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 |
|---|---|---|---|---|---|---|---|---|
3. The first 24 digits of $\pi$ are
3. 14159265358979323846264 .
Let $a_{1}, a_{2}, \cdots, a_{24}$ be any permutation of these 24 digits. Then
$$
\begin{array}{l}
\left(a_{1}-a_{2}\right)\left(a_{3}-a_{4}\right) \cdots\left(a_{23}-a_{24}\right) \\
\equiv \quad(\bmod 2) .
\end{array}
$$
$(\bmod 2)$. | 3. 0 .
In the first 24 digits of $\pi$, there are 13 odd numbers. When these 13 odd numbers are placed into 12 parentheses, by the pigeonhole principle, there must be two in the same parenthesis, making the difference in this parenthesis even. Thus, the product $\left(a_{1}-a_{2}\right)\left(a_{3}-a_{4}\right) \cdots\... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $a, b, c \in \mathbf{R}$, and $a+b+c=3$. Then the minimum value of $3^{a} a+3^{b} b+3^{c} c$ is $\qquad$ | 3. 9 .
Since the function $y=3^{x}$ is an increasing function on $(-\infty,+\infty)$, by the property of increasing functions, for any $x_{1}$, $x_{2}$, we have $\left(x_{1}-x_{2}\right)\left(3^{x_{1}}-3^{x_{2}}\right) \geqslant 0$, so,
$$
(a-1)\left(3^{a}-3^{1}\right) \geqslant 0,
$$
which means $3^{a} a-3^{\circ} \... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 12 The number of positive integer solutions to the system of equations $\left\{\begin{array}{l}x y+y z=63, \\ x z+y z=23\end{array}\right.$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | From the second equation, we get $z(x+y)=23$, and since 23 is a prime number, we can obtain $\left\{\begin{array}{l}z=1, \\ x+y=23\end{array}\right.$ or $\left\{\begin{array}{l}z=23, \\ x+y=1 \text {. }\end{array}\right.$ According to the problem, $x+y=1$ is not valid, so we discard it. Substituting $z=1$ into the give... | 2 | Algebra | MCQ | Yes | Yes | cn_contest | false |
8. Let $P_{n}(k)$ denote the number of permutations of $\{1,2, \cdots, n\}$ with $k$ fixed points. Let $a_{t}=\sum_{k=0}^{n} k^{t} P_{n}(k)$. Then
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1}-2 a_{0} \\
=
\end{array}
$$ | 8. 0 .
Notice
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1} \\
=\sum_{k=0}^{n}\left(k^{5}-10 k^{4}+35 k^{3}-50 k^{2}+25 k\right) P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] . \\
\frac{n!}{(n-k)!k!} P_{n-k}(0) \\
=\sum_{k=0}^{n}\left[\frac{n!... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
265 It is known that for all positive integers $n$,
$$
\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right) \geqslant \frac{k}{2} \sqrt[3]{19 n+8}
$$
always holds. Try to find the maximum value of $k$. | Let $T_{n}=\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right)$. Then
$$
\frac{T_{n+1}}{T_{n}}=1+\frac{1}{3 n+2}=\frac{3 n+3}{3 n+2} \text {. }
$$
Given $T_{n} \geqslant \frac{k}{2} \sqrt[3]{19 n+8}$, we have $k \leqslant \frac{2 T_{n}}{\sqrt[3]{19 n+8}}$.
Let $f(n)=\frac{2 T_{n}}{\sqrt[3]{19 n+8}}$. Then
$$
\begin{array}{l... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Given real numbers $a, b \neq 0$, let
$$
x=\frac{a}{|a|}+\frac{b}{|b|}+\frac{a b}{|a b|} \text {. }
$$
Then the sum of the maximum and minimum values of $x$ is $\qquad$ [1] | Question 1 Original Solution ${ }^{[1]}$ According to the number of negative numbers in $a$ and $b$, there are three cases:
(1) If $a$ and $b$ are both positive, then $x=3$;
(2) If $a$ and $b$ are one positive and one negative, then $x=-1$;
(3) If $a$ and $b$ are both negative, then $x=-1$.
In summary, the maximum valu... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $a, b, c$ satisfy
$$
\frac{a b}{a+b}=\frac{1}{3}, \frac{b c}{b+c}=\frac{1}{4}, \frac{c a}{c+a}=\frac{1}{5} \text {. }
$$
then $a b+b c+c a=$ $\qquad$ | $$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b}=3, \frac{1}{b}+\frac{1}{c}=4, \frac{1}{c}+\frac{1}{a}=5 \\
\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6 \\
\Rightarrow \frac{1}{c}=3, \frac{1}{a}=2, \frac{1}{b}=1 \\
\Rightarrow a=\frac{1}{2}, b=1, c=\frac{1}{3} \\
\Rightarrow a b+b c+c a=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$ are six different positive integers, taking values from $1, 2, 3, 4, 5, 6$. Let
$$
\begin{aligned}
S= & \left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\left|x_{3}-x_{4}\right|+ \\
& \left|x_{4}-x_{5}\right|+\left|x_{5}-x_{6}\right|+\left|x_{6}-x_{1}\right| .
\end{alig... | 4. 10 .
Since equation (1) is a cyclic expression about $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$, we can assume $x_{1}=6, x_{j}=1(j \neq 1)$. Then
$$
\begin{array}{l}
S \geqslant\left|\left(6-x_{2}\right)+\left(x_{2}-x_{3}\right)+\cdots+\left(x_{j-1}-1\right)\right|+ \\
\quad\left|\left(x_{j+1}-1\right)+\left(x_{j+... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the value of $\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}+15^{1998}}{7^{1998}+35^{1998}}}$. | $\begin{array}{l}\text { Solve the original expression }=\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}\left(1+5^{1998}\right)}{7^{1998}\left(1+5^{1998}\right)}} \\ =\left(\frac{7}{3}\right)^{999} \times\left(\frac{3}{7}\right)^{999}=\left(\frac{7}{3} \times \frac{3}{7}\right)^{999}=1 .\end{array}$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Person A writes down the positive integers $1, 2, \cdots$, 2009 on the blackboard, then turns away from the blackboard, and asks Person B to erase some of these numbers and then add the remainder of the sum of the erased numbers when divided by 7. After several such operations, only two numbers remain on the blackbo... | 3.5.
Since $1+2+\cdots+2009 \equiv 0(\bmod 7)$, therefore, the single digit $a$ satisfies $100+a \equiv 0(\bmod 7)$. Hence $a \equiv 5(\bmod 7)$. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. The number of solutions to the equation $\pi^{x-1} x^{2}+\pi^{x^{2}} x-\pi^{x^{2}}=x^{2}+x-1$ is $\qquad$ ( $\pi$ is the ratio of a circle's circumference to its diameter). | 4. 2 .
The original equation is transformed into
$$
x^{2}\left(\pi^{x-1}-1\right)+(x-1)\left(\pi^{x^{2}}-1\right)=0 \text {. }
$$
When $x \neq 0,1$, $x^{2}\left(\pi^{x-1}-1\right)$ and $(x-1)\left(\pi^{x^{2}}-1\right)$ have the same sign.
Therefore, $x=0,1$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (15 points) Given a sequence of positive numbers $\left\{a_{n}\right\}(n \geqslant 0)$ that satisfies $a_{n}=\frac{a_{n-1}}{m a_{n-2}}(n=2,3, \cdots, m$ is a real parameter $)$. If $a_{2009}=\frac{a_{0}}{a_{1}}$, find the value of $m$. | 10. Since $a_{2}=\frac{a_{1}}{m a_{0}}, a_{3}=\frac{\frac{a_{1}}{m a_{0}}}{m a_{1}}=\frac{1}{m^{2} a_{0}}$,
$$
\begin{array}{l}
a_{4}=\frac{\frac{1}{m^{2} a_{0}}}{m \frac{a_{1}}{m a_{0}}}=\frac{1}{m^{2} a_{1}}, a_{5}=\frac{\frac{1}{m^{2} a_{1}}}{m \frac{1}{m^{2} a_{0}}}=\frac{a_{0}}{m a_{1}}, \\
a_{6}=\frac{\frac{a_{0}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $x=2 \frac{1}{5}, y=-\frac{5}{11}, z=-2 \frac{1}{5}$. Then $x^{2}+x z+2 y z+3 x+3 z+4 x y+5=$ | Solve: From the given, we have
$$
x y=-1, x+z=0, y z=1 \text {. }
$$
Therefore, the expression to be found is
$$
\begin{array}{l}
=x(x+z)+2 y z+3(x+z)+4 x y+5 \\
=2 \frac{1}{5} \times 0+2 \times 1+3 \times 0+4 \times(-1)+5=3 .
