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Five, (20 points) For the parabola $y^{2}=2 p x(p>0)$ with focus $F$, does there exist an inscribed isosceles right triangle such that one of its legs passes through $F$? If it exists, how many are there? If not, explain the reason.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
As shown in Figure 4, let \( A\left(x_{A}, y_{A}\right) \), \( B\left(x_{B}, y_{B}\right) \), \( C\left(x_{C}, y_{C}\right) \), and the inclination angle of \( AC \) be \( \theta \). Clearly, \( \theta \neq 0 \), \( \frac{\pi}{2} \). Suppose \( \theta \in \left(0, \frac{\pi}{2}\right) \). By the focal chord length formula, we have
\[
\begin{array}{l}
|A C|=\frac{2 p}{\sin ^{2} \theta} . \\
\because\left\{\begin{array}{l}
y_{A}-y_{C}=|A C| \sin \theta=\frac{2 p}{\sin \theta}, \\
y_{A}+y_{C}=2 p \frac{x_{A}-x_{C}}{y_{A}-y_{C}}=2 p \cot \theta,
\end{array}\right.
\end{array}
\]
\[
\left\{\begin{array}{l}
y_{C}-y_{B}=|C B| \cos \theta=|A C| \cos \theta=\frac{2 p \cos \theta}{\sin ^{2} \theta}, \\
y_{C}+y_{B}=2 p \frac{x_{C}-x_{B}}{y_{C}-y_{B}}=-2 p \tan \theta .
\end{array}\right.
\]
From (1), we get \( y_{C}=p \frac{\cos \theta-1}{\sin \theta} \),
From (2), we get \( y_{C}=p \frac{\cos ^{2} \theta-\sin ^{3} \theta}{\sin ^{2} \theta \cos \theta} \),
\(\therefore \cos\) squared minus \(\sin ^{3} \theta\),
which means \(\cos \theta+\sin \theta=\tan \theta\),
or equivalently, \(\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right)=\tan \theta\).
The problem reduces to whether (3) has a solution in \(\left(0, \frac{\pi}{2}\right)\). From the graphs of trigonometric functions, (3) has exactly one intersection point in \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), indicating that there exists an isosceles right triangle that satisfies the given conditions. By symmetry, when \(\theta \in \left(\frac{\pi}{2}, \pi\right)\), there is also one intersection point.
In summary, there exist exactly two isosceles right triangles that satisfy the given conditions.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $a$ is a natural number, there exists a linear polynomial with integer coefficients and $a$ as the leading coefficient, which has two distinct positive roots less than 1. Then, the minimum value of $a$ is $\qquad$ .
|
2.5.
Let $f(x)=a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, $00$. Then from $f(0)$ and $f(1)$ being positive integers, we get $f(0) f(1) \geqslant 1$, that is, $a^{2} x_{1} x_{2}\left(1-x_{1}\right)\left(1-x_{2}\right) \geqslant 1$.
Also, $x(1-x) \leqslant \frac{1}{4}$, with equality when $x=\frac{1}{2}$. Since $x_{1} \neq x_{2}$, we have $x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right)4$.
When $a=5$, the equation $5 x^{2}-5 x+1=0$ has two distinct positive roots in the interval $(0,1)$. Therefore, the minimum value of $a$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (20 points) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=3, a_{n}=$ $3^{a_{n-1}}(n \geqslant 2)$. Find the last digit of $a_{n}(n \geqslant 2)$.
|
Let's prove, when $n \geqslant 2$, we have
$$
a_{n}=4 m+3, m \in \mathbf{N} \text {. }
$$
Using mathematical induction:
(i) When $n=2$, $a_{2}=3^{3}=4 \times 6+3$, so equation (1) holds.
(ii) Assume when $n=k(k \geqslant 2)$, equation (1) holds, i.e.,
$$
a_{k}=4 m+3, m \in \mathbf{N} \text {. }
$$
Then $a_{k+1}=3^{a_{k}}=3^{4 m+3}=(4-1)^{4 m+3}$
$$
\begin{aligned}
= & \mathrm{C}_{4 m+3}^{0} \cdot 4^{4 m+3}(-1)^{0}+\mathrm{C}_{4 m+3}^{1} 4^{4 m+2}(-1)^{1} \\
& +\cdots+\mathrm{C}_{4 m+3}^{4 m+2} 4^{1}(-1)^{4 m+2} \\
& +\mathrm{C}_{4 m+3}^{4 m+3} 4^{0}(-1)^{4 m+3} \\
= & 4 t-1=4(t-1)+3 .
\end{aligned}
$$
Here, $t=\mathrm{C}_{4 m+3}^{0} 4^{4 m+2}+\mathrm{C}_{4 m+3}^{1} 4^{4 m+1}(-1)^{1}+\cdots+$ $\mathrm{C}_{4 m+3}^{4 m+2}(-1)^{4 m+2}$ is an integer, and since $\left\{a_{n}\right\}$ is an increasing sequence, we have $a_{k+1} \geqslant a_{k} \geqslant \cdots \geqslant a_{2}=27$, i.e., $4 t-1 \geