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 |
|---|---|---|---|---|---|---|---|---|---|
Example 1. Find $\lim _{x \rightarrow 0} \frac{\ln (1+\alpha x)}{\sin \beta x}$. | Since $\ln (1+\alpha x) \sim \alpha x, \sin \beta x \sim$ $\beta x(x \rightarrow 0$ when $)$, the original expression $=\lim _{x \rightarrow 0}-\frac{\sim x}{\beta^{3} x}=\frac{\alpha}{\beta}$. | \frac{\alpha}{\beta} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,848 |
8. For each proper divisor of $1,000,000$, take the logarithm to the base 10, add these logarithmic values together, to get the sum $S$, find the integer closest to $S$.
(Proper divisors of a natural number $n$ are the divisors of $n$ that are neither 1 nor $n$.) | 8. Since $1000000=2^{\circ} \cdot 5^{\circ}$ has $(6+1)$ - $(6+1)=49$ factors (since its factors must be of the form $2^{i} \cdot 5$, where $0 \leqslant i, j \leqslant 6$), excluding 1000, the remaining 48 factors form 24 pairs, each with a product of $1000000=10^{6}$, so the product of all factors of 1000000 is $1000 ... | 141 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,850 |
9. As shown in the figure, in $\triangle A B C$, $A B$ $=425, B C=450, C A=510, P$ is inside the triangle, $D E 、 F G 、 H I$ all pass through $P$, have the same length $d$, and are parallel to $A B 、 B C 、 C A$ respectively. Find $d$. | 9. As shown in the figure,
$$
\begin{array}{l}
E H=B C-(B E+H C)=B C-(F P \\
+P G)=450-d .
\end{array}
$$
Similarly, we get $G D=510-d$,
From the similarity of $\triangle D P G$ and $\triangle A B C$, we have
$$
\begin{aligned}
D P= & \frac{A B}{C A} \cdot G D=\frac{425}{510}(510-d) \\
& =425-\frac{5}{6} d
\end{aligne... | 306 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,851 |
10. In a game, the "magician" asks a person to think of a three-digit number $(a b c)$ (where $a, b, c$ are the digits of the number in base 10), and asks this person to select 5 numbers $(a c b), (b a c), (b c a), (c a b)$, and $(c b a)$, and to find the sum $N$ of these 5 numbers, and tell the sum $N$ to the magician... | 10. Add (abc) to $N$, since $a$, $b$, and $c$ each appear exactly twice in every position, we get
$$
N+(a b c)=222(a+b+c) .
$$
Therefore, we should look for a multiple of 222, $222k$, which should be greater than $N$ (since $(abc) \neq 0$) and less than $N+1000$ (since $(abc)$ is a three-digit number), and make equati... | 358 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,852 |
11. Let the polynomial $1-x+x^{2}-x^{8}+\cdots+x^{16}$ $-x^{17}$ be written as $a_{0}+a_{1} y+a_{2} y^{2}+a_{3} y^{3}+\cdots$ $+a_{10} y^{16}+a_{17} y^{17}$,
where $y=x+1$, and all $a_{\mathrm{i}}$ are constants, find $a_{2}$. | $$
\begin{array}{l}
\text { 11. Replace } x \text { with } y-1 \text {, the polynomial becomes } \\
\text { 1- }(y-1)+(y-1)^{2}-(y-1)^{3} \\
+\cdots+(y-1)^{10}-(y-1)^{17}, \\
\quad 1+(1-y)+(1-y)^{2}+(1-y)^{3} \\
+\cdots+(1-y)^{10}+(1-y)^{17} .
\end{array}
$$
Notice that the coefficient of $y^{2}$ in $(1-y)^{\mathrm{k}... | 816 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,853 |
12. The sum of a set of numbers is the sum of all its elements. Let $S$ be a set composed of positive integers not exceeding 15, such that the sums of any two disjoint subsets of $S$ are not equal, and among all sets with the above property, the sum of $S$ is the largest. Find the set $S$ and its sum. | 12. First, $S$ can contain at most 5 elements, because otherwise, $S$ would have at least 6 elements, and the number of subsets with elements not exceeding 4 would be at least $C_{8}^{1} + C_{8}^{2} + C_{8}^{3} + C_{8}^{4} = 56$. The sum of each such subset is at most 54 (since $15 + 14 + 13 + 12 = 54$). By the pigeonh... | 61 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 702,854 |
15. Let $\triangle A B C$ be a right triangle on the $x y$ plane, with $C$ as the right angle, and the chord $A B$ has a length of 60. The medians drawn from $A$ and $B$ lie on the lines $y=x+3$ and $y=2 x+4$, respectively. Find the area of $\triangle A B C$. | 15. In this problem, $r=\frac{60}{2}=30$.
From figure (1), it is easy to calculate $\operatorname{tg} \theta=\frac{1}{3}, \sin \theta=\frac{1}{\sqrt{10}}$. As shown in figure (2), taking the midpoints of the three sides as $M, N, O$, with $D$ as the incenter, $p, q, s, t$ are defined as shown in the figure, $\angle \t... | 400 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,857 |
Example 3. Find $\lim _{x \rightarrow 0} \frac{\ln \left(\sin ^{2} x+e^{x}\right)-x}{\ln \left(x^{2}+e^{2 x}\right)-2 x}$. | $$
\begin{array}{l}
\text { Solve the original expression }=\lim _{x \rightarrow 0} \ln \left(\sin ^{2} x+e^{x}\right)-\ln e^{x}\left(x^{2}+e^{2 x}\right)-\ln e^{2 x} \\
=\lim _{x \rightarrow 0} \frac{\ln \left(\frac{\sin ^{2} x+e^{x}}{e^{x}}\right)}{\ln \left(\frac{x^{2}+e^{2 x}}{e^{2 x}}\right)} \\
=\lim _{x \rightar... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,859 |
Example 4. Find $\lim _{y \rightarrow 0} \frac{e^{y}+\sin y-1}{\ln (1+y)}$. | Since $\lim _{y \rightarrow 0} \frac{e^{y}-1}{y}=1, e^{y}-1 \sim y$
$$
\lim _{\substack{(y \rightarrow 0 \\(y \rightarrow 0 \text { when) }}} \frac{e^{y}-1+\sin y}{2 y}=1, e^{y}-1+\sin y \sim 2 y
$$
Therefore, the original expression $=\lim _{y \rightarrow 0} \frac{2 y}{y}=2$.
It is worth noting that when using equiva... | 2 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,860 |
Example 1. In the triangular prism $BCD-B_{1}C_{1}D_{1}$, $D D_{1}=D_{1} C_{1}, \angle B_{1} D_{1} D=\angle B_{1} D_{1} C_{1}$.
Prove: $B_{1} C \perp D C_{1}$.
(Figure 1) | Prove that by supplementing the triangular prism to form a parallelepiped $A C_{1}$, because
$$
D D_{1}=D_{1} C_{1}
$$
$$
\angle B_{1} D_{1} D=\angle B_{1} D_{1} C_{1}, \triangle D_{1} O C_{1} \cong \triangle D_{1} O D
$$
Thus, $D O=O C_{1}=A_{1} O, \angle A_{1} D C_{1}=90^{\circ}$,
$$
A_{1} D \perp D C_{1} \text {. }... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,861 |
Example: Find the value of $\arcsin \frac{1}{\sqrt{10}}+\arccos \frac{5}{\sqrt{26}}$ $+\operatorname{arctg} \frac{1}{7}+\arccos \frac{8}{\sqrt{65}}$. | Let the four terms on the left side of the equation be denoted as $\alpha, \beta, \gamma, \delta$, clearly they all belong to $\left(0, \frac{\pi}{4}\right)$.
From $\sin \alpha=\frac{1}{\sqrt{10}}$ and $\alpha \in\left(0, \frac{\pi}{4}\right)$, it is easy to get $\operatorname{tg} \alpha = \frac{1}{3}, a=\operatorname... | \frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,862 |
$\begin{array}{l} \text { Example } 6 . \text { Given } 0<a<b<1, \text { prove: } \\ \sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+(1-b)^{2}} \\ +\sqrt{(1-a)^{2}+b^{2}}+\sqrt{(1-a)^{2}+(1-b)^{2}} \\ \geqslant 2 \sqrt{2} .\end{array}$ | When solving this problem, observe that the expression under each square root on the left side of the inequality is the sum of squares of two numbers, which can guide students to think of proving it using the following methods: (1) using the basic inequality, (2) using the property of the modulus of complex numbers, (3... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,864 |
Example 7. If $x+y+z=xyz$, prove:
$$
\begin{array}{l}
\frac{2 x}{1-x^{2}}+\frac{2 y}{1-y^{2}}+\frac{2 z}{1-z^{2}} \\
=\frac{8 x y z}{\left(1-x^{2}\right)\left(1-y^{2}\left(1-z^{2}\right)\right.} .
