problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
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10. 1 Given? three quadratic trinomials; $P_{1}(x)=x^{2}+p_{1} x$ $+q_{1}, P_{2}(x)=x^{2}-p_{2} x+q_{2}$ and $P_{3}(x)=x^{2}+p_{3} x+q_{3}$. Prove: The equation $\left|P_{1}(x)\right|+\left|P_{2}(x)\right|=\left|P_{3}(x)\right|$ has at most 8 real roots. | 10.1 Each root of the original equation should be a root of a quadratic trinomial of the form $\pm p_{1}: p_{2} \pm p_{3}$, and with different choices of signs, there are 8 such quadratic trinomials. Since the coefficient of $x^{2}$ has the form $\pm 1 \pm 1 \pm 1$, the coefficient of $x^{2}$ will never be 0 regardless... | 8 | Algebra | proof | Yes | Yes | cn_contest | false |
3. Let $a$ be the decimal part of $\sqrt{3}$, $b$ be the decimal part of $\sqrt{2}$: $\frac{a}{(a-b) b}$ has an integer part of $\qquad$ | $$
\begin{array}{l}
\text { 3. } \frac{a}{(a-b) b}=\frac{\sqrt{3}-1}{(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)} \\
=\frac{(\sqrt{3}-\sqrt{2})+(\sqrt{2}-1)}{(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)} \\
=\frac{1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}-\sqrt{2}} \\
=\sqrt{3}+2 \sqrt{2}+1 \\
\approx 5.56 .
\end{array}
$$
The integer part is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $p, q, \frac{2 p-1}{q}, \frac{2 q-1}{p}$ all be integers, and $p>1, q>$
1. Then $p+q=$ $\qquad$ . | 6. $p+q=8$.
From $\frac{2 q-1}{p}, \frac{2 p-1}{q}$ both being positive integers, we know that one of them must be less than 2. Otherwise, $\frac{2 q-1}{p}>2, \frac{2 p-1}{q}>2$, leading to $2 p-1+2 q-1>2 p+2 q$, which is a contradiction. Assume $0<\frac{2 q-1}{p}<2$, then it must be that $2 q-1=p$. It is easy to see ... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. As shown in the figure, in pentagon $A B C D E$, $\angle B=\angle A E D=$ $90^{\circ}, A B=C D=A E=B C+D E=1$. Find the area of this pentagon. (1992, Beijing Junior High School Grade 2 Mathematics Competition) | Analysis We immediately think of connecting $A C, A D$, because there are two right-angled triangles. But we also find that it is not easy to directly calculate the area of each triangle. Given the condition $B C+D E=1$, we wonder if we can join $B C$ to one end of $D E$, making $E F=B C$. Connect $A F$. Then we find $... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. As shown in the figure, let $D$ and $E$ be on the sides $AC$ and $AB$ of $\triangle ABC$, respectively. $BD$ and $CE$ intersect at $F$, $AE = EB$, $\frac{AD}{DC} = \frac{2}{3}$, and $S_{\triangle ABC} = 40$. Find $S_{\text{quadrilateral AEFD}}$. (6th National Junior High School Mathematics Correspondence Com... | Analyzing, first connect $A F$ and then set:
$$
S_{\triangle A E P}=x, S_{\triangle A P D}=y, S_{\triangle C D F}=z, S_{\triangle B E P}=t\left(S_{\triangle B C F}\right.
$$
$=u$ ), to find $S_{\text {quadrilateral } A E F D}$, we only need to find the values of $x$ and $y$.
We easily get:
$$
\begin{array}{l}
x+y+z=20,... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $S=\{1,2, \cdots, 10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$ that satisfy: (1) $\left|A_{i}\right|=5, i=1,2, \cdots, k_{i}$ (2) $\mid A_{i} \cap$ $A_{j} \mid \leqslant 2, i, j=1,2, \cdots, k, i \neq j$. Find the maximum value of $k$. (6th test, 1st question, provided by Wu Chang) | First, prove that each element in $S$ belongs to at most 3 of the sets $A_{1}, A_{2}, \cdots, A_{k}$.
By contradiction, suppose there is an element in $S$ that belongs to at least 4 subsets. Without loss of generality, assume $1 \in A_{1}, A_{2}, A_{3}, A_{4}$.
(1) If each of the other elements in $S$ belongs to at mo... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $x$ be a positive real number, then the minimum value of the function $y=x^{2}-x+\frac{1}{x}$ is $\qquad$ . | 3. 1.
The formula yields
$$
\begin{aligned}
y & =(x-1)^{2}+x+\frac{1}{x}-1 \\
& =(x-1)^{2}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+1 .
\end{aligned}
$$
When $x=1$, both $(x-1)^{2}$ and $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}$ simultaneously take the minimum value of 0, so the minimum value of $y=x^{2}-x+\f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The "Monkey Cycling" performance in the circus is carried out by 5 monkeys using 5 bicycles, with each monkey riding at least once, but no monkey can ride the same bicycle more than once. After the performance, the 5 monkeys rode the bicycles $2, 2, 3, 5, x$ times respectively, and the 5 bicycles were ridden $1, 1, ... | ,- 1 (B).
Every time the monkey rides a bike, the monkey and the bike each get one count, so the number of times the monkey rides a bike = the number of times the bike is ridden, i.e., $2+2+3+5+x=1+1+2+4+y$. Simplifying, we get $x+4=y$. From $x \geqslant 1$, we get $y \geqslant 5$. Also, from the problem, $y \leqslant ... | 6 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
4. The ice numbers $a_{1}, u_{2}, \cdots, a_{9}$ can only take two different values, +1 or -1. Then, the maximum value of the expression $a_{1} a_{5} a_{9}-a_{1} a_{6} a_{8}+a_{2} a_{8} a_{7}-$ $a_{2} a_{4} a_{9}+a_{3} a_{4} a_{8}-a_{3} a_{5} a_{7}$ is $\qquad$. | 4. 4 .
Since each term in the sum can only be 11 or -1, the sum is even (the parity of $a-b$ and $a+b$ is the same), so the maximum value of the sum is at most 6. It is easy to see that the sum cannot be 6, because in this case $a_{1} a_{5} a_{9}, a_{2} a_{6} a_{7}, a_{3} a_{6} a_{8}$ should all be +1, making their pr... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $f(x)=a x^{2}+b x+c(a, b, c \in R, a \neq 0)$. If for $|x| \leqslant 1$, $|f(x)| \leqslant 1$, then for $|x| \leqslant 1$, the maximum value of $|2 a x+b|$ is $\qquad$ | 4. 4 .
Given $f(0)=c, f(1)=a+b+c, f(-1)=a-b+c$,
we solve to get $a=\frac{f(1)+f(-1)-2 f(0)}{2}, b=\frac{f(1)-f(-1)}{2}$.
For $|x| \leqslant 1$, $|f(x)| \leqslant 1$, we have
$$
\begin{array}{l}
|2 a x+b| \\
=\left|[f(1)+f(-1)-2 f(0)] x\right. \\
\left.+\frac{f(1)-f(-1)}{2} \right| \\
=\left|\left(x+\frac{1}{2}\right)... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
(4) (Total 20 points) In the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $M$ is a point on $AB$, and $AM^2 + BM^2 + CM^2 = 2AM + 2BM + 2CM - 3$. If $P$ is a moving point on the segment $AC$, and $\odot O$ is the circle passing through points $P, M, C$, and a line through $P$ parallel to $AB$ intersects $\o... | (1) From the given, we have
$$
(A M-1)^{2}+(B M-1)^{2}+(C M-1)^{2}=0,
$$
which implies $A M=B M=C M=1$, meaning $M$ is the midpoint of $A B$.
(2) From $\left\{\begin{array}{l}M A=M C \Rightarrow \angle A=\angle M C A, \\ P D / / A B: \angle A=\angle C P D\end{array}\right.$
$\Rightarrow \angle M C A=\angle C P D$, thu... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. In the table below, the average of any three adjacent small squares in the upper row is $1(x \neq 0)$, and the square of any four adjacent small squares in the lower row is also 1. Then the value of $\frac{x^{2}+y^{2}+z^{2}-32}{y z+16}$ is $ـ$. $\qquad$ | ii. 3 .
