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Five. (Full marks: 15 points) From the 91 natural numbers $1,2,3, \cdots, 90,91$, select $k$ numbers such that there must be two natural numbers $p, q$ satisfying $\frac{2}{3} \leqslant \frac{q}{p} \leqslant \frac{3}{2}$. Determine the minimum value of the natural number $k$, and explain your reasoning. | Five, divide the 91 natural numbers from 1 to 91 into nine groups, such that the ratio of any two natural numbers in each group is no less than $\frac{2}{3}$ and no more than $\frac{3}{2}$, and the division is as follows:
$$
\begin{array}{l}
A_{1}=\{1\}, A_{2}=\{2,3\}, A_{3}=\{4,5,6\}, \\
A_{4}=\{7,8,9,10\}, \\
A_{5}=\... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 If $x, y, z$ are real numbers, and
$$
\begin{aligned}
(y-z)^{2} & +(z-x)^{2}+(x-y)^{2} \\
= & (y+z-2 x)^{2}+(z+x-2 y)^{2} \\
& +(x+y-2 z)^{2},
\end{aligned}
$$
find the value of $M=\frac{(y z+1)(z x+1)(x y+1)}{\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right)}$. | Solution: The condition can be simplified to
$$
x^{2}+y^{2}+z^{2}-x y-y z-z x=0 .
$$
Then $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0$,
which implies $x=y=z$.
Therefore, $M=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five, in a chess tournament, there are an odd number of participants, and each participant plays one game against every other participant. The scoring system is as follows: 1 point for a win, 0.5 points for a draw, and 0 points for a loss. It is known that two of the participants together scored 8 points, and the avera... | Five, suppose there are $(n+2)$ players in total, except for 2 people who get 8 points, $n$ people on average get $k$ points each ($k$ is an integer).
$\because$ Each person plays one match with everyone else, and there are $(n+2)$ people,
$\therefore$ A total of $\frac{(n+1)(n+2)}{2}$ matches are played.
Since each ma... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 As shown, in Rt $\triangle ABC$, the hypotenuse $AB=5, CD \perp AB$. It is known that $BC, AC$ are the two roots of the quadratic equation $x^{2}-(2 m-1) x+4(m-1)=0$. Then the value of $m$ is $\qquad$. | Solution: Let $A C=b$, $B C=a$. By Vieta's formulas, we get $a+b=2 m-$
$$
\begin{array}{l}
1, a b=4(m-1) . \\
\begin{aligned}
\therefore A B^{2} & =a^{2}+b^{2}=(a+b)^{2}-2 a b \\
& =(2 m-1)^{2}-2 \times 4(m-1)=5^{2},
\end{aligned}
\end{array}
$$
i.e., $m^{2}-3 m-4=0$.
$$
\therefore m=4 \text { or } m=-1 \text {. }
$$
... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. As shown in the figure, in the right trapezoid $A B C D$, the base $A B=13$, $C D=8, A D \perp A B$, and $A D=$ 12. Then the distance from $A$ to the side $B C$ is ( ).
(A) 12
(B) 13
(C) 10
(D) $\frac{12 \times 21}{13}$ | 6. (A).
Draw $C C^{\vee} \perp A B$, then
$$
B C=13-8=5 \text {. }
$$
Let the distance from $A$ to $B C$ be $h$,
Connect AC.
$$
\begin{aligned}
& \because S_{\triangle A C D}+S_{\triangle A B C} \\
& =S_{\text {UEABCD }}, \\
\therefore & \frac{1}{2} \times 8 \times 12+\frac{1}{2} \times h \times 13=\frac{1}{2}(8+13) ... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false |
2. Given $|x| \leqslant 2$, the sum of the maximum and minimum values of the function $y=x-|1+x|$ is $\qquad$ . | 2. -4 .
$$
\begin{array}{l}
\because|x| \leqslant 2, \text { i.e., }-2 \leqslant x \leqslant 2, \\
\therefore y=x-|1+x| \\
\quad=\left\{\begin{array}{ll}
2 x+1, & \text { when }-2 \leqslant x<-1, \\
-1, & \text { when }-1 \leqslant x \leqslant 2.
\end{array}\right.
\end{array}
$$
As shown in the figure.
Indeed, when $... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $M=\cos 5^{\circ} \sin 15^{\circ} \sin 25^{\circ} \sin 35^{\circ}, N=$ $\sin 5^{\circ} \cos 15^{\circ} \cos 25^{\circ} \cos 35^{\circ}$. Then $\frac{M}{N}=$ $\qquad$ . | $=1.1$.
$$
\begin{aligned}
\frac{M}{N} & =\frac{\frac{1}{2}\left(\sin 20^{\circ}+\sin 10^{\circ}\right) \cdot \frac{1}{2}\left(\cos 10^{\circ}-\cos 60^{\circ}\right)}{\frac{1}{2}\left(\sin 20^{\circ}-\sin 10^{\circ}\right) \cdot \frac{1}{2}\left(\cos 10^{\circ}+\cos 60^{\circ}\right)} \\
& =\frac{\left(\sin 20^{\circ}+... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $a+\lg a=10, b+10^{b}=10$. Then $a+b$ | 4. 10.
Thought 1: From the given information,
$$
\begin{array}{l}
a=10^{10-a}, \\
10-b=10^{b} .
\end{array}
$$
Subtracting, we get $10-a-b=10^{b}-10^{10-a}$.
If $10-a-b>0$, then $10-a>b$. By the monotonicity of the exponential function, we get $10^{10-a}>10^{b}$. Substituting into (1), we get
$$
010-a-b=10^{b}-10^{10... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $a, b$ be unequal real numbers, and $a^{2}+2 a-5$ $=0, b^{2}+2 b-5=0$. Then $a^{2} b+a b^{2}=$ $\qquad$ . | （Answer:10） | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For example, 610 people go to the bookstore to buy books, it is known that
(1) each person bought three books;
(2) any two people have at least one book in common.
How many people at most bought the book that was purchased by the fewest people? | Solution: Let the number of people who bought the most popular book be $x$. Among the 10 people, person A bought three books. Since the other 9 people each have at least one book in common with A, and $9 \div 3=3$, it follows that among A's three books, the most popular one must have been bought by at least 4 people, s... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given that $x$, $y$, $z$ are positive integers, and $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(y+z)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Explanation: This is a typical example of constructing a geometric figure to solve a problem. As shown in Figure 1, construct $\triangle ABC$, where the lengths of the three sides are
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z, \\
c=z+x .
\end{array}\right.
$$
Then its area is
$$
\begin{aligned}
\triangle & =\sqrt{p(p... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$ . | Solution: Since $m \neq n$, by the definition of roots, $m, n$ are two distinct roots of the equation $x^{2}-x-1=0$. By Vieta's formulas, we have
$$
\begin{array}{l}
m+n=1, m n=-1 . \\
\because m^{2}+n^{2}=(m+n)^{2}-2 m n \\
\quad=1^{2}-2 \times(-1)=3, \\
\quad m^{3}+n^{3}=(m+n)^{3}-3 m n(m+n) \\
\quad=1^{3}-3 \times(-... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, let $S=\{1,2,3,4\}, n$ terms of the sequence: $a_{1}$, $a_{2}, \cdots, a_{n}$ have the following property, for any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted as $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
| Three, the minimum value of $n$ is 8.
First, prove that each number in $S$ appears at least 2 times in the sequence $a_{1}, a_{2}, \cdots, a_{n}$. This is because, if a number in $S$ appears only once in this sequence, since there are 3 two-element subsets containing this number, but in the sequence, the adjacent pairs... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in the figure, the side lengths of $\triangle A B C$ are $A B=14, B C$ $=16, A C=26, P$ is a point on the angle bisector $A D$ of $\angle A$, and $B P \perp A D, M$ is the midpoint of $B C$. Find the value of $P M$ $\qquad$ | 3. 6 .
From the figure, take $B^{\prime}$ on $A C$ such that $A B^{\prime}=A B=14$, then $B^{\prime} C=12$. Since $\triangle A B B^{\prime}$ is an isosceles triangle, we know that the intersection point of $B B^{\prime}$ and $A D$ is $P$ (concurrency of five lines), so $P$ is the midpoint of $B B^{\prime}$. | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $x, y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1 \\
(y-1)^{3}+1997(y-1)=1 .
\end{array}\right.
$$
Then $x+y=$ $\qquad$ (Proposed by the Problem Group) | $=、 1.2$.
The original system of equations is transformed into
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1, \\
(1-y)^{3}+1997(1-y)=-1 .
\end{array}\right.
$$
Since $f(t)=t^{3}+1997 t$ is monotonically increasing on $(-\infty,+\infty)$, and $f(x-1)=f(1-y)$, it follows that $x-1=1-y$, i.e., $x+y=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$, a line $l$ intersects the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$ such that there are exactly 3 lines $l$, then $\lambda=$
(Proposed by the Problem Committee) | 2. 4 .
First, note the following conclusion: For a chord passing through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ and intersecting the right branch at two points, the chord attains its minimum length $\frac{2 b^{2}}{a}=4$ if and only if the chord is perpendicular to the $x$-axis. (In fact, the polar ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $x=\frac{-\sqrt[1]{17}+\sqrt{\sqrt{17}+4 \sqrt{15}}}{2 \sqrt{3}}$. Then, $3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5=$ | 2. 0 .
From the given, we have
$$
2 \sqrt{3} x+\sqrt[4]{17}=\sqrt{\sqrt{17}+4 \sqrt{15}} \text {. }
$$
Squaring and rearranging, we get
$$
\begin{array}{l}
\sqrt{3} x^{2}+\sqrt[4]{17} x-\sqrt{5}=0 . \\
\text { Also, } 3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5 \\
=\left(3 x^{4}-2 \sqrt{3} \cdot \sqrt{5} x^{2}-5\right)-1... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that the integer $n$ is not a multiple of 5. Then the remainder when $n^{4}+4$ is divided by 5 is $\qquad$ . | 4. 0 .
$$
\text { Given } \begin{aligned}
& n^{4}+4=\left(n^{4}-1\right)+5 \\
= & \left(n^{2}+1\right)\left(n^{2}-1\right)+5 \\
= & \left(n^{2}-4\right)\left(n^{2}-1\right)+5\left(n^{2}-1\right)+5 \\
= & (n+2)(n-2)(n+1)(n-1) \\
& +5\left(n^{2}-1\right)+5,
\end{aligned}
$$
and $n$ is not a multiple of 5, so among $n+2,... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. For a cube wooden block $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with an edge length of 1, take points $P, Q, R$ on the three edges passing through vertex $A_{1}$, such that $A_{1} P=A_{1} Q=A_{1} R$. After cutting off the tetrahedron $A_{1}-P Q R$, use the section $\triangle P Q R$ as the base to drill a triangular pri... | 6. (B).
Through point $R$, a line parallel to $A_{1} C$ intersects $A C$ at a point $R^{\prime}$. Through point $Q$, a line parallel to $A_{1} C$ intersects $B_{1} C$ at a point $Q^{\prime}$. A plane through $R Q$ and parallel to $A_{1} C$ intersects the side face of the triangular prism at a point $C_{0}$ on the edge... | 6 | Geometry | MCQ | Yes | Yes | cn_contest | false |
1. Let $n=\underbrace{111 \cdots 11}_{1999 \uparrow 1}, f(n)=90 n^{2000}+20 n+$ 1997. Then the remainder when $f(n)$ is divided by 3 is | Ni.1.1.
A natural number $a$ has the same remainder when divided by 3 as its digit sum $S(a)$ when divided by 3. Therefore, $n=\underbrace{11 \cdots 111}_{1999 \uparrow 1}$ has the same remainder when divided by 3 as 1999, which is 1.
$90 n^{2000}$ has a remainder of 0 when divided by 3, and $20 n$ has a remainder of 2... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $a$, $b$, and $c$ are the lengths of the three sides of a right triangle, and for a natural number $n$ greater than 2, the following holds:
$$
\left(a^{n}+b^{n}+c^{n}\right)^{2}=2\left(a^{2 n}+b^{2 n}+c^{2 n}\right) .
$$
Then $n=$ | 6. 4 .
Let $x=a^{\frac{n}{2}}, y=b^{\frac{n}{2}}, z=c^{\frac{\pi}{2}}$, then
$$
\begin{aligned}
0= & 2\left(a^{2 n}+b^{2 n}+c^{2 n}\right)-\left(a^{n}+b^{n}+c^{n}\right)^{2} \\
= & 2\left(x^{4}+y^{4}+z^{4}\right)-\left(x^{2}+y^{2}+z^{2}\right)^{2} \\
= & x^{4}+y^{4}+z^{4}-2 x^{2} y^{2}-2 x^{2} z^{2}-2 y^{2} z^{2} \\
=... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Six. (12 points) On the hypotenuse $AB$ of the right triangle $\triangle ABC$, color the points $P$ that satisfy $PC^{2} = PA \cdot PB$ in red. How many red points are there at least, and at most, on the hypotenuse? | Six, as shown, let $B P=x$, then $A P=c-x$. Draw the altitude $C H$ on the hypotenuse, then $C H=\frac{a b}{c}, B H$ $=\frac{a^{2}}{b}$. Therefore,
$$
\left.A P=\frac{a^{2}}{c}-x \right| \, \text { ( } P \text { can be between } A \text { and } Y \text {, also please consider } H \text { and }
$$
$B$ points.
In the rig... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. In the Cartesian coordinate system, points of the form $\left(m, n^{2}\right)$ are painted red (where $m, n$ are integers), referred to as red points, and their surrounding points are not colored. Then, the parabola $y=x^{2}-$ $196 x+9612$ has $\qquad$ red points. | 2. 2 .
Let $\left(m, n^{2}\right)$ be on the parabola $y=x^{2}-196 x+9612$, then $n^{2}=m^{2}-196 m+9612$.
Completing the square and factoring, we get
$$
\begin{array}{l}
(n+m-98)(n-m+98) \\
=8=2 \times 4=(-2) \times(-4) .
\end{array}
$$
$\because m, n$ are integers,
$\therefore n+m-98$ and $n-m+98$ are of the same pa... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given, as shown in the figure, a semicircle $O$ with a diameter of $20 \mathrm{~cm}$ has two points $P$ and $Q$, $P C \perp A B$ at $C, Q D$ $\perp A B$ at $D, Q E \perp$ $O P$ at $E, A C=4 \mathrm{~cm}$. Then $D E=$ | $3.8 \mathrm{~cm}$.
Take the midpoint $M$ of $O P$ and the midpoint $N$ of $O Q$, and connect $C M, D N$, and $E N$. Then
$$
\begin{array}{c}
M C=P E=\frac{1}{2} O P \\
=\frac{1}{2} O Q=E N=D N, \\
\angle P M C=\angle M C O+ \\
\angle M O C=2 \angle M O C .
\end{array}
$$
Since $O, D, Q, E$ are concyclic, with $N$ as t... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Let $a_{1}, a_{2}, \cdots, a_{n}$ represent any permutation of the integers 1, 2, $\cdots, n$. Let $f(n)$ be the number of such permutations that satisfy: (1) $a_{1}=1$; (2) $\left|a_{i}-a_{i+1}\right| \leqslant 2, i=1$, $2, \cdots, n-1$. Determine whether $f(1996)$ is divisible by 3. | Solution: It is easy to find that $f(1)=f(2)=1, f(3)=2$.
When $n \geqslant 4$, we have $a_{1}=1, a_{2}=2$ or 3.
