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1. Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies
$$
f(f(x)+f(y))=f(x)+y(\forall x, y \in \mathbf{R}) \text {. }
$$
then $f(2011)=$ $\qquad$ . | $-1.2011$.
Let $x=0$ in equation (1), we get
$$
f(f(0)+f(y))=f(0)+y,
$$
which shows that $f$ is surjective.
Let $x$ be the zero root $a$ of $f(x)$ in equation (1). Then $f(f(y))=y$.
Replace $x$ with $f(x)$ in equation (1) to get
$$
f(x+f(y))=x+y \text {. }
$$
Let $y=a$ in equation (2) to get $f(x)=x+a$.
Substitute in... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a_{1}, a_{2}, \cdots, a_{2011}$ be positive real numbers, $S=\sum_{i=1}^{2011} a_{i}$, and
$$
\begin{array}{l}
(S+2011)\left(S+a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2011} a_{1}\right) \\
=4\left(\sqrt{a_{1} a_{2}}+\sqrt{a_{2} a_{3}}+\cdots+\sqrt{a_{2011} a_{1}}\right)^{2} .
\end{array}
$$
Then $S=$ . $\qquad$ | 3. 2011.
By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
& (S+2011)\left(S+a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2011} a_{1}\right) \\
= & \left(a_{1}+a_{2}+\cdots+a_{2011}+\frac{1+\cdots+1}{2011}\right) . \\
& \left(a_{2}+a_{3}+\cdots+a_{1}+a_{1} a_{2}+\cdots+a_{2011} a_{1}\right) \\
\geqslant & {\left[2\lef... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. From the set $\{1,2, \cdots, 2011\}$, randomly select 1005 different numbers such that their sum is 1021035. Then, there are at least some odd numbers among them. | 6.5.
From 1 to 2011, taking 1005 even numbers, their sum is $\frac{1005(2+2010)}{2}=1011030$.
Replacing $2,4,6,8,10$ with 2003, 2005, $2007,2009,2011$, increases the sum by
$$
2001 \times 5=10005,
$$
making the total sum 1021035.
Thus, there are at least 5 odd numbers. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Inside a large sphere with a radius of 4, 24 cubes with edge lengths of 1 have been placed arbitrarily. Prove: At least 4 small spheres with a radius of $\frac{1}{2}$ can still be placed inside the large sphere, such that these small spheres and the cubes are all within the large sphere and do not overl... | 11. To place a small ball with a radius of $\frac{1}{2}$ completely inside the large ball $D(O, 4)$, the distance from its center to the surface of the large ball should be no less than $\frac{1}{2}$, meaning the center of the small ball should be within the ball $D\left(0,4-\frac{1}{2}\right)$. The volume of this ball... | 4 | Geometry | proof | Yes | Yes | cn_contest | false |
Three, (50 points) Try to find the last two non-zero digits of 2011!.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let the last two non-zero digits of a positive integer $k$ be denoted as $f(k)$. Below, we discuss under modulo 100.
Clearly, $5 \times f(k!)$, and if $5 \times f(k), 5 \times f(l)$, then $f(k l) \equiv f(k) f(l)$. Therefore,
$$
\begin{array}{l}
f(2011!) \equiv f\left(\frac{2011!}{2000!}\right) f(2000!) \\
\equiv 68 f(... | 44 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For any positive integer $n(n \geqslant 2)$, try to find:
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right]+ \\
\frac{1}{2} \log _{\frac{3}{2}}\left[1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}\right]
\end{array}
$$
the value. | Notice
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right] \\
=\sum_{k=2}^{n} \log _{\frac{3}{2}} \frac{k^{3}+1}{k^{3}-1}=\log _{\frac{3}{2}} \prod_{k=2}^{n} \frac{k^{3}+1}{k^{3}-1} \\
=\log _{\frac{3}{2}} \prod_{k=2}^{n} \frac{(k+1)\left(k^{2}-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. In $\triangle A B C$, it is known that $A B=A C, \angle C$'s bisector $C D$ intersects $A B$ at point $D, B D, B C, C D$ are three consecutive integers. Find the perimeter of $\triangle A B C$. | Let $A B=A C=b, B C=a$.
By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
B D=\frac{a b}{a+b}, \\
B C-B D=a-\frac{a b}{a+b}=\frac{a^{2}}{a+b} \\
=k \in\{1,2\} .
\end{array}
$$
Following Example 1, we get
$$
C D^{2}=B D(B D+B C)=\frac{a^{2} b(a+2 b)}{(a+b)^{2}} \text {. }
$$
If $k=1$, then $b=a^{2}-a$.
Thus, ... | 45 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Assuming the Earth rotates once around the axis connecting the North Pole and the South Pole in 23 hours 56 minutes 4 seconds, and the Earth's equatorial radius is $6378.1 \mathrm{~km}$. Then, when you stand on the equator and rotate with the Earth, the linear velocity around the axis is $\qquad$ meters/second (roun... | 1. 465 | 465 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (20 points) The numbers $1,2, \cdots, 666$ are written on a blackboard. In the first step, the first eight numbers: $1,2, \cdots, 8$, are erased, and the sum of these numbers, 36, is written after 666; in the second step, the next eight numbers: $9,10, \cdots, 16$, are erased, and the sum of these numbers, 100, ... | (1) Since each step reduces the numbers by seven, after $\frac{666-1}{7}=95$ steps, only one number remains.
(2) From $666-512=154$, it follows that after $\frac{154}{7}=22$ steps, there are 512 numbers left.
In 22 steps, a total of $22 \times 8=176$ numbers are crossed out, and their sum is
$1+2+\cdots+176=88 \times ... | 904020 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the inverse function of $y=f(x+1)$ is
$$
\begin{array}{c}
y=f^{-1}(x+1) \text {, and } f(1)=4007 \text {. Then } \\
f(1998)=
\end{array}
$$ | II, 7.2010.
From $y=f^{-1}(x+1)$, we get $x+1=f(y)$, which means $x=f(y)-1$.
Thus, the inverse function of $y=f^{-1}(x+1)$ is $y=f(x)-1$. Therefore, $f(x+1)-f(x)=-1$.
Let $x=1,2, \cdots, 1997$, add up all the equations and simplify to get $f(1998)-f(1)=-1997$.
Hence, $f(1998)=2010$. | 2010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given the quadratic function
$$
y=a x^{2}+b x+c \geqslant 0(a<b) \text {. }
$$
Then the minimum value of $M=\frac{a+2 b+4 c}{b-a}$ is $\qquad$ | 11. 8.
From the conditions, it is easy to see that $a>0, b^{2}-4 a c \leqslant 0$.
Notice that
$$
\begin{array}{l}
M=\frac{a+2 b+4 c}{b-a}=\frac{a^{2}+2 a b+4 a c}{a(b-a)} \\
\geqslant \frac{a^{2}+2 a b+b^{2}}{a(b-a)} .
\end{array}
$$
Let $t=\frac{b}{a}$. Then $t>1$. Thus,
$$
\begin{array}{l}
M \geqslant \frac{a^{2}+2... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
16. (15 points) As shown in Figure 2, it is known that the ellipse $C$ passes through the point $M(2,1)$, with the two foci at $(-\sqrt{6}, 0)$ and $(\sqrt{6}, 0)$. $O$ is the origin, and a line $l$ parallel to $OM$ intersects the ellipse $C$ at two different points $A$ and $B$.
(1) Find the maximum value of the area o... | 16. (1) Let the equation of the ellipse $C$ be
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$
From the problem, we have
$$
\left\{\begin{array} { l }
{ a ^ { 2 } - b ^ { 2 } = 6 , } \\
{ \frac { 4 } { a ^ { 2 } } + \frac { 1 } { b ^ { 2 } } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a^{2... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
17. (15 points) Given the function
$$
f(x)=\frac{1}{2} m x^{2}-2 x+1+\ln (x+1)(m \geqslant 1) \text {. }
$$
(1) If the curve $C: y=f(x)$ has a tangent line $l$ at point $P(0,1)$ that intersects $C$ at only one point, find the value of $m$;
(2) Prove that the function $f(x)$ has a decreasing interval $[a, b]$, and find ... | 17. (1) Note that the domain of the function $f(x)$ is $(-1,+\infty)$,
$$
f^{\prime}(x)=m x-2+\frac{1}{x+1}, f^{\prime}(0)=-1 .
$$
Therefore, the slope of the tangent line $l$ at the point of tangency $P(0,1)$ is -1.
