problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
1. If the quadratic equation with integer coefficients
$$
x^{2}+(a+3) x+2 a+3=0
$$
has one positive root $x_{1}$ and one negative root $x_{2}$, and $\left|x_{1}\right|<\left|x_{2}\right|$, then
$$
a=
$$
$\qquad$ | $=、 1 .-2$.
Since the two roots of the equation are not equal, we have $\Delta>0$, that is
$$
(a+3)^{2}>4(2 a+3) \text {. }
$$
Solving this, we get $a>3$ or $a-3, a<-\frac{3}{2}$, which means $-3<a<-\frac{3}{2}$.
Since $a$ is an integer, then $a=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Given the quadratic function
$$
y=x^{2}+2 m x-n^{2} \text {. }
$$
(1) If the graph of this quadratic function passes through the point $(1,1)$, and let the larger of the two numbers $m, n+4$ be $P$, find the minimum value of $P$;
(2) If $m, n$ vary, these functions represent different parabolas. If eac... | (1) From the quadratic function passing through the point $(1,1)$, we get $m=\frac{n^{2}}{2}$.
Notice that
$$
\begin{array}{l}
m-(n+4)=\frac{n^{2}}{2}-(n+4) \\
=\frac{1}{2}\left(n^{2}-2 n-8\right)=\frac{1}{2}(n-4)(n+2),
\end{array}
$$
Therefore, $P=\left\{\begin{array}{ll}\frac{n^{2}}{2}, & n \leqslant-2 \text { or } ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In pentagon $A B C D E$, $\angle A=\angle C=90^{\circ}$, $A B=B C=D E=A E+C D=3$. Then the area of this pentagon is $(\quad)$.
(A) 9
(B) 10.5
(C) 12
(D) 13.5 | 3. A.
As shown in Figure 3, extend $DC$ to point $F$ such that $CF = AE$, and connect $BE$, $BD$, and $BF$. Then,
$$
DF = DE = 3.
$$
Also, $\triangle BCF \cong \triangle BAE$, so
$$
BE = BF.
$$
Since $BD = BD$, we have
$$
\triangle BED \cong \triangle BFD.
$$
Therefore, $S_{\triangle BRD} = S_{\triangle BFD}$. Henc... | 9 | Geometry | MCQ | Yes | Yes | cn_contest | false |
11. Given $a+b+c=0, a^{2}+b^{2}+c^{2}=4$. Then, the value of $a^{4}+b^{4}+c^{4}$ is $\qquad$ . | 11.8.
From the known equations, we get $a+b=-c, a^{2}+b^{2}=4-c^{2}$.
$$
\begin{array}{l}
\text { Also, } a b=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right] \\
=\frac{1}{2}\left[(-c)^{2}-\left(4-c^{2}\right)\right]=c^{2}-2 .
\end{array}
$$
Therefore, $a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$ and $b$ are non-negative real numbers, and
$$
a^{2005}+b^{2005}=1, a^{2006}+b^{2006}=1 \text {. }
$$
then $a^{2007}+b^{2007}=$ $\qquad$ | Ni, 1.1.
If $a$ or $b$ is greater than 1, then $a^{2005} + b^{2005} > 1$; if $0 < a, b < 1$, then $a^{2005} > a^{2006}$, $b^{2005} > b^{2006}$. Therefore, $a^{2000} + b^{2000} > a^{2006} + b^{2000}$, which means $1 > 1$, a contradiction. Hence, one of $a$ or $b$ is 0, and the other is 1. Thus, $a^{200} + b^{20 n} = 1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Let the function $f(x)=x^{2}+a x+b(a, b \in$
$\mathbf{R})$. If there exists a real number $m$, such that
$$
|f(m)| \leqslant \frac{1}{4} \text {, and }|f(m+1)| \leqslant \frac{1}{4},
$$
find the maximum and minimum values of $\Delta=a^{2}-4 b$. | Solution 1: If $\Delta=a^{2}-4 b$
$$
\frac{-a-\sqrt{\Delta+1}}{2} \leqslant x \leqslant \frac{-a-\sqrt{\Delta-1}}{2}
$$
or $\frac{-a+\sqrt{\Delta-1}}{2} \leqslant x \leqslant \frac{-a+\sqrt{\Delta+1}}{2}$.
If $|f(m)| \leqslant \frac{1}{4}$, and $|f(m+1)| \leqslant \frac{1}{4}$, then it must be true that
$$
\frac{-a+\s... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given positive numbers $a, b, c$ satisfying $a+b+c=3$.
Prove:
$$
\begin{array}{l}
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+ \\
\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant 5 .
\end{array}
$$ | $$
\begin{array}{l}
\text { 5. } \frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}=\frac{a^{2}+9}{2 a^{2}+(3-a)^{2}} \\
=\frac{1}{3}\left(1+\frac{2 a+6}{a^{2}-2 a+3}\right)=\frac{1}{3}\left[1+\frac{2 a+6}{(a-1)^{2}+2}\right] \\
\leqslant \frac{1}{3}\left(1+\frac{2 a+6}{2}\right)=\frac{1}{3}(4+a) .
\end{array}
$$
Similarly, $\frac{b^{... | 5 | Inequalities | proof | Yes | Yes | cn_contest | false |
II. (25 points) As shown in Figure 4, in $\triangle ABC$, $\angle BAC=90^{\circ}$, $AB=AC$, points $D_{1}$ and $D_{2}$ are on $AC$, and satisfy $AD_{1}=CD_{2}$, $AE_{1} \perp BD_{1}$, $AE_{2} \perp BD_{2}$, intersecting $BC$ at points $E_{1}$ and $E_{2}$, respectively. Prove that: $\frac{CE_{2}}{BE_{2}}+\frac{CE_{1}}{B... | $$
\begin{array}{l}
\frac{C E_{1}}{B E_{1}}=\frac{A F}{A B}, \\
\angle E_{1} A C=\angle F C A . \\
\text { Also, } \angle A B D_{1}=90^{\circ}-\angle B A E_{1}=\angle E_{1} A C, \text { then } \\
\angle A B D_{1}=\angle A C F .
\end{array}
$$
Since $\angle B A D_{1}=\angle C A F=90^{\circ}, A B=A C$, therefore, $\tria... | 1 | Geometry | proof | Yes | Yes | cn_contest | false |
10. There are four numbers, among which the sum of every three numbers is $24, 36, 28, 32$. Then the average of these four numbers is $\qquad$ . | 10.10 .
The sum of these four numbers is $(24+36+28+32) \div 3=40$, so the average of these four numbers is 10 . | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. If $a^{4}+b^{4}=a^{2}-2 a^{2} b^{2}+b^{2}+6$, then $a^{2}+b^{2}=$ $\qquad$ . | 11.3.
Given $a^{4}+b^{4}=a^{2}-2 a^{2} b^{2}+b^{2}+6$, we have $\left(a^{2}+b^{2}\right)^{2}-\left(a^{2}+b^{2}\right)-6=0$.
Therefore, $a^{2}+b^{2}=3$ or -2.
Since $a^{2}+b^{2} \geqslant 0$, we have $a^{2}+b^{2}=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. If real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x y+x+y+7=0, \\
3 x+3 y=9+2 x y,
\end{array}\right.
$$
then $x^{2} y+x y^{2}=$ | 13.6.
From $\left\{\begin{array}{l}x y+x+y+7=0, \\ 3 x+3 y=9+2 x y,\end{array}\right.$ we get $x y=-6, x+y=-1$.
Therefore, $x^{2} y+x y^{2}=6$. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. A person's 5 trips to work (unit: $\mathrm{min}$) are $a, b, 8, 9, 10$. It is known that the average of this set of data is 9, and the variance is 2. Then the value of $|a-b|$ is $\qquad$. | 15.4.
Since the average of $a, b, 8, 9, 10$ is 9 and the variance is 2, it follows that $a, b, 8, 9, 10$ are 5 consecutive integers, so $a=7, b=11$ or $a=11, b=7$. Therefore, the value of $|a-b|$ is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
16. If the integer $m$ makes the equation
$$
x^{2}-m x+m+2006=0
$$
have non-zero integer roots, then the number of such integers $m$ is
$\qquad$. | 16.5.
