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values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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9. (16 points) As shown in Figure 1, let the base of the pyramid $E-ABCD$ be a rhombus, and $\angle ABC=60^{\circ}, AB=EC=2$, $AE=BE=\sqrt{2}$.
(1) Prove: Plane $EAB \perp$ Plane $ABCD$;
(2) Find the cosine value of the dihedral angle $A-EC-D$.
保留了原文的换行和格式。 | (1) Take the midpoint $O$ of $AB$, and connect $EO$, $CO$. Since $AE=EB=\sqrt{2}, AB=2$, we know that $\triangle AEB$ is an isosceles right triangle.
Thus, $EO \perp AB, EO=1$.
Also, $AB=BC, \angle ABC=60^{\circ}$, so $\triangle ACB$ is an equilateral triangle.
Therefore, $CO=\sqrt{3}$.
Given $EC=2$,
$EC^{2}=EO^{2}+CO^... | \frac{2 \sqrt{7}}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,432 |
10. (20 points) Given the function
$$
f(x)=\frac{\ln (1+x)}{x} \text {. }
$$
(1) When $x>0$, prove: $f(x)>\frac{2}{x+2}$.
(2) When $x>-1$, and $x \neq 0$, the inequality
$$
f(x)<\frac{1+k x}{1+x}
$$
holds, find the value of the real number $k$. | 10. (1) Let $h(x)=\ln (1+x)-\frac{2 x}{x+2}$. Then $h^{\prime}(x)=\frac{x^{2}}{(1+x)(2+x)^{2}}$.
When $x>0$, it is easy to see that $h^{\prime}(x)>0$.
Thus, $h(x)$ is an increasing function on $(0,+\infty)$.
Therefore, $h(x)>h(0)=0$, which means
$\ln (1+x)-\frac{2 x}{x+2}>0$
$\Rightarrow \ln (1+x)>\frac{2 x}{x+2}$.
Sin... | k \leqslant \frac{1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 726,433 |
11. (20 points) In the sequence $\left\{x_{n}\right\}$, it is known that
$$
x_{1}=1 \text {, and } x_{n+1}=1+\frac{1}{x_{n}+1} \text {. }
$$
(1) If $a_{n}=\frac{1}{x_{n}+\sqrt{2}}$, find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) If $b_{n}=\left|x_{n}-\sqrt{2}\right|$, and the sum of the firs... | 11. (1) Notice,
$$
\begin{array}{l}
a_{n+1}=\frac{1}{x_{n+1}+\sqrt{2}} \\
=\frac{1}{1+\frac{1}{x_{n}+1}+\sqrt{2}} \\
=\frac{x_{n}+1}{x_{n}+2+\sqrt{2} x_{n}+\sqrt{2}} \\
=\frac{\left(x_{n}+\sqrt{2}\right)+(1-\sqrt{2})}{\left(x_{n}+\sqrt{2}\right)(1+\sqrt{2})} \\
=\frac{1}{1+\sqrt{2}}+\frac{1-\sqrt{2}}{1+\sqrt{2}} \cdot ... | S_{n}<\frac{\sqrt{2}}{2} | Algebra | proof | Yes | Yes | cn_contest | false | 726,434 |
12. (20 points) Given the ellipse $\frac{x^{2}}{4}+y^{2}=1$, and $P$ is any point on the circle $x^{2}+y^{2}=16$. Tangents $PA$ and $PB$ are drawn from $P$ to the ellipse, touching the ellipse at points $A$ and $B$ respectively. Find the maximum and minimum values of $\overrightarrow{P A} \cdot \overrightarrow{P B}$. | 12. Let point $P(m, n), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then the tangent equations are
$$
\begin{array}{l}
l_{A P}: \frac{x_{1} x}{4}+y_{1} y=1, \\
l_{P B}: \frac{x_{2} x}{4}+y_{2} y=1 .
\end{array}
$$
Since the tangents $P A$ and $P B$ both pass through point $P$, we have $\frac{x_{1} m}{4}+y... | \frac{33}{4} \text{ and } \frac{165}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,435 |
1. Given $x-\frac{2008}{x}=5$. Then $\frac{(x-2)^{3}-(x-1)^{2}+1}{x-2}=(\quad)$.
(A) 2009
(B) 2010
(C) 2011
(D) 2012 | -.1. D.
From the given, $x^{2}-5 x=2008$.
$$
\begin{array}{l}
\text { Then } \frac{(x-2)^{3}-(x-1)^{2}+1}{x-2} \\
=\frac{(x-2)^{3}-x(x-2)}{x-2} \\
=x^{2}-5 x+4=2012 .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,436 |
2. The equation
$$
6 y^{4}-35 y^{3}+62 y^{2}-35 y+6=0
$$
has ( ) integer solutions.
(A) 1
(B) 2
(C) 3
(D) 4 | 2. B.
From the given, we have
$$
\begin{array}{l}
6\left(y+\frac{1}{y}\right)^{2}-35\left(y+\frac{1}{y}\right)+50=0 \\
\Rightarrow y+\frac{1}{y}=2+\frac{1}{2} \text { or } y+\frac{1}{y}=3+\frac{1}{3} \\
\Rightarrow y_{1}=2, y_{2}=\frac{1}{2}, y_{3}=3, y_{4}=\frac{1}{3} .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,437 |
3. In Rt $\triangle A B C$, $D$ is the midpoint of the hypotenuse $B C$, and $E, F$ are points on $A B, A C$ respectively. Let the perimeter of $\triangle D E F$ be $l$. Then $(\quad)$.
(A) $l>B C$
(B) $l=B C$
(C) $l<B C$
(D) cannot be determined | 3. A.
As shown in Figure 4, construct the symmetric points $G$ and $H$ of $D$ with respect to $AB$ and $AC$, respectively, and connect $AH$, $AG$, $AD$, $HF$, and $GE$.
Then, $DE = GE$, $DF = HF$, and $AD = AG = AH$.
It is easy to prove that points $H$, $A$, and $G$ are collinear, and $\triangle HDG$ is a right triang... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,438 |
4. Let $[x]$ denote the integer part of the real number $x$. Then $[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{49}]=(\quad)$.
(A) 146
(B) 161
(C) 210
(D) 365 | 4. C.
$$
\begin{array}{l}
{[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{49}]} \\
=1 \times 3+2 \times 5+3 \times 7+4 \times 9+5 \times 11+6 \times 13+7 \\
=210
\end{array}
$$ | 210 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,439 |
5. As shown in Figure 1, given a cyclic quadrilateral $A B E C, A E \perp$ $B C$ at point $D$, and $B D=2$, $C D=3, \angle B E C=135^{\circ}$. Then the area of quadrilateral $A B E C$ is ( ).
(A) 12
(B) 18
(C) $\frac{35}{2}$
(D) $\frac{25}{2}$ | 5. C.
Notice that, $\angle B A C=45^{\circ}$.
Let $A \dot{D}=x$.
Since $A D \perp B C$
$$
\begin{array}{l}
\Rightarrow A B^{2}=x^{2}+4, A C^{2}=x^{2}+9 \\
\Rightarrow S_{\triangle A B C}^{2}=\left(\frac{1}{2} A B \cdot A C \sin 45^{\circ}\right)^{2} . \\
\text { Also } S_{\triangle B C}^{2}=\left(\frac{1}{2} B C \cdot... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,440 |
Example 1 From the 205 positive integers $1,2, \cdots, 205$, what is the maximum number of integers that can be selected such that for any three selected numbers $a, b, c (a<b<c)$, we have
$$
a b \neq c ?^{[1]}
$$
(2005, (Casio Cup) National Junior High School Mathematics Competition) | Estimate first.
Since $14 \times 15=210>205$, then $14,15, \cdots$, 205 satisfy that for any three numbers $a 、 b 、 c(a<b<c)$, we have $a b \neq c$.
Because 1 multiplied by any number equals the number itself, $1,14,15, \cdots, 205$ satisfy the condition.
Therefore, there are $205-14+1+1=193$ numbers in total.
If we se... | 193 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,441 |
Example 2: 9 judges score 12 athletes participating in a bodybuilding competition. Each judge gives 1 point to the athlete they consider to be in 1st place, 2 points to the athlete in 2nd place, ..., and 12 points to the athlete in 12th place. The final scoring shows: the difference between the highest and lowest score... | Explanation: It is impossible for 9 judges to give 1 point to five or more athletes, because among five or more athletes, at least one athlete must be rated no less than 5 by a judge. However, according to the problem, each of these five athletes is rated no more than 4 by each judge, which is a contradiction.
Therefo... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,442 |
Example 2 In $\triangle A B C$, it is known that $\angle A: \angle B: \angle C = 4: 2: 1, \angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively.
(1) Prove: $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$;
(2) Find the value of $\frac{(a+b-c)^{2}}{a^{2}+b^{2}+c^{2}}$. | (1) Proof As shown in Figure 3, construct the angle bisector of $\angle ABC$, intersecting the circumcircle of $\triangle ABC$ at point $M$, and construct the angle bisector of $\angle BAC$, intersecting the circumcircle of $\triangle ABC$ at point $N$.
Then $AM=MC=AB=c, AM \parallel BC$;
$$
CN=NB=AC=b, AB \parallel CN... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,443 |
4. Given the equation in $x$
$$
(a-1) x^{2}+2 x-a-1=0
$$
has roots that are all integers. Then the number of integer values of $a$ that satisfy this condition is
$\qquad$. | 4.5.
When $a=1$, $x=1$.
When $a \neq 1$, it is easy to see that $x=1$ is an integer root of the equation.
Furthermore, from $1+x=\frac{2}{1-a}$ and $x$ being an integer, we know
$$
1-a= \pm 1, \pm 2 \text {. }
$$
Therefore, $a=-1,0,2,3$.