From the problem, we know that in the above row, the numbers in every two small squares separated by one are equal, thus we can deduce
$$
\frac{x+7-y}{3}=1 \text {. }
$$
Similarly, for the row below, we get
$$
\begin{array}{l}
\frac{z-x+9.5-1.5}{4}=1 . \\
\begin{array}{l}
\therefore y=x+4, z=x-4, \text { then... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. In a tennis tournament, $n$ women and $2 n$ men participate, and each player plays against all other players exactly once. If there are no ties, the ratio of the number of games won by women to the number of games won by men is 7:5. Then $n=$
| 6. $n=3$.
Let $k$ represent the number of times the woman wins over Luo Ziyue.
Here, we set $\frac{k+\mathrm{C}_{0}^{2}}{\mathrm{C}_{\mathrm{jn}}^{2}}=\frac{7}{12}, k=\frac{7}{12} \mathrm{C}_{3 n}-\mathrm{C}_{n}^{2}$. From $k \leq 2 n^{2}$, we can solve to get $n \leq 3$.
When $n=1$, we get $k=\frac{7}{4}$; when $n=... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at $D$, $P$ is the midpoint of $AD$, $BP$ intersects $AC$ at $E$, $EF \perp BC$ at $F$. Given $AE=3$, $EC=$ 12. Find the length of $EF$. | Analysis: Consider $\triangle A B C$ as its machine triangle, given that $E$, $P$ are the fixed ratio points of $A C, A D$ respectively, hence from (*) we get $\left(1+\frac{C D}{D B}\right) \cdot \frac{3}{12}=1$, i.e., $\frac{C D}{D B}=3$.
From the projection theorem, we can find $B D=\frac{5}{2} \sqrt{3}, D C$ $=\fr... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Among the 35 numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 35^{2}$, the numbers with an odd digit in the tens place are $\qquad$ in total. | 3. 7. Among $1^{2}, 2^{2}, \cdots, 9^{2}$, those with an odd tens digit are only $4^{2}=16, 6^{2}=36$.
The square of a two-digit number can be expressed as
$$
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2} \text {. }
$$
It is evident that, when the unit digit is $A$ and $\hat{C}$, the tens digit of the square is odd. That is, on... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\frac{a}{a^{3}+a^{2} b+6 b^{2}+b^{3}}+\frac{b}{a^{3}-a^{2} b+a b^{2}-b^{3}} \\
+\frac{1}{a^{2}-b^{2}}-\frac{1}{b^{2}+a^{2}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} .
\end{array}
$$ | \[
\begin{aligned}
=、 \text { Original expression } & =\frac{a}{(a+b)\left(a^{2}+b^{2}\right)} \\
& +\frac{b}{(a-b)\left(a^{2}+b^{2}\right)}+\frac{2 b^{2}}{a^{4}-b^{4}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} \\
= & \frac{a^{2}+b^{2} .}{\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)}-\frac{a^{2}+b^{2}}{a^{4}-b^{4}} \\
&... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the equation $2 x^{2}+2 k x-13 k+1=0$ has two real roots whose squares sum to 13. Then $k=$ $\qquad$ . | 1.1. By the relationship between roots and coefficients, we solve to get $k=1$ or $k=$ -14. Substituting these values back into the original equation, we find that $k=-14$ does not satisfy the conditions, so it is discarded; $k=1$ satisfies the conditions. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that the area of $\triangle A B C$ is $1, D$ is the midpoint of $B C, E$, $F$ are on $A C, A B$ respectively, and $S_{\triangle B D P}=\frac{1}{5}, S_{\triangle C D E}=\frac{1}{3}$. Then $S_{\triangle D S P}=$ $\qquad$ | (7)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems there was a misunderstanding in your request. You asked me to translate a text, but the text you provided is just a number in parentheses. If you need a ... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four, (18 points) In a round-robin football tournament with $n$ teams (i.e., each team must play a match against every other team), each match awards 2 points to the winning team, 1 point to each team in the case of a draw, and 0 points to the losing team. The result is that one team has more points than any other team... | Let the team with the highest score be Team $A$, and assume Team $A$ wins $k$ games, draws $m$ games, and thus Team $A$'s total points are $2k + m$.
From the given conditions, every other team must win at least $k+1$ games, and their points must be no less than $2(k+1)$. Therefore, we have $2k + m > 2(k+1)$, which sim... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. On the diagonal $BD$ of square $ABCD$, take two points $E$ and $F$, such that the extension of $AE$ intersects side $BC$ at point $M$, and the extension of $AF$ intersects side $CD$ at point $N$, with $CM = CN$. If $BE = 3$, $EF = 4$, what is the length of the diagonal of this square? | 3. From the condition, we get
$\triangle A^{\prime} B M I \triangle \triangle A C M$.
$\therefore \angle B E=\angle D A E$.
Also, $A B=A D$,
$\angle A B E=\angle A D F$,
$\therefore \triangle A B E \simeq \triangle A D F$.
Therefore, $D F=B E=3$.
$\therefore A C=B D=10$. | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. 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 $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. 3 .
From $[\lg x] \leqslant \lg x$, we get $\lg ^{2} x-\lg x-2 \leqslant 0$, which means $-1 \leqslant \lg x \leqslant 2$.
When $-1 \leqslant \lg x<0$, we have $[\lg x]=-1$. Substituting into the original equation, we get $\lg x= \pm 1$, but $\lg x=1$ is not valid, so $\lg x=-1, x_{1}=\frac{1}{10}$.
When $0 \leqs... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. 21 people participate in an exam, with the test paper containing 15 true/false questions. It is known that for any two people, there is at least one question that both of them answered correctly. What is the minimum number of people who answered the most questions correctly? Please explain your reasoning. | For the $i$-th problem, $a_{i}$ people answered correctly, thus there are exactly
$$
b_{i}=C_{s_{i}}^{z}
$$
pairs of people who answered the $i$-th problem correctly $(i=1,2, \cdots, 15)$.
Below, we focus on the sum $\sum_{i=1}^{15} b_{i}$.
Let $a=\max \left\{a_{1}, a_{2}, \cdots, a_{15}\right\}$, then we have
$$
15 C... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $S=\left\{A=\left(a_{1}, \cdots, a_{8}\right) \mid a_{i}=0\right.$ or $1, i=1, \cdots$, 8\}. For two elements $A=\left(a_{1}, \cdots, a_{8}\right)$ and $B=\left(b_{1}\right.$, $\cdots, b_{8}$ ) in $S$, denote
$$
d(A, B)=\sum_{i=1}^{\delta}\left|a_{i}-b_{i}\right|,
$$
and call it the distance between $A$ and $B$... | Solution One
(I) First, we point out that the sum of the weights of any two codewords in $\mathscr{D}$ does not exceed 11; otherwise, if
$$
\omega(X)+\omega(Y) \geqslant 12,
$$
then because $12-8=4$, these two codewords must share at least four positions with 1s, and the distance between them
$$
a^{\prime}\left(X, y^{... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}(n>4)$ where all interior angles are integer multiples of $15^{\circ}$, and $\angle A_{1}+\angle A_{2}+\angle A_{3}=$ $285^{\circ}$. Then, $n=$ | 2. 10 (Hint: The sum of the $n-3$ interior angles other than the male one is $(n-2) \cdot$ $180^{\circ}-285^{\circ}$, it can be divided by $n-3$, and its quotient should also be an integer multiple of $15^{\circ}$.） | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. For real numbers $x, y$, define a new operation: $x * y=a x+b y+$ $c$, where $a, b, c$ are constants, and the right side of the equation is the usual addition and multiplication operations. It is known that $3 * 5=15,4 * 7=28$. Then, 1*1= $\qquad$ | 7. -11
Brief solution: According to the definition, we know that $3 * 5=3a+5b+c=15, 4 * 7$ $=4a+7b+c=28$, and $1 * 1=a+b+c=3(3a+5b+$c) $-2(4a+7b+c)=3 \times 15-2 \times 28=-11$. | -11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (12 points) Given points $P_{1}\left(x_{1}, 1994\right), P_{2}\left(x_{2}, 1994\right)$ are two points on the graph of the quadratic function $y=a x^{2}+b x+7(a \neq 0)$. Try to find the value of the quadratic function $y=x_{1}+x_{2}$. | Three, since $P_{1}, P_{2}$ are two points on the graph of a quadratic function, we have
$$
\begin{array}{l}
a x_{1}^{2}+b x_{1}+7=1994, \\
a x_{2}^{2}+b x_{2}+7=1994 .