Thus, the equation of the tangent line is $y=-x+1$.
Since the tangent line $l$ intersects the curve $C... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given real numbers $a, b (a \neq b)$, and they satisfy
$$
\begin{array}{l}
(a+1)^{2}=3-3(a+1), \\
3(b+1)=3-(b+1)^{2} .
\end{array}
$$
Then the value of $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}$ is ( ).
(A) 23
(B) -23
(C) -2
(D) -13 | 【Analysis】Transform the known two equations into
$$
\begin{array}{l}
(a+1)^{2}+3(a+1)-3=0, \\
(b+1)^{2}+3(b+1)-3=0 .
\end{array}
$$
It can be seen that $a$ and $b$ are the two roots of the equation with respect to $x$
$$
(x+1)^{2}+3(x+1)-3=0
$$
Solution: From equation (1) in the analysis, simplifying and rearranging ... | -23 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 2 Let real numbers $s, t$ satisfy
$$
\begin{array}{l}
19 s^{2}+99 s+1=0, \\
t^{2}+99 t+19=0(s t \neq 1) . \\
\text { Find the value of } \frac{s t+4 s+1}{t} \text { . }
\end{array}
$$
(1999, National Junior High School Mathematics Competition) | 【Analysis】Transform the first equation of the known equations, and combine it with the second equation to find that $\frac{1}{s} 、 t(s t \neq 1)$ are the two roots of the quadratic equation
$$
x^{2}+99 x+19=0
$$
Since $s \neq 0$, the first equation can be transformed into
$$
\left(\frac{1}{s}\right)^{2}+99\left(\frac{... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, for a $7 \times 7$ grid of small squares, how many of the small squares can be shaded such that no two shaded squares are adjacent? Adjacent means they share a side.
How many of the small squares can be shaded? | Four, color at most 26 small squares.
The coloring in Figure 2 satisfies the conditions.
The following proves: At most 26 small squares can be colored.
First, according to the problem, for any $2 \times 2$ square grid, at most two of the small squares can be colored; for a $3 \times 3$ square grid, at most 5 of the s... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given 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? | Prompt: Example 6. From the problem, we know that $x$ and $y$ are the two real roots of the equation
$$
t^{2}-(z-1) t+z^{2}-7 z+14=0
$$
By the discriminant $\Delta \geqslant 0$, we get
$$
3 z^{2}-26 z+55 \leqslant 0 \text {. }
$$
Solving this, we get $\frac{11}{3} \leqslant z \leqslant 5$.
$$
\begin{array}{l}
\text {... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
the integer part.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Let
then $2<\sqrt[3]{23}=a_{1}<3, a_{n+1}=\sqrt[3]{23+a_{n}}$.
By mathematical induction, it is easy to prove
$$
2<a_{n}<3\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
In fact,
the integer part is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given real numbers $x, y$ satisfy
$$
x^{2}+3 y^{2}-12 y+12=0 \text {. }
$$
then the value of $y^{x}$ is $\qquad$ | Hint: Treat the known equation as a quadratic equation in $y$ (the main variable)
$$
3 y^{2}-12 y+\left(12+x^{2}\right)=0 \text {. }
$$
From $\Delta=-12 x^{2} \geqslant 0$, and since $x^{2} \geqslant 0$, then $x=0$.
Thus, $y=2$. Therefore, $y^{x}=2^{0}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given real numbers $a, b, c$ satisfy
$$
a=2 b+\sqrt{2}, a b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 \text {. }
$$
Then $\frac{b c}{a}=$ $\qquad$ . | Hint: Eliminate $a$ from the two known equations, then treat $c$ as a constant, to obtain a quadratic equation in $b$
$$
2 b^{2}+\sqrt{2} b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 .
$$
From the discriminant $\Delta \geqslant 0$, we get $c=0$.
Therefore, $\frac{b c}{a}=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If real numbers $x, y$ satisfy
$$
\begin{array}{l}
\frac{x}{3^{3}+4^{3}}+\frac{y}{3^{3}+6^{3}}=1, \\
\frac{x}{5^{3}+4^{3}}+\frac{y}{5^{3}+6^{3}}=1,
\end{array}
$$
then $x+y=$ $\qquad$
(2005, National Junior High School Mathematics Competition) | 【Analysis】It is easy to notice that the denominators of the two given equations contain $3^{3}$ and $5^{3}$ respectively. Therefore, we can consider $3^{3}$ and $5^{3}$ as the roots of a certain equation.
Solution From the given conditions, it is easy to see that $3^{3}$ and $5^{3}$ are the roots of the equation in ter... | 432 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given
$\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$, and $a \neq 0$.
Then $\frac{b+c}{a}=$ $\qquad$ | From the given equation, we have
$$
(b-c)^{2}-4(a-b)(c-a)=0 \text {. }
$$
When $a \neq b$, by the discriminant of a quadratic equation, we know that the quadratic equation in $x$
$$
(a-b) x^{2}+(b-c) x+(c-a)=0
$$
has two equal real roots.
$$
\text { Also, }(a-b)+(b-c)+(c-a)=0 \text {, thus, }
$$
$x=1$ is a root of eq... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (15 points) Can 2010 be written as the sum of squares of $k$ distinct prime numbers? If so, try to find the maximum value of $k$; if not, please briefly explain the reason. | If 2010 can be written as the sum of squares of $k$ prime numbers, taking the sum of the squares of the smallest 10 distinct primes, then
$$
\begin{array}{l}
4+9+25+49+121+169+ \\
289+361+529+841 \\
=2397>2010 .
\end{array}
$$
Therefore, $k \leqslant 9$.
It is easy to see that there is only one even prime number 2, an... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. If the function $f(x)=\ln \frac{\mathrm{e} x}{\mathrm{e}-x}$, then $\sum_{k=1}^{2010} f\left(\frac{k e}{2011}\right)=$ $\qquad$ . | 13.2010.
Notice
$$
\begin{array}{l}
f(x)+f(\mathrm{e}-x) \\
=\ln \left[\frac{\mathrm{e} x}{\mathrm{e}-x} \cdot \frac{\mathrm{e}(\mathrm{e}-x)}{\mathrm{e}-(\mathrm{e}-x)}\right]=2 . \\
\text { Therefore } \sum_{k=1}^{2010} f\left(\frac{k \mathrm{e}}{2011}\right) \\
=\sum_{k=1}^{1005}\left(f\left(\frac{k \mathrm{e}}{201... | 2010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
19. (15 points) In a certain grade, $n$ students participate in Chinese and
Mathematics exams, with scores ranging from 0 to 100 for each subject. Suppose no two students have exactly the same scores (i.e., at least one subject score is different). Additionally, “A is better than B” means that student A’s scores in bo... | 19. Establish a Cartesian coordinate system $x O y$.
If a student's Chinese score is $i$ points, and mathematics score is $j$ points, let it correspond to the integer point $(i, j)$ on the plane, called a "score point". Thus, the examination results of $n$ students are mapped to $n$ score points within the range
$$
0 ... | 401 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given points $M(0,2)$ and $N(-3,6)$, the distances from these points to line $l$ are $1$ and $4$, respectively. The number of lines $l$ that satisfy these conditions is . $\qquad$ | 2.3.
It is easy to get $M N=5$.
Then the circle $\odot M$ with radius 1 is externally tangent to the circle $\odot N$ with radius 4. Therefore, there are 3 lines $l$ that satisfy the condition. | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{n+1}=\frac{1+a_{n}}{1-a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Let $T_{n}=a_{1} a_{2} \cdots a_{n}$. Then $T_{2010}=$ $\qquad$ | 5. -6 .