Let the two integer roots of the equation be $\alpha, \beta$, then
$$
\begin{array}{l}
\alpha+\beta=m, a \beta=m+2006, \\
\text { i.e., } \alpha \beta-(\alpha+\beta)+1=2006+1 \\
=2007=(\alpha-1)(\beta-1) .
\end{array}
$$
Thus, $\alpha-1= \pm 1, \pm 3, \pm 9$;
$$
\beta-1= \pm 2007, \pm 669, \pm 223 \text {. }
$$... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $x, y$ be real numbers, the algebraic expression
$$
5 x^{2}+4 y^{2}-8 x y+2 x+4
$$
has a minimum value of | 9.3.
$$
\begin{array}{l}
\text { Since } 5 x^{2}+4 y^{2}-8 x y+2 x+4 \\
=5\left(x-\frac{4}{5} y+\frac{1}{5}\right)^{2}+\frac{4}{5}(y+1)^{2}+3,
\end{array}
$$
Therefore, when $y=-1, x=-1$, the original expression has a minimum value of 3. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. In a $3 \times 3$ grid filled with the numbers $1 \sim 9$, the largest number in each row is colored red, and the smallest number in each row is colored green. Let $M$ be the smallest number among the three red squares, and $m$ be the largest number among the three green squares. Then $M-m$ can have $\qquad$ differ... | 14.8.
From the conditions, it is known that the values of $m$ and $M$ are positive integers from 3 to 7 (inclusive of 3 and 7).
Obviously, $M \neq m$, so the value of $M-m$ is
$$
1,2,3,4,-1,-2,-3,-4 \text {. }
$$
Therefore, $M-m$ can have 8 different values. | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
13. Given $\frac{\sin (\alpha+2 \beta)}{\sin \alpha}=3$, and $\beta \neq \frac{1}{2} k \pi$, $\alpha+\beta \neq n \pi+\frac{\pi}{2}(n, k \in \mathbf{Z})$. Then the value of $\frac{\tan (\alpha+\beta)}{\tan \beta}$ is $\qquad$ | $\begin{array}{l}13.2 \text {. } \\ \frac{\tan (\alpha+\beta)}{\tan \beta}=\frac{\sin (\alpha+\beta) \cdot \cos \beta}{\cos (\alpha+\beta) \cdot \sin \beta} \\ =\frac{\frac{1}{2}[\sin (\alpha+2 \beta)+\sin \alpha]}{\frac{1}{2}[\sin (\alpha+2 \beta)-\sin \alpha]}=\frac{\frac{\sin (\alpha+2 \beta)}{\sin \alpha}+1}{\frac{... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Given a finite set of planar vectors $M$, for any three elements chosen from $M$, there always exist two elements $\boldsymbol{a}, \boldsymbol{b}$ such that $\boldsymbol{a}+\boldsymbol{b} \in M$. Try to find the maximum number of elements in $M$.
| Three, the maximum number of elements in the set $M$ is 7.
Let points $A, B, C$ be any three points on a plane. Consider the 7-element set $M=\{\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C A}, \boldsymbol{B A}, \boldsymbol{C B}, \boldsymbol{A C}, \mathbf{0}\}$, which clearly satisfies the conditions. We will now p... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. The number of positive integers $n$ such that $n+1$ divides $n^{2006}+2006$ is $\qquad$.
Makes the positive integer $n$ such that $n+1$ can divide $n^{2006}+2006$ total $\qquad$.
Note: The second sentence seems to be a repetition or a different phrasing of the first. If it's meant to be a different statement, plea... | 5. 5 .
From the problem, we have
$$
\begin{array}{l}
n^{2006}+2006 \equiv(-1)^{2006}+2006=2007 \\
\equiv 0(\bmod (n+1)) .
\end{array}
$$
Since $2007=3 \times 3 \times 223$, then
$$
n+1=3,9,223,669,2007 \text {. }
$$
Therefore, $n=2,8,222,668,2006$. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given $n(n>1)$ integers (which can be the same) $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2007 .
$$
Then the minimum value of $n$ is $\qquad$ | 9. 5 .
Given $a_{1} a_{2} \cdots a_{n}=2007$, we know that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. Also, $a_{1}+a_{2}+\cdots+a_{n}=2007$ is an odd number, so $n$ is odd.
If $n=3$, i.e., $a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3}=2007$, without loss of generality, assume $a_{1} \geqslant a_{2} \geqslant a_{3}$, t... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8.3. There are 19 cards. Can a non-zero digit be written on each card so that these 19 cards can be arranged to form a 19-digit number divisible by 11? | 8.3. Answer: Yes.
Write a 2 on each of 10 cards, and a 1 on each of the remaining cards. It is well known that a positive decimal integer is divisible by 11 if and only if the difference $S$ between the sum of the digits in the odd positions and the sum of the digits in the even positions is a multiple of 11. Under al... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of positive integer pairs $(x, y)$ that satisfy the equation
$$
\begin{array}{l}
x \sqrt{y}+y \sqrt{x}-\sqrt{2006 x}-\sqrt{2006 y}+\sqrt{2006 x y} \\
\quad=2006
\end{array}
$$ | 3.8 .
Given the equation can be transformed into
$$
(\sqrt{x}+\sqrt{y}+\sqrt{2006})(\sqrt{x y}-\sqrt{2006})=0 \text {. }
$$
Since $\sqrt{x}+\sqrt{y}+\sqrt{2006}>0$, we have
$$
\sqrt{x y}=\sqrt{2006} \text {, }
$$
which means $\square$
$$
\begin{array}{l}
x y=2006=1 \times 2006=2 \times 1003 \\
=17 \times 118=34 \tim... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. A three-digit number $\overline{x y z}, 1 \leqslant x \leqslant 9,0 \leqslant y, z \leqslant 9$, and $x!+y!+z!=\overline{x y z}$. Then the value of $x+y+z$ is $\qquad$ | 5.10.
From $\overline{x y z}=x!+y!+z!$, we get
$$
100 x+10 y+z=x!+y!+z!.
$$
It is easy to see that $x, y, z \leqslant 6$. Otherwise,
$$
x!+y!+z!\geqslant 7!>1000.
$$
If $x=6$, then the left side of equation (1) is $700$, which is a contradiction.
Therefore, $x \leqslant 5$.
Similarly, $y \leqslant 5, z \leqslant 5$.... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. If the graph of the inverse proportion function $y=\frac{k}{x}$ intersects the graph of the linear function $y=a x+b$ at points $A(-2, m)$ and $B(5, n)$, then the value of $3 a+b$ is
保留了源文本的换行和格式。 | Ni.7.0.
Since points $A(-2, m)$ and $B(5, n)$ lie on the graph of an inverse proportion function, we have
$$
\left\{\begin{array}{l}
m=-\frac{k}{2}, \\
n=\frac{k}{5} .
\end{array}\right.
$$
Since points $A(-2, m)$ and $B(5, n)$ also lie on the line $y=a x+b$, we get
$$
\left\{\begin{array}{l}
m=-2 a+b \\
n=5 a+b
\end... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Given that $a$, $b$, and $c$ are positive integers, and the graph of the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from points $A$ and $B$ to the origin $O$ are both less than 1, find the minimum value of $a+b+c$. | 14. Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, where $x_{1}$ and $x_{2}$ are the roots of the equation $a x^{2}+b x+c=0$.
Given that $a$, $b$, and $c$ are positive integers, we have
$x_{1}+x_{2}=-\frac{b}{a} < 0$.
Thus, the equation $a x^{2}+b x+c=0$ has two negative real roots, i.e., $x_{1} < 0$.
Solvi... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. At a party, 9 celebrities performed $n$ "trio dance" programs. If in these programs, any two people have collaborated exactly once, then $n=$ $\qquad$ | 12.12.
Consider 9 people as 9 points. If two people have performed in the same group, then a line segment is drawn between the corresponding two points. Thus, every pair of points is connected, resulting in a total of 36 line segments. Each trio dance corresponds to a triangle, which has three sides. When all sides ap... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Find the smallest positive real number $k$ such that for any 4 distinct real numbers $a, b, c, d$ not less than $k$, there exists a permutation $p, q, r, s$ of $a, b, c, d$ such that the equation $\left(x^{2}+p x+q\right)\left(x^{2}+r x+s\right)=0$ has 4 distinct real roots. (Feng Zhigang) | 2. On one hand, if $k4(d-a)>0, c^{2}-4 b>4(c-b)$ $>0$, therefore, equations (1) and (2) both have two distinct real roots.