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (15 points) Divide the sides of an equilateral $\triangle ABC$ with side length 3 into three equal parts, and draw lines parallel to the other two sides through each division point. The 10 points where the sides of $\triangle ABC$ and these parallel lines intersect are called grid points. If $n$ grid points are cho... | 11. $n_{\min }=5$.
Let the two equal division points on side $AB$ from point $A$ to $B$ be $D$ and $E$, the two equal division points on side $BC$ from point $B$ to $C$ be $F$ and $G$, and the two equal division points on side $CA$ from point $C$ to $A$ be $H$ and $I$, with the central grid point being $K$.
If the mi... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Let a tangent line of the circle $x^{2}+y^{2}=1$ intersect the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Then the minimum value of $|AB|$ is $\qquad$ . | 3. 2 .
By symmetry, without loss of generality, assume the point of tangency is
$$
P(\cos \theta, \sin \theta)\left(0<\theta<\frac{\pi}{2}\right) \text {. }
$$
Then $|P A|=\tan \theta,|P B|=\cot \theta$.
Thus, $|A B|=|P A|+|P B|=\tan \theta+\cot \theta \geqslant 2$.
Equality holds if and only if $\tan \theta=\cot \th... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $f(x)$ be an odd function defined on $\mathbf{R}$, $f(1)=2$, when $x>0$, $f(x)$ is an increasing function, and for any $x, y \in \mathbf{R}$, we have $f(x+y)=f(x)+f(y)$. Then the maximum value of the function $f(x)$ on the interval $[-3,-2]$ is | 5. -4 .
Since $f(x)$ is an odd function and is increasing on $(0,+\infty)$, therefore, $f(x)$ is also increasing on $(-\infty, 0)$. Thus, $f(-3) \leqslant f(x) \leqslant f(-2)$.
$$
\begin{array}{l}
\text { Also, } f(2)=f(1)+f(1)=4, \text { then } \\
f(-2)=-f(2)=-4 .
\end{array}
$$
Therefore, the maximum value of the ... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. As shown in Figure $2, A B$ is the diameter of the semicircle $\odot O$, and $C, D$ are two moving points on the semicircle, with $C D / / A B$. If the radius of the semicircle $\odot O$ is 1, then the maximum value of the perimeter of trapezoid $A B C D$ is $\qquad$ . | 7.5.
In Figure 2, connect $A C$, and draw $C H \perp A B$ at point $H$. Let $\angle A B C=\theta\left(0<\theta<\frac{\pi}{2}\right)$. Then $A D=B C=A B \cos \theta=2 \cos \theta$.
Thus, $B H=B C \cos \theta=2 \cos ^{2} \theta$.
Therefore, $C D=A B-2 B H=2-4 \cos ^{2} \theta$.
Hence, the perimeter of trapezoid $A B C D... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 3, in $\triangle A B C$, $A B=3, A C=5$. If $O$ is the circumcenter of $\triangle A B C$, then the value of $\overrightarrow{A O} \cdot \overrightarrow{B C}$ is | 8. 8 .
Let $D$ be the midpoint of side $B C$, and connect $O D, A D$. Then $O D \perp B C$. Therefore,
$$
\begin{array}{l}
\overrightarrow{A O} \cdot \overrightarrow{B C}=(\overrightarrow{A D}+\overrightarrow{D O}) \cdot \overrightarrow{B C} \\
=\overrightarrow{A D} \cdot \overrightarrow{B C}+\overrightarrow{D O} \cdo... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Calculate:
$$
\frac{2009^{2}-2008^{2}}{19492009^{2}-19492008 \times 19492010+2 \times 2008} .
$$ | Let $19492009=a, 2008=b$. Then the original expression $=\frac{(b+1)^{2}-b^{2}}{a^{2}-(a-1)(a+1)+2 b}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given real numbers $x, y, z$ satisfy $x-y=8$, $xy+z^{2}=-16$. Then $x+y+z=$ $\qquad$ . | Let $x=a+b, y=a-b$. Then $x-y=2b \Rightarrow b=4$.
Since $xy+z^{2}=-16$, we have $(a+b)(a-b)+z^{2}+16=0$. Also, $-b^{2}=-16$, then $a^{2}+z^{2}=0$. By the property of non-negative numbers, we know $a=0, z=0$. Therefore, $x=4, y=-4$. Hence, $x+y+z=4-4+0=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In the Cartesian coordinate system, a point whose both horizontal and vertical coordinates are integers is called an "integer point". The number of integer points on the graph of the function $y=\frac{x+12}{2 x-1}$ is $\qquad$ | 2. 6 .
Notice that $2 y=1+\frac{25}{2 x-1}$.
Since $y$ is an integer, $\frac{25}{2 x-1}$ must also be an integer. Therefore, $2 x-1= \pm 1, \pm 5, \pm 25$.
Solving for $x$ gives $x=-12,-2,0,1,3,13$.
Correspondingly, $y=0,-2,-12,13,3,1$.
Thus, there are 6 integer points. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $a$ and $b$ are integers. Then the number of ordered pairs $(a, b)$ that satisfy $a + b + ab = 2008$ is $\qquad$ groups. | 4. 12 .
$$
\begin{array}{r}
\text { Since }(a+1)(b+1) \\
=2009=41 \times 7^{2} \text {, and given }
\end{array}
$$
that $a, b$ take integer values, so, $a+1, b+1$ have 12 factor combinations, i.e., $1 \times 2009, 7 \times 287, 41 \times 49$ and their negatives as well as swaps. | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let convex quadrilateral $ABCD$ satisfy $AB=AD=1$, $\angle A=160^{\circ}, \angle C=100^{\circ}$. Then the range of the length of diagonal $AC$ is $\qquad$ . | 2. $\{1\}$.
Since $\angle C=180^{\circ}-\frac{\angle A}{2}$, therefore, points $B$, $C$, and $D$ lie on a circle with $A$ as the center and a radius of 1. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Simplify:
$$
\frac{(\sqrt{x}-\sqrt{y})^{3}+2 x \sqrt{x}+y \sqrt{y}}{x \sqrt{x}+y \sqrt{y}}+\frac{3 \sqrt{x y}-3 y}{x-y} .
$$ | Let $\sqrt{x}=a+b, \sqrt{y}=a-b$. Then
$$
\sqrt{x y}=a^{2}-b^{2}, \sqrt{x}+\sqrt{y}=2 a, \sqrt{x}-\sqrt{y}=2 b \text {. }
$$
Therefore, the original expression is
$$
\begin{aligned}
= & \frac{(2 b)^{3}+2(a+b)^{3}+(a-b)^{3}}{(a+b)^{3}+(a-b)^{3}}+ \\
& \frac{3\left(a^{2}-b^{2}\right)-3(a-b)^{2}}{4 a b} \\
= & \frac{3\le... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. (22 points) Suppose a set of planar points $S$ has the following properties:
(1) No three points are collinear;
(2) The distance between any two points is unique.
For two points $A$ and $B$ in $S$, if there exists a point $C \in S$ such that $|A C|<|A B|<|B C|$, then $A B$ is called a "middle edge" of $S$. For thre... | 4. Color all middle edges of $S$ red, and color other edges blue.
When $n \geqslant 6$, according to Ramsey's theorem, there must exist a monochromatic triangle, which must have a middle edge, and it must be a middle edge triangle.
The following set of five points with properties (1) and (2) does not have a middle ed... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. The line $l: x+y=t$ intersects the circle $\odot O: x^{2}+y^{2}=20$ at points $A$ and $B$, and $S_{\triangle O A B}$ is an integer. Then the number of all positive integer values of $t$ that satisfy the condition is $\qquad$ . | 7.2.
Let $\angle A O B=2 \alpha, O C \perp A B$, with the foot of the perpendicular being $C$. Then $O C=\sqrt{20} \cos \alpha=\frac{t}{\sqrt{1+1}}=\frac{t}{\sqrt{2}}$ $\Rightarrow \cos \alpha=\frac{t}{2 \sqrt{10}}$.
Also, $S_{\triangle O A B}=10 \sin 2 \alpha \leqslant 10$, so $\sin 2 \alpha=\frac{k}{10}(k \in\{1,2, ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. If the polynomial $f(x)=x^{3}-6 x^{2}+a x+a$ has three roots $x_{1}, x_{2}, x_{3}$ that satisfy
$$
\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0 \text {, }
$$
then the value of the real number $a$ is .. $\qquad$ | 6. -9 .