It is easy to get $a_{1}=2, a_{2}=-3, a_{3}=-\frac{1}{2}, a_{4}=\frac{1}{3}$, $a_{1} a_{2} a_{3} a_{4}=1$.
Also, $a_{5}=2=a_{1}$, by induction it is easy to know $a_{n+4}=a_{n}\left(n \in \mathbf{N}_{+}\right)$.
Therefore, $T_{2010}=T_{4 \times 502+2}=a_{1} a_{2}=-6$. | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$ | Given $a=\frac{-1+\sqrt{5}}{2}$, we can obtain the corresponding quadratic equation $a^{2}+a-1=0$.
Thus, the original expression is
$$
\begin{array}{l}
=\frac{\left(a^{2}+a-1\right)\left(a^{3}-a\right)-2(a-1)}{a\left(a^{2}-1\right)} \\
=\frac{-2}{a^{2}+a}=-2 .
\end{array}
$$ | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Given that $a$ and $b$ are integers, and satisfy $a-b$ is a prime number, $ab$ is a perfect square. If $a \geqslant 2011$, find the minimum value of $a$.
| Three, let $a-b=m$ (where $m$ is a prime number) and $ab=n^2$ (where $n$ is a positive integer).
$$
\begin{array}{l}
\text { From }(a+b)^{2}-4ab=(a-b)^{2} \\
\Rightarrow(2a-m)^{2}-4n^{2}=m^{2} \\
\Rightarrow(2a-m+2n)(2a-m-2n)=m^{2} \times 1 .
\end{array}
$$
Since $2a-m+2n$ and $2a-m-2n$ are both positive integers, and... | 2025 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the sequence $\left\{x_{n}\right\}$ :
$$
1,3,3,3,5,5,5,5,5, \cdots
$$
formed by all positive odd numbers arranged from smallest to largest, and each odd number $k(k=1,3,5, \cdots)$ appears consecutively $k$ times. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$, then $a+b+c+d=$ $\qqu... | 4.3.
For $k^{2}+1 \leqslant n \leqslant(k+1)^{2}$, $x_{n}=2 k+1, k=[\sqrt{n-1}]$,
therefore, $x_{n}=2[\sqrt{n-1}]+1$.
Thus, $(a, b, c, d)=(2,1,-1,1)$.
So $a+b+c+d=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. $\sum_{i=0}^{50} \sum_{j=0}^{50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}$ modulo 31 is
$\qquad$ . | 6.1.
Original expression $=\sum_{i=0}^{50} C_{50}^{i} \cdot \sum_{j=0}^{50} C_{50}^{j}=\left(2^{50}\right)^{2}=2^{100}$.
And $2^{5} \equiv 1(\bmod 31)$, so
Original expression $=2^{100} \equiv 1(\bmod 31)$.
Therefore, the remainder is 1. | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. There are 20 teams participating in the national league. Question: What is the minimum number of matches that must be played so that in any group of three teams, at least two teams have played against each other? | Let team $A$ have the minimum number of matches (which is $k$ matches). Then
(1) There are $k$ teams that have played against team $A$, and each of these teams has played at least $k$ matches;
(2) There are $19-k$ teams that have not played against team $A$, and these teams must all have played against each other, othe... | 90 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Among the integers $1,2, \cdots, 2011$, the number of integers that can be expressed in the form $[x[x]]$ is $\qquad$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. | 2. 1.990.
Let $x=k+a, k=[x], 0 \leqslant a<1$. Then $[x[x]]=[k(k+a)]=k^{2}+[k a]$.
Therefore, $k^{2}, k^{2}+1, \cdots, k^{2}+k-1(k=1,2$, $\cdots, 44)$ can all be expressed in the form of $[x[x]]$. Hence, the number of integers that meet the requirement is
$$
1+2+\cdots+44=990 \text { (numbers). }
$$ | 990 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. In a certain kingdom, there are 32 knights, some of whom are servants to other knights. Each servant can have at most one master, and each master must be richer than any of his servants. If a knight has at least four servants, he is ennobled as a noble. If it is stipulated that a servant of $A$'s servant is not a se... | 2.7.
According to the problem, the richest knight is not a servant of any other knight, so at most 31 knights can be servants of other knights.
Since each noble has at least four servants, there can be at most 7 nobles.
Number the 32 knights as $1,2, \cdots, 32$, with wealth decreasing as the number increases, then
... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $x_{1}=x_{2011}=1$, $\left|x_{n+1}\right|=\left|x_{n}+1\right|(n=1,2, \cdots, 2010)$.
Then $x_{1}+x_{2}+\cdots+x_{2010}=$ $\qquad$ | 3. -1005 .
From the given, it is easy to obtain
$$
x_{n+1}^{2}=x_{n}^{2}+2 x_{n}+1(n=1,2, \cdots, 2010) \text {. }
$$
By summing up and organizing these 2010 equations, we get
$$
\begin{array}{l}
x_{2011}^{2}=2\left(x_{1}+x_{2}+\cdots+x_{2010}\right)+2011 . \\
\text { Therefore, } x_{1}+x_{2}+\cdots+x_{2010}=-1005 .
... | -1005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Among the positive integers not greater than 100, the ordered integer pairs $(m, n)$ that satisfy
$$
\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n}
$$
are $\qquad$ pairs. | 4. 170 .
Notice that $\sqrt{2} n-1<m<\sqrt{2}(n+1)$.
For each $n$, the number of $m$ is given by
$$
\begin{array}{l}
{[\sqrt{2}(n+1)]-[\sqrt{2} n-1]} \\
=[\sqrt{2}(n+1)]-[\sqrt{2} n]+1
\end{array}
$$
Since $100<71 \sqrt{2}<101<72 \sqrt{2}$, we have $n \leqslant 71$.
But when $n=71$, $m=100$.
Therefore, the number of ... | 170 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) When $1<x<5$, find the minimum value of the algebraic expression
$$
\frac{\sqrt{(7 x+1)(5-x)(x-1)(7 x+5)}}{x^{2}}
$$ | $$
\begin{array}{l}
\frac{\sqrt{(7 x+1)(5-x)(x-1)(7 x+5)}}{x^{2}} \\
=\sqrt{\frac{\left(7 x^{2}-2 x-5\right)\left(-7 x^{2}+34 x+5\right)}{x^{4}}} \\
=\sqrt{\left(7 x-\frac{5}{x}-2\right)\left(-7 x+\frac{5}{x}+34\right)} \text {. } \\
\end{array}
$$
Let $t=7 x-\frac{5}{x}$. Then
$$
\begin{array}{l}
\text { Equation (1)... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) Given 15 quadratic equations $x^{2}-p_{i} x+q_{i}=0(i=1,2, \cdots, 15)$ with coefficients $p_{i} 、 q_{i}$ taking values from $1,2, \cdots, 30$, and these coefficients are all distinct. If an equation has a root greater than 20, it is called a "good equation." Find the maximum number of good equations. | Second, if there exists an equation $x^{2}-p_{i} x+q_{i}=0$ with two roots $x_{1}, x_{2}$, then from $x_{1}+x_{2}=p_{i}>0, x_{1} x_{2}=q_{i}>0$, we know that both roots are positive.
Next, if a certain equation has a root greater than 20, then $p_{i}=x_{1}+x_{2}>20$. However, among the numbers $1,2, \cdots, 30$, there... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) As shown in Figure 2, given points $A$ and $B$ are two distinct points outside circle $\odot O$, point $P$ is on $\odot O$, and $PA$, $PB$ intersect $\odot O$ at points $D$ and $C$ respectively, different from point $P$, and $AD \cdot AP = BC \cdot BP$.
(1) Prove: $\triangle OAB$ is an isosceles tria... | (1) Draw $A T$ tangent to $\odot O$ at point $T$, and connect $O T$.
Let the radius of $\odot O$ be $R$. Then
$$
A D \cdot A P=A T^{2}=O A^{2}-R^{2} \text {. }
$$
Similarly, $B C \cdot B P=O B^{2}-R^{2}$.
