problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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values | synthetic bool 1
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Example 3 Proof:
$$
(1+2+\cdots+9) \mid\left(1^{5}+2^{5}+\cdots+9^{5}\right) .
$$ | Let $s=1^{5}+2^{5}+\cdots+9^{5}$. Then
$$
\begin{array}{l}
2 s=\left(1^{5}+9^{5}\right)+\left(2^{5}+8^{5}\right)+\cdots+\left(9^{5}+1^{5}\right) \\
\Rightarrow 10 \mid 2 s, \\
2 s=\left(0^{5}+9^{5}\right)+\left(1^{5}+8^{5}\right)+\cdots+\left(9^{5}+0^{5}\right) \\
\Rightarrow 9 \mid 2 s . \\
\text { Also, }(9,10)=1, \t... | 45|s | Number Theory | proof | Yes | Yes | cn_contest | false | 725,355 |
1. Find all positive integer pairs $(x, y)$ that satisfy the equation
$$
2 x^{2}+5 y^{2}=11(x y-11)
$$ | ```
By rearranging and factoring, we get
\[
\begin{array}{l}
(2 x-y)(x-5 y)=-121 . \\
\text { The solution is }(x, y)=(14,27) .
\end{array}
\]
``` | (x, y)=(14,27) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,356 |
2. Find the integer solutions of $\frac{1}{x^{2}}+\frac{1}{x y}+\frac{1}{y^{2}}=1$. | The equation is transformed into $x^{2}+x y+y^{2}=x^{2} y^{2}$.
Therefore, $(x+y)^{2}=x^{2} y^{2}+x y$, $4(x+y)^{2}+1=(2 x y+1)^{2}$.
Solving yields $(x, y)=(1,-1),(-1,1)$. | (1,-1),(-1,1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,357 |
3. Prove: The equation $y^{2}=x^{3}-5$ has no integer solutions. | First, if $x$ is even, modulo 4 we get
$$
y^{2}=3(\bmod 4),
$$
which is impossible;
if $x$ is odd, when $x \equiv 3(\bmod 4)$, from the original equation we get $y^{2} \equiv 2(\bmod 4)$, hence it is impossible.
Therefore, $x \equiv 1(\bmod 4)$.
Further classification and discussion can also show that there is no sol... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,358 |
4. Prove: When $n>1$, there do not exist an odd prime $p$ and a positive integer $m$ such that $p^{n}+1=2^{m}$ holds. | If $n$ is odd and $n>1$, then $p^{n}+1=2^{m}$ has no prime solution $p$, thus, $n$ is even.
Let $n=2 s(s \geqslant 1)$. Therefore, $m>2$.
Hence, $p^{2 s}+1 \equiv 0(\bmod 4)$.
Since the odd prime $p=4 k \pm 1$, we have,
$$
p^{2 s}+1 \equiv( \pm 1)^{2 s}+1 \equiv 2(\bmod 4) \text {, }
$$
contradiction. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,359 |
As shown in Figure 1, BM and CN are the angle bisectors of $\triangle ABC$. Point $P$ is inside $\triangle ABC$, and perpendiculars are drawn from $P$ to $BC$, $AC$, and $AB$, with $D$, $E$, and $F$ being the feet of the perpendiculars, respectively. The necessary and sufficient condition for point $P$ to lie on segmen... | Proof. As shown in Figure 1, draw $M H \perp B C$ at point $H$, $M I \perp A B$ at point $I$, draw $N G \perp B C$ at point $G$, $N J \perp A C$ at point $J$.
Necessity.
Assume point $P$ is on line segment $M N$.
In $\triangle M J N$, let $\frac{P M}{P N}=\frac{m}{n}$.
By $P E / / N J \Rightarrow P E=\frac{m}{m+n} N J$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,360 |
Question As shown in Figure 1, in the inscribed $\triangle A B C$, $\angle A$ is the largest angle, and points $D$ and $E$ on the arc $\overparen{B C}$ not containing point $A$ are the midpoints of arcs $\overparen{A B C}$ and $\overparen{A C B}$, respectively. Let the circle passing through points $A$ and $B$ and tang... | Proof 1 As shown in Figure 1, extend $CA$ to point $X$, and connect $AE$, $PB$, $PE$, $BE$. Let the intersection of line $BD$ and $CE$ be point $K$.
Since points $D$ and $E$ are the midpoints of arcs $\overparen{ABC}$ and $\overparen{ACB}$, respectively, $BD$ and $CE$ are the external angle bisectors of $\angle B$ and... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,362 |
Question 1 Given that $f(x)$ is a quadratic function, and $a$, $f(a)$, $f(f(a))$, $f(f(f(a)))$ form a geometric sequence. Prove: $f(a)=a$.
保留源文本的换行和格式，直接输出翻译结果如下：
Question 1 Given that $f(x)$ is a quadratic function, and $a$, $f(a)$, $f(f(a))$, $f(f(f(a)))$ form a geometric sequence. Prove: $f(a)=a$. | Proof: Let $f(x)=m x^{2}+n x+l(m \neq 0)$, and the common ratio of a geometric sequence be $q$. Then
$$
\begin{array}{l}
f(a)=m a^{2}+n a+l=q a, \\
f(f(a))=m f^{2}(a)+n f(a)+l=a f(a), \\
f(f(f(a))) \\
=m f^{2}(f(x))+n f(f(a))+l \\
=a f(f(a)) .
\end{array}
$$
Therefore, $f(x)-q x=0$ has roots $a$, $a q$, and $a q^{2}$,... | f(a)=a | Algebra | proof | Yes | Yes | cn_contest | false | 725,363 |
Question 2 In the acute triangle $\triangle ABC$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c (a>b>c)$, respectively. Prove: The side length of the largest inscribed square in $\triangle ABC$ is
$$
\frac{a c \sin B}{c+a \sin B} .
$$ | Proof: Let the inscribed square of $\triangle ABC$ be $DEFG$.
It is easy to know that the square must have two vertices on two sides of $\triangle ABC$, and the other two vertices on the same side.
Assume point $D$ is on side $AC$, point $G$ is on side $BC$, and points $E, F$ are on side $AB$, with the side length of ... | \frac{a c \sin B}{c+a \sin B} | Geometry | proof | Yes | Yes | cn_contest | false | 725,364 |
Question 3: From point $O$, draw two rays $l_{1}$ and $l_{2}$. A moving line $l$ intersects $l_{1}$ and $l_{2}$ at points $A$ and $B$, respectively. The midpoint of segment $AB$ is $X$, and the trajectory of the moving point $X$ is $\Gamma$. The rays $l_{1}$ and $l_{2}$, and the line $l$ form $\triangle OAB$ with a con... | Prove that by taking the bisector of the angle formed by rays $l_{1}$ and $l_{2}$ as the $x$-axis, and the line passing through point $O$ and perpendicular to the $x$-axis as the $y$-axis, we establish a Cartesian coordinate system.
Let the moving point $X\left(x_{0}, y_{0}\right)$,
ray $l_{1}: y=k x(k>0)$,
$$
\begin{a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,365 |
Example 4 Let $n$ be a natural number. Prove:
$$
A=3237^{n}-632^{n}-855^{n}+235^{n}
$$
can be divided by 1985. | Notice that,
$$
1985=397 \times 5 \text {, and }(397,5)=1 \text {. }
$$
It suffices to prove that $51 \mid A$ and $397 \mid A$.
Since $51 \mid \left(3237^{n}-632^{n}\right)$, it follows that $51 \mid A$.
Also, $A=\left(3237^{n}-855^{n}\right)-\left(632^{n}-235^{n}\right)$,
and $3237^{n}-855^{n}$ is divisible by
$$
323... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,366 |
Question 4 Given $a_{i}>0(i=1,2, \cdots, 10)$, and
$$
\sum_{i=1}^{10} a_{i}=30, \prod_{i=1}^{10} a_{i}<21 .
$$
Prove: $a_{i}$ contains at least one number less than 1. | Proof by contradiction.
Assume $a_{i} \geqslant 1(i=1,2, \cdots, 10)$.
$$
\begin{array}{l}
\text { By } \sum_{i=1}^{10} a_{i}=30, \text { we know } \sum_{i=1}^{10} a_{i} \leqslant 29 . \\
\text { Also, } a_{i} \geqslant 1, a_{i} a_{i+1} \cdots a_{10} \geqslant 1(i=1,2, \cdots, 9), \\
\text { then } 9<\sum_{i=1}^{10} a_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,367 |
Prove: For any $n \in \mathbf{N}_{+}, (\sqrt{2}+1)^{n}$ can be written in the form $\sqrt{m}+\sqrt{m-1}\left(m \in \mathbf{N}_{+}\right)$. Find the analytical expression of $m$ in terms of $n$. | The simplest method is to consider the dual form of $(\sqrt{2}+1)^{n}$, which is $(\sqrt{2}-1)^{n}$.
