problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
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8. The constant term in the expansion of $\left(x^{2}+x-\frac{1}{x}\right)^{6}$ is $\qquad$ (answer with a specific number). | 8. -5 .
From the conditions, the constant term is
$$
C_{6}^{2}(-1)^{4}+C_{6}^{3}(-1)^{3}=-5 .
$$ | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Let the sequence $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, satisfying $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$. Then the value of $S_{20}$ is $\qquad$. | 9. 0 .
From the condition, we know $a_{1}=\frac{\left(a_{1}+1\right)^{2}}{4}$, solving this gives $a_{1}=1$.
When $n \geqslant 2$, from $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$, we know
$$
S_{n-1}=\frac{\left(a_{n-1}+1\right)^{2}}{4} \text {. }
$$
Then $a_{n}=\frac{1}{4}\left(a_{n}+1\right)^{2}-\frac{1}{4}\left(a_{... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Let the function $f(x)=\sin x+\sqrt{3} \cos x+1$.
(1) Find the maximum and minimum values of the function $f(x)$ on $\left[0, \frac{\pi}{2}\right]$;
(2) If real numbers $a$, $b$, and $c$ satisfy
$$
a f(x)+b f(x-c)=1
$$
for any $x \in \mathbf{R}$, find the value of $\frac{b \cos c}{a}$. | Three, 13. (1) From the given conditions,
$$
f(x)=2 \sin \left(x+\frac{\pi}{3}\right)+1 \text {. }
$$
From $0 \leqslant x \leqslant \frac{\pi}{2} \Rightarrow \frac{\pi}{3} \leqslant x+\frac{\pi}{3} \leqslant \frac{5 \pi}{6}$.
Thus, $\frac{1}{2} \leqslant \sin \left(x+\frac{\pi}{3}\right) \leqslant 1$.
Therefore, when ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $\triangle A B C$ is an isosceles right triangle, $\angle A$ $=90^{\circ}$, and $\overrightarrow{A B}=a+b, \overrightarrow{A C}=a-b$.
If $a=(\cos \theta, \sin \theta)(\theta \in \mathbf{R})$, then $S_{\triangle A B C}$ $=$ . $\qquad$ | 4. 1.
From the problem, we know that $A B \perp A C,|A B|=|A C|$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
(a+b) \cdot(a-b)=0, \\
|a+b|=|a-b|
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
|a|=|b|, \\
a \cdot b=0 .
\end{array}\right.
\end{array}
$$
Since $|a|=1$, we have $|b|=1$.
According ... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. In a regular tetrahedron $ABCD$, $AO \perp$ plane $BCD$, with the foot of the perpendicular being $O$. Let $M$ be a point on the line segment $AO$ such that $\angle BMC=90^{\circ}$. Then $\frac{AM}{MO}=$ $\qquad$ | 5. 1.
As shown in Figure 3, connect $O B$. Let the edge length of the regular tetrahedron $A B C D$ be $a$. Then
$$
\begin{array}{c}
O B=\frac{\sqrt{3}}{3} a, \\
M B=\frac{\sqrt{2}}{2} a . \\
\text { Therefore, } M O=\sqrt{M B^{2}-O B^{2}} \\
=\frac{\sqrt{6}}{6} a=\frac{1}{2} A O=A M . \\
\text { Hence, } \frac{A M}{M... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $p$ and $q$ be two different prime numbers. Then the remainder when $p^{q-1}+q^{p-1}$ is divided by $p q$ is $\qquad$ | 8. 1 .
Since $p$ and $q$ are different prime numbers, by Fermat's Little Theorem, we have
$$
p^{q-1} \equiv 1(\bmod q) \text {. }
$$
Also, $q^{p-1} \equiv 0(\bmod q)$, thus
$$
p^{q-1}+q^{p-1} \equiv(\bmod q) \text {. }
$$
Similarly, $p^{q-1}+q^{p-1} \equiv 1(\bmod p)$.
Therefore, $p^{q-1}+q^{p-1} \equiv 1(\bmod p q)... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 4, in the "dart-shaped" quadrilateral $ABCD$, $AB=4\sqrt{3}$, $BC=8$, $\angle A=\angle B=\angle C=30^{\circ}$. Then the distance from point $D$ to $AB$ is $\qquad$ | 3. 1.
As shown in Figure 9, extend $A D$ to intersect $B C$ at point $E$. Then $A E=B E$.
Draw $E F \perp A B$ at point $F$.
It is easy to see that $A F=B F=2 \sqrt{3}$,
$$
E F=2, A E=4 \text {. }
$$
Thus, $C E=4$.
Also, $\angle A D C=90^{\circ}$
$\Rightarrow D E=2$
$\Rightarrow D$ is the midpoint of $A E$
$\Rightarr... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Given that $f(x)$ is a function defined on the set of real numbers $\mathbf{R}$, $f(0)=2$, and for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(5+2 x)=f(-5-4 x), \\
f(3 x-2)=f(5-6 x) .
\end{array}
$$
Find the value of $f(2012)$. | In equation (1), let $x=\frac{1}{2} y-\frac{3}{2}(y \in \mathbf{R})$, we get $f(2+y)=f(1-2 y)$.
In equation (2), let $x=\frac{1}{3} y+\frac{2}{3}(y \in \mathbf{R})$, we get $f(y)=f(1-2 y)$.
Thus, $f(2+y)=f(y)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2012)=f(2 \times 1006+0)=f(0)=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let $x, y$ be real numbers. Then the minimum value of the algebraic expression
$$
2 x^{2}+4 x y+5 y^{2}-4 x+2 y-5
$$
is $\qquad$ [1]
(2005, National Junior High School Mathematics League Wuhan CASIO Cup Selection Competition) | $$
\begin{array}{l}
\text { Original expression }= x^{2}+4 x y+4 y^{2}+x^{2}-4 x+4+ \\
y^{2}+2 y+1-10 \\
=(x+2 y)^{2}+(x-2)^{2}+(y+1)^{2}-10 .
\end{array}
$$
Therefore, when $x=2, y=-1$, the minimum value of the required algebraic expression is -10. | -10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $m>0$. If the function
$$
f(x)=x+\sqrt{100-m x}
$$
has a maximum value of $g(m)$, find the minimum value of $g(m)$.
(2011, National High School Mathematics League Sichuan Province Preliminary Contest) | First, use the discriminant method to find $g(m)$.
Notice that the original function is $y-x=\sqrt{100-m x}$. Squaring both sides and rearranging, we get
$$
x^{2}+(m-2 y) x+y^{2}-100=0 \text {. }
$$
By $\Delta \geqslant 0$, we have $y \leqslant \frac{m}{4}+\frac{100}{m}$.
Thus, $g(m)=\frac{m}{4}+\frac{100}{m}$
$$
\geq... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given that $a$ and $b$ are real numbers, and
$$
a^{2}+a b+b^{2}=3 \text {. }
$$
If the maximum value of $a^{2}-a b+b^{2}$ is $m$, and the minimum value is $n$, find the value of $m+n$. ${ }^{[3]}$
$(2008$, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | Let $a^{2}-a b+b^{2}=k$.
From $\left\{\begin{array}{l}a^{2}+a b+b^{2}=3 \\ a^{2}-a b+b^{2}=k,\end{array}\right.$, we get $a b=\frac{3-k}{2}$.
Thus, $(a+b)^{2}=\left(a^{2}+a b+b^{2}\right)+a b$
$=3+\frac{3-k}{2}=\frac{9-k}{2}$.
Since $(a+b)^{2} \geqslant 0$, then $\frac{9-k}{2} \geqslant 0$, which means $k \leqslant 9$.... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 If real numbers $x, y$ satisfy $|x|+|y| \leqslant 1$, then the maximum value of $x^{2}-x y+y^{2}$ is $\qquad$ [4]
(2010, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | Solve: Completing the square for $x^{2}-x y+y^{2}$ yields
$$
\begin{array}{l}
x^{2}-x y+y^{2}=\frac{1}{4}(x+y)^{2}+\frac{3}{4}(x-y)^{2} . \\
\text { Also, }|x \pm y| \leqslant|x|+|y| \leqslant 1, \text { then } \\
x^{2}-x y+y^{2} \leqslant \frac{1}{4}+\frac{3}{4}=1 .