From the given, it is easy to see that $O A=O B$.
(2) From the proof in (1), we know
$$
p(2 p+1)=(m-1)^{2}-9 \text... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the inequality
$$
3 x+4 \sqrt{x y} \leqslant a(x+y)
$$
holds for all positive numbers $x$ and $y$. Then the minimum value of the real number $a$ is . $\qquad$ | -1.4 .
From the problem, we know that $a \geqslant\left(\frac{3 x+4 \sqrt{x y}}{x+y}\right)_{\text {max }}$.
$$
\text { Also, } \frac{3 x+4 \sqrt{x y}}{x+y} \leqslant \frac{3 x+(x+4 y)}{x+y}=4 \text {, }
$$
with equality holding if and only if $x=4 y>0$.
Therefore, the minimum value of $a$ is 4. | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the equation in $x$
$$
x^{2}+(a-2010) x+a=0 \quad (a \neq 0)
$$
has two integer roots. Then the value of the real number $a$ is $\qquad$. | 3.4024.
Let the roots of the equation be $x_{1} 、 x_{2}\left(x_{1} \leqslant x_{2}\right)$.
By Vieta's formulas, we have
$$
x_{1}+x_{2}=-(a-2010), x_{1} x_{2}=a \text {. }
$$
Then $x_{1} x_{2}+x_{1}+x_{2}=2010$, which means
$$
\left(x_{1}+1\right)\left(x_{2}+1\right)=2011 \text {. }
$$
Since 2011 is a prime number, ... | 4024 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. $[x]$ is the greatest integer not exceeding the real number $x$. It is known that the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{3}{2}, a_{n+1}=a_{n}^{2}-a_{n}+1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then $m=\left[\sum_{k=1}^{2011} \frac{1}{a_{k}}\right]$ is . $\qquad$ | 4.1.
$$
\begin{array}{l}
\text { Given } a_{n+1}=a_{n}^{2}-a_{n}+1 \\
\Rightarrow a_{n+1}-1=a_{n}\left(a_{n}-1\right) \\
\Rightarrow \frac{1}{a_{n+1}-1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}} . \\
\text { Then } \frac{1}{a_{n}}=\frac{1}{a_{n}-1}-\frac{1}{a_{n+1}-1} . \\
\text { Therefore } m=\left[\frac{1}{a_{1}-1}-\frac{1}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) In a certain drill activity, the drill leader arranged $n$ students, numbered $1 \sim n (n>3)$, in a circular formation, and they performed a $1-2-3$ cyclic counting. The drill leader recorded the numbers of the students who reported, and required students who reported 1 and 2 to leave the formation,... | Three, the numbers recorded by the leader are $a_{1}$, $a_{2}, \cdots, a_{n}, a_{n+1}, \cdots, a_{m}$, with their sum being $S_{n}$. Thus,
(1) After each 1-2-3 counting cycle, the total number of students decreases by 2, but the sum of the numbers remains unchanged.
(2) If $n=3^{k}$, after $3^{k-1}$ 1-2-3 counting cycl... | 16011388 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given $a, b, c > 0, a^{2}+b^{2}+c^{2}=14$. Prove: $a^{5}+\frac{b^{5}}{8}+\frac{c^{5}}{27} \geqslant 14$. | Prove the construction of the $3 \times 5$ matrix
$$
\left(\begin{array}{ccccc}
a^{5} & a^{5} & 1 & 1 & 1 \\
\frac{b^{5}}{8} & \frac{b^{5}}{8} & 4 & 4 & 4 \\
\frac{c^{5}}{27} & \frac{c^{5}}{27} & 9 & 9 & 9
\end{array}\right) .
$$
Using Carleman's inequality, we get
$$
\begin{array}{l}
{\left[\left(a^{5}+\frac{b^{5}}{8... | 14 | Inequalities | proof | Yes | Yes | cn_contest | false |
4. If the product of the first 2011 positive integers
$$
1 \times 2 \times \cdots \times 2011
$$
can be divided by $2010^{k}$, then the maximum value of the positive integer $k$ is | 4.30.
$$
2010=2 \times 3 \times 5 \times 67 \text {. }
$$
In $1 \times 2 \times \cdots \times 2011$, the exponent of 67 is
$$
\left[\frac{2011}{67}\right]+\left[\frac{2011}{67^{2}}\right]+\cdots=30 \text {. }
$$
Obviously, the exponents of $2,3,5$ in $1 \times 2 \times \cdots \times 2011$ are all greater than 30.
The... | 30 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given integers $p$ and $q$ satisfy $p+q=2010$, and the quadratic equation $67 x^{2}+p x+q=0$ has two positive integer roots. Then $p=$ $\qquad$ . | 7. -2278 .
Let the two positive integer roots of the equation be $x_{1}, x_{2}\left(x_{1} \leqslant x_{2}\right)$.
Then $x_{1}+x_{2}=-\frac{p}{67}, x_{1} x_{2}=\frac{q}{67}$.
Thus, $x_{1} x_{2}-x_{1}-x_{2}=\frac{p+q}{67}=\frac{2010}{67}=30$
$$
\begin{array}{l}
\Rightarrow\left(x_{1}-1\right)\left(x_{2}-1\right)=31 \\
... | -2278 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Place the nine digits $1,2, \cdots, 9$ into the nine small squares in Figure 4, so that the seven three-digit numbers $\overline{a b c} 、 \overline{d e f} 、 \overline{g h i} 、 \overline{a d g} 、 \overline{b e h} 、 \overline{c f i}$ and $\overline{a e i}$ are all divisible by 11. Find the maximum value of the three-... | 12. According to the problem, for modulo 11 we have
$$
\begin{array}{l}
a+c \equiv b, d+f \equiv e, g+i \equiv h, a+g \equiv d, \\
b+h \equiv e, c+i \equiv f, a+i \equiv e . \\
\text { Then }(a+c)+(d+f)+(g+i)+(b+h)+e \\
\equiv b+h+3 e \equiv 4 e(\bmod 11) .
\end{array}
$$
The left side of the above equation is
$$
1+2+... | 734 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. A natural number is called a "good number" if it is exactly 2007 more than the sum of its digits. Then the sum of all good numbers is $\qquad$ . | 5.20145.
Let $f(n)=n-S(n)$ ($S(n)$ is the sum of the digits of the natural number $n$). Then the function $f(n)$ is a non-strictly increasing function, and
$$
\begin{array}{l}
f(2009)<f(2010) \\
=f(2011)=\cdots=f(2019)=2007 \\
<f(2020) .
\end{array}
$$
Therefore, there are only 10 natural numbers that satisfy the con... | 20145 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the sum of 12 distinct positive integers is 2010. Then the maximum value of the greatest common divisor of these positive integers is . $\qquad$ | $-1.15$.
Let the greatest common divisor be $d$, and the 12 numbers be $a_{1} d$, $a_{2} d, \cdots, a_{12} d$, where $\left(a_{1}, a_{2}, \cdots, a_{12}\right)=1$.
Let $S=\sum_{i=1}^{12} a_{i}$. Then $2010=S d$.
To maximize $d$, $S$ should be minimized.
Since $a_{1}, a_{2}, \cdots, a_{12}$ are distinct, then
$S \geqsla... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n}+a_{n+1}=-n^{2} \text {. }
$$
then $a_{15}=$ $\qquad$ | 7. -104 .
Rewrite $a_{n}+a_{n+1}=-n^{2}$ as
$$
\left(a_{n}+\frac{n^{2}}{2}-\frac{n}{2}\right)+\left[a_{n+1}+\frac{(n+1)^{2}}{2}-\frac{n+1}{2}\right]=0 \text {. }
$$
Let $b_{n}=a_{n}+\frac{n^{2}}{2}-\frac{n}{2}$. Then
$$
b_{1}=1 \text {, and } b_{n+1}=-b_{n} \text {. }
$$
Therefore, $b_{2 k-1}=1, b_{2 k}=-1(k=1,2, \c... | -104 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. If a four-digit number $n$ contains at most two different digits among its four digits, then $n$ is called a "simple four-digit number" (such as 5555 and 3313). Then, the number of simple four-digit numbers is | 8. 576.