From the problem, we have
$$
\begin{array}{l}
g(t)=f(t+3) \\
=(t+3)^{3}-6(t+3)^{2}+a(t+3)+a \\
=t^{3}+3 t^{2}+(a-9) t+4 a-27
\end{array}
$$
The three roots \( t_{1}, t_{2}, t_{3} \) satisfy \( t_{1}^{3}+t_{2}^{3}+t_{3}^{3}=0 \).
By the relationship between roots and coefficients, we get
$$
\begin{array}{l}
\l... | -9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given an ellipse centered at the origin, with foci on the $x$-axis, the length of the major axis is twice the length of the minor axis, and it passes through the point $M(2,1)$. A line $l$ parallel to $OM$ has a $y$-intercept of $m (m<0)$, and intersects the ellipse at two distinct points $A$ and $B$. F... | 10. Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
Then $\left\{\begin{array}{l}a=2 b, \\ \frac{4}{a^{2}}+\frac{1}{b^{2}}=1\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=8, \\ b^{2}=2 \text {. }\end{array}\right.\right.$
Therefore, the equation of the ellipse is $\frac{x^{2}... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}, x y z>0$,
and
$$
\begin{array}{l}
\sqrt[3]{\frac{2010}{x^{2}}+\frac{2011}{y^{2}}+\frac{2012}{z^{2}}} \\
=\sqrt[3]{2010}+\sqrt[3]{2011}+\sqrt[3]{2012} .
\end{array}
$$
Find the value of $x+y+z$. | Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}=k$, obviously $k \neq 0$.
Then $2010=x^{3} k, 2011=y^{3} k, 2012=z^{3} k$.
From the given, we have
$$
\sqrt[3]{x k+y k+z k}=\sqrt[3]{x^{3} k}+\sqrt[3]{y^{3} k}+\sqrt[3]{z^{3} k},
$$
which means $\sqrt[3]{k} \sqrt[3]{x+y+z}=\sqrt[3]{k}(x+y+z)$.
Since $k \neq... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Find all positive integers $n$, such that $2^{n-1} n+1$ is a perfect square.
(2004, Slovenia IMO National Team Selection Test | Let $2^{n-1} n+1=m^{2}\left(m \in \mathbf{N}_{+}\right)$. Then $2^{n-1} n=(m+1)(m-1)$.
When $n=1,2,3,4$, $2^{n-1} n+1$ is not a perfect square.
Therefore, $n \geqslant 5, 16 \mid (m+1)(m-1)$.
Since $m+1$ and $m-1$ have the same parity, both $m+1$ and $m-1$ are even, and $m$ is odd.
Let $m=2 k-1\left(k \in \mathbf{N}_{+... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Proof: There exist positive integers $a_{i}(1 \leqslant i \leqslant 8)$, such that
$$
\begin{array}{l}
\sqrt{\sqrt{a_{1}}-\sqrt{a_{1}-1}}+\sqrt{\sqrt{a_{2}}-\sqrt{a_{2}-1}}+\cdots+ \\
\sqrt{\sqrt{a_{8}}-\sqrt{a_{8}-1}}=2 .
\end{array}
$$ | Prove the construction of the identity:
$$
\begin{array}{l}
\sqrt{(2 i+1)^{2}}-\sqrt{(2 i+1)^{2}-1} \\
=2 i+1-2 \sqrt{i(i+1)} \\
=(\sqrt{i+1}-\sqrt{i})^{2} .
\end{array}
$$
Take \( a_{i}=(2 i+1)^{2} \) for \( i=1,2, \cdots, 8 \), then
$$
\begin{array}{l}
\sqrt{\sqrt{a_{1}}-\sqrt{a_{1}-1}}+\sqrt{\sqrt{a_{2}}-\sqrt{a_{2... | 2 | Algebra | proof | Yes | Yes | cn_contest | false |
7. Given that $a$ is a root of the equation $x^{2}-3 x+1=0$. Then the value of the fraction $\frac{2 a^{6}-6 a^{4}+2 a^{5}-a^{2}-1}{3 a}$ is $\qquad$ - | II. 7. -1.
According to the problem, we have $a^{2}-3 a+1=0$.
$$
\begin{array}{l}
\text { Original expression }=\frac{2 a^{3}\left(a^{2}-3 a+1\right)-\left(a^{2}+1\right)}{3 a} \\
=-\frac{a^{2}+1}{3 a}=-1 .
\end{array}
$$ | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Robots A and B simultaneously conduct a $100 \mathrm{~m}$ track test at a uniform speed, and the automatic recorder shows: when A is $1 \mathrm{~m}$ away from the finish line, B is $2 \mathrm{~m}$ away from the finish line; when A reaches the finish line, B is $1.01 \mathrm{~m}$ away from the finish line. After calc... | 8. 1 .
Let the actual length of the track be $x \mathrm{~m}$, and the speeds of robots 甲 and 乙 be $v_{\text {甲 }}$ and $v_{\text {乙 }}$, respectively. When 甲 is $1 \mathrm{~m}$ away from the finish line, the time spent is $t$. Then $v_{\text {乙 }} t=x-2$.
Thus, $\frac{v_{\text {甲 }}}{v_{\text {乙 }}}=\frac{x-1}{x-2}$.
... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. The graphs of the functions $f(x)=2 x^{2}-2 x-1$ and $g(x)=$ $-5 x^{2}+2 x+3$ intersect at two points. The equation of the line passing through these two points is $y=a x+b$. Find the value of $a-b$. | 6. The graphs of the functions $f(x)$ and $g(x)$ intersect at two points, which are the solutions to the system of equations $\left\{\begin{array}{l}y=2 x^{2}-2 x-1, \\ y=-5 x^{2}+2 x+3\end{array}\right.$. Therefore,
$$
\begin{array}{l}
7 y=5\left(2 x^{2}-2 x-1\right)+2\left(-5 x^{2}+2 x+3\right) \\
=-6 x+1 .
\end{arra... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { Three. (20 points) (1) Prove: } \\
\left(4 \sin ^{2} x-3\right)\left(4 \cos ^{2} x-3\right) \\
=4 \sin ^{2} 2 x-3(x \in \mathbf{R}) ;
\end{array}
$$
(2) Find the value: $\prod_{t=0}^{2^{8}}\left(4 \sin ^{2} \frac{t \pi}{2^{9}}-3\right)$. | $$
\begin{array}{l}
=(1)\left(4 \sin ^{2} x-3\right)\left(4 \cos ^{2} x-3\right) \\
=16 \sin ^{2} x \cdot \cos ^{2} x-12\left(\sin ^{2} x+\cos ^{2} x\right)+9 \\
=4 \sin ^{2} 2 x-3
\end{array}
$$
(2) Let $A_{k}=\prod_{i=0}^{2 k-1}\left(4 \sin ^{2} \frac{t \pi}{2^{k}}-3\right)$.
When $k \geqslant 2$, by (1) we have
$$
... | -3 | Algebra | proof | Yes | Yes | cn_contest | false |
Three. (20 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$ (for example, $f(3)$ is the sum of the digits of $3 \times 3^{2}+3+1=31$, i.e., $f(3)=4$).
(1) Prove that for any positive integer $n, f(n) \neq 1$, and $f(n) \neq 2$;
(2) Try to ... | (1) Since $3 n^{2}+n+1$ is an odd number greater than 3, hence $f(n) \neq 1$.
If $f(n)=2$, then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e., $3 n^{2}+n+1=10^{k}+1$ (where $k$ is an integer greater than 1).
Thus, $n(3 n+1)=2^{k} \times 5^{k}$.
Since $(n, ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a three-digit number $x y z(1 \leqslant x \leqslant 9,0 \leqslant y, z$ $\leqslant 9)$. If $x y z=x!+y!+z!$, then the value of $x+y+z$ is | 7. 10 .
Since $6!=720$, we have $0 \leqslant x, y, z \leqslant 5$.
And $1!=1, 2!=2, 3!=6, 4!=24, 5!=120$.
By observation, we get $x=1, y=4, z=5$. Therefore,
$$
x+y+z=10 \text {. }
$$ | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) Given that $a$ and $b$ are integers, the equation $a x^{2} + b x + 2 = 0$ has two distinct negative real roots greater than -1. Find the minimum value of $b$.
| Let the equation $a x^{2}+b x+2=0(a \neq 0)$ have two distinct negative real roots $x_{1} 、 x_{2}\left(x_{1}<x_{2}<0\right)$.