\end{array}
$$
When $x$ and $y$ are such that one i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given that $x_{1}, x_{2}, \cdots, x_{6}$ are six different positive integers, taking values from $1,2, \cdots, 6$. Let
$$
\begin{aligned}
S= & \left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\left|x_{3}-x_{4}\right|+ \\
& \left|x_{4}-x_{5}\right|+\left|x_{5}-x_{6}\right|+\left|x_{6}-x_{1}\right| .
\end{alig... | Since equation (1) is a cyclic expression about $x_{1}, x_{2}, \cdots, x_{6}$, we can assume $x_{1}=6, x_{j}=1(j \neq 1)$. Then
$$
\begin{aligned}
S \geqslant & \left|\left(6-x_{2}\right)+\left(x_{2}-x_{3}\right)+\cdots+\left(x_{j-1}-1\right)\right|+ \\
& \left|\left(x_{j+1}-1\right)+\left(x_{j+2}-x_{j+1}\right)+\cdots... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the function $S=|x-2|+|x-4|$.
(1) Find the minimum value of $S$;
(2) If for any real numbers $x, y$ the inequality
$$
S \geqslant m\left(-y^{2}+2 y\right)
$$
holds, find the maximum value of the real number $m$. | (1) Using the absolute value inequality, the answer is 2.
(2) From the problem, we know that for any real number $y$,
$$
m\left(-y^{2}+2 y\right) \leqslant 2
$$
holds. It is easy to find that the maximum value of $-y^{2}+2 y$ is 1. Therefore, $0 \leqslant m \leqslant 2$, and the maximum value of $m$ is 2. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) A fly and $k$ spiders are placed at some intersections of a $2012 \times 2012$ grid. An operation consists of the following steps: first, the fly moves to an adjacent intersection or stays in place, then each spider moves to an adjacent intersection or stays in place (multiple spiders can occupy the s... | The minimum value of $k$ is 2.
(1) First, prove that a single spider cannot catch the fly.
Establish a Cartesian coordinate system, then the range of the spider and fly's movement is
$$
\{(x, y) \mid 0 \leqslant x, y \leqslant 2012, x, y \in \mathbf{N}\}.
$$
For each point in the above set, there are at least two poi... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given the sets
$$
\begin{array}{l}
M=\{(x, y) \mid x(x-1) \leqslant y(1-y)\}, \\
N=\left\{(x, y) \mid x^{2}+y^{2} \leqslant k\right\} .
\end{array}
$$
If $M \subset N$, then the minimum value of $k$ is $\qquad$ .
(2007, Shanghai Jiao Tong University Independent Admission Examination) | Notice that,
$$
M=\left\{(x, y) \left\lvert\,\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2}\right.\right\}
$$
represents a disk with center $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{\sqrt{2}}{2}$.
By $M \subset N \Rightarrow \sqrt{k} \geqslant \sqrt{2} \times \frac{... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Find the largest positive integer $k$ such that the set of positive integers can be partitioned into $k$ subsets $A_{1}, A_{2}, \cdots, A_{k}$, so that for all integers $n(n \geqslant 15)$ and all $i \in\{1,2, \cdots, k\}$, there exist two distinct elements in $A_{i}$ whose sum is $n$. | 4. The largest positive integer $k$ is 3.
When $k=3$, let
$A_{1}=\{1,2,3\} \cup\{3 m \mid m \in \mathbf{Z}$, and $m \geqslant 4\}$,
$A_{2}=\{4,5,6\} \cup\{3 m-1 \mid m \in \mathbf{Z}$, and $m \geqslant 4\}$,
$A_{3}=\{7,8,9\} \cup\{3 m-21 m \in \mathbf{Z}$, and $m \geqslant 4\}$.
Then the sum of two different elements ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. (50 points) Let the sequence $\left\{x_{n}\right\}$ satisfy
$$
x_{1}=1, x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \text {. }
$$
Find the units digit of $x_{2012}$. | 7. Clearly, $x_{2}=7$, and for any positive integer $n, x_{n}$ is a positive integer.
By the property of the floor function, we have
$$
\begin{array}{l}
4 x_{n}+\sqrt{11} x_{n}>x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \\
>4 x_{n}+\sqrt{11} x_{n}-1 .
\end{array}
$$
Multiplying both sides of the above inequality by ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A country has $n(n \geqslant 3)$ cities and two airlines. There is exactly one two-way flight between every pair of cities, and this two-way flight is operated exclusively by one of the airlines. A female mathematician wants to start from a city, pass through at least two other cities (each city is visited only once... | 6. Solution 1 Consider each city as a vertex, each flight route as an edge, and each airline as a color. Then, the country's flight network can be seen as a complete graph with $n$ vertices whose edges are colored with two colors.
From the condition, we know that any cycle contains edges of both colors, meaning that t... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. At the beginning, there are 111 pieces of clay of equal weight on the table. Perform the following operations on the clay: First, divide a part or all of the clay into several groups, with the same number of pieces in each group, then knead the clay in each group into one piece. It is known that after $m$ operations... | 1. 2 .
Obviously, one operation can result in at most two different weights of clay blocks.
Below, we show that two operations can achieve the goal.
Assume without loss of generality that each block of clay initially weighs 1.
In the first operation, select 74 blocks and divide them into 37 groups, with two blocks in ... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let $f(x)$ represent a quartic polynomial in $x$. If $f(1)=f(2)=f(3)=0, f(4)=6$, $f(5)=72$, then the last digit of $f(2010)$ is $\qquad$. ${ }^{3}$
(2010, International Cities Mathematics Invitational for Youth) | 【Analysis】It is given in the problem that $1,2,3$ are three zeros of the function $f(x)$, so we consider starting from the zero point form of the function.
Solution From the problem, we know
$$
f(1)=f(2)=f(3)=0 \text {. }
$$
Let the one-variable quartic polynomial function be
$$
f(x)=(x-1)(x-2)(x-3)(a x+b) \text {, }
... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $f(x)$ be a polynomial with integer coefficients, $f(0)=$
11. There exist $n$ distinct integers $x_{1}, x_{2}, \cdots, x_{n}$, such that $f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010$.
Then the maximum value of $n$ is $\qquad$ (6)
$(2010$, Xin Zhi Cup Shanghai High School Mathem... | Let $g(x)=f(x)-2010$.
$$
\begin{array}{l}
\text { By } f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010, \text { we have } \\
g\left(x_{1}\right)=g\left(x_{2}\right)=\cdots=g\left(x_{n}\right)=0,
\end{array}
$$
That is, $x_{1}, x_{2}, \cdots, x_{n}$ are $n$ distinct integer roots of the integer-... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. On the edge $AS$ of the tetrahedron $S-ABC$, mark points $M, N$ such that $AM=MN=NS$. If the areas of $\triangle ABC$, $\triangle MBC$, and $\triangle SBC$ are $1$, $2$, and $\sqrt{37}$, respectively, find the area of $\triangle NBC$. | 8. Let the areas of $\triangle A B C$, $\triangle M B C$, $\triangle N B C$, $\triangle S B C$ be $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}$, and let $h_{1}$, $h_{2}$, $h_{3}$, $h_{4}$ be the heights from these triangles to the common base $B C$, as shown in Figure 2.
Points $A^{\prime}$, $B^{\prime}$, $C^{\prime}$, $S^{\prim... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. Given trapezoid $A B C D$ with bases $A D=3, B C=1$, the diagonals intersect at point $O$, two circles intersect base $B C$ at points $K, L$, these two circles are tangent at point $O$, and are tangent to line $A D$ at points $A, D$ respectively. Find $A K^{2}+D L^{2}$. | 10. As shown in Figure 4, draw the common tangent of the two circles through point $O$, intersecting the lower base $AD$ at point $P$. By the properties of tangents, we have
$$
A P=O P=D P \text {. }
$$
This indicates that $\triangle A O D$ is a right triangle and is similar to $\triangle C D B$, with the similarity r... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) Arrange all positive integers that satisfy the following conditions in descending order, denoted as $M$, and the $k$-th number as $b_{k}$: each number's any three consecutive digits form a non-zero perfect square. If $b_{16}-b_{20}=2^{n}$, find $n$.