Secondly, if equations (1) and (2) have a common real root $\beta$, then
$$
\left\{\begin{array}{l}
\beta^{2}+d \beta+a=0, \\
\beta^{2}+\phi \beta+b=0 .
\end{array}\right.
$$
Subtracting the two e... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $a$, $b$, $c$ be positive integers, and satisfy
$$
a^{2}+b^{2}+c^{2}-a b-b c-c a=19 \text {. }
$$
Then the minimum value of $a+b+c$ is $\qquad$ | 4. 10 .
By the cyclic symmetry of $a, b, c$, without loss of generality, assume $a \geqslant b \geqslant c$.
Let $a-b=m, b-c=n$, then $a-c=m+n$, and
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[m^{2}+n^{2}+(m+n)^{2}\right],
\end{array}
$$
which is $m^{2}+m n+n^{2}-19=0$.
For the equation in ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 1, label the six vertices of a regular hexagon with the numbers $0,1,2,3,4,5$ in sequence. An ant starts from vertex 0 and crawls counterclockwise, moving 1 edge on the first move, 2 edges on the second move, $\cdots \cdots$ and $2^{n-1}$ edges on the $n$-th move. After 2005 moves, the value at th... | Ni.1.1.
According to the problem, after 2005 movements, the ant has crawled over
$$
1+2+\cdots+2^{2004}=2^{2005}-1
$$
edges.
Let $x=2^{2005}-1$. Consider the remainder of $x$ modulo 6.
Since $x \equiv 1(\bmod 2), x \equiv 1(\bmod 3)$, we have
$$
3 x \equiv 3(\bmod 6), 2 x \equiv 2(\bmod 6), 5 x \equiv 5(\bmod 6) \text... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $x \in\left(0, \frac{\pi}{2}\right), \sin ^{2} x, \sin x \cdot \cos x$, and $\cos ^{2} x$ cannot form a triangle, and the length of the interval of $x$ that satisfies the condition is $\arctan k$. Then $k$ equals $\qquad$ (Note: If $x \in(a, b), b>a$, then the length of such an interval is $b-a$). | 2.2.
First, find the range of $x$ that can form a triangle.
(1) If $x=\frac{\pi}{4}$, then $\sin ^{2} \frac{\pi}{4} 、 \cos ^{2} \frac{\pi}{4} 、 \sin \frac{\pi}{4} \cdot \cos \frac{\pi}{4}$ can form a triangle.
(2) If $x \in\left(0, \frac{\pi}{4}\right)$, at this time,
$$
\sin ^{2} x\cos ^{2} x$, that is,
$$
\tan ^{2} ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
4 y^{3}+\frac{1}{2} \sin 2 y+a=0 .
\end{array}\right.
$$
then the value of $\cos (x+2 y)$ is | 8.1.
Let $f(t)=t^{3}+\sin t$. Then $f(t)$ is monotonically increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
From the original system of equations, we get $f(x)=f(-2 y)=2 a$, and since $x$ and $-2 y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, it follows that $x=-2 y$, hence $x+2 y=0$. Therefore, $\cos (... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If the six edges of a tetrahedron are $2,3,4,5$, 6,7, then there are $\qquad$ different shapes (if two tetrahedra can be made to coincide by appropriate placement, they are considered the same shape). | 6.10.
Let the edge of length $k$ be denoted as $l_{k}(k \in\{2,3,4,5,6,7\})$. Consider $l_{2}$ and $l_{3}$.
(1) If $l_{2}$ and $l_{3}$ are coplanar, then the other side of the plane must be $l_{4}$.
(i) If $l_{2}$, $l_{3}$, and $l_{4}$ form a triangle in a clockwise direction (all referring to the direction of the thr... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. (16 points) Does there exist a smallest positive integer $t$, such that the inequality
$$
(n+t)^{n+t}>(1+n)^{3} n^{n} t^{t}
$$
holds for any positive integer $n$? Prove your conclusion. | $$
\text { Three, 13. Take }(t, n)=(1,1),(2,2),(3,3) \text {. }
$$
It is easy to verify that when $t=1,2,3$, none of them meet the requirement.
When $t=4$, if $n=1$, equation (1) obviously holds.
If $n \geqslant 2$, we have
$$
\begin{array}{l}
4^{4} n^{n}(n+1)^{3}=n^{n-2}(2 n)^{2}(2 n+2)^{3} \times 2^{3} \\
\leqslant\... | 4 | Inequalities | proof | Yes | Yes | cn_contest | false |
15. (22 points) Let $A$ and $B$ be the common left and right vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Let $P$ and $Q$ be moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
... | 15. (1) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}}.
$$
Similarly,
$$
k_{3}+k_{4}=-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}}.
$$
Let $O$ be the... | 8 | Geometry | proof | Yes | Yes | cn_contest | false |
3. Points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the parabola $y=2 x^{2}$ are symmetric with respect to the line $y=x+m$. If $2 x_{1} x_{2}=-1$, then the value of $2 m$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 3. A.
$$
\begin{array}{l}
\text { Given } \frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-1, \\
\frac{y_{1}+y_{2}}{2}=\frac{x_{1}+x_{2}}{2}+m, \\
2 x_{1} x_{2}=-1 \text { and } y_{1}=2 x_{1}^{2}, y_{2}=2 x_{2}^{2},
\end{array}
$$
we get
$$
\begin{array}{l}
x_{2}-x_{1}=y_{1}-y_{2}=2\left(x_{1}^{2}-x_{2}^{2}\right) \Rightarrow x_{1}+x... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
7. If $\frac{1-\cos \theta}{4+\sin ^{2} \theta}=\frac{1}{2}$, then
$$
\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=
$$ | ニ、7.9.
From the condition, we get $2-2 \cos \theta=4+\sin ^{2} \theta$, then
$$
\begin{array}{l}
(\cos \theta-1)^{2}=4 \Rightarrow \cos \theta=-1 . \\
\text { Therefore, }\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=9 .
\end{array}
$$ | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. The sequence $\left\{x_{n}\right\}: 1,3,3,3,5,5,5,5,5, \cdots$ is formed by arranging all positive odd numbers in ascending order, and each odd number $k(k=1,3,5, \cdots)$ appears consecutively $k$ times. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$ (where $[x]$ denotes the greatest inte... | 8.3.
Given $x_{k^{2}+1}=x_{k^{2}+2}=\cdots=x_{(k+1)^{2}}=2 k+1$, that is, when $k^{2}+1 \leqslant n \leqslant(k+1)^{2}$, we have $x_{n}=2 k+1(k=[\sqrt{n-1}])$.
Therefore, $x_{n}=2[\sqrt{n-1}]+1$.
Hence, $(a, b, c, d)=(2,1,-1,1), a+b+c+d=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. (12 points) Find the area of the figure formed by the set of points on the right-angle coordinate plane $b O a$
$$
S=\left\{(b, a) \mid f(x)=a x^{3}+b x^{2}-3 x\right.
$$
is a monotonic function on $\mathbf{R}$, and $a \geqslant-1\}$. | Three, 15. When $a=0$, by $f(x)$ being monotonic on $\mathbf{R}$, we know $b=0$. $f(x)$ being monotonic on $\mathbf{R}$ $\Leftrightarrow f^{\prime}(x)$ does not change sign on $\mathbf{R}$.
Since $f^{\prime}(x)=3 a x^{2}+2 b x-3$, therefore, by $\Delta=4 b^{2}+36 a \leqslant 0$,
we get, $a \leqslant-\frac{1}{9} b^{2}$.... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The number of prime pairs $(p, q)$ that satisfy $\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=q$ is $\qquad$ . | 5.2.
(1) When $p=2$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=1$, which is not a prime number, contradiction.
(2) When $p=3$, $q=\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=1+1=2$ is a prime number.
(3) When $p=5$, $q=\left[\frac{p}{2}\right]+\left[\fr... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $f(x)$ is a function defined on $\mathbf{R}$, $f\left(\frac{\pi}{4}\right)=0$, and for any $x, y \in \mathbf{R}$, we have
$$
f(x)+f(y)=2 f\left(\frac{x+y}{2}\right) f\left(\frac{x-y}{2}\right) .
$$
Then $f\left(\frac{\pi}{4}\right)+f\left(\frac{3 \pi}{4}\right)+f\left(\frac{5 \pi}{4}\right)+\cdots+f\left... | 6.0 .