\end{array}
$$ | Inspired by the guidance, students are not hard to think of setting $x=\operatorname{tg} A$, $y=\operatorname{tg} B, z=\operatorname{tg} C$.
Solution: Let $x=\operatorname{tg} A, y=\operatorname{tg} B, z=\operatorname{tg} C$, and $A+B+C=\pi$, then $\operatorname{tg} 2 A=\frac{2 x}{1-x^{2}}$, $\operatorname{tg} 2 B=\fra... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,865 |
Example 8. Find the value of $2 \operatorname{arctg} \frac{1}{5}+\operatorname{arctg} \frac{1}{4}$. | This problem is to find the inverse function, but using the tangent makes the calculation easier.
Slightly solving, let $\operatorname{arctg} \frac{1}{5}=\alpha$, $\operatorname{arc} \operatorname{tg} \frac{1}{4}=\beta$, then $\operatorname{tg} \alpha=\frac{1}{5}, \operatorname{tg} \beta=\frac{1}{4}$.
$\operatorname{t... | \operatorname{arctg} \frac{32}{43} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,866 |
Example 10. Solve the equation
$$
\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+x}}}}=x .
$$ | When solving this problem, you can guide students to use conjecture and recursive methods to explore the solution path. By observation, one can conjecture: Is $x$ $=\sqrt{3+x}$? After several iterations, we get:
$$
\begin{aligned}
x & =\sqrt{3+x}=\sqrt{3+\sqrt{3+x}} \\
& =\sqrt{3+\sqrt{3+\sqrt{3+x}}} \\
& =\sqrt{3+\sqr... | x=\sqrt{3+x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,868 |
Example 11. Prove that each term in the sequence 16, 1156, 111556, 11115556, $\cdots$ is a perfect square. | First, use the square root operation to check the first few terms, and we get:
$$
16=4^{2}, 1156=34^{2}, 111556=334^{2} \text {, }
$$
By observing and analyzing the pattern of the first few terms, we can conjecture that
$$
11 \underbrace{\cdots}_{n} \underbrace{55 \cdots 56}_{n}=(\underbrace{33 \cdots 33}_{n}+1)^{2} \... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 702,869 |
Example 12. For $>0, \frac{1}{b}-\frac{1}{a}=1$. Compare the sizes of the following four numbers: $\sqrt{1+a}, \frac{1}{1-\frac{b}{2}}, 1+\frac{a}{2}$,
$$
\frac{1}{\sqrt{1-b}}
$$ | For these four numbers, if we use the method of finding differences to compare their sizes, we need to perform $C_{4}^{2}=6$ comparisons to get the answer. Can we first estimate the size relationship of these four numbers? We can use the method of substitution to test it.
$\because a>0$, let's assume $a=1$, from $\frac... | \frac{1}{1-\frac{b}{2}}<\frac{1}{\sqrt{1-b}}=\sqrt{1+a}<1+\frac{a}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,870 |
Example 13. There are $n$ parabolas on a plane, where each two intersect at only two points, and no three intersect at the same point. Find how many parts these $n$ parabolas divide the plane into. | Because the number of parts into which the plane is divided is related to the number of parabolas, we can conjecture that they have a functional relationship, and then attempt to construct this function.
Solution If there are already $n-1$ parabolas satisfying the conditions on the plane, dividing it into $f(n-1)$ par... | f(n)=n^{2}+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,871 |
Example 1. The numbers $a, b, c$ belong to the open interval $\left(0, \frac{\pi}{2}\right)$ and satisfy the following equations
$$
\begin{array}{l}
\cos a=a, \\
\sin \cos b=b, \\
\cos \sin c=c .
\end{array}
$$
Arrange these three numbers in ascending order. | Solution 1. First, separately plot the functions
$$
\begin{array}{l}
y=\cos x, \\
y=\sin \cos x, \\
y=\cos \sin x
\end{array}
$$
(see Figure 1), then the x-coordinates of their intersections with the line $y=x$ are $a$, $b$, and $c$, respectively. From the graph, it can be seen that the order should be
$$
b<a<c.
$$
T... | b<a<c | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,872 |
Example 2. If $\frac{\pi}{4} \leqslant \alpha \leqslant \frac{\pi}{2}$, prove:
$$
\frac{1+\sin \alpha-\cos \alpha}{\sin \alpha} \geqslant \sqrt{2}
$$ | Proof 1. As shown in Figure 3, first construct a unit square $ABCD$, then draw a ray from $A$ making an angle $\alpha$ with $AD$ to intersect $BC$ at $E$. Then we have
$$
\begin{array}{l}
A E=\csc \alpha, \\
E C=1-\operatorname{ctg} \alpha \\
\text { and } A E+E Q>A C,
\end{array}
$$
Assume $\csc \alpha+1$
$-\operator... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,873 |
Example 4. If $0<x<1$, prove:
$$
\begin{array}{l}
\sqrt{1+x^{2}}+\sqrt{1+(1-x)^{2}} \\
<2 \sqrt{2-1} \text { (b) }
\end{array}
$$ | Proof 1. Take a unit
square $ABCD$,
on $BC$ take $M$, let
$BM=x$, then connect
$AM, MD$ (as shown in Figure 7),
then we have
$$
\begin{array}{l}
\sqrt{1+x^{2}}+(1-x) \\
=AM+MC>AC \\
=\sqrt{2},
\end{array}
$$
$$
\begin{array}{l}
x+\sqrt{1+(1-x)^{2}}=DM+MD \\
>BD=\sqrt{2}.
\end{array}
$$
Adding the two inequalities and ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,876 |
$$
a^{2}+b^{2}+c^{2}=1 \text {. }
$$
Prove: $\sqrt{1-a^{2}}+\sqrt{1-b^{2}}$
$$
+\sqrt{1-c^{2}}>3-(a+b+c) .
$$ | Proof 1. Construct a rectangular parallelepiped $AC'$ with edges $a, b, c$, and set $AB=a, AD=b, AA'=c$ (as shown in Figure 8). Then,
$$
\begin{array}{l}
BC'=\sqrt{b^{2}+c^{2}}=\sqrt{1-a^{2}}, \\
A'B=\sqrt{a^{2}+c^{2}}=\sqrt{1-b^{2}}, \\
A'C'=\sqrt{a^{2}+b^{2}}=\sqrt{1-c^{2}} .
\end{array}
$$
From
$$
\begin{array}{l}
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,877 |
Example 6. Given $x_{1}, x_{2}, x_{3}>0$, and $x_{1}+x_{2}+x_{3}=1$. Find
$$
u=\sqrt{1-x_{1}}+\sqrt{1-x_{2}}+\sqrt{1-x_{3}}
$$
the upper and lower bounds. | Consider three points in the Cartesian coordinate system:
$$
\begin{array}{l}
A\left(x_{1}, \sqrt{1-x_{1}}\right), B\left(x_{2}, \sqrt{1-x_{2}}\right), \\
C\left(x_{3}, \sqrt{1-x_{3}}\right).
\end{array}
$$
It is evident that they all lie on the parabola $y=\sqrt{1-x}$. As shown in Figure 9, let the centroid of $\tria... | 2 < u < \sqrt{6} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 702,878 |
Example 2. Given: $\left\{\begin{array}{l}a^{2}+b^{2}=1, \\ x^{2}+y^{2}=1, \\ a x+b y=0\end{array}\right.$
Prove: $\left\{\begin{array}{l}a^{2}+x^{2}=1, \\ b^{2}+y^{2}=1, \\ a b+x y=0\end{array}\right.$ | Consider: From $a^{2}+b^{2}=x^{2}+y^{2}$, construct two right triangles with a common hypotenuse.