Arrange all positive integers that satisfy the follow... | II. Notice that, among three-digit numbers, the perfect squares are $100, 121, 144, 169, 196, 225, 256, 289$, $324, 361, 400, 441, 484, 529, 576, 625$, $676, 729, 784, 841, 900, 961$.
For each number in $M$, consider its leftmost three digits.
(1) The six-digit number with the leftmost three digits 100 is 100169, the f... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. If five vertices of a regular nonagon are colored red, then there are at least $\qquad$ pairs of congruent triangles (each pair of triangles has different vertex sets) whose vertices are all red. | 8. 4 .
A triangle with both vertices colored red is called a "red triangle". Thus, there are $\mathrm{C}_{5}^{3}=10$ red triangles. For a regular nonagon, the triangles formed by any three vertices are of only seven distinct types (the lengths of the minor arcs of the circumcircle of the regular nonagon corresponding ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (40 points) Find the largest positive real number $\lambda$ such that for all positive integers $n$ and positive real numbers $a_{i} (i=1,2, \cdots, n)$, we have
$$
1+\sum_{k=1}^{n} \frac{1}{a_{k}^{2}} \geqslant \lambda\left[\sum_{k=1}^{n} \frac{1}{\left(1+\sum_{s=1}^{k} a_{s}\right)^{2}}\right] .
$$ | II. Define $S_{k}=\sum_{i=1}^{k} a_{i}+1$, and supplement the definition $S_{0}=1$. First, prove a lemma.
Lemma For any positive integer $k \geqslant 1$, we have $\frac{1}{a_{k}^{2}}+\frac{1}{S_{k-1}^{2}} \geqslant \frac{8}{S_{k}^{2}}$.
Proof Let $S_{k-1}=a_{k} t_{k}$. Then $S_{k}=a_{k}\left(t_{k}+1\right)$.
Thus, equa... | 7 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: For $n$ consecutive positive integers, if each number is written in its standard prime factorization form, and each prime factor is raised to an odd power, such a sequence of $n$ consecutive positive integers is called a "consecutive $n$ odd group" (for example, when $n=3$, $22=2^{1} \times 11^{1}$, $23=23^{... | 【Analysis】Notice that, in a connected $n$-singular group, if there exists a multiple of 4, then by the definition of a connected $n$-singular group, it must be a multiple of 8.
Let this number be $2^{k} A\left(k, A \in \mathbf{N}_{+}, k \geqslant 3, A\right.$ is an odd number). Then $2^{k} A+4$ and $2^{k} A-4$ are bot... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 The number of prime pairs \((a, b)\) that satisfy the equation
$$
a^{b} b^{a}=(2 a+b+1)(2 b+a+1)
$$
is \qquad (2]
(2011, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | 【Analysis】If $a$ and $b$ are both odd, then
the left side of equation (1) $\equiv 1 \times 1 \equiv 1(\bmod 2)$,
the right side of equation (1) $\equiv(2 \times 1+1+1)(2 \times 1+1+1)$ $\equiv 0(\bmod 2)$.
Clearly, the left side is not congruent to the right side $(\bmod 2)$, a contradiction.
Therefore, at least one of... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the smallest positive integer $n$, such that there exist rational-coefficient polynomials $f_{1}, f_{2}, \cdots, f_{n}$, satisfying
$$
x^{2}+7=f_{1}^{2}(x)+f_{2}^{2}(x)+\cdots+f_{n}^{2}(x) .
$$
(51st IMO Shortlist) | 【Analysis】For the case $n=5$,
$$
x^{2}+7=x^{2}+2^{2}+1+1+1 \text {, }
$$
it meets the requirement.
Now we prove that $n \leqslant 4$ does not meet the requirement.
Assume there exist four rational coefficient polynomials $f_{1}(x)$, $f_{2}(x)$, $f_{3}(x)$, $f_{4}(x)$ (which may include the zero polynomial), such that ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Find all real numbers $x$ such that $4 x^{5}-7$ and $4 x^{13}-7$ are both perfect squares. ${ }^{[6]}$
(2008, German Mathematical Olympiad) | 【Analysis】Let
$$
4 x^{5}-7=a^{2}, 4 x^{13}-7=b^{2}(a, b \in \mathbf{N}) \text {. }
$$
Then $x^{5}=\frac{a^{2}+7}{4}>1$ is a positive rational number, and $x^{13}=\frac{b^{2}+7}{4}$ is a positive rational number.
Therefore, $x=\frac{\left(x^{5}\right)^{8}}{\left(x^{13}\right)^{3}}$ is a positive rational number.
Let $x... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $a, b, c, d$ satisfy: for any real number $x$,
$a \cos x + b \cos 2x + c \cos 3x + d \cos 4x \leq 1$.
Find the maximum value of $a + b - c + d$ and the values of the real numbers $a, b, c, d$ at that time.
(Supplied by Li Shenghong) | 4. Let $f(x)=a \cos x+b \cos 2 x+$ $c \cos 3 x+d \cos 4 x$.
From $f(0)=a+b+c+d$,
$f(\pi)=-a+b-c+d$,
$f\left(\frac{\pi}{3}\right)=\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2}$,
then $a+b-c+d$
$$
=f(0)+\frac{2}{3} f(\pi)+\frac{4}{3} f\left(\frac{\pi}{3}\right) \leqslant 3 .
$$
Equality holds if and only if $f(0)=f(\pi)=f\left... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $m$ be a positive integer, $n=2^{m}-1$, and the set of $n$ points on the number line be $P_{n}=\{1,2, \cdots, n\}$.
A grasshopper jumps on these points, each step moving from one point to an adjacent point. Find the maximum value of $m$ such that for any $x, y \in P_{n}$, the number of ways to jump from point $... | 8. When $m \geqslant 11$, $n=2^{m}-1>2013$.
Since there is only one way to jump from point 1 to point 2013 in 2012 steps, this is a contradiction, so $m \leqslant 10$.
We will now prove that $m=10$ satisfies the condition.
We use mathematical induction on $m$ to prove a stronger proposition: $\square$
For any $k \geq... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Can 2010 be written as the sum of squares of $k$ distinct prime numbers? If so, find the maximum value of $k$; if not, please briefly explain the reason. | 提示: As the sum of the squares of the smallest 10 distinct prime numbers is
$$
\begin{array}{l}
4+9+25+49+121+169+289+361+529+841 \\
=2397>2010,
\end{array}
$$
thus, $k \leqslant 9$.
By analyzing the parity and the fact that the square of an odd number is congruent to 1 modulo 8, it is easy to prove that $k \neq 8, k \... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a$ is a constant, and real numbers $x, y, z$ satisfy
$$
(x-1)^{2}+(y-\sqrt{5})^{2}+(z+1)^{2}=a
$$
when, $-8 \leqslant 4 x-\sqrt{5} y+2 z \leqslant 2$. Then $a=$ $\qquad$ | 3. 1.
Let $4 x-\sqrt{5} y+2 z=k$. Then $-8 \leqslant k \leqslant 2$.
From the given equation, we have
$$
\frac{(4 x-4)^{2}}{16 a}+\frac{(-\sqrt{5} y+5)^{2}}{5 a}+\frac{(2 z+2)^{2}}{4 a}=1 \text {. }
$$
Using the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
(16 a+5 a+4 a)\left[\frac{(4 x-4)^{2}}{16 a}+\frac{(... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $x, y, z \in (0, \sqrt{2})$, and satisfying
$$
\left(2-x^{2}\right)\left(2-y^{2}\right)\left(2-z^{2}\right)=x^{2} y^{2} z^{2} \text{. }
$$
Then the maximum value of $x+y+z$ is | 4.3.
$$
\begin{array}{l}
\text { Let } x=\sqrt{2} \cos \alpha, y=\sqrt{2} \cos \beta, \\
z=\sqrt{2} \cos \gamma\left(\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)\right) \text {. }
\end{array}
$$
Then the given equation transforms to
$$
\tan \alpha \cdot \tan \beta \cdot \tan \gamma=1 \text {. }
$$
Assume wi... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $p$ and $5 p^{2}-2$ both be prime numbers: Find the value of $p$.
(2012, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | It is easy to prove that when $3 \times p$, $3 \mid \left(5 p^{2}-2\right)$.
Since $5 p^{2}-2>3$, thus $5 p^{2}-2$ is not a prime number, which contradicts the given condition.