Let $\frac{x-y}{2}=\frac{\pi}{4}$, then
$$
\begin{array}{l}
f(x)+f\left(x-\frac{\pi}{2}\right)=2 f\left(x-\frac{\pi}{4}\right) f\left(\frac{\pi}{4}\right)=0 . \\
\text { Therefore } f\left(\frac{\pi}{4}\right)+f\left(\frac{3 \pi}{4}\right)+f\left(\frac{5 \pi}{4}\right)+\cdots+f\left(\frac{2007 \pi}{4}\right)=0 .... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (50 points) Find the maximum value of a prime number $p$ with the following property: there exist two permutations (which can be the same) of $1, 2, \cdots, p$, $a_{1}, a_{2}, \cdots, a_{p}$ and $b_{1}, b_{2}, \cdots, b_{p}$, such that the remainders of $a_{1} b_{1}$, $a_{2} b_{2}, \cdots, a_{p} b_{p}$ when divided... | Lemma (Wilson's Theorem): For any prime $p$, we have $(p-1)! \equiv -1 \pmod{p}$.
Proof of the Lemma: When $p=2,3$, equation (1) is obviously true.
When $p>3$, we prove: For any $2 \leqslant k \leqslant p-2$, there must exist $k^{\prime}\left(2 \leqslant k^{\prime} \leqslant p-2, k^{\prime} \neq k\right)$, such that $... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Square $A B C D$ and square $A B E F$ are in planes that form a $120^{\circ}$ angle, $M$ and $N$ are points on the diagonals $A C$ and $B F$ respectively, and $A M=F N$. If $A B=1$, then the maximum value of $M N$ is $\qquad$ | 5.1.
As shown in Figure 2, draw $M P \perp A B$ at $P$, and connect $P N$. It can be proven that $P N \perp A B$. Thus, $\angle M P N=120^{\circ}$.
Let $A M=F N=x$.
Therefore, $\frac{M P}{1}=\frac{A M}{\sqrt{2}}$,
which means $M P=\frac{\sqrt{2}}{2} x$.
Hence, $P N=\frac{\sqrt{2}-x}{\sqrt{2}}\left(\frac{P N}{1}=\frac... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $\cos \beta+\sin \beta \cdot \cot \theta=\tan 54^{\circ}$, $\cos \beta-\sin \beta \cdot \cot \theta=\tan 18^{\circ}$.
Then the value of $\tan ^{2} \theta$ is $\qquad$ | 6.1
From the given conditions, we have
$\cos \beta=\frac{1}{2}\left(\tan 54^{\circ}+\tan 18^{\circ}\right)=\frac{1}{2 \cos 54^{\circ}}$,
$\sin \beta=\frac{\tan \theta}{2}\left(\tan 54^{\circ}-\tan 18^{\circ}\right)=\frac{\tan \theta}{2 \cos 18^{\circ}}$.
Since $\sin ^{2} \beta+\cos ^{2} \beta=1$, we have
$\frac{1}{\co... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { 2. Let } M=\frac{2 \cos 34^{\circ}-\cos 22^{\circ}}{\cos 14^{\circ}} \text {, } \\
N=\sin 56^{\circ} \cdot \sin 28^{\circ} \cdot \sin 14^{\circ} \text {. } \\
\text { Then } \frac{M}{N}=
\end{array}
$$ | 2.8.
$$
\begin{array}{l}
M=\frac{2 \cos 34^{\circ}-\sin 68^{\circ}}{\cos 14^{\circ}}=\frac{2 \cos 34^{\circ}\left(1-\sin 34^{\circ}\right)}{\cos 14^{\circ}} \\
=\frac{2 \cos 34^{\circ}\left(1-\cos 56^{\circ}\right)}{\cos 14^{\circ}}=\frac{4 \cos 34^{\circ} \cdot \sin ^{2} 28^{\circ}}{\cos 14^{\circ}} \\
=8 \cos 34^{\ci... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. From the 6 face diagonals of the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, two tetrahedra $A-B_{1} C D_{1}$ and $A_{1}-B C_{1} D$ can be formed. If the combined volume of the two tetrahedra is 1 (overlapping parts are counted only once), then the volume of the rectangular prism is $\qquad$ . | 3.2.
Let the length, width, and height of the rectangular prism be $a$, $b$, and $c$ respectively, then the volume of the rectangular prism $V_{0}=a b c$. The volumes of the two tetrahedra (as shown in Figure 4) are both
$$
\begin{array}{l}
V_{1}=V_{0}-V_{A_{1}-A B D}-V_{A_{1}-B B C_{1}}-V_{A_{1}-D_{1} D C_{1}}-V_{C_{... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $x=\frac{1}{2-\sqrt{5}}$. Then $x^{3}+3 x^{2}-5 x+1=$ | From $x=\frac{1}{2-\sqrt{5}}=-2-\sqrt{5}$, we get $x+2=-\sqrt{5}$. Therefore, $(x+2)^{2}=5$, which means $x^{2}+4 x=1$.
Then $x^{3}+3 x^{2}-5 x+1=\left(x^{2}+4 x-1\right)(x-1)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let for any natural numbers $m$, $n$ satisfying $\frac{m}{n}<\sqrt{7}$, the inequality $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\lambda}{n^{2}}$ always holds. Then the maximum value of $\lambda$ is $\qquad$. | 5.3.
The original inequality $\Leftrightarrow 7 n^{2}-m^{2} \geqslant \lambda$.
Since $7 n^{2} \equiv 0(\bmod 7), m^{2} \equiv 0,1,2,4(\bmod 7)$, then $\left(7 n^{2}-m^{2}\right)_{\text {min }}=3$, so, $\lambda_{\text {max }}=3$. | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) Given the semi-ellipse $\frac{x^{2}}{4}+y^{2}=1(y>0)$, two perpendicular lines are drawn through a fixed point $C(1,0)$ intersecting the ellipse at points $P$ and $Q$, respectively. Here, $O$ is the origin, and $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse.
(1) Find the minimum value ... | (1) Since $P O$ is the median of $\triangle P F_{1} F_{2}$, we have
$$
P F_{1}+P F_{2}=2 P O, \left|P F_{1}+P F_{2}\right|=2|P O| \text {. }
$$
Therefore, when $P$ is at the vertex of the minor axis, $\left|P F_{1}+P F_{2}\right|$ achieves its minimum value of 2.
(2) From the problem analysis, the slopes of the lines ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
204 Given that $a, b, c$ are positive numbers satisfying $a+b+c=1$. Prove:
$$
\frac{a^{3}+b}{a(b+c)}+\frac{b^{3}+c}{b(c+a)}+\frac{c^{3}+a}{c(a+b)} \geqslant 5 .
$$ | Prove: $\frac{a^{3}+b}{a(b+c)}=\frac{a^{2}(1-b-c)+b}{a(b+c)}$
$$
\begin{array}{l}
=\frac{a^{2}+b}{a(b+c)}-a=\frac{a(1-b-c)+b}{a(b+c)}-a \\
=\frac{a+b}{a(b+c)}-1-a .
\end{array}
$$
Similarly, $\frac{b^{3}+c}{b(c+a)}=\frac{b+c}{b(c+a)}-1-b$,
$$
\frac{c^{3}+a}{c(a+b)}=\frac{c+a}{c(a+b)}-1-c \text {. }
$$
Adding the abov... | 5 | Inequalities | proof | Yes | Yes | cn_contest | false |
Example 3 For all $a, b, c \in \mathbf{R}_{+}$, find
$$
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}}
$$
the minimum value. | Explanation: Make the substitution
$$
\begin{array}{l}
x=\frac{a}{\sqrt{a^{2}+8 b c}}, y=\frac{b}{\sqrt{b^{2}+8 a c}}, \\
z=\frac{c}{\sqrt{c^{2}+8 a b}} .
\end{array}
$$
Then $x, y, z \in (0, +\infty)$.
Thus, $x^{2}=\frac{a^{2}}{a^{2}+8 b c}$, which means $\frac{1}{x^{2}}-1=\frac{8 b c}{a^{2}}$.