(a) When $a_{2}=2$, we proceed as follows: delete $a_{1}$, $b_{1}=a_{2}-1, b_{2}=a_{3}-1, \cdots, b_{n-1}=a_{n}-1$, to get a permutation $b_{1}, \cdots, b_{n-1}$ that meets the conditions, and the number of s... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Find the largest integer $n$, such that all non-zero solutions of the equation $(z+1)^{n}=z^{n}+1$ lie on the unit circle. | $$
\begin{array}{l}
\text { Solution: Using the binomial theorem, the equation can be transformed into } \\
z\left(C_{n}^{1} z^{n-2}+C_{n}^{2} z^{n-3}+C_{n}^{3} z^{n-4}+\cdots+C_{n}^{n-2} z+\right. \\
\left.C_{n}^{n-1}\right)=0(n>3) .
\end{array}
$$
Let the non-zero solutions of the equation be $z_{i}(i=1,2, \cdots, n... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $S=1^{2}-2^{2}+3^{2}-1^{2}+\cdots+99^{2}-$ $100^{2}+101^{2}$. Then the remainder when $S$ is divided by 103 is $\qquad$ | 2. 1 .
$$
\begin{aligned}
S= & 1+\left(3^{2}-2^{2}\right)+\left(5^{2}-4^{2}\right)+\cdots+\left(99^{2}-98^{2}\right) \\
& +\left(101^{2}-100^{2}\right) \\
= & 1+2+3+\cdots+100+101 \\
= & \frac{101 \times 102}{2}=5151=103 \times 50+1 .
\end{aligned}
$$
Therefore, the required remainder is 1. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Locations A and B are respectively upstream and downstream on a lake. Each day, there is a boat that departs on time from both locations and travels at a constant speed towards each other, usually meeting at 11:00 AM on the way. ... Due to a delay, the boat from location A left 40 minutes late, and as a result, the ... | 3. 6 .
As shown in the figure, the two ships usually meet at point $A$. On the day when ship B is late, they meet at point $B$. According to the problem, ship A takes $15^{\prime}$ to navigate segment $AB$, and ship C takes
$$
\begin{array}{l}
(40-15)^{\prime}=25^{\prime} . \\
\therefore \frac{15}{60}(44+v)=\frac{25}{... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $a$ is an integer, the two real roots of the equation $x^{2}+(2 a-1) x+$ $a^{2}=0$ are $x_{1}$ and $x_{2}$. Then $\left|\sqrt{x_{1}}-\sqrt{x_{2}}\right|=$ | 4. 1.
From the problem, the discriminant $\Delta=-4 a+1 \geqslant 0$, so $a \leqslant 0$.
$$
\begin{array}{l}
\left(\sqrt{x_{1}}-\sqrt{x_{2}}\right)^{2}=\left(x_{1}+x_{2}\right)-2 \sqrt{x_{1} x_{2}} \\
=1-2 a+2 a=1 \\
\therefore\left|\sqrt{x_{1}}-\sqrt{x_{2}}\right|=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the equation $\left(a^{2}-1\right) x^{2}-2(5 a+1) x+24=0$ has two distinct negative integer roots. Then the integer value of $a$ is $\qquad$ .
（The 1st Zu Chongzhi Cup Junior High School Mathematics Competition） | (Solution: $a=-2$ ) | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $k$ is an integer, and the equation $\left(k^{2}-1\right) x^{2}-$ $3(3 k-1) x+18=0$ has two distinct positive integer roots. Then $k=$ $\qquad$
（4th Hope Forest Junior High School Mathematics Competition） | (Solution: $k=2$ ) | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $a$ is an integer, the equation $x^{2}+(2 a+1) x+a^{2}=0$ has integer roots $x_{1} 、 x_{2}, x_{1}>x_{2}$. Try to find $\sqrt[4]{x_{1}^{2}}-\sqrt[4]{x_{2}^{2}}$.
(1991, Nanchang City Junior High School Mathematics Competition) | (Given $a>0$, we know $0>x_{1}>x_{2}$, the result is -1) | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 For any $a>1, b>1$,
we have $\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant 8$.
(1992, 26th Commonwealth Mathematics Olympiad (10th grade)) | To prove for $a:=1+t_{1}, b:=1+t_{2}$, where $t_{1}, t_{2}>0$. The inequality to be proven becomes
$$
\begin{array}{l}
\frac{\left(1+t_{1}\right)^{2}}{t_{2}}+\frac{\left(1+t_{2}\right)^{2}}{t_{1}} \geqslant 8 . \\
\text { The left side of the above equation } \geqslant \frac{2\left(1+t_{1}\right)\left(1+t_{2}\right)}{\... | 8 | Inequalities | proof | Yes | Yes | cn_contest | false |
5. Product $\prod_{k=1}^{7}\left(1+2 \cos \frac{2 k \pi}{7}\right)=$ $\qquad$ | 5.3 .
$$
\begin{array}{l}
\text { Let } \omega=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7} \text {. Then } \omega^{7}=1 . \\
\omega^{k}=\cos \frac{2 k \pi}{7}+i \sin \frac{2 k \pi}{7}, \\
\omega^{-k}=\cos \frac{2 k \pi}{7}-i \sin \frac{2 k \pi}{7}, \\
\therefore \omega^{k}+\omega^{-k}=2 \cos \frac{2 k \pi}{7} . \\
\tex... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, on the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$ there are 16 points, sequentially $P_{1}, P_{2}, \cdots, P_{16}, F$ is the left focus, and the angles between each adjacent pair of points and $F$ are equal $\left(\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\cdots=\angle P_{16} F P_{1}\right)$. Let the dis... | $$
\begin{array}{l}
a=5, b \\
=4, c=3 . \text { Let } \angle X F P_{1} \\
=\alpha \cdot, \angle P_{1} F P_{2}= \\
\angle P_{2} F P_{3}=\cdots= \\
\angle P_{16} F P_{1}=\frac{\pi}{8} . \\
F M=d:-P_{i} F \cos \left[(i-1) \frac{\pi}{8}+\alpha\right] \\
\quad=\frac{a^{2}}{c}-c=\frac{16}{3} .
\end{array}
$$
By the definiti... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
63. Let $a_{1}=1997^{1997^{1997}}{ }^{.197}$ (1997 sevens in total), the sum of the digits in the decimal representation of $a_{1}$ is $a_{2}$, the sum of the digits of $a_{2}$ is $a_{3}$, and so on. Find $a_{2000}$. | Solution: Let $x_{n}=10000^{10000^{10000}}$ (with $n$ 10000s), $n=1,2, \cdots$, then for $n \geqslant 2$, we have
$$
\begin{array}{l}
x_{n}=10000^{x_{n-1}}=10^{4 x_{n-1}} . \\
\therefore a_{1}<x_{1997}=10^{4 x_{1996}}, \\
a_{2} \leqslant 9 \times 4 x_{1996}<100 x_{1996} \\
=100 \cdot 10^{4 x 1995}=10^{4 x_{1995}+2}, \... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $m=\sqrt{5}+1$. Then the integer part of $m+\frac{1}{m}$ is $\qquad$ . | \begin{array}{l}\text { 1. } m=\sqrt{5}+1, \frac{1}{m}=\frac{1}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{4}, \\ \therefore m+\frac{1}{m}=\frac{5}{4} \sqrt{5}+\frac{3}{4},\left[m+\frac{1}{m}\right]=3 .\end{array} | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $x^{2}-x-1=0$. Then, the value of the algebraic expression $x^{3}$ $-2 x+1$ is $\qquad$ . | $\begin{array}{l} \text { 3. } x^{3}-2 x+1 \\ \quad=\left(x^{3}-x^{2}-x\right)+\left(x^{2}-x-1\right)+2 \\ =x\left(x^{2}-x-1\right)+\left(x^{2}-x-1\right)+2=2\end{array}$
The translation is as follows:
$\begin{array}{l} \text { 3. } x^{3}-2 x+1 \\ \quad=\left(x^{3}-x^{2}-x\right)+\left(x^{2}-x-1\right)+2 \\ =x\left(x... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $m$ and $n$ are rational numbers, and the equation $x^{2}+$ $m x+n=0$ has a root $\sqrt{5}-2$. Then the value of $m+n$ is $\qquad$ . | 4. Since $m, n$ are rational, the other root is $-\sqrt{5}-2$, thus by Vieta's formulas,
$$
\begin{array}{l}
w:=4, n=-1 . \\
\therefore m+n=3 .