Therefore, $3 \mid p$. Then $p=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-2 x-m=0$, and $2 x_{1}+x_{2}=0$. Then the value of $m$ is $\qquad$ . | By the relationship between roots and coefficients, we know $x_{1}+x_{2}=2$.
Also, $2 x_{1}+x_{2}=0$, then
$$
\begin{array}{l}
x_{1}+2=0 \\
\Rightarrow x_{1}=-2 \\
\Rightarrow(-2)^{2}-2 \times(-2)-m=0 \\
\Rightarrow m=8 .
\end{array}
$$ | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a, b$ be positive real numbers, $m$ be a positive integer, and satisfy
$$
\left\{\begin{array}{l}
a+b \leqslant 14, \\
a b \geqslant 48+m .
\end{array}\right.
$$
Then the value of $m$ is | 3.1.
Notice,
$$
\begin{array}{l}
14 \geqslant a+b \geqslant 2 \sqrt{a b} \geqslant 2 \sqrt{48+m} \\
\geqslant 2 \sqrt{48+1}=14 .
\end{array}
$$
Therefore, all equalities hold. Hence \( m=1 \). | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. In a ball game competition, there are eight teams participating, and each pair of teams has to play a match. A team gets 2 points for a win, 1 point for a draw, and 0 points for a loss. If a team wants to ensure it enters the top four (i.e., its points must exceed those of at least four other teams), then the minimu... | 4. 11.
Since there are eight teams, there will be $\frac{8 \times 7}{2}=28$ matches, totaling $28 \times 2=56$ points.
If the top five teams draw with each other and each win against the bottom three teams, and the bottom three teams draw with each other, then there will be five teams each with 10 points, and the oth... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $P$ be any point on the graph of the function $y=x+\frac{2}{x}(x>0)$, and draw perpendiculars from $P$ to the line $y=x$ and the $y$-axis, with the feet of the perpendiculars being $A$ and $B$ respectively. Then $\overrightarrow{P A} \cdot \overrightarrow{P B}=$ $\qquad$ | - 1. -1 .
Solution 1 Let $P\left(x_{0}, x_{0}+\frac{2}{x_{0}}\right)$.
Then $l_{P A}: y-\left(x_{0}+\frac{2}{x_{0}}\right)=-\left(x-x_{0}\right)$, which is $y=-x+2 x_{0}+\frac{2}{x_{0}}$.
Solving the above equation with $y=x$ yields point $A\left(x_{0}+\frac{1}{x_{0}}, x_{0}+\frac{1}{x_{0}}\right)$.
Also, point $B\lef... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $\triangle A B C$ have interior angles $\angle A, \angle B, \angle C$ with opposite sides $a, b, c$ respectively, and satisfy
$$
\begin{array}{l}
a \cos B-b \cos A=\frac{3}{5} c . \\
\text { Then } \frac{\tan A}{\tan B}=
\end{array}
$$ | 2.4.
Solution 1 From the given and the cosine rule, we have
$$
\begin{array}{l}
a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{5} c \\
\Rightarrow a^{2}-b^{2}=\frac{3}{5} c^{2} . \\
\text { Therefore, } \frac{\tan A}{\tan B}=\frac{\sin A \cdot \cos B}{\sin B \cdot \cos A}=\fra... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. For the parabola $y^{2}=2 p x(p>0)$, the focus is $F$, and the directrix is $l$. Points $A$ and $B$ are two moving points on the parabola, and they satisfy $\angle A F B=\frac{\pi}{3}$. Let $M$ be the midpoint of segment $A B$, and let $N$ be the projection of $M$ on $l$. Then the maximum value of $\frac{|M N|}{|A B... | 4. 1 .
Solution 1 Let $\angle A B F=\theta\left(0<\theta<\frac{2 \pi}{3}\right)$. Then by the Law of Sines, we have
$$
\frac{|A F|}{\sin \theta}=\frac{|B F|}{\sin \left(\frac{2 \pi}{3}-\theta\right)}=\frac{|A B|}{\sin \frac{\pi}{3}} .
$$
Thus, $\frac{|A F|+|B F|}{\sin \theta+\sin \left(\frac{2 \pi}{3}-\theta\right)}=... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let two regular tetrahedra $P-ABC$ and $Q-ABC$ be inscribed in the same sphere. If the dihedral angle between a lateral face and the base of the regular tetrahedron $P-ABC$ is $45^{\circ}$, then the tangent value of the dihedral angle between a lateral face and the base of the regular tetrahedron $Q-ABC$ is . $\qqua... | 5.4.
As shown in Figure 6, connect $P Q$. Then $P Q \perp$ plane $A B C$, with the foot of the perpendicular $H$ being the center of the equilateral $\triangle A B C$, and $P Q$ passing through the center of the sphere $O$.
Connect $C H$ and extend it to intersect $A B$ at point $M$. Then $M$ is the midpoint of side ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. The function $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)+\arcsin x$. Then the solution set of $f(x)+f\left(2-x^{2}\right) \leqslant 0$ is $\qquad$ | $-、 1 .\{-1\}$.
It is known that the function $f(x)$ is a monotonically increasing odd function defined on $[-1,1]$.
$$
\begin{array}{l}
\text { Given } f(x)+f\left(2-x^{2}\right) \leqslant 0 \\
\Rightarrow f(x) \leqslant f\left(x^{2}-2\right) \\
\Rightarrow-1 \leqslant x \leqslant x^{2}-2 \leqslant 1 \\
\Rightarro... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For the expression $\frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{2013}$, when written as a decimal, find the digit before the decimal point. | Let $a_{n}=\frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{n}-\frac{\sqrt{5}}{5}\left(\frac{1-\sqrt{5}}{2}\right)^{n}$.
Then $a_{1}=a_{2}=1, a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3)$.
The last digits are
$$
\begin{array}{l}
1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7 \\
7,4,1,5,6,1,7,8,5,3,8,1,9,0,9,9 \\
8,7,5,2,7,9,6,5,1,6,7,3,0... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 The family of sets $\Omega$ consists of 11 five-element sets $A_{1}, A_{2}$, $\cdots, A_{11}$, where the intersection of any two sets is not empty. Let $A=\bigcup_{i=1}^{11} A_{i}=\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, for any $x_{i} \in A$, the number of sets in $\Omega$ that contain the element $x_{i}... | It is known that $\sum_{i=1}^{n} k_{i}=55$.
Notice that, the $k_{i}$ sets containing $x_{i}$ form
$$
\mathrm{C}_{k_{i}}^{2}=\frac{k_{i}\left(k_{i}-1\right)}{2}
$$
pairs of sets. Since the intersection of any two sets is not empty, the sum $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2}$ includes all pairs of sets, with some re... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 When $n$ is any real number and $k$ is a certain specific integer, the equation
$$
n(n+1)(n+2)(n+3)+1=\left(n^{2}+k n+1\right)^{2}
$$
holds. Then $k=$ $\qquad$ . [1]
(2010, Taiyuan Junior High School Mathematics Competition) | 【Analysis】Since the left side of the given equation is a polynomial and the right side is in the form of a product, we only need to factorize the left side. The method of factorization is to use the overall idea and the complete square formula to handle it.
Solution Note that,
$$
\begin{array}{l}
n(n+1)(n+2)(n+3)+1 \\
... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 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:
(1) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}$;
(2... | 【Analysis】The condition equation in this problem is a fractional equation, which is relatively complex. The key to solving this problem is to simplify the relationship between the letters $a, b, c$. First, eliminate the denominator and rearrange the condition equation into the form $f(a, b, c)=0$, then factorize $f(a, ... | 2 | Algebra | proof | Yes | Yes | cn_contest | false |
3. The graph of the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at two points $A$ and $B$, with the vertex at $C$. If $\triangle A C B$ is a right triangle, then the value of the discriminant is $\qquad$. | 3. 4 .
As shown in Figure 4.
From the problem, we know $\Delta=b^{2}-4 a c>0$.
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right), C\left(-\frac{b}{2 a}, \frac{4 a c-b^{2}}{4 a}\right)$.
By Vieta's formulas, we have
$x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}$.