Prove: Construct two right triangles $A B C$ and $A D C$ such that $A C=1 ; A B=$
$$
\begin{array}{l}
|a|, B C=|b|, \\
A D=|x|, D C \\
=|y| .
\end{array}
$$
From $\frac{a}{b}=-\frac{y}{x}$ we get: $-\frac{|a|}{|b|}=-\frac... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,880 |
Example 3. Let $a>0, b>0, 2c>a+b$. Prove: $c^{2}>ab$, and $c-\sqrt{c^{2}-ab}<a<c+\sqrt{c^{2}-ab}$. | Consider: Let $x=\sqrt{a b}, c^{2}-a b$, a right-angled triangle is good.
Let a semicircle $O$ have diameter $A B=2 c$. On $A B$, take $A C=a, C D$ $=b$ successively; construct a semicircle with $A D$ as the diameter, and draw $C E \perp$ $A D$ intersecting the circle at $E$. From $2 c$ $>a+b$ we know the two circles ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,881 |
Example 4. If $|a|<1, \quad|b|<1$, prove: $\left|a b \pm \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}\right| \leqslant 1$.
untranslated text is kept as is, only the problem statement is translated. | Consider: From $|a| \cdot|b|+\sqrt{1-a^{2}} - \sqrt{1-b^{2}} \leqslant 1 \cdot 1$, construct a right triangle and a cyclic quadrilateral.
Proof: As shown in the figure, the diameter
$A B=1$, chord $A C=|a|, B D=|b|$.
Connect $B C, A D, D C$, then
$$
\begin{array}{l}
B C=\sqrt{1-a^{2}}, \\
A D=\sqrt{1-b^{2}} .
\end{arra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,882 |
Example 5. If $a, b, c, d$ are all positive numbers, prove that
$$
\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \geqslant(a c+b d)^{2} \text{. }
$$ | Prove that without loss of generality,
$$
a^{2}+b^{2} \geqslant c^{2}+d^{2} \text {. }
$$
Construct a right $\triangle ABC$,
such that $AB=a, BC=$
b, $\angle B=90^{\circ}$. Also
construct a right $\triangle ADE$,
such that point $E$ lies on $AC$, $AD=d, DE=c$. Draw $EF \perp AB$ intersecting $AB$ at $F$, and connect $... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,883 |
Example 6. $a$, $b$, $c$, $d$ are all positive numbers, $\frac{a}{b}=\frac{c}{d}$, and $a$ is the largest. Prove: $a+d>b+c$.
| Consider: From $a d=b c$, construct two secants of a circle.
Proof: Take a straight line $A B C$, such that $A C=$ $a, A B=d$, and construct a semicircle $O$ with $B C$ as its diameter. Without loss of generality, let $b \geqslant c$, and draw the secant $A D=b$, intersecting the circle at $E$. Draw the perpendicular $... | a+d>b+c | Algebra | proof | Yes | Yes | cn_contest | false | 702,884 |
Example. Let $y=f(u)=\lg u, u \in R^{+}$, $u=\varphi(x)=\sin x, x \in R$. Try to discuss the domain of the composite function $y=f[\varphi(x)]$. | To make the function $y=f(u)=\lg u$ defined, it is necessary to have $u>0$, i.e., $\sin x>0$, which means $x \in(2 k \pi, (2 k+1) \pi), k \in \mathbb{Z}$.
Therefore, the domain of the composite function $y=f[\varphi(x)]=\lg \sin x$ is: $\{x \mid 2 k \pi < x < (2 k+1) \pi, k \in \mathbb{Z}\}$.
The domain of $u=\varphi(... | \{x \mid 2 k \pi < x < (2 k+1) \pi, k \in \mathbb{Z}\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,885 |
Example 1. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>$
0 ), find the locus of the midpoints of parallel chords with slope $k$. | Let the endpoints of the chord be \( P_{\mathrm{j}}\left(x_{\mathrm{j}}, y_{\mathrm{j}}\right) \) \((j=1,2)\), and the midpoint be \( P(x, y) \). Then we have
\[
\left\{\begin{array}{l}
\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1, \\
\frac{x_{2}^{2}}{a^{2}}+\frac{y_{2}^{2}}{b^{2}}=1, \\
x_{1}+x_{2}=2 x, \\
y_{1}+... | b^{2} x + a^{2} k y = 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,886 |
Example 2. Draw a chord through one of the foci of a hyperbola, and prove that the locus of the midpoints of all such chords is also a hyperbola, with the same eccentricity as the original hyperbola. | Proof: Let the equation of the hyperbola be: \(\frac{x^{2}}{c^{2}} - \frac{y^{2}}{b^{2}} = 1\), with one focus at \(F(c, 0)\) and \(a > 0, b > 0\), \(c = \sqrt{a^{2} + b^{2}}\). Let the endpoints of the chord be \(P_{1}(x_{1}, y_{1})\) and the midpoint be \(P(x, y)\). According to the problem, we have:
\[
\left\{\begin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,887 |
Example 1. For the right triangle $R_{\mathrm{t}} \triangle A B C$ with hypotenuse $A B$ as the diameter, draw a circle. The tangent line through $C$ intersects the circles with diameters $A C$ and $B C$ at $D$ and $E$ respectively. Prove that $C D = C E$.
| $$
\begin{array}{l}
=\frac{A C \cos \angle 3}{B C \cos \angle 4} \\
=\frac{A B \cos \angle 4 \cos \angle 3}{A B \cos \angle 3 \cos \angle 4} \\
=1 . \quad \therefore C D=C E .
\end{array}
$$
Proof: We have $\frac{C D}{C E}=\frac{A C \cos \angle 1}{B C \cos \angle 2}$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,889 |
Example 3. $P$ is a point on the arc $\widehat{A B}$ of the circumcircle of square $A B C D$. Prove that:
$$
P D^{2}-P B^{2}=2 P A \cdot P C .
$$ | As shown in the figure, let the radius of the circle be $R$, and $\angle P D A = a$, then
$$
\begin{aligned}
& \frac{P D^{2}-P B^{2}}{2 P A \cdot P C} \\
= & \frac{4 R^{2} \sin \left(45^{\circ}+\alpha\right)-4 R^{2} \sin \left(45^{\circ}-a\right)}{2 \cdot 2 R \sin \alpha \cdot 2 R \cos C} \\
= & \frac{2 \cdot \sin 45^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,891 |
Given $a x^{3}=b y^{3}=c z^{3}(a, b, c$ are constants), and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.
Find the value of $\sqrt[3]{a x^{2}+b y^{2}+c z^{2}}$. | $$
\begin{array}{l}
\frac{a x^{2}}{1 / x}=\frac{b y^{2}}{1 / y}=\frac{c z^{2}}{1 / z}=\frac{a x^{2}+b y^{2}+c z^{2}}{1 / x+1 / y+1 / z} \\
=a x^{2}+b y^{2}+c z^{2}, \\
\therefore \sqrt[3]{a x^{2}+b y^{2}+c z^{2}}=\frac{\sqrt[3]{a}}{1 / x}=\frac{\sqrt[3]{b}}{1 / y} \\
=\frac{\sqrt[3]{c}}{1 / z}=\frac{\sqrt[3]{a}+\sqrt[3... | \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,892 |
$\begin{array}{l}\text { Example 3. If } \frac{a}{x^{2}-yz}=\frac{b}{y^{2}-zx} \\ =\frac{c}{z^{2}-xy}, \text { and } xyz \neq 0, \text { prove: } \\ ax+by+cz=(a+b+c)(x+y+z)\end{array}$ | Prove that $\begin{array}{l}\frac{a}{x^{2}-y z}=\frac{b}{y^{2}-z x}=\frac{c}{z^{2}-x y} \\ =\frac{a+b+c}{x^{2}+y^{2}+z^{2}-x y-y z-z x} \\ =\frac{(a+b+c)(x+y+z)}{x^{3}+y^{3}+z^{3}-3 x y z} \text { and } \\ \frac{a x}{x^{3}-x y z}=\frac{b y}{y^{3}-x y z}=\frac{c z}{z^{3}-x y z} \\ =\frac{a x+b y+c z}{x^{3}+y^{3}+z^{3}-3... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 702,893 |
Example 4. For a geometric sequence $\left\{a_{\mathrm{n}}\right\}$ with a common ratio of $q$, find the formula for the sum of the first $n$ terms.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $\begin{array}{l}\text { Sol } q=\frac{a_{2}}{a_{1}}=\cdots=\frac{a_{\mathrm{n}}}{a_{\mathrm{n}-1}}=\frac{S_{\mathrm{n}}-a_{1}}{S_{\mathrm{n}}-a_{1} q^{\mathrm{n}-1}} \\ \therefore S_{\mathrm{n}}=\left\{\begin{array}{l}\frac{a_{1}\left(1-q^{\mathrm{n}}\right)}{1-q}, \quad q \neq 1 ; \\ n a_{1}, \quad q=1 .\end{array}\r... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,894 |
Example 3. Let $A$ be a fixed point inside a fixed circle $O$, and $B$ be any point on the circle. Line $l$ passes through $B$ and $l \perp A B$. Prove that line $l$ is a tangent to a certain ellipse, first determine the ellipse, then prove that $l$ is its tangent. (Indicate the foci and the length of the major axis of... | Proof: Let $A^{\prime}$ be the point symmetric to point $A$ with respect to the center $O$ of the circle, and $C$ be the point symmetric to point $A$ with respect to point $B$ (i.e., with respect to line $l$). Connect $A^{\prime} C$ and let it intersect line $l$ at point $P$. Then the trajectory of point $P$ is an elli... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,896 |
Example 2. A person at point $A$ on a straight road $A E$ observes two objects $M$ and $N$ on the same side of the road, with viewing angles of $60^{\circ}$ for both; advancing 10 meters to point $B$, it is measured that $\angle M B N=60^{\circ}$; advancing another 10 meters to point $C$, it is found that $\angle M C N... | Solving the problem is evident:
$$
M, B, N, C
$$
are concyclic, $\angle 1=\angle 2$.
