M A N}=\frac{1}{2} M N \cdot A H=\frac{1}{2} M N \cdot A B$, thus, $\left.\frac{S_{\text {square } A B C D}}{S_{\triangle M A N}}=\frac{A B^{2}}{\frac{1}{2} M N \cdot A B}=\frac{2 A B}{M N}.\right)$ | \frac{2 A B}{M N} | Geometry | proof | Yes | Yes | cn_contest | false | 721,377 |
Example 1 As shown in Figure $1, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$.
(2008, National High School Mathematics Competition) | Explanation: By the symmetry of the parabola, we can assume
$$
P\left(2 t^{2}, 2 t\right)(t>0) \text {. }
$$
Since the equation of the circle is $x^{2}+y^{2}-2 x=0$, the equation of the chord of contact $M N$ is
$$
2 t^{2} x+2 t y-\left(x+2 t^{2}\right)=0,
$$
which simplifies to $\left(2 t^{2}-1\right) x+2 t y-2 t^{2... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,378 |
Example 2 As shown in Figure 2, through point $P$ on the line $l: 5 x-7 y-70=0$, draw the tangents $P M$ and $P N$ to the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, with the points of tangency being $M$ and $N$, respectively, and connect $M N$.
(1) When point $P$ moves on the line $l$, prove that the line $M N$ alwa... | Explanation: (1) Take any point on the line $l$
$$
P(7 t+7,5 t-5),
$$
Thus, the equation of the chord of tangents $M N$ of the ellipse with respect to $P$ is
$$
\begin{array}{l}
\frac{7 t+7}{25} \cdot x+\frac{5 t-5}{9} \cdot y-1=0 \\
\Leftrightarrow\left(\frac{7}{25} x+\frac{5}{9} y\right) t+\frac{7}{25} x-\frac{5}{9}... | Q\left(\frac{25}{14},-\frac{9}{10}\right) | Geometry | proof | Yes | Yes | cn_contest | false | 721,379 |
Example 3 Given that the parabola $y=-x^{2}+b x+c$ is tangent to the parabola $y=x^{2}$. Find the geometric position of the vertex of the parabola.
(17th All-Russian Mathematical Olympiad) | Explanation: Let the point of tangency of the two parabolas be $Q\left(x_{1}, y_{1}\right)$. Then their equations at the point of tangency are
$$
\begin{aligned}
\frac{y_{1}+y}{2} & =-x_{1} x+b \cdot \frac{x_{1}+x}{2}+c, \\
\frac{y_{1}+y}{2} & =x_{1} x .
\end{aligned}
$$
Since the two parabolas are tangent to each oth... | \left(\frac{b}{2}, \frac{b^{2}}{8}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,380 |
Property 4 In the acute triangle $\triangle A M N$, $\angle A=45^{\circ}$, $A H \perp M N$ at point $H$. Then
$$
\frac{A H^{2}+M H^{2}}{A H^{2}+N H^{2}}=\frac{A H+M H}{A H+N H} .
$$ | Proof of Property 4: As shown in Figure 2, let $A H = h, M H =$
$$
\begin{array}{l}
m, H N = n, \angle M A H = \alpha. \text{ Then } \\
\angle N A H = 45^{\circ} - \alpha, m = h \tan \alpha, \\
n = h \tan \left(45^{\circ} - \alpha\right). \\
\text{ Therefore, } \frac{A H^{2} + M H^{2}}{A H^{2} + N H^{2}} = \frac{h^{2} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,381 |
Example 4 As shown in Figure 3, given the parabola $x^{2}=4 y$ and a fixed point $P(0,8)$, $A$ and $B$ are two moving points on the parabola, and $\overrightarrow{A P}=\lambda \overrightarrow{P B}$ $(\lambda>0)$. Tangent lines to the parabola are drawn through points $A$ and $B$, and their intersection point is $M$.
(1... | (1) From $\overrightarrow{A P}=\lambda \overrightarrow{P B}(\lambda>0)$, we know that points $A$, $P$, and $B$ are collinear, and point $P$ lies between $A$ and $B$:
Let $M\left(x_{0}, y_{0}\right)$. Then the equation of the chord of contact $A B$ is
$$
x_{0} x=2\left(y_{0}+y\right) \text {. }
$$
Substituting the coor... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,382 |
Example 5 As shown in Figure 4, $AB$ is the major axis of an ellipse, $O$ is the center, $F$ is a focus, $P$ is a point on the ellipse, $CD$ is a chord passing through $O$ and parallel to the tangent at $P$, and the line $PF$ intersects $CD$ (or its extension) at point $Q$. Prove or disprove that $PQ=OA=OB$. | Given the equation of an ellipse:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$
Let $P(a \cos \theta, b \sin \theta)$. The equation of the tangent line to the ellipse at point $P$ is
$$
\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1,
$$
which can be rewritten as $x b \cos \theta+y a \sin \theta... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,383 |
Example 6 As shown in Figure 5, let $M$ be the midpoint of a chord $UV$ of an ellipse. Draw chords $AB$ and $CD$ through $M$. Let $AC$ and $BD$ intersect $UV$ at points $P$ and $Q$ respectively. Prove that $M$ is also the midpoint of segment $PQ$.