If the four digits of a four-digit number are all the same, then there are nine such four-digit numbers.
If the four digits of a four-digit number have two different values, the first digit \( a \in \{1,2, \cdots, 9\} \) has 9 possible choices. After choosing \( a \), select \( b \in \{0,1, \cdots, 9\} \) and... | 576 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and
$$
f(x)=f(1-x) \text {. }
$$
Then $f(2010)=$ $\qquad$ . | $$
\begin{array}{l}
f(0)=0, \\
f(x+1)=f(-x)=-f(x) .
\end{array}
$$
From the conditions, we have $f(x+2)=f(x)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2010)=f(0)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Given real numbers $x, y$ satisfy
$$
3|x+1|+2|y-1| \leqslant 6 \text {. }
$$
Then the maximum value of $2 x-3 y$ is $\qquad$ . | 10.4.
The figure determined by $3|x+1|+2|y-1| \leqslant 6$ is the quadrilateral $ABCD$ and its interior, where,
$$
A(-1,4) 、 B(1,1) 、 C(-1,-2) 、 D(-3,1) \text {. }
$$
By the knowledge of linear programming, the maximum value of $2x-3y$ is 4.
The maximum value can be achieved when $x=-1, y=-2$. | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
12. The capacity of a set refers to the sum of its elements. Then the total capacity of all non-empty sets $A$ that satisfy the condition “ $A \subseteq\{1,2, \cdots, 7\}$, and if $a \in A$ then $8-a \in A$ ” is
(Answer with a specific number).
| 12.224.
First, find the single-element and two-element sets that satisfy the conditions:
$$
A_{1}=\{4\}, A_{2}=\{1,7\}, A_{3}=\{2,6\}, A_{4}=\{3,5\} .
$$
Then, any combination of elements from these four sets also meets the requirements.
Therefore, the total sum of elements in all sets $A$ that satisfy the condition... | 224 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3: There is an $8 \times 8$ chessboard, and at the start, each of the 64 small squares contains a "castle" chess piece. If a castle chess piece can attack an odd number of other castle chess pieces still on the board, it is removed. Question: What is the maximum number of castle chess pieces that can be removed (a cast... | First, prove that the pieces on the four corners of the chessboard will not be taken away. Since a castle on a corner can attack at most two other castles, if this castle is taken away, it means that the castle can only attack one. Without loss of generality, we can assume that the castle in the upper left corner is th... | 59 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the sequence $\left\{8 \times\left(-\frac{1}{3}\right)^{n-1}\right\}$ have the sum of its first $n$ terms as $S_{n}$. Then the smallest integer $n$ that satisfies the inequality
$$
\left|S_{n}-6\right|<\frac{1}{125}
$$
is | 2. 7 | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. For $0<x<1$, if the complex number
$$
z=\sqrt{x}+\mathrm{i} \sqrt{\sin x}
$$
corresponds to a point, then the number of such points inside the unit circle is $n=$ | 6. 1 .
From the point on the unit circle, we have
$$
x+\sin x=1(00(x \in(0,1))$, which means $\varphi(x)$ is a strictly increasing function.
Also, $\varphi(0)=-10$, so the equation $x+\sin x=1$ has exactly one real root in $(0,1)$.
Therefore, $n=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010}$.
Then $m=$ $\qquad$ $(\bmod 2011)$. | 8. 0 .
Notice
$$
\begin{array}{l}
\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010} \\
=\left(\frac{1}{1}+\frac{1}{2010}\right)+\left(\frac{1}{2}+\frac{1}{2009}\right)+\cdots+ \\
\left(\frac{1}{1005}+\frac{1}{1006}\right) \\
= \frac{2011}{1 \times 2010}+\frac{2011}{2 \times 2009}+\cdots+\frac{2011}{1005 \times 1006} \\... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Take the subset $A_{i}=\left\{a_{i}, a_{i+1}, \cdots, a_{i+59}\right\}(i=1,2, \cdots$, 70 ) of the set $S=\left\{a_{1}, a_{2}, \cdots, a_{70}\right\}$, where $a_{70+i}=a_{i}$. If there exist $k$ sets among $A_{1}, A_{2}, \cdots, A_{70}$ such that the intersection of any seven of them is non-empty, fin... | Given:
$$
A_{1} \cap A_{2} \cap \cdots \cap A_{60}=\left\{a_{60}\right\} \text {, }
$$
Furthermore, the intersection of any seven sets from $A_{1}, A_{2}, \cdots, A_{\infty}$ is non-empty, hence $k \geqslant 60$.
We will now prove that if $k>60$, it cannot be guaranteed that the intersection of any seven sets from th... | 60 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Consider the matrix
$$
\left(a_{i j}\right)_{n \times n}\left(a_{i j} \in\{1,2,3\}\right) \text {. }
$$
If $a_{i j}$ is such that its row $i$ and column $j$ both contain at least three elements (including $a_{i j}$) that are equal to $a_{i j}$, then the element $a_{i j}$ is called "good". If the matrix $\left(a_{i j}\... | The minimum value of $n$ is 7.
When $n=6$, the matrix
$\left(\begin{array}{llllll}1 & 1 & 2 & 2 & 3 & 3 \\ 1 & 1 & 3 & 3 & 2 & 2 \\ 2 & 2 & 1 & 1 & 3 & 3 \\ 2 & 2 & 3 & 3 & 1 & 1 \\ 3 & 3 & 2 & 2 & 1 & 1 \\ 3 & 3 & 1 & 1 & 2 & 2\end{array}\right)$
has no good elements, so $n \geqslant 7$.
Below, we use proof by contrad... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
410 translators are invited to an international mathematics conference. Each translator is proficient in exactly two of the five languages: Greek, Slovenian, Vietnamese, Spanish, and German, and no two translators are proficient in the same pair of languages. The translators are to be assigned to five rooms, with two t... | Construct a graph $G$, where points $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ represent five languages, and any two points (such as $x_{1}, x_{2}$) are connected by an edge $(x_{1} x_{2})$ representing a translator proficient in these two languages.
By the problem statement, graph $G$ is a simple complete graph.
Next, orient... | 144 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. A. Let $a=\sqrt{7}-1$. Then the value of the algebraic expression $3 a^{3}+12 a^{2}-$ $6 a-12$ is ( ).
(A) 24
(B) 25
(C) $4 \sqrt{7}+10$
(D) $4 \sqrt{7}+12$ | ,- 1. A. A.
Notice
$$
a=\sqrt{7}-1 \Rightarrow a^{2}+2 a-6=0 \text {. }
$$
Then $3 a^{3}+12 a^{2}-6 a-12$
$$
=\left(a^{2}+2 a-6\right)(3 a+6)+24=24 .
$$ | 24 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. B. Given that $\angle A, \angle B$ are two acute angles, and satisfy
$$
\begin{array}{l}
\sin ^{2} A+\cos ^{2} B=\frac{5}{4} t, \\
\cos ^{2} A+\sin ^{2} B=\frac{3}{4} t^{2} .
\end{array}
$$
Then the sum of all possible real values of $t$ is ().
(A) $-\frac{8}{3}$
(B) $-\frac{5}{3}$
(C) 1
(D) $\frac{11}{3}$ | 3. B. C.
Adding the two equations yields $3 t^{2}+5 t=8$.
Solving gives $t=1, t=-\frac{8}{3}$ (discard).
When $t=1$, $\angle A=45^{\circ}, \angle B=30^{\circ}$ satisfies the given equations.
Therefore, the sum of all possible values of the real number $t$ is 1. | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false |
5. A. Let $S=\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{99^{3}}$. Then the integer part of $4 S$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 5. A. A.