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. The number of integer pairs $(m, n)$ that satisfy $1998^{2}+m^{2}=1997^{2}+n^{2}(0<m$ $<n<1998)$ is $\qquad$.
| 6. $n^{2}-m^{2}=3995=5 \times 17 \times 47,(n-m)(n+m)=5 \times 17 \times 47$, obviously any integer factorization of 3995 can yield $(m, n)$, given the condition $(0<m<n<1998)$, thus there are 3 integer pairs $(m, n)$ that satisfy the condition. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $a, b$ be real numbers. Then the minimum value of $a^{2}+a b+b^{2}-$ $a-2 b$ is $\qquad$. | 11 .
$$
\begin{array}{l}
a^{2}+a b+b^{2}-a-2 b \\
=a^{2}+(b-1) a+b^{2}-2 b \\
=\left(a+\frac{b-1}{2}\right)^{2}+\frac{3}{4} b^{2}-\frac{3}{2} b-\frac{1}{4} \\
=\left(a+\frac{b-1}{2}\right)^{2}+\frac{3}{4}(b-1)^{2}-1 \geqslant-1 .
\end{array}
$$
When $a+\frac{b-1}{2}=0, b-1=0$,
i.e., $a=0, b=1$, the equality in the abo... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Every book has an international book number:
ABCDEFGHIJ
where $A B C D E F G H I$ are composed of nine digits, and $J$ is the check digit.
$$
\text { Let } \begin{aligned}
S= & 10 A+9 B+8 C+7 D+6 E+5 F \\
& +4 G+3 H+2 I,
\end{aligned}
$$
$r$ is the remainder when $S$ is divided by 11. If $r$ is not 0 or 1, then $J... | $$
\text { 15. } \begin{aligned}
S=9 \times 10 & +6 \times 9+2 \times 8+y \times 7+7 \times 6 \\
& +0 \times 5+7 \times 4+0 \times 3+1 \times 2
\end{aligned}
$$
$\therefore S$ the remainder when divided by 11 is equal to the remainder when $7 y+1$ is divided by 11.
From the check digit, we know that the remainder when... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given that the line $y=-2 x+3$ intersects the parabola $y=$ $x^{2}$ at points $A$ and $B$, and $O$ is the origin. Then, the area of $\triangle O A B$ is $\qquad$ . | 7.6 .
As shown in the figure, the line $y=$
$-2 x+3$ intersects the parabola $y$ $=x^{2}$ at points $A(1,1), B(-3,9)$.
Construct $A A_{1}, B B_{1}$ perpendicular to the $x$-axis, with feet of the perpendiculars at
$$
\begin{array}{l}
A_{1} 、 B_{1} . \\
\therefore S_{\triangle O A B}=S_{\text {trapezoid } A A_{1} B_{1}... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $n$ be a positive integer. $0<x \leqslant 1$. In $\triangle A B C$, if $A B=n+x, B C=n+2 x, C A=n+3 x, B C$ has a height $A D=n$. Then, the number of such triangles is ( ).
(A) 10
(B) 11
(C) 12
(D) infinitely many | 6. (C).
Let $\triangle A B C$ be the condition; then $c=n+1, a=n+2 x$, $b=n+3 x, p=\frac{1}{2}(a+b+c)=\frac{1}{2}(3 n+6 x)$. By Heron's formula, we get
$$
\begin{array}{l}
\sqrt{\frac{3 n+6 x}{2} \cdot \frac{n+4 x}{2} \cdot \frac{n+2 x}{2} \cdot \frac{n}{2}}=\frac{n(n+2 x)}{2} . \\
\therefore \frac{3 n(n+2 x)^{2}(n+4 ... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false |
Example 2 Let $[x]$ denote the greatest integer not greater than the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is $\qquad$
(1995, National High School Mathematics Competition) | Analysis: The difficulty of this problem lies in the uncertainty of $[\lg x]$. Since $[\lg x] \leqslant \lg x$, the original problem is first transformed into finding the values of $x$ that satisfy the inequality $\lg ^{2} x-\lg x-2 \leqslant 0$.
Solving this, we get $-1 \leqslant \lg x \leqslant 2$.
Therefore, the num... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. For all real numbers $x, y$, if the function $f$ satisfies:
$$
f(x y)=f(x) \cdot f(y)
$$
and $f(0) \neq 0$, then $f(1998)=$ $\qquad$ . | $\approx 、 1.1$.
Let $f(x y)=f(x) \cdot f(y)$, set $y=0$, we have $f(x \cdot 0)=$ $f(x) \cdot f(0)$, which means $f(0)=f(x) \cdot f(0)$, and since $f(0) \neq 0$, it follows that $f(x)=1$, thus $f(1998)=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Quadrilateral $ABCD$ is inscribed in a circle, $BC=CD=4$, $AC$ and $BD$ intersect at $E$, $AE=6$, and the lengths of $BE$ and $DE$ are both integers. Then the length of $BD$ is $\qquad$ | 4.7.
From $\overparen{B C}=\overparen{C D}, \angle B D C=\angle C A D, \angle A C D=\angle A C D$, we know $\triangle D C E \backsim \triangle A C D$, so $\frac{C E}{C D}=\frac{C D}{A C}$, which means $C E \cdot A C=C D^{2}$. Therefore, $C E(C E+6)=16$, solving this gives $C E=2$. Also, from $B E \cdot D E=A E \cdot C... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (Full marks 25 points) As shown in the figure, $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at $M$, and the angle between their two external common tangents is $60^{\circ}$. The line connecting the centers intersects $\odot O_{1}$ and $\odot O_{2}$ at $A$ and $B$ (different from $M$), respectively. A line... | From the given conditions, we have $O_{1} E \perp P E, O_{1} F \perp P F, O_{1} E= O_{1} F$, thus $O_{1}$ lies on the bisector of $\angle E P F$. Similarly, $O_{2}$ lies on the bisector of $\angle E P F$. Therefore, $P A$ is the bisector of $\angle E P F$. Since $O_{2} Q / / P E$, we have $\angle Q O_{2} O_{1}=\angle E... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $f(x)=x^{10}+2 x^{9}-2 x^{8}-2 x^{7}+x^{6}$ $+3 x^{2}+6 x+1$, then $f(\sqrt{2}-1)=$ | $=1.4$.
Let $x=\sqrt{2}-1$, then $x+1=\sqrt{2} \Rightarrow (x+1)^{2}=2 \Rightarrow x^{2}+2x-1=0$. That is, $x=\sqrt{2}-1$ is a root of $x^{2}+2x-1=0$. But $f(x)=(x^{8}-x^{6}+3)(x^{2}+2x-1)+4$, so $f(\sqrt{2}-1)=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If
$$
\dot{z}=\frac{(1+i)^{2000}(6+2 i)-(1-i)^{1998}(3-i)}{(1+i)^{1996}(23-7 i)+(1-i)^{1994}(10+2 i)} \text {, }
$$
then $|z|=$ . $\qquad$ | 3.1.
$$
\begin{array}{l}
\text { When }(1+i)^{2}=2 i, (1-i)^{2}=-2 i \text { and } X \in \mathbb{Z}, \\
i^{4 k+1}=i, i^{4 k+2}=-1, i^{4 k+3}=-i, i^{4 k}=1. \\
\text { Therefore, }(1+i)^{2000}=(2 i)^{1000}=2^{1000}, \\
(1-i)^{1998}=(-2 i)^{999}=2^{999} \cdot i, \\
(1+i)^{1996}=(2 i)^{998}=-2^{998}, \\
(1-i)^{1994}=(-2 i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-3 x-\right.$ $3)^{2001}$ is expanded and like terms are combined. The sum of the coefficients of the odd powers of $x$ in the resulting expression is $\qquad$. | 4. -1 .
$$
\text { Let } \begin{aligned}
f(x)= & \left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-3 x-3\right)^{2001} \\
= & A_{0}+A_{1} x+A_{2} x^{2}+\cdots+A_{4001} x^{4001} \\
& +A_{4002} x^{4002} .