(24th Putnam Mathematical Competition) | Let $M\left(x_{0}, y_{0}\right)$. Then the equation of the ellipse is
$$
\begin{array}{l}
a x^{2}+b y^{2}=c\left(a, b, c \in \mathbf{N}_{+}\right), \\
l_{A B}: a_{1}\left(x-x_{0}\right)+b_{1}\left(y-y_{0}\right)=0, \\
l_{C D}: a_{2}\left(x-x_{0}\right)+b_{2}\left(y-y_{0}\right)=0,
\end{array}
$$
The equation of the ch... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,384 |
Example 7 As shown in Figure 6, a broken line with a finite number of segments is inscribed in a parabola, with its starting point coinciding with the vertex of the parabola. Any two segments of the broken line sharing a common vertex form equal angles with the tangent line of the parabola at that point. Prove: Such a ... | Suppose the parabola is $y=a x^{2}(a>0)$. Take three consecutive vertices $A_{i}\left(x_{i}, a x_{i}^{2}\right)$ $(i=n, n+1, n+2, n \in \mathbf{N})$ on the broken line.
From the tangent line equation of the parabola at point $A_{n+1}$ (or by finding the derivative), we know its slope
$$
\begin{array}{l}
k=2 a x_{n+1},... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,385 |
1. For a point $P$ on an ellipse, let $d$ be the distance from the center of the ellipse to the tangent line through point $P$. Prove: When $P$ moves along the ellipse, $P F_{1} \cdot P F_{2} d^{2}$ is a constant, where $P F_{1}$ and $P F_{2}$ are the distances from point $P$ to the foci $F_{1}$ and $F_{2}$ of the elli... | (Hint: From the ellipse equation $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$ $(a>b>0)$ and the tangent line equation, using the point-to-line distance formula and the distance formula between two points, we know $P F_{1} \cdot P F_{2} d^{2}=$ $a^{2} b^{2}$ (a constant). ) | a^{2} b^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 721,386 |
2. If $A, B$ are two different points on the parabola $y^{2}=4 x$, and the perpendicular bisector of the chord $A B$ (not parallel to the $y$-axis) intersects the $x$-axis at point $P$, then the chord $A B$ is called a "related chord" of point $P$. It is known that when $x>2$, point $P(x, 0)$ has infinitely many relate... | (Hint: (1) Use the equation of the chord through the midpoint to find the equation of the perpendicular bisector of the related chord $AB$. It is known that the x-coordinate of the midpoint $Q(x_1, y_1)$ of all related chords of point $P(x_0, 0)$ is the same and is $x_0 - 2$.
(2) Substitute the equation of the related ... | 2x_0 - 2 | Algebra | proof | Yes | Yes | cn_contest | false | 721,387 |
3. Find the equation of the parabola that is tangent to the $x$-axis and $y$-axis at points $(1,0)$ and $(0,2)$, respectively, and find the axis of symmetry and the coordinates of the vertex of the parabola. | (The origin $(0,0)$ is related to the chord of tangents of the parabola, which is the line $l: 2 x+y-2=0$ passing through the tangent points $(1,0)$ and $(0,2)$. It is easy to find that the equation of the parabola is
$$
(2 x-y)^{2}-4(2 x+y)+4=0 \text {. }
$$
Therefore, the axis of symmetry of the parabola is $10 x-5 ... | (2 x-y)^{2}-4(2 x+y)+4=0, 10 x-5 y-6=0, \left(\frac{16}{25}, \frac{2}{25}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,388 |
4. Let the two tangents to the parabola $y^{2}=4 a x$ intersect at an angle $\theta$ (where $\theta$ is a constant). Find the equation of the locus of the vertex of the angle.
| Let the vertex of the angle be \( P\left(x_{0}, y_{0}\right) \), and the points of tangency be \( A\left(a t^{2}, 2 a t\right) \) and \( B\left(a s^{2}, 2 a s\right) \). Substituting these into the equation of the chord of contact \( A B \), we get \( y_{0} y=2 a\left(x_{0}+x\right) \).