Thus, $\angle N B C=$
$$
\angle N C B, B N=
$$
$C N$, construct $N D \perp$
$B C$ at $D, A N=15 / \cos 60^{\circ}=30$, by the secant
theorem: $A M=A B \cdot A C / A N=20 / 3$, then
$$
M N=23 \frac{1}{3}(m) \text {. }
$$ | 23 \frac{1}{3}(m) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,898 |
Example 1. $A D$ is the median of $\triangle A B C$, prove: $A D^{2}=\frac{1}{2}\left(b^{2}+c^{2}-\frac{a^{2}}{2}\right)$. (Junior High School Algebra Volume 4, p. 2(6)). | Prove: Let $A E \perp$ line $B C$ at $E$, set $F D=t$, then $E B, E C$ are respectively $\frac{a}{2}+t$ or $\left|\frac{a}{2}-t\right|$, $A E^{2}=A D^{2}-2$, by the Pythagorean theorem,
$$
\begin{aligned}
b^{2}+c^{2}= & A \bar{D}^{2}-t^{2}+\left(\frac{a}{2}+t\right)^{2}+A D^{2} \\
& -t^{2}+\left(\frac{a}{2}-t\right)^{2... | A D^{2}=\frac{1}{2}\left(b^{2}+c^{2}-\frac{a^{2}}{2}\right) | Geometry | proof | Yes | Yes | cn_contest | false | 702,899 |
Example 2. In $\triangle A B C$, the incircle $I$ touches the sides at $D, E$, and $F$ (as shown in the figure), and $A C \cdot B C=2 A D \cdot D B$. Prove: $A C \perp B C$. | Prove: As shown in the figure, let the parameters $x$, $y$, and $z$, then
$$
(x+z)(z+y)=2 x y \text{. }
$$
Solving for $z=\frac{1}{2}\left(\sqrt{(x+y)^{2}+4 x y}\right.$
$-(x+y))$, then
$$
\begin{array}{c}
A B+B C+C A=A B+B D+C E \\
+C E+A D \\
=2(A B+C E)=2(x+y)+ \\
\sqrt{(x+y)^{2}+2(x+z)(y+z)}-(x+y),
\end{array}
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,900 |
Example 3. $P$ is a point inside square $ABCD$, and $PA : PB : PC = 1 : 2 : 3$. Prove that $\angle APB = 135^{\circ}$. | Solve as shown in the figure, set the parameters, then apply the sine and cosine laws:
$$
\begin{array}{l}
\sin \alpha=\frac{k \sin \theta}{a}, \\
a^{2}=k^{2}+4 k^{2} \\
-4 k^{2} \cos \theta, \\
9 k^{2}=4 k^{2}+a^{2}-4 a k \cos \left(5^{\circ}-\alpha\right) .
\end{array}
$$
Find $\operatorname{tg} \theta=1, \theta=135... | 135^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 702,901 |
Given: $A, B$ are any two points on a sphere. Construct a great circle passing through $A, B$, obtaining the minor arc $\overparen{A C B}$. Construct any small circle passing through $A, B$, obtaining the minor arc $\overparen{A D}$. Prove: $A C B < A D B$.
---
The translation maintains the original text's line break... | Proof: It can be divided into three steps. First, let the center of the large circle be $O$, the radius of the large circle be $R$, the center of the small circle be $O^{\prime}$, and the radius of the small circle be $r$. Connect $A B, A O$, $B O, A O^{\prime}, B O^{\prime}$, and denote $\angle A O B=\alpha, \angle A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,902 |
Example 1. Suppose in a plane there are any two determined line segments $AB$ and $CD$ of unequal lengths, $X$ represents the set of points on line segment $AB$, and $Y$ represents the set of points on line segment $CD$. Prove that $X \sim Y$.
| Proof: The key to the proof lies in establishing a one-to-one correspondence $f$ between the elements of $X$ and $Y$.
We place line segments $AB$ and $CD$ as shown in Figure 1, and connect $CA$ and $DB$, extending their lines to intersect at point $O$. We establish the following correspondence:
$$
f:\left\{\begin{arra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,903 |
Example 3. Prove: $\{$ natural numbers $\}$ and $\{$ real numbers $\}$ are not equivalent, i.e., $N$ is not equivalent to $R$. | Proof:
(i) First, prove: $N \not\sim (0,1)$. (Using proof by contradiction)
Assume $N \sim (0,1)$, then $(0,1)$ can be written in the form $\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{3}, \cdots\right\}$. Write them in decimal form (including finite decimals, infinite repeating decimals, and infinite non-repeating decimals)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 702,905 |
For example, prove that the set of rational numbers is an infinite set. | Proof: Let $Q^{\prime}$ be the set of all rational numbers remaining after removing “1” from the set of rational numbers $Q$. We know $Q^{\prime} \supset Q$, i.e., $Q^{\prime}$ is a proper subset of $Q$.
From $Q^{\prime}$, take $\{2,3,4, 6, \cdots\}$, and from $Q$, take $\{1,2,3,4,5, \cdots\}$. Establish the correspon... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 702,906 |
Example 1. The quadratic function $f(x)$ satisfies the relation $f(2+x)=$ $f(2-x)$, compare $f\left(\frac{\sqrt{2}}{2}\right)$ and $f(\pi)$. | From the structure of $f(2+x)=f(2-x)$, we can see that the graph of the quadratic function $f(x)$ is symmetric about the line $x=2$.
Also, $\left|2-\frac{\sqrt{2}}{2}\right|>|2-\pi|$,
so when $a>0$, $f\left(\frac{\sqrt{2}}{2}\right)>f(\pi)$,
when $a<0$, $f\left(\frac{\sqrt{2}}{2}\right)<f(\pi)$. | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,907 |
$$
\begin{array}{l}
\text { 1. Let } p>0, x_{1}>0, \quad a_{i}>0(i=1,2, \\
\cdots, n), a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}=S \text { (constant} \\
\end{array}
$$
value), and $a_{1}+a_{2}+\cdots+a_{n}=1$, then
$$
\begin{array}{c}
\left(1+\frac{1}{x_{1}^{p}}\right)^{a_{1}}\left(1+\frac{1}{x_{2}^{p}}\right)^{a_{2}... | Consider the function $f(x)=\ln \left(1+\frac{1}{x^{\mathrm{p}}}\right)$.