By the binomial theorem, we have
$$
\begin{array}{l}
(\sqrt{2}+1)^{n}=\sqrt{a}+\sqrt{b} \\
\Rightarrow(\sqrt{2}-1)^{n}=\sqrt{a}-\sqrt{b}\left(a, b \in \mathbf{N}_{+}\right) \\
\Rightarrow(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sq... | a_{n}=\frac{1}{4}(3+2 \sqrt{2})^{n}+\frac{1}{4}(3-2 \sqrt{2})^{n}+\frac{1}{2} | Number Theory | proof | Yes | Yes | cn_contest | false | 725,368 |
1. If the perimeter of a triangle is even, and the lengths of two of its sides are 2 and 5, then the shape of this triangle must be ( ).
(A) right triangle
(B) isosceles triangle
(C) isosceles right triangle
(D) cannot be determined | -、1. B.
Let the third side be $c$. Then $3<c<7$. Also, $c$ must be an odd number, so $c=5$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,369 |
2. Xiao Wang's hourly wage in the first week is $a$ dollars, working $b$ hours, in the second week the hourly wage increases by $10 \%$, and the total working hours decrease by $10 \%$. Then, compared to the total wage of the first week, the total wage of the second week ( ).
(A) increases by $1 \%$
(B) decreases by $1... | 2. B.
$$
\begin{array}{l}
\frac{a(1+10\%) b(1-10\%)-a b}{a b} \times 100\% \\
=-1\% .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,370 |
3. Given that $a$, $b$, and $c$ are any real numbers. Then $(a-b)^{2}-4(a-c)(c-b)$ is always ( ).
(A) greater than 0
(B) equal to 0
(C) less than 0
(D) greater than or equal to 0 | 3. D.
$$
(a-b)^{2}-4(a-c)(c-b)=(a+b-2 c)^{2} \text{. }
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,371 |
4. Given that $x$ and $y$ are not equal to 0. If
$$
|x+y|=|| x|-| y|| \text {, }
$$
then ( ).
(A) $x>0, y>0$
(B) $x<0, y<0$
(C) $x y<0$
(D) Any case holds | 4. C.
From the problem, we have
$$
x+y=-(|x|-|y|)=-|x|+|y|
$$
or $x+y=|x|-|y|$.
Therefore, $x<0, y>0$ or $x>0, y<0$, which means $x$ and $y$ have opposite signs. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,372 |
5. As shown in Figure 1, if $P$
is a point inside the square
$A B C D$, and $S_{\triangle P A B}=5$,
$S_{\triangle P A D}=2$, then $S_{\triangle P A C}$ $=$ ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 5. C.
Let $S_{\triangle P A H}=m, S_{\triangle P C H}=n$.
Then $2+S_{\triangle B H C}+n=S_{\triangle P A D}+S_{\triangle B P C}$ $=\frac{1}{2} S_{\square A B C D}=S_{\triangle A B C}=S_{\triangle A B H}+S_{\triangle B H C}$ $=5-m+S_{\triangle B H C}$.
Therefore, $S_{\triangle P A C}=m+n=3$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,373 |
6. If $a=\frac{1}{2}$, then the integer part of the algebraic expression
$$
\frac{1}{1+a}+\frac{2}{1+a^{2}}+\frac{4}{1+a^{4}}
$$
is ( ).
(A) 6
(B) 8
(C) 10
(D) 12 | 6. A.
Notice that,
$$
\begin{aligned}
& \frac{1}{1+a}+\frac{2}{1+a^{2}}+\frac{4}{1+a^{4}} \\
= & \frac{1}{1-a}+\frac{1}{1+a}+\frac{2}{1+a^{2}}+\frac{4}{1+a^{4}}-\frac{1}{1-a} \\
= & \frac{8}{1-a^{8}}-\frac{1}{1-a}=6+\frac{8}{255} .
\end{aligned}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,374 |
1. As shown in Figure 2, pentagon $A B C D I$ is a regular pentagon, and hexagon DEFGHI is a regular hexagon. Then the size of $\angle I A H$ is $\qquad$ | $$
=1.24^{\circ} \text {. }
$$
From the interior angles of a regular pentagon and a regular hexagon, we know that
$$
\angle A I D=108^{\circ}, \angle H I D=120^{\circ} \text {. }
$$
Therefore, $\angle A I H=132^{\circ}$.
Since $\triangle A I H$ is an isosceles triangle, we have
$$
\angle I A H=24^{\circ}
$$ | 24^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,375 |
2. If $\frac{1}{a}+\frac{1}{b}=\frac{5}{a+b}$, then $\frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}=$ | 2. 7 .
$$
\begin{array}{l}
\text { Given } \frac{1}{a}+\frac{1}{b}=\frac{5}{a+b} \Rightarrow \frac{b}{a}+\frac{a}{b}=3 \\
\Rightarrow \frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}=\left(\frac{b}{a}+\frac{a}{b}\right)^{2}-2=7 .
\end{array}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,376 |
Example 5 Let $a, b, c$ be three distinct positive integers. Prove: Among $a^{3} b-a b^{3}, b^{3} c-b c^{3}, c^{3} a-c a^{3}$, at least one is divisible by 10.
(1986, National Junior High School Mathematics Competition) | Prove that because $a^{3} b$ and $a b^{3}$ have the same parity, $a^{3} b - a b^{3}$ is an even number.
Similarly, the other two numbers are also even.
Now, classify them by their remainders when divided by 5.
When one of $a$, $b$, and $c$ is a multiple of 5,
$$
a b\left(a^{2}-b^{2}\right), b c\left(b^{2}-c^{2}\right),... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,377 |
3. If $x-1$ is a factor of $x^{3}+a x^{2}+1$, then the value of $a$ is $\qquad$ . | 3. -2 .
Let $x^{3}+a x^{2}+1=(x-1)\left(x^{2}-m x-1\right)$, that is
$$
\begin{array}{l}
x^{3}+a x^{2}+1 \\
=x^{3}-(m+1) x^{2}+(m-1) x+1 .
\end{array}
$$
By comparing, we get $m-1=0, a=-(m+1)$.
Thus, $m=1, a=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,378 |
4. As shown in Figure 3, in the square paper piece $A B C D$, $E$ is the midpoint of $B C$. Fold the square so that point $A$ coincides with $E$, and flatten it with the crease being $M N$. Then the ratio of the area of trapezoid $A D M N$ to the area of trapezoid $B C M N$ is $\qquad$ | 4. $\frac{3}{5}$.
As shown in Figure 5, let the side length of the square be 2. Then $A E=\sqrt{5}$.
Draw a perpendicular from point $M$ to $A B$, intersecting at point $P$.
It is easy to see that $M N$ is the perpendicular bisector of $A E$.
Thus, $\triangle M P N \cong \triangle$
$A B E$
$$
\Rightarrow P N=B E=1 \... | \frac{3}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,379 |
Three. (20 points) Given
$$
(3 a+5 b-1)^{2}+|a+3 b+1|=0 \text {. }
$$
Find the solution set of the inequality with respect to $x$
$$
a x-b>\frac{x}{3}+6
$$ | Three, from the conditions we know
$$
\left\{\begin{array} { l }
{ 3 a + 5 b - 1 = 0 , } \\
{ a + 3 b + 1 = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=2, \\
b=-1 .
\end{array}\right.\right.
$$
Therefore, $2 x+1>\frac{x}{3}+6 \Rightarrow \frac{5}{3} x>5 \Rightarrow x>3$.
Thus, the solution set is $x>3$. | x>3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,380 |
Four, (25 points) Let the two intersection points of the functions $y=2x$ and $y=\frac{4}{x}$ be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\left(x_{1}>x_{2}\right)$, and point $C(\sqrt{2},-2 \sqrt{2})$. Find the area of $\triangle ABC$. | From $\left\{\begin{array}{l}y=2 x, \\ y=\frac{4}{x},\end{array}\right.$ eliminating $y$ we get
$$
2 x=\frac{4}{x} \Rightarrow x^{2}=2 \Rightarrow x= \pm \sqrt{2} \text {. }
$$
Therefore, $A(\sqrt{2}, 2 \sqrt{2}), B(-\sqrt{2},-2 \sqrt{2})$.
Since point $C(\sqrt{2},-2 \sqrt{2})$, $\triangle A B C$ is a right triangle, ... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,381 |
Five. (25 points) As shown in Figure 4, in quadrilateral $\square B C D$, $A E \perp$ $B C, A F \perp C D, H$ is the orthocenter of $\triangle A E F$. Prove:
$$
A C^{2}=A H^{2}+E F^{2} \text {. }
$$ | Five, as shown in Figure 6, connect $E H$ and $H F$.
Since $E H \perp A F, A F \perp C D$, therefore, $E H \parallel C D$.
Similarly, $H F \parallel B C$.
Thus, quadrilateral $E C F H$ is a parallelogram.
Translate $\triangle A E H$ along $\overrightarrow{H F}$ to get $\triangle A^{\prime} C F$.