In summary, there are 5 integer values of $a$ that satisfy the condition. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,444 |
II. (25 points) As shown in Figure 4, point $P$ is inside quadrilateral $A B C D$, and satisfies $\angle P A B=\angle P C B$. Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the circumcenters of $\triangle P A B, \triangle P B C, \triangle P C D, \triangle P D A$, respectively. Prove: The area of quadrilateral $O_{1} O_{2} O_{3} O... | As shown in Figure 7, let $P A, P B, P C, P D$ intersect $\mathrm{O}_{4} \mathrm{O}_{1}, \mathrm{O}_{1} \mathrm{O}_{2}, \mathrm{O}_{2} \mathrm{O}_{3}, \mathrm{O}_{3} \mathrm{O}_{4}$ at points $\mathrm{H}, E, F, G$ respectively. Construct quadrilateral $H E F G$, and then draw $P Q \perp B C$, connecting $Q D, Q C$.
It ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,445 |
Three. (25 points) Let the quadratic function
$$
f(x)=a x^{2}+b x+c
$$
satisfy the following conditions:
(i) When $x$ is a real number, its minimum value is 0, and
$$
f(x-1)=f(-x-1)
$$
holds;
(ii) There exists a real number $m (m>1)$, such that there exists a real number $t$, as long as $1 \leqslant x \leqslant m$, t... | (1) In condition (ii), let $x=1$, we get $f(1)=1$.
From condition (i), we know that the quadratic function opens upwards and is symmetric about $x=-1$, so we can assume the quadratic function to be
$$
f(x)=a(x+1)^{2}(a>0) .
$$
Substituting $f(1)=1$ into the above equation, we get $a=\frac{1}{4}$.
Therefore, $f(x)=\fra... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,446 |
1. Let $m, n \in \mathbf{N}_{+}$, and $m>n$, sets $A, B, C$ satisfy
$$
\begin{array}{l}
A=\{1,2, \cdots, m\}, B=\{1,2, \cdots, n\}, \\
C \subseteq A, B \cap C \neq \varnothing .
\end{array}
$$
Then the number of sets $C$ that meet the conditions is $\qquad$ | $$
-1.2^{m-n}\left(2^{n}-1\right) \text {. }
$$
From the condition, we know that the elements of set $C$ partly come from the non-empty subsets of set $B$, which have $2^{n}-1$ ways of selection; the other part comes from the set $\{n+1, n+2, \cdots, m\}$, which has $2^{m-n}$ ways of selection. Therefore, there are a ... | 2^{m-n}\left(2^{n}-1\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,447 |
2. Given the general term of the sequence $\left\{a_{n}\right\}$
$$
a_{n}=\frac{(n+1)^{4}+n^{4}+1}{(n+1)^{2}+n^{2}+1} \text {. }
$$
Then the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, $S_{n}=$ $\qquad$ | $\begin{array}{l}\text { 2. } \frac{1}{3} n\left(n^{2}+3 n+5\right) . \\ S_{n}=\sum_{k=1}^{n}\left(k^{2}+k+1\right) \\ =\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}+n \\ =\frac{1}{3} n\left(n^{2}+3 n+5\right) .\end{array}$ | \frac{1}{3} n\left(n^{2}+3 n+5\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,448 |
3. Given a geometric sequence $\left\{a_{n}\right\}$ satisfies
$$
\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\cdots+a_{n}\right)=-2 \text {. }
$$
Then the range of values for $a_{1}$ is $\qquad$ | 3. $(-4,-2) \cup(-2,0)$.
Let the common ratio be $q$. According to the problem, we have
$$
\left\{\begin{array}{l}
0<|q|<1, \\
\frac{a_{1}}{1-q}=-2 .
\end{array}\right.
$$
Then $\left|a_{1}+2\right|=|2 q| \in(0,2)$.
Therefore, $a_{1} \in(-4,-2) \cup(-2,0)$. | (-4,-2) \cup(-2,0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,449 |
5. As shown in Figure 1, given a regular tetrahedron $A B C D$ with edge length $1, M$ is the midpoint of $A C$, and $P$ lies on the line segment $D M$. Then the minimum value of $A P + B P$ is $\qquad$ | 5. $\sqrt{1+\frac{\sqrt{6}}{3}}$.
Let $\angle B D M=\theta$. In $\triangle B D M$, $B D=1, B M=M D=\frac{\sqrt{3}}{2}$.
It is easy to get $\cos \theta=\frac{\sqrt{3}}{3}, \sin \theta=\frac{\sqrt{6}}{3}$.
As shown in Figure 3, rotate $\triangle B D M$ around $D M$ so that $\triangle B D M$ lies in the plane $A C D$,
at... | \sqrt{1+\frac{\sqrt{6}}{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,451 |
6. When rolling a die twice, the numbers obtained are $m, n$, and the vector $a=(m, n)$ is formed. Then the probability that the angle between $a$ and $b=(1,-1)$ is an angle in a right triangle is $\qquad$
6. $\frac{7}{12}$.
Since $m, n$ can both take values from $1 \sim 6$, the vector $a$ has | $6 \times 6=36$ ways to choose.
And $\cos \langle\boldsymbol{a}, \boldsymbol{b}\rangle=\frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}||\boldsymbol{b}|}=\frac{m-n}{\sqrt{2} \cdot \sqrt{m^{2}+n^{2}}}$.
Therefore, $\langle a, b\rangle$ being a right angle or an acute angle is equivalent to $m \geqslant n$.
Suc... | \frac{7}{12} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,452 |
Example 3 Given that $C$ is the midpoint of line segment $A B$, the circle $\odot O_{1}$ passing through points $A$ and $C$ intersects the circle $\odot O_{2}$ passing through points $B$ and $C$ at points $C$ and $D$. $P$ is the midpoint of arc $\overparen{A D}$ on $\odot O_{1}$ (excluding point $C$), and $Q$ is the mi... | Proof As shown in Figure 4, connect $P A, P D, Q D, Q B$, and let $P C$ intersect $A D$ at point $E$, and $Q C$ intersect $B D$ at point $F$.
By Property 4, we know
$$
P D^{2}=P E \cdot P C, Q D^{2}=Q F \cdot Q C,
$$
and $C E \cdot C P=C A \cdot C D=C B \cdot C D=C F \cdot C Q$.
Therefore, $P D^{2}-Q D^{2}=P E \cdot P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,453 |
7. From the arithmetic sequence $2,5,8, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$ ـ. | 7.8.
First, let's take $x_{1}, x_{2}, \cdots, x_{k}$ from the known sequence such that
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{k}}=1 .
$$
Let $y_{i}=\frac{x_{1} x_{2} \cdots x_{k}}{x_{i}}$. Then
$$
y_{1}+y_{2}+\cdots+y_{k}=x_{1} x_{2} \cdots x_{k} \text {. }
$$
It is easy to see that for any $n$, $x_{n... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,454 |
8. Given $x, y, z \in \mathbf{R}_{+}$, and $\sqrt{x^{2}+y^{2}}+z=1$. Then the maximum value of $x y+2 x z$ is $\qquad$ . | 8. $\frac{\sqrt{3}}{3}$.
$$
\begin{array}{l}
\text { Given } \sqrt{x^{2}+y^{2}}+z=1 \\
\Rightarrow x^{2}+y^{2}=(1-z)^{2} \\
\Rightarrow x^{2}=(1-z-y)(1-z+y) .
\end{array}
$$
Then $(x y+2 x z)^{2}=x^{2}(y+2 z)^{2}$
$$
\begin{array}{l}
=\frac{1}{3}(3-3 z-3 y)(1-z+y)(y+2 z)(y+2 z) \\
\leqslant \frac{1}{3}\left[\frac{(3-3... | \frac{\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,455 |
9. (16 points) Given non-zero real numbers $a, b, c, t$ satisfy
$$
\left\{\begin{array}{l}
a=t b+c, \\
b=c\left(1+t+t^{2}\right) .
\end{array}\right.
$$
(1) Prove: The quadratic equation
$$
x^{2}+c(b-2 c) x-\left(b^{2}+c^{2}\right)(b-c)=0
$$
must have real roots;
(2) When $a=15, b=7$, find $c, t$. | (1) From $\left\{\begin{array}{l}a=t b+c, \\ b=c\left(1+t+t^{2}\right),\end{array}\right.$ eliminating $t$ yields $b=c\left[1+\frac{a-c}{b}+\left(\frac{a-c}{b}\right)^{2}\right]$.
Rearranging and simplifying gives
$$
c a^{2}+c(b-2 c) a-\left(b^{2}+c^{2}\right)(b-c)=0 .
$$
Therefore, the equation
$$
c x^{2}+c(b-2 c) x-... | c=1, t=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,456 |
10. (20 points) Given the parabola $x^{2}=2 p y(p>0)$ and the line $y=b(b<0)$, point $P(t, b)$ moves on the line. Two tangent lines are drawn from $P$ to the parabola, touching it at points $A$ and $B$ respectively, and the midpoint of segment $AB$ is $M$. Find
(1) the locus of point $M$;
(2) the minimum value of $|AB|... | 10. (1) From the problem, we have
$$
y=\frac{1}{2 p} x^{2} \Rightarrow y^{\prime}=\frac{1}{p} x \text {. }
$$
Let $A\left(x_{1}, \frac{1}{2 p} x_{1}^{2}\right), B\left(x_{2}, \frac{1}{2 p} x_{2}^{2}\right)$.
Thus, $k_{P A}=\frac{x_{1}}{p}, k_{P B}=\frac{x_{2}}{p}$.