\end{array}
$$
Subtracting the two equations and simplifying, we get
$$
\left(x_{1}-x_{2}\right)\left[a\left(x_{1}+x_{2}\right)+b\right]=0 .
$$
Since... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. (India) The sequence of positive integers $\left\{f_{n}\right\}_{m=1}^{m}$ is defined as follows:
$f(1)=1$, and for $n \geqslant 2$,
$$
f(n)=\left\{\begin{array}{l}
f(n-1)-n, \text { when } f(n-1)>n ; \\
f(n-1)+n, \text { when } f(n-1) \leqslant n .
\end{array}\right.
$$
Let $S=\{n \in \mathbb{N} \mid f(n)=1993\}$.... | (i) We point out that if $f(n)=1$, then $f(3n+3)=1$. From $f(n)=1$, we can sequentially obtain according to the definition:
$$
\begin{array}{l}
f(n+1)=n+2, f(n+2)=2n+4, \\
f(n+3)=n+1, f(n+4)=2n+5, \\
f(n+5)=n, \quad f(n+6)=2n+6 \\
f(n+7)=n-1, \cdots
\end{array}
$$
It is evident that the sequence $f(n+1), f(n+3), f(n+5... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $[x]$ denotes the greatest integer not exceeding $x$, $a, b, c \in$ $R^{+}, a+b+c=1$, let $M=\sqrt{3 a+1}+\sqrt{3 b+1}+$ $\sqrt{3 c+1}$. Then the value of $[M]$ is $($. | 5. B.
Given $a \in(0,1), b \in(0,1), c \in(0,1) \Rightarrow a^{2} & \sqrt{a^{2}+2 a+1}+\sqrt{b^{2}+2 b+1} \\
& +\sqrt{c^{2}+2 c+1} \\
= & (a+b+c)+3=4 .
\end{aligned}
$$
And $M=\sqrt{(3 a+1) \cdot 1}+\sqrt{(3 b+1) \cdot 1}$
$$
\begin{aligned}
& +\sqrt{(3 c+1) \cdot 1} \\
& \frac{(3 a+1)+1}{2}+\frac{(3 b+1)+1}{2}+\frac... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the functions $y=2 \cos \pi x(0 \leqslant x \leqslant 2)$ and $y=2(x \in$ $R$ ) whose graphs enclose a closed plane figure. Then the area of this figure is $\qquad$ . | 2. As shown in the figure, by symmetry, the area of $CDE$ = the area of $AOD$ + the area of $BCF$, which means the shaded area is equal to the area of square $OABF$, so the answer is 4. | 4 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
1. If the natural numbers $a, x, y$ satisfy $\sqrt{a-2 \sqrt{6}}=\sqrt{x} -\sqrt{y}$, then the maximum value of $a$ is $\qquad$. | $$
=, 1, a_{\max }=7 \text {. }
$$
Squaring both sides of the known equation yields $a-2 \sqrt{6}=x+y-$ number, it must be that $x+y=\alpha$ and $x y=6$. From the known equation, we also know that $x>y$, so it can only be $x=6, y=1$ or $x=3, y=2$. Therefore, $a=7$ or 2, the maximum value is 7. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Person A and Person B start from the same point $A$ on a circular track at the same time and run in opposite directions. Person A's speed is $5 \mathrm{~m}$ per second, and Person B's speed is $7 \mathrm{~m}$ per second. They stop running when they meet again at point $A$ for the first time. During this period, they... | 3. 12 .
Two $\lambda \mu A$ start from $A$ and meet once (not at point $A$). At the moment of their first meeting, the two have run a certain distance. From the ratio of their speeds, it can be deduced that person A has run $\frac{5}{12}$ of a lap, and person B has run $\frac{7}{12}$ of a lap. Therefore, after the $n$... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
4. Consider the hyperbola \((x-2)^{2}-\frac{y^{2}}{2}=1\). A line \(l\) is drawn through the right focus of the hyperbola, intersecting the hyperbola at points \(A\) and \(B\). If \(|A B|=4\), then the number of such lines is \(\qquad\). | 4. 3 lines.
The coordinates of the right focus are $(2+\sqrt{3}, 0)$. The length of the chord perpendicular to the real axis through the right focus is exactly 4. The length of the chord $d$ formed by passing through the right focus and intersecting both branches of the hyperbola has the range $d>2$, so there are two ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. Write the number $1234567802011 \cdots 19941995$ on the blackboard, forming the integer $N_{1}$. Erase the digits of $N_{1}$ that are in even positions, leaving the remaining digits to form the integer $N_{2}$. Remove the digits of $N_{2}$ that are in odd positions, leaving the remaining digits to form the integer ... | $$
=、 9 \times 1+90 \times 2+900 \times 3+996 \times 4=6873 \text {. }
$$
The integer $N_{1}$ is composed of 6873 digits, which are sequentially numbered as $1,2,3, \cdots, 6873$. We still examine the set of numbers $\{1,2, \cdots, 6873\}$. Removing the digits at even positions from $N_{1}$ is equivalent to removing $... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that the inverse function of $y=f(x)$ is $\varphi(x)$, and $\varphi(x)$ . $=\log _{m a_{0}{ }^{2}}\left(\frac{1996}{x}-\sin ^{2} 0\right), 0 \in\left(0, \frac{\pi}{2}\right)$. Then the solution to the equation $f(x)=$ 1996 is | $=, 1 .-1$.
Since $f(x)$ and $\varphi(x)$ are inverse functions of each other, solving $f(x) = 1996$ is equivalent to finding the value of $\varphi(1996)$.
$$
\varphi(1996)=\log _{\sec ^{2} \theta}\left(1-\sin ^{2} \theta\right)=-1 .
$$ | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
When the expression changes, 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 $y=\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$.
(*)
Rearranging (*), we get
$$
(y-6) x^{2}+(2 y-12) x+2 y-10=0 \text {. }
$$
Since $x$ is a real number, we have
$$
\Delta=(2 y-12)^{2}-4(y-6)(2 y-10) \geqslant 0 \text {, }
$$
which simplifies to $y^{2}-10 y+24 \leqslant 0$.
Solving this, we get $4 \leqslant y \le... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. Let $x$ be a positive real number, then the minimum value of the function $y=x^{2}-x+\frac{1}{x}$ is $\qquad$ - (1995, National Junior High School Competition) | - $y=x^{2}-x+\frac{1}{x}$ $=(x-1)^{2}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+1$.
From the above equation, it is known that when $x-1=0$ and $\sqrt{x}-\frac{1}{\sqrt{x}}$ $=0$, i.e., $x=1$, $y$ takes the minimum value of 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 13. Let $a, b, c$ be distinct integers from 1 to 9. What is the largest possible value of $\frac{a+b+c}{a b c}$? (1992, 1st Dannevirke-Shanghai Friendship Correspondence Competition) | Let $P=\frac{a+b+c}{a b c}$.
In equation (1), let $a, b$ remain unchanged temporarily, and only let $c$ vary, where $c$ can take any integer from 1 to 9. Then, from $P=\frac{a+b+c}{a b c}=\frac{1}{a b}+\frac{a+b}{a b c}$, we know that when $c=1$, $P$ reaches its maximum value. Therefore, $c=1$.