Then point $A^{\prime}$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,382 |
4. Given $a>0, b>0$. If $x$ represents the smallest of the three numbers $1$, $a$, $\frac{b}{a^{2}+b^{2}}$, when $a$, $b$ vary, the maximum value of $x$ is $\qquad$ | 4. $\frac{\sqrt{2}}{2}$ | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,386 |
Example 6 Proof: All integers of the form
$$
10017, 100117, 1001117, \cdots
$$
can be divided by 53.
(1995, Moscow Mathematical Olympiad (8th grade)) | Prove that from $10017=53 \times 189$, we know 53110017.
Assume $100 \underbrace{11 \cdots 17}_{n \uparrow}$ can be divided by 53.
Since $1001 \underbrace{1 \cdots 17}_{n+1 \uparrow}=10011 \underbrace{1 \cdots 17}_{n \uparrow}+901 \underbrace{00 \cdots 0}_{n+1 \uparrow}$, and $901=53 \times 17$, thus, $100 \underbrace... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,388 |
7. In a regular tetrahedron $ABCD$ with edge length $1 \mathrm{~m}$, a small insect starts crawling from vertex $A$ according to the following rules: at each vertex, it chooses one of the three edges connected to that vertex with equal probability, and crawls all the way to the end of that edge. Let the probability tha... | 7. $\frac{7}{27}$ | \frac{7}{27} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,390 |
(16 points) Let $a>0$, and the function $f(x)=x^{3}-a x$ is monotonic on $[1,+\infty)$.
(1) Find the range of values for $a$;
(2) If $x_{0} \geqslant 1$, and satisfies
$$
f\left(x_{0}\right) \geqslant 1, f\left(f\left(x_{0}\right)\right)=x_{0} \text {, }
$$
Prove: $f\left(x_{0}\right)=x_{0}$. | (1) When $a>0$, from
$$
f^{\prime}(x)=3 x^{2}-a=0 \text {, }
$$
we get $x= \pm \sqrt{\frac{a}{3}}$.
To make $f(x)=x^{3}-a x$ a monotonic function on $[1,+\infty)$, it is only necessary that $\sqrt{\frac{a}{3}} \leqslant 1$, i.e., $a \leqslant 3$.
Therefore, the range of $a$ is $(0,3]$.
(2) From the conditions, we know... | f\left(x_{0}\right)=x_{0} | Algebra | proof | Yes | Yes | cn_contest | false | 725,392 |
Three. (20 points) As shown in Figure 2, let the pyramid be $P$ $A B C D$, with the base $A B C D$ being a rhombus, $P A \perp$ plane $A B C D, \angle A B C=60^{\circ}$, and $E, F$ being the midpoints of $B C, P C$ respectively.
(1) Prove: $A E \perp P D$;
(2) If $H$ is a moving point on $P D$, and the maximum angle fo... | (1) Since quadrilateral $ABCD$ is a rhombus, $\angle ABC = 60^{\circ}$, we know that $\triangle ABC$ is an equilateral triangle.
Since $E$ is the midpoint of $BC$, $AE \perp BC$.
Also, $BC \parallel AD$, so $AE \perp AD$.
Since $PA \perp$ plane $ABCD$ and $AE \subset$ plane $ABCD$, we have $PA \perp AE$. Since $PA \sub... | \frac{\sqrt{15}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,393 |
In the Cartesian coordinate system $x O y$, with the origin $O$ as the center, two circles are drawn with radii $a$ and $b$ ($a > b > 0$). Point $Q$ is the intersection of the radius $O P$ of the larger circle with the smaller circle. A perpendicular line $A N \perp O x$ is drawn from point $P$, with the foot of the pe... | (1) Let point $M(x, y)$, take $\angle x O P$ as parameter $\varphi$. Then
$$
\left\{\begin{array}{l}
x=a \cos \varphi, \\
y=b \sin \varphi .
\end{array}\right.
$$
Eliminating parameter $\varphi$ yields $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, which is the equation of curve $E$.
(2) Let $A\left(x_{1}, y_{1}\right),... | \frac{3 \sqrt{3} a b}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,394 |
Five. (20 points) Given $\tan \alpha=\sqrt{2}-1$, the function
$$
f(x)=x^{2} \tan 2 \alpha+x \sin \left(2 \alpha+\frac{\pi}{4}\right),
$$
where $\alpha \in\left(0, \frac{\pi}{2}\right)$.
(1) Find the analytical expression of $f(x)$;
(2) If the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{2}, a_{n+1}=f\l... | (1) From $\tan \alpha=\sqrt{2}-1$, we get $\tan 2 \alpha=1$.
Since $\alpha$ is an acute angle, thus $2 \alpha=\frac{\pi}{4}$.
Therefore, $\sin \left(2 \alpha+\frac{\pi}{4}\right)=1$.
Hence, $f(x)=x^{2}+x$.
(2) (i) From the condition, we know $a_{n+1}=a_{n}^{2}+a_{n}$.
Since $a_{1}=\frac{1}{2}>0$, thus $a_{n}>0$.
Also, ... | 1<g(n)<2 | Algebra | proof | Yes | Yes | cn_contest | false | 725,395 |
1. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1} \leqslant \frac{a_{n+2}+a_{n}}{2}, a_{1}=1, a_{403}=2011 \text {. }
$$
Then the maximum value of $a_{5}$ is $\qquad$ | - 1.21.
Obviously, the sequence of points $\left(n, a_{n}\right)$ is arranged in a convex function. When the sequence of points is distributed on the line determined by the points $(1,1)$ and $(403,2011)$, $a_{5}$ takes the maximum value 21. | 21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,396 |
2. If $x$ and $y$ are both positive integers, and the value of $x^{5}-y^{5}$ is exactly a four-digit number composed of one 2, one 0, and two 1s, then all the four-digit numbers that satisfy the condition are $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and ... | 2.2101.
When $x \geqslant 6$,
$$
x^{5}-y^{5} \geqslant 6^{5}-5^{5}>2110 \text {. }
$$
Thus, $x \leqslant 5$.
It can be verified that the only solution is $5^{5}-4^{5}=2101$. | null | Number Theory | proof | Yes | Yes | cn_contest | false | 725,397 |
3. Given $a^{2}+b^{2}+c^{2}=1$. Then the range of $a b+b c+a c$ is . $\qquad$ | 3. $\left[-\frac{1}{2}, 1\right]$.
Obviously, $a b+b c+a c \leqslant a^{2}+b^{2}+c^{2}=1$.
On the other hand,
$$
\begin{array}{l}
a b+b c+a c=a(b+c)+b c \\
\geqslant-\frac{a^{2}+(b+c)^{2}}{2}+b c \\
=-\frac{a^{2}+b^{2}+c^{2}}{2}=-\frac{1}{2},
\end{array}
$$
Equality holds only when $a+b+c=0$, and $a^{2}+b^{2}+c^{2}=1... | \left[-\frac{1}{2}, 1\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,398 |
Example 7 Let $n \in \mathbf{N}_{+}$. Prove:
$$
512 \mid\left(3^{2 n}-32 n^{2}+24 n-1\right) .
$$ | Let $f(n)=3^{2 n}-32 n^{2}+24 n-1$.
Since $f(1)=0$, we have $512 \mid f(1)$.
$$
\begin{array}{l}
\text { Hence } f(n+1)-f(n) \\
= {\left[3^{2(n+1)}-32(n+1)^{2}+24(n+1)-1\right]-} \\
{\left[3^{2 n}-32 n^{2}+24 n-1\right] } \\
= 8\left(3^{2 n}-8 n-1\right) .
\end{array}
$$
Since $512=8 \times 64$, it suffices to prove:... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,399 |
4. Labeled as $1,2, \cdots, 13$, there are 4 different colored cards totaling 52 cards, plus two blank cards, which are evenly distributed into three different boxes. If a box contains two blank cards, four 1s, and one each of the $2,3, \cdots, 13$ cards, it is called a "super box". What is the probability of a super b... | 4. $\left(\frac{1}{3}\right)^{5} \frac{\left(\mathrm{C}_{4}^{1}\right)^{12}}{\mathrm{C}_{46}^{12}}$.
First, consider that a blank card must be placed in one of the boxes. The probability that the second card is also placed in the same box is $\frac{1}{3}$, and the probability that all 4 ones are placed in the same box... | null | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 725,400 |
5. Given $a_{1}=1, a_{2}=3$,
$$
a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n} \text {, }
$$
when $m \geqslant n$, $a_{m}$ is divisible by 9. Then the minimum value of $n$ is $\qquad$ . | 5.5.
Notice,
$$
\begin{array}{l}
a_{n+2}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right) \\
=\cdots=(n+2)(n+1) n \cdots \cdots \cdot 3\left(a_{2}-a_{1}\right) \\
=(n+2)!.
\end{array}
$$
Thus, $a_{n}=a_{1}+\sum_{i=1}^{n-1}\left(a_{i+1}-a_{i}\right)=\sum_{i=1}^{n} i!$.