Thus, \( t \) and \( s \) are t... | x^{2} \sin^{2} \theta + 2 a\left(1 + \cos^{2} \theta\right) x - y^{2} \cos^{2} \theta + a^{2} \sin^{2} \theta = 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,389 |
5. There is a moving line $y=\frac{2}{t}(x-t-1)$ (where $t$ is a parameter), and a parabola $y^{2}=4 x$. What is the locus of the midpoint of the line segment joining the two points of intersection of the line and the parabola? Prove that this locus is tangent to the line $x+y+1=0$.
---
The translation is provided as... | (Tip: Let the midpoint be $P\left(x_{0}, y_{0}\right)$. Then the equation of the chord with midpoint $P$ for the parabola $y^{2}=4 x$ is
$$
y_{0} y-2\left(x_{0}+x\right)=y_{0}^{2}-4 x_{0} \text {. }
$$
This line is the same as the given moving line. By comparing the coefficients of corresponding terms and eliminating ... | 2 x-1=(y+1)^{2} | Algebra | proof | Yes | Yes | cn_contest | false | 721,390 |
Example 6 Given that when $x \in[0,1]$, the inequality
$$
x^{2} \cos \theta-x(1-x)+(1-x)^{2} \sin \theta>0
$$
always holds. Try to find the range of $\theta$.
(1999, National High School Mathematics Competition) | Let
$$
f(x)=x^{2} \cos \theta-x(1-x)+(1-x)^{2} \sin \theta \text {. }
$$
To ensure that $f(x)>0$ for all $x \in[0,1]$, it is sufficient that $f(x)_{\text {min }}>0$ for $x \in[0,1]$.
Notice that
$$
\begin{array}{c}
f(x)=(1+\sin \theta+\cos \theta) x^{2}- \\
(1+2 \sin \theta) x+\sin \theta \\
=(1+\sin \theta+\cos \thet... | 2 k \pi+\frac{\pi}{12}<\theta<2 k \pi+\frac{5 \pi}{12}(k \in \mathbf{Z}) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,391 |
Property 5 As shown in Figure 3, in the acute triangle $\triangle A M N$, $\angle A=$ $45^{\circ}$, draw $M E \perp A N$ at point $E$, $N F \perp A M$ at point $F$. Then $E F$ bisects the area of $\triangle A M N$, that is, $S_{\triangle A E F}=S_{\text {quadrilateral MNEF }}$. | Proof of Property 5: From $\angle M A N=45^{\circ}$, we get
$$
\angle A M E=45^{\circ}, \angle A N F=45^{\circ} \text {. }
$$
Thus, $A E=E M, A F=F N$, and the acute angle formed by $M E$ and $F N$ is $45^{\circ}$. Therefore,
$$
\begin{array}{l}
S_{\triangle A E F}=\frac{1}{2} A E \cdot A F \sin 45^{\circ} \\
=\frac{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,392 |
Example 7 Let the function $f(x)=a x^{2}+8 x+3(a<0)$. For a given negative number $a$, there is a largest positive number $l(a)$ such that the inequality $|f(x)| \leqslant 5$ holds for the entire interval $[0, l(a)]$. Question: For what value of $a$ is $l(a)$ the largest? Find this largest $l(a)$ and prove your conclus... | Notice that
$$
f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}
$$
has the axis of symmetry at $x=-\frac{4}{a}$.
Therefore, when $x=-\frac{4}{a}$, $f(x)_{\text {max }}=3-\frac{16}{a}$.
Combining the graph, we can consider two cases: $3-\frac{16}{a}>5$ and $3-\frac{16}{a} \leqslant 5$, and find the maximum positive ... | \frac{\sqrt{5}+1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,393 |
Example 8 Quadratic Function
$$
f(x)=a x^{2}+b x+c(a, b \in \mathbf{R} \text {, and } a \neq 0)
$$
satisfies the conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and
$$
f(x) \geqslant x \text {; }
$$
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\math... | Explanation: First, derive the analytical expression of $f(x)$ from the given conditions, then classify and discuss $m$ and $t$ to determine the value of $m$.
Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about the line $x=-1$. Therefore,
$$
-\frac{b}{2 a}=-1 \Rightarrow b=2 a \text {. }
$$
From condi... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,394 |
Example 9 Given the function $f(x)=a x^{2}+b x+c$, when $x \in[-1,1]$, $|f(x)| \leqslant 1$. Prove: When $x \in$ $[-1,1]$, $|f(x)| \leqslant \frac{5}{4}$. | Explanation: Notice that in $|f(x)|=\left|a x^{2}+b x+c\right|$, there are three variables $a$, $b$, and $c$, and the problem only provides one condition: "when $x \in [-1,1]$, $|f(x)| \leqslant 1$". However, we can select appropriate values in $[-1,1]$, express the corresponding function values in terms of $a$, $b$, a... | \frac{5}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 721,395 |
1. Given points $A(0,4)$ and $B(4,0)$. If the parabola $y=x^{2}-m x+m+1$ intersects the line segment $A B$ (excluding endpoints $A$ and $B$) at two distinct points, then the range of values for $m$ is $\qquad$.