\end{aligned}
$$
$$
\begin{array}{l}
\text { Then } A_{0}+A_{1}+A_{2}+\cdots+A_{4001}+A_{4002} \\
\quad=f(1)=0, \\
A_{0}-A_... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (Full marks 20 points) An arithmetic sequence with a common difference of 4 and a finite number of terms, the square of its first term plus the sum of the rest of the terms does not exceed 100. Please answer, how many terms can this arithmetic sequence have at most?
保留源文本的换行和格式，所以翻译结果如下：
Four, (Full marks 20 po... | Let the arithmetic sequence be $a_{1}, a_{2}, \cdots, a_{n}$, with common difference $d=4$. Then
$$
\begin{array}{l}
a_{1}^{2}+a_{2}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+\frac{2 a_{1}+4 n}{2}(n-1) \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0 .
\end{a... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Initial 65. Given a real-coefficient polynomial function $y=a x^{2}+b x+c$, for any $|x| \leqslant 1$, it is known that $|y| \leqslant 1$. Try to find the maximum value of $|a|+|b|+|c|$. | Proof: First, we prove the following auxiliary proposition.
Proposition: Let real numbers $A, B$ satisfy $|A| \leqslant 2, |B| \leqslant 2$. Then $|A+B| + |A-B| \leqslant 4$.
In fact, without loss of generality, let $|A| \geqslant |B|$.
From $A^2 \geqslant B^2$, we have $(A+B)(A-B) \geqslant 0$,
thus $|A+B| + |A-B| = |... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 If any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>$ 0 , to make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is $\qquad$
(1993, National High School Mathematics Competi... | Analysis: The inequality can be transformed into
$$
\begin{array}{l}
k \leqslant-\frac{\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993}{\log _{\frac{x_{0}}{x_{3}}} 1993} \\
=f\left(x_{0}, x_{1}, x_{2}, x_{3}\right) \text {. } \\
\end{array}
$$
Thus,
$$
\{k\}_{\text {m... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 12 Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\sin y \cos y+a=$ 0. Then $\cos (x+2 y)=$ $\qquad$
(1994, National High School Mathematics Competition) | Analysis: Since $2 a=x^{3}+\sin x=(-2 y)^{3}+$ $\sin (-2 y)$, if we let $f(t)=t^{3}+\sin t$, then we have $f(x)$ $=f(-2 y)$.
And $f(t)$ is strictly increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so by monotonicity we have
$$
x=-2 y \text {, hence } x+2 y=0 \text {. }
$$
Thus $\cos (x+2 y)=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, let integers $a, b, c$ satisfy $1 \leqslant a<b<c \leqslant 5, a^{3}$, $b^{3}, c^{3}$ have unit digits $x, y, z$ respectively. When $(x-y)(y-z)(z-x)$ is the smallest, find the maximum value of the product $a b c$.
Let integers $a, b, c$ satisfy $1 \leqslant a<b<c \leqslant 5, a^{3}$, $b^{3}, c^{3}$ have unit di... | $$
\text { Three, } \because 1^{3}=1,2^{3}=8,3^{3}=27,4^{3}=64,5^{3}=125 \text {. }
$$
$\therefore(x, y, z)$ has the following 10 possibilities:
(1) $(1,8,7) ;(2)(1,8,4) ;(3)(1,8,5)$;
(4) $(1,7,4) ;(5)(1,7,5) ;(6)(1,4,5)$;
(7) $(8,7,4) ;(8)(8,7,5) ;(9)(8,4,5)$;
$(10)(7,4,5)$.
Then the values of $(x-y)(y-z)(z-x)$ are
$$... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, on a circular road, there are 4 middle schools arranged clockwise: $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$. They have 15, 8, 5, and 12 color TVs, respectively. To make the number of color TVs in each school the same, some schools are allowed to transfer color TVs to adjacent schools. How should the TVs be transferred ... | Let $A_{1}$ High School transfer $x_{1}$ color TVs to $A_{2}$ High School (if $x_{1}$ is negative, it means $A_{2}$ High School transfers $|x_{1}|$ color TVs to $A_{1}$ High School. The same applies below), $A_{2}$ High School transfer $x_{2}$ color TVs to $A_{3}$ High School, $A_{3}$ High School transfer $x_{3}$ color... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: Xiao Zhang is riding a bicycle on a road next to a double-track railway. He notices that every 12 minutes, a train catches up with him from behind, and every 4 minutes, a train comes towards him from the opposite direction. If the intervals between each train are constant, the speeds are the same, and both t... | Solution: Let the trains depart from the station ahead and behind Xiao Zhang every $x$ minutes, Xiao Zhang's cycling speed be $v_{1}$, and the train speed be $v_{2}$, then
$$
\left\{\begin{array}{l}
4\left(v_{1}+v_{2}\right)=x v_{2}, \\
12\left(v_{2}-v_{1}\right)=x v_{2} .
\end{array}\right.
$$
Solving, we get $x=6$ (... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 A person walks from place A to place B, and there are regular buses running between A and B, with equal intervals for departures from both places. He notices that a bus going to A passes by every 6 minutes, and a bus going to B passes by every 12 minutes. How often do the buses depart from their respective st... | Analysis: Let the distance between two consecutive buses in the same direction be $s$, and the time be $t$ minutes, then the speed of the bus is $\frac{s}{t}$.
(1) A person's speed relative to the bus going to location A is $\frac{s}{6}$, so the person's speed is $\frac{s}{6}-\frac{s}{t}$.
(2) A person's speed relative... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
One, (Full marks 20 points) Given that $x$, $y$, $z$ are positive integers, and satisfy $x^{3}-y^{3}-z^{3}=3 x y z$, $x^{2}=2(y+z)$. Find the value of $x y+y z+z x$.
---
The translation is provided as requested, maintaining the original format and line breaks. | $$
\begin{aligned}
- & \because x^{3}-y^{3}-z^{3}-3 x y z \\
= & x^{3}-(y+z)^{3}-3 x y z+3 y^{2} z+3 y z^{2} \\
= & (x-y-z)\left(x^{2}+x y+x z+y^{2}+2 y z+z^{2}\right) \\
& -3 y z(x-y-z) \\
= & (x-y-z)\left(x^{2}+y^{2}+z^{2}-x y-y z+x z\right),
\end{aligned}
$$
X. $x^{3}-y^{3}-z^{3}=3 x y z$,
$$
\begin{array}{l}
\there... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Full marks 25 points) On the blackboard, all natural numbers from 1 to 1997 are written. Students $A$ and $B$ take turns to perform the following operations: Student $A$ subtracts the same natural number from each number on the blackboard (the number subtracted can be different in different operations); Student... | Three, because after each operation by student $A$ and student $B$, the number of numbers written on the blackboard decreases by 1. Since students $A$ and $B$ take turns operating, when $B$ completes the last operation, only one number remains on the blackboard, and both have performed 1996 operations. Let $d_{k} (k=1,... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. From the center of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, two perpendicular chords $A C$ and $B D$ are drawn. Connecting $A, B, C, D$ in sequence forms a quadrilateral. Then, the maximum value of the area $S$ of quadrilateral $A B C D$ is | 3. 12.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Given $O A \perp O B$, we have
$$
\begin{array}{l}
x_{1} x_{2}+y_{1} y_{2}=0 . \\
\begin{aligned}
\therefore S & =4 S_{\triangle A O B}=2|O A| \cdot|O B| \\
& =2 \sqrt{\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)} \\
& =2 \sqrt{\l... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Given a pyramid $S-ABCD$ inscribed in a sphere with the base being a rectangle $ABCD$, and $SA=4, SB=8, SD=7$, $\angle SAC=\angle SBC=\angle SDC$. Then the length of $BD$ is | 5. $B D=9$.