Given $a_{1}=1, a_{2}=3$, we have
$$
a_{3}=9, a_{4}=33, a_... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,401 |
6. The graph of the function $f(x)=\sum_{k=1}^{2011} \frac{x+k-1}{x+k}$ is symmetric about the center at | 6. $(-1006,2011)$.
Notice that,
$f(x)=\sum_{k=1}^{2011} \frac{x+k-1}{x+k}=2011-\sum_{k=1}^{2011} \frac{1}{x+k}$.
Let $g(x)=\sum_{k=1}^{2011} \frac{1}{x+k}$. Then
$$
g(x-1006)=\sum_{k=-1005}^{1005} \frac{1}{x+k}
$$
is an odd function.
Thus, the graph of $g(x)$ is symmetric about $(-1006,0)$.
Therefore, the graph of $f... | (-1006,2011) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,402 |
7. Six college graduates apply to three employers. If each employer hires at least one of them, the number of different hiring scenarios is $\qquad$ .
| 7. 2100 .
The number of ways for three people to be hired is $\mathrm{A}_{6}^{3}=120$; the number of ways for four people to be hired is $\mathrm{C}_{6}^{4} \mathrm{C}_{4}^{2} \mathrm{~A}_{3}^{3}=15 \times 6 \times 6=540$ (ways); the number of ways for five people to be hired is $C_{6}^{5}\left(C_{5}^{3} A_{3}^{3}+\fr... | 2100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,403 |
8. Let $O$ be the origin, $B(4,0), C(5,0)$, and draw a perpendicular line to the $x$-axis through point $C$. Let $M$ be a moving point on this perpendicular line. Construct a circle with $O$ as the center and $OB$ as the radius. $M T_{1}$ and $M T_{2}$ are tangents to the circle. Then the equation of the locus of the o... | 8. $\left(x-\frac{16}{5}\right)^{2}+y^{2}=\left(\frac{16}{5}\right)^{2}(x>0)$.
The equation of the circle with center $O$ and radius $O B$ is
$$
x^{2}+y^{2}=16 \text {. }
$$
As shown in Figure 2, connect $O T_{1}$ and $O T_{2}$. Let $H(x, y)$ be the orthocenter of $\triangle M T_{1} T_{2}$, $M(5, b)$, and $N$ be the ... | \left(x-\frac{16}{5}\right)^{2}+y^{2}=\left(\frac{16}{5}\right)^{2}(x>0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,404 |
9. (12 points) Solve the inequality
$$
\sqrt{x+\frac{1}{x^{2}}}-\sqrt{x-\frac{1}{x^{2}}}<\frac{1}{x} .
$$ | From the problem, we have
$$
x+\frac{1}{x^{2}}>x-\frac{1}{x^{2}} \geqslant 0 \Rightarrow \frac{x^{3}-1}{x^{2}} \geqslant 0 \Rightarrow x \geqslant 1 \text {. }
$$
Therefore, the original inequality is equivalent to
$$
\begin{array}{l}
\sqrt{x^{3}+1}-\sqrt{x^{3}-1} < \frac{1}{2} \Rightarrow x^{3}>\frac{5}{4} \Rightarro... | x>\frac{\sqrt[3]{10}}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,405 |
10. (12 points) As shown in Figure 1, given that $A$ and $B$ are the two intersection points of $\odot O$: $x^{2}+y^{2}=4$ with the $x$-axis, $P$ is a moving point on the line $l$: $x=4$, and $PA$, $PB$ intersect $\odot O$ at another points $M$ and $N$ respectively. Prove: the line $MN$ passes through a fixed point. | 10. Let $P\left(4, y_{0}\right), M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. Then
$$
\begin{aligned}
& k_{B P}=\frac{y_{0}}{2}=3 \cdot \frac{y_{0}}{6}=3 k_{A P} \\
\Rightarrow & \frac{3 y_{1}}{x_{1}+2}=\frac{y_{2}}{x_{2}-2} \\
\Rightarrow & \frac{9\left(4-x_{1}^{2}\right)}{\left(x_{1}+2\right)^{2}}=\frac{4... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,406 |
11. (12 points) Prove: When $n \geqslant 23$, we always have
$$
2<1+\frac{1}{\sqrt{2^{3}}}+\frac{1}{\sqrt{3^{3}}}+\cdots+\frac{1}{\sqrt{n^{3}}}<3
$$
holds. | 11. First, prove: $1+\frac{1}{\sqrt{2^{3}}}+\cdots+\frac{1}{\sqrt{n^{3}}}2(n \geqslant 23)$.
Similarly, we can get
$$
1+\frac{1}{\sqrt{2^{3}}}+\cdots+\frac{1}{\sqrt{n^{3}}}>1+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{n+1}}\right) \text {. }
$$
Therefore, it suffices to have $1+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{n... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,407 |
12. (12 points) Given
$$
\begin{array}{l}
f(x, y) \\
=x^{3}+y^{3}+x^{2} y+x y^{2}-3\left(x^{2}+y^{2}+x y\right)+3(x+y),
\end{array}
$$
and $x, y \geqslant \frac{1}{2}$. Find the minimum value of $f(x, y)$. | 12. When $x \neq y$, multiplying both sides of the function by $x-y$ gives
$$
\begin{array}{l}
(x-y) f(x, y) \\
=\left(x^{4}-y^{4}\right)-3\left(x^{3}-y^{3}\right)+3\left(x^{2}-y^{2}\right) . \\
\text { Let } g(x)=x^{4}-3 x^{3}+3 x^{2} .
\end{array}
$$
Then $f(x, y)=\frac{g(x)-g(y)}{x-y}$ is the slope of the line segm... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,408 |
13. (15 points) (1) In $\triangle A B C$, $\angle B C A=$ $90^{\circ}$, then $A C^{2}+B C^{2}=A B^{2}$; by analogy in three-dimensional space, what conclusion can you draw? Please provide a proof.
(2) In $\triangle A B C$, $\angle B C A=90^{\circ}$, if the distance from point $C$ to $A B$ is $h$, and the inradius of $\... | 13. (1) Conclusion: In the tetrahedron $S-ABC$, if the side edges $SA, SB, SC$ are pairwise perpendicular (let's call it a right-angled tetrahedron), then
$S_{\triangle ABC}^{2}=S_{\triangle SBC}^{2}+S_{\triangle SBA}^{2}+S_{\triangle SAC}^{2}$.
Proof: Let $SA=a, SB=b, SC=c$.
Draw $SD \perp BC$ at point $D$, and connec... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,409 |
Example 8 Given that $a, b$ are integers, $a^{2}+b^{2}$ is divisible by 3. Prove: $a, b$ are both divisible by 3. | Proof Assume that $a$ and $b$ are not both divisible by 3, then there are the following two cases.
(1) Exactly one of $a$ and $b$ is divisible by 3. Without loss of generality, let $3 \mid a$, and $3 \nmid b$.
Then $3 \mid a^{2}, 3 \nmid b^{2} \Rightarrow 3 \nmid (a^{2}+b^{2})$, which is a contradiction.
(2) Neither $a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,410 |
14. (15 points) Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=2 p, a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{p^{2}}{a_{n}}\right), \\
b_{n}=\frac{a_{n}+p}{a_{n}-p}\left(n \in \mathbf{N}_{+}, p>0\right) .
\end{array}
$$
(1) Find the general term of the sequence $\left... | 14. (1) From the given, we have
$$
\begin{array}{l}
b_{n+1}=\frac{a_{n+1}+p}{a_{n+1}-p}=\frac{a_{n}+\frac{p^{2}}{a_{n}}+2 p}{a_{n}+\frac{p^{2}}{a_{n}}-2 p} \\
=\frac{\left(a_{n}+p\right)^{2}}{\left(a_{n}-p\right)^{2}}=b_{n}^{2}>0 .
\end{array}
$$
Then $\lg b_{n+1}=2 \lg b_{n}$
$$
\Rightarrow \lg b_{n}=2^{n-1} \lg 3 \R... | S_{n}<\left(n+\frac{23}{18}\right) p | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,411 |
1. Given $\mathrm{i}$ is the imaginary unit. Then the complex number $\frac{1+2 \mathrm{i}}{\mathrm{i}-2}=$ ( ).
(A)i
(B) $-\mathrm{i}$
(C) $-\frac{4}{5}-\frac{3}{5} i$
(D) $-\frac{4}{5}+\frac{3}{5}$ i | \begin{array}{l}\text {-1. B. } \\ \frac{1+2 \mathrm{i}}{\mathrm{i}-2}=\frac{(1+2 \mathrm{i})(\mathrm{i}+2)}{(\mathrm{i}-2)(\mathrm{i}+2)}=-\mathrm{i}\end{array}
The translation is:
\begin{array}{l}\text {-1. B. } \\ \frac{1+2 \mathrm{i}}{\mathrm{i}-2}=\frac{(1+2 \mathrm{i})(\mathrm{i}+2)}{(\mathrm{i}-2)(\mathrm{i}+... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,412 |
3. Given that $a$ and $b$ are both unit vectors, and their included angle is $\theta$. Then the proposition $p:|a-b|>1$ is the ( ) condition for the proposition $q: \theta \in$ $\left[\frac{\pi}{2}, \frac{5 \pi}{6}\right)$.