(1997, Shanghai High School Mathematics Competition) | Answer: $3<m<\frac{17}{3}$. | 3<m<\frac{17}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,396 |
A right angle rotates around its fixed vertex, which is inside a circle. Then, the intersection point of the two tangents to the circle passing through the intersection points of the sides of the angle with the circle traces a circle ${ }^{[1]}$. | First, provide the construction method:
(1) As shown in Figure 1, take the midpoint $O_{1}$ of $O M$, and draw a circle $\odot O_{1}$ with $O_{1}$ as the center and $O_{1} N$ as the radius.
(2) Extend
$P N$ to intersect $\odot O$ at point $Q$, connect $O_{1} Q$, and draw $P Q_{2} \parallel O_{1} Q$ to intersect the ex... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,397 |
Example 1 A $6 \times 6$ chessboard has 36 squares. Placing chess pieces in the squares, if there are 4 pieces of the same color in a straight line (horizontal, vertical, or at a $45^{\circ}$ diagonal), it is called a "four-in-a-row". Player A places white pieces, and Player B places black pieces. If A goes first, what... | Solution: At least 10 chess pieces need to be placed.
This is because, to prevent the four-in-a-row formation by the black pieces placed subsequently by player B, player A must place at least one white piece in each $1 \times 4$ rectangle, meaning that at least 8 white pieces must be placed in the $2 \times 4$ rectangl... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,398 |
Example 2 As shown in Figure 4, in a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected". Now, some of the 56 chess pieces are removed so that no 5... | Analysis: The difficulty ratio of this problem has significantly increased from 1, mainly due to its "apparent" asymmetry (the number of rows and columns are inconsistent). Therefore, the key to solving the problem is to find the symmetry hidden behind this asymmetry.
Inspired by Example 1, we will also start from the ... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,399 |
3. Let $S \subseteq \mathbf{R}$ be a set of real numbers, and call a pair of functions $(f, g)$ from $S$ to $S$ a "Spanish combination" if it satisfies the following conditions:
(i) Both functions are strictly increasing, i.e., for any $x, y \in S$, and $x<y$, we have
$$
f(x)<f(y), g(x)<g(y) \text {; }
$$
(ii) For any ... | 3. (1) Does not exist.
Let $g_{0}(x)=x, g_{k}(x)=\underbrace{g(g(\cdots g(x) \cdots))}_{k \uparrow}$.
Assuming in the set $\mathbf{N}_{+}$, there exists a Spanish pair $(f, g)$. From condition (i), we have $f(x) \geqslant x$, $g(x) \geqslant x$ for any $x \in \mathbf{N}_{+}$.
We will prove: For any non-negative integ... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,400 |
5. Given positive real numbers $a, b, c, d$ satisfying $a b c d=1$,
$$
a+b+c+d>\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \text { . }
$$
Prove: $a+b+c+d<\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}$. | 5. First prove: If $a b c d=1$, then $a+b+c+d$ does not exceed the weighted average of $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$ and $\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}$.
By the AM-GM inequality, we have
$$
\begin{array}{l}
a=\sqrt[4]{\frac{a^{4}}{a b c d}}=\sqrt[4]{\frac{a}{b} \cdot \frac{a}{b} \c... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,402 |
Property 6 In the acute triangle $\triangle A M N$, draw $M E \perp A N$ at point $E$, $N F \perp A M$ at point $F$, and let $K$ be the circumcenter of $\triangle A M N$. Then $\angle A=45^{\circ}$ if and only if $K$ is the orthocenter of $\triangle A E F$. | Proof of Property 6: As shown in Figure 4, let $M E$ and $F N$ intersect at point $C$. Then $C$ is the orthocenter of $\triangle A M N$. Connect $A C$, $A K$, $M K$, and $N K$. Then $A$, $F$, $C$, and $E$ are concyclic.