Also, from $P(t, b)$, we have
$$
\frac{\frac{x_{1}^{2... | 2 \sqrt{-2 p b} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,457 |
11. (20 points) Let \( a, b, c \) be distinct positive numbers. Prove:
$$
\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \geqslant a+b+c+\frac{4(a-b)^{2}}{a+b+c} .
$$ | 11. Notice that,
$$
\begin{aligned}
\frac{a^{2}}{b}+b-2 a & =\frac{(a-b)^{2}}{b} \\
\frac{b^{2}}{c}+c-2 b & =\frac{(b-c)^{2}}{c}, \\
\frac{c^{2}}{a}+a-2 c & =\frac{(c-a)^{2}}{a} .
\end{aligned}
$$
Adding the above three equations, we get
$$
\begin{array}{l}
\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \\
=(a+b+c)+\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,458 |
一、(40 points) As shown in Figure 2, given that $\triangle ABC$ is a non-equilateral acute triangle, $P$ and $Q$ are the midpoints of sides $AB$ and $BC$, respectively, and $H$ is the orthocenter. Extend $PH$ and $QH$ to intersect the circumcircle $\odot O$ at points $M$ and $N$. Prove that points $P$, $Q$, $M$, and $N$... | Extend $A O$ to intersect $\odot O$ at point $A^{\prime}$. Then $A^{\prime} C \perp A C$. Since $H$ is the orthocenter, we know $B H \perp A C$. Therefore, $B H / / A^{\prime} C$. Similarly, $C H / / A^{\prime} B$.
Thus, quadrilateral $B H C A^{\prime}$ is a parallelogram.
Furthermore, since $Q$ is the midpoint of $B C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,459 |
II. (40 points) Given the sequence of real numbers $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{3}, a_{n+1}=2 a_{n}-\left[a_{n}\right],
$$
where $[x]$ denotes the greatest integer less than or equal to the real number $x$.
$$
\text { Find } \sum_{i=1}^{2012} a_{i} \text {. }
$$ | $$
\text { II. } a_{1}=\frac{1}{3}, a_{2}=\frac{2}{3}, a_{3}=\frac{4}{3}, a_{4}=\frac{5}{3} .
$$
By mathematical induction, it is easy to prove
$$
\left\{\begin{array}{l}
a_{2 k+1}=\frac{3 k+1}{3}, \\
a_{2 k+2}=\frac{3 k+2}{3} .
\end{array}\right.
$$
Then $a_{2 k+1}+a_{2 k+2}=2 k+1$.
Therefore, $\sum_{i=1}^{2012} a_{... | 1012036 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,460 |
Three. (50 points) Given the function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ satisfies for any $m, n \in \mathbf{N}_{+}$,
$$
\left(f^{2}(m)+f(n)\right) \mid\left(m^{2}+n\right)^{2} .
$$
Prove: For any $n$, $f(n)=n$. | Let $m=n=1$. Then $\left(f^{2}(1)+f(1)\right) \mid 4$. Since $f(1) \in \mathbf{N}_{+}$, it can only be that $f(1)=1$.
Let $m=1, n=p-1$ (where $p$ is any prime number), we get $(f(p-1)+1) \mid p^{2}$.
Assume $f(p-1)+1=p^{2}$. Let $m=p-1, n=1$.
Thus, $\left(f^{2}(p-1)+1\right) \mid\left[(p-1)^{2}+1\right]^{2}$.
Therefore... | f(n)=n | Number Theory | proof | Yes | Yes | cn_contest | false | 726,461 |
Four, (50 points) Let $n \in \mathbf{N}_{+}$. Prove:
$$
\frac{1}{2} \sum_{k=1}^{n}\left[\frac{n}{k}\right]\left(\left[\frac{n}{k}\right]+1\right)=\sum_{k=1}^{n} k\left[\frac{n}{k}\right] .
$$ | For the region
$$
G=\{(x, y) \mid x y \leqslant n, x>0, y>0\}
$$
the sum $S$ of the x-coordinates of the integer points within the region is calculated in two ways.
On one hand, let $k \in \mathbf{N}_{+}$, for $1 \leqslant k \leqslant n$, it is easy to see that the line $x=k$ has $\left[\frac{n}{k}\right]$ integer poi... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,462 |
Solve the system of equations in the real number range
$$
\left\{\begin{array}{l}
a^{2}-b c=-5, \\
b^{2}-c a=1, \\
c^{2}-a b=7 .
\end{array}\right.
$$ | Solving the system of equations by subtracting each pair of the three equations, we get
$$
\begin{array}{l}
(a-b)(a+b+c)=-6, \\
(b-c)(a+b+c)=-6, \\
(c-a)(a+b+c)=12 .
\end{array}
$$
Adding the three equations in the system, we get
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a=3 . \\
(1)^{2}+(2)^{2}+(3)^{2} \text { ... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,463 |
Example 4 Convex quadrilateral $A B C D$ is inscribed in circle $\Gamma$, and a circle that intersects side $B C$ is internally tangent to circle $\Gamma$, and is tangent to $B D$ and $A C$ at points $P$ and $Q$, respectively. Prove that the incenter of $\triangle A B C$ and the incenter of $\triangle D B C$ both lie o... | Proof As shown in Figure 5, let two circles be internally tangent at point $T$, and the lines $TP$ and $TQ$ intersect circle $\Gamma$ at points $E$ and $F$ respectively.
By the properties of internally tangent circles, we know that $EF \parallel PQ$.
By Corollary $1(1)$, we know that $E$ and $F$ are the midpoints of ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,464 |
Initial 348 As shown in Figure 3, given two parallel lines $l_{1}$ and $l_{2}$, $A, B, C$ are three points on $l_{1}$, and $D, E, F$ are three points on $l_{2}$. The line $AE$ intersects $CF$ at point $G$, $AD$ intersects $BF$ at point $H$, and $BE$ intersects $CD$ at point $K$. Prove: $G, H, K$ are collinear. | Proof As shown in Figure 3, let $A E$ and $C D$ intersect at point $S$. Draw a line through $S$ parallel to $l_{1}$, intersecting $B E$ and $C F$ at points $L$ and $T$ respectively.
By $A C \parallel D E \Rightarrow \frac{A E}{E S}=\frac{D C}{S D}$
$$
\Rightarrow \frac{A E}{E S} \cdot \frac{S D}{D C}=1 \text {, }
$$
a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,465 |
Given $x, y, z \in \mathbf{R}_{+}$, and satisfying $x y z=1$, $\alpha \geqslant 0$. Prove:
$$
\sum \frac{x^{\alpha+3}+y^{\alpha+3}}{x^{2}+x y+y^{2}} \geqslant 2,
$$
where, “$\sum$” denotes the cyclic sum. | To prove:
\[
\begin{array}{l}
\frac{x^{3}}{x^{2}+x y+y^{2}} \geqslant \frac{2}{3} x-\frac{1}{3} y \\
\Leftrightarrow 3 x^{3} \geqslant(2 x-y)\left(x^{2}+x y+y^{2}\right) \\
\Leftrightarrow x^{3}+y^{3} \geqslant x^{2} y+x y^{2} \\
\Leftrightarrow(x-y)^{2}(x+y) \geqslant 0 .
\end{array}
\]
The last inequality is obviousl... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,466 |
348 Proof:
(1) There does not exist an integer-point regular hexagon;
(2) There exists a regular hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ and integer points $B_{1}, B_{2}, \cdots, B_{6}$, such that for each $k(k=1,2, \cdots, 6)$, we have $A_{k} B_{k}<\frac{1}{2013}$. | Prove (1) If there exists an integer-point regular hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$, let $A_{k}\left(x_{k}, y_{k}\right)(k=1,2, \cdots, 6)$, where $x_{k} 、 y_{k} \in \mathbf{Z}$, and $A_{1}, A_{2}, \cdots, A_{6}$ are arranged counterclockwise. Then
$$
\overrightarrow{A_{2} A_{1}}=\overrightarrow{A_{2} A_{3... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,467 |
Example 5 Let $D$ be the midpoint of arc $\overparen{B C}$ on the circumcircle $\Gamma$ of acute $\triangle A B C$, and let point $X$ be on arc $\overparen{B D}$. $E$ is the midpoint of arc $\overparen{A B X}$, and $S$ is a point on arc $\overparen{A C}$. Line $S D$ intersects $B C$ at point $R$, and $S E$ intersects $... | Proof As shown in Figure 6.
From $R T // D E$, we know that $\triangle D E S$ and $\triangle R T S$ are homothetic.
Therefore, the circumcircle of $\triangle T R S$ is internally tangent to circle $\Gamma$ at point $S$.
Draw the tangent line $E Z$ to circle $\Gamma$ through point $E$.
Since $E$ is the midpoint of arc $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,468 |
Example 6 In the inscribed $\triangle A B C$, $\angle A$ is the largest angle, and points $D, E$ on the arc $\overparen{B C}$ not containing point $A$ are the midpoints of arcs $\overparen{A B C}$ and $\overparen{A C B}$, respectively. Let the circle passing through points $A, B$ and tangent to $A C$ be $\odot O_{1}$, ... | Proof As shown in Figure 7, let the lines $B D$ and $C E$ intersect $\odot O_{1}$ and $\odot O_{2}$ at points $N$ and $M$, respectively, and connect $N A$, $A M$, $B P$, $P E$, and $B E$.
Since $D$ and $E$ are the midpoints of arcs $A B C$ and $A C B$, respectively, by property $4(2)$, we know that lines $B D$ and $C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,469 |
Example 1 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=a, a_{2}=b, 2 a_{n+2}=a_{n+1}+a_{n} \text {. }
$$
(1) If $b_{n}=a_{n+1}-a_{n}$, prove: for $a \neq b$, $\left\{b_{n}\right\}$ is a geometric sequence;
(2) If $\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\cdots+a_{n}\right)=4$, find the values of $... | (1) Proof From the problem, we have
$$
\begin{array}{l}
2\left(a_{n+2}-a_{n+1}\right)=-\left(a_{n+1}-a_{n}\right) \\
\Rightarrow b_{n+1}=-\frac{1}{2} b_{n}, \text { and } b_{1}=a_{2}-a_{1}=b-a \neq 0 .