\end{array}\right.
$$
Solving, we get $a>0, b>0$.
Since $a$ and $b$ are both integers, it follows that $a$ and $b$ are both positive integers.
Let $y=a x^{2}+b x+2$. Then this parabola opens up... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 2, fold $\triangle A B C$ along the dotted line $D E$ to get a heptagon $A D E C F G H$. If the area ratio of the heptagon to the original triangle is $2: 3$, and the area of the overlapping part after folding is 4, then the area of the original $\triangle A B C$ is | 3. 12 .
Let the area of the non-overlapping part after folding be $x$. Then the area of the original triangle is $8+x$, and the area of the heptagon is $4+x$. From the given condition, we have
$$
\begin{array}{l}
(8+x):(4+x)=3: 2 \\
\Rightarrow 16+2 x=12+3 x \Rightarrow x=4 .
\end{array}
$$
$$
\text { Hence } S_{\tria... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ The value is
The value of $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ is | 4. 4 .
Let $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}=x$.
Cubing both sides and simplifying, we get
$$
x^{3}-6 x-40=0 \text {. }
$$
By observation, 4 is a root of the equation. Therefore,
$$
(x-4)\left(x^{2}+4 x+10\right)=0 \text {. }
$$
Since $\Delta=4^{2}-4 \times 10=-24<0$, the equation $x^{2}+4 x+10=0$ h... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Given the polynomial
$$
\begin{array}{l}
(1+x)+(1+x)^{2}+\cdots+(1+x)^{n} \\
=b_{0}+b_{1} x+\cdots+b_{n} x^{n},
\end{array}
$$
and $b_{1}+b_{2}+\cdots+b_{n}=1013$.
Then a possible value of the positive integer $n$ is | 13. 9 .
Let $x=0$, we get $b_{0}=n$.
Let $x=1$, we get
$$
2+2^{2}+\cdots+2^{n}=b_{0}+b_{1}+\cdots+b_{n},
$$
which is $2\left(2^{n}-1\right)=n+1013$.
Solving this, we get $n=9$. | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 4, given $\angle A O M=60^{\circ}$, there is a point $B$ on ray $O M$ such that the lengths of $A B$ and $O B$ are both integers, thus $B$ is called an "olympic point". If $O A=8$, then the number of olympic points $B$ in Figure 4 is $\qquad$ | 4. 4 .
As shown in Figure 7, draw $A H \perp O M$ at point $H$.
Since $\angle A O M$
$=60^{\circ}$, and $O A$
$=8$, therefore,
$O H=4$,
$A H=4 \sqrt{3}$.
Let $A B=$
$m, H B=n$
($m, n$ are positive integers). Clearly, in the right triangle $\triangle A H B$, we have
$$
m^{2}-n^{2}=(4 \sqrt{3})^{2} \text {, }
$$
which ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. If $2n+1, 20n+1 \left(n \in \mathbf{N}_{+}\right)$ are powers of the same positive integer, then all possible values of $n$ are | 6. 4 .
According to the problem, we know that $(2 n+1) \mid(20 n+1)$. Then $(2 n+1) \mid[10(2 n+1)-(20 n+1)]=9$. Therefore, $n \in\{1,4\}$. Upon verification, $n=4$. | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: There is an electronic flea jumping back and forth on a number line, and it lights up a red light when it jumps to a negative number, but does not light up when it jumps to a positive number. The starting point is at the point representing the number -2 (record one red light), the first step is to jump 1 uni... | (1) Notice
$$
\begin{array}{l}
S=\mid-2-1+2-3+4-5+\cdots-9+101 \\
=1-2+1 \times 5 \mid=3 .
\end{array}
$$
Therefore, the negative numbers appear as $-2,-3,-1,-4,-5$, $-6,-7$, a total of seven times. So, the red light will flash seven times.
(2) In fact, after the electronic flea jumps four times, it returns to the ori... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 If real numbers $a, b$ satisfy the conditions
$$
a^{2}+b^{2}=1,|1-2 a+b|+2 a+1=b^{2}-a^{2} \text {, }
$$
then $a+b=$ $\qquad$
(2009, National Junior High School Mathematics Joint Competition) | Given $a^{2}+b^{2}=1$, we have
$$
b^{2}=1-a^{2} \text {, and }-1 \leqslant a \leqslant 1,-1 \leqslant b \leqslant 1 \text {. }
$$
From $|1-2 a+b|+2 a+1=b^{2}-a^{2}$, we get
$$
\begin{array}{l}
|1-2 a+b|=b^{2}-a^{2}-2 a-1 \\
=\left(1-a^{2}\right)-a^{2}-2 a-1=-2 a^{2}-2 a .
\end{array}
$$
Thus, $-2 a^{2}-2 a \geqslant ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Let the lengths of the two legs of a right triangle be $a$ and $b$, and the length of the hypotenuse be $c$. If $a$, $b$, and $c$ are all integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition. | 12. By the Pythagorean theorem, we have
$$
c^{2}=a^{2}+b^{2} \text {. }
$$
Also, $c=\frac{1}{3} a b-(a+b)$, substituting into equation (1) gives
$$
\begin{array}{l}
a^{2}+b^{2} \\
=\frac{1}{9}(a b)^{2}-\frac{2}{3} a b(a+b)+a^{2}+2 a b+b^{2} .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
a b-6(a+b)+18=0 \\
\... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. In $\triangle A B C$, it is known that $A B=b^{2}-1, B C=$ $a^{2}, C A=2 a$, where $a$ and $b$ are both integers greater than 1. Then the value of $a-b$ is $\qquad$ | 2.0.
Since $a$ is an integer greater than 1, we have $a^{2} \geqslant 2 a$. By the triangle inequality, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
a^{2}+2 a>b^{2}-1, \\
2 a+b^{2}-1>a^{2}
\end{array}\right. \\
\Leftrightarrow(a-1)^{2}<b^{2}<(a+1)^{2} .
\end{array}
$$
Given $a, b \in \mathbf{N}_{+}$, hence $b=a... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the equation in $x$
$$
x^{4}+2 x^{3}+(3+k) x^{2}+(2+k) x+2 k=0
$$
has real roots. If the product of all real roots is -2, then the sum of the squares of all real roots is $\qquad$ . | 4.5.
Factorizing the left side of the equation, we get
$$
\left(x^{2}+x+2\right)\left(x^{2}+x+k\right)=0 \text {. }
$$
Since $x^{2}+x+2=0$ has no real roots, the equation $x^{2}+x+k=0$ has two real roots $x_{1} 、 x_{2}$.
According to the problem, $x_{1} x_{2}=k=-2$.
$$
\begin{array}{l}
\text { Also, } x_{1}+x_{2}=-1 ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 The function $y=|x+1|+|x+2|+|x+3|$. When $x=$ $\qquad$, $y$ has its minimum value, and the minimum value is $\qquad$
(2007, National Junior High School Mathematics Competition Zhejiang Regional Finals) | It is known that $|x+1|+|x+2|+|x+3|$ represents the sum of the distances (lengths of segments) from point $x$ to points $-1$, $-2$, and $-3$ on the number line. It is easy to see that when $x=-2$, this sum of distances is minimized, i.e., the value of $y$ is the smallest, at which point, $y_{\min }=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For example, the minimum value of $7 y=2 x^{2}+4|x|-1$ is
$\qquad$
(2007, National Junior High School Mathematics Competition, Zhejiang Province Re-test) | Let $t=|x| \geqslant 0$. Then
$$
y=2 t^{2}+4 t-1=2(t+1)^{2}-3(t \geqslant 0) \text{.}
$$
It is easy to see that when $t=|x|=0$, i.e., $x=0$, $y$ is minimized at -1. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. If for any real number $x$, the function
$$
f(x)=x^{2}-2 x-|x-1-a|-|x-2|+4
$$
is always a non-negative real number, then the maximum value of the real number $a$ is | 7. 1 .
From the conditions, we have $\left\{\begin{array}{l}f(0)=-|1+a|+2 \geqslant 0, \\ f(1)=-|a|+2 \geqslant 0 .\end{array}\right.$
Solving this, we get $-2 \leqslant a \leqslant 1$.