By the properties of the orthocenter and circumcenter of a triangle, we know
$$
\angle F A K = \ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,403 |
6. Let the function $f: \mathbf{R} \rightarrow \mathbf{N}_{+}$ satisfy that for any $x, y \in$ $\mathbf{R}$, we have
$$
f\left(x+\frac{1}{f(y)}\right)=f\left(y+\frac{1}{f(x)}\right) \text {. }
$$
Prove: there exists a positive integer that does not belong to the range of $f$. | 6. Assume the conclusion is not correct. Then $f(\mathbf{R})=\mathbf{N}_{+}$.
To derive a contradiction, we first prove some properties of the function $f$.
We can assume $f(0)=1$.
Indeed, let $a \in \mathbf{R}$ such that $f(a)=1$.
Consider the function $g(x)=f(x+a)$.
Substitute $x+a$ and $y+a$ for $x$ and $y$ in equa... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,404 |
7. Prove: For any positive real numbers $a, b, c, d$, we have
$$
\begin{array}{l}
\frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+ \\
\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b} \geqslant 0,
\end{array}
$$
and determine the conditions under which equality holds. | 7. Let \( A = \frac{(a-b)(a-c)}{a+b+c}, B = \frac{(b-c)(b-d)}{b+c+d} \),
\[
C = \frac{(c-d)(c-a)}{c+d+a}, D = \frac{(d-a)(d-b)}{d+a+b} \text{. }
\]
Then \( 2A = A' + A'' \), where,
\[
A' = \frac{(a-c)^2}{a+b+c}, A'' = \frac{(a-c)(a-2b+c)}{a+b+c} \text{. }
\]
Similarly, we have
\[
2B = B' + B'', 2C = C' + C'', 2D = D'... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,405 |
1. Given non-zero real numbers $a$ and $b$ satisfy
$$
|2 a-4|+|b+2|+\sqrt{(a-3) b^{2}}+4=2 a \text {. }
$$
then $a+b$ equals ( ).
(A) -1
(B) 0
(C) 1
(D) 2 | - 1. C
From the given, we know $a \geqslant 3$, so the given equation is
$$
|b+2|+\sqrt{(a-3) b^{2}}=0 \Rightarrow a+b=1 \text {. }
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,406 |
2. As shown in Figure 1, the side length of rhombus $A B C D$ is $a$, and $O$ is a point on the diagonal $A C$, with $O A=a, O B=$ $O C=O D=1$. Then $a$ equals ( ).
(A) $\frac{\sqrt{5}+1}{2}$
(B) $\frac{\sqrt{5}-1}{2}$
(C) 1
(D) 2 | 2. A. Since $\triangle B O C \sim \triangle A B C$, we have $\frac{B O}{A B}=\frac{B C}{A C}$, which means
$$
\frac{1}{a}=\frac{a}{a+1} \Rightarrow a^{2}-a-1=0 .
$$
Given $a>0$, solving yields $a=\frac{1+\sqrt{5}}{2}$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,407 |
3. A fair cube die, with faces numbered $1, 2, 3, 4, 5, 6$, is thrown twice. Let the number on the first throw be $a$, and the number on the second throw be $b$. Then, the probability that the system of equations
$$
\left\{\begin{array}{l}
a x+b y=3, \\
x+2 y=2
\end{array}\right.
$$
has only positive solutions for $x$... | 3. D.
When $2 a-b=0$, the system of equations has no solution.
When $2 a-b \neq 0$, the solution to the system of equations is
$$
(x, y)=\left(\frac{6-2 b}{2 a-b}, \frac{2 a-3}{2 a-b}\right) \text {. }
$$
From the given information, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{6-2 b}{2 a-b}>0, \\
\frac{2... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,408 |
4. As shown in Figure 2, in the right trapezoid $A B C D$, $A B / /$ $D C, \angle B=90^{\circ}$, a moving point $P$ starts from point $B$ and moves along the sides of the trapezoid in the order $B \rightarrow C \rightarrow D \rightarrow A$. Let the distance traveled by $P$ be $x$, and the area of $\triangle A B P$ be $... | 4. B.
According to the image, we get $B C=4, C D=5, D A=5$. Then we find $A B=8$. Therefore, $S_{\triangle A B C}=\frac{1}{2} \times 8 \times 4=16$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,409 |
5. The number of integer solutions $(x, y)$ for the equation $x^{2}+x y+2 y^{2}=29$ is $(\quad)$.
(A) 2
(B) 3
(C) 4
(D) infinitely many | 5. C.
Transform the original equation into a quadratic equation in terms of $x$, which is $x^{2}+y x+\left(2 y^{2}-29\right)=0$.