Similarly, $\frac{1}{y... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Find
$$
\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}
$$
the minimum value. | Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}(x, y, z \in \mathbf{R}_{+})$, then
$$
\begin{array}{l}
\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1} \\
=\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{[y(y+2 x)+z(z+2 y)+x(x+2 z)] \cdot}... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a, b, c \in \mathbf{R}_{+}$, and $a+b+c=1$. Find
$$
\frac{3 a^{2}-a}{1+a^{2}}+\frac{3 b^{2}-b}{1+b^{2}}+\frac{3 c^{2}-c}{1+c^{2}}
$$
the minimum value. | ( Hint: First prove $\frac{3 a^{2}-a}{1+a^{2}} \geqslant \frac{9}{10}\left(a-\frac{1}{3}\right)$. Similarly, obtain the other two inequalities. Then add the three inequalities, to get the minimum value of 0. ) | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 The inequality $x^{2}+|2 x-4| \geqslant p$ holds for all real numbers $x$. Find the maximum value of the real number $p$.
保留了原文的换行和格式，如上所示。 | Explanation 1: Transform the original inequality into
$|x-2| \geqslant-\frac{x^{2}}{2}+\frac{p}{2}$.
Let $y_{1}=|x-2|, y_{2}=-\frac{x^{2}}{2}+\frac{p}{2}$.
Thus, solving inequality (1) is equivalent to determining the range of $x$ values for which the graph of the function $y_{1}=|x-2|$ is above the graph of the functi... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
6. As shown in Figure 1, in the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $CA=4$, $P$ is the midpoint of the semicircular arc $\overparen{AC}$, connect $BP$, the line segment $BP$ divides the figure $APCB$ into two parts. The absolute value of the difference in the areas of these two parts is $\qquad$. | Ni.6.4.
As shown in Figure 8, let $A C$ and $B P$ intersect at point $D$, and the point symmetric to $D$ with respect to the circle center $O$ is denoted as $E$. The line segment $B P$ divides the figure $A P C B$ into two parts, and the absolute value of the difference in the areas of these two parts is the area of $\... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12.B. The real numbers $a, b, c$ satisfy $a \leqslant b \leqslant c$, and $ab + bc + ca = 0, abc = 1$. Find the largest real number $k$ such that the inequality $|a+b| \geqslant k|c|$ always holds. | 12. B. When $a=b=-\sqrt[3]{2}, c=\frac{\sqrt[3]{2}}{2}$, the real numbers $a, b, c$ satisfy the given conditions, at this time, $k \leqslant 4$.
Below is the proof: The inequality $|a+b| \geqslant 4|c|$ holds for all real numbers $a, b, c$ that satisfy the given conditions.
From the given conditions, we know that $a, ... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11.B. Given the parabolas $C_{1}: y=-x^{2}-3 x+4$ and $C_{2}: y=x^{2}-3 x-4$ intersect at points $A$ and $B$, point $P$ is on parabola $C_{1}$ and lies between points $A$ and $B$; point $Q$ is on parabola $C_{2}$, also lying between points $A$ and $B$.
(1) Find the length of segment $A B$;
(2) When $P Q \parallel y$-ax... | 11. B. (1) Solve the system of equations $\left\{\begin{array}{l}y=-x^{2}-3 x+4, \\ y=x^{2}-3 x-4,\end{array}\right.$ to get
$$
\left\{\begin{array} { l }
{ x _ { 1 } = - 2 , } \\
{ y _ { 1 } = 6 , }
\end{array} \left\{\begin{array}{l}
x_{2}=2, \\
y_{2}=-6 .
\end{array}\right.\right.
$$
Therefore, $A(-2,6)$ and $B(2,... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $a b=1$, and $\frac{1}{1-2^{x} a}+\frac{1}{1-2^{y+1} b}=1$, then the value of $x+y$ is $\qquad$. | 8. -1 .
Given that both sides of the equation are multiplied by $\left(1-2^{x} a\right)\left(1-2^{y+1} b\right)$, we get
$$
2-2^{x} a-2^{y+1} b=1-2^{x} a-2^{y+1} b+2^{x+y+1} a b \text {, }
$$
which simplifies to $1=2 \times 2^{x+y}$.
Therefore, $x+y=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. As shown in Figure 2, in Pascal's Triangle, the numbers above the diagonal form the sequence: $1,3,6,10, \cdots$, let the sum of the first $n$ terms of this sequence be $S_{n}$. Then, as $n \rightarrow +\infty$, the limit of $\frac{n^{3}}{S(n)}$ is $\qquad$ | 10.6 .
It is known that $S_{n}=\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\cdots+\mathrm{C}_{n+1}^{2}=\mathrm{C}_{\mathrm{n}+2}^{3}$. Therefore, $\lim _{n \rightarrow+\infty} \frac{n^{3}}{S(n)}=\lim _{n \rightarrow+\infty} \frac{n^{3} \times 6}{(n+2)(n+1) n}=6$. | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Given sets $A_{1}, A_{2}, \cdots, A_{n}$ are different subsets of the set $\{1,2, \cdots, n\}$, satisfying the following conditions:
(i) $i \notin A_{i}$ and $\operatorname{Card}\left(A_{i}\right) \geqslant 3, i=1,2, \cdots, n$;
(ii) $i \in A_{j}$ if and only if $j \notin A_{i}(i \neq j$, $i, j=1,2, ... | (1) Let $A_{1}, A_{2}, \cdots, A_{n}$ be $n$ sets, and let $r_{i}$ be the number of sets that contain element $i$ for $i=1,2, \cdots, n$. Then,
$$
\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)=\sum_{i=1}^{n} r_{i} \text {. }
$$
Assume the $r_{1}$ sets that contain element 1 are
$$
A_{2}, A_{3}, \cdots, A_{r_{1}... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $S$ be a subset of $\{1,2, \cdots, 9\}$ such that the sum of any two distinct elements of $S$ is unique. How many elements can $S$ have at most?
(2002, Canadian Mathematical Olympiad) | (When $S=\{1,2,3,5,8\}$, $S$ meets the requirements of the problem. If $T \subseteq\{1,2, \cdots, 9\},|T| \geqslant 6$, then since the sum of any two different numbers in $T$ is between 3 and 17, at most 15 different sum numbers can be formed. And choosing any two numbers from $T$, there are at least $\mathrm{C}_{6}^{2... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $x=\frac{1}{\sqrt{2}-1}, a$ be the fractional part of $x$, and $b$ be the fractional part of $-x$. Then $a^{3}+b^{3}+3 a b=$ $\qquad$ . | ニ、1. 1 .
Since $x=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$, and $2<\sqrt{2}+1<3$, therefore,
$$
\begin{array}{l}
a=x-2=\sqrt{2}-1 . \\
\text { Also }-x=-\sqrt{2}-1, \text { and }-3<-\sqrt{2}-1<-2 \text {, so, } \\
b=-x-(-3)=2-\sqrt{2} .
\end{array}
$$
Then $a+b=1$.
Thus $a^{3}+b^{3}+3 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+3 a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given a right trapezoid $A B C D$ with side lengths $A B=2, B C=C D=10, A D=6$, a circle is drawn through points $B$ and $D$, intersecting the extension of $B A$ at point $E$ and the extension of $C B$ at point $F$. Then the value of $B E-B F$ is $\qquad$ | 3.4 .
As shown in Figure 3, extend $C D$ to intersect $\odot O$ at point $G$. Let the midpoints of $B E$ and $D G$ be $M$ and $N$, respectively. It is easy to see that $A M = D N$. Since $B C = C D = 10$, by the secant theorem, it is easy to prove that
$$
\begin{aligned}
B F = & D G. \text{ Therefore, } \\
& B E - B F... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let positive integers $a, b, k$ satisfy $\frac{a^{2}+b^{2}}{a b-1}=k$. Prove: $k=5$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | For every $k$ that satisfies the above equation, let $a_{0}, b_{0}$ satisfy $\frac{a_{0}^{2}+b_{0}^{2}}{a_{0} b_{0}-1}=k$, and $a_{0}+b_{0}$ is the smallest pair. Without loss of generality, assume $a_{0} \geqslant b_{0}$.
(1) If $a_{0}=b_{0}$, then
$$
k=\frac{a_{0}^{2}+b_{0}^{2}}{a_{0} b_{0}-1}=2+\frac{2}{a_{0}^{2}-1}... | 5 | Number Theory | proof | Yes | Yes | cn_contest | false |
Three. (25 points) Let $a$ be a positive integer, the quadratic function $y=x^{2}+(a+17) x+38-a$, and the reciprocal function $y=\frac{56}{x}$. If the intersection points of the two functions are all integer points (points with both coordinates as integers), find the value of $a$.