\end{array}
$$
Therefore, $\left\{b_{n}\right\}$ is a geometric sequence with the first term $b-a$ and the common rat... | (a, b)=(6,-3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,470 |
Example 2 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{n}=2^{2 n} a_{n-1}+2^{n^{2}} n \quad (n=2,3, \cdots) \text {. }
$$
Find the general term $a_{n} \ (n=1,2, \cdots)$.
(Sixth Northern Mathematical Olympiad Invitational Competition) | Let $b_{n}=\frac{a_{n}}{2^{n^{2}}}$. Then $b_{n}=2 b_{n-1}+n$, i.e., $b_{n}+n=2\left(b_{n-1}+n-1\right)+2, b_{1}=1$.
Let $c_{n}=b_{n}+n$. Then $c_{n}=2 c_{n-1}+2$, i.e.,
$$
\begin{array}{l}
c_{n}+2=2\left(c_{n-1}+2\right), c_{1}=2 \\
\Rightarrow c_{n}+2=2^{n-1}\left(c_{1}+2\right)=2^{n+1} \\
\Rightarrow c_{n}=2^{n+1}-2... | a_{n}=2^{n^{2}}\left(2^{n+1}-n-2\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,471 |
Example $\mathbf{3}$ Given the Fibonacci sequence $\left\{a_{n}\right\}$ satisfies
$$
\left\{\begin{array}{l}
a_{1}=a_{2}=1, \\
a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3) .
\end{array}\right.
$$
Find the general term formula $a_{n}$ of the sequence $\left\{a_{n}\right\}$. | Solve from $x^{2}=x+1$, we get
$$
x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=\frac{1-\sqrt{5}}{2} \text {. }
$$
Let $a_{n}=A\left(\frac{1+\sqrt{5}}{2}\right)^{n}+B\left(\frac{1-\sqrt{5}}{2}\right)^{n}$.
From $a_{1}=a_{2}=1$, we get
$$
\left\{\begin{array}{l}
\frac{1+\sqrt{5}}{2} A+\frac{1-\sqrt{5}}{2} B=1, \\
A\left(\frac{1+\s... | a_{n}=\frac{\sqrt{5}}{5}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,472 |
Example 4 Let the function $f(x)=\frac{x+m}{x+1}$, and there exists a function $s=\varphi(t)=a t+b\left(t>\frac{1}{2}, a \neq 0\right)$, satisfying
$$
f\left(\frac{2 t-1}{t}\right)=\frac{2 s+1}{s} .
$$ | Prove: (1) There exists a function $t=g(s)=c s+d(s>0)$, satisfying $f\left(\frac{2 s+1}{s}\right)=\frac{2 t-1}{t}$;
(2) If $x_{1}=3, x_{n+1}=f\left(x_{n}\right)(n=1,2, \cdots)$, then $\left|x_{n}-2\right| \leqslant \frac{1}{3^{n+1}}$.
(2010, Joint Autonomous Admissions Examination of Tsinghua University and Other Unive... | \left|x_{n}-2\right| \leqslant \frac{1}{3^{n+1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,473 |
Example 5 Let $\left\{a_{n}\right\}$ be a sequence of real numbers, satisfying the relation for all $n$
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=\frac{1}{a_{1} a_{2} \cdots a_{n}} .
$$
(1) Write down the relation between $a_{n}$ and $a_{n+1}$ for $(n \geqslant 2)$;
(2) Prove that if $a_{1} \in(0,1)$, t... | (1) Solution: From $\frac{1}{a_{1}}+\frac{1}{a_{2}}=\frac{1}{a_{1} a_{2}}$, we get $a_{2}=1-a_{1}$.
Assume that when $n=k \geqslant 2$, we have
$$
\begin{array}{l}
a_{k}=1-a_{1} a_{2} \cdots a_{k-1}. \\
\text { By } \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{k}}=\frac{1}{a_{1} a_{2} \cdots a_{k}} \\
\Rightarrow... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,474 |
Example 3 Can 2010 be written as the sum of squares of $k$ distinct positive integers? If so, try to find the maximum value of $k$; if not, please briefly explain the reason. ${ }^{[2]}$
(2010, Beijing Middle School Mathematics Competition (Grade 8)) | Estimate the approximate range of $k$ first, then discuss by classification.
Let $p_{i}$ be a prime number.
If 2010 can be written as the sum of squares of $k$ prime numbers, then by
$$
\begin{array}{l}
2^{2}+3^{2}+5^{2}+7^{2}+11^{2}+13^{2}+17^{2}+19^{2}+23^{2}+29^{2} \\
=2397>2010,
\end{array}
$$
we know $k \leqslant... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,475 |
1. A. Let $n$ be a positive integer, and denote $n!=1 \times 2 \times \cdots \times n$. Then the last digit of $1!+2!+\cdots+10!$ is ( ).
(A) 0
(B) 1
(C) 3
(D) 5 | One, 1. A. C.
Notice that, $1!=1, 2!=2, 3!=6, 4!=24$.
It is also known that the last digit of $5!, 6!, 7!, 8!, 9!, 10!$ is 0, and the last digit of $1!+2!+3!+4!$ is 3, so the last digit of $1!+2!+\cdots+10!$ is 3. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,476 |
B. Let non-zero real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+2 b+3 c=0, \\
2 a+3 b+4 c=0 .
\end{array}\right. \\
\text { Then } \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}=(\quad \text {. }
\end{array}
$$
(A) $-\frac{1}{2}$
(B) 0
(C) $\frac{1}{2}$
(D) 1 | B. A.
From the given we have
$$
\begin{array}{l}
a+b+c=(2 a+3 b+4 c)-(a+2 b+3 c)=0 \\
\Rightarrow(a+b+c)^{2}=0 \\
\Rightarrow a b+b c+c a=-\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right) \\
\Rightarrow \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}=-\frac{1}{2} .
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,477 |
2. A. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
\frac{2 x+5}{3}-x>-5, \\
\frac{x+3}{2}-t<x
\end{array}\right.
$$
has exactly five integer solutions. Then the range of $t$ is ( ).
(A) $-6<t<-\frac{11}{2}$
(B) $-6 \leqslant t<-\frac{11}{2}$
(C) $-6<t \leqslant-\frac{11}{2}$
(D) $-6 \leqslant ... | 2. A. C.
From the system of inequalities, we get
$$
3-2 t<x<20 \text {. }
$$
Since the system of inequalities has exactly five integer solutions, the integer solutions can only be $15, 16, 17, 18, 19$.
Thus, $14 \leqslant 3-2 t<15 \Rightarrow-6<t \leqslant-\frac{11}{2}$. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 726,478 |
B. Given that $a$, $b$, and $c$ are real constants, the quadratic equation in $x$
$$
a x^{2} + b x + c = 0
$$
has two non-zero real roots $x_{1}$ and $x_{2}$. Then, among the following quadratic equations in $x$, the one that has $\frac{1}{x_{1}^{2}}$ and $\frac{1}{x_{2}^{2}}$ as its two real roots is ( ).
(A) $c^{2} ... | B. B.
From the problem, we know that $a \neq 0$, and
$$
x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a} \neq 0 \text {. }
$$
Thus, $c \neq 0$,
$$
\begin{array}{l}
\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{x_{1}^{2} x_{2}^{2}}=\frac{b^{2}-2 a c}{c^{2}}, \\
\frac{1}{x_{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,479 |
3. A. Given the equation in terms of $x$
$$
\frac{x}{x-2}+\frac{x-2}{x}=\frac{a-2 x}{x^{2}-2 x}
$$
has exactly one real root. Then the number of real values of $a$ that satisfy this condition is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 3. A. C.
The original equation can be transformed into
$$
2 x^{2}-2 x+4-a=0 \text {. }
$$
(1) When $\Delta=4(2 a-7)=0$, i.e., $a=\frac{7}{2}$, $x=\frac{1}{2}$ satisfies the condition;
(2) When $\Delta=4(2 a-7)>0$, i.e., $a>\frac{7}{2}$, the two distinct real roots of $2 x^{2}-2 x+4-a=0$ must include one extraneous roo... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,480 |
B. As shown in Figure 1, in the right triangle $\triangle ABC$, it is known that $O$ is the midpoint of the hypotenuse $AB$, $CD \perp AB$ at point $D$, and $DE \perp OC$ at point $E$. If the lengths of $AD$, $DB$, and $CD$ are all rational numbers, then among the following line segments, the length that is not necessa... | B. D.
Given that the lengths of $A D$, $D B$, and $C D$ are all rational numbers, we have
$$
O A=O B=O C=\frac{A D+B D}{2}
$$
which are rational numbers.
Thus, $O D=O A-A D$ is a rational number.
By the similarity of Rt $\triangle D O E \backsim \mathrm{Rt} \triangle C O D$, we know
$$
O E=\frac{O D^{2}}{O C}, D E=\f... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,481 |
4. A. As shown in Figure 2, the area of $\triangle A B C$ is 24, point $D$ is on line segment $A C$, and point $F$ is on the extension of line segment $B C$, with $B C=4 C F$. If quadrilateral $D C F E$ is a parallelogram, then the area of the shaded part in the figure is ( ).
(A) 3
(B) 4
(C) 6
(D) 8 | 4. A. C.
As shown in Figure 7, in quadrilateral DCFE, $DE // CF, EF // DC$:
Connect $CE$ and $AF$.