( $x>0, p>0$ )
Since $f^{\prime}(x)=-\frac{p}{x\left(1+x^{p}\right)}$,
$$
f^{\prime \prime}(x)=p \frac{1+(1+p) x^{p}}{x^{2}\left(1+x^{p}\right)^{2}}>0,
$$
Therefore, $f(x)$ is a convex function on $(0,+\infty)$, and by Jensen's inequality for co... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,909 |
Theorem 2 $\frac{p_{1 \mathrm{a}}^{2}+p_{1 \mathrm{~b}}^{2}+p_{1 \mathrm{c}}^{2}}{a^{2}+b^{2}+c^{2}}=\frac{7}{9}$. | $$
\begin{array}{r}
\text { Prove } p_{1 a}^{2}=A B^{2} \frac{C D}{B C}+A C^{2} \frac{B D}{B C} \\
-B C^{2} \frac{B D}{B C} \cdot \frac{C D}{B C} \\
=c^{2} \cdot \frac{2}{3}+b^{2} \cdot \frac{1}{3}-a^{2} \frac{1}{3} \cdot \frac{2}{3}-\frac{1}{9}\left(\ell c^{2}+3 b^{2}-2 a^{2}\right)
\end{array}
$$
Prove $p_{1 \mathrm... | \frac{7}{9} | Algebra | proof | Yes | Yes | cn_contest | false | 702,911 |
Theorem 4 Let $R, R^{\prime}$ be the circumradii of triangle $A B C$ and triangle $G C F$ respectively. Then
$$
\frac{R p_{1 \mathrm{a}} p_{1 \mathrm{~b}} p_{1 \mathrm{c}}}{R^{\prime} a b c}=\frac{7}{9} .
$$ | $\begin{array}{c}\text { Prove } \because R=\frac{a b c}{4 S}, R^{\prime}=\frac{p_{1 \Delta} p_{1 \mathrm{~b}} p_{1 \mathrm{c}}}{4 S 1}, \\ \therefore \frac{R^{\prime}}{R}=\frac{p_{1 \mathrm{~A}} p_{1 \mathrm{~b}} p_{1 \mathrm{c}}}{4 S_{1}} / \frac{a b c}{4 S} \\ -\frac{S}{S_{1}} \frac{p_{1 \mathrm{a}} p_{1 \mathrm{~b}... | \frac{7}{9} | Geometry | proof | Yes | Yes | cn_contest | false | 702,913 |
Example 4. Determine the non-negative roots of the equation $\sin x=\frac{1}{100} x$ | The roots of the equation are the x-coordinates of the intersection points of the graphs of $y=\sin x$ and $y=\frac{1}{100} x$. Therefore, we only need to find the number of intersection points of the two functions in the first quadrant. This can be solved by observing the graphs.
Solution: The curve $y=\sin x$ and the... | 32 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 702,918 |
Example 3. In an arithmetic sequence with the first term $a_{1}>0$, the sum of the first 9 terms is equal to the sum of the first 17 terms. How many terms have the maximum sum? | This is a common problem, usually solved by completing the square or solving a system of inequalities, which can be cumbersome. Let's first analyze the implicit conditions in the problem: (1) It is a decreasing sequence (2) $a_{10}+a_{11}+a_{12}+a_{13}+a_{14}+a_{1 \mathrm{~s}}+a_{16}+a_{17}=0$. By the properties of an ... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,919 |
Example 5. If $A=20^{\circ}, B=25^{\circ}$, find $(1+\operatorname{tg} A)$
- $(1+\operatorname{tg} B)$.
untranslated part:
- $(1+\operatorname{tg} B)$ 的值.
Note: The untranslated part is kept as is because "tg" is a notation for the tangent function, and the phrase "的值" means "the value of" in this context. However,... | $$
\text{Solve } \begin{aligned}
& (1+\tan A) \cdot(1+\tan B) \\
= & \left(1+\tan 20^{\circ}\right) \cdot\left(1+\tan 25^{\circ}\right) \\
= & 1+\tan 20^{\circ}+\tan 25^{\circ}+\tan 20^{\circ} \cdot \tan 25^{\circ} .
\end{aligned}
$$
Using the tangent addition formula in reverse $1=\tan 45^{\circ}$
$$
\begin{array}{l}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,920 |
There are 13 playing cards, all face down. Now, each time 6 cards are flipped. Can all the cards be face up after several flips?
. | How to solve this problem? Most students might try this way: he takes 13 cards and arranges them according to the requirements (for example, $\times$ indicates the back side, ○ indicates the front side facing up).
$x \times \times \times \times \times \times \times \times \times \times \times \times$
The first time, he... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 702,921 |
Example 2. Put a positive or negative sign in front of each of the numbers $1,2,3,4, \cdots, 1985,1986$
one thousand nine hundred and eighty-six numbers, and ask whether the sum obtained is odd or even? | This problem seems intractable at first glance, because we do not know which numbers should have a positive sign and which should have a negative sign in front of them, and different signs can lead to different results. However, we are only concerned with the parity of the final result. Since the sum and difference of ... | odd | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,922 |
Example 1 Proof: A positive decimal integer with 3 digits being 1 and all other digits being 0 is not a perfect square. | We know that such positive integers are numerous, for example, 111, 101010, 106001, 19000011, ... etc. If we verify them, we can find that they are indeed not perfect squares. However, no matter how many such positive integers we verify, it cannot be considered a proof of the proposition, because there are still many s... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 702,924 |
Example 2. Prove that there exist infinitely many positive integers with the following properties: (i) none of their digits is 0; (ii) they are all divisible by the sum of their digits. | We know that there are many such positive integers, for example: $111,225,322,414,1236,122112, \cdots$. However, no matter how many we list, it still cannot prove that there are infinitely many such positive integers. What should we do? Someone came up with the following ingenious proof:
Consider all positive integers... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 702,925 |
Example 3. There are $2 n$ points on a plane, where no three points are collinear, and $n$ of them are red points, $n$ are blue points. Prove: they can be connected into $n$ non-intersecting segments, each connecting a red point and a blue point.
| Using mathematical induction is an effective way to prove this problem. However, without using mathematical induction, one can also arrive at the solution by observing special cases. The following
Obviously, there are ways to connect these points into $n$ line segments, each connecting one red and one blue point. The ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,926 |
Example 4. There are 25 points on a plane, among any 3 points, the distance between at least two of them does not exceed 1. Prove: it is always possible to cover at least 13 of these points with a circle of radius 1. | First, let's consider the "obvious" case - among the 25 points, the distance between any two points does not exceed 1. In this case, by taking any one of the points as the center and drawing a circle with a radius of 1, all 25 points can be covered, and the conclusion naturally holds.
Next, let's examine the "usual" c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,927 |
Example 5. Proof: Two equilateral triangles with side lengths less than 1 cannot cover an equilateral triangle with a side length of 1.
untranslated text remains the same as it is already in the target language. | Obviously, each equilateral triangle with a side length less than 1 can only cover one vertex of an equilateral triangle with a side length of 1. Therefore, when using two equilateral triangles with side lengths less than 1 to cover an equilateral triangle with a side length of 1, at least one vertex cannot be covered.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,928 |
Example 6. Prove: Two circular paper pieces with radii less than 1 cannot cover a circular paper piece with a radius of 1.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems the translation request was already... | Quan Ran, each circular paper piece with a radius less than 1 can at most cover a minor arc on the circumference of a circular paper piece with a radius of 1, therefore two circular paper pieces cannot cover its entire circumference, QED. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,929 |
Example 1. Given $a, b, c \in R^{+}$. Prove:
$$
\frac{b^{2} c^{2}+c^{2} a^{2}+a^{2} b^{2}}{a+b+c} \geqslant a b c .
$$
(Algebra for Key High Schools in the Six-Year System, Second Edition 95) Problem 13) | Assume $a \geqslant b \geqslant c>0$, then $b c \leqslant c a \leqslant a b$,
H. $\quad \frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c}$.
By the rearrangement inequality, we get
$$
\frac{b c}{a}+\frac{c a}{b}+\frac{a b}{c} \geqslant \frac{b c}{c}+\frac{c a}{a}+\frac{a b}{b},
$$
which means $\frac{b^{2} c^{2}+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,930 |
Example 2 $a>0, b>0, c>0$. Prove that;
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{a^{8}+b^{8}+c^{8}}{a^{3} b^{3} c^{3}} .