As shown in the figure, $\because \angle S A C=\angle S B C=\angle S D C$,
$$
\therefore \frac{S C}{\sin \angle S A C}=\frac{S C}{\sin \angle S B C}=\frac{S C}{\sin \angle S D C} \text {. }
$$
Therefore, the circumcircles of $\triangle S A C$,
$\triangle S B C, \triangle S D C$ have equal diameters, meani... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (Full marks 20 points) The complex numbers $z_{1}, z_{2}, z_{3}, z_{4}, z_{5}$ satisfy
$$
\left\{\begin{array}{l}
\left|z_{1}\right| \leqslant 1, \\
\left|z_{2}\right| \leqslant 1, \\
\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right|, \\
\left|2 z_{4}-\left(z_{1}+z_{2}\right)\right|... | $$
\begin{array}{l}
\text { 3, } \because\left|z_{1}-z_{2}\right| \geqslant\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \\
\quad \geqslant|2| z_{3}|-| z_{1}+z_{2}||, \\
\therefore\left|z_{1}+z_{2}\right|-\left|z_{1}-z_{2}\right| \leqslant 2\left|z_{3}\right| \\
\quad \leqslant\left|z_{1}+z_{2}\right|+\left|z_{1}-z_{2}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 13 Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ all be natural numbers, and $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5}$. Try to find the maximum value of $x_{5}$. | Solution: Without loss of generality, let $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}$.
$$
\because x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} \text {. }
$$
$x_{4} x_{5}$,
$$
\begin{aligned}
\therefore 1= & \frac{1}{x_{2} x_{3} x_{4} x_{5}}+\frac{1}{x_{1} x_{3} x_{4} x_{5}} \\
& +\frac{1... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If $x y=1$, then the minimum value of the algebraic expression $\frac{1}{4}+\frac{1}{4 y^{4}}$ is $\qquad$ | Solution: From $a+b \geqslant 2 \sqrt{a b}$ we know $\frac{1}{x^{4}}+\frac{1}{4 y^{4}} \geqslant 2 \sqrt{\frac{1}{x^{4} \cdot 4 y^{4}}}=1$. Therefore, the minimum value of $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example: $3 A, B, C, D, E$ five people participate in an exam, with 7 questions, all of which are true or false questions. The scoring rule is: for each question, a correct answer earns 1 point, a wrong answer deducts 1 point, and no answer neither earns nor deducts points. Figure 1 records the answers of $A, B, C, D, ... | Let: Assign $k=1,2, \cdots, 7$. When the conclusion of the $k$-th question is correct, i.e., $x_{k}:=1$, if it is judged as correct (i.e., marked with the symbol “$\checkmark$”), then $x_{k}$ points are scored; if it is judged as incorrect (i.e., marked with the symbol “$X$”), then $-x_{k}$ points are scored. When the ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given that $x, y, z$ are all positive numbers, and $x y z \cdot (x+y+z)=1$. Then, the minimum value of $(x+y)(y+z)$ is . $\qquad$ | Analysis: To find the minimum value of the original expression, that is, to find the minimum value of $x y+y^{2}+$ $x z+y z$, we can achieve this by appropriately combining terms so that the sum becomes the sum of two terms, and the product of these two terms should be a constant.
$$
\text { Solution: } \begin{array}{l... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the complex number $z=\cos \theta+i \sin \theta\left(0^{\circ} \leqslant \theta \leqslant\right.$ $\left.180^{\circ}\right)$, and the complex numbers $z, (1+i)z, 2\bar{z}$ correspond to three points $P, Q, R$ in the complex plane. When $P, Q, R$ are not collinear, the fourth vertex of the parallelogram formed by... | 2.3.
Let the complex number $w$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have
$$
w+z=2 \bar{z}+(1+i) z \text{, i.e., } w=2 \bar{z}+i z \text{. }
$$
Therefore, $|w|^{2}=(2 \bar{z}+i z)(2 z-i \bar{z})$
$$
\begin{array}{l}
=4+1+2 i\left(z^{2}-\bar{z}^{2}\right) \\
=5-4 \sin 2 A<5 \div 4=9 .
\end{a... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In an arithmetic sequence with real number terms, the common difference is 4, and the square of the first term plus the sum of the remaining terms does not exceed 100. Such a sequence can have at most terms. | 4.8 .
Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence with a common difference of 4, then
$$
\begin{aligned}
& a_{1}^{2}+a_{2}+a_{3}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow & a_{1}^{2}+\frac{\left(a_{1}+4\right)+\left[a_{1}+4(n-1)\right]}{2} \cdot(n-1) \\
& \leqslant 100 \\
\Leftrightarrow & a_{1}^{2... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 When $x$ varies, the minimum value of the fraction $\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$
(1993, National Junior High School Competition) | Let $y=\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$. Then,
$$
\left(3-\frac{1}{2} y\right) x^{2}+(6-y) x+(5-y)=0 \text {. }
$$
Since $x$ is a real number, $\Delta \geqslant 0$, so,
$$
y^{2}-10 y+24 \leqslant 0 \text {. }
$$
Thus, $4 \leqslant y \leqslant 6$.
When $y=4$, $x=1$.
Therefore, when $x=1$, the minimum value... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Four cities each send 3 political advisors to participate in $k$ group inspection activities (each advisor can participate in several groups), with the rules: (1) advisors from the same city are not in the same group; (2) any two advisors from different cities exactly participate in one activity together. Then the m... | 2.9 Group.
First, consider the CPPCC members from two cities, Jia and Yi. Let the members from Jia city be $A_{1}, A_{2}, A_{3}$, and the members from Yi city be $B_{1}, B_{2}, B_{3}$. They can form 9 pairs: $A_{1} B_{1}, A_{1} B_{2}, A_{1} B_{3}, A_{2} B_{1}, A_{2} B_{2}, A_{2} B_{3}$, $A_{3} B_{1}, A_{3} B_{2}, A_{3... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. In $\triangle A B C$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given $a^{2}+b^{2}-7 c^{2}=0$, then $\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=$ | 6.3 .
$$
\begin{array}{l}
\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}=\frac{\cos C \sin A \sin B}{\sin (A+B) \sin C} \\
=\cos C \cdot \frac{\sin A \sin B}{\sin ^{2} C}=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \cdot \frac{a b}{c^... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{c}
\text { II. (50 points) }(1) \text { For } 0 \leqslant x \leqslant 1, \text { find the range of the function } \\
h(x)=(\sqrt{1+x}+\sqrt{1-x}+2) . \\
\left(\sqrt{1-x^{2}}+1\right)
\end{array}
$$
(2) Prove: For $0 \leqslant x \leqslant 1$, there exists a positive number $\beta$ such that the inequal... | $$
\begin{array}{l}
=(1) 00$, the inequality $\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{a}}{\beta}(x \in[0,1])$ does not hold. Conversely, i.e., $-\frac{2 x^{2}}{h(x)} \leqslant-\frac{x^{0}}{\beta}$, which means $x^{2 \cdots} \geqslant \frac{h(x)}{2 \beta}$ holds.
Since $2-\alpha>0$, let $x \rightarrow 0$, we get
$$
0... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
For a constant $p \in N$, if the indeterminate equation $x^{2}+y^{2}=$ $p(x y-1)$ has positive integer solutions, prove that $p=5$ must hold. | Proof: Let $x_{0}, y_{0}$ be positive integer solutions of the equation. If $x_{0}=y_{0}$, substituting gives $p=(p-2) x_{0}^{2}$.
$$
\therefore x_{0}^{2}=\frac{p}{p-2}=1+\frac{2}{p-2} \in \mathbb{N} \text { : }
$$
Then $x_{0}^{2}=2$ or 3, which contradicts $x_{0} \in \mathbb{N}$.