Since $DE // CF$, which means $DE // BF$, therefore,
$S_{\triangle DEB} = S_{\triangle DEC}$.
Thus, the area of the shaded part is equal to the area of $\triangle ACE$.
Similarly, $S_{\triangle CEF} = S_{\triangle MCF}$.... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,482 |
B. Let the function be
$$
y=(\sqrt{4+x}+\sqrt{4-x}+1)\left(\sqrt{16-x^{2}}+2\right) \text {. }
$$
Then the range of $y$ is . .
(A) $2 \leqslant y \leqslant 20$
(B) $2 \leqslant y \leqslant 30$
(C) $4 \sqrt{2}+2 \leqslant y \leqslant 20$
(D) $4 \sqrt{2}+2 \leqslant y \leqslant 30$ | B. D.
Notice,
$$
\begin{array}{l}
y=\left[\sqrt{(\sqrt{4+x}+\sqrt{4-x})^{2}}+1\right]\left(\sqrt{16-x^{2}}+2\right) \\
=\left(\sqrt{8+2 \sqrt{16-x^{2}}}+1\right)\left(\sqrt{16-x^{2}}+2\right)
\end{array}
$$
Therefore, when $x=4$, $y_{\text {min }}=\sqrt{2}+2$;
when $x=0$, $y_{\max }=30$.
Hence, the range of $y$ is $4... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,483 |
5.A. Given nine balls labeled with numbers $2,3, \cdots, 10$, two balls are randomly drawn, and their labels are noted. The probability that the larger label is divisible by the smaller label is ( ).
(A) $\frac{1}{4}$
(B) $\frac{2}{9}$
(C) $\frac{5}{18}$
(D) $\frac{7}{36}$ | 5. A. B.
Let the binary array $(a, b)$ represent the labels of the two balls drawn, where $a$ and $b$ are the smaller and larger labels, respectively. Then, the pairs $(a, b)$ that satisfy the condition are as follows:
$$
\begin{array}{l}
(2,4),(2,6),(2,8),(2,10), \\
(3,6),(3,9),(4,8),(5,10) .
\end{array}
$$
The tota... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,484 |
B. For any real numbers $x, y, z$, the operation “*” is defined as
$$
x * y=\frac{3 x^{3} y+3 x^{2} y^{2}+x y^{3}+45}{(x+1)^{3}+(y+1)^{3}-60} \text {, }
$$
and $x * y * z=(x * y) * z$.
Then $2013 * 2012 * \cdots * 2=$ ( ).
(A) $\frac{607}{967}$
(B) $\frac{1821}{967}$
(C) $\frac{5463}{967}$
(D) $\frac{16389}{967}$ | B. C.
$$
\begin{array}{l}
\text { Let } 2013 * 2012 * \cdots * 4 = m \text {. Then } \\
(2013 * 2012 * \cdots * 4) * 3 = m * 3 \\
=\frac{3 m^{3} \times 3 + 3 m^{2} \times 9 + m \times 27 + 45}{m^{3} + 3 m^{2} + 3 m + 1 + 64 - 60} = 9 \text {. } \\
\text { Therefore, } (2013 * 2012 * \cdots * 3) * 2 \doteq 9 * 2 \\
=\fr... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,485 |
Example 4 Choose $k$ numbers from $1,2, \cdots, 2004$, such that among the chosen $k$ numbers, there are definitely three numbers that can form the side lengths of a triangle (the three numbers must be distinct). What is the minimum value of $k$ that satisfies this condition? | When selecting three numbers from 1 to 2004 to form the sides of a triangle, there are too many possibilities. Instead, let's approach it from the opposite direction and list all sets of three numbers that cannot form the sides of a triangle.
First, 1, 2, 3 cannot form the sides of a triangle. Adding 5, the set \(1, 2... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,486 |
6. A. Let $a=\sqrt[3]{3}, b$ be the fractional part of $a^{2}$. Then $(b+2)^{3}=$ $\qquad$ | ニ、6. A.9.
From $2<a^{2}<3$, we know $b=a^{2}-2=\sqrt[3]{9}-2$. Therefore, $(b+2)^{3}=(\sqrt[3]{9})^{3}=9$. | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,487 |
B. Let $a=\sqrt[3]{3}, b, c$ be the fractional parts of $a, a^{2}$, respectively. Then $b(b+c+4)=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | B. 2 .
From $1<a<2<a^{2}<3$, we know $b=a-1, c=a^{2}-2$.
Thus $b(b+c+4)$
$$
\begin{array}{l}
=(a-1)\left(a-1+a^{2}-2+4\right) \\
=(a-1)\left(a^{2}+a+1\right)=a^{3}-1=2 .
\end{array}
$$ | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,488 |
7. A. A uniform cube is marked with the numbers $1,2, \cdots, 6$ on its six faces. If the cube is rolled three times, the probability that the sum of the numbers on the top faces is a multiple of 3 is . $\qquad$ | 7. A. $\frac{1}{3}$.
Notice that, the sums of the numbers on the top faces of a cube rolled three times that are multiples of 3 are $3, 6, 9, 12, 15, 18$, and
$$
\begin{array}{l}
3=1+1+1, \\
6=1+1+4=1+2+3=2+2+2, \\
9=1+2+6=1+3+5 \\
=1+4+4=2+2+5 \\
=2+3+4=3+3+3, \\
12=1+5+6=2+4+6 \\
=2+5+5=3+3+6 \\
=3+4+5=4+4+4, \\
15=... | \frac{1}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,489 |
B. As shown in Figure 3, given that $D$ and $E$ are points on the sides $AC$ and $AB$ of $\triangle ABC$, respectively, and line $BD$ intersects $CE$ at point $F$. If the areas of $\triangle CDF$, $\triangle BFE$, and $\triangle BCF$ are $3$, $4$, and $5$, respectively, then the area of quadrilateral $AEFD$ is $\qquad$ | B. $\frac{204}{13}$.
Connect $A F$. Then
$$
\begin{array}{l}
\frac{S_{\triangle A E F}+4}{S_{\triangle A F D}}=\frac{S_{\triangle A E F}+S_{\triangle B F E}}{S_{\triangle A F D}} \\
=\frac{B F}{F D}=\frac{S_{\triangle A C F}}{S_{\triangle C D F}}=\frac{5}{3}, \\
\frac{S_{\triangle A F D}+3}{S_{\triangle A E F}}=\frac{... | \frac{204}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,490 |
8. A. Given positive integers $a$, $b$, $c$ satisfy
$$
\begin{array}{l}
a+b^{2}-2 c-2=0, \\
3 a^{2}-8 b+c=0 .
\end{array}
$$
Then the maximum value of $a b c$ is $\qquad$ | 8. A. 2013.
The two equations are simplified and rearranged to get
$$
(b-8)^{2}+6 a^{2}+a=66 \text {. }
$$
Given that $a$ is a positive integer and $6 a^{2}+a \leqslant 66$, we have $1 \leqslant a \leqslant 3$. If $a=1$, then $(b-8)^{2}=59$, which has no positive integer solutions; if $a=2$, then $(b-8)^{2}=40$, whic... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,491 |
B. If $k$ numbers are chosen from 2, $, 8, \cdots, 101$ these 34 numbers, where the sum of at least two of them is 43, then the minimum value of $k$ is: $\qquad$ | B. 28.
Divide the 14 numbers less than 43 into the following seven groups: $(2,41),(5,38),(8,35),(11,32)$, $(14,29),(17,26),(20,23)$.
The sum of the two numbers in each group is 43. After selecting one number from each group and then taking all numbers greater than 43, a total of 27 numbers are selected. The sum of a... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,492 |
9. A. Given real numbers $a, b, c, d$ satisfy: the quadratic equation $x^{2}+c x+d=0$ has roots $a, b$, and the quadratic equation $x^{2}+a x+b=0$ has roots $c, d$. Then all sets of $(a, b, c, d)$ that meet the conditions are $\qquad$ . | 9. A. $(1,-2,1,-2),(t, 0,-t, 0)(t$ is any real number).
By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
a+b=-c, \\
a b=d, \\
c+d=-a, \\
c d=b .
\end{array}\right.
$$
It is easy to see that $b=-a-c=d$.
If $b=d \neq 0$, then
$$
\begin{array}{l}
a=\frac{d}{b}=1, c=\frac{b}{d}=1 \\
\Rightarrow b=d=-a-c=-2 ;
\end{a... | (1,-2,1,-2),(t, 0,-t, 0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,493 |
B. The system of equations in $x, y, z$
$$
\left\{\begin{array}{l}
x y+y z+z x=1, \\
5 x+8 y+9 z=12
\end{array}\right.
$$
all real solutions $(x, y, z)$ are $\qquad$ | $\begin{array}{l}\text { B. }\left(1, \frac{1}{2}, \frac{1}{3}\right) . \\ \quad \text { Substitute } z=\frac{1}{9}(12-5 x-8 y) \text { into } x y+y z+z x \\ =1 \text {, we get } \\ \quad 5 x^{2}+8 y^{2}+4 x y-12 x-12 y+9=0 \\ \Rightarrow(4 y+x-3)^{2}+9(x-1)^{2}=0 \\ \Rightarrow x-1=0,4 y+x-3=0 \\ \Rightarrow x=1, y=\f... | B | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,494 |
10. A. Xiaoming volunteered to sell pens at a stationery store one day. Pencils were sold at 4 yuan each, and ballpoint pens at 7 yuan each. At the beginning, it was known that he had a total of 350 pencils and ballpoint pens. Although he did not sell them all that day, his sales revenue was 2013 yuan. Then he sold at ... | 10. A. 207.
Let $x$ and $y$ represent the number of pencils and ballpoint pens sold, respectively. Then
\[
\begin{array}{l}
\left\{\begin{array}{l}
4 x+7 y=2013 ; \\
x+y=204
\end{array}\right.