---
The above text has been translat... | Three, by eliminating $y$ from the two equations, we get
$$
x^{2}+(a+17) x+38-a=\frac{56}{x} \text {, }
$$
which simplifies to $x^{3}+(a+17) x^{2}+(38-a) x-56=0$.
Factoring, we get
$$
(x-1)\left[x^{2}+(a+18) x+56\right]=0 \text {. }
$$
Clearly, $x_{1}=1$ is a root of equation (1), and $(1,56)$ is one of the intersect... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given that for all real numbers $x$, we have
$$
|x+1|+\sqrt{x-1} \geqslant m-|x-2|
$$
always holds. Then the maximum value that $m$ can take is $\qquad$ . | When $-1 \leqslant x \leqslant 2$, the minimum value of $|x+1|+|x-2|$ is
3. Since $\sqrt{x-1} \geqslant 0$, when $x=1$,
$$
|x+1|+\sqrt{x-1}+|x-2|
$$
the minimum value is 3, so, $3 \geqslant m$, hence the maximum value of $m$ is 3. | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $x$ and $y$ are real numbers, and $x^{2}+x y+y^{2}=3$. Let the maximum and minimum values of $x^{2}-x y+y^{2}$ be $m$ and $n$, respectively. Then the value of $m+n$ is $\qquad$ | $=, 1.10$.
Given the equation is symmetric about $x, y$, let $x=a+b, y=a-b$, then $3=x^{2}+xy+y^{2}=3a^{2}+b^{2}$.
Therefore, $3a^{2}=3-b^{2}$.
So, $0 \leqslant b^{2} \leqslant 3$.
$$
\begin{array}{l}
\text { Also, } x^{2}-xy+y^{2}=a^{2}+3b^{2}=\frac{1}{3}\left(3a^{2}+b^{2}\right)+\frac{8}{3}b^{2} \\
=\frac{1}{3} \time... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $a$ and $b$ be integers, and one root of the equation $x^{2} + a x + b = 0$ is $\sqrt{4 - 2 \sqrt{3}}$. Then the value of $a + b$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2 | 4.B.
Notice that $\sqrt{4-2 \sqrt{3}}=\sqrt{3}-1$.
According to the problem, we have $(\sqrt{3}-1)^{2}+a(\sqrt{3}-1)+b=0$, which simplifies to $(a-2) \sqrt{3}+4-a+b=0$.
Thus, $a-2=0$ and $4-a+b=0$.
Solving these, we get $a=2, b=-2$. Therefore, $a+b=0$. | 0 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. As shown in Figure 1, in the right trapezoid $A B C D$, $A B=B C=4$, $M$ is a point on the leg $B C$, and $\triangle A D M$ is an equilateral triangle. Then $S_{\triangle C D M}$ : $S_{\triangle A B M}=$ $\qquad$ . | 2.2.
As shown in Figure 6, draw $A E \perp$ $C D$ intersecting the extension of $C D$ at point $E$, then quadrilateral $A B C E$ is a square.
It is easy to prove
$\mathrm{Rt} \triangle A B M \cong \mathrm{Rt} \triangle A E D$.
Therefore, $B M=D E$.
Thus, $C M=C D$.
Let this value be $x$, then
$$
\begin{array}{l}
x^{2}... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. The quality requirements of a product are divided into four different levels from low to high, labeled as $1,2,3,4$. If the working hours remain unchanged, the workshop can produce 40 units of the lowest level (i.e., level 1) product per day, with a profit of 16 yuan per unit; if the level is increased by one, the p... | 3.3.
Let the profit obtained from producing products of the $x$-th grade in the workshop be $y$. According to the problem, we have
$$
\begin{array}{l}
y=[40-2(x-1)][16+(x-1)] \\
=-2 x^{2}+12 x+630=-2(x-3)^{2}+648 .
\end{array}
$$
Therefore, when $x=3$, the profit $y$ is maximized, at 648 yuan. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. The number of all different integer solutions to the equation $2 x^{2}+5 x y+2 y^{2}=2007$ is $\qquad$ groups.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note above is not part of the translation but is provided to clarify the instruction. The actual translation is above this note. | 4.4 .
Let's first assume $x \geqslant y$, the original equation can be transformed into $(2 x+y)(x+2 y)=2007$.
Since $2007=2007 \times 1=669 \times 3=223 \times 9$
$$
\begin{array}{l}
=(-1) \times 2007=(-3) \times(-669) \\
=(-9) \times(-223),
\end{array}
$$
Therefore, $\left\{\begin{array}{l}2 x+y=2007, \\ x+2 y=1 .\... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
9. For any 4 vertices $A, B, C, D$ of a cube that do not lie on the same plane, the number of cosine values of the dihedral angle $A-BC-D$ that are less than $\frac{1}{2}$ is
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 9.4 .
Among the cosines of the dihedral angles with the body diagonal as the edge,
$$
\begin{array}{l}
\cos \angle A_{1} O_{1} C_{1} \\
=\cos 120^{\circ}=-\frac{1}{2} .
\end{array}
$$
Among the cosines of the dihedral angles with the edge as the edge, only 0 is less than $\frac{1}{2}$.
Among the cosines of the dihedr... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. There are 5 rooms $A, B, C, D, E$ arranged in a circular pattern, with the number of people living in them being $17, 9, 14, 16, 4$ respectively. Now, adjustments are to be made so that the number of people in each room is the same, and it is stipulated that people can only move to the adjacent left or right room. H... | (Tip: Let the number of people moved from room $A$ to room $B$ be $x_{B}$, and so on, then we have
$$
\begin{array}{l}
9+x_{B}-x_{C}=14+x_{C}-x_{D}=16+x_{D}-x_{E} \\
=4+x_{E}-x_{A}=17+x_{A}-x_{B} \\
=\frac{1}{5}(17+9+14+16+4)=12 .
\end{array}
$$
This is transformed into finding
$$
\begin{array}{l}
y=\left|x_{B}-5\righ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
2. The number of triangles with integer side lengths and a perimeter of 20 is $\qquad$ . | 2.8.
Let the three sides of a triangle be $a, b, c$, and $a \geqslant b \geqslant c$, $a+b+c=20$, then $a \geqslant 7$.
Also, from $b+c>a$, we get $2 a<a+b+c=20 \Rightarrow a<10$. Therefore, $7 \leqslant a \leqslant 9$. We can list
$$
\begin{array}{l}
(a, b, c)=(9,9,2),(9,8,3),(9,7,4),(9,6,5), \\
(8,8,4),(8,7,5),(8,6,... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange the four numbers $1, 2, 3, 4$ to form a four-digit number, such that this number is a multiple of 11. Then the number of such four-digit numbers is $\qquad$.
| 4.8.
Since $1+4=2+3$, we can place 1 and 4 in the even positions, and 2 and 3 in the odd positions, which gives us four arrangements; placing 2 and 3 in the even positions, and 1 and 4 in the odd positions also gives us four arrangements. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the sequence
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{2}=\frac{1}{3}+\frac{2}{3}, \cdots, \\
a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1}, \cdots .
\end{array}
$$
Let $S_{n}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}}$. Then the integer closest to $S_{2006}$ is $(\qua... | 3.C.
Since $a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1}=\frac{n}{2}$, we have
$$
\begin{array}{l}
S_{n}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}} \\
=\frac{4}{1 \times 2}+\frac{4}{2 \times 3}+\cdots+\frac{4}{n(n+1)} \\
=4\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\fr... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and the graph of $y=f(x)$ is symmetric about the line $x=\frac{1}{2}$. Then
$$
f(1)+f(2)+\cdots+f(2006)=
$$
$\qquad$ | 2.0.
Given that the graph of $y=f(x)$ is symmetric about the line $x=\frac{1}{2}$, we know that $f(x)=f(1-x)$.
Also, since $f(x)$ is an odd function defined on $\mathbf{R}$, it follows that $f(1-x)=-f(x-1)$.
Therefore, $f(x)+f(x-1)=0$.
Thus, $f(1)+f(2)+\cdots+f(2006)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $\triangle A B C$ with three sides $A B=\sqrt{34}, B C$ $=5 \sqrt{10}, C A=2 \sqrt{26}$. Then the area of $\triangle A B C$ is $\qquad$ | 5.10.