Thus, $P=\frac{a+b+1}{a ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If 20 points divide a circle into 20 equal parts, then the number of regular polygons that can be formed with vertices only among these 20 points is ( ) .
(A) 4
(B) 8
(C) 12
(D) 24 | 6. (C).
Let the regular $k$-sided polygon satisfy the condition, then the 20 $-k$ points other than the $k$ vertices are evenly distributed on the minor arcs opposite to the sides of the regular $k$-sided polygon.
Thus, $\frac{20-k}{k}=\frac{20}{k}-1$ is an integer, so $k \mid 20$.
But $k \geqslant 3$,
$\therefore k=4... | 12 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
Three, (25 points) Given that $a, b, c$ are positive integers, and the parabola $y=$ $a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A, B$. If the distances from $A, B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
Translate the above text into English, please retain the origin... | Three, let the coordinates of $A, B$ be $\left(x_{1}, 0\right),\left(x_{2}, 0\right)$ and $x_{1}<0, x_{2}>0$,
\[
\begin{array}{l}
\therefore x_{1} x_{2}=\frac{c}{a}<0, \\
\therefore a c<0, \text { i.e., } a>0, c<0 .
\end{array}
\]
\[
\begin{array}{l}
\because x_{1}+x_{2}=-\frac{b}{a}>0, \\
\therefore b<0 .
\end{array}
... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) As shown in the figure, given that $AB, CD$ are perpendicular chords in a circle $\odot O$ with radius 5, intersecting at point $P$. $E$ is the midpoint of $AB$, $PD=AB$, and $OE=3$. Try to find the value of $CP + CE$.
---
The translation is provided as requested, maintaining the original text's form... | By the intersecting chords theorem, we have
$$
C P \cdot P D=A P \cdot P B \text {. }
$$
Also, $A E=E B=\frac{1}{2} A B$,
$$
\begin{aligned}
\because C P \cdot P D & =\left(\frac{1}{2} A B+E P\right)\left(\frac{1}{2} A B-E P\right) \\
& =\frac{1}{4} A B^{2}-E P^{2} .
\end{aligned}
$$
And $P D=A B$,
$$
\therefore \qua... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. The Dao sequence $\left\{\begin{array}{l}x_{1}=x_{2}=1, \\ x_{n+2}=a x_{n+1}+b x_{n}(n \in N) .\end{array}\right.$
If $T=1996$ is the smallest natural number such that $x_{T+1}=x_{T+2}=1$, then $\sum_{i=1}^{1006} x_{i}=$ $\qquad$ . | 2. 0 .
From $T=1996$ being the smallest natural number that makes $x_{1+1}=x_{T+2}=1$, we know that $a+b \neq 1$. Otherwise, if $a+b=1$, then $x_{3}=a x_{2}+b x_{1}=a+b=1$. This would mean that when $T=1$, $x_{T+1}=x_{T+2}=1$, which contradicts the fact that $T=1996$ is the smallest value. Furthermore, we have
$$
\beg... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. In the plane $\alpha$ there is a $\triangle A B C, \angle A B C=105^{\circ}$, $A C=2(\sqrt{6}+\sqrt{2})$. On both sides of the plane $\alpha$, there are points $S, T$, satisfying $S A=S B=S C=\sqrt{41}, T A=T B=T C=$ 5. Then $S T=$ $\qquad$. | 3. 8 .
From Figure 4, since $S A=S B=S C$, we know that the projection $D$ of $S$ on plane $\alpha$ is the circumcenter of $\triangle A B C$. Similarly, the projection of $T$ on plane $\alpha$ is also the circumcenter of $\triangle A B C$. By the uniqueness of the circumcenter, we have $S T \perp a$.
Connect $A D$. By... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given functions $f(x)$ and $g(x)$ are defined on $R$, and $f(x-y)=f(x) g(y)-g(x) f(y)$. If $f(1)=f(2) \neq$ 0, then $g(1)+g(-1)=$ $\qquad$ . | 4. 1 .
For $x \in R$, let $x=u-v$, then we have
$$
\begin{array}{l}
f(-x)=f(v-u) \\
=f(v) g(u)-g(v) f(u) \\
=-[f(u) g(v)-g(u) f(v)] \\
=-f(u-v)=-f(x) .
\end{array}
$$
Therefore, $f(x)$ is an odd function, and we have
$$
\begin{array}{l}
f(2)=f[1-(-1)] \\
=f(1) g(-1)-g(1) f(-1) \\
=f(1) g(-1)+g(1) f(1) .
\end{array}
$... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ have side lengths $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$, respectively, and their areas be $S$ and $S^{\prime}$. If for all values of $x$, $a x^{2}+b x+c=3\left(a^{\prime} x^{2}+b^{\prime} x+c^{\prime}\right)$ always holds, then $\fra... | II. 1. From the given, $\left(a-3 a^{\prime}\right) x^{2}+\left(b-3 b^{\prime}\right) x+(c$ $\left.-3 c^{\prime}\right)=0$ is an identity, hence $\left(a-3 a^{\prime}\right)=0,\left(b-3 b^{\prime}\right)=0$, $\left(c-3 c^{\prime}\right)=0$, which gives $\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}=3$,... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. In the interior angles of an $n(n \geqslant 3)$-sided polygon, the maximum number of acute angles is $\qquad$ . | II, 1.3.
Let a polygon with $n$ sides have $k$ acute angles, then
$$
(n-2) \times 180^{\circ}<k \times
$$
$90^{\circ}+(n-k) \times 180^{\circ}$.
Thus, $k<4$.
As shown in the figure, construct a $60^{\circ}$ sector $A_{1} A_{2} A_{1}$. When $n=3$, $\triangle A_{1} A_{2} A_{3}$ has three interior angles, all of which are... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. There are two roads $O M, O N$ intersecting at a $30^{\circ}$ angle. Along the direction of road $O M$, 80 meters from $A$ is a primary school. When a tractor travels along the direction of $O N$, areas within 50 meters on both sides of the road will be affected by noise. Given that the speed of the tractor is 18 ki... | 3. 12 .
As shown in the figure, when the tractor reaches point $B$, it starts to have an impact, and when it reaches point $C$, the impact ceases. Thus, $A B = A C = 50$ meters. Draw $A D \perp$ $O N$ from $A$. Given that $\angle A O D = 30^{\circ}$, we know that $A D = \frac{1}{2} O A = 40$ meters. Therefore, $B D = ... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let real numbers $x, y$ satisfy the equation $9 x^{2}+4 y^{2}-3 x+2 y=0$. Then the maximum value of $z=3 x+2 y$ is $\qquad$ . | From $9 x^{2}+4 y^{2}=3 x-2 y$ we know $3 x-2 y \geqslant 0$ and $\frac{9 x^{2}+4 y^{2}}{3 x-2 y}$ $=1$,
we have $z=3 x+2 y=\frac{(3 x)^{2}-(2 y)^{2}}{3 x-2 y} \leqslant \frac{(3 x)^{2}+(2 y)^{2}}{3 x-2 y}$ $=1$ (equality holds when $y=0$). | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 17. Given as shown, in quadrilateral $ABCD$, $AD=DC=1, \angle DAB=$ $\angle DCB=90^{\circ}, BC, AD$ extended intersect at $P$. Find the minimum value of $AB \cdot S_{\triangle PAB}$.
(1994, Sichuan Province Junior High School Mathematics League Competition) | Let $DP = x$, then $PC = \sqrt{x^2 - 1}$.
Since $\triangle PCD \sim \triangle PAB$,
$\therefore CD \perp AB = PC : PA$.
$\therefore AB = \frac{CD \cdot PA}{PC} = \frac{x + 1}{\sqrt{x^2 - 1}}$.
Let $y = AB \cdot S_{\triangle PAB}$, then
$$
y = \frac{1}{2} AB^2 \cdot PA = \frac{(x + 1)^3}{2(x^2 - 1)}.
$$
Eliminating the... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. If $a=\sqrt{17}-1$, find the value of $\left(a^{5}+2 a^{4}-17 a^{3}\right.$ $\left.-a^{2}+18 a-17\right)^{1993}$.