(A) Sufficient but not necessary
(B) Necessary but not sufficient
(C) Sufficient and necessary
(... | 3. From the geometric meaning of vectors, we know that
$$
\begin{array}{l}
|a-b|>1 \Rightarrow \theta \in\left(\frac{\pi}{3}, \pi\right) . \\
\theta \in\left[\frac{\pi}{2}, \frac{5 \pi}{6}\right) \Rightarrow|a-b|>1 .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,414 |
4. Given the sets
$$
\begin{array}{l}
P=\{x \mid 1 \leqslant x \leqslant 2\}, \\
M=\{x \mid 2-a \leqslant x \leqslant 1+a\} .
\end{array}
$$
If $P \cap M=P$, then the range of the real number $a$ is ( ).
(A) $(-\infty, 1]$
(B) $[1,+\infty)$
(C) $[-1,1]$
(D) $[-1,+\infty)$ | 4. B.
$$
\begin{array}{l}
\text { Given } P \cap M=P \Rightarrow P \subset M \\
\Rightarrow 2-a \leqslant 1 \\
\Rightarrow a \geqslant 1 .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,415 |
8. Let $i$ and $j$ represent the unit vectors on the $x$-axis and $y$-axis of a Cartesian coordinate system, respectively, and
$$
|a-i|+|a-2 j|=\sqrt{5} \text {. }
$$
Then the range of $|a+2 i|$ is ( ).
(A) $[2 \sqrt{2}, 3]$
(B) $\left[\frac{6 \sqrt{5}}{5}, 2 \sqrt{2}\right]$
(C) $[\sqrt{5}, 4]$
(D) $\left[\frac{6 \sq... | 8. D.
Since the endpoint of vector $a$ lies on the line segment $2 x+y-2=0(0 \leqslant x \leqslant 1)$ that satisfies $|a-i|+|a-2 j|=\sqrt{5}$, the maximum value of $|a+2 i|$ is the distance between the point $(-2,0)$ and $(1,0)$, which is 3, and the minimum value is the distance from the point $(-2,0)$ to the line se... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,419 |
9. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: \frac{x^{2}}{9}-\frac{y^{2}}{27}=1$, and let point $A\left(\frac{9}{2}, \frac{\sqrt{135}}{2}\right)$. Then the coordinates of point $M$, where the angle bisector of $\angle F_{1} A F_{2}$ intersects the $x$-axis, are ( ).
(A) $(2,0)$
(B) $(-2,0)... | 9. A.
Notice,
$$
\begin{array}{l}
F_{1}(-6,0), F_{2}(6,0) \\
\Rightarrow \frac{\left|A F_{1}\right|}{\left|A F_{2}\right|}=2 \Rightarrow \frac{\left|M F_{1}\right|}{\left|M F_{2}\right|}=2 \Rightarrow M(2,0) .
\end{array}
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,420 |
1. $p$ is a prime number greater than 3. Prove: The square of $p$ leaves a remainder of 1 when divided by 24. | 提示: 只需证 $24 \mid\left(p^{2}-1\right)$. 由于 $p$ 是大于 3 的质数, 则 $p-1 、 p+1$ 为两个连续的偶数, $p-1 、 p 、 p+1$ 为三个连续的整数, 于是,
$$
\begin{array}{c}
8 \text { I }(p-1)(p+1), 3 \mid(p-1)(p+1) . \\
\text { 又 }(3,8)=1, \text { 故 } 24 \mid\left(p^{2}-1\right) .
\end{array}
$$
Hint: It is sufficient to prove $24 \mid (p^2 - 1)$. Since $p$ i... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,421 |
10. Let $f(x)=x^{2}+b x+c$. If the equation $f(x)=x$ has no real roots, then the equation $f(f(x))=x(\quad)$.
(A) has four distinct real roots
(B) has two distinct real roots
(C) has one real root
(D) has no real roots | 10. D.
From $f(x)=x$ having no real roots, we know that the graph of the quadratic function $f(x)=x^{2}+b x+c$ is above the line $y=x$, i.e., $f(x)>x$.
Therefore, $f(f(x))>f(x)>x$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,422 |
11. Let the line $y=a x-4$ be symmetric to the line $y=8 x-b$ with respect to the line $y=x$. Then $a=$ $\qquad$ ,$b=$ $\qquad$ | II. 11. $\frac{1}{8},-32$.
It is easy to know that the equation of the line symmetric to $y=a x-4$ with respect to $y=x$ is
$$
\begin{array}{l}
y=\frac{1}{a} x+\frac{4}{a} . \\
\text { Therefore, } \frac{1}{a}=8, \frac{4}{a}=-b \Rightarrow a=\frac{1}{8}, b=-32 .
\end{array}
$$ | a=\frac{1}{8}, b=-32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,423 |
12. Let $\frac{1-|\cos x|}{1+|\cos x|}=\sin x$. Then $x=$ | $\begin{array}{l}\text { 12. } k \pi, 2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z}) . \\ \text { From } \frac{1-|\cos x|}{1+|\cos x|}=\sin x \\ \Rightarrow \sin ^{2} x=\sin x(1+|\cos x|)^{2} \\ \Rightarrow \sin x=0 \text { or } \cos x=0, \text { and } \sin x=1 \\ \Rightarrow x=k \pi, 2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z}) .... | x=k \pi, 2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,424 |
14. Let real numbers $a, b, c, d$ satisfy $ab=c^{2}+d^{2}=1$. Then the minimum value of $(a-c)^{2}+(b-d)^{2}$ is $\qquad$ | 14. $3-2 \sqrt{2}$.
It is easy to see that the minimum value of $(a-c)^{2}+(b-d)^{2}$ is the square of the minimum distance between a point on the hyperbola $ab=1$ and a point on the circle $c^{2}+d^{2}=1$, which is the square of the distance between the points $(1,1)$ and $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}... | 3-2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,426 |
15. Let $\left\{a_{n}\right\}$ be a geometric sequence, and each term is greater than
1. Then
$$
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=
$$
$\qquad$ | 15.2011.
When the common ratio is 1, $\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=2011$.
When the common ratio is $q \neq 1$,
$$
\begin{array}{l}
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}} \\
=\frac{\lg a_{1} \cdot \lg a_{2012}}{\lg q} ... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,427 |
16. Figure 3 is a defective $3 \times 3$ magic square, in which the sum of the three numbers in each row, each column, and each diagonal is equal. Then the value of $x$ is . $\qquad$ | $$
\begin{array}{l}
\text { Magic Square } \\
4017+2012 \\
=x-2003+x \\
\Rightarrow x=4016 \text {. } \\
\end{array}
$$ | 4016 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 725,428 |
17. Let $x>0$. Then
$$
f(x)=\frac{\left(x+\frac{1}{x}\right)^{4}-\left(x^{4}+\frac{1}{x^{4}}\right)}{\left(x+\frac{1}{x}\right)^{3}-\left(x^{3}+\frac{1}{x^{3}}\right)}
$$
the minimum value of $f(x)$ is | 17. $\frac{7}{3}$.
Let $t=x+\frac{1}{x}$. Then $t \geqslant 2$,
$$
\begin{array}{l}
x^{4}+\frac{1}{x^{4}}=t^{4}-4 t^{2}+2, \\
x^{3}+\frac{1}{x^{3}}=t^{3}-3 t .
\end{array}
$$
Thus $f(x)=g(t)=\frac{4}{3} t-\frac{2}{3} t^{-1}$
$$
\geqslant g(2)=\frac{4}{3} \times 2-\frac{2}{3} \times \frac{1}{2}=\frac{7}{3}=f(1) \text ... | \frac{7}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,429 |
18. Given real numbers $x_{1}, x_{2}, \cdots, x_{10}$ satisfy $\sum_{i=1}^{10}\left|x_{i}-1\right| \leqslant 4, \sum_{i=1}^{10}\left|x_{i}-2\right| \leqslant 6$. Find the average $\bar{x}$ of $x_{1}, x_{2}, \cdots, x_{10}$. | $$
\begin{array}{l}
10=\left|\sum_{i=1}^{10}\left[\left(x_{i}-1\right)-\left(x_{i}-2\right)\right]\right| \\
\leqslant \sum_{i=1}^{10}\left|x_{i}-1\right|+\sum_{i=1}^{10}\left|x_{i}-2\right| \leqslant 10 . \\
\text { Then } \sum_{i=1}^{10}\left|x_{i}-1\right|=4, \sum_{i=1}^{10}\left|x_{i}-2\right|=6, \text { and } \\
1... | 1.4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,430 |
19. Let $P$ be a moving point on the major axis of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$. A line passing through $P$ with slope $k$ intersects the ellipse at points $A$ and $B$. If $|P A|^{2}+|P B|^{2}$ depends only on $k$ and not on $P$, find the value of $k$. | 19. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and the equation of the line passing through point $P$ with slope $k$ be $y=k(x-a)$.