When $k=2,3, \cdots, 99$, we have
$$
\frac{1}{k^{3}}<\frac{1}{k\left(k^{2}-1\right)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right] \text {, }
$$
Thus, $1<S=1+\sum_{k=2}^{29} \frac{1}{k^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99 \times 100}\right)<\frac{5}{4} \text {. }
$$
Therefore, $... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
6. B. Given that the lengths of the two legs are integers $a$ and $b$ $(b<2011)$. Then the number of right triangles with the hypotenuse length $b+1$ is | 6. B. 31.
By the Pythagorean theorem, we have
$$
a^{2}=(b+1)^{2}-b^{2}=2 b+1 \text{. }
$$
Given $b<2011$, we know that $a$ is an odd number in the interval $(1, \sqrt{4023})$, so $a$ must be $3, 5, \cdots, 63$.
Therefore, there are 31 right-angled triangles that satisfy the conditions. | 31 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. A. As shown in Figure 1, points $A$ and $B$ lie on the line $y=x$. Parallel lines to the $y$-axis through $A$ and $B$ intersect the hyperbola $y=\frac{1}{x} (x>0)$ at points $C$ and $D$. If $BD=2AC$, then the value of $4OC^2-OD^2$ is $\qquad$ | 8. A. 6 .
Let points $C(a, b), D(c, d)$.
Then points $A(a, a), B(c, c)$.
Since points $C, D$ are on the hyperbola $y=\frac{1}{x}$, we have $a b=1, c d=1$.
Given $B D=2 A C$
$$
\begin{array}{l}
\Rightarrow|c-d|=2|a-b| \\
\Rightarrow c^{2}-2 c d+d^{2}=4\left(a^{2}-2 a b+b^{2}\right) \\
\Rightarrow 4\left(a^{2}+b^{2}\rig... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. A. As shown in Figure 3, in the right triangle $\triangle ABC$, the hypotenuse $AB$ is 35 units long, and the square $CDEF$ is inscribed in $\triangle ABC$ with a side length of 12. Then the perimeter of $\triangle ABC$ is $\qquad$ | 10. A. 84.
Let $BC = a, AC = b$. Then,
$$
a^{2} + b^{2} = 35^{2} = 1225.
$$
Since Rt $\triangle AFE \sim \text{Rt} \triangle ACB$, we have,
$$
\frac{FE}{CB} = \frac{AF}{AC} \Rightarrow \frac{12}{a} = \frac{b-12}{b}.
$$
Thus, $12(a + b) = ab$.
From equations (1) and (2), we get
$$
\begin{array}{l}
(a + b)^{2} = a^{2}... | 84 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. B. Let the four-digit number
$\overline{a b c d}$ satisfy
$$
a^{3}+b^{3}+c^{3}+d^{3}+1=10 c+d .
$$
Then the number of such four-digit numbers is $\qquad$ | 10. B. 5 .
From $d^{3} \geqslant d$ we know $c^{3}+1 \leqslant 10 c \Rightarrow 1 \leqslant c \leqslant 3$.
If $c=3$, then $a^{3}+b^{3}+d^{3}=2+d$.
Thus, $d=1$ or 0. Therefore, $a=b=1$, which gives us $1131, 1130$ as solutions.
If $c=2$, then $a^{3}+b^{3}+d^{3}=11+d$. Thus, $d \leqslant 2$.
When $d=2$, $a^{3}+b^{3}=5$... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. B. If five pairwise coprime distinct integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are randomly selected from $1,2, \cdots, n$, and one of these integers is always a prime number, find the maximum value of $n$. | 13. B. When $n \geqslant 49$, take the integers $1, 2^{2}, 3^{2}, 5^{2}, 7^{2}$. These five integers are five pairwise coprime distinct integers, but none of them are prime.
When $n=48$, in the integers $1, 2, \cdots, 48$, take any five pairwise coprime distinct integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. If $a_{1},... | 48 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $f(x)$ be a polynomial with integer coefficients, $f(0)=11$, and there exist $n$ distinct integers $x_{1}, x_{2}, \cdots, x_{n}$, such that
$$
f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010 .
$$
Then the maximum value of $n$ is | 7.3.
Let $g(x)=f(x)-2010$. Then $x_{1}, x_{2}, \cdots, x_{n}$ are all roots of $g(x)=0$.
Thus, $g(x)=\prod_{i=1}^{n}\left(x-x_{i}\right) \cdot q(x)$,
where $q(x)$ is a polynomial with integer coefficients. Therefore,
$$
\begin{array}{l}
g(0)=11-2010=-1999 \\
=\prod_{i=1}^{n}\left(-x_{i}\right) q(0) .
\end{array}
$$
... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. (14 points) As shown in Figure 2, the corridor is 3 m wide, the angle between the corridors is 120°, the ground is level, and the ends of the corridor are sufficiently long. Question: What is the maximum length of a horizontal rod (neglecting its thickness) that can pass through the corridor? | II. 9. As shown in Figure 4, draw any horizontal line through the inner vertex $P$ of the corridor corner, intersecting the outer sides of the corridor at points $A$ and $B$. The length of a wooden rod that can pass through the corridor in a horizontal position is less than or equal to $AB$.
Let $\angle B A Q=\alpha$. ... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. (16 points) Given a positive integer $n$ that satisfies the following condition: for each positive integer $m$ in the open interval $(0,2009)$, there always exists a positive integer $k$, such that
$$
\frac{m}{2009}<\frac{k}{n}<\frac{m+1}{2010} \text {. }
$$
Find the minimum value of such $n$. | 12. Notice
$$
\begin{array}{l}
\frac{m}{2009}2010 k
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
m n+1 \leqslant 2009 k, \\
m n+n-1 \geqslant 2010 k
\end{array}\right. \\
\Rightarrow 2009(m n+n-1) \geqslant 2009 \times 2010 k \\
\geqslant 2010(m n+1) \\
\Rightarrow 2009 m n+2009 n-2009 \\
\geqslant 2010 m ... | 4019 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $k_{1}<k_{2}<\cdots<k_{n}$ be non-negative integers, satisfying $2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{n}}=227$.
Then $k_{1}+k_{2}+\cdots+k_{n}=$ $\qquad$ | - 1. 19.
Notice that
$$
\begin{array}{l}
227=1+2+32+64+128 \\
=2^{0}+2^{1}+2^{5}+2^{6}+2^{7} .
\end{array}
$$
Therefore, $k_{1}+k_{2}+\cdots+k_{n}=0+1+5+6+7=19$. | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a>0$, the graphs of the functions $f(x)=|x+2a|$ and $g(x)=|x-a|$ intersect at point $C$, and they intersect the $y$-axis at points $A$ and $B$ respectively. If the area of $\triangle ABC$ is 1, then $a=$ $\qquad$ . | 2. 2 .
From the graphs of $f(x)$ and $g(x)$, we know that $\triangle ABC$ is an isosceles right triangle with base $a$, so its area is $\frac{a^{2}}{4}=1$. Therefore, $a=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the function $y=x^{3}$, the tangent line at $x=a_{k}$ intersects the $x$-axis at point $a_{k+1}$. If $a_{1}=1, S_{n}=\sum_{i=1}^{n} a_{i}$, then $\lim _{n \rightarrow \infty} S_{n}$ $=$ . $\qquad$ | 4.3.
It is known that $y^{\prime}=3 x^{2}$.
Therefore, the equation of the tangent line to $y=x^{3}$ at $x=a_{k}$ is $y-a_{k}^{3}=3 a_{k}^{2}\left(x-a_{k}\right)$.
Thus, the above equation intersects the $x$-axis at the point $\left(a_{k+1}, 0\right)$. Hence, $-a_{k}^{3}=3 a_{k}^{2}\left(a_{k+1}-a_{k}\right)$.
From th... | 3 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
5. The function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies for all $x, y, z \in \mathbf{R}$
$$
f(x+y)+f(y+z)+f(z+x) \geqslant 3 f(x+2 y+z) .
$$
Then $f(1)-f(0)=$ $\qquad$ | 5.0.