$$
(Additional Questions for Zhejiang High School Mathematics Textbook, Volume 2, P26, Question 51) | Given $a \geqslant b \geqslant c>0$, then
$$
\frac{1}{b c} \geq \frac{1}{c a}>\frac{1}{a b} \text {, }
$$
Thus, $\frac{a^{8}+b^{8}+c^{8}}{a^{3} b^{3} c^{3}}=\frac{a^{5}}{b^{3} c^{3}}+\frac{b^{5}}{c^{3} a^{3}}+\frac{c^{8}}{a^{3} b^{5}}$
$$
\begin{array}{l}
>\frac{a^{5}}{c^{3}}+\frac{b^{5}}{a^{3} b^{3}}+\frac{c^{5}}{b^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,931 |
Example 3. If $a, b, c$ are positive numbers, prove:
$$
a^{2a} \cdot b^{2b} \cdot c^{2c} \geqslant a^{b+c} \cdot b^{c+a} \cdot c^{a+b} \cdot
$$
(1978 Shanghai Middle School Mathematics Competition Question) | Given $a \geqslant b \geqslant c>0$, then
$$
\lg a \geqslant \lg b \geqslant \lg c .
$$
By the rearrangement inequality, we have
$$
\begin{array}{l}
a \lg a + b \lg b + c \lg c \geqslant b \lg a + c \lg b + a \lg c, \\
a \lg a + b \lg b + c \lg c \geqslant c \lg a + a \lg b + b \lg c
\end{array}
$$
Therefore, $2 a \l... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,932 |
Let's start with a number-filling game. Please fill in the blanks with positive integers:
$$
\begin{array}{l}
\square^{2}+17^{2}+8^{2}=\square^{2}, \\
\square^{2}+18^{2}+12^{2}=\square^{2}, \\
\square^{2}+7^{2}+9^{2}=\square^{2} .
\end{array}
$$ | How should we fill in these numbers? Let's analyze it calmly.
For example, in (1), 17 is an odd number, and 8 is an even number, so $17^{2}+8^{2}$ is an odd number, which we can write as $2k+1$; then, what number can we add to make it a perfect square? By the multiplication formula
$$
k^{2}+2k+1=(k+1)^{2}
$$
we know ... | 176^{2}+17^{2}+8^{2}=177^{2}, 116^{2}+18^{2}+12^{2}=118^{2}, \text{no solution for the third equation} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,933 |
Example 5. Let $a, b, c$ be the lengths of the three sides of a triangle. Prove:
$$
\begin{array}{l}
a^{2}(b+c-a)+b^{2}(c+a-b) \\
+c^{2}(a+b-c) \leqslant 3 a b c .
\end{array}
$$
(From the 6th International Mathematical Olympiad) | Assume without loss of generality that $a \geqslant b \geqslant c$,
then
$$
\begin{array}{l}
b(c+a-b)-a(b+c-a) \\
=(a-b)(a+b-c) \geqslant 0, \\
c(a+b-c)-b(c+a-b) \\
=(b-c)(2+a-c) \geqslant 0,
\end{array}
$$
Therefore, $a(b+c-c) \leqslant b(c+a-b)$
$$
\leqslant c(a+b-c) \text {. }
$$
Since $a \geqslant b \geqslant c$,... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,935 |
Example 6. Let $a, b, c$ be the lengths of the sides of a triangle. Prove:
$$
\begin{array}{l}
a^{2} b(a-b)+b^{2} c(b-c) \\
+c^{2} a(c-a) \geqslant 0 .
\end{array}
$$
(24th International Mathematical Olympiad Problem) | Assume without loss of generality that $a \geqslant b \geqslant c$. From the proof of Example 5, we know that
$$
\begin{array}{l}
a(b+c-a) \leqslant b(c+a-b) \\
\leqslant c(a+b-c).
\end{array}
$$
But $\frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c}$,
so by the rearrangement inequality, we have
$$
\begin{array}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,936 |
Example 7. Let $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$,
$$
y_{1} \leqslant y_{2} \leqslant \cdots \leqslant y_{n} \text {, }
$$
and $z_{1}, z_{2}, \cdots, z_{n}$ be a permutation of $y_{1}, y_{2}, \cdots, y_{n}$. Prove that:
$$
\sum_{i=1}^{n}\left(x_{i}-y_{i}\right)^{2} \leqslant \sum_{i=1}^{n}\left(x... | Prove that by the sorting principle,
i.e., $-\sum_{i=1}^{n} 2 x_{i} y_{i} \leqslant -\sum_{i=1}^{n} 2 x_{i} z_{i}$.
But $\sum_{i=1}^{n}\left(x_{i}^{2}+y_{i}^{2}\right)=\sum_{i=1}^{n}\left(x_{i}^{2}+z_{i}^{2}\right)$,
Therefore, $\sum_{i=1}^{n}\left(x_{i}^{2}-2 x_{i} y_{i}+y_{i}^{2}\right)$
$$
\leqslant \sum_{i=1}^{n}\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,937 |
Example 8. Given that $a_{1}, a_{2}, \cdots, a_{n}$ are distinct positive integers. Prove that:
$$
\sum_{k=1}^{n} \frac{a_{k}}{k^{2}} \geqslant \sum_{k=1}^{n} \frac{1}{k} .
$$
(20th International Mathematical Olympiad Problem) | To prove that for $a_{1}, a_{2}, \cdots, a_{n}$ arranged in order of magnitude, let them be
$$
a^{\prime}\frac{1}{2^{2}}>\cdots>\frac{1}{n^{2}},
$$
By the rearrangement inequality, we have
$$
\sum_{k=1}^{n} \frac{a_{k}}{k^{2}} \geqslant \sum_{i=1}^{n} \frac{a_{k}^{\prime}}{l^{2}} .
$$
Since $a_{1}^{\prime}, a_{2}^{\p... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,938 |
Given: Triangular pyramid $S-A B C$, the base is an equilateral triangle with side length $4 \sqrt{2}$, the edge $S C$ has a length of 2 and is perpendicular to the base, $E$ and $D$ are the midpoints of $B C$ and $A B$ respectively. Find: The distance between $C D$ and $S E$.
---
Translate the above text into Englis... | [3D Geometry Solution] Since $CD$ and $SE'$ are two skew lines, it is not obvious where their common perpendicular intersects these two lines. Therefore, the problem needs to be transformed into whether a plane can be found that passes through $SE$ and is parallel to $CD$. $E$ is the midpoint of $BC$. Find the midpoint... | \frac{2\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,940 |
In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ $=1: 2$. Find the dihedral angle formed by the plane $B_{1} E F$ and the base $A_{1} B_{1} C_{1} D_{1}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 〔Solution One〕, see "Intermediate Mathematics" 1985, No. 6, P 45 ~ P 46
This method of finding the plane angle of a dihedral angle is the most basic, but when the plane angle is not obvious, it is difficult to add auxiliary lines.
〔Solution Two〕 Draw a perpendicular line $BO$ from $B$ to the plane $B_{1}EF$, with $O$... | \operatorname{tg}^{-1} \frac{\sqrt{37}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,941 |
Let $A B C$ be a triangle, and a circle passing through vertices $A J C$, and line segments $A B$ and $B C$ intersect at points $K$ and $N$, respectively, with $K$ and $N$ being distinct. The circumcircles of $\triangle A B C$ and $\triangle K B N$ intersect at $B$ and another point $M$. Prove that $\angle O M B$ is a ... | Analysis: The conditions of this problem include three circumferences and seven points. These three circumferences connect six of the points into one piece, with only point $O$ appearing isolated, so it can be temporarily ignored. Among the remaining six points, point $M$ is the most important, as it is the vertex of t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,942 |
In the 1986 National Junior High School Mathematics Competition, the Third Question is: Let $P, Q$ be two fixed points on the line segment $BC$, and $BP = CQ$. $A$ is a moving point outside of $BC$ (as shown in Figure 1). When point $A$ moves to make $\angle BAP = \angle CAQ$,
what kind of triangle is $\triangle ABC$? ... | $$
\begin{array}{l}
\text { Proof 1 } \because S \triangle \triangle A B=S \triangle A C Q, \\
\therefore \quad-\frac{A B \cdot A P \sin \angle B A P}{A C \cdot A Q \sin \angle C A Q} . \\
= \frac{A B \cdot A P}{A C \cdot A Q}=1 .