When $a=1$, we have
$$
f(x)=x^{2}-2 x-2|x-2|+4,
$$
which is $f(x)=\left\{\begin{array}{ll}x^{2}, & x \leqslant 2 ; \\ x^{2}-4 x+8, &... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Find the minimum value of the function
$$
f(x)=\max \left\{|x+1|,\left|x^{2}-5\right| \right\}
$$
and find the value of the independent variable $x$ when $f(x)$ takes the minimum value. | Solve As shown in Figure 1, in the same Cartesian coordinate system, draw the graphs of $f_{1}(x)=|x+1|$ and $f_{2}(x)=\left|x^{2}-5\right|$. The two graphs intersect at four points, $A$, $B$, $C$, and $D$, whose x-coordinates can be obtained by solving the equation $|x+1|=\left|x^{2}-5\right|$.
Removing the absolute v... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Real numbers $a, b$ satisfy
$$
|a+1|+|a+3|+|b+2|+|b-5|=9 \text {. }
$$
Let the maximum and minimum values of $ab + a + b$ be $m, n$ respectively. Then the value of $m+n$ is ( ).
(A) -12 (B) -10 (C) -14 (D) -8 | -1. A.
$$
\begin{array}{l}
\text { Since } 9=|a+1|+|a+3|+|b+2|+|b-5| \\
=(|a+1|+|-a-3|)+(|b+2|+|5-b|) \\
\geqslant|1-3|+|2+5|=9 .
\end{array}
$$
Thus $-3 \leqslant a \leqslant-1$, and $-2 \leqslant b \leqslant 5$
$$
\begin{array}{l}
\Rightarrow-2 \leqslant a+1 \leqslant 0, \text { and }-1 \leqslant b+1 \leqslant 6 \\
... | -12 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 9 On the blackboard, there are two positive integers, one is 2002, and the other is a number less than 2002. If the average of these two numbers $m$ is an integer, then the following operation can be performed: one of the numbers is erased and replaced by $m$. How many times can such an operation be performed a... | Without loss of generality, let these two positive integers be $a, b(a>b)$.
Consider the absolute value of the difference between the two numbers.
Initially, it is $|a-b|$. After the first operation, the absolute value of the difference between the two numbers is $\left|a-\frac{a+b}{2}\right|=\left|\frac{a-b}{2}\right|... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. If the equation
$$
n \sin x+(n+1) \cos x=n+2
$$
has two distinct real roots in $0<x<\pi$, then the minimum value of the positive integer $n$ is $\qquad$. | 4.4.
From the given, we have $\frac{1}{n}=-1-\frac{-1-\sin x}{2-\cos x}$.
And $\frac{-1-\sin x}{2-\cos x}$ represents the slope $k$ of the line connecting a moving point $P(\cos x, \sin x)$ on the upper half of the unit circle (excluding endpoints) and a fixed point $Q(2,-1)$.
To satisfy the problem, the line $PQ$ mu... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The sum of the x-coordinates of the points where the graph of the function $y=x^{2}-2009|x|+2010$ intersects the x-axis is $\qquad$ . | 3. 0 .
The original problem can be transformed into finding the sum of all real roots of the equation
$$
x^{2}-2009|x|+2010=0
$$
If a real number $x_{0}$ is a root of equation (1), then its opposite number $-x_{0}$ is also a root of equation (1).
Therefore, the sum of all real roots of the equation is 0, that is, th... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given
$$
A=17^{2012 n}+4 \times 17^{4 n}+7 \times 19^{7 n}(n \in \mathbf{N})
$$
can be expressed as the product of $k(k \in \mathbf{N}, k>1)$ consecutive integers. Then $n+k=$ $\qquad$ . | 4. 2 .
If $k \geqslant 4$, then $81 A$.
$$
\begin{array}{l}
\text { But } A=17^{2012 n}+4 \times 17^{4 n}+7 \times 19^{7 n} \\
\equiv 1+4-3^{7 n} \equiv 5-3^{n} \not \equiv 0(\bmod 8),
\end{array}
$$
Contradiction.
If $k=3$, let $A=m\left(m^{2}-1\right)(m \in \mathbf{Z})$.
But $A \equiv 2^{2012 n}-2^{4 n}+2 \times(-1... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A photographer took some photos of eight people at a party, with any two people (there are 28 possible combinations) appearing in exactly one photo. Each photo can be a duo or a trio. How many photos did the photographer take at least?
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。 | 6. 12 .
Let the number of group photos with three people be $x$, and the number of group photos with two people be $y$. Then $3 x+y=28$.
Thus, when $x$ is maximized, the total number of photos taken is minimized.
When $x>8$, by $\frac{3 x}{8}>3$ we know that there is a person $A$ who appears 4 times, and in the 4 grou... | 12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
10. For $i=2,3, \cdots, k$, the remainder when the positive integer $n$ is divided by $i$ is $i-1$. If the smallest value of $n$, $n_{0}$, satisfies $2000<n_{0}<3000$, then the smallest value of the positive integer $k$ is | 10.9.
Since $n+1$ is a multiple of $2,3, \cdots, k$, the smallest value of $n$, $n_{0}$, satisfies
$$
n_{0}+1=[2,3, \cdots, k],
$$
where $[2,3, \cdots, k]$ represents the least common multiple of $2,3, \cdots, k$.
$$
\begin{array}{l}
\text { Since }[2,3, \cdots, 8]=840, \\
{[2,3, \cdots, 9]=2520,} \\
{[2,3, \cdots, 1... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. B. Color the five sides and five diagonals of the convex pentagon $A B C D E$, such that any two segments sharing a common vertex are of different colors. Find the minimum number of colors needed. | 14. B. Since vertex $A$ is the common point of four segments $A B$, $A C$, $A D$, and $A E$, at least four colors are needed.
If only four colors are used, let's assume they are red, yellow, blue, and green. Then, the four segments extending from each vertex must include one each of red, yellow, blue, and green. There... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. If the equation about $x$
$$
x^{3}+a x^{2}+b x-4=0\left(a 、 b \in \mathbf{N}_{+}\right)
$$
has a positive integer solution, then $|a-b|=$ | 7.1.
Solution 1 Let $m$ be a positive integer solution of the equation.
If $m \geqslant 2$, then $a m^{2}+b m=4-m^{3}<0$, which contradicts that $a$ and $b$ are both positive integers.
Therefore, only $m=1$. Substituting it in, we get $a+b=3$.
Since $a, b \in \mathbf{N}_{+}$, $\{a, b\}=\{1,2\}$.
Thus, $|a-b|=1$.
Solut... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $|x| \leqslant 1,|y| \leqslant 1$, and
$$
k=|x+y|+|y+1|+|2 y-x-4| \text {. }
$$
Then the sum of the maximum and minimum values of $k$ is $\qquad$ | 2. 10 .
From the given conditions, we know
$$
-1 \leqslant x \leqslant 1, -1 \leqslant y \leqslant 1 \text{. }
$$
Thus, $y+1 \geqslant 0, 2y-x-40$ when,
$$
k=x+y+y+1-(2y-x-4)=2x+5 \text{. }
$$
Therefore, $3 \leqslant k \leqslant 7$.
Hence, the maximum value of $k$ is 7, and the minimum value is 3. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given real numbers $x, y, z$ satisfy $x+y=5$ and $z^{2}=x y+y-9$.
Then $x+2 y+3 z=$ $\qquad$ . | Solve: From $x+y=5$, we can set $x=\frac{5}{2}+t, y=\frac{5}{2}-t$.
Substitute into $z^{2}=x y+y-9$ and simplify to get $z^{2}+\left(t+\frac{1}{2}\right)^{2}=0$.
Thus, $z=0, t=-\frac{1}{2}$.
Therefore, $x=2, y=3$.
Hence, $x+2 y+3 z=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Let real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x+y=z-1, \\
x y=z^{2}-7 z+14 .
\end{array}\right.
$$
Question: What is the maximum value of $x^{2}+y^{2}$? For what value of $z$ does $x^{2}+y^{2}$ achieve its maximum value? | Solve: From $x+y=z-1$, we can set
$$
x=\frac{z-1}{2}+t, y=\frac{z-1}{2}-t \text {. }
$$
Substituting into $x y=z^{2}-7 z+14$ and simplifying, we get
$$
3 z^{2}-26 z-55=-4 t^{2} \text {. }
$$
Therefore, $3 z^{2}-26 z-55 \leqslant 0$.
Solving this, we get $\frac{11}{3} \leqslant z \leqslant 5$.
$$
\begin{array}{l}
\tex... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given an equilateral $\triangle A B C$ with side length $2, P$ is a point inside $\triangle A B C$, and the distances from point $P$ to the three sides $B C, A C, A B$ are $x, y, z$ respectively, and their product is $\frac{\sqrt{3}}{9}$. Then the sum of the squares of $x, y, z$ is | 2. 1 .