Let $x=-y=z$, we get
$$
f(2 x) \geqslant f(0) \Rightarrow f(1) \geqslant f(0) \text {. }
$$
Let $x=y=-z$, we get
$$
f(0) \geqslant f(2 x) \Rightarrow f(0) \geqslant f(1) \text {. }
$$
Thus, $f(1)-f(0)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. (18 points) Let $S$ be a set of distinct quadruples $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$, where $a_{i}=0$ or 1 $(i=1,2,3,4)$. It is known that the number of elements in $S$ does not exceed 15, and satisfies:
if $\left(a_{1}, a_{2}, a_{3}, a_{4}\right) 、\left(b_{1}, b_{2}, b_{3}, b_{4}\right) \in S$, then $\lef... | 12. Clearly, there are 16 possible quadruples. Since at least one quadruple is not in $S$, it follows that $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, $(0,0,0,1)$ must have at least one that is not in $S$. Otherwise, by the given conditions, all quadruples would be in $S$.
Assume $(1,0,0,0) \notin S$.
In this case, by the g... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let the sequence of rational numbers $\left\{a_{n}\right\}$ be defined as follows:
$a_{k}=\frac{x_{k}}{y_{k}}$, where $x_{1}=y_{1}=1$, and
if $y_{k}=1$, then $x_{k+1}=1, y_{k+1}=x_{k}+1$;
if $y_{k} \neq 1$, then $x_{k+1}=x_{k}+1, y_{k+1}=y_{k}-1$.
How many terms in the first 2011 terms of this sequence ... | 11. The sequence is
$$
\frac{1}{1}, \frac{1}{2}, \frac{2}{1}, \frac{1}{3}, \frac{2}{2}, \frac{3}{1}, \cdots, \frac{1}{k}, \frac{2}{k-1}, \cdots, \frac{k}{1}, \cdots \text {. }
$$
Group it as follows:
$$
\begin{array}{l}
\left(\frac{1}{1}\right),\left(\frac{1}{2}, \frac{2}{1}\right),\left(\frac{1}{3}, \frac{2}{2}, \fra... | 213 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) A dance troupe has $n(n \geqslant 5)$ actors, and they have arranged some performances, each of which is performed by four actors on stage. In one performance, they found that: it is possible to appropriately arrange several performances so that every two actors in the troupe perform on stage togethe... | Three, use $n$ points to represent $n$ actors.
If two actors have performed on the same stage once, then connect the corresponding points with an edge. Thus, the condition of this problem is equivalent to:
Being able to partition the complete graph $K_{n}$ of order $n$ into several complete graphs $K_{4}$ of order 4, ... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. From the numbers $1,2, \cdots, 2014$, what is the maximum number of numbers that can be selected such that none of the selected numbers is 19 times another? | According to the problem, if $k$ and $19 k$ cannot both appear in $1,2, \cdots, 2014$, and since $2014=19 \times 106$, and $106=5 \times 19+11$, then choose
$$
1,2,3,4,5,106,107, \cdots, 2013 \text {, }
$$
These 1913 numbers satisfy the requirement.
The numbers not chosen are $6,7, \cdots, 105,2014$, a total of 101 nu... | 1913 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $T \subseteq\{1,2, \cdots, 25\}$. If for any two distinct elements $a, b (a \neq b)$ in $T$, their product $ab$ is not a perfect square, find the maximum number of elements in $T$, and the number of subsets $T$ that satisfy this condition. | Prompt: Divide the set $\{1,2, \cdots, 25\}$ into several subsets such that the product of any two elements in the same subset is a perfect square, and the product of any two elements in different subsets is not a perfect square.
To maximize the number of elements in $T$, let
$$
\begin{array}{l}
A_{1}=\{1,4,9,16,25\}, ... | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Find the smallest positive integer $n$ such that for any $n$ integers, there exist at least two numbers whose sum or difference is divisible by 1991.
(1991, Australian Mathematical Olympiad) | Let $M=\left\{a_{i} \mid a_{i}=0,1, \cdots, 995\right\}$.
Since $a_{i}+a_{j} \leqslant 995+994=1989<1991$, $0<\left|a_{i}-a_{j}\right| \leqslant 995$,
then the sum and difference of any two numbers in $M$ are not multiples of 1991.
Therefore, $n \geqslant 997$.
Let $a_{1}, a_{2}, \cdots, a_{997}$ be any 997 integers.
I... | 997 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Find the units digit of $\left(7^{2004}+36\right)^{818}$.
$(2004$, Shanghai Jiao Tong University Independent Recruitment Examination) | $$
\begin{array}{l}
\text { Solution: Since } 7^{4} \equiv 1(\bmod 10) \\
\Rightarrow 7^{2004} \equiv 1(\bmod 10) \\
\Rightarrow 7^{2004}+36 \equiv 7(\bmod 10) .
\end{array}
$$
$$
\text { Therefore, the original expression } \equiv 7^{818} \equiv 7^{2} \equiv 9(\bmod 10) \text {. }
$$
Thus, the unit digit is 9. | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Let $u$ be a root of the equation
$$
x^{3}-3 x+10=0
$$
Let $f(x)$ be a quadratic polynomial with rational coefficients, and
$$
\alpha=\frac{1}{2}\left(u^{2}+u-2\right), f(\alpha)=u .
$$
Find $f(0)$.
(2010, Five Schools Joint Examination for Independent Enrollment) | From the given, we have $u^{3}-3 u+10=0$. Then
$$
\begin{aligned}
\alpha^{2} &=\frac{1}{4}\left(u^{2}+u-2\right)^{2} \\
&=\frac{1}{4}\left(u^{4}+2 u^{3}-3 u^{2}-4 u+4\right) \\
&=\frac{1}{4}\left[u(3 u-10)+2(3 u-10)-3 u^{2}-4 u+4\right] \\
&=-4-2 u .
\end{aligned}
$$
Let $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{Q}, a \... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given real numbers $x, y, z$ satisfy
$$
x y z=32, x+y+z=4 \text {. }
$$
Then the minimum value of $|x|+|y|+|z|$ is $\qquad$ .
(2010, Hubei Province High School Mathematics Competition) | Answer: 12.
The text above has been translated into English, maintaining the original text's line breaks and format. Here is the direct output of the translation result. | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given positive integers $a, b$ satisfy
$$
|b-2|+b-2=0,|a-b|+a-b=0,
$$
and $a \neq b$. Then the value of $a b$ is $\qquad$ . | 2.1.2.
From the given conditions, we know that $a>0, b-2 \leqslant 0, a-b \leqslant 0$. Therefore, $a<b \leqslant 2$. Hence, $a=1, b=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the side lengths of a trapezoid are $3,4,5,6$. Then the area of this trapezoid is $\qquad$ . | 3. 18.
First, determine the lengths of the two bases. As shown in Figure 7, let trapezoid $ABCD$ have $AD$ and $BC$ as the upper and lower bases, respectively. Draw $AE \parallel CD$. Then $BE$ is the difference between the upper and lower bases.
In $\triangle ABE$, $AB - AE = AB - CD < BE = BC - AD$. Therefore, $AB ... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 3, in $\triangle ABC$, it is given that $D$ is a point on side $BC$ such that $AD = AC$, and $E$ is the midpoint of side $AD$ such that $\angle BAD = \angle ACE$. If $S_{\triangle BDE} = 1$, then $S_{\triangle ABC}$ is $\qquad$. | 4.4.
Notice that
$$
\begin{array}{l}
\angle E C D=\angle A C D-\angle A C E \\
=\angle A D C-\angle B A D=\angle A B C .
\end{array}
$$
Therefore, $\triangle E C D \backsim \triangle A B C$.
$$
\text { Then } \frac{C D}{B C}=\frac{E D}{A C}=\frac{E D}{A D}=\frac{1}{2} \text {. }
$$
Thus, $D$ is the midpoint of side ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (20 points) For a certain project, Team A alone needs 12 days to complete, and Team B alone needs 9 days to complete. If the two teams are scheduled to work in whole days, how many schemes are there to ensure that the project is completed within 8 days? | Three, let teams A and B work for $x, y$ days respectively to just meet the requirements. Then we have
$$
\left\{\begin{array}{l}
\frac{x}{12}+\frac{y}{9}=1, \\
0 \leqslant x \leqslant 8, \\
0 \leqslant y \leqslant 8,
\end{array}\right.
$$
and $x, y$ are both integers.