Since this equation has integer roots, the discriminant $\Delta \geqslant 0$, and it must be a perfect square, i.e.,
$$
\Delta=y^{2}-4\left(2 y^{2}-29\right)=-7 y^{2}+116 \geqslant 0 \text {... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,410 |
6. A bicycle tire, if installed on the front wheel, will wear out after the bicycle has traveled $5000 \mathrm{~km}$; if installed on the rear wheel, it will wear out after the bicycle has traveled $3000 \mathrm{~km}$. After traveling a certain distance, the front and rear tires can be swapped. If the front and rear ti... | II. 6.3750.
Let the total wear of each new tire when it is scrapped be $k$. Then the wear per $1 \mathrm{~km}$ for a tire installed on the front wheel is $\frac{k}{5000}$, and the wear per $1 \mathrm{~km}$ for a tire installed on the rear wheel is $\frac{k}{3000}$. Suppose a pair of new tires travels $x \mathrm{~km}$ b... | 3750 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,411 |
7. Given that the midpoint of line segment $A B$ is $C$, a circle is drawn with point $A$ as the center and the length of $A B$ as the radius. On the extension of line segment $A B$, take point $D$ such that $B D = A C$; then, with point $D$ as the center and the length of $D A$ as the radius, draw another circle, whic... | 7. $\frac{1}{3}$.
As shown in Figure 7, extend $A D$ to intersect $\odot D$ at point $E$, and connect $A F, E F$.
From the given conditions, we have
$$
\begin{array}{l}
A C=\frac{1}{3} A D, \\
A B=\frac{1}{3} A E .
\end{array}
$$
In $\triangle F H A$ and $\triangle E F A$,
$$
\begin{array}{l}
\angle E F A=\angle F H ... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,412 |
8. Given that $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five distinct integers satisfying the condition
$$
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=9
$$
If $b$ is an integer root of the equation
$$
\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)\left(x-a_{4}\right)\left(x-a_{5}\right)=2009
$$
then the value of $b$ is... | 8. 10 .
Notice
$$
\left(b-a_{1}\right)\left(b-a_{2}\right)\left(b-a_{3}\right)\left(b-a_{4}\right)\left(b-a_{5}\right)=2009,
$$
and $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five different integers, so all $b-a_{1}, b-a_{2}, b-a_{3}, b-a_{4}, b-a_{5}$ are also five different integers.
$$
\begin{array}{l}
\text { Also, ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,413 |
Example 1 Let $P$ be any point on the hypotenuse $AB$ of an isosceles right $\triangle ACB$, $PE \perp AC$, $PF \perp BC$ intersect at points $E$ and $F$ respectively, $PG \perp EF$ at point $G$, extend $GP$ and take a point $D$ on its extension such that $PD = PC$. Prove:
$$
BC \perp BD \text{, and } BC = BD \text{. }... | Proof: As shown in Figure 5, from the given conditions, we have
$$
\begin{array}{l}
\angle E P G=\angle E F P \\
=\angle C P F .
\end{array}
$$
Then $\angle D P B$
$$
\begin{aligned}
= & \angle A P G \\
& =45^{\circ}+\angle E P G \\
= & 45^{\circ}+\angle C P F \\
= & \angle B P F+\angle C P F \\
= & \angle B P C .
\en... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,414 |
9. As shown in Figure 4, in $\triangle A B C$, $C D$ is the altitude, and $C E$ is the angle bisector of $\angle A C B$. If $A C$ $=15, B C=20, C D$ $=12$, then the length of $C E$ is | 9. $\frac{60 \sqrt{2}}{7}$.
By the Pythagorean theorem, we know $A D=9, B D=16$.
Therefore, $A B=A D+B D=25$.
Hence, by the converse of the Pythagorean theorem, $\triangle A C B$ is a right triangle, and $\angle A C B=90^{\circ}$.
As shown in Figure 4, draw $E F \perp B C$, with the foot of the perpendicular at $F$. L... | \frac{60 \sqrt{2}}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,415 |
11. A. For the function $y=x^{2}+(2 k-1) x+k^{2}$, are the two intersection points with the $x$-axis both to the left of the line $x=1$? If so, please explain the reason; if not, find the range of values for $k$ when both intersection points are to the right of the line $x=1$. | Three, 11. A. Not necessarily.
For example, when $k=0$, the intersections of the function's graph with the $x$-axis are at $(0,0)$ and $(1,0)$, not both to the right of the line $x=1$.
Let the $x$-coordinates of the two intersections of the function with the $x$-axis be $x_{1}$ and $x_{2}$. Then $x_{1}+x_{2}=-(2 k-1),... | k<-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,417 |
11. B. Given the parabola $y=x^{2}$ and the moving line $y=(2 t-1) x-c$ have common points $\left(x_{1}, y_{1}\right) 、\left(x_{2}, y_{2}\right)$, and $x_{1}^{2}+x_{2}^{2}=t^{2}+2 t-3$.