\end{array}
\]
Thus, $y_{\text {min }}=207$, at which point, $x=141$. | 207 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,495 |
$\begin{array}{l}\text { B. } \frac{\sqrt{1^{4}+2^{4}+1}}{1^{2}+2^{2}-1}+\frac{\sqrt{2^{4}+3^{4}+1}}{2^{2}+3^{2}-1}+\cdots+\frac{\sqrt{99^{4}+100^{4}+1}}{99^{2}+100^{2}-1} \\ =\end{array}$ | B. $\frac{9999 \sqrt{2}}{200}$.
Let $k>0$. Then
$$
\begin{array}{l}
\frac{\sqrt{k^{4}+(k+1)^{4}+1}}{k^{2}+(k+1)^{2}-1}=\frac{\sqrt{2\left(k^{2}+k+1\right)^{2}}}{2\left(k^{2}+k\right)} \\
=\frac{\sqrt{2}}{2}\left[1+\frac{1}{k(k+1)}\right]=\frac{\sqrt{2}}{2}\left(1+\frac{1}{k}-\frac{1}{k+1}\right) .
\end{array}
$$
Summ... | \frac{9999 \sqrt{2}}{200} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,496 |
Example 5 Let $x_{1}, x_{2}, \cdots, x_{100}$ be positive integers, and $x_{1}<x_{2}<\cdots<x_{100}$. If $x_{1}+x_{2}+\cdots+x_{100}=7001$, then the maximum value of $x_{1}+x_{2}+\cdots+x_{50}$ is ( ).
(A) 2225
(B) 2226
(C) 2227
(D) 2228 | Because
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{50} \\
\leqslant 50 x_{50}-(1+2+\cdots+49) \\
=50 x_{50}-1225,
\end{array}
$$
Therefore, we need to determine the value of $x_{50}$.
To maximize the sum $x_{1}+x_{2}+\cdots+x_{50}$, the latter part should be minimized.
$$
\begin{array}{l}
\text { Hence } x_{51}+x_{52}+... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,497 |
11. A. As shown in Figure 4, the parabola $C: y=a x^{2}+b x-3$, with vertex $E$, intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. It is given that $O B=O C=3 O A$. The line $y=-\frac{1}{3} x+1$ intersects the $y$-axis at point $D$. Find $\angle D B C-\angle C B E$. | Three, 11. A.
It is easy to know $D(0,1), C(0,-3)$.
Then $B(3,0), A(-1,0)$.
Obviously, the line $y=-\frac{1}{3} x+1$ passes through point $B$.
Substitute the coordinates of point $C(0,-3)$ into
$$
y=a(x+1)(x-3),
$$
we get $a=1$.
Therefore, the vertex of the parabola $y=x^{2}-2 x-3$ is $E(1,-4)$.
By the Pythagorean the... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,498 |
B. In the Cartesian coordinate system $x O y$, it is known that $O$ is the origin, point $A(10,100), B\left(x_{0}, y_{0}\right)$, where $x_{0} 、 y_{0}$ are integers, and points $O 、 A 、 B$ are not collinear. For all points $B$ that satisfy the above conditions, find the minimum area of $\triangle O A B$. | B. Draw a line parallel to the $x$-axis through point $B$, intersecting line $O A$ at point $C\left(\frac{y_{0}}{10}, y_{0}\right)$.
Thus, $B C=\left|x_{0}-\frac{y_{0}}{10}\right|$.
Since points $O$, $A$, and $B$ are not collinear, we have
$$
\begin{array}{l}
S_{\triangle O A B}=\frac{1}{2} B C \times 100=50\left|x_{0}... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,499 |
12. A. As shown in Figure 5, given that $A B$ is the diameter of $\odot O$, $C$ is a point on the circumference, and $D$ is a point on the line segment $O B$ (not at the endpoints), satisfying $C D \perp A B$ and $D E \perp C O$ at point $E$. If $C E = 10$, and the lengths of $A D$ and $D B$ are both positive integers,... | 12. A. Connect $A C$ and $B C$, then $\angle A C B=90^{\circ}$.
From $\mathrm{Rt} \triangle C D E \backsim \mathrm{Rt} \triangle C O D$, we know $C E \cdot C O=C D^{2}$.
From $\mathrm{Rt} \triangle A C D \backsim \mathrm{Rt} \triangle C B D$, we know $C D^{2}=A D \cdot B D$.
Therefore, $C E \cdot C O=A D \cdot B D$.
L... | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,500 |
B. Let the circumcenter and orthocenter of $\triangle A B C$ be $O$ and $H$ respectively. If $B, C, H, O$ are concyclic, find all possible measures of $\angle B A C$ for all $\triangle A B C$.
The text is translated while preserving the original formatting and line breaks. | B. Discuss in three cases.
(1) $\triangle A B C$ is an acute triangle (as shown in Figure 8).
From $\angle B H C=180^{\circ}-\angle A, \angle B O C=2 \angle A$,
. Then $\angle B H C=\angle B O C$
$$
\begin{array}{l}
\Rightarrow 180^{\circ}-\angle A=2 \angle A \\
\Rightarrow \angle A=60^{\circ} .
\end{array}
$$
(2) $\t... | 60^{\circ} \text{ and } 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,501 |
13. A. Let $a, b, c$ be prime numbers, and denote
$$
x=b+c-a, y=c+a-b, z=a+b-c \text {. }
$$
Question: When $z^{2}=y, \sqrt{x}-\sqrt{y}=2$, can $a, b, c$ form the three sides of a triangle? Prove your conclusion. | 13. A. No.
From the problem, we have
$$
a=\frac{1}{2}(y+z), b=\frac{1}{2}(x+z), c=\frac{1}{2}(x+y) \text {. }
$$
Since \( y=z^{2} \), we have
$$
a=\frac{1}{2}(y+z)=\frac{1}{2}\left(z^{2}+z\right)=\frac{z(z+1)}{2} \text {. }
$$
Given that \( z \) is an integer and \( a \) is a prime number, then \( z=2 \) or \( -3 \)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,502 |
B. As shown in Figure 6, given that point $D$ is on the circumcircle of $\triangle A B C$ and is the midpoint of arc $\overparen{B C}$, point $X$ is on arc $\overparen{B D}$, and $E$ is the midpoint of arc $\overparen{A X}$. A line $R T \parallel D E$ is drawn through the incenter $I$ of $\triangle A B C$, intersecting... | B. As shown in Figure 10, let $D R$ intersect the circumcircle of $\triangle A B C$ at point $S^{\prime}$, and $A X$ intersect $S^{\prime} E$ at point $T^{\prime}$. Connect $S^{\prime} C$, $C D$, $S^{\prime} A$, $A E$, and $A D$.
Since $D$ is the midpoint of arc $\overparen{B C}$, we know that $A$, $I$, and $D$ are co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,503 |
14. A. If placing the positive integer $M$ to the left of the positive integer $m$ results in a new number that is divisible by 7, then $M$ is called the "magic number" of $m$ (for example, placing 86 to the left of 415 results in the number 86415, which is divisible by 7, so 86 is called the magic number of 415). Find... | 14. A. If $n \leqslant 6$, take $m=1,2, \cdots, 7$. By the pigeonhole principle, there must be a positive integer $M$ among $a_{1}, a_{2}, \cdots, a_{n}$ that is a common magic number of $i$ and $j (1 \leqslant i<j \leqslant 7)$, i.e.,
$$
7 \mid(10 M+i), 7 \mid(10 M+j) \text {. }
$$
Then $7 \mid (j-i)$. But $0<j-i \le... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,504 |
B. Let $n$ be a non-negative integer, $x=\frac{1+\sqrt{5}}{2}$. If there exist integers $k_{0}<k_{1}<\cdots<k_{n}$, such that $x^{k_{0}}+x^{k_{1}}+\cdots+x^{k_{n}}=13$.
Prove: $k_{0} \leqslant-2$. | B. Assume $k_{0} \geqslant-1$.
From $x=\frac{1+\sqrt{5}}{2}$, we get $x^{2}=x+1$.
Thus, $x^{-1}=x-1, x^{3}=2 x+1, x^{4}=3 x+2$, $x^{5}=5 x+3, x^{6}=8 x+5$.
From $x^{50} \text {. }$
$$
Therefore, the left side of equation (1) is an irrational number, leading to a contradiction.
(2) If $-1 \leqslant k_{n} \leqslant 0$,... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,505 |
1. Given sets
$$
A=\{1,2,3,4,5\}, B=\{2,3,4,5,6\} \text {. }
$$
Then the set
$C=\{(a, b) \mid a \in A, b \in B$, and the equation $x^{2}+2 a x+b^{2}=0$ has real roots $\}$ has how many elements? $(\quad$.
(A) 7
(B) 8
(C) 9
(D) 10 | -,1.D.
When $a>0, b>0$, the equation $x^{2}+2 a x+b^{2}=0$ has real roots if and only if $a \geqslant b$.
Thus, $C=\{(a, b) \mid a, b \in A, a \geqslant b\}$.
Therefore, the number of elements in set $C$ is equal to 10. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,506 |
2. Given $\sqrt{24-a}-\sqrt{8-a}=2$. Then $\sqrt{24-a}+\sqrt{8-a}=(\quad)$.
(A) 7
(B) 8
(C) 9
(D) 10 | $\begin{array}{l}\text { 2. B. } \\ \sqrt{24-a}+\sqrt{8-a} \\ =\frac{(\sqrt{24-a})^{2}-(\sqrt{8-a})^{2}}{\sqrt{24-a}-\sqrt{8-a}}=8 .\end{array}$
The translation is as follows:
$\begin{array}{l}\text { 2. B. } \\ \sqrt{24-a}+\sqrt{8-a} \\ =\frac{(\sqrt{24-a})^{2}-(\sqrt{8-a})^{2}}{\sqrt{24-a}-\sqrt{8-a}}=8 .\end{array... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,507 |
Example 6 Given $n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=2008$. Find the maximum value of the product $x_{1} x_{2} \cdots x_{n}$. ${ }^{[3]}$
(2008, National Junior High School Mathematics Competition, Tianjin Preliminary) | Let the maximum value of $x_{1} x_{2} \cdots x_{n}$ be $M$.