Then $\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. $a_{1}, a_{2}, a_{3}, \cdots$ is an arithmetic sequence, where $a_{1}>0, s_{n}$ represents the sum of the first $n$ terms. If $S_{3}=S_{11}$, in $S_{1}, S_{2}, S_{3}, \cdots$ the largest number is $S_{k}$, then $k=$ $\qquad$ . | -1.7 .
Let the common difference be $d$. Then
$$
\begin{array}{l}
a_{n}=a_{1}+(n-1) d . \\
\text { By } S_{3}=S_{11} \Rightarrow d=-\frac{2}{13} a_{1}<0 . \\
\text { Hence } a_{n}=a_{1}+(n-1)\left(-\frac{2}{13} a_{1}\right) \\
=\frac{a_{1}}{13}(15-2 n),
\end{array}
$$
and the largest positive integer $n$ for which $a_... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Consider a tangent line to the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$, which intersects the two symmetry axes of the ellipse at points $A$ and $B$. Then the minimum length of segment $AB$ is $\qquad$ . | 2. 8 .
Let the point of tangency be \( P(5 \cos \theta, 3 \sin \theta) \). Then the equation of the tangent line to the ellipse at point \( P \) is
$$
\frac{\cos \theta}{5} x + \frac{\sin \theta}{3} y = 1,
$$
which intersects the \( x \)-axis and \( y \)-axis at
$$
A\left(\frac{5}{\cos \theta}, 0\right), B\left(0, \f... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. In rectangle $A B C D$, it is known that $A B=2, B C=3$, $E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. Rotate $\triangle F A B$ $90^{\circ}$ around $E F$ to $\triangle F A^{\prime} B^{\prime}$. Then the volume of the tetrahedron $A^{\prime} B^{\prime} C D$ is $\qquad$ . | 3. 2 .
It is known that $E F=B C=3$, and the plane of $\triangle F A^{\prime} B^{\prime}$ divides the tetrahedron $A^{\prime} B^{\prime} C D$ into two tetrahedrons of equal volume, $C F A^{\prime} B^{\prime}$ and $D F A^{\prime} B^{\prime}$. Their heights are $C F=D F=1$, and $S_{\triangle F A^{\prime} B^{\prime}}=3$.... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find all real roots of the equation
$$
x^{2}-x+1=\left(x^{2}+x+1\right)\left(x^{2}+2 x+4\right)
$$
All real roots. ${ }^{[4]}$
(2011, International Invitational Competition for Young Mathematicians in Cities) | 【Analysis】The most basic method to solve higher-degree equations is to convert them into lower-degree equations for solving, that is, to handle them as linear or quadratic equations. Factorization is the most powerful tool to achieve such a transformation. The preferred method for factoring higher-degree polynomials is... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. For any $x, y \in [0,1]$, the function
$$
f(x, y)=x \sqrt{1-y}+y \sqrt{1-x}
$$
has a maximum value of $\qquad$ . | 7. 1.
Since $x, y \in [0,1]$, then $x \leqslant \sqrt{x}, y \leqslant \sqrt{y}$.
Let $x=\sin ^{2} \alpha, y=\sin ^{2} \beta\left(\alpha, \beta \in\left[0, \frac{\pi}{2}\right]\right)$.
Thus, $f(x, y)=x \sqrt{1-y}+y \sqrt{1-x}$
$$
\begin{array}{l}
\leqslant \sqrt{x(1-y)}+\sqrt{y(1-x)} \\
=\sin \alpha \cdot \cos \beta+\... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. In the tetrahedron $ABCD$, it is known that $AD=2\sqrt{3}$, $\angle BAC=60^{\circ}$, $\angle BAD=\angle CAD=45^{\circ}$. The radius of the sphere that passes through $D$ and is tangent to the plane $ABC$ and internally tangent to the circumscribed sphere of the tetrahedron is 1, then the radius of the circumscribed... | 12.3.
As shown in Figure 3, draw a perpendicular from point $D$ to plane $ABC$, with the foot of the perpendicular being $H$. Draw $DE \perp AB$ and $DF \perp AC$, with the feet of the perpendiculars being $E$ and $F$ respectively.
Then $HE \perp AB$, $HF \perp AC$, and
$AE = AF = AD \cos 45^{\circ} = \sqrt{6}$.
From ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 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. ${ }^{(6)}$
(2010, National Junior High School Mathematics Competiti... | 【Analysis】In a right-angled triangle, the three sides satisfy the Pythagorean theorem. Given the condition $c=\frac{1}{3} a b-(a+b)$, one unknown can be eliminated to obtain a quadratic equation in two variables, which is generally difficult to solve. However, the problem of integer solutions to a quadratic equation in... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. If the equation with respect to $x$
$$
x^{2}+2(m+3) x+m^{2}+3=0
$$
has two real roots $x_{1}$ and $x_{2}$, then the minimum value of $\left|x_{1}-1\right|+\left|x_{2}-1\right|$ is $\qquad$. | 2.6 .
According to the problem, we have
$$
\begin{array}{l}
\Delta=[2(m+3)]^{2}-4\left(m^{2}+3\right) \geqslant 0 \\
\Rightarrow m \geqslant-1 .
\end{array}
$$
Then $x_{1}+x_{2}=-2(m+3)<0$.
When $x=1$, the left side of the equation is greater than 0, thus, $x_{1}$ and $x_{2}$ are on the same side of 1.
$$
\begin{arra... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) As shown in Figure 2, in the isosceles right triangle $\triangle ABC$, $\angle C=90^{\circ}$, points $D$ and $E$ are on side $BC$, and point $F$ is on the extension of $AC$, such that $BE=ED=CF$. Find the tangent value of $\angle CEF + \angle CAD$.
---
The translation preserves the original text's li... | $\begin{array}{l}\text { I. Draw } D G \perp A B \text { at point } G . \\ \text { Let } A C=B C=x, \\ B E=E D=C F=y\left(0<y<\frac{x}{2}\right) . \\ \text { Then } G D=\sqrt{2} y \Rightarrow A G=\sqrt{2}(x-y)=\sqrt{2} C E \text {. } \\ \text { Therefore, } \mathrm{Rt} \triangle E C F \backsim \mathrm{Rt} \triangle A G... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $\frac{m}{n}\left(m, n \in \mathbf{N}_{+},(m, n)=1\right)$ has a segment of digits $\overline{2012}$ in its decimal part, where $n$ is the smallest number satisfying the condition. Then $\left[\frac{m}{\sqrt{n}}\right]=$ $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 8. 2 .
Let $\frac{m}{n}=\overline{A . B 2012 C}$, where $A, B, C$ are digit strings, and the lengths of $A, B$ are $k, l (k, l \in \mathbf{N})$. Then
$$
\frac{10^{6}(m-n A)-n B}{n}=\overline{0.2012 C} \triangleq \frac{a}{b},
$$
where $(a, b)=1, 1 \leqslant b \leqslant n$.
Assume $\frac{m}{n}=\overline{0.2012 C}$.
$$
... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given that $a$ and $b$ are real numbers, and $a^{2} + ab + b^{2} = 3$. If the maximum value of $a^{2} - ab + b^{2}$ is $m$, and the minimum value is $n$, find the value of $m + n$. ${ }^{\text {[2] }}$ | Let $a^{2}-a b+b^{2}=t$.