Assume $x_{0}>y_{0} \geqslant 2$. Con... | 5 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 15 Find the minimum value of $\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Construct right triangles $\triangle P A C$ and $\triangle P B I$ as shown in Figure 1, such that
$$
\begin{array}{l}
A C=1, B D=2, P C \\
=x, C I=4, \text { and }
\end{array}
$$
$P C$ and $P D$ lie on line $l$. Then the problem of finding the minimum value is converted to "finding a point $I$ on line $l$ suc... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 If the fractional parts of $9+\sqrt{13}$ and $9-\sqrt{13}$ are $a$ and $b$ respectively, then $a b-4 a+3 b-2=$ $\qquad$ | Solution: $\because 3<\sqrt{13}<4$,
$\therefore 9+\sqrt{13}$ has an integer part of 12, and a decimal part
$$
\begin{array}{l}
a=\sqrt{13}-3 . \\
\because-4<-\sqrt{13}<-3,
\end{array}
$$
i.e., $0<4-\sqrt{13}<1$,
$\therefore 9-\sqrt{13}$ has an integer part of 5, and a decimal part $b$ $=4-\sqrt{13}$.
Thus, $a b-4 a+3... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Simplify $\frac{2 b-a-c}{(a-b)(b-c)}$
$$
+\frac{2 c-a-b}{(b-c)(c-a)}+\frac{2 a-b-c}{(c-a)(a-b)} .
$$ | $\begin{array}{l}\text { Solution: Original expression }=\frac{(b-c)-(a-b)}{(a-b)(b-c)} \\ \quad+\frac{(c-a)-(b-c)}{(b-c)(c-a)}+\frac{(a-b)-(c-a)}{(c-a)(a-b)} \\ =\frac{1}{a-b}-\frac{1}{b-c}+\frac{1}{b-c}-\frac{1}{c-a} \\ \quad+\frac{1}{c-a}-\frac{1}{a-b}=0 .\end{array}$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 If $x=\sqrt{19-8 \sqrt{3}}$. Then the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15}=$
Preserve the original text's line breaks and format, output the translation result directly. | Given: From the above, $x=\sqrt{(4-\sqrt{3})^{2}}=4-\sqrt{3}$, thus $x-4=-\sqrt{3}$.
Squaring and simplifying the above equation yields
$$
x^{2}-8 x+13=0 \text {. }
$$
Therefore, the denominator $=\left(x^{2}-8 x+13\right)+2=2$.
By long division, we get
the numerator $=\left(x^{2}-8 x+13\right)\left(x^{2}+2 x+1\right)... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. For a $4 n+2$-sided polygon $A_{1} A_{2} \cdots A_{4 n-2}$ (where $n$ is a natural number), each interior angle is an integer multiple of $30^{\circ}$, and $\angle A_{1}=\angle A_{2}=\angle A_{3}=90^{\circ}$, then the possible values of $n$ are $\qquad$ | 4. 1 .
From the problem, we get
$$
\begin{array}{l}
90^{\circ} \times 3+(4 n+2-3) \times 30^{\circ} \mathrm{k} \\
=(4 n+2-2) \times 180^{\circ},
\end{array}
$$
where $k \geqslant 4 n-1$.
Simplifying, we get $(4 n-1) k=24 n-9$,
which means $24 n-9 \geqslant(4 n-1)^{2}$.
Thus, $8 n^{2}-16 n+5 \leqslant 0$.
Clearly, $0... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. The solution set of the inequality $\frac{1}{x-1}+\frac{2}{x-2} \geqslant \frac{3}{2}$, is the union of some non-overlapping intervals with a total length of $\qquad$. | 5.2.
For $g(x)=(x-1)(x-2)>($ or $\text { (or } \leqslant \text { ) } 0 \text {. }$
The graph of $y=f(x)$ is a parabola opening upwards, and the graph of $y=$ $g(x)$ is a parabola opening upwards.
For $f(x):$ when $x=1$, $y0$.
Therefore, the graph intersects the
$x$-axis at two points
$$
\begin{array}{l}
\left(x_{1}, 0... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Four, for what real number $x$ does $y=x^{2}-x+1+$ $\sqrt{2(x+3)^{2}+2\left(x^{2}-5\right)^{2}}$ have a minimum value? What is the minimum value? | As shown in Figure 5, let \( P\left(x, x^{2}\right) \) be a point on the parabola \( y=x^{2} \). The distance from \( P \) to the line \( y=x-1 \) is
\[
\begin{aligned}
|P Q| & =\frac{\left|x^{2}-x+1\right|}{\sqrt{(-1)^{2}+1^{2}}} \\
& =\frac{\sqrt{2}}{2} \left| x^{2}-x+1 \right| \\
& =\frac{\sqrt{2}}{2}\left(x^{2}-x+1... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For example, 610 people go to the bookstore to buy books, it is known that
(1) each person bought three books;
(2) any two people have at least one book in common.
How many people at most could have bought the book that was purchased by the fewest people?
(1993, China Mathematical Olympiad) | Solution: Let's assume that person A bought three books, and since A has at least one book in common with each of the other 9 people, among A's three books, the book purchased by the most people is bought by no less than 4 people.
If the book purchased by the most people is bought by 4 people, then all three books bou... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $S=\{1,2,3,4\}$. An $n$-term sequence: $q_{1}, q_{2}, \cdots, q_{n}$ has the following property: For any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted by $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
(1997, Shan... | Solution: First, each number in $S$ appears at least twice in the sequence $q_{1}, q_{2}$, $\cdots, q_{n}$, otherwise, since there are 3 binary subsets containing a certain number, but in the sequence, the number of adjacent pairs containing this number is at most 2, therefore, $n \geqslant 8$.
Moreover, the 8-term se... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For $n \in \mathbf{N}$, 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 $a_{1}+a_{2}+\cdots+a_{n}=17$. If there exists a unique $n$ such that $S_{n}$ is also an integer, find the value of $n$. | Solution: Based on the structure of $\sqrt{(2 k-1)^{2}+a_{k}^{2}}$, we construct the complex number $z_{k}=(2 k-1)+\sigma_{k} i, k=1,2, \cdots, n$.
Then $\sqrt{(2 k-1)^{2}+a_{k}^{2}}=\left|(2 k-i)+a_{k} i\right|$.
Thus $\sum_{k=1}^{n} \vee \sqrt{(2 k-1)^{2}+a_{k}^{2}}$
$=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} i\right|$
$=\... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 The elements of set $A$ are all integers, the smallest of which is 1, and the largest is 100. Except for 1, each element is equal to the sum of two numbers (which can be the same) in set $A$. Find the minimum number of elements in set $A$.
| Solution: Construct a set with as many elements as possible to satisfy the conditions, such as $\{1,2,3,5,10,20,25,50, 9\}$.
Extend $\left\{1,2, x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, 100\right\}$ to also satisfy the conditions, then $x_{1} \leqslant 4, x_{2} \leqslant 8, x_{3} \leqslant 16, x_{4} \leqslant 32, x_{5}$ $\l... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 For a finite set $A$, there exists a function $f: N \rightarrow$ $A$, with the following property: if $i, j \in N$, and $|i-j|$ is a prime number, then $f(i) \neq f(j)$. How many elements must the set $A$ have at least? | Solution: Since the absolute value of the difference between any two numbers among $1,3,6,8$ is a prime number, according to the problem, $f(1)$, $f(3)$, $f(6)$, and $f(8)$ are four distinct elements in $A$, thus $|A| \geqslant 4$.
On the other hand, if we let $A=\{0,1,2,3\}$, and the mapping $f: N \rightarrow A$ is d... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The solution set of the equation $\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$ is $\qquad$ . | Let $y=\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$, then we have
$$
\left\{\begin{array}{l}
5^{y}=3^{x}+4^{x}, \\
4^{y}=5^{x}-3^{x}
\end{array}\right.
$$
(1) + (2) gives $5^{y}+4^{y}=5^{x}+4^{x}$.
However, the function $f(x)=5^{x}+4^{x}$ is an increasing function, so from (3) we have
$$
f(y)=f(... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that the pure imaginary numbers $x_{1}, x_{2}, \cdots, x_{1999}$ have a modulus of 1. Then the remainder when $x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{1998} x_{1999}+x_{1999} x_{1}$ is divided by 4 is $\qquad$ | 4.1.
Obviously, if $x_{k}$ takes $i$ or $-i$, we have
$$
\begin{array}{l}
\left(i+x_{k}\right)\left(i-x_{k+1}\right) \\
=\left\{\begin{array}{l}
0, \text{ when } x_{k}=-i \text{ or } x_{k+1}=i; \\
-4, \text{ when } x_{k}=i \text{ and } x_{k+1}=-i
\end{array}\right.
\end{array}
$$
Both are multiples of 4, so the sum i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Using $1,2, \cdots, n$ to form an $n$-digit number without repeating digits, where 2 cannot be adjacent to 1 or 3, a total of 2400 different $n$-digit numbers are obtained. Then $n=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation ... | 6.7 .
Obviously, $n \geqslant 4$, otherwise 2 must be adjacent to 1 or 3.
When 2 is adjacent to 1, there are $2 \cdot(n-1)$! arrangements, and when 2 is adjacent to 3, there are also $2 \cdot(n-1)$! arrangements. When 2 is adjacent to both 1 and 3, there are $2 \cdot(n-2)$! arrangements. Therefore, the number of arran... | 7 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Four. (Full marks 20 points) As shown in Figure 8, point $O$ represents the sun, $\triangle A B C$ represents a triangular sunshade, and points $A$ and $B$ are two fixed points on the ground in the north-south direction. The sunlight $O C D$ from the due west direction casts the shadow of the sunshade onto the ground, ... | (1) As shown in Figure 17, draw the perpendicular line $O H$ from $O$ to the ground, connect $H D$ to intersect $A B$ at $E$, and connect $C E$. Then $H D$ is the projection of the oblique line $O D$ on the ground, and $\angle C D E = \theta$. Given that $A B$ is in the north-south direction and $C D$ is in the east-we... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$ and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ | $$
\begin{array}{l}
\because(b-c)^{2}=4(a-b)(c-a), \\
b^{2}-2 b c+c^{2}=4 a c-4 b c+4 a b-4 a^{2}, \\
\therefore(b+c)^{2}-4 a(b+c)+4 a^{2}=0 .
\end{array}
$$
Therefore, $[2 a-(b+c)]^{2}=0$, which means $2 a=b+c$.
$$
\therefore \frac{b+c}{a}=2 \text {. }
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a$ and $b$ are integers, and satisfy
$$
\left(\frac{\frac{1}{a}}{\frac{1}{a}-\frac{1}{b}}-\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}\right)\left(\frac{1}{a}-\frac{1}{b}\right) \frac{1}{\frac{1}{a^{2}}+\frac{1}{b^{2}}}=\frac{2}{3} \text {. }
$$
Then $a+b=$ $\qquad$ . | 3.3.
$$
\begin{array}{l}
\text { Left side }=\frac{a b}{a-b}=\frac{2}{3}, \\
\therefore(3 b-2)(3 a-2)=4 .
\end{array}
$$
Given that $a \neq b$ and they are integers, hence $3 b-2, 3 a-2$ can only take the values 1, 4 or -1, -4.
(1) Suppose $3 b-2=1, 3 a-2=4$.
Solving gives $b=1, a=2$. Therefore, $a+b=3$.
(2) Suppose $... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If the 5th term of the expansion of $\left(x \sqrt{x}-\frac{1}{x}\right)^{6}$ is $\frac{15}{2}$, then $\lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-a}\right)=$ | $\begin{array}{l}\text { II, 1.1. } \\ \because T_{5}=C_{6}^{4}(x \sqrt{x})^{2} \cdot\left(-\frac{1}{x}\right)^{4}=\frac{15}{x}, \\ \text { from } \frac{15}{x}=\frac{15}{2} \text {, we get } x^{-1}=\frac{1}{2} . \\ \therefore \lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-n}\right) \\ =\frac{\frac{1}{2}}{1-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. If the functions $f(x)$ and $g(x)$ are defined on $\mathbf{R}$, and
$$
\begin{array}{l}
f(x-y)=f(x) g(y)-g(x) f(y), f(-2) \\
=f(1) \neq 0, \text { then } g(1)+g(-1)=\ldots
\end{array}
$$
(Answer with a number). | 4. -1 .
$$
\begin{array}{l}
\because f(x-y)=f(x) g(y)-g(x) f(y), \\
\therefore f(y-x)=f(y) g(x)-g(y) f(x) \\
\quad=-[f(x) g(y)-g(x) f(y)]
\end{array}
$$
There is $f(x-y)=-f(y-x)$
$$
=-f[-(x-y)] \text {. }
$$
Then $f(-x)=-f(x)$, i.e., $f(x)$ is an odd function.
$$
\text { Hence } \begin{aligned}
f(1) & =f(-2)=f(-1-1) ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2: In a certain year, the total coal production of a coal mine, apart from a certain amount of coal used for civilian, export, and other non-industrial purposes each year, the rest is reserved for industrial use. According to the standard of industrial coal consumption of a certain industrial city in that year,... | Solution: Let the total coal production of the mine for the year be $x$, the annual non-industrial coal quota be $y$, and the industrial coal consumption of each industrial city for the year be $z$. Let $p$ be the number of years the coal can supply only one city. According to the problem, we have the system of equatio... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (Full marks 12 points) At the foot of the mountain is a pond, the scene: a steady flow (i.e., the same amount of water flows into the pond from the river per unit time) continuously flows into the pond. The pond contains a certain depth of water. If one Type A water pump is used, it will take exactly 1 hour to p... | Three, let the amount of water flowing into the pond from the spring every minute be $x$ m$^{3}$, each water pump extracts $y$ m$^{3}$ of water per minute, and the pond originally contains $z$ m$^{3}$ of water. It takes $t$ minutes for three water pumps to drain the pond. According to the problem, we have
$$
\left\{\be... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given two points $A(0,1), B(6,9)$. If there is an integer point $C$ (Note: A point with both coordinates as integers is called an integer point), such that the area of $\triangle A B C$ is minimized. Then the minimum value of the area of $\triangle A B C$ is $\qquad$ | 5.1.
Since the slope of line $AB$ is $k=\frac{3}{4}$, the line passing through point $C$ and parallel to $AB$ can be expressed as $3 y-4 x=m$. Let the coordinates of point $C$ be $\left(x_{0}, y_{0}\right)$, where $x_{0}, y_{0}$ are integers. Therefore, we have
$$
3 y_{0}-4 x_{0}=m \text {. }
$$
Thus, $m$ is an integ... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 20 points) Given $0<\alpha_{i}<\frac{\pi}{4}(i=1$, $2,3,4)$, and $\sum_{i=1}^{4} \sin ^{2} \alpha_{i}=1$. Prove: $\sum_{i=1}^{4} \frac{\sin ^{2} \alpha_{i}}{\cos 2 \alpha_{i}} \geqslant$ 2. | Let $a_{i}=\sin ^{2} \alpha_{i}$, then $0<a_{i}<\frac{1}{2}$,
$$
\begin{array}{l}
\cos 2 \alpha_{i}=1-2 \sin ^{2} \alpha_{i}=1-2 a_{i} . \\
\frac{\sin ^{2} \alpha_{i}}{\cos 2 a_{i}}=\frac{a_{i}}{1-2 a_{i}}=2\left(\frac{a_{i}^{2}}{1-2 a_{i}}+\frac{a_{i}}{2}\right) \\
=\frac{2}{1-2 a_{i}}\left(a_{i}-\frac{1-2 a_{i}}{2}\r... | 2 | Inequalities | proof | Yes | Yes | cn_contest | false |
7. Given $x=\frac{1}{\sqrt{3}+\sqrt{2}}, y=\frac{1}{\sqrt{3}-\sqrt{2}}$. Then, $x^{2}+y^{2}$ is
untranslated part: 轩隹
Note: The term "轩隹" does not have a clear meaning in this context and has been left untranslated. If you can provide more context or clarify the term, I can attempt to translate it accurately. | $=, 7.10$.
Let $x=\sqrt{3}-\sqrt{2}, y=\sqrt{3}+\sqrt{2}$, so,
$$
x^{2}+y^{2}=(\sqrt{3}-\sqrt{2})^{2}+(\sqrt{3}+\sqrt{2})^{2}=10 .
$$ | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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