(1) Find the range of real number $t$;
(2) For what value of $t$ does $c$ attain its minimum value, and find the minimum value of $c$. | 11. B. (1) By combining $y=x^{2}$ and $y=(2 t-1) x- c$, eliminating $y$ yields the quadratic equation
$$
x^{2}-(2 t-1) x+c=0
$$
with real roots $x_{1}$ and $x_{2}$, then $x_{1}+x_{2}=2 t-1, x_{1} x_{2}=c$.
Thus, $c=x_{1} x_{2}=\frac{1}{2}\left[\left(x_{1}+x_{2}\right)^{2}-\left(x_{1}^{2}+x_{2}^{2}\right)\right]$
$$
\b... | \frac{11-6 \sqrt{2}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,418 |
12. A. In the Cartesian coordinate system $x O y$, points with integer abscissas and ordinate values that are perfect squares are called "good points". Find the coordinates of all good points on the graph of the quadratic function $y=(x-90)^{2}-4907$. | 12. A. Let $y=m^{2},(x-90)^{2}=k^{2}(m, k$ are non-negative integers). Then,
$$
k^{2}-m^{2}=7 \times 701=1 \times 4907,
$$
i.e., $(k-m)(k+m)=7 \times 701=1 \times 4907$.
$$
\text { Then }\left\{\begin{array} { l }
{ k + m = 7 0 1 , } \\
{ k - m = 7 ; }
\end{array} \left\{\begin{array}{l}
k+m=4907, \\
k-m=1 .
\end{arr... | (444,120409),(-264,120409),(2544,6017209),(-2364,6017209) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,419 |
12. B. Let the positive integer $a$ satisfy $192 \mid\left(a^{3}+191\right)$, and $a<2$ 009. Find the sum of all possible positive integers $a$ that meet the condition. | 12. B. From the problem, we have $192 \mid\left(a^{3}-1\right)$.
$$
\begin{array}{l}
\text { Also, } 192=3 \times 2^{6}, \text { and } \\
a^{3}-1=(a-1)[a(a+1)+1] \\
=(a-1) a(a+1)+(a-1) .
\end{array}
$$
Since $a(a+1)+1$ is odd, we have
$$
2^{6}\left|\left(a^{3}-1\right) \Leftrightarrow 2^{6}\right|(a-1) \text {. }
$$
... | 10571 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,420 |
13. A. As shown in Figure 6, given an acute triangle $\triangle ABC$, with $BC < CA$, $AD$ and $BE$ are two altitudes of the triangle. A tangent line $l$ is drawn through point $C$ to the circumcircle of $\triangle ABC$. Perpendiculars are drawn from points $D$ and $E$ to line $l$, with the feet of the perpendiculars b... | 13. A. As shown in Figure 8, connect $D E$.
Since $\angle A D B$
$$
=\angle A E B=90^{\circ} \text {, }
$$
Therefore, $A, B, D, E$
are concyclic. Hence
$$
\begin{array}{l}
\angle C E D \\
=\angle A B C .
\end{array}
$$
Also, $l$ is the tangent line of $\odot O$ at point $C$, then
$$
\angle A C G=\angle A B C \text {.... | D F=E G | Geometry | proof | Yes | Yes | cn_contest | false | 721,421 |
13. B. Given that $AB$ is the diameter of $\odot O$, chord $DC \parallel AB$, and $DO$ is connected. A perpendicular line to $DO$ is drawn through point $D$, intersecting the extension of $BA$ at point $E$. A line parallel to $AC$ is drawn through point $E$, intersecting $CD$ at point $F$. A line parallel to $AC$ is dr... | 13. B. As shown in Figure 9, connect $A D$ and $B C$.
Since quadrilateral $A E F C$ is a parallelogram, we have
$$
A E = F C.
$$
Given that
$$
\begin{array}{c}
A D = C B, \\
\angle D A E = \angle B C F,
\end{array}
$$
therefore, $\triangle D A E \cong \triangle B C F$.
Thus, $\angle A D E = \angle C B F$.