Combining this with the given equation, we get
$$
a b=\frac{3-t}{2}, a+b= \pm \sqrt{\frac{9-t}{2}} .
$$
Thus, $a$ and $b$ are the two real roots of the quadratic equation in $x$:
$$
x^{2} \pm \sqrt{\frac{9-t}{2}} x+\frac{3-t}{2}=0
$$
Therefore, $\Delta=\left( \pm \sqrt{\frac{9-t}{2}}\right)^{... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Initially 340 Given two points $A$ and $B$ on a straight line $l$, the distance between them is $10000 \mathrm{~cm}$. At points $A$ and $B$, there are two movable barriers, designated as Barrier 1 and Barrier 2, respectively. Assume there is a ping-pong ball between $A$ and $B$, moving along the straight line $l$ at a ... | Let's assume that when the ping-pong ball contacts the 2nd board for the $n$-th time, the position of the 2nd board after it moves quickly is $B_{n}$; when the ping-pong ball contacts the 1st board for the $n$-th time, its position is $A_{n}$, and we set
$$
\begin{array}{l}
B_{0}=B, A_{n} B_{n-1}=x_{n}, A_{n} B_{n}=y_{... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Try to find the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
This line is not translated as it seems to be an instruction and not part of the text to be translated. If you need this line translated as well, please l... | Notice,
$$
(\sqrt{2}+\sqrt{3})^{2012}=(5+2 \sqrt{6})^{1000} \text {. }
$$
Consider the integer-coefficient quadratic equation $x^{2}-10 x+1=0$ with $5+2 \sqrt{6}$ as one of its roots, the other root being $5-2 \sqrt{6}$.
$$
\begin{array}{l}
\text { Let } a=5+2 \sqrt{6}, b=5-2 \sqrt{6}, \\
u_{n}=a^{n}+b^{n}(n=1,2, \cdo... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let $x$ be a real number. Then
$$
|x-1|+|x+1|+|x+5|
$$
the minimum value is $\qquad$ (s) | Let $y=|x-1|+|x+1|+|x+5|$. When $x<-5$,
$$
y=1-x-x-1-x-5=-3 x-5 \text{; }
$$
When $-5 \leqslant x<-1$,
$$
y=1-x-x-1+x+5=-x+5 \text{; }
$$
When $-1 \leqslant x<1$,
$$
y=1-x+x+1+x+5=x+7 \text{; }
$$
When $x \geqslant 1$,
$$
y=x-1+x+1+x+5=3 x+5 \text{. }
$$
By the properties of linear functions, we know that when $x=-... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the vectors $\overrightarrow{O A}=(1,0), \overrightarrow{O B}=(1,1)$, and $O$ be the origin. A moving point $P(x, y)$ satisfies
$$
\left\{\begin{array}{l}
0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O A} \leqslant 1, \\
0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O B} \leqslant 2
\end{array... | 2. 2 .
From the problem, the point $P(x, y)$ satisfies
$$
\begin{array}{l}
\left\{\begin{array}{l}
0 \leqslant x \leqslant 1, \\
0 \leqslant x+y \leqslant 2 .
\end{array}\right. \\
\text { Let }\left\{\begin{array}{l}
x+y=u, \\
y=v .
\end{array}\right.
\end{array}
$$
Then the point $Q(u, v)$ satisfies
$$
\left\{\begi... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $(a, b)$ be real numbers. Then the minimum value of $a^{2}+a b+b^{2}-a-2 b$ is . $\qquad$ | Hint: Use the method of completing the square. The original expression can be transformed into
$$
\left(a+\frac{1}{2} b-\frac{1}{2}\right)^{2}+\frac{3}{4}(b-1)^{2}-1 \geqslant-1 \text {. }
$$
When $a+\frac{1}{2} b-\frac{1}{2}=0, b-1=0$, i.e., $a=0$, $b=1$, the original expression can achieve the minimum value of -1. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let positive real numbers $a, b, c$ satisfy $\frac{2}{a}+\frac{1}{b}=\frac{\sqrt{3}}{c}$. Then the minimum value of $\frac{2 a^{2}+b^{2}}{c^{2}}$ is $\qquad$ . | 4. 9 .
From the given, we have $\frac{2 c}{a}+\frac{c}{b}=\sqrt{3}$.
By the Cauchy-Schwarz inequality and the AM-GM inequality, we get
$$
\begin{array}{l}
\frac{2 a^{2}+b^{2}}{c^{2}} \\
=\frac{1}{3}(2+1)\left[2\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}\right] \\
\geqslant \frac{1}{3}\left(\frac{2 a}{c}+... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the equation in $x$
$$
x^{3}-4 x^{2}+5 x+a=0(a \in \mathbf{R})
$$
has three real roots $x_{1}, x_{2}, x_{3}$. Then the maximum value of $\max \left\{x_{1}, x_{2}, x_{3}\right\}$ is $\qquad$ . | 6.2.
Assume $x_{3}=\max \left\{x_{1}, x_{2}, x_{3}\right\}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=4, \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=5
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x_{1}+x_{2}=4-x_{3}, \\
x_{1} x_{2}=5-x_{3}\left(x_{1}+x_{2}\right... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a>1$. Then when the graphs of $y=a^{x}$ and $y=\log _{a} x$ are tangent, $\ln \ln a=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. -1.
Since the two functions are inverse functions of each other and are symmetric about the line $y=x$, the point of tangency lies on $y=x$.
Let the point of tangency be $\left(x_{0}, y_{0}\right)$. Then
$$
\begin{array}{l}
x_{0}=a^{x_{0}}, \\
a^{x_{0}} \ln a=1 .
\end{array}
$$
Substituting equation (1) into equat... | -1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Question 3 Given $a b c=-1, a^{2} b+b^{2} c+c^{2} a=t$, $\frac{a^{2}}{c}+\frac{b}{c^{2}}=1$. Try to find the value of $a b^{5}+b c^{5}+c a^{5}$. | 【Analysis】Using $a b c=-1$ can make common substitutions
$$
a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x} \text {. }
$$
This problem involves three letters, and can also be solved by the method of elimination.
Solution 1 Let $a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x}$.
$$
\begin{array}{l}
\text { Then } \frac{a^... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $\tan \alpha+\tan \beta+\tan \gamma=\frac{17}{6}$, $\cot \alpha+\cot \beta+\cot \gamma=-\frac{4}{5}$, $\cot \alpha \cdot \cot \beta+\cot \beta \cdot \cot \gamma+\cot \gamma \cdot \cot \alpha=-\frac{17}{5}$. Then $\tan (\alpha+\beta+\gamma)=$ $\qquad$ | 3. 11 .
Let $\tan \alpha=x, \tan \beta=y, \tan \gamma=z$.
Then the given equations can be written as
$$
\begin{array}{l}
x+y+z=\frac{17}{6}, \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{4}{5}, \\
\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=-\frac{17}{5} . \\
\text { From (1) } \div \text { (3) we get } \\
x y z=-\frac{... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $a, b, c$ satisfy
$$
a+b+c=a^{2}+b^{2}+c^{2} \text {. }
$$
Then the maximum value of $a+b+c$ is $\qquad$ | 4.3.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
3(a+b+c) \\
=\left(1^{2}+1^{2}+1^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)
\end{array}
$$
$$
\begin{array}{l}
\geqslant(a+b+c)^{2} \\
\Rightarrow a+b+c \leqslant 3 .
\end{array}
$$
Equality holds if and only if \(a=b=c=1\). | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that the base edge length of a regular tetrahedron is 6, and the side edge is 4. Then the radius of the circumscribed sphere of this regular tetrahedron is $\qquad$ | 3. 4 .
From the problem, we know that the distance from point $A$ to the base $B C D$ is 2, and the radius of its circumscribed sphere is $R$. Then
$$
R^{2}-12=(R-2)^{2} \Rightarrow R=4 \text {. }
$$ | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. In the arithmetic sequence $\left\{a_{n}\right\}$, if $S_{4} \leqslant 4, S_{5} \geqslant 15$, then the minimum value of $a_{4}$ is $\qquad$ . | 4. 7 .
Let the common difference be $d$. From the given conditions, we have
$$
\begin{array}{l}
2 a_{4} \leqslant 2+3 d, d \leqslant a_{4}-4 \\
\Rightarrow 2 a_{4} \leqslant 2+3 d \leqslant 2+3\left(a_{4}-3\right) \\
\Rightarrow a_{4} \geqslant 7 .
\end{array}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 As shown in Figure 3, in the acute triangle $\triangle ABC$, it is known that $BE \perp AC$ at point $E$, $CD \perp AB$ at point $D$, $BC=25$, $CE=7$, $BD=15$. If $BE$ and $CD$ intersect at point $H$, connect $DE$, and construct a circle with $DE$ as the diameter, which intersects $AC$ at another point $F$. F... | Solve for DF. From the given conditions, we have
$$
\begin{array}{l}
\cos B=\frac{3}{5}, \cos C=\frac{7}{25} \\
\Rightarrow \sin B=\frac{4}{5}, \sin C=\frac{24}{25} \\
\Rightarrow \sin A=\sin B \cdot \cos C+\sin C \cdot \cos B=\frac{4}{5}=\sin B \\
\Rightarrow \angle A=\angle B \Rightarrow AC=BC=25 .