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore, } A B \cdot A P=A C \cdot A Q . \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,943 |
Example 4. Is there any such numbers $a, \nu, c, d$ that satisfy the system of equations
$$
\begin{array}{l}
a b c d - a = 1 \underbrace{198519851985}_{1985 \text { ones } 1985}, \\
a b c d - b = \underbrace{135135 \cdots 135}_{135 \text { ones } 135}, \\
a b c d - c = \underbrace{13571357 \cdots 1357}_{1357 \text { on... | From (1), $a(b c d-1)=$ odd number, so $a$ is odd.
Similarly, from (2), (3), and (4), we get $b$, $c$, and $d$ are odd.
Since $a, b, c, d$ are odd, from (1) we have
$a b c d-a=$ odd number - odd number $=$ even number $\neq$ odd number.
Therefore, it is impossible for integers $a, b, c, d$ to simultaneously satisfy a... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,944 |
1. What is the last digit of $11^{6}+14^{6}+16^{8}$? | 1. The last digit of the given sum can be determined by adding the last digits of each power: the last digit of any power of a number ending in 1 and 6 remains 1 and 6, the last digit of an even power of a number ending in 4 is 6. Therefore, the last digit of the given sum is 3. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,945 |
2. Prove: The square of any odd number is always congruent to 1 modulo 8.
保留了源文本的换行和格式。 | 2. $(2 k+1)^{2}=4 k(k+1)+1$. Among two consecutive natural numbers $k, k+1$, one must be even, hence the first term on the right is divisible by 8. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 702,946 |
3 . Proof: The number $\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}}$ is less than 3.
| 3. By contradiction, assume the given number is not less than 3. In this case, its square is not less than 9. That is,
$$
\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}} \geqslant 3 .
$$
Continuing with this method, we get $\sqrt{6} \geqslant 3$, which is a contradiction. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 702,947 |
$$
\begin{array}{l}
\sqrt{5 \frac{5}{24}}=5 \sqrt{\frac{5}{24}} ; \\
\sqrt{12-\frac{12}{143}}=12 \sqrt{\frac{12}{143}} .
\end{array}
$$
4. Generalize the following correct equations into a general rule:
$$
\begin{array}{l}
\sqrt{5 \frac{5}{24}}=5 \sqrt{\frac{5}{24}} ; \\
\sqrt{12-\frac{12}{143}}=12 \sqrt{\frac{12}{143... | 4. The two known equations are special cases of the following general equation:
$$
\begin{array}{l}
\sqrt{a+\frac{a}{a^{2}-1}}=\sqrt{\frac{a^{3}}{a^{2}-1}} \\
=a \sqrt{\frac{a}{a-2}} \quad(a>0, a \neq 1)
\end{array}
$$ | a \sqrt{\frac{a}{a-2}} \quad(a>0, a \neq 1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,948 |
5 . Find the root of the equation $x^{x^{3}}=3$.
| 5. $\sqrt[3]{3}$. , (Let $x^{3}=y^{\prime}$, we get $x^{y}=3$, $x^{3y}=3^{3}$. That is, $y^{y}=3^{3}$. The equation clearly has a root $y^{\prime}=2$, i.e., $x=\sqrt[3]{3}$. Let's see if there are any other roots: if $y>3$, then $y^{3}>y^{3}>3^{3}$; if $0<y<3$, then $y^{y}<y^{3}<3^{3}$. Therefore, the only root of this... | \sqrt[3]{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,949 |
6. Explain the following strange equation:
$$
\lg \left(16+\frac{16}{15}\right)=\lg 16+\lg \frac{16}{15}
$$ | For positive numbers $a, b$, we have:
\[
\begin{aligned}
& \lg (a+b)=\lg a+\lg b \Leftrightarrow a+b=a b \\
\Leftrightarrow & a(b-1)=b \Leftrightarrow b>1, a=\frac{b}{b-1} .
\end{aligned}
\] | a = \frac{b}{b-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,950 |
7. What is the last digit of $1! +2!+3!+\cdots+8!+9!$? | 7. The last digit of this sum is 3. Because from 5! onwards, the last digit of each term is 0, and $1!+$ $2!+3!+4!=1+2+6+24=33$ | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 702,951 |
9 . Solve the equation $\cos ^{58} x+\sin ^{40} x=1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\text { 9. } \cos ^{58} x \leqslant \cos ^{2} x, \sin ^{40} x \leqslant \sin ^{2} x
$$
Therefore, $\cos ^{58} x+\sin ^{40} x \leqslant \cos ^{2} x+\sin ^{2} x$, so $\cos ^{58} x+\sin ^{40} x \leqslant 1$.
The equality in (2) holds if and only if the two equalities in (1) hold simultaneously, and in (1), the first e... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,953 |
10. For what value of $x$ does the equation $\sin \frac{\pi}{1+x^{2}}$ $=0$ hold? | 10. Let $1+x^{2}=\frac{1}{k}$, where $k$ is an integer, $k \neq 0$, then
$$
x= \pm \sqrt{\frac{1}{k}-1} .
$$
Only when $k=1$, $x$ is a real number, so $x=0$. | x=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,954 |
Example 1. Simplify the quadratic curve equation:
$$
\begin{array}{l}
9 x^{2}-24 x y+16 y^{2}-20 x-14 y \\
+200=0,
\end{array}
$$
and find its focus coordinates and directrix equation. | Solve $\Delta=B^{2}-4 A C=0$, the curve is a non-central quadratic curve. From (2.4), the equation of the axis of symmetry of the curve (2.5) can be obtained as $l_{1}: 3 x-4 y+10=0$.
Substituting $l_{1}$ into (2.5) yields the intersection points of $l_{1}$ and the curve as $O\left(\frac{-2}{5}, \frac{11}{5}\right)$. ... | \vec{F}\left(\frac{2}{5}, \frac{14}{5}\right), \, 4 x+3 y=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,956 |
Example 2. Simplify the quadratic curve equation
$$
\begin{aligned}
14 x^{2} & +24 x y+21 y^{2}-4 x+18 y \\
& -139=0 .
\end{aligned}
$$ | Solve $\triangle=B^{2}-4 A C=-600<0$, then the equation represents an elliptical quadratic curve. By
$$
\begin{array}{l}
f_{x}^{\prime}(x, y)=14 x+12 y-2=0, \\
f_{y}^{\prime}(x, y)=12 x+21 y+S=0,
\end{array}
$$
we get the center of this curve as $O(1,-1)$. The parametric equations of the line system are:
$$
\left(\beg... | \frac{x^{\prime 2}}{5}+\frac{y^{\prime 2}}{30}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,957 |
Example 4. Simplify the quadratic curve equation
$$
x^{2}-3 x y+y^{2}+10 x-10 y+21=0 \text {, }
$$
and find its eccentricity, vertex, foci, directrix equation, and asymptote equation in the original coordinate system. | From $\Delta=B^{2}-4 A C=5>0$, we know that the equation is of hyperbolic type.
From (3.8), the symmetry axis equations of line (3.9) can be obtained as
$$
l_{1}: x+y=0, l_{2}: x-y+4=0 .
$$
According to the normal direction of $l_{1}, l_{2}$, determine the positive direction of these two lines as coordinate axes. Sin... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,959 |
Example 5. If an arc in the Euclidean plane is part of a regular quadratic curve, try to determine what type of curve it represents. | Solve: According to Corollary 1 and Corollary 2, construct two lines
$$
f_{x}^{\prime}(x, y)=0, f_{y}^{\prime}(x, y)=0 .
$$
(1) If the two lines are parallel, then the quadratic curve is a parabola;
(2) If the intersection point $O$ of the two lines is inside the concave segment of the curve, then the quadratic curve i... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,960 |
Example 6. Determine the range of the curve
$$
2 x^{2}-2 x y+y^{2}-6 x-4 y+27=0
$$
in the coordinate plane. | Solve $A=B^{2}-4 A C=-i<0$, this curve represents an ellipse.
The derivatives are $f_{x}^{\prime}(x, y)=4 x-2 y-6=0$,
$$
f_{y}^{\prime}(x, y)=-2 x+2 y-4=0
$$
Substituting into the curve equation $(5 \cdot 2)$, we find that the intersection points of the line $f_{x}^{\prime}(x, y)=0$ and the curve $(5 \cdot 2)$ are $P_... | x=2+\sqrt{5}, x=2-\sqrt{5}, y=5, y=9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,961 |
Example 6. Let $a, b, c, d$ be odd integers, $0<a<b$ $<c<d$, and $ad=bc$. Prove: If $a+b=2^{k}$ $b+c=2^{m}, k$ and $m$ are integers, then $a=1$.