(Adapted from the 1987 Chongqing Junior High School Mathematics Invitational Competition) | Given $a+1=\sqrt{17}$, squaring both sides yields $a^{2}+2 a+1=17$.
Then
$$
\begin{array}{l}
\left(a^{5}+2 a^{4}-17 a^{3}-a^{2}+18 a-17\right)^{1993} \\
=\left[a^{5}+2 a^{4}-\left(a^{2}+2 a+1\right) a^{3}-a^{2}+\left(a^{2}\right.\right. \\
\left.+2 a+1+1) a-\left(a^{2}+2 a+1\right)\right]^{1993} \\
=(-1)^{1993}=-1 \tex... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. If $m^{2}=m+1, n^{2}=n+1$, then what is the value of $m^{5}+n^{5}$?
(1989, Jiangsu Province High School Mathematics Competition) | Let $S_{k}=m^{k}+n^{k}$, construct the recurrence relation for $S_{k}$.
$$
\begin{array}{l}
\because m^{2}=m+1, n^{2}=n+1, \\
\therefore m^{k}=m^{k-1}+m^{k-2}, n^{k}=n^{k-1}+n^{k-2} .
\end{array}
$$
Adding the two equations gives $S_{k}=S_{k-1}+S_{k-2}$.
$$
\begin{aligned}
\therefore S_{5} & =S_{4}+S_{3}=\left(S_{3}+S... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 10. Given $a^{2}+2 a-5=0, b^{2}+2 b-5=0$, and $a \neq b$. Then the value of $a b^{2}+a^{2} b$ is what?
(1989, Sichuan Province Junior High School Mathematics Competition) | Given that $a, b$ are the roots of the quadratic equation $x^{2}+$ $2 x-5=0$, then $a+b=-2, ab=-5$. Therefore, $ab^{2}+a^{2}b=ab(a+b)=10$. Note: Here, the definition of the roots is used to construct the equation. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a+b+c=0, a^{3}+b^{3}+c^{3}=0$. Find the value of $a^{15}+b^{15}+c^{15}$. | 2. From $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right.$ $-a b-b c-a c)$, we have $a b c=0$. Then at least one of $a, b, c$ is zero, for example, $c=0$, then it is known that $a, b$ are opposites, i.e., $a^{15}+b^{15}+c^{15}$ $=0$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. Find a natural number $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square.
(2nd All-Russian High School Mathematics Olympiad) | Let $2^{4}=x$, then $2^{8}=x^{2}, 2^{11}=2^{7} \cdot x$. Therefore, $2^{8}+2^{11}+2^{n}=x^{2}+2^{7} x+2^{n}$.
According to the condition for a quadratic trinomial to be a perfect square, we get $\Delta=\left(2^{7}\right)^{2}-4 \times 2^{n}=0$.
Solving for $n$ yields $n=12$. | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $\frac{x}{m}+\frac{y}{n}+\frac{z}{p}=1, \frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0$. Calculate the value of $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}}$. | 6. From $\frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0$, we have $\frac{m y z+n x z+p x y}{x y z}=0$.
Since $x, y, z$ are not zero, then $m y z+n x z+p x y=0$. Also,
$$
\begin{aligned}
& \frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}} \\
= & \left(\frac{x}{m}+\frac{y}{n}+\frac{z}{p}\right)^{2}-2\left(\frac{x y}{m ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that $x, y, z$ are three non-negative rational numbers, and satisfy $3 x$ $+2 y+z=5, x+y-z=2$. If $S=2 x+y-z$, then what is the sum of the maximum and minimum values of $S$? | 9. $\left\{\begin{array}{l}3 x+y+z=5, \\ x+y-z=2, \\ 2 x+y-z=S .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}x=S-2, \\ y=\frac{15-4 S}{3}, \\ z=\frac{3-S}{3} .\end{array}\right.$ Since $x, y, z$ are all non-negative, we have
$$
\left\{\begin{array} { l }
{ S - 2 \geqslant 0 , } \\
{ \frac { 1 5 - 4 S } ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In $\triangle A B C$, $G$ is the centroid, and $I$ is the intersection of the angle bisectors of $\angle B$ and $\angle C$. If $I G / / B C$, and $B C=5$, then $A B+A C$ $=$ . $\qquad$ | 3. Connect $A G, B G$ and $C G$, and extend $A G$ to intersect $B C$ at $M$, then $S_{\triangle G B C}: S_{\triangle A B C}=G M: A M=1: 3$.
Connect $A I$, then by $I G$ $/ / B C$, we know
$$
\begin{array}{l}
S_{\triangle I B C}=S_{\triangle G B C}=\frac{S_{\triangle A B C}}{3}, \\
\therefore S_{\triangle I A B}+S_{\tr... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 20 points, Question (1) 8 points, Question (2) 12 points) As shown in the figure, there is a cube-shaped wire frame, and the midpoints of its sides $I, J, K, L$ are also connected with wire.
(1) There is an ant that wants to crawl along the wire from point A to point G. How many shortest routes are th... | (1) "The shortest path" means that the ant can neither move left nor down, otherwise it would be taking a "detour" rather than the "shortest" path. The ant can crawl on face $ABCGFE$ or on $ADCGHE$. The number of "shortest paths" on these two faces is the same, and the possible paths can be represented by the following... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. 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 such that
(i) $a_{1}=1$;
(ii) $\left|a_{i}-a_{i+1}\right| \leqslant 2 . i=1,2, \cdots, n-1$.
Determine whether $f(1996)$ is divisible by 3. | 3. Verify that $f(1)=f(2)=1$ and $f(3)=2$.
Let $n \geqslant 4$. Then it must be that $a_{1}=1, a_{2}=2$ or 3.
For $a_{2}=2$, the number of permutations is $f(n-1)$, because by deleting the first term and reducing all subsequent terms by 1, we can establish a one-to-one correspondence of sequences.
If $a_{2}=3$, then ... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. If $360^{x}=3, 360^{y}=5$, then $72^{\frac{1-2 x-y}{3(1-y)}}=$ | \begin{array}{l}=1.2. \\ \because 72=\frac{360}{5}=\frac{360}{360^{y}}=360^{1-y}, \\ \therefore \text { original expression }=\left(360^{1-y}\right)^{\frac{1}{3(2 x-y)}} \\ =(360)^{\frac{1}{3}(1-2 x-y)}-\left(360^{1-2 x-y}\right)^{\frac{1}{3}} \\ =\left(\frac{360}{\left(360^{x}\right)^{2} \cdot 360^{5}}\right)^{\frac{1... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If $x=\frac{1}{2}-\frac{1}{4 x}$, then $1-2 x+2^{2} x^{2}-2^{3} x^{3}+2^{4} x^{4}$ $-\cdots-2^{1995} x^{1995}$ is. $\qquad$. | 2. 1 .
From $x=\frac{1}{2}-\frac{1}{4 x}$ we can get $1-2 x+(2 x)^{2}=0$.
$$
\begin{array}{c}
\therefore \text { the original expression }=1-2 x\left[1-2 x+(2 x)^{2}\right]+(2 x)^{4}[1-2 x \\
\left.+(2 x)^{2}\right]-\cdots-(2 x)^{1993}\left[1-2 x+(2 x)^{2}\right]=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. If $x=\sqrt{19-8 \sqrt{3}}$, then the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15}=$ $\qquad$ | Solve: From $x=\sqrt{19-2 \sqrt{48}}=4-\sqrt{3}$, we get $x-4=-\sqrt{3}$. Squaring both sides and rearranging, we obtain
$$
x^{2}-8 x+13=0 \text {. }
$$
Therefore,
$$
\begin{aligned}
\text { Original expression } & =\frac{\left(x^{2}-8 x+13\right)\left(x^{2}+2 x+1\right)+10}{\left(x^{2}-8 x+13\right)+2} . \\
& =5 .
\e... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a, b$ be positive integers, and $a+b \sqrt{2}$ $=(1+\sqrt{2})^{100}$. Then the units digit of $a b$ is $\qquad$ | 6.4.