As shown in Figure 4, with $BC$ as the hypotenuse, construct a right triangle $\triangle BCD$ on one side of $\triangle ABC$ such that $\angle BDC = 90^\circ$, $BD = 5$, and $CD = 15$.
Then construct a rectangle $DEA'F$ such that $DE = 2$ and $DF = 5$. Thus, $BE = 3$ and $CF = 10$. At this point,
$$
\begin{arra... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. The area of the figure enclosed by the two curves $y=x^{3}$ and $x=y^{3}$ on the Cartesian plane is $\qquad$ . | 5.1.
Since the two curves are symmetric with respect to the origin, it is only necessary to calculate the area $A$ of the figure enclosed by the two curves in the first quadrant.
When $x>1$, $x^{3}>\sqrt[3]{x}$;
When $0<x<1$, $x^{3}<\sqrt[3]{x}$.
Therefore, the two curves have a unique intersection point $(1,1)$ in th... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
8.3. On the side $BC$ of the rhombus $ABCD$, take a point $M$. Draw perpendiculars from $M$ to the diagonals $BD$ and $AC$, intersecting the line $AD$ at points $P$ and $Q$. If the lines $PB$ and $QC$ intersect $AM$ at the same point, find the ratio $\frac{BM}{MC}$. | 8.3. As shown in Figure 1, let the intersection of lines $P B$, $Q C$, and $A M$ be $R$.
From the given conditions, $P M \parallel A C$ and $M Q \parallel B D$, thus quadrilaterals $P M C A$ and $Q M B D$ are both parallelograms. Therefore,
$$
\begin{aligned}
M C &= P A, B M = D Q, \text{ and } \\
P Q &= P A + A D + D... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8.7. For a natural number $n(n>3)$, we use “$n$ ?" to denote the product of all prime numbers less than $n$. Solve the equation
$$
n ?=2 n+16
$$ | 8.7. The given equation is
$$
n ?-32=2(n-8) \text {. }
$$
Since $n$ ? cannot be divisible by 4, it follows from equation (1) that $n-8$ is odd.
Assume $n>9$, then $n-8$ has an odd prime factor $p$.
Also, $p2 \times 9+16$.
When $n=7$, it is clearly a root of the equation:
However, when $n=5$, we have $n ?=6<16$.
Theref... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. In the arithmetic sequence $\left\{a_{n}\right\}$, if $a_{2}+a_{4}+a_{6}+$ $a_{8}+a_{10}=80$, then $a_{7}-\frac{1}{2} a_{8}=(\quad)$.
(A) 4
(B) 6
(C) 8
(D) 10 | 4.C.
Since $a_{2}+a_{4}+a_{6}+a_{8}+a_{10}=5 a_{6}=80$, therefore, $a_{6}=16$.
Thus, $a_{7}-\frac{1}{2} a_{8}=a_{6}+d-\frac{1}{2}\left(a_{6}+2 d\right)=\frac{1}{2} a_{6}=8$. | 8 | Algebra | MCQ | Yes | Yes | cn_contest | false |
11. If the three medians $A D$, $B E$, $C F$ of $\triangle A B C$ intersect at point $M$, then $M A+M B+M C=$ | Ni, 11.0.
Diagram (omitted). Let the midpoint of $A B$ be $D$. By the parallelogram rule, we have
$$
\begin{array}{l}
M A+M B=2 M D=-M C \text {. } \\
\text { Therefore, } M A+M B+M C=0 \text {. } \\
\end{array}
$$ | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. In the sequence $\left\{a_{n}\right\}$, $a_{1}=2, a_{n}+a_{n+1}=1$ $\left(n \in \mathbf{N}_{+}\right)$, let $S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then
$$
\mathrm{S}_{2007}-2 \mathrm{~S}_{2006}+\mathrm{S}_{2000}=
$$
$\qquad$ | 13.3.
When $n$ is even, we have
$$
a_{1}+a_{2}=a_{3}+a_{4}=\cdots=a_{n-1}+a_{n}=1 .
$$
Thus, $S_{n}=\frac{n}{2}$.
When $n$ is odd, we have
$$
a_{1}=2, a_{2}+a_{3}=a_{4}+a_{5}=\cdots=a_{n-1}+a_{n}=1 \text {. }
$$
Thus, $S_{n}=2+\frac{n-1}{2}=\frac{n+3}{2}$.
Therefore, $S_{2007}-2 S_{2006}+S_{2005}$
$$
=1005-2 \times ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. (12 points) In $\triangle A B C$, it is known that $\sin A \cdot \cos ^{2} \frac{C}{2}+\sin C \cdot \cos ^{2} \frac{A}{2}=\frac{3}{2} \sin B$. Find the value of $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}$. | Three, 15. From the given,
$\sin A \cdot \frac{1+\cos C}{2}+\sin C \cdot \frac{1+\cos A}{2}=\frac{3}{2} \sin B$.
Then, $\sin A+\sin C+\sin A \cdot \cos C+\cos A \cdot \sin C$ $=3 \sin B$.
Thus, $\sin A+\sin C+\sin (A+C)=3 \sin B$, which means $\sin A+\sin C=2 \sin B$.
Therefore, $2 \sin \frac{A+C}{2} \cdot \cos \frac{A... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For example, $\triangle ABC$ is an equilateral triangle with side length 8, $M$ is a point on side $AB$, $MP \perp AC$ at point $P$, $MQ \perp BC$ at point $Q$, and connect $PQ$.
(1) Find the minimum length of $PQ$;
(2) Find the maximum area of $\triangle CPQ$. | Solution: (1) Let the height of $\triangle ABC$ be $h$, then $h=4 \sqrt{3}$.
From $S_{\triangle C M}+S_{\triangle B C M}=S_{\triangle B B C}$, we get
$$
M P+M O=h=4 \sqrt{3} \text {. }
$$
As shown in Figure 11, draw perpendiculars from points $P$ and $Q$ to side $AB$, with the feet of the perpendiculars being $P_{1}$ ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2-1 Given that $x, y, z$ are positive numbers, and $xyz(x+y+z)=1$.
Find the minimum value of $(x+y)(y+z)$.
(1989, All-Soviet Union Mathematical Competition) | Solution 1: $(x+y)(y+z)$
$$
\begin{array}{l}
=x z+y(x+y+z) \\
\geqslant 2 \sqrt{x y z(x+y+z)}=2 .
\end{array}
$$
When $x=z=1, y=\sqrt{2}-1$, $y(x+y+z) = xz$, $(x+y)(y+z)$ takes the minimum value 2.
Solution 2: As shown in Figure 1, construct
$\triangle ABC$, with side lengths
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2-3 Let quadrilateral $A B C D$ be a rectangle with an area of 2, $P$ a point on side $C D$, and $Q$ the point where the incircle of $\triangle P A B$ touches side $A B$. The product $P A \cdot P B$ varies with the changes in rectangle $A B C D$ and point $P$. When $P A \cdot P B$ is minimized,
(1) Prove: $A B ... | Solution 1: (1) From the given information,
$S_{\triangle P A B}=\frac{1}{2} S_{\text {rectangle } A B C D}=1$.
Thus, $P A \cdot P B=\frac{2 S_{\triangle P A B}}{\sin \angle A P B} \geqslant 2$.
Equality holds if and only if $\angle A P B=90^{\circ}$. At this time, point $P$ lies on the circle with $A B$ as its diamete... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $p$ be a positive odd number. Then the remainder of $p^{2}$ divided by 8 is $\qquad$ . | Because $p$ is an odd positive integer, let $p=2k-1\left(k \in \mathbf{N}_{+}\right)$, so
$$
p^{2}=(2k-1)^{2}=4k^{2}-4k+1=4(k-1)k+1 \text{. }
$$
Since $(k-1)k$ is even, therefore, $p^{2}$ leaves a remainder of 1 when divided by 8. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given an isosceles triangle $\triangle A B C$ with side lengths $a$, $b$, and $c$ all being integers, and satisfying $a+b c+b+c a=24$. Then the number of such triangles is $\qquad$. | 4.3.
$$
\begin{array}{l}
\text { Since } a+b c+b+a a \\
=(a+b)(c+1)=24=12 \times 2=8 \times 3=6 \times 4, \text { and }
\end{array}
$$
$\triangle A B C$ is an isosceles triangle, so the length of the base can only be $c$.