From the given equation, we know that each $x_{i}>1$ $(i=1,2, \cdots, 2008)$. If there is an $x_{i} \geqslant 4$, we can split $x_{i}$ into $x_{i}-2$ and 2, and consider their product, we have
$$
\left(x_{i}-2\right) \times 2=x_{i}+\left(x_{i}-4\right) \geqsla... | 2^{2} \times 3^{668} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,508 |
3. As shown in Figure 1, the diagonal $BD$ of rectangle $ABCD$ passes through the origin $O$, and the sides of the rectangle are parallel to the coordinate axes. Point $C$ lies on the graph of the inverse proportion function $y=\frac{3k+1}{x}$. If $A(-2,-2)$, then $k=(\quad)$.
(A) 2
(B) 1
(C) 0
(D) -1 | 3. B.
Since the diagonal of a rectangle bisects the rectangle, therefore,
$S_{\text {partCHOG }}=S_{\text {partOFAE }}=|-2| \times|-2|=4$.
Thus $3 k+1=O G \cdot G C=4 \Rightarrow k=1$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,509 |
4. The even function $f(x)$ defined on $\mathbf{R}$ satisfies $f(x+1)=-f(x)$,
and is increasing in the interval $[-1,0]$, then ().
(A) $f(3)<f(\sqrt{3})<f(2)$
(B) $f(2)<f(3)<f(\sqrt{3})$
(C) $f(3)<f(2)<f(\sqrt{3})$
(D) $f(2)<f(\sqrt{3})<f(3)$ | 4. A.
From the problem, we know
$$
\begin{array}{l}
f(x)=-f(x+1)=f(x+2) . \\
\text { Then } f(3)=f(1)=f(-1), \\
f(2)=f(0), f(\sqrt{3})=f(\sqrt{3}-2) .
\end{array}
$$
And $-1<\sqrt{3}-2<0$, $f(x)$ is increasing in the interval $[-1,0]$
so, $f(3)<f(\sqrt{3})<f(2)$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,510 |
5. The product of the first $n$ positive integers starting from 1 is denoted as $n!=1 \times 2 \times \cdots \times n$, for example, $3!=1 \times 2 \times 3=6$. Then $\sum_{n=2}^{8} \frac{n-1}{n!}=(\quad$.
( A) $\frac{719}{720}$
(B) $\frac{5039}{5040}$
(C) $\frac{40319}{40320}$
(D) $\frac{40321}{40320}$ | 5. C.
Notice,
$$
\begin{array}{l}
\frac{n-1}{n!}=\frac{n}{n!}-\frac{1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!} . \\
\text { Hence } \sum_{n=2}^{8} \frac{n-1}{n!}=\sum_{n=2}^{8}\left[\frac{1}{(n-1)!}-\frac{1}{n!}\right] \\
=1-\frac{1}{8!}=1-\frac{1}{40320}=\frac{40319}{40 \cdot 320} .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,511 |
6. As shown in Figure 2, square $A B C D$ is inscribed in $\odot O, P$ is a point on the minor arc $\overparen{C D}$, $P A$ intersects $B D$ at point $M$, $P B$ intersects $A C$ at point $N$, and let $\angle P A C=\theta$. If $M N \perp P A$, then $2 \cos ^{2} \theta-\tan \theta=(\quad)$.
(A) 1
(B) $\frac{\sqrt{2}}{2}$... | 6. A.
Given that quadrilateral $A B C D$ is a square, we know
$$
\begin{array}{l}
\angle A C B=45^{\circ}, D B \perp A C \\
\Rightarrow \angle A P B=\angle A C B=45^{\circ} . \\
\text { Also, } M N \perp P A \\
\Rightarrow \angle M N P=\angle A P B=45^{\circ} \\
\Rightarrow M P=M N .
\end{array}
$$
Since $A C$ is the... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,512 |
$$
\begin{array}{l}
\frac{\sin ^{2} 30^{\circ}+\sin ^{2} 355^{\circ}+\sin ^{2} 40^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 50^{\circ}+\sin ^{2} 55^{\circ}+\sin ^{2} 60^{\circ}}{\tan 36^{\circ} \cdot \tan ^{3} 39^{\circ} \cdot \tan ^{5} 42^{\circ} \cdot \tan ^{7} 45^{\circ} \cdot \tan ^{5} 48^{\circ} \cdot \tan ^{3} 55^{\... | 2, 1.3.5.
Notice,
$$
\sin ^{2} \alpha+\sin ^{2}\left(90^{\circ}-\alpha\right)=\sin ^{2} \alpha+\cos ^{2} \alpha=1 \text {, }
$$
and $\sin ^{2} 45^{\circ}=\frac{1}{2}$.
When $n$ is a positive integer,
$$
\tan ^{n} \alpha \cdot \tan ^{n}\left(90^{\circ}-\alpha\right)=\tan ^{n} \alpha \cdot \cot ^{n} \alpha=1 \text {, }
... | 3.5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,513 |
2. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x \geqslant 0$,
$f(x)=2^{x}+2 x+b$ ( $b$ is a constant).
Then $f(-10)=$ $\qquad$ . | 2. -1043 .
From the given condition, we easily know that
$$
f(0)=2^{0}+2 \times 0+b=0 \text {. }
$$
Solving for $b$ yields $b=-1$.
By the property of odd functions $f(-x)=-f(x)$, we have
$$
\begin{array}{l}
f(-10)=-f(10)=-2^{10}-2 \times 10+1 \\
=-1043 .
\end{array}
$$ | -1043 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,514 |
3. If real numbers $x, y, z$ satisfy the equation
$$
\sqrt{x+9+\sqrt{x-7}}+\frac{|x+y-z|}{4}=4 \text {, }
$$
then the units digit of $(5 x+3 y-3 z)^{2013}$ is $\qquad$ | 3. 4 .
It is known that $x \geqslant 7$, then
$$
\left\{\begin{array}{l}
\sqrt{x+9+\sqrt{x-7}} \geqslant 4, \\
\frac{|x+y-z|}{4} \geqslant 0 .
\end{array}\right.
$$
Combining the given equations, we have
$$
\left\{\begin{array}{l}
\sqrt{x+9+\sqrt{x-7}}=4, \\
\frac{|x+y-z|}{4}=0 .
\end{array}\right.
$$
Therefore, $x=... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,515 |
4. As shown in Figure 3, square $ABCD$ is divided into 8 triangles of equal area. If $AG=\sqrt{50}$, then the area $S$ of square $ABCD$ is $ \qquad $. | 4. 128 .
As shown in Figure 5, draw $K L / / D C$ through point $F$, take the midpoint $N$ of $A B$, and connect $G N$ with $A H$ intersecting at point $P$.
Let the side length of the square $A B C D$ be $a$.
Given $S_{\triangle D C I}=S_{\triangle M B H}=\frac{1}{8} S$, we know
$C I=B H=\frac{1}{4} B C=\frac{a}{4}$.
... | 128 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,516 |
5. Given real numbers $m, n$ satisfy $m-n=\sqrt{10}$, $m^{2}-3 n^{2}$ is a prime number. If the maximum value of $m^{2}-3 n^{2}$ is $a$, and the minimum value is $b$, then $a-b=$ $\qquad$ | 5.11 .
Let $m^{2}-3 n^{2}=p$ (where $p$ is a prime number).
From $m-n=\sqrt{10}$, we get
$$
m=\sqrt{10}+n \text {. }
$$
Substituting equation (2) into equation (1) and simplifying, we get
$$
\begin{array}{l}
2 n^{2}-2 \sqrt{10} n+p-10=0 \\
\Rightarrow \Delta=40-8 p+80 \geqslant 0 \\
\Rightarrow p \leqslant 15 \\
\Rig... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,517 |
6. As shown in Figure 4, on side $BC$ of $\triangle ABC$, there is a point $D$, and $\angle ADB$ is an acute angle. $P$ and $Q$ are the circumcenters of $\triangle ABD$ and $\triangle ACD$, respectively, and the area of quadrilateral $APDQ$ is $\frac{3}{4}$ of the area of $\triangle ABC$. Then $\sin \angle ADB=$ $\qqua... | 6. $\frac{\sqrt{6}}{3}$.
As shown in Figure 6, connect $P Q$ and $C Q$.
It is easy to prove that $\triangle A Q P \cong \triangle D Q P$.
Then $\frac{S_{\triangle P P}}{S_{\triangle A B C}}=\frac{3}{8}$.
Also, $\triangle A P Q \backsim \triangle A B C$
$$
\Rightarrow \frac{S_{\triangle A P Q}}{S_{\triangle A B C}}=\le... | \frac{\sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,518 |
Example 7 Let $x_{1}, x_{2}, \cdots, x_{2006}$ be integers, and satisfy the following conditions:
$$
\begin{array}{l}
-1 \leqslant x_{n} \leqslant 2 \quad (n=1,2, \cdots, 2006), \\
x_{1}+x_{2}+\cdots+x_{2006}=200, \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{2006}^{2}=2006 .
\end{array}
$$
Find the minimum and maximum values of $... | Given the problem, let there be $a$ numbers of $-1$, $b$ numbers of $0$, $c$ numbers of $1$, and $d$ numbers of $2$. Then,
$$
\begin{array}{c}
a+b+c+d=2006, \\
-a+c+2 d=200, \\
a+c+4 d=2006 .