From $\frac{x}{12}+\frac{y}{9}=1$, we get $x=12-y... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the polynomial $f(x)$ satisfy: for any $x \in \mathbf{R}$, we have
$$
f(x+1)+f(x-1)=2 x^{2}-4 x .
$$
Then the minimum value of $f(x)$ is $\qquad$ | - 1. - -2 .
Since $f(x)$ is a polynomial, we know that $f(x+1)$ and $f(x-1)$ have the same degree as $f(x)$. Therefore, $f(x)$ is a quadratic polynomial (let's assume $f(x) = ax^2 + bx + c$). Then,
$$
\begin{array}{l}
f(x+1) + f(x-1) \\
= 2ax^2 + 2bx + 2(a + c) = 2x^2 - 4x.
\end{array}
$$
Thus, $a = 1, b = -2, c = -1... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (25 points) For a set $M=\left\{p_{1}, p_{2}, \cdots, p_{2_{n}}\right\}$ consisting of $2 n$ prime numbers, its elements can be paired to form $n$ products, resulting in an $n$-element set. If
$$
\begin{aligned}
A & =\left\{a_{1} a_{2}, a_{3} a_{4}, \cdots, a_{2 n-1} a_{2 n}\right\} \\
\text { and } \quad B & =\lef... | 11. Six elements can form fifteen different "slips of paper," listed as follows:
$$
\begin{array}{l}
\{a b, c d, e f\},\{a b, c e, d f\},\{a b, c f, d e\}, \\
\{a c, b d, e f\},\{a c, b e, d f\},\{a c, b f, d e\}, \\
\{a d, b c, e f\},\{a d, b e, c f\},\{a d, b f, c e\}, \\
\{a e, b c, d f\},\{a e, b d, c f\},\{a e, b ... | 60 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. If real numbers $x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
(x-1)^{2011}+(x-1)^{2009}+2010 x=4020, \\
(y-1)^{2011}+(y-1)^{2009}+2010 y=0,
\end{array}\right.
$$
then $x+y=$ $\qquad$ . | 2. 2 .
Transform the original system of equations into
$$
\begin{array}{l}
\left\{\begin{array}{l}
(x-1)^{2011}+(x-1)^{2009}+2010(x-1)=2010, \\
(y-1)^{2011}+(y-1)^{2009}+2010(y-1)=-2010 .
\end{array}\right. \\
\text { Let } f(t)=t^{2011}+t^{2009}+2010 t(t \in \mathbf{R}) .
\end{array}
$$
Then the function $f(t)$ is a... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $a, b, c \in \mathbf{R}$, and
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \text {, }
$$
then there exists an integer $k$, such that the following equations hold for
$\qquad$ number of them.
(1) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1... | 3. 2 .
From the given equation, we have
$$
\begin{array}{l}
\frac{(b c+a c+a b)(a+b+c)-a b c}{a b c(a+b+c)}=0 . \\
\text { Let } P(a, b, c) \\
=(b c+a c+a b)(a+b+c)-a b c \\
=(a+b)(b+c)(c+a) .
\end{array}
$$
From $P(a, b, c)=0$, we get
$$
a=-b \text { or } b=-c \text { or } c=-a \text {. }
$$
Verification shows that... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let two fixed points in the plane be $A(-3,0)$ and $B(0,-4)$, and let $P$ be any point on the curve $y=\frac{12}{x}(x>0)$. Draw $PC \perp x$-axis and $PD \perp y$-axis, with the feet of the perpendiculars being $C$ and $D$, respectively. Then the minimum value of $S_{\text{quadrilateral } ACD}$ is | 4. 24.
Notice that
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \\
=2\left(x+\frac{9}{x}\right)+12 \geqslant 24 .
\end{array}
$$
The equality holds if and only if $x=3$. Therefore, the minimum value of $S_{\text {quadrilateral } A B C D}$ is 24. | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $k_{1}, k_{2}, \cdots, k_{n}$ are $n$ distinct positive integers, and satisfy $\sum_{i=1}^{n} k_{i}^{3}=2024$. Then the maximum value of the positive integer $n$ is $\qquad$ . | 6. 8 .
From the problem, we know that when $n \geqslant 9$, we have
$$
\sum_{i=1}^{n} k_{i}^{3} \geqslant \sum_{i=1}^{n} i^{3} \geqslant \sum_{i=1}^{9} i^{3}=2025>2024,
$$
which is a contradiction. Therefore, $n \leqslant 8$.
Also, $2^{3}+3^{3}+\cdots+9^{3}=2024$, so the maximum value of the positive integer $n$ is 8... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $m, n$ be given positive integers. In each square of an $m \times n$ chessboard, fill a number according to the following rule: first, fill the numbers in the 1st row and the 1st column arbitrarily, then, for any other square, let the number filled in it be $x$, and the number in the 1st row in the same column a... | 6.1.
Let the cell at the $i$-th row and $j$-th column of the chessboard be denoted as $a_{i j}$, and the number filled in cell $a_{i j}$ is also represented by $a_{i j}$.
Consider any rectangle on the chessboard, and let the four corner cells of this rectangle be $a_{i j}, a_{i t}, a_{s j}, a_{s t} (i<s, j<t)$.
By th... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 There are three types of goods, A, B, and C. If you buy 3 pieces of A, 7 pieces of B, and 1 piece of C, it costs a total of 315 yuan; if you buy 4 pieces of A, 10 pieces of B, and 1 piece of C, it costs a total of 420 yuan. Question: How much would it cost to buy one piece each of A, B, and C? | Let the unit prices of A, B, and C be $x$, $y$, and $z$ yuan, respectively. Then, according to the problem, we have
$$
\left\{\begin{array}{l}
3 x+7 y+z=315, \\
4 x+10 y+z=420 .
\end{array}\right.
$$
The problem actually only requires finding the value of $x+y+z$, without necessarily solving for $x$, $y$, and $z$ indi... | 105 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $p$ is a prime number greater than 5, and $m$ is the smallest non-negative remainder when $\left(p^{2}+5 p+5\right)^{2}$ is divided by 120. Then the units digit of $2009^{m}$ is $\qquad$ | Notice that
$$
\begin{array}{l}
\left(p^{2}+5 p+5\right)^{2}=\left[\left(p^{2}+5 p+5\right)^{2}-1\right]+1 \\
=\left(p^{2}+5 p+6\right)\left(p^{2}+5 p+4\right)+1 \\
=(p+1)(p+2)(p+3)(p+4)+1 .
\end{array}
$$
Let $M=(p+1)(p+2)(p+3)(p+4)$.
Clearly, $5! \mid p M$.
Since $p$ is a prime number greater than 5, 120 divides $M$... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. For $\triangle A B C$, squares are constructed outward on its three sides $a, b, c$, with their areas denoted as $S_{a}, S_{b}, S_{c}$ respectively. If $a+b+c=18$, then the minimum value of $S_{a}+S_{b}+S_{c}$ is $\qquad$ | 2. 108.
$$
\begin{array}{l}
\text { Given }(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0 \\
\Rightarrow 2\left(a^{2}+b^{2}+c^{2}\right) \geqslant 2(a b+b c+c a) \\
\Rightarrow 3\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2} \\
\Rightarrow a^{2}+b^{2}+c^{2} \geqslant \frac{18^{2}}{3}=108 . \\
\text { Therefore, }\left(... | 108 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. If the real numbers $x, y, z, w$ satisfy
$$
\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}=1, \\
\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}
$$
then $x^{2... | 3. 36 .
It is known that $2^{2}$ and $4^{2}$ are the two roots of the equation with respect to $t$:
$$
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}=1
$$
which means they are the two roots of the equation:
$$
t^{2}-\left(1^{2}+3^{2}+x^{2}+y^{2}\right) t+1^{2} \times 3^{2}+3^{2} x^{2}+1^{2} \times y^{2}=0
$$
Therefore,... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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