By the binomial theorem, we have
$$
a-b \sqrt{2}=(1-\sqrt{2})^{100} \text {. }
$$
Therefore, $a=\frac{1}{2}\left((1+\sqrt{2})^{100}+(1-\sqrt{2})^{100}\right)$,
$$
\begin{array}{l}
b=\frac{1}{2 \sqrt{2}}\left((1+\sqrt{2})^{100}-(1-\sqrt{2})^{100}\right) . \\
\text { Hence } \left.a b=\frac{1}{4 \sqrt{2}}(1+\sqrt{... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 13. Given $4 x-3 y-6 z=0, x+2 y-7 z$ $=0, x y z \neq 0$. Then the value of $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}$ is equal to | Solve the system of equations by treating $\approx$ as a constant,
$$
\left\{\begin{array}{l}
4 x-3 y=6 z, \\
x-2 y=7 z,
\end{array}\right.
$$
we get
$$
\begin{array}{l}
x=3 z, y=2 z . \\
\text { Therefore, the original expression }=\frac{2 \cdot(3 z)^{2}+3 \cdot(2 z)^{2}+6 z^{2}}{(3 z)^{2}+5 \cdot(2 z)^{2}+7 z^{2}}=1... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 14. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq$ $n$, then $m^{5}+n^{5}=$ $\qquad$ | $$
\begin{array}{l}
\text { Sol } \because(a-b)\left(a^{n}+b^{n-1}\right) \\
=a^{n}+b^{n}+a b\left(a^{n-2}+b^{n-2}\right), \\
\therefore a^{n}+b^{n}=(a+b)\left(a^{n-1}+b^{n-1}\right) \\
-a b\left(a^{n-2}+b^{n-2}\right) \text {. } \\
\end{array}
$$
Let $S_{n}=a^{n}-b^{n}$, we get the recursive formula
$$
S_{n}=(a+b) S_... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the equation $a x^{2}+b x+c=0(a \neq 0)$, the sum of the roots is $s_{1}$, the sum of the squares of the roots is $s_{2}$, and the sum of the cubes of the roots is $s_{3}$. Then the value of $a s_{3}+$ $\left\langle s_{2}\right.$ $+c s_{1}$ is . $\qquad$ | (Tip: Let the two roots of the equation be $x_{1}, x_{2}$.
Then by definition, we have $a x_{1}^{2}+b x_{1}+c=0, a x_{2}^{2}+b x_{2}+c=0$. The original expression $=0$.) | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. If $a$ is a root of $x^{2}-3 x+1=0$, try to find the value of $\frac{2 a^{5}-5 a^{4}+2 a^{3}-8 a^{2}}{a^{2}+1}$. | From the definition of the root, we know that $a^{2}-3 a+1=0$.
Thus, $a^{2}+1=3 a$ or $a^{2}-3 a=-1$ or
$$
\begin{array}{l}
a^{3}=3 a^{2}-a \\
\therefore \text { the original expression }=\frac{a\left[2 a^{2}\left(a^{2}+1\right)-5 a^{3}-8 a\right]}{3 a} \\
=\frac{1}{3}\left(a^{3}-8 a\right)=\frac{1}{3}\left(3 a^{2}-9 a... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 8. Let } x>\frac{1}{4} \text {. Simplify } \sqrt{x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}} \\ -\sqrt{x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1}}=\end{array}$ | Let $a$
$$
\begin{array}{l}
=x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}, b=x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1} . \text { Original } \\
\text { expression }=1 .)
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. The value of $x$ that satisfies the following equations is
$$
\begin{array}{l}
(123456789) x+9=987654321, \\
(12345678) x+8=98765432 . \\
(1234567) x+7=9876543 . \\
\cdots \cdots .
\end{array}
$$ | Observe the numerical changes on both sides of each equation, it is easy to know that the last equation should be $x+1=9$, i.e., $x=8$. Upon verification, $x=8$ is the solution. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. Let $a^{2}+2 a-1=0, b^{4}-2 b^{2}-1=0$ and $1-a b^{2} \neq 0$. Then the value of $\left(\frac{a b^{2}+b^{2}+1}{a}\right)^{1990}$ is $\qquad$. | From the given, we have $\left(\frac{1}{a}\right)^{2}-\frac{2}{a}-1=0$, $\left(b^{2}\right)^{2}-2 b^{2}-1=0$. By the definition of roots, $\frac{1}{a}, b^{2}$ are the roots of the equation $x^{2}-2 x-1=0$. Then
$$
\begin{aligned}
\frac{1}{a}+b^{2} & =2, \frac{1}{a} \cdot b^{2}=-1 . \\
\therefore \text { the original ex... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Glue the bases of two given congruent regular tetrahedra together, precisely to form a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$ | 4. 3.
As shown in the figure, construct $CE \perp AD$, connect $EF$, it is easy to prove that $EF \perp AD$. Therefore, $\angle CEF$ is the plane angle of the dihedral angle formed by plane $ADF$ and plane $ACD$.
Let $G$ be the midpoint of $CD$. Similarly, $\angle AGB$ is the plane angle of the dihedral angle formed ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. In the Cartesian coordinate plane, the number of integer points (i.e., points with both coordinates as integers) on the circumference of a circle centered at $(199,0)$ with a radius of 199 is $\qquad$ . | 6. 4 .
Let $A(x, y)$ be an integer point on circle $O$. As shown in the figure, the equation of circle $O$ is $y^{2}+(x-199)^{2}=$ $199^{2}$.
$$
\begin{array}{l}
\text { Clearly, } x=0, y=0 ; \\
x=199, y=199 ; \\
x=199, y=-199 ; x=389, y=0
\end{array}
$$
These are 4 solutions to the equation. However, when $y \neq 0,... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 8. Calculate
$$
\begin{array}{l}
\sqrt{3633 \times 3635 \times 3639 \times 3641+36} \\
-3636 \times 3638=
\end{array}
$$ | Let $3637=a$. Then
the original expression $=$
$$
\begin{array}{l}
\sqrt{(a-4)(a-2)(a+2)(a+4)+36} \\
-(a+1)(a-1) \\
=\sqrt{\left(a^{2}-10\right)^{2}}-\left(a^{2}-1\right) \\
=a^{2}-10-a^{2}+1=-9 . \\
\end{array}
$$ | -9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $a, b$ be unequal real numbers, and $a^{2}+2 a=$ $5 \cdot b^{2}+2 b=5$. Then $a^{2} b+a b^{2}=$ | 4. (10).
It is known that $a, b$ are two unequal real numbers which are exactly the two real roots of the equation $x^{2}+2 x-5=0$. Therefore, $a+b=-2, ab=-5$.
Thus, $a^{2} b+ab^{2}=ab(a+b)=(-2) \cdot(-5)=$
10. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. If $m=1996^{3}-1995^{3}+1994^{3}-1993^{3}$ $+\cdots+4^{3}-3^{3}+2^{3}-1^{3}$, then the last digit of $m$ is | 5. (0).
$$
2^{3}-1^{3}=7,4^{3}-3^{3}=37,6^{3}-5^{3}=91,8^{3}-7^{3}=
$$
$169,10^{3}-9^{3}=271$. Therefore, the sum of the last digits of the first 10 numbers $10^{3}-9^{3}+8^{3}-7^{3}$ $+6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}$ is $1+9+1+$ $7+7=25$, and the last digit of 25 is 5. The last digit of the algebraic sum of every... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let 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 Competition) | Solution: According to the problem, all logarithms are positive. Therefore, from (*) we get
$$
\begin{aligned}
\text { LHS } & =\frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}} \\
& \geqslant \frac{(1+1+1)^{2}}{\log _{1993}\left(\frac{x_{0}... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
10.3. $x, y$ are natural numbers. $k$ is a natural number of size 1. Find all natural numbers $n$ that satisfy: $3^{n}=x^{k}+y^{k}$, and provide a proof. | 10. 3. Answer: $n=2$.