Thus, there are 3 triangles that satisfy the conditions:
$$
c=1, a=b=6 ; c=2, a=b=4 ; c=3, a=b=3 \... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) Given that the graph of the linear function $y=a x+b$ passes through the points $A(\sqrt{3}, \sqrt{3}+2), B(-1, \sqrt{3})$, and $C(c, 2-c)$. Find the value of $a-b+c$.
---
The above text has been translated into English, preserving the original text's line breaks and format. | Three, from $\left\{\begin{array}{l}\sqrt{3}+2=\sqrt{3} a+b, \\ \sqrt{3}=-a+b\end{array} \Rightarrow\left\{\begin{array}{l}a=\sqrt{3}-1, \\ b=2 \sqrt{3}-1 .\end{array}\right.\right.$
Therefore, $2-c=a c+b=(\sqrt{3}-1) c+(2 \sqrt{3}-1)$.
Solving for $c$ gives $c=\sqrt{3}-2$.
Thus, $a-b+c=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (50 points)
As shown in Figure 3, in a $7 \times 8$
rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected." Now, some of the 56 chess pieces are to be removed ... | Second, at least 11 chess pieces must be taken out to possibly meet the requirement.
The reason is as follows:
If a square is in the $i$th row and the $j$th column, then this square is denoted as $(i, j)$.
Step 1 Proof: If any 10 chess pieces are taken, then the remaining chess pieces must have a five-in-a-row, i.e., ... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Let $x, y, a, m, n$ be positive integers, and $x+y=a^{m}, x^{2}+y^{2}=a^{n}$. Find how many digits $a^{30}$ has.
保留源文本的换行和格式，直接输出翻译结果。 | Three, from the known we get
$$
a^{2 m}=x^{2}+y^{2}+2 x y=a^{n}+2 x y \text {. }
$$
From the problem and equation (1), we know that $a^{2 m}>a^{n}$. Therefore, $2 m>n$.
Dividing both sides of equation (1) by $a^{n}$, we get
$$
a^{2 m-n}=1+\frac{2 x y}{a^{n}} \text {. }
$$
Since the left side of equation (2) is a posi... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) As shown in Figure 5, there are 12 equally spaced points on the circumference of a clock face, marked with the numbers $1, 2, \cdots, 12$ in sequence. Please use these equally spaced points as vertices to form 4 triangles (dividing these 12 equally spaced points into 4 groups) such that the following... | Three, let 4 triangles be $\left(a_{i}, b_{i}, c_{i}\right), i=1,2, 3,4$, where $a_{i}=b_{i}+c_{i}, b_{i}>c_{i}$.
Let $a_{1}=27$.
Thus, $10 \leqslant a_{3} \leqslant 11$.
If $a_{3}=10$, then from $a_{1}+a_{2}=17, a_{2}a_{1}+a_{2}=17$, we get $a_{2}=9, a_{1}=8$, that is
$$
\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(8,9,10... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. In a $3 \times 3$ grid, the numbers $1,2,3,4,5,6,7,8,9$ are filled in, with one number per cell. Now, the cells containing the maximum number in each row are colored red, and the cells containing the minimum number in each row are colored green. Let $M$ be the smallest number in the red cells, and $m$ be the largest... | 2.8
Obviously, $3 \leqslant m, M \leqslant 7$, and $m \neq M$. Therefore,
$$
M-m \in\{-4,-3,-2,-1,1,2,3,4\} \text {. }
$$
As shown in Figure 7, all 8 values can be obtained
\begin{tabular}{|l|l|l|}
\hline 7 & 9 & 8 \\
\hline 6 & 5 & 4 \\
\hline 3 & 2 & 1 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 6 & 9 &... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of positive integers $m$ that make $m^{2}+m+7$ a perfect square is $\qquad$ . | 3.2.
It has been verified that when $m=1$, $m^{2}+m+7=9$ is a perfect square;
when $m=2,3,4,5$, $m^{2}+m+7$ are not perfect squares;
when $m=6$, $m^{2}+m+7=49$ is a perfect square.
When $m>6$, $m^{2}+m+7$ are not perfect squares.
Therefore, there are only 2 positive integers $m$ that meet the condition. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (20 points) For any real numbers $x, y$, we have
$$
|x-2|+|x-4| \geqslant m\left(-y^{2}+2 y\right)
$$
Determine the maximum value of the real number $m$. | Three, by the geometric meaning of absolute value, $|x-2|+|x-4|$ has a minimum value of 2 when $x \in [2,4]$.
And $-y^{2}+2y=-(y-1)^{2}+1$ has a maximum value of 1 when $y=1$.
From the condition, $2 \geqslant m \times 1$, then $m \leqslant 2$.
Therefore, the maximum value of $m$ is 2. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Five. (25 points) Given the system of equations in $x$ and $y$
$$
\left\{\begin{array}{l}
x^{2}-y^{2}=p, \\
3 x y+p(x-y)=p^{2}
\end{array}\right.
$$
has integer solutions $(x, y)$. Find the prime number $p$ that satisfies the condition. | Five, from $p=x^{2}-y^{2}=(x-y)(x+y)$ and $p$ being a prime number, we have
$\left\{\begin{array}{l}x+y=p, \\ x-y=1\end{array}\right.$ or $\quad\left\{\begin{array}{l}x+y=-p, \\ x-y=-1\end{array}\right.$
or $\left\{\begin{array}{l}x+y=1, \\ x-y=p\end{array}\right.$ or $\left\{\begin{array}{l}x+y=-1, \\ x-y=-p .\end{ar... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Add the same integer $a(a>0)$ to the numerator and denominator of $\frac{2008}{3}$, making the fraction an integer. Then the integer $a$ added has $\qquad$ solutions. | $=1.3$.
From the problem, we know that $\frac{2008+a}{3+a}$ should be an integer, which means $\frac{2008+a}{3+a}=\frac{2005}{3+a}+1$ should be an integer. Therefore, $(3+a) \mid 2005=5 \times 401$.
Since 2005 has 4 divisors, at this point, $a$ can take the values $2002$, $398$, $2$, and one divisor corresponds to a va... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $|a|>1$, simplify
$$
\left(a+\sqrt{a^{2}-1}\right)^{4}+2\left(1-2 a^{2}\right)\left(a+\sqrt{a^{2}-1}\right)^{2}+3
$$
the result is $\qquad$ . | Let $x_{0}=a+\sqrt{a^{2}-1}$. Clearly, $x_{0}$ is a root of the equation $x^{2}-2 a x+1=0$, and thus it is also a root of the equation
$$
\left(x^{2}-2 a x+1\right)\left(x^{2}+2 a x+1\right)=0
$$
Then $\left(x_{0}^{2}+1\right)^{2}-4 a^{2} x_{0}^{2}=0$, which means
$$
x_{0}^{4}+2\left(1-2 a^{2}\right) x_{0}^{2}+1=0 \te... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given
$$
\frac{y+z-x}{x+y+z}=\frac{z+x-y}{y+z-x}=\frac{x+y-z}{z+x-y}=p \text {. }
$$
Then $p^{3}+p^{2}+p=$ $\qquad$ . | Notice
$$
\begin{array}{l}
p^{2}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x}=\frac{z+x-y}{x+y+z}, \\
p^{3}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x} \cdot \frac{x+y-z}{z+x-y} \\
=\frac{x+y-z}{x+y+z}, \\
p^{3}+p^{2}+p \\
=\frac{x+y-z}{x+y+z}+\frac{z+x-y}{x+y+z}+\frac{y+z-x}{x+y+z} \\
=\frac{x+y+z}{x+y+z}=1 .
\end{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that the two roots of the equation $x^{2}+x-1=0$ are $\alpha, \beta$. Then the value of $\frac{\alpha^{3}}{\beta}+\frac{\beta^{3}}{\alpha}$ is $\qquad$ | Let $A=\frac{\alpha^{3}}{\beta}+\frac{\beta^{3}}{\alpha}, B=\frac{\alpha^{3}}{\alpha}+\frac{\beta^{3}}{\beta}=\alpha^{2}+\beta^{2}$.
From the given information,
$$
\begin{array}{l}
\alpha+\beta=-1, \alpha \beta=-1 . \\
\text { Hence } B=(\alpha+\beta)^{2}-2 \alpha \beta=1+2=3 . \\
\text { Also, } \alpha^{3}+\beta^{3}=(... | -7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.