\end{array}
$$
Fr... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let two ellipses be
$$
\frac{x^{2}}{t^{2}+2 t-2}+\frac{y^{2}}{t^{2}+t+2}=1
$$
and $\frac{x^{2}}{2 t^{2}-3 t-5}+\frac{y^{2}}{t^{2}+t-7}=1$
have common foci. Then $t=$ $\qquad$ . | 5.3.
Given that the two ellipses have a common focus, we have
$$
\begin{array}{l}
t^{2}+2 t-2-\left(t^{2}+t+2\right) \\
=2 t^{2}-3 t-5-\left(t^{2}+t-7\right) \\
\Rightarrow t=3 \text { or } 2 \text { (rejected). }
\end{array}
$$
Therefore, $t=3$. | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Calculate:
$$
\sum_{k=0}^{2013}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2014}^{k}}=
$$
$\qquad$ | 5. 0 .
Let $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2 n}^{k}}$.
And $\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n}-k-1}$, so $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{\mathrm{C}_{2 n}^{2 n... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Let $a, b, c > 0$, and $a+b+c=3$. Find the maximum value of $a^{2} b+b^{2} c+c^{2} a+a b c$.
| Let's assume $a \leqslant b \leqslant c$ or $c \leqslant b \leqslant a$.
Then $c(b-a)(b-c) \leqslant 0$
$$
\begin{aligned}
\Rightarrow & b^{2} c+c^{2} a \leqslant a b c+c^{2} b \\
\Rightarrow & a^{2} b+b^{2} c+c^{2} a+a b c \\
& \leqslant a^{2} b+c^{2} b+2 a b c \\
& =b(a+c)^{2}=b(3-b)^{2} \\
& \leqslant \frac{1}{2}\le... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50) Find the maximum value of $n$ such that there are $n$ points in the plane, where among any three points, there must be two points whose distance is 1. | If there exist $n(n \geqslant 8)$ points satisfying the conditions of the problem, let $V=\left\{v, v_{1}, v_{2}, \cdots, v_{7}\right\}$ represent any eight of these points. When and only when the distance between two points is 1, connect an edge between these two points, forming a graph $G$.
If there exists a point (... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given $x, y$ are positive real numbers, $n \in \mathrm{N}$, and $n \geqslant 2$. Prove:
$$
\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}}+\sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}} \geqslant 4 \text {. }
$$ | $$
\begin{array}{l}
\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}}+\sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}} \\
\geqslant 2 \sqrt{\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}} \cdot \sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}}} \\
=2 \sqrt[2 n]{\frac{\left(2^{n}-1\right)\left(x^{2}+y^{2}\right)+\left(2^{2 n}-2^{n+1}+2\... | 4 | Inequalities | proof | Yes | Yes | cn_contest | false |
Example 6 Let the functions $f(x)=\ln x, g(x)=\frac{1}{2} x^{2}$. If $x_{1}>x_{2}>0$, for what value of $m(m \in \mathbf{Z}, m \leqslant 1)$ is it always true that
$$
m\left(g\left(x_{1}\right)-g\left(x_{2}\right)\right)>x_{1} f\left(x_{1}\right)-x_{2} f\left(x_{2}\right)
$$
holds. | Introduce an auxiliary function
$$
t(x)=m g(x)-x f(x)=\frac{m}{2} x^{2}-x \ln x \quad (x>0) \text {. }
$$
By the problem, $x_{1}>x_{2}>0$.
Therefore, if the original inequality always holds for $x>0$, i.e., the function $t(x)$ is monotonically increasing, then
$$
t^{\prime}(x)=m x-\ln x-1 \geqslant 0
$$
always holds.... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
3 x-3 \geqslant 6 x+a, \\
x \geqslant 1
\end{array}\right.
$$
the solution is $1 \leqslant x \leqslant 3$. Then $a=$ | $=, 1 .-12$.
From the given, we have $1 \leqslant x \leqslant \frac{1}{3}(-a-3)$. Then $\frac{1}{3}(-a-3)=3 \Rightarrow a=-12$. | -12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $x$ and $y$ be two distinct non-negative integers, and satisfy $x y + 2x + y = 13$. Then the minimum value of $x + y$ is $\qquad$ | 3. 5 .
From the problem, we know that $(x+1)(y+2)=15$.
$$
\begin{array}{l}
\text { Then }(x+1, y+2) \\
=(15,1),(5,3),(3,5),(1,15) \\
\Rightarrow(x, y)=(14,-1),(4,1),(2,3),(0,13) .
\end{array}
$$
Therefore, the minimum value of $x+y$ is 5. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. In the arithmetic sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{20} \simeq \frac{1}{a}, a_{201}=\frac{1}{b}, a_{2012}=\frac{1}{c} \text {. }
$$
Then $1992 a c-1811 b c-181 a b=$ | 12. 0 .
Let the common difference of the arithmetic sequence be $d$. Then, according to the problem, we have
$$
\begin{array}{l}
a_{201}-a_{20}=\frac{a-b}{a b}=181 d, \\
a_{2012}-a_{201}=\frac{b-c}{b c}=1811 d, \\
a_{2012}-a_{20}=\frac{a-c}{a c}=1992 d .
\end{array}
$$
Therefore, $1992 a c-1811 b c-181 a b$
$$
=\frac... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Given constants $a, b$ satisfy $a, b>0, a \neq 1$, and points $P(a, b), Q(b, a)$ are both on the curve $y=\cos (x+c)$, where $c$ is a constant. Then $\log _{a} b=$ $\qquad$ | 13. 1 .
Given points $P(a, b), Q(b, a)$ are both on the curve
$$
y=\cos (x+c)
$$
we know
$$
\begin{array}{l}
a-b=\cos (b+c)-\cos (a+c) \\
=2 \sin \left(\frac{a+b+2 c}{2}\right) \cdot \sin \frac{a-b}{2} .
\end{array}
$$
Without loss of generality, assume $a \geqslant b$.
If $a>b$, then
$$
\left|\sin \frac{a+b+2 c}{2}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $m$ is an integer, the equation
$$
x^{2}-m x+3-n=0
$$
has two distinct real roots, the equation
$$
x^{2}+(6-m) x+7-n=0
$$
has two equal real roots, and the equation
$$
x^{2}+(4-m) x+5-n=0
$$
has no real roots.
$$
\text { Then }(m-n)^{2013}=
$$ | 3. According to the problem, we have
$$
\left\{\begin{array}{l}
m^{2}-4(3-n)>0, \\
(6-m)^{2}-4(7-n)=0, \\
(4-m)^{2}-4(5-n)\frac{5}{3}, m<3 \text {. }
\end{array}\right.
$$
Then $m=2$. Consequently, $n=3$.
Therefore, $(m-n)^{2013}=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) In a dormitory of a school, there are several students. On New Year's Day, each student in the dormitory gives a greeting card to every other student, and each student also gives a greeting card to each dormitory administrator, who in turn gives a card back to each student. In this way, a total of 51 g... | Let there be $x$ students and $y$ administrators living in the dormitory $\left(x, y \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
x(x-1)+x y+y=51 \\
\Rightarrow y=\frac{51+x-x^{2}}{x+1}=\frac{49}{x+1}-x+2 .
\end{array}
$$
Since $x, y$ are positive integers, we know that $(x, y)=(6,3)$.
Therefore, there are 6 s... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. In $\triangle A B C$, it is known that $O$ is the circumcenter, the three altitudes $A D, B E, C F$ intersect at point $H$, line $E D$ intersects $A B$ at point $M$, and $F D$ intersects $A C$ at point $N$. Then $\overrightarrow{O H} \cdot \overrightarrow{M N}=$ $\qquad$ | 3. 0 .
It is known that, $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{O H}$.