$$
\begin{array}{l}
\text { From }\left\{\begin{array}{l}
y=k(x-a), \\
\frac{x^{2}}{25}+\frac{y^{2}}{16}=1,
\end{array}\right. \text { we get } \\
x_{1}+x_{2}=\frac{50 a... | k= \pm \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,431 |
2. Prove: $2011 \mid \left(2012^{2012}-2010^{2010}\right)$. | Hint: $2012^{2012}-2010^{2010}$
$$
=\left(2012^{2012}-2010^{2012}\right)+\left(2010^{2012}-2010^{2010}\right) \text {. }
$$ | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,432 |
20. Let $p, q \in \mathbf{Z}_{+}$, and $q \leqslant p^{2}$. Prove: For $n \in$ $\mathbf{Z}_{+}$, there exists $N \in \mathbf{Z}_{+}$, such that
$$
\left(p-\sqrt{p^{2}-q}\right)^{n}=N-\sqrt{N^{2}-q^{n}},
$$
and $\left(p+\sqrt{p^{2}-q}\right)^{n}=N+\sqrt{N^{2}-q^{n}}$. | 20. Let $x_{1}=\left(p-\sqrt{p^{2}-q}\right)^{n}$,
$$
x_{2}=\left(p+\sqrt{p^{2}-q}\right)^{n} \text {. }
$$
Then $x_{1} x_{2}=q^{n}$.
Let $N=\frac{1}{2}\left(x_{1}+x_{2}\right)$
$$
=\sum_{j=0}^{\left[\frac{n}{2}\right]}(-1)^{j} \mathrm{C}_{n}^{2 j} p^{n-2 j}\left(p^{2}-q\right)^{j} \text {. }
$$
Clearly, $N \in \math... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,433 |
21. Let $\odot O_{4}$ be externally tangent to $\odot O_{1}$, $\odot O_{1}$ be externally tangent to $\odot O_{2}$, $\odot O_{2}$ be externally tangent to $\odot O_{3}$, and $\odot O_{3}$ be externally tangent to $\odot O_{4}$ at points $P_{1}$, $P_{2}$, $P_{3}$, and $P_{4}$, respectively. Prove:
(1) $P_{1}$, $P_{2}$, ... | (1) From the theorem that the angle between a tangent and a chord is half the central angle, we get
$$
\begin{array}{l}
\angle P_{1} P_{2} P_{3}=\frac{1}{2}\left(\angle O_{4} O_{1} O_{2}+\angle O_{1} O_{2} O_{3}\right), \\
\angle P_{3} P_{4} P_{1}=\frac{1}{2}\left(\angle O_{2} O_{3} O_{4}+\angle O_{3} O_{4} O_{1}\right... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,434 |
22. Let $i_{1}, i_{2}, \cdots, i_{10}$ be a permutation of $1,2, \cdots, 10$, and let
$$
S=\left|i_{1}-i_{2}\right|+\left|i_{3}-i_{4}\right|+\cdots+\left|i_{9}-i_{10}\right| .
$$
Find all possible values of $S$. | 22. Note that,
$$
\begin{array}{l}
S \geqslant 1+1+1+1+1=5, \\
S \leqslant 6+7+\cdots+10-(1+2+\cdots+5)=25,
\end{array}
$$
and $S \equiv \sum_{k=1}^{10} k(\bmod 2) \equiv 1(\bmod 2)$.
We will prove below that: $S$ can take all odd numbers from $5 \sim 25$. Let
$$
f\left(i_{2}, i_{4}, i_{6}, i_{8}, i_{10} ; i_{1}, i_{3... | 5,7,9,11,13,15,17,19,21,23,25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,435 |
1. As shown in Figure 1, in the acute triangle $\triangle ABC$, $\angle A > 60^{\circ}$, $H$ is the orthocenter of $\triangle ABC$, points $M$ and $N$ are on sides $AB$ and $AC$ respectively, such that $\angle HMB = \angle HNC = 60^{\circ}$, $O$ is the circumcenter of $\triangle HMN$, point $D$ is on the same side of l... | 1. Proof 1 As shown in Figure 2, let $T$ be the orthocenter of $\triangle H M N$, extend $H M$ and $C A$ to intersect at point $P$, and extend $H N$ and $B A$ to intersect at point $Q$.
It is easy to know that points $N, M, P, Q$ are concyclic.
From $\angle T H M = \angle O H N$, we have
$\angle P Q H - \angle O H N = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,436 |
3. Find the smallest real number $c$ such that:
For any real-coefficient polynomial of degree 2012 with leading coefficient 1,
$$
P(x)=x^{2012}+a_{2011} x^{2011}+a_{2010} x^{2010}+\cdots+a_{0} \text {, }
$$
some of its coefficients can be multiplied by -1, while the others remain unchanged, so that each root $z$ of t... | 3. First, prove: $c \geqslant \cot \frac{\pi}{4022}$.
Consider the polynomial $P(x)=x^{2012}-x$.
By changing the signs of the coefficients of $P(x)$, we obtain four polynomials $P(x), -P(x), Q(x)=x^{2012}+x$, and $-Q(x)$.
Notice that, $P(x)$ and $-P(x)$ have the same roots, one of which is
$$
z_{1}=\cos \frac{1006}{2... | \cot \frac{\pi}{4022} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,438 |
4. Given an integer $n \geqslant 4$, let $A, B \subseteq \{1, 2, \cdots, n\}$. It is known that for any $a \in A, b \in B, ab + 1$ is a perfect square. Prove:
$$
\min \{|A|,|B|\} \leqslant \log _{2} n .
$$
(Xiong Bin) | 4. First, prove a lemma.
Lemma: Let $a, a', b, b'$ be positive integers such that $a < a'$ and $b < b'$.
Proof: Notice that,
$$
\begin{aligned}
& (a b+1)\left(a^{\prime} b^{\prime}+1\right)>\left(a b^{\prime}+1\right)\left(a^{\prime} b+1\right) \\
(\Leftrightarrow & \left.\left(a^{\prime}-a\right)\left(b^{\prime}-b\ri... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,439 |
6. A square chessboard is composed of $2012 \times 2012$ unit squares, and some of these squares are occupied by beetles, with at most one beetle per square. At a certain moment, all the beetles fly up and then land again on the squares of the chessboard, with at most one beetle per square. The vector from the center o... | 6. Establish a Cartesian coordinate system parallel to the grid lines with the center of the chessboard as the origin. Let the set of centers of all small squares be denoted as $S$, and the set of centers of the small squares initially occupied by beetles be denoted as $M_{1} \subseteq S$. After the beetles move again,... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,441 |
1. Arrange the numbers $1, 2, 3, 4, 5$ in a row, with the last number being even, and ensure that the sum of any three consecutive numbers is divisible by the first of these three numbers. How many arrangements satisfy this condition?
(A) 8
(B) 9
(C) 10
(D) 11 | - 1. C.
From the problem, we know that in three consecutive numbers, the first number can divide the sum of the last two numbers. Therefore, we first arrange the last digit and then test one by one from back to front. Thus, the numbers that meet the requirements are $54132, 14532, 35142, 43152, 23514$, $15324, 32154, ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 725,442 |
Example 1 As shown in Figure 1, in square $A B C D$, $E$ and $F$ are points on sides $B C$ and $C D$ respectively, with $\angle E A F=$ $45^{\circ}$. Prove:
$$
B E+D F=E F \text {. }
$$ | Prove: As shown in Figure 1, draw a perpendicular line from point $A$ to $A F$ intersecting line $C B$ at point $G$. Given $\angle E A F=45^{\circ}$, we know $\angle G A E=45^{\circ}$. Therefore, $\angle B A E+\angle D A F=45^{\circ}$ $\Rightarrow \angle B A G=\angle D A F$. Since $A B=A D$
$$
\begin{array}{l}
\Rightar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,443 |
Example 2 In $\triangle A B C$, it is known that $\angle A=120^{\circ}$ and $A D, B E$ are the angle bisectors of $\triangle A B C$. Prove:
$$
A B+A E=A D+B D .
$$ | Proof As shown in Figure 2, let the circumcircle of $\triangle ABE$ intersect side $BC$ at point $G$. Take a point $F$ on the extension of $BA$ such that $BF = BG$. Connect $EF$, $ED$, and $EG$.
Obviously, point $F$ is symmetric to point $G$ with respect to line $BE$. Therefore,
$$
\begin{array}{l}
\angle F = \angle E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,444 |
Example 6 Find the positive integer $n$ that satisfies the following conditions: $\varphi(n)>2$, and for any positive integers $a>b>c$ not exceeding $n$, if $(a, n)=(b, n)=(c, n)=1$, then $(a+b+c, n)=1$. | 【Analysis】 $\varphi(n)>2$, so $n=5$ or $n \geqslant 7$.
When $n=5$, by $1+2+3+4=10$, we know that the sum after removing any number is coprime with 5, so $n=5$ satisfies the condition.