\end{array}
$$
Thus, $b=3 d \geqslant 0$,
$$
\begin{array}{l}
c=1103-3 d \geqslant 0, \\
a=903-d \geqslant 0 .
\end{array}
$$
... | 200 \text{ and } 2402 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,519 |
7. Let $S(x)$ denote the sum of the digits of the natural number $x$. Then the solution set of the equation
$$
x+S(x)+S(S(x))=2013
$$
is $\qquad$ . | 7. $\{1979,1985,1991,2003\}$.
Obviously, $x<2013$.
And $S(x)$ is at most $28$, $S(S(x))$ is at most 10, so $x$ is at least
$$
2013-38=1975 \text {. }
$$
Thus, $1975 \leqslant x<2013$.
Upon inspection, when $x=2003$,
$$
\begin{array}{l}
S(2003)=5, S(S(2003))=5 \\
\Rightarrow 2003+5+5=2013 ;
\end{array}
$$
When $x=19... | \{1979,1985,1991,2003\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,520 |
8. In Rt $\triangle A B C$, the incircle $\odot O$ touches the hypotenuse $A B$ at point $D$, and touches $B C, C A$ at points $E, F$ respectively. Draw $D K \perp A C$ at point $K$, and $D P \perp B C$ at point $P$. Given $A D$ $=m, B D=n$. Express the area $S$ of rectangle $C K D P$ in terms of $m, n$ as $\qquad$ | 8. $\frac{2 m^{2} n^{2}}{(m+n)^{2}}$.
Let the inradius be $r$. As shown in Figure 7, connect $O D$, $O E$, and $O F$.
Then $O D=O E=O F=r$.
By the tangent segment theorem,
$$
A D=A F=m, B D=B E=n, C E=C F=r \text {. }
$$
Let the area of $\triangle A B C$ be $S_{1}$. Then
$$
\begin{aligned}
S_{1}= & \frac{(r+m)(r+n)}{... | \frac{2 m^{2} n^{2}}{(m+n)^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,521 |
1. Let $A$ and $B$ be two non-empty finite sets, and the universal set $U$
$$
\begin{aligned}
= & A \cup B, \text{ and } |U|=m. \text{ If } \\
& \left|\left(\complement_{U} A\right) \cup\left(\complement_{U} B\right)\right|=n,
\end{aligned}
$$
then $|A \cap B|=$ | $$
-1 . m-n \text {. }
$$
Notice that,
$$
\left(\complement_{U} A\right) \cup\left(\complement_{U} B\right)=\complement_{U}(A \cap B) .
$$
From the Venn diagram, we know $|A \cap B|=m-n$. | m-n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,522 |
2. In $\triangle A B C$, it is known that the three interior angles $\angle A$, $\angle B$, $\angle C$ are opposite to the sides $a$, $b$, $c$ respectively, and satisfy $a \sin A \cdot \sin B + b \cos ^{2} A = \sqrt{2} a$. Then $\frac{b}{a}=$ . $\qquad$ | 2. $\sqrt{2}$.
From the given and the Law of Sines, we have
$$
\begin{array}{l}
\sqrt{2} a=a \sin A \cdot \sin B+b\left(1-\sin ^{2} A\right) \\
=b+\sin A(a \sin B-b \sin A)=b . \\
\text { Therefore, } \frac{b}{a}=\sqrt{2} .
\end{array}
$$ | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,523 |
3. In the Cartesian coordinate system $x O y$, it is known that there are three points $A(a, 1), B(2, b), C(3,4)$.
If the projections of $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are the same, then $3 a-4 b=$ $\qquad$ | 3. 2 .
Solution 1 The projections of vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are $\frac{\overrightarrow{O A} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}, \frac{\overrightarrow{O B} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}$, respectively.
A... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,524 |
4. In a regular tetrahedron $P-ABC$, it is known that the angle between the lateral edge and the base is $45^{\circ}$. Then the cosine value of the dihedral angle between two adjacent lateral faces is $\qquad$ | 4. $\frac{1}{5}$.
As shown in Figure 2, let the base edge length of the regular tetrahedron $P-ABC$ be $a$, and $E$ be the midpoint of $AB$.
Then $\angle PCE$ is the angle between the lateral edge $PC$ and the base $ABC$, i.e., $\angle PCE=45^{\circ}$.
Draw $AF \perp PC$ at point $F$ from point $A$.
By symmetry,
$BF ... | \frac{1}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,525 |
5. Given three distinct integers $x, y, z$ whose sum lies between 40 and 44. If $x, y, z$ form an arithmetic sequence with a common difference of $d$, and $x+y, y+z, z+x$ form a geometric sequence with a common ratio of $q$, then $d q=$ $\qquad$ | 5. 42 .
$$
\begin{array}{l}
\text { Given } x=y-d, z=y+d \\
\Rightarrow x+y=2 y-d, y+z=2 y+d \\
\Rightarrow z+x=2 y . \\
\text { Also, }(x+y)(z+x)=(y+z)^{2} \\
\Rightarrow 2 y(2 y-d)=(2 y+d)^{2} \\
\Rightarrow d(d+6 y)=0 .
\end{array}
$$
Since $d \neq 0$, we have $d=-6 y$.
$$
\begin{array}{l}
\text { Also, } 40<x+y+z=... | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,526 |
6. Let points $P$ and $Q$ move on the lines
$$
3 x-y+5=0 \text{ and } 3 x-y-13=0
$$
respectively, the midpoint of segment $P Q$ is $M\left(x_{0}, y_{0}\right)$, and $x_{0}+y_{0} \geqslant 4$. Then the range of $\frac{y_{0}}{x_{1}}$ is $\qquad$ | 6. $[1,3)$.
Solution 1 Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}+x_{2}=2 x_{0}, y_{1}+y_{2}=2 y_{0} \text {, }
$$
and $3 x_{1}-y_{1}+5=0,3 x_{2}-y_{2}-13=0$.
Adding the two equations, we get
$$
\begin{array}{l}
3\left(x_{1}+x_{2}\right)-\left(y_{1}+y_{2}\right)-8=0 \\
\Rightarrow 3 x... | [1,3) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,527 |
7. If three points are randomly taken on a circle, the probability that the triangle formed by these three points is an acute triangle is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 7. $\frac{1}{4}$.
Assume $\triangle A B C$ is any inscribed triangle in a circle with radius 1, and the arc lengths opposite to $\angle A$ and $\angle B$ are $x$ and $y$, respectively. Then,
$$
\left\{\begin{array}{l}
0<x<2 \pi, \\
0<y<2 \pi, \\
0<x+y<2 \pi .
\end{array}\right.
$$
This system of inequalities represen... | \frac{1}{4} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 726,528 |
8. Let $M=1^{4}+2^{4}+\cdots+2013^{4}$. Then the unit digit of $M$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 8. 1 .
Let $a, b$ be positive integers. Then, $(10 a+b)^{4} \equiv b^{4}(\bmod 10)$.
Thus, $1^{4}+2^{4}+\cdots+10^{4}$
$$
\begin{array}{l}
=1+6+1+6+5+6+1+6+1+0 \\
=3(\bmod 10)
\end{array}
$$
Therefore, $M=1^{4}+2^{4}+3^{4}+201 \times 3$
$$
=1(\bmod 10)
$$ | null | Other | math-word-problem | Yes | Yes | cn_contest | false | 726,529 |
For example, $8 n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$
and the arithmetic mean of any $n-1$ different numbers among $a_{1}, a_{2}, \cdots, a_{n}$ is a positive integer. Find the maximum value of $n$. [4]
(2009, "Mathematics Weekly Cup" National Junior High Sch... | Let $a_{1}, a_{2}, \cdots, a_{n}$ be such that removing $a_{i} (i=1, 2, \cdots, n)$ leaves the arithmetic mean of the remaining $n-1$ numbers as a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$
Thus, for any $1 \leqslant i<j \leqslant n$, we have
$$
b_{i}-b_{j}=\f... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,530 |
9. If for any $x \in\left(-\frac{1}{2}, 1\right)$, we have
$$
\frac{x}{1+x-2 x^{2}}=\sum_{k=0}^{\infty} a_{k} x^{k},
$$
then $a_{3}+a_{4}=$ . $\qquad$ | 9. -2 .
In equation (1), let $x=0$, we get $a_{0}=0$. Then $\frac{1}{1+x-2 x^{2}}=\sum_{k=1}^{\infty} a_{k} x^{k-1}$. Substituting $x=0$ into the above equation, we get $a_{1}=1$. Then $\frac{-1+2 x}{1+x-2 x^{2}}=\sum_{k=2}^{\infty} a_{k} x^{k-2}$. Substituting $x=0$ into the above equation, we get $a_{2}=-1$. Similar... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,531 |
10. If $0 \leqslant x_{i} \leqslant 1(i=1,2, \cdots, 5)$, then
$$
M=x_{1}-x_{2}^{3}+x_{2}-x_{3}^{3}+x_{3}-x_{4}^{3}+x_{4}-x_{5}^{3}+x_{5}-x_{1}^{3}
$$
the maximum value is $\qquad$ | 10.4.
If there exists a positive integer $j$, such that
$$
x_{j}=x_{j+1}\left(j=1,2, \cdots, 5, x_{6}=x_{1}\right) \text {, }
$$
then $M \leqslant 4$.
When $x_{1}=0, x_{2}=1, x_{3}=0, x_{4}=1, x_{5}=0$, the equality holds.
If for any positive integer $i$, we have $x_{i} \neq x_{i+1}$ $(i=1,2, \cdots 5)$, then either... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,532 |
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