As shown in Figure 4, connect
$$
P A, P B, P C \text{. }
$$
It is easy to see that, in the equilateral $\triangle A B C$, we have
$$
\begin{array}{l}
x+y+z \\
=\frac{\sqrt{3}}{2} A B=\sqrt{3} .
\end{array}
$$
Thus, $\sqrt{3}=x+y+z$
$$
\geqslant 3 \sqrt[3]{x y z}=3 \sqrt[3]{\frac{\sqrt{3}}{9}}=\sqrt{3} \text{.... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the quadratic function
$$
y=3 a x^{2}+2 b x-(a+b) \text {, }
$$
when $x=0$ and $x=1$, the value of $y$ is positive. Then, when $0<x<1$, the parabola intersects the $x$-axis at $\qquad$ points. | 3. 2 .
From the given, we have $\left\{\begin{array}{l}-(a+b)>0, \\ 3 a+2 b-(a+b)>0 .\end{array}\right.$
Solving, we get $a>0$.
Let $x=\frac{1}{2}$, then
$$
y=\frac{3 a}{4}+b-(a+b)=-\frac{a}{4}<0 .
$$
By the Intermediate Value Theorem, $3 a x^{2}+2 b x-(a+b)=0$ has one root when $0<x<\frac{1}{2}$, and one root when $... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $x, y, z \in \mathbf{R}_{+}$. Then the minimum value of $\frac{\left(x^{2}+y^{2}\right)^{3}+z^{6}}{2 x^{3} y^{3}+y^{3} z^{3}+z^{3} x^{3}}$ is $\qquad$ . | 3.2.
Let $x=y=1, z=\sqrt[3]{2}$, we get
$$
\frac{\left(x^{2}+y^{2}\right)^{3}+z^{6}}{2 x^{3} y^{3}+y^{3} z^{3}+z^{3} x^{3}}=2 \text {. }
$$
Notice that
$$
\begin{array}{l}
\left(x^{2}+y^{2}\right)^{3}+z^{6} \\
=\left(x^{6}+y^{6}+3 x^{4} y^{2}+3 x^{2} y^{4}\right)+z^{6} \\
=\left(2 x^{4} y^{2}+2 x^{2} y^{4}\right)+\le... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. Fill $1,2, \cdots, n^{2}$ into an $n \times n$ chessboard, with each cell containing one number, and each row forming an arithmetic sequence with a common difference of 1. If any two of the $n$ numbers on the chessboard are neither in the same row nor in the same column, then the sum of these $n$ numbers is called a... | 5.1.
The numbers in the $k$-th row of the number table are
$$
(k-1) n+1,(k-1) n+2, \cdots,(k-1) n+n \text {. }
$$
Let the number taken from the $k$-th row be
$$
(k-1) n+a_{k} \text {. }
$$
Since these numbers are from different columns, $a_{1}, a_{2}, \cdots, a_{n}$ is a permutation of $1,2, \cdots, n$.
Therefore, t... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 2 of the hopscotch game: a person can only enter the first
square from outside; in the squares, each time they can
jump forward 1 square or 2 squares.
Then, the number of ways a person can jump
from outside to the sixth square is $\qquad$. | Solution 1 (Enumeration Method) Transform the problem as follows:
Enter the 6th grid from the 1st grid, walking 5 grids. Represent the number 5 as the sum of several 1s or 2s, where different orders of 1s and 2s represent different methods (e.g., $5=1+1+1+1+1$). For convenience, abbreviate the expression to $5=11111$. ... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 The condition for three line segments to form a triangle is: the sum of the lengths of any two line segments is greater than the length of the third line segment. There is a wire of length $144 \mathrm{~cm}$, which is to be cut into $n$ $(n>2)$ small segments, with each segment being no less than $1 \mathrm{~... | Since the necessary and sufficient condition for forming a triangle is that the sum of any two sides is greater than the third side, the condition for not forming a triangle is that the sum of any two sides is less than or equal to the largest side. The shortest piece of wire is 1, so we can place 2 ones, and the third... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. A certain linear function graph is parallel to the line $y=\frac{5}{4} x+\frac{95}{4}$, and intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and passes through the point $(-1,-25)$. Then on the line segment $AB$ (including $A$ and $B$), the number of points with both integer coordinates is $... | 3.5.
Let the linear function be $y=\frac{5}{4} x+b$.
Given that it passes through the point $(-1,-25)$, we get $b=-\frac{95}{4}$.
Therefore, $y=\frac{5}{4} x-\frac{95}{4}$.
Thus, $A(19,0)$ and $B\left(0,-\frac{95}{4}\right)$.
From $y=\frac{5}{4} x-\frac{95}{4}(0 \leqslant x \leqslant 19)$, taking $x=3,7,11$,
15,19, $y$... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the function $f(x)$ satisfies for all real numbers $x, y$,
$$
\begin{array}{l}
f(x)+f(2 x+y)=f(3 x-y)+x-2010 \text {. } \\
\text { Then } f(2010)=
\end{array}
$$ | 4.0.
Let $x=2010, y=1005$, then
$$
\begin{array}{l}
f(2010)+f(5025) \\
=f(5025)+2010-2010 .
\end{array}
$$
Therefore, $f(2010)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given real numbers $a, b$ satisfy
$$
a^{2}+a b+b^{2}=1 \text {, and } t=a b-a^{2}-b^{2} \text {. }
$$
Then the product of the maximum and minimum values of $t$ is $\qquad$ | 8. 1.
Let $a=x+y, b=x-y$. Then
$$
(x+y)^{2}+(x+y)(x-y)+(x-y)^{2}=1 \text {. }
$$
Simplifying, we get $y^{2}=1-3 x^{2}$.
Since $y^{2} \geqslant 0$, we have $0 \leqslant x^{2} \leqslant \frac{1}{3}$.
Thus, $t=a b-a^{2}-b^{2}$
$$
\begin{array}{l}
=(x+y)(x-y)-(x+y)^{2}-(x-y)^{2} \\
=-x^{2}-3 y^{2}=8 x^{2}-3 .
\end{array}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (12 points) As shown in Figure 3, in $\triangle ABC$, it is given that $AB=9, BC=8, AC=7$, and $AD$ is the angle bisector. A circle is drawn with $AD$ as a chord, tangent to $BC$, and intersecting $AB$ and $AC$ at points $M$ and $N$, respectively. Find the length of $MN$. | II. As shown in Figure 3, connect DM. Given
$\angle B D M=\angle B A D=\angle C A D=\angle D M N$
$\Rightarrow M N / / B C \Rightarrow \triangle A M N \backsim \triangle A B C$.
It is easy to know that $B D=\frac{9}{2}$.
Then $B M \cdot B A=B D^{2} \Rightarrow B M \cdot 9=\left(\frac{9}{2}\right)^{2}$
$\Rightarrow B M=... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=a(a \neq 0$, and $a \neq 1)$, the sum of the first $n$ terms is $S_{n}$, and $S_{n}=\frac{a}{1-a}\left(1-a_{n}\right)$. Let $b_{n}=a_{n} \lg \left|a_{n}\right|\left(n \in \mathbf{N}_{+}\right)$. When $a=-\frac{\sqrt{7}}{3}$, does there exist a positive integ... | 2. When $n \geqslant 2$,
$$
S_{n}=\frac{a}{1-a}\left(1-a_{n}\right), S_{n-1}=\frac{a}{1-a}\left(1-a_{n-1}\right) .
$$
Then $a_{n}=S_{n}-S_{n-1}$
$$
\begin{array}{l}
=\frac{a}{1-a}\left[\left(1-a_{n}\right)-\left(1-a_{n-1}\right)\right] \\
=\frac{a}{1-a}\left(a_{n-1}-a_{n}\right),
\end{array}
$$
i.e., $a_{n}=a a_{n-1}... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9.1. Distribute 24 pencils of four colors (6 pencils of each color) to 6 students, with each student getting 4 pencils. It is known that no matter how the pencils are distributed, there will always be $n$ students such that the $4 n$ pencils they have are of four colors. Find the minimum value of $n$.
| 9.1. The minimum value of $n$ is 3.
First, we prove: There are always 3 students who have pencils of 4 different colors.
In fact, there are 6 pencils of each color, and each student has 4 pencils, so there exists a student who has at least two colors of pencils. Clearly, any other color must be owned by at least one ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
A4. Find the smallest positive integer $v$, such that for any two-coloring of the edges of the complete graph $K_{v}$, there are two monochromatic triangles that share exactly one vertex. | The solution to A4 is $v=9$.