保留源文本的换行和格式，直接输出翻译结果。 | Prove that $ac-b$.
Since $ad=bc$, we have
$$
\begin{aligned}
(2^k)^2 & = (a+d)^2 = (a-d)^2 + 4ad \\
& > (c-b)^2 + 4bc = (b+c)^2 = (2^m)^2.
\end{aligned}
$$
Therefore, $k > m$.
From $d = 2^k - a$ and $c = 2^m - b$, we get
$$
\begin{aligned}
a(2^k - a) & = b(2^m - b), \\
2^k a - 2^m b & = a^2 - b^2,
\end{aligned}
$$
$$... | a = 1 | Number Theory | proof | Yes | Yes | cn_contest | false | 702,962 |
Example 7. Can the numbers $1,1,2,2, \cdots, 1986,1986$ be arranged in a row so that there is one number between the two 1s, two numbers between the two 2s, ..., and one thousand nine hundred and eighty-six numbers between the two 1986s?
Please prove your conclusion. | Let's first imagine arranging these 3972 numbers in a row, with each of the 3972 numbers occupying a position, and we number each position.
First, consider the even numbers. Suppose there are two even numbers $m$, then there are $m$ numbers between the two $m$s. If one $m$ is in an odd-numbered position, then the othe... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 702,963 |
Toe 1. The three sides of a triangle are three consecutive integers, and one angle is twice that of another. Find the lengths of the sides of this triangle.
Note: The original text seems to have some typographical errors or unconventional notations, such as "趾" (toe) and "笛" (flute), which do not make sense in the con... | Let $\triangle A B C$ have three sides $A C=n, A B=n-1, B C=n+1$, then we have $A>B>C$. Since one angle is twice another, we have $4 C\cos 45^{\circ}=\frac{\sqrt{2}}{2}$,
which means
$$
\begin{aligned}
\cos C & =\frac{n^{2}+(n+1)^{2}-(n-1)^{2}}{2 n(n+1)} \\
& =\frac{n+4}{2(n+1)}>\frac{\sqrt{2}}{2} .
\end{aligned}
$$
... | 4,5,6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 702,964 |
Example 2. Prove: There exist 1986 points in the plane, not all collinear, such that the distance between any two points is an integer.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Example 2. Prove: There exist 1986 ... | Analysis: As shown in the figure, on the x-axis of the Cartesian coordinate system $\mathrm{xOy}$, take point $O$ and another 1984 points, and take point $P$ on the $Oy$ axis. It is only required that point $P$ and any point $A_{1}$ on the $X$ axis form a right-angled triangle $\triangle P O A_{1}$ with integer sides.
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 702,965 |
Example 3. Prove: On a circle with radius 1, there exist 1975 points, such that the straight-line distance between any two points is a rational number. (17th IMO, Problem 5) | Analysis: According to the formula $\left(\frac{2 u v}{u^{2}+v^{2}}\right)^{2} +\left(\frac{u^{2}-v^{2}}{u^{2}+v^{2}}\right)^{2}=1$, it is not difficult to find 1973 points, $A_{1}, A_{2}, \cdots, A_{1 \ominus 7 s}$, on a semicircle with a diameter of 1, excluding the endpoints $R, S$.
Let the diameter $R S=1$ of the ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 702,966 |
Example 1. Which of the following expressions are meaningful? Among the meaningful radicals, which are arithmetic roots? Express the radicals that are not arithmetic roots using arithmetic roots;
(1) $\sqrt[3]{\frac{1}{8}}$,
(2) $\sqrt{-9}$
(3) $\sqrt[3]{-8}$ | Solution: (1) is an arithmetic root;
(2) is meaningless;
(3) is meaningful, but not an arithmetic root,
$$
\sqrt[3]{-8}=-\sqrt[3]{8}=-2 .
$$ | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,967 |
Example 2. Answer the following questions:
(1) $\sqrt{6^{2}}=6$, is this correct?
(2) $\sqrt{(-6)^{2}}=-6$, is this correct? | Answer: (1) Yes, (2) No.
Example 2 can be easily generalized to the general case: the square root of a number $a$ squared is the absolute value of the number $a$, that is,
$$
\sqrt{a^{2}}=|a|=\left(\begin{array}{ll}
a ; & (a>0) \\
0, & (a=0) \\
-a, & (a<0)
\end{array}\right.
$$
With this general conclusion, the soluti... | \sqrt{a^{2}}=|a| | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,968 |
3. Determine the parity of the function $f(x)=\log _{a}\left(x+\sqrt{1+x^{2}}\right)$. | Solution: $f(-x)=\log _{a}\left(-x+\sqrt{1+x^{2}}\right)$,
$$
\because f(-x) \neq-f(x), f(-x) \neq f(x) \text {, }
$$
$\therefore f(x)$ has no odd or even properties.
Analysis: This problem makes a hasty judgment that $f(x)$ has no odd or even properties based on the superficial result without necessary equivalent tran... | f(x) \text{ is an odd function} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,969 |
Example 4. Determine the odd or even nature of $f(x)=\frac{1+\sin x-\cos x}{1+\sin x+\cos x}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Example 4. Determine the odd or even nature of $f(x)=\frac{1+\sin x-\cos x}{1+\sin ... | Solution: It is difficult to directly compare the relationship between $f(x)$ and $f(-x)$, so we need to simplify $f(x)$,
$$
\begin{array}{l}
f(x)=\frac{1+\sin x-\cos x}{1+\sin x+\cos x} \\
=\frac{2 \sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}} \... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 702,970 |
Example 1. If the diagonals of a cyclic quadrilateral are perpendicular to each other, then the segment from the intersection of the diagonals to the midpoint of one side is equal to the distance from the center of the circle to the opposite side (i.e., prove $E N$ $=O M$). | Proof: Let the radius of $\odot O$ be 1,
in $\triangle C B D$, $O M=\cos \alpha$.
in $\triangle A C B$, $A B=2 \sin \beta$,
and $\cos \alpha=\sin \beta$,
$$
\therefore O M=\cos \alpha=\sin \beta=\frac{1}{2} A B=E N \text {. }
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,971 |
Example 2. Given that $AB$ is the diameter of $\odot O$, $AC$ is tangent to $\odot O$ at $A$, $BC$ intersects $\odot O$ at $D$, and a tangent line through $D$ to $\odot O$ intersects $AC$ at $E$. Prove: $DE^2 = \frac{1}{4} BC \cdot DC$. | $$
\begin{array}{l}
\because B C \cdot D C \\
=A C^{2}=A B^{2} \cdot \tan^{2} B \\
= 4 \tan^{2} B . \text { and } D E^{2} \\
= A E^{2}=\tan^{2} B, \\
\therefore D E^{2}=\frac{1}{4} B C \cdot D C .
\end{array}
$$ | D E^{2}=\frac{1}{4} B C \cdot D C | Geometry | proof | Yes | Yes | cn_contest | false | 702,972 |
Example 4. In $\triangle A B C$, prove that $\sin 2 A$
$$
\begin{array}{l}
+\sin 2 B+\sin 2 C \\
=4 \sin A \sin B \sin C
\end{array}
$$ | Figure 11
Proof: Without loss of generality, let the radius of the circumcircle of $\triangle ABC$ be 1.
(1) If $\triangle ABC$ is an acute triangle, as shown in Figure 11,
then $S_{\triangle OBC} = \frac{1}{2} BC \cdot OM$
$$
= \frac{1}{2} \cdot 2 \sin A \cos A = \frac{1}{2} \sin 2A.
$$
Similarly, $S_{\triangle OCA} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 702,974 |
Example 5. Solve the inequality: $\operatorname{tg} x>\frac{\sqrt{3}}{3}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: Draw a rough sketch of the unit circle and the tangent line $A T=\overline{\sqrt{3}} \frac{3}{3}$ (Figure 12). It can be seen that the solution within $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ is: $\frac{\pi}{6}<x<\frac{\pi}{2}$.
$\therefore$ The general solution is $n \pi+\frac{\pi}{6}<x<n \pi+\frac{\pi... | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 702,975 |
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