Since $D E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,422 |
14. A. $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the following conditions:
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$
and the arithmetic mean of any $n-1$ different numbers among $a_{1}, a_{2}, \cdots, a_{n}$ is a positive integer. Find the maximum value of $n$. | 14. A. Let in $a_{1}, a_{2}, \cdots, a_{n}$, after removing $a_{i}(i=1$, $2, \cdots, n)$, the arithmetic mean of the remaining $n-1$ numbers is a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$
Therefore, for any $i, j(1 \leqslant i<j \leqslant n)$, we have
$$
b_{i... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,423 |
14. B. Given positive integers $x, y$ such that $\frac{4 x y}{x+y}$ is an odd number. Prove: There exists a positive integer $k$, such that $4 k-1$ divides $\frac{4 x y}{x+y}$. | 14. B. Let \( x=2^{s} a, y=2^{t} b \) (where \( s, t \) are non-negative integers, and \( a, b \) are odd numbers), without loss of generality, assume \( s \geqslant t \). Then
\[
\frac{4 x y}{x+y}=\frac{2^{s+t+2} a b}{2^{t}\left(2^{s-t} a+b\right)}=\frac{2^{s+2} a b}{2^{s-t} a+b} \text{. }
\]
If \( s > t \), then the... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,424 |
Example 2 As shown in Figure 6, in isosceles right $\triangle A B C$, it is known that $A B=1, \angle A=$ $90^{\circ}, E$ is the midpoint of $A C$, and point $F$ is on the base $B C$, with $F E$ $\perp B E$. Find the area of $\triangle C E F$.
(1998, National Junior High School Mathematics Competition) | Solution: As shown in Figure 6, draw $C D \perp C E$ intersecting the extension of $E F$ at point $D$.
Since $\angle A B E + \angle A E B = 90^{\circ}$,
$\angle C E D + \angle A E B = 90^{\circ}$,
thus, $\angle A B E = \angle C E D$.
Therefore, Rt $\triangle A B E \backsim \mathrm{Rt} \triangle C E D$.
Hence, $\frac{S_... | \frac{1}{24} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,425 |
1. If the function $f(x)=\lg \left(a x^{2}-4 x+a-3\right)$ has a range of $\mathbf{R}$, then the range of real number $a$ is ( ).
(A) $(4,+\infty)$.
(B) $[0,4]$
(C) $(0,4)$
(D) $(-\infty,-1) \cup(4,+\infty)$ | $-1 . B$.
To make the range of $f(x)$ be $\mathbf{R}$, the argument $a x^{2}-$ $4 x+a-3$ should be able to take all positive values. Therefore, either $a=0$; or $a>0$, and $4^{2}-4 a(a-3) \geqslant 0$.
Solving this, we get $0 \leqslant a \leqslant 4$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,426 |
2. Let $a^{2}+b^{2}=1(b \neq 0)$. If the line $a x+b y$ $=2$ and the ellipse $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$ have a common point, then the range of $\frac{a}{b}$ is ( ).
(A) $\left[-\frac{1}{2}, \frac{1}{2}\right]$
(B) $[-1,1]$
(C) $(-\infty,-1] \cup[1,+\infty)$
(D) $[-2,2]$ | 2. C.
Substitute $y=\frac{2-a x}{b}$ into the ellipse equation and rearrange to get $\left(3 a^{2}+b^{2}\right) x^{2}-12 a x+12-6 b^{2}=0$. Since the line and the ellipse have a common point, the discriminant $\Delta=(12 a)^{2}-4\left(3 a^{2}+b^{2}\right)\left(12-6 b^{2}\right) \geqslant 0$. Using $a^{2}+b^{2}=1$, sim... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,427 |
3. The six edges of the tetrahedron $ABCD$ are 7, 13, 18, 27, 36, 41, and $AB=41$. Then $CD=(\quad)$.
(A) 7
(B) 13
(C) 18
(D) 27 | 3. B.
In a tetrahedron, except for $CD$, all other edges are adjacent to $AB$. If the edge of length 13 is adjacent to $AB$, let's assume $BC=13$. According to the triangle inequality, we have
$$
\begin{array}{l}
AC \notin\{7,18,27\} \Rightarrow AC=36 \Rightarrow BD=7 \\
\Rightarrow\{AD, CD\}=\{18,27\} .
\end{array}
$... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,428 |
4. For all real numbers $x$, we have
$$
\sin ^{k} x \cdot \sin k x+\cos ^{k} x \cdot \cos k x=\cos ^{k} 2 x,
$$
then $k=(\quad)$.
(A) 6
(B) 5
(C) 4
(D) 3 | 4. D.
$$
\text { Let } f(x)=\sin ^{k} x \cdot \sin k x+\cos ^{k} x \cdot \cos k x-\cos ^{k} 2 x \text {. }
$$
Given that $f(x)$ is always 0, taking $x=\frac{\pi}{2}$, we get $\sin \frac{h \pi}{2}$ $=(-1)^{k}$. Therefore, $k$ is an odd number.
Let $k=2 n-1$. Then the equation becomes
$$
\sin \left(n \pi-\frac{\pi}{2}\r... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,429 |
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