On one hand, take two red complete graphs $K_{4}$, and color all the edges between them blue. The resulting complete graph $K_{8}$ has no two monochromatic triangles with exactly one common vertex.
On the other hand, consider any two-coloring of the complete graph $K_{9}$. By problem $\mat... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. A set of $n$ points $P_{1}, P_{2}, \cdots, P_{n}$ in the plane, no three of which are collinear, is denoted as $D$. A line segment is drawn between any two points, and the lengths of all these line segments are distinct. In a triangle, the side that is neither the longest nor the shortest is called the "middle side"... | 4. The minimum value of $n$ is 11.
When $n \geqslant 11$, regardless of how $l$ is chosen, there always exists a subset, let's assume it is $D_{1}$, such that $D_{1}$ contains at least six points.
Consider all the middle edges of the triangles in $D_{1}$, and color them red, then color the other edges blue.
By Ramse... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. 1. Distribute 40 pencils of four colors (10 pencils of each color) to 10 students, with each student getting 4 pencils. It is known that no matter how the pencils are distributed, there will always be $n$ students such that the $4 n$ pencils they have are of four colors. Find the minimum value of $n$. | 10.1. The minimum value of $n$ is 3.
First, we prove: There are always 3 students who have pencils of 4 different colors.
Since there are 10 pencils of each color, and each student has 4 pencils, there must be a student who has at least two different colors of pencils.
Obviously, any other color of pencils must be o... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. In two boxes, Jia and Yi, each contains the same number of pieces of jewelry. After transferring a piece of jewelry worth 50,000 yuan from Jia to Yi, the average value of the jewelry in Jia decreases by 10,000 yuan, and the average value of the jewelry in Yi increases by 10,000 yuan. Then the total value of the jewe... | 4. B.
Let the original number of jewels in box A be $n(n>1)$, with a total value of $x$ million yuan, and the total value in box B be $y$ million yuan. Then $\left\{\begin{array}{l}\frac{x-5}{n-1}+1=\frac{x}{n}, \\ \frac{y+5}{n+1}-1=\frac{y}{n}\end{array} \Rightarrow\left\{\begin{array}{l}x=-n^{2}+6 n, \\ y=-n^{2}+4 n... | 12 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. Set $A=\left\{(x, y) \left\lvert\,\left\{\begin{array}{l}y=\sqrt{1-x}, \\ y=1-x^{2}\end{array}\right\}\right.\right.$ has the number of subsets as $\qquad$ | 3. 8 .
Notice that the number of elements in set $A$ is the number of solutions to the system of equations
$$
\left\{\begin{array}{l}
y=\sqrt{1-x}, \\
y=1-x^{2}
\end{array}\right.
$$
Substituting equation (1) into equation (2) and simplifying, we get
$$
x(x-1)\left(x^{2}+x-1\right)=0 \text {. }
$$
Solving this, we g... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\text { Three. (50 points) Given the sequence }\left\{a_{n}\right\}: 1,3,5,7, \cdots \text {, }
$$
starting from the 5th term, $a_{n+4}$ is the unit digit of $a_{n}+a_{n+3}$. Find:
$$
a_{2008}^{2}+a_{2009}^{2}+a_{2010}^{2}+a_{2011}^{2}
$$
Can it be divisible by 4? | Consider the sequence $\left\{a_{n}\right\}$ and its modulo 2 residue sequence $\left\{b_{n}\right\}$.
Given that $a_{n+4}$ is the unit digit of $a_{n}+a_{n+3}$, we have $b_{n+4} \equiv b_{n}+b_{n+3}(\bmod 2)$.
By recursion, we get
$$
\begin{array}{l}
b_{n+10} \equiv b_{n+6}+b_{n+9} \\
\equiv\left(b_{n+2}+b_{n+5}\right... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
279 Given positive real numbers $x, y, z$ satisfying $(\sqrt{3}+1) x y + 2 \sqrt{3} y z + (\sqrt{3}+1) z x = 1$.
(1) Find the minimum value of $x + y + z$;
(2) Find the minimum value of $\frac{\sqrt{3} x y}{z} + \frac{(8-4 \sqrt{3}) y z}{x} + \frac{\sqrt{3} z x}{y}$. | (1) By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{(\sqrt{3}+1) x^{2}}{2}+(\sqrt{3}-1) y^{2} \geqslant 2 x y, \\
2 y^{2}+2 z^{2} \geqslant 4 y z, \\
(\sqrt{3}-1) z^{2}+\frac{(\sqrt{3}+1) x^{2}}{2} \geqslant 2 x z .
\end{array}
$$
Adding the above three inequalities, we get
$$
\begin{array}{l}
(\sqrt{3}+1)\... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$
(Fourth Jiangsu Province Junior High School Mathematics Competition) | Given that $m$ and $n$ are the roots of the equation $x^{2}=x+1$, i.e., $x^{2}-x-1=0$.
Let $a_{k}=m^{k}+n^{k}$. Then
$$
a_{k}=a_{k-1}+a_{k-2}(k \geqslant 3) \text {. }
$$
By $a_{1}=m+n=1$,
$$
a_{2}=(m+n)^{2}-2 m n=3 \text {, }
$$
we know $a_{3}=a_{2}+a_{1}=4, a_{4}=a_{3}+a_{2}=7$.
Therefore, $m^{5}+n^{5}=a_{5}=a_{4}+... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $m$ and $n$ are rational numbers, and the equation
$$
x^{2}+m x+n=0
$$
has a root $\sqrt{5}-2$. Then $m+n=$ $\qquad$ .
(2001, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | $$
\begin{array}{l}
(\sqrt{5}-2)^{2}+m(\sqrt{5}-2)+n=0 \\
\Rightarrow(n-2 m+9)=(4-m) \sqrt{5} \\
\Rightarrow\left\{\begin{array} { l }
{ n - 2 m + 9 = 0 , } \\
{ 4 - m = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=4, \\
n=-1
\end{array}\right.\right. \\
\Rightarrow m+n=3 \text {. } \\
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) Let positive integers $a, b, c (a \geqslant b \geqslant c)$ be the lengths of the sides of a triangle, and satisfy
$$
a^{2}+b^{2}+c^{2}-a b-a c-b c=13 \text {. }
$$
Find the number of triangles that meet the conditions and have a perimeter not exceeding 30. | Given the known equation:
$$
(a-b)^{2}+(b-c)^{2}+(a-c)^{2}=26 \text {. }
$$
Let $a-b=m, b-c=n$.
Then $a-c=m+n(m, n$ are natural numbers $)$.
Thus, equation (1) becomes
$$
m^{2}+n^{2}+m n=13 \text {. }
$$
Therefore, the pairs $(m, n)$ that satisfy equation (2) are:
$$
(m, n)=(3,1),(1,3) \text {. }
$$
(1) When $(m, n)=... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given the quadratic function $y=x^{2}+b x-c$ whose graph passes through two points $P(1, a)$ and $Q(2,10 a)$.
(1) If $a$, $b$, and $c$ are all integers, and $c<b<8 a$, find the values of $a$, $b$, and $c$;
(2) Let the graph of the quadratic function $y=x^{2}+b x-c$ intersect the $x$-axis at points $A... | Three, given that points $P(1, a)$ and $Q(2, 10a)$ are on the graph of the quadratic function $y = x^2 + bx - c$, we have
$$
1 + b - c = a, \quad 4 + 2b - c = 10a.
$$
Solving these equations, we get $b = 9a - 3$ and $c = 8a - 2$.
(1) From $c < b < 8a$, we know
$$
8a - 2 < 9a - 3 < 8a \Rightarrow 1 < a < 3.
$$
Since $a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Let $p$ be a prime number greater than 2, and $k$ be a positive integer. If the graph of the function $y=x^{2}+p x+(k+1) p-4$ intersects the $x$-axis at two points, at least one of which has an integer coordinate, find the value of $k$.
---
The function is given by:
\[ y = x^2 + px + (k+1)p - 4 \]
... | From the problem, we know that the equation
$$
x^{2}+p x+(k+1) p-4=0
$$
has at least one integer root among its two roots $x_{1}, x_{2}$.
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}=-p, x_{1} x_{2}=(k+1) p-4 . \\
\text { Hence }\left(x_{1}+2\right)\left(x_{2}+2\right) \\... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let $a, b$ be positive integers, and satisfy $2\left(\sqrt{\frac{1}{a}}+\sqrt{\frac{15}{b}}\right)$ is an integer. Then the number of such ordered pairs $(a, b)$ is $\qquad$ pairs.