At the time: Let $3^{n}=x^{k}+y^{k}$, where $x$ and $y$ are prime (assuming $x > y$), $k>1$, and $n$ is a natural number. Of course, neither $x$ nor $y$ can be divisible by 3.
If $k$ is even, then $x^{k}$ and $y^{k}$ leave a remainder of 1 when divided by 3. Thus, the sum of $x^{k}$ and $y^{k}$ ... | 2 | Number Theory | proof | Yes | Yes | cn_contest | false |
Example 1 Find the positive integer root of the equation
$$
\frac{2}{n}+\frac{3}{n+1}+\frac{4}{n+2}=\frac{133}{60}
$$
(1990, Shanghai Junior High School Mathematics Competition) | Solution: Since $n$ is a natural number, by the properties of natural numbers and
$$
\begin{array}{l}
\frac{1}{n}>\frac{1}{n+1}>\frac{1}{n+2} \\
\therefore \frac{2+3+4}{n+2}<\frac{2}{n}+\frac{3}{n+1}+\frac{4}{n+2} \\
<\frac{2+3+4}{n} .
\end{array}
$$
That is, $\frac{9}{n+2}<\frac{133}{60}<\frac{9}{n}$.
Solving this, ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Solve the equation
\[
\begin{array}{l}
\sqrt{x^{2}+5 x-14}+\sqrt{x+7}+\sqrt{2-x}+x-5 \\
=0 .
\end{array}
\] | Solution: By the properties of quadratic radicals, we get
$$
\left\{\begin{array}{l}
x^{2}+5 x-14 \geqslant 0, \\
x+7 \geqslant 0, \\
2-x \geqslant 0 .
\end{array}\right.
$$
Solving, we get $-7 \leqslant x \leqslant-7$ or $2 \leqslant x \leqslant 2$.
That is, $x=-7$ or $x=2$.
Upon verification, $x=2$ is a root of the ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 For a finite set $A$, function $f: N \rightarrow A$ has only the following property: if $i, j \in N, |H| i-j |$ is a prime number, then $f(i) \neq f(j)$. How many elements does set $A$ have at least? | Solution: Let $|A|$ denote the number of elements in the finite set $A$, and estimate the lower bound of $|A|$.
Since the absolute value of the difference between any two numbers among $1, 3, 6, 8$ is a prime number, by the problem's condition: $f(1)$, $f(3)$, $f(6)$, $f(8)$ are four distinct elements in $A$. Therefor... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Five, let $A=\{1,2,3, \cdots, 17\}$. For any function $f: A \rightarrow A$, denote
$$
f^{[1]}(x)=f(x), f^{[k+1]}(x)=f\left(f^{[k]}(x)\right)
$$
$(k \in \mathbb{N})$. Find the natural number $M$ such that:
$$
\begin{array}{l}
\text { (1) When } m<M, 1 \leqslant i \leqslant 16, \text { we have } \\
f^{[m]}(i+1)-f^{[m]}(i... | Five, the required $M_{0}=8$.
First, prove $M_{0} \geqslant 8$.
In fact, we can define the mapping $f(i) \equiv 3 i-2(\bmod 17)$, where $i \in A, f(i) \in A$.
If $f(i) \equiv f(j)(\bmod 17)$,
then $3 i-2 \equiv 3 j-2(\bmod 17)$,
we have $i \equiv j(\bmod 17)$,
$$
\therefore i=j \text {. }
$$
The mapping $f$ is a mappi... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. Solve the equation
$$
\begin{array}{l}
\sqrt{x-1}+\sqrt{2 x-3}+\sqrt{3 x-5} \\
+\sqrt{4 x-7}=5 x-6 .
\end{array}
$$
(First Yangtze Cup Correspondence Competition for Junior High School Students) | Solution: By applying the method of completing the square, the original equation can be transformed into
$$
\begin{array}{l}
(\sqrt{x-1}-1)^{2}+(\sqrt{2 x-3}-1)^{2} \\
+(\sqrt{3 x-5}-1)^{2}+(\sqrt{4 x-7}-1)^{2}=0 . \\
\therefore \sqrt{x-1}-1=\sqrt{2 x-3}-1 \\
\quad=\sqrt{3 x-5}-1=\sqrt{4 x-7}-1=0 .
\end{array}
$$
Solv... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
24. (15 points) Given the parabola $y^{2}=\operatorname{tar}(0<a<1)$ with focus $\vec{F}$, a semicircle is constructed above the $x$-axis with center $A(a+4,0)$ and radius $|A F|$, intersecting the parabola at two distinct points $M$ and $N$. Let $P$ be the midpoint of segment $M N$.
(1) Find the value of $|M F|+|N F|$... | 24. (1) From the given, we have $F(a, 0)$, and the semicircle is
$$
[x-(a+4)]^{2}+y^{2}=16(y \geqslant 0) \text {. }
$$
Substituting $y^{2}=4 a x$ into the equation, we get
$$
x^{2}-2(4-a) x+a^{2}+8 a=0 \text {. }
$$
Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. Then, by the definition of the parabola... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given that $x, y, z$ are two non-negative rational numbers, and satisfy $3x+2y+z=5, x+y-z=2$. If $S=2x+y-z$, then what is the sum of the maximum and minimum values of $S$?
(1996, Tianjin City Junior High School Mathematics Competition) | Analysis: By representing the algebraic expression $S$ with one letter and determining the range of this letter, we can find the maximum or minimum value of $S$.
From the given, we solve $y=\frac{7-4 x}{3}, z=\frac{1-x}{3}$, and.
$$
\left\{\begin{array}{l}
\frac{7-4 x}{3} \geqslant 0, \\
\frac{1-x}{3} \geqslant 0, \\
x... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
$(1996$, National Junior High School Mathematics Leagu... | $$
\begin{array}{l}
\text{Analysis: Let } x_{1} \text{ and } x_{2} \text{ be the x-coordinates of the points where the parabola intersects the x-axis, then } x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}. \text{ It is easy to know that } -\frac{b}{a}<0, \text{ and } -116 c(a-b+c). \\
\because a-b+c>0, c \geqslant 1... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Natural numbers $m, n$ satisfy $8 m+10 n>9 m n$. Then
$$
m^{2}+n^{2}-m^{2} n^{2}+m^{4}+n^{4}-m^{4} n^{4}=
$$
$\qquad$ | 3. 2 .
It is known that $\frac{8}{n}+\frac{10}{m}>9$. It is easy to see that one of the two natural numbers $m, n$ must be 1 (otherwise $\frac{8}{m}+\frac{10}{m} \leqslant \frac{8}{2}+\frac{10}{2}=9$ which is a contradiction). Without loss of generality, let $m=1$, then
$$
\begin{array}{l}
m^{2}+n^{2}-m^{2} n^{2}+m^{4... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $(-1)^{n} 2^{n} \equiv b_{n}(\bmod 9), 0 \leq b_{n} \leq 8$. Then the range of values for $b_{n}$ is \{ $\qquad$ \}. | $\begin{array}{l}\text { II.1.1. } \\ \text { \| }(-1)^{n} 2^{3 n}=(-8)^{n}=(1-9)^{n} \\ \quad=C_{n}^{n}-C_{n}^{1} 9+C_{n}^{2} 9^{2}+\cdots+(-1)^{n} C_{n}^{n} 9^{n}, \\ \text { know }(-1)^{n} 2^{3 n} \equiv 1(\bmod 9) .\end{array}$ | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
The struggle of the land, by 1996, the forest coverage rate of the whole county had reached 30% (becoming an oasis). From 1997 onwards, each year will see such changes: 16% of the original desert area will be converted, and 4% of it will be eroded, turning back into desert.
(1) Assuming the total area of the county is ... | Let the accumulated area be $b_{n+1}$. Then
$$
a_{1}+b_{1}=1, \quad a_{n}+b_{n}=1 .
$$
According to the problem, $a_{n+1}$ consists of two parts: one part is the remaining area of the original oasis $a_{n}$ after being eroded by $\frac{4}{100} \cdot a_{n}$, which is
$$
a_{n}-\frac{4}{100} a_{n}=\frac{96}{100} a_{n} .
... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Initial 52. Given real numbers $a, b, x, y$ satisfy $a x^{n}+b y^{n}=1+2^{n+1}$ for any natural number $n$. Find the value of $x^{a}+y^{b}$. | Let $S_{n}=a x^{\prime \prime}+b y^{\prime \prime}$. Then
$$
S_{1}=a x+b y=5 \text {, }
$$
$$
\begin{array}{l}
S_{0}=a x^{2}+b y^{2}=9 . \\
S=a x^{3}+b y^{3}=17 . \\
S_{1}=a x^{4}+b y^{\prime}=33 .
\end{array}
$$
Let $x+y=A, x y=B$. Then $x, y$ are the roots of the quadratic equation $t^{2}-A t+B=0$. Therefore,
$$
\be... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{c}
\text { Example } 1 \text { Calculate } \frac{a}{a^{3}+a^{2} b+a b^{2}+b^{3}}+ \\
\frac{b}{a^{3}-a^{2} b+a b^{2}-b^{3}}+\frac{1}{a^{2}-b^{2}}-\frac{1}{b^{2}+a^{2}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} .
\end{array}
$$
(1995, Tianjin City, Grade 7 "Mathematics Competition) | $\begin{array}{l}\text { Solution: Original expression }=\frac{a}{(a+b)\left(a^{2}+b^{2}\right)} \\ +\frac{b}{(a-b)\left(a^{2}+b^{2}\right)}+\frac{2 b^{2}}{a^{4}-b^{4}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} \\ =\frac{a^{2}+b^{2}}{\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)}-\frac{a^{2}+b^{2}}{a^{4}-b^{4}} \\ =\frac{1}... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 If $a+b+c=a b c \neq 0$, find the value of $\frac{\left(1-a^{2}\right)\left(1-b^{2}\right)}{a b}+\frac{\left(1-b^{2}\right)\left(1-c^{2}\right)}{b c}+$ $\frac{\left(1-c^{2}\right)\left(1-a^{2}\right)}{a c}$.
(1990, Wuhan City Mathematics Competition) | $\begin{array}{l}\text { Solution: Original expression }=\frac{1-a^{2}-b^{2}+a^{2} b^{2}}{a b} \\ +\frac{1-b^{2}-c^{2}+b^{2} c^{2}}{b c}+\frac{1-a^{2}-c^{2}+a^{2} c^{2}}{a c} \\ =\frac{1}{a b}-\frac{a}{b}-\frac{b}{a}+a b+\frac{1}{b c}-\frac{b}{c}-\frac{c}{b}+b c \\ +\frac{1}{a c}-\frac{a}{c}-\frac{c}{a}+a c \\ =\left(\... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Full marks 20 points) Given the equation $x^{2}+m x-m+1=0$ (where $m$ is an integer) has two distinct positive integer roots. Find the value of $m$.
---
The translation maintains the original format and line breaks as requested. | Three, let the two distinct positive integer roots be $\alpha, \beta(\alpha<\beta)$. By Vieta's formulas, we have $\left\{\begin{array}{l}\alpha+\beta=-m, \\ \alpha \beta=-m+1 .\end{array}\right.$ Eliminating $m$, we get $\alpha \beta-\alpha-\beta=1$.
That is, $(\alpha-1)(\beta-1)=2$.
Then $\left\{\begin{array}{l}\alph... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a, b, c$ are distinct. Find the value of $\frac{2 a-b-c}{(a-b)(a-c)}$ $+\frac{2 b-c-a}{(b-c)(b-a)}+\frac{2 c-a-b}{(c-a)(c-b)}$. | (Answer: 0) | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Real numbers $a, b, c$ satisfy $a+b+c=0, a b c>0$. If $x=$
$$
\begin{array}{l}
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}, y=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}\right. \\
\left.+\frac{1}{b}\right) \text {. Then, } x+2 y+3 x y=
\end{array}
$$
. | 1. 2
The above text has been translated into English, maintaining the original text's line breaks and format. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $X=\{-1,0,1\}, Y=\{-2,-1,0,1,2\}$, and for all elements $x$ in $X$, $x+f(x)$ is even. Then the number of mappings $f$ from $X$ to $Y$ is ( ).
(A) 7
(B) 10
(C) 12
(D) 15. | 4. (C).
$x$ and $f(x)$ are both even, or both odd.
When $x=0$, there are three possibilities: $-2, 0, 2$;
When $x=-1, 1$, there are two possibilities each for -1 and 1, making 4 possibilities.
Therefore, the total is $3 \times 4=12$. | 12 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 1. } \lim _{n \rightarrow \infty} \frac{1}{\sqrt[3]{n}}\left(\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+ \\ \frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}+ \\ \left.\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}}\right)=\end{array}$ | $$
\approx .1 .1 \text {. }
$$
Since
$$
\begin{array}{l}
\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+\cdots+ \\
\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}} \\
=\frac{\sqrt[3]{2}-1}{2-1}+\cdots+\frac{\sqrt[3]{n}-\sqrt[3]{n-1}}{n-(n-1)} . \\
=\sqrt[3]{n}-1,
\end{array}
$$
Therefore, the required limit is $\l... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
9. Simplify $\frac{(x+b)(x+c)}{(a-b)(a-c)}+\frac{(x+c)(x+a)}{(b-c)(b-a)}$ $+\frac{(x+a)(x+b)}{(c-a)(c-b)}$. | (Answer: 1).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $x=b y+c z, y=c z+a x, z=a x$ $+b y$. Find the value of $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}$. | Solution: $\frac{a}{a+1}=\frac{a x}{a x+x}=\frac{a x}{a x+b y+c z}$.
Similarly, $\frac{b}{b+1}=\frac{b y}{a x+b y+c z}$,
$$
\frac{c}{c+1}=\frac{c z}{a x+b y+c z} \text {. }
$$
Adding them up, the value of the desired expression is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 If $\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=$ $\frac{t}{x+y+z}$, let $f=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}$. Prove: $f$ is an integer.
(1990, Hungarian Mathematical Competition) | Proof: If $x+y+z+t \neq 0$, by the property of proportion, we have
$$
\begin{array}{l}
\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y} \\
=\frac{t}{x+y+z}=\frac{x+y+z+t}{3(x+y+z+t)}=\frac{1}{3} . \\
\text { Then we have }\left\{\begin{array}{l}
y+z+t=3 x, \\
z+t+x=3 y, \\
t+x+y=3 z, \\
x+y+z=3 t .
\end{array}\right.
\e... | -4 | Algebra | proof | Yes | Yes | cn_contest | false |
Example 7 Given $4 x-3 y-6 z=0, x+2 y-7 z$ $=0(x y z \neq 0)$. Find the value of $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}$.
(1992, Sichuan Province Junior High School Mathematics League Preliminary) | Solution: Let $y=k_{1} x, z=k_{2} x$. Then
$$
\left\{\begin{array} { l }
{ 4 - 3 k _ { 1 } - 6 k _ { 2 } = 0 , } \\
{ 1 + 2 k _ { 1 } - 7 k _ { 2 } = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
k_{1}=\frac{2}{3}, \\
k_{2}=\frac{1}{3} .
\end{array}\right.\right.
$$
Then $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given the theorem: "If three prime numbers greater than 3, $a, b$, and $c$, satisfy the equation $2a + 5b = c$, then $a + b + c$ is a deficient number of the integer $n$." What is the maximum possible value of the integer $n$ in the theorem? Prove your conclusion. | Three, the maximum possible value of $n$ is 9.
First, prove that $a+b+c$ can be divided by 3.
In fact, $a+b+c=a+b+2a+5b=3(a+2b)$, so $a+b+c$ is a multiple of 3.
Let the remainders when $a$ and $b$ are divided by 3 be $r_{a}$ and $r_{b}$, respectively, then $r_{a} \neq 0, r_{b} \neq 0$.
If $r_{a} \neq r_{b}$, then $r_{a... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false |
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