When $n=7$, by $(1+2+4,7) \neq 1$, we know the proposition does not hold.
When $n=8$, by $1+3+5+7=16$, we know that the sum after rem... | 5,10,12,2^{k}(k \geqslant 3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,445 |
Four. (50 points) There are 2013 representatives from different countries attending a conference, and each representative speaks several languages. It is known that any four representatives can communicate without the help of any other representatives, meaning that any two of the four can communicate directly in a comm... | First, prove two lemmas.
Lemma 1: In a 4-vertex connected graph $G$, there are three points forming a connected subgraph.
Proof of Lemma 1: By the connectivity of $G$, it has a spanning tree. Removing a pendant vertex from this tree results in a connected subgraph with three points.
Lemma 2: If in a 6-vertex graph $G... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,446 |
Given real numbers $a, b, c, d$ satisfy
$$
2 a^{2}+3 c^{2}=2 b^{2}+3 d^{2}=(a d-b c)^{2}=6 \text{. }
$$
Prove: $a^{2}+b^{2}=3, c^{2}+d^{2}=2$. | Proof: Let $m=2 a^{2}+2 b^{2}, n=3 c^{2}+3 d^{2}$.
Then $m+n=2 a^{2}+3 c^{2}+2 b^{2}+3 d^{2}=12$.
And $4 m n \leqslant(m+n)^{2}=12^{2}$, hence
$m n \leqslant 36$.
$$
\begin{array}{l}
\text { Also } m n=\left(2 a^{2}+2 b^{2}\right)\left(3 c^{2}+3 d^{2}\right) \\
=6\left(a^{2} d^{2}+b^{2} c^{2}+a^{2} c^{2}+b^{2} d^{2}\ri... | a^{2}+b^{2}=3, c^{2}+d^{2}=2 | Algebra | proof | Yes | Yes | cn_contest | false | 725,447 |
Given $S=\{1,2, \cdots, 16\}$. Can we select 20 four-element subsets of $S$ and divide them into 5 groups such that the following conditions are met:
(1) the four subsets in each group are pairwise disjoint;
(2) any two subsets from different groups have exactly one common element? | The answer is affirmative.
20 quadruple subsets can be constructed as follows:
$$
\begin{array}{l}
\{1,2,3,4\},\{5,8,11,14\}, \\
\{6,9,12,15\},\{7,10,13,16\}, \\
\{1,5,6,7\},\{2,10,12,14\}, \\
\{3,9,11,16\},\{4,8,13,15\}, \\
\{1,8,9,10\},\{2,7,11,15\}, \\
\{3,6,13,14\},\{4,5,12,16\}, \\
\{1,11,12,13\},\{2,6,8,16\}, \\
... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,449 |
Let $n$ be a positive integer. Prove:
$$
\sum_{k=1}^{n} \frac{2 k-1}{(2 k+1)(2 k+3)} \geqslant \frac{n^{3}}{(n+2)\left(2 n^{2}+4 n-1\right)} .
$$ | Proof: Let
$$
b_{k}=(2 k-1)(2 k+1)(2 k+3)(2 k+5) \text {, }
$$
where, $k \in \mathbf{N}_{+}$.
$$
\begin{array}{l}
\text { Then } b_{k}-b_{k-1} \\
=(2 k-1)(2 k+1)(2 k+3)(2 k+5)- \\
(2 k-3)(2 k-1)(2 k+1)(2 k+3) \\
= 8(2 k-1)(2 k+1)(2 k+3) .
\end{array}
$$
Thus, $b_{k}-b_{k-1}$
Let $k=1,2, \cdots, n$, and sum up:
$$
\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,450 |
Question As shown in figures $1$ and $2$, for the scalene $\triangle ABC$, the circumcenter is $O$, and the centroid is $G$. $A_{1}, B_{1}, C_{1}$ are the midpoints of sides $BC, CA$, and $AB$, respectively. Draw perpendiculars $l_{b}$ and $l_{c}$ from points $B$ and $C$ to $OG$. $l_{b}$ intersects $A_{1}C_{1}$ at poin... | Proof: Let the lines $B E$ and $C A$ intersect at point $M$, and the lines $C F$ and $B A$ intersect at point $P$, as shown in figures 3 and 4. Draw $M N \parallel A E$ intersecting the line $A B$ at point $N$, and draw $P Q \parallel A F$ intersecting the line $A C$ at point $Q$.
Since $B C_{1} = C_{1} A$, we have $B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,451 |
1. The sum of all roots of the equation $|2 x-4|=5$ is ( ).
(A) -0.5
(B) 4.5
(C) 5
(D) 4 | - 1. D.
$$
\begin{array}{l}
\text { Given }|2 x-4|=5 \Rightarrow 2 x-4= \pm 5 \\
\Rightarrow x_{1}=4.5, x_{2}=-0.5 \\
\Rightarrow x_{1}+x_{2}=4.5-0.5=4 .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,452 |
2. In the Cartesian coordinate system $x O y$, the line $y=a x+24$ forms a triangle with the positive axes of the two coordinates, and the area of this triangle is 72. Then the coordinates of the point not on the line are ( ).
(A) $(3,12)$
(B) $(1,20)$
(C) $(-0.5,26)$
(D) $(-2.5,32)$ | 2. D.
The line $y=a x+24$ intersects the positive x-axis at the point $\left(-\frac{24}{a}, 0\right)$, and the positive y-axis at the point $(0,24)$. This gives the lengths of the two legs of the right triangle as 24 and $-\frac{24}{a}$, and the area is
$$
\frac{1}{2} \times 24 \times\left(-\frac{24}{a}\right)=-\frac{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,453 |
3. The arithmetic mean of two positive numbers is $2 \sqrt{3}$, and the arithmetic square root of their product is $\sqrt{3}$. Then the larger number is greater than the smaller number by ( ).
(A) 4
(B) $2 \sqrt{3}$
(C) 6
(D) $3 \sqrt{3}$ | 3. C.
Let these two positive numbers be $x, y$. According to the conditions, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+y=4 \sqrt{3}, \\
x y=3
\end{array}\right. \\
\Rightarrow|x-y|=\sqrt{(x-y)^{2}}=6 .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,454 |
4. In $\triangle A B C$, it is known that $M$ is the midpoint of $A B$, $N$ is a point on side $B C$, and $C N=2 B N$. Connecting $A N$ and $M C$ intersects at point $O$. The area of quadrilateral $B M O N$ is 14. Then the area of $\triangle A B C$ is ( ).
(A) 56
(B) 60
(C) 64
(D) 68 | 4. B.
As shown in Figure 1, connect $B O$.
Let $S_{\triangle BMO}=x$,
$$
S_{\triangle BNO}=y \text{. }
$$
$$
\text{Then } S_{\triangle MO}=x \text{, }
$$
$$
\begin{array}{l}
S_{\triangle CNO}=2 y, \\
S_{\triangle ACO}=3 y . \\
\text{Therefore, } \frac{2 x}{y}=\frac{S_{\triangle ABO}}{S_{\triangle BON}}=\frac{AO}{ON}=\... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,455 |
5. When $a=1.67, b=1.71, c=0.46$,
$$
\begin{array}{l}
\frac{1}{a^{2}-a c-a b+b c}+\frac{2}{b^{2}-a b-b c+a c}+\frac{1}{c^{2}-a c-b c+a b} \\
=\text { ( ). }
\end{array}
$$
(A) 20
(B) 15
(C) 10
(D) 5.55 | 5. A.
Notice,
$$
\begin{array}{l}
\frac{1}{a^{2}-a c-a b+b c}+\frac{2}{b^{2}-a b-b c+a c}+\frac{1}{c^{2}-a c-b c+a b} \\
=\frac{1}{(a-b)(a-c)}+\frac{2}{(b-a)(b-c)}+\frac{1}{(c-a)(c-b)} \\
=\frac{(c-b)+2(a-c)+(b-a)}{(a-b)(b-c)(c-a)} \\
=\frac{a-c}{(a-b)(b-a)(c-a)} \\
=-\frac{1}{(a-b)(b-c)} \\
=20 .
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,456 |
$$
\begin{array}{l}
1 \times 2 - 3 \times 4 + 5 \times 6 - 7 \times 8 + \cdots + \\
2009 \times 2010 - 2011 \times 2012 \\
= \quad .
\end{array}
$$ | $$
\text { II, 1. }-2025078 \text {. }
$$
Notice that,
$$
\begin{array}{l}
(n+2)(n+3)-n(n+1)=4 n+6 \\
=n+(n+1)+(n+2)+(n+3) .
\end{array}
$$
Therefore, the original expression is
$$
=-(1+2+\cdots+2012)=-2025078 .
$$ | -2025078 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,457 |
2. The ten positive integers from $1 \sim 10$ are written in a row in some order, denoted as $a_{1}, a_{2}, \cdots, a_{10}, S_{1}=a_{1}, S_{2}=a_{1}+a_{2}$, $\cdots, S_{10}=a_{1}+a_{2}+\cdots+a_{10}$. Then, among $S_{1}, S_{2}, \cdots, S_{10}$, the maximum number of primes that can occur is . | 2. 7 .