$(2009$, National Junior High School Mathematics League) | First, guess that $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are both rational numbers.
The following is an attempt to prove this.
Let $\frac{15}{a}=A, \frac{15}{b}=B$. According to the problem, we can assume
$\sqrt{A}+\sqrt{B}=C\left(A, B, C \in \mathbf{Q}_{+}\right)$.
Thus, $\sqrt{A}=C-\sqrt{B} \Rightarrow A=(C-... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If real numbers $a, b$ satisfy
$$
\left(a-\sqrt{a^{2}+2010}\right)\left(b+\sqrt{b^{2}+2010}\right)+2010=0 \text {, }
$$
then $a \sqrt{b^{2}+2011}-b \sqrt{a^{2}+2011}=$ $\qquad$ | From the given, we have
$$
\begin{array}{l}
a-\sqrt{a^{2}+2010}=-\frac{2010}{b+\sqrt{b^{2}+2010}} \\
=b-\sqrt{b^{2}+2010} .
\end{array}
$$
Similarly,
$$
b+\sqrt{b^{2}+2010}=a+\sqrt{a^{2}+2010} \text {. }
$$
Subtracting (2) from (1) and simplifying, we get $a=b$.
Therefore, the answer is 0. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $P$ is a point inside the circle $\odot O$ with radius 15, among all the chords passing through point $P$, 24 chords have integer lengths. Then $O P=$ $\qquad$ . | 2. 12 .
Among the chords of circle $\odot O$ passing through point $P$, the diameter is the longest chord, and there is only one; the shortest chord is the one perpendicular to $OP$, and there is also only one. Therefore, there is only one chord each with lengths of $30$ and $18$, and two chords each with lengths of $... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Prove: $\frac{(2+\sqrt{3})^{2010}+(2-\sqrt{3})^{2010}}{2}$ is an integer, and find its remainder when divided by 4. | Notice
$$
\begin{array}{l}
(2+\sqrt{3})+(2-\sqrt{3})=4, \\
(2+\sqrt{3})(2-\sqrt{3})=1 .
\end{array}
$$
Therefore, $2+\sqrt{3}$ and $2-\sqrt{3}$ are the two roots of $x^{2}-4 x+1=0$.
Let $a_{n}=\frac{1}{2}(2+\sqrt{3})^{n}+\frac{1}{2}(2-\sqrt{3})^{n}$. Then
$$
a_{n}-4 a_{n-1}+a_{n-2}=0
$$
That is, $a_{n}=4 a_{n-1}-a_{n... | 3 | Number Theory | proof | Yes | Yes | cn_contest | false |
4. $F$ is the right focus of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$, and $P$ is a moving point on the ellipse. For the fixed point $A(-2, \sqrt{3}),|P A|+$ $2|P F|$ the minimum value is $\qquad$ . | 4. 10 .
It is known that the eccentricity of the ellipse is $\frac{1}{2}$, and the equation of the right directrix $l$ is $x=8$. The distance from point $A$ to $l$ is 10.
Let the projection of point $P$ onto $l$ be $H$. Then
$\frac{|P F|}{|P H|}=\frac{1}{2}$.
Thus, $|P A|+2|P F|=|P A|+|P H| \geqslant 10$.
The minimum ... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Given $0<a<1$, and
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a$ ] equals $\qquad$
$(2009$, Beijing Mathematical Competition (Grade 8)) | Notice that
$0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2$.
Then $\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right]$ equals
0 or 1.
By the problem statement, 18 of these are equal to 1. Therefore,
$\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}\right]=\cdots=\lef... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Let
$$
\begin{aligned}
S & =\frac{1}{\left[\frac{(10 \times 11-1)^{2}}{10 \times 11}\right]}+\frac{1}{\left[\frac{(11 \times 12-1)^{2}}{11 \times 12}\right]}+\cdots+ \\
& \frac{1}{\left[\frac{(49 \times 50-1)^{2}}{49 \times 50}\right]}
\end{aligned}
$$
Then $[30 S]=(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 0
(2002, "F... | When $n \geqslant 10$, and $n$ is an integer,
$$
\begin{array}{l}
{\left[\frac{[n(n+1)-1]^{2}}{n(n+1)}\right]=\left[n(n+1)-2+\frac{1}{n(n+1)}\right]} \\
=n(n+1)-2=(n-1)(n+2) .
\end{array}
$$
Thus, $S=\frac{1}{9 \times 12}+\frac{1}{10 \times 13}+\cdots+\frac{1}{48 \times 51}$.
Notice that $\frac{1}{k(k+3)}=\frac{1}{3}\... | 2 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Let $n$ ($n<100$) be a positive integer, and there exists a positive integer $k$, such that $1 \leqslant k \leqslant n-1$, satisfying
$$
\frac{4 k+1}{2 n}+\frac{1-2 k^{2}}{n^{2}}=\frac{1}{2} \text {. }
$$
How many values of $n$ satisfy the condition? Prove your conclusion. | 1. From equation (1), we get $(2 k-n)^{2}=n+2$.
Solving, we get $k=\frac{1}{2}(n \pm \sqrt{n+2})$.
Since $k$ is an integer, therefore, $n=m^{2}-2(m \in \mathbf{N})$.
Thus, $k=\frac{1}{2}\left(m^{2} \pm m-2\right)$.
Also, $1 \leqslant k \leqslant n-1$, so
$$
1 \leqslant \frac{m^{2} \pm m-2}{2} \leqslant m^{2}-3 \text {... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If real numbers $x, y$ satisfy $|x|+|y| \leqslant 1$, then the maximum value of $x^{2}-$ $xy+y^{2}$ is $\qquad$ | 2. 1 .
Notice that
$$
\begin{array}{l}
x^{2}-x y+y^{2}=\frac{1}{4}(x+y)^{2}+\frac{3}{4}(x-y)^{2} . \\
\text { Also }|x \pm y| \leqslant|x|+|y| \leqslant 1 \text {, then } \\
x^{2}-x y+y^{2} \leqslant \frac{1}{4}+\frac{3}{4}=1 .
\end{array}
$$
When $x$ and $y$ take 0 and 1 respectively, the equality holds. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
6. In trapezoid $A B C D$, $A D / / B C, E F$ is the midline, the area ratio of quadrilateral $A E F D$ to quadrilateral $E B C F$ is $\frac{\sqrt{3}+1}{3-\sqrt{3}}$, and the area of $\triangle A B D$ is $\sqrt{3}$. Then the area of trapezoid $A B C D$ is | 6. 2 .
Let $A D=a, B C=b$, and the height be $2 h$. The area of trapezoid $A B C D$ is $S$. Then
$$
\begin{array}{l}
E F=\frac{a+b}{2}, \frac{a}{b}=\frac{S_{\triangle A B D}}{S_{\triangle B C D}}=\frac{\sqrt{3}}{S-\sqrt{3}} . \\
\text { Also, } \frac{\sqrt{3}+1}{3-\sqrt{3}}=\frac{S_{\text {quadrilateral } A E F D}}{S_... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Five. (25 points) Let positive real numbers $a$, $b$, $c$ satisfy
$$
(a+2 b)(b+2 c)=9 \text {. }
$$
Prove: $\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \geqslant 3$.
(Zhang Lei) | Because $a, b, c > 0$, so,
$$
\begin{array}{l}
\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \\
\geqslant \frac{a+b}{2}+2 \cdot \frac{b+c}{2} \\
=\frac{1}{2}[(a+2 b)+(b+2 c)] \\
\geqslant \sqrt{(a+2 b)(b+2 c)}=3 .
\end{array}
$$ | 3 | Inequalities | proof | Yes | Yes | cn_contest | false |
6. As shown in Figure 1, let $G$ and $H$ be the centroid and orthocenter of $\triangle ABC$ respectively, $F$ be the midpoint of segment $GH$, and the circumradius of $\triangle ABC$ be $R=1$. Then
$$
\begin{array}{l}
|\overrightarrow{A F}|^{2}+|\overrightarrow{B F}|^{2}+|\overrightarrow{C F}|^{2} \\
= \\
.
\end{array}... | 6.3.
Let the circumcenter $O$ of $\triangle ABC$ be the origin of a Cartesian coordinate system. Then,
$$
\begin{array}{l}
\overrightarrow{O H}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}, \\
\overrightarrow{O G}=\frac{1}{3}(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C})
\end{array}... | 3 | Geometry | 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.