Then $\overrightarrow{O H} \cdot \overrightarrow{M N}=(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \cdot \overrightarrow{M N}$
$$
\begin{aligned}
= & \overrightarrow{O A} \cdot(\overright... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the real number pair $(x, y)$ satisfies the equation $(x-2)^{2}+y^{2}=3$, let the minimum and maximum values of $\frac{y}{x}$ be $m$ and $n$ respectively. Then $m+n=$ | $\begin{array}{l}\text { II. 1.0. } \\ \text { Let } y=t x \text {. Then }\left(1+t^{2}\right) x^{2}-4 x+1=0 \text {. } \\ \text { By } \Delta=(-4)^{2}-4\left(1+t^{2}\right) \geqslant 0 \\ \Rightarrow-\sqrt{3} \leqslant t \leqslant \sqrt{3} \\ \Rightarrow m=-\sqrt{3}, n=\sqrt{3} \\ \Rightarrow m+n=0 \text {. }\end{arra... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 2, in
rhombus $A B C D$, it is
known that $\angle A B C=60^{\circ}$, line
$E F$ passes through point $D$, and
intersects the extensions of
$B A$ and $B C$ at points $E$ and $F$, respectively. $M$
is the intersection of $C E$ and $A F$. If $C M=4, E M=5$, then $C A=$ $\qquad$ | 3. 6 .
It is easy to prove $\triangle E A D \backsim \triangle D C F$.
$$
\begin{array}{l}
\text { Therefore, } \frac{E A}{A D}=\frac{D C}{C F} \Rightarrow \frac{E A}{A C}=\frac{A C}{C F} \\
\Rightarrow \triangle E A C \backsim \triangle A C F \\
\Rightarrow \angle A E C=\angle C A F \\
\Rightarrow C A^{2}=C E \cdot C... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Let $x, y (x>y)$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}, \cdots, a_{n}$, satisfying $x-y \geqslant \frac{xy}{31}$. Find the maximum value of the number of elements $n$ in this set of natural numbers. | Let's assume $a_{1}1 \\
\Rightarrow d_{6} \geqslant 2 \Rightarrow a_{7}=a_{6}+d_{6} \geqslant 8 \\
\Rightarrow d_{7} \geqslant \frac{8^{2}}{31-8}>2 \Rightarrow d_{7} \geqslant 3 . \\
\Rightarrow a_{8}=a_{7}+d_{7} \geqslant 11 \\
\Rightarrow d_{8} \geqslant \frac{11^{2}}{31-11}>6 \Rightarrow d_{8} \geqslant 7 \\
\Righta... | 10 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $a, b, x \in \mathbf{N}_{+}$, and $a \leqslant b$. $A$ is the solution set of the inequality
$$
\lg b - \lg a < \lg x < \lg b + \lg a
$$
It is known that $|A|=50$. When $ab$ takes its maximum possible value,
$$
\sqrt{a+b}=
$$ | 4. 6 .
It is easy to know, $\frac{b}{a}<x<a b, a \neq 1$.
Therefore, $a \geqslant 2$,
$$
\begin{array}{l}
50 \geqslant a b-\frac{b}{a}-1=a b\left(1-\frac{1}{a^{2}}\right)-1 \geqslant \frac{3}{4} a b-1 \\
\Rightarrow a b \leqslant 68 .
\end{array}
$$
Upon inspection, when and only when $a=2, b=34$, the equality holds.... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The function defined on the domain $R$
$$
f(x)=|\lg | x-2||-1 \text{. }
$$
If $b<0$, then the equation concerning $x$
$$
f^{2}(x)+b f(x)=0
$$
has $\qquad$ distinct real roots. | 5. 8 .
From the problem, we know that the graph of $y=\lg x$ is first symmetrical about the $y$-axis, then the part below the $x$-axis is flipped up, followed by a rightward shift of 2 units and a downward shift of 1 unit to form the graph of $f(x)$. The zeros of $g(x)=f^{2}(x)+b f(x)$ are the roots of the equation. I... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $z$ be a complex number with modulus 2. Then the sum of the maximum and minimum values of $\left|z-\frac{1}{z}\right|$ is $\qquad$ [2] | Given $|z|=2$, we know
$$
\begin{array}{l}
|z+1|^{2}=(z+1)(\bar{z}+1) \\
=z \bar{z}+z+\bar{z}+1=5+2 \operatorname{Re} z, \\
|z-1|^{2}=(z-1)(\bar{z}-1) \\
=z \bar{z}-z-\bar{z}+1=5-2 \operatorname{Re} z .
\end{array}
$$
Therefore, $\left|z-\frac{1}{z}\right|=\left|\frac{z^{2}-1}{z}\right|$
$$
=\frac{|z+1||z-1|}{|z|}=\fr... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For a natural number $n$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and their sum is 17. If there exists a unique $n$ such that $S_{n}$ is also an integer, find $n .{ }^{(4]}$ | Solve: Regarding $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the modulus of the complex number $(2 k-1)+a_{k} \mathrm{i}$.
$$
\begin{array}{l}
\text { Hence } \sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}} \\
=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} \mathrm{i}\right| \\
\geqslant\left|\sum_{k=1}^{n}\left[(2 k-1)+a_{k} \mat... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In rectangle $A B C D$, it is known that $A B=5, B C=9$, points $E, F, G, H$ are on sides $A B, B C, C D, D A$ respectively, such that $A E=C G=3, B F=D H=4, P$ is a point inside the rectangle. If the area of quadrilateral $A E P H$ is 15, then the area of quadrilateral $P F C G$ is $\qquad$ | 4. 11.
As shown in Figure 3, let the distances from $P$ to $AB$ and $AD$ be $a$ and $b$, respectively. Then the distances from $P$ to $BC$ and $CD$ are $5-b$ and $9-a$, respectively. Given that $S_{\text {quadrilateral } AEPH}=\frac{1}{2}(3a+5b)=15$, we have
$$
\begin{array}{l}
S_{\text {quadrilateral PFCG }}=\frac{1}... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. The number of integers $n$ that make $n^{4}-3 n^{2}+9$ a prime number is $\qquad$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 5.4.
If $n=0$, then $n^{4}-3 n^{2}+9=9$ is not a prime number.
If $n>0$, then
$$
\begin{array}{l}
n^{4}-3 n^{2}+9=\left(n^{2}+3\right)^{2}-(3 n)^{2} \\
=\left(n^{2}-3 n+3\right)\left(n^{2}+3 n+3\right) .
\end{array}
$$
Notice that, $n^{2}+3 n+3>3$.
Therefore, $n^{2}-3 n+3=1$.
Solving this gives $n=1$ or 2 (correspond... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given any positive integer $a$, define the integer sequence $x_{1}, x_{2}$, $\cdots$, satisfying
$$
x_{1}=a, x_{n}=2 x_{n-1}+1(n \geqslant 1) .
$$
If $y_{n}=2^{x_{n}}-1$, determine the maximum integer $k$ such that there exists a positive integer $a$ for which $y_{1}, y_{2}, \cdots, y_{k}$ are all prime numbers. | 1. If $y_{i}$ is a prime number, then $x_{i}$ is also a prime number.
Otherwise, if $x_{i}=1$, then $y_{i}=1$ is not a prime number; if $x_{i}=m n($ integers $m, n>1)$, then
$\left(2^{m}-1\right) \mid\left(2^{x_{i}}-1\right)$, i.e., $x_{i}$ and $y_{i}$ are composite numbers.
Below, we prove by contradiction: For any ... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies
$$
\begin{array}{l}
f(x+1)=f(-x), \\
f(x)=\left\{\begin{array}{ll}
1, & -1<x \leqslant 0 \\
-1, & 0<x \leqslant 1 .
\end{array}\right.
\end{array}
$$
Then $f(f(3.5))=$ $\qquad$ | 2. -1 .
From $f(x+1)=-f(x)$, we know $f(x+2)=f(x)$.
Then $f(3.5)=f(-0.5)=1$.
Therefore, $f(f(3.5))=f(1)=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $P$ is a moving point on the line $l$:
$$
k x+y+4=0(k>0)
$$
$P A$ and $P B$ are the two tangents from $P$ to the circle $C$:
$$
x^{2}+y^{2}-2 y=0
$$
with points of tangency $A$ and $B$ respectively. If the minimum area of quadrilateral $P A C B$ is 2, then $k=$ $\qquad$ | 6.2.
$$
\begin{array}{l}
\text { Given } S_{\text {quadrilateral } P A C B}=P A \cdot A C=P A \\
=\sqrt{C P^{2}-C A^{2}}=\sqrt{C P^{2}-1},
\end{array}
$$
we know that the area is minimized when $|C P|$ is minimized, i.e., when $C P \perp l$. Also, $\sqrt{C P^{2}-1}=2$, then $C P=\sqrt{5}$. Using the point-to-line dist... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
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