Adding an odd number changes the sum to the opposite parity, and among even numbers, only 2 is a prime. Let $b_{i}$ be the $i$-th ($i=1,2,3,4,5$) odd number in this row. Then, when adding $b_{2}$ and $b_{4}$, the sums $S_{k}$ and $S_{n}$ are even numbers greater than 2. Therefore, $S_{k} \backslash S_{n}$ and $... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,458 |
3. In $\triangle A B C$, it is known that $A B=12, A C=9, B C$ $=13$. From $A$, perpendiculars are drawn to the angle bisectors of $\angle C$ and $\angle B$, with the feet of the perpendiculars being $M$ and $N$, respectively. Connect $M N$. Then $\frac{S_{\triangle A M N}}{S_{\triangle A B C}}=$ $\qquad$ . | 3. $\frac{2}{13}$.
As shown in Figure 2, extend $A M$ to intersect $B C$ at point $D$, and extend $A N$ to intersect $B C$ at point $E$. Then, $\triangle A C D$ and $\triangle A B E$ are both isosceles triangles.
Thus, $B E=A B=12, C D=A C=9$. Since $B C=13$, we have $D E=12+9-13=8$. Therefore, $\frac{S_{\triangle A D... | \frac{2}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,459 |
Example 3 Given an acute triangle $\triangle ABC, \angle BAC=60^{\circ}$, $AB=c, AC=b(b>c)$, the orthocenter and circumcenter of $\triangle ABC$ are $H$ and $O$ respectively, and $OH$ intersects sides $AB$ and $AC$ at points $X$ and $Y$. Prove: $OH=b-c .{ }^{[1]}$ | Prove as shown in Figure 3, connect $H A, O A, O B, O C$.
Obviously, $\angle A X Y=\angle A B H+\angle X H B=60^{\circ}$.
Given $\angle B A C=60^{\circ}$
$\Rightarrow \triangle A X Y$ is an equilateral triangle
$\Rightarrow A Y=A X$.
Given $\angle O A C=90^{\circ}-\angle A B C=\angle B A H$
$\Rightarrow X H=Y O$
$\Righ... | OH=b-c | Geometry | proof | Yes | Yes | cn_contest | false | 725,460 |
4. Real numbers $x, y$ satisfy
$$
x^{2}+12 x y+52 y^{2}-8 y+1=0 \text {. }
$$
Then $x^{2}-y^{2}=$ $\qquad$ | 4. $\frac{35}{16}$.
From the problem, we get
$$
\begin{array}{l}
(x+6 y)^{2}+(4 y-1)^{2}=0 \\
\Rightarrow x+6 y=4 y-1=0 \\
\Rightarrow y=\frac{1}{4}, x=-\frac{3}{2} \\
\Rightarrow x^{2}-y^{2}=\frac{35}{16} .
\end{array}
$$ | \frac{35}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,461 |
5. Let $P$ be a point inside an equilateral $\triangle A B C$, with $A P=3$, $B P=4$, $C P=5$. Then the area of quadrilateral $A B C P$ is $\qquad$ | $$
\text { 5. } 6+4 \sqrt{3} \text {. }
$$
As shown in Figure 3, with point $B$ as the center of rotation, $\triangle A P B$ is rotated $60^{\circ}$ clockwise to the position of $\triangle B Q C$. Connect $P Q$.
It is easy to see that $\triangle B P Q$ is
an equilateral triangle, i.e.,
$$
\begin{array}{l}
P Q=B Q=B P... | 6+4 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,462 |
Three, (10 points) Prove: For any three distinct numbers $a, b, c$, we have
$$
\frac{(a+b-c)^{2}}{(a-c)(b-c)}+\frac{(b+c-a)^{2}}{(b-a)(c-a)}+\frac{(c+a-b)^{2}}{(c-b)(a-b)}
$$
is a constant. | $$
\begin{array}{l}
S=\frac{(a+b-c)^{2}}{(a-c)(b-c)}+\frac{(b+c-a)^{2}}{(b-a)(c-a)}+\frac{(c+a-b)^{2}}{(c-b)(a-b)}, \\
p=a^{2}+b^{2}+c^{2}-2 a b-2 b c-2 c a . \\
\text { Then } (a+b-c)^{2}=p+4 a b, \\
(b+c-a)^{2}=p+4 b c, \\
(c+a-b)^{2}=p+4 c a . \\
\text { Hence } S=\frac{p+4 a b}{(a-c)(b-c)}+\frac{p+4 b c}{(b-a)(c-a)... | 4 | Algebra | proof | Yes | Yes | cn_contest | false | 725,463 |
Four, (15 points) It is known that a positive integer $n$ can be expressed as the sum of 2011 identical natural numbers, and also as the sum of 2012 identical natural numbers. Determine the minimum value of $n$.
| Let $a_{1}, a_{2}, \cdots, a_{2011}$ be 2011 natural numbers with the same digit sum, and $b_{1}, b_{2}, \cdots, b_{2012}$ be 2012 natural numbers with the same digit sum, and they satisfy
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2011}=n \\
=b_{1}+b_{2}+\cdots+b_{2012} .
\end{array}
$$
Since each of the numbers $a_{1... | 10055 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,464 |
Five. (15 points) In $\triangle A B C$, it is known that $\angle A B C=$ $\angle B A C=70^{\circ}, P$ is a point inside the triangle, $\angle P A B=40^{\circ}$, $\angle P B A=20^{\circ}$. Prove: $P A+P B=P C$. | Because $\angle ABC = \angle BAC = 70^{\circ}$, therefore, $AC = BC$, $\angle ACB = 40^{\circ}$.
As shown in Figure 4, take a point $N$ on the extension of $BP$ such that $PN = PA$, and connect $AN$ and $CN$. It is easy to see that $\angle APB = 120^{\circ}$.
Then $\angle APN = 60^{\circ}$.
Therefore, $\triangle APN... | PA + PB = PC | Geometry | proof | Yes | Yes | cn_contest | false | 725,465 |
1. Let the sets be
$$
\begin{array}{l}
S=\left\{x \mid x^{2}-5 x-6<0\right\}, \\
T=\{x|| x+2 \mid \leqslant 3\} .
\end{array}
$$
Then $S \cap T=(\quad)$.
(A) $\{x \mid-5 \leqslant x<-1\}$
(B) $\{x \mid-5 \leqslant x<5\}$
(C) $\{x \mid-1<x \leqslant 1\}$
(D) $\{x \mid 1 \leqslant x<5\}$ | -1. C.
Notice,
$$
\begin{array}{l}
S=\{x \mid-1<x<6\}, \\
T=\{x \mid-5 \leqslant x \leqslant 1\} .
\end{array}
$$
Therefore, $S \cap T=\{x \mid-1<x \leqslant 1\}$. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 725,466 |
2. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the angle formed by $B C_{1}$ and the section $B B_{1} D_{1} D$ is ( ).
(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{3}$
(D) $\frac{\pi}{2}$ | 2. A.
As shown in Figure 2, draw $C_{1} O \perp B_{1} D_{1}$ at point $O$, and connect $B O$. Then $\angle C_{1} B O$ is the required angle.
Given $C_{1} O=\frac{1}{2} B C_{1}$,
we know $\angle C_{1} B O=\frac{\pi}{6}$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,467 |
3. Given $f(x)=x^{2}-2 x+3, g(x)=k x-1$, then “ $|k| \leqslant 2$ ” is the ( ) condition for “ $f(x) \geqslant g(x)$ to hold for all $x \in \mathbf{R}$ ”.
(A) Sufficient but not necessary
(B) Necessary but not sufficient
(C) Sufficient and necessary
(D) Neither sufficient nor necessary | 3. A.
Given $f(x) \geqslant g(x)$ always holds
$\Rightarrow x^{2}-(2+k) x+4 \geqslant 0$ always holds
$\Rightarrow(2+k)^{2}-4 \times 4 \leqslant 0$
$\Rightarrow-6 \leqslant k \leqslant 2$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,468 |
4. Let the area of the equilateral triangle $\triangle_{1}$ be $S_{1}$, construct the incircle of $\triangle_{1}$, then construct the inscribed equilateral triangle $\triangle_{2}$ in the incircle, with area $S_{2}$, and so on, creating a series of equilateral triangles $\triangle_{3}$, $\triangle, \cdots$, with corres... | 4. B.
Notice that the series of triangles obtained are similar triangles, with a similarity ratio of $\frac{1}{2}$, so the areas form a geometric sequence with a common ratio of $\frac{1}{4}$. Therefore,
$$
T_{n}=\frac{1-\left(\frac{1}{4}\right)^{n}}{1-\frac{1}{4}} \text {. }
$$
Thus, $\lim _{n \rightarrow+\infty} T_... | \frac{4}{3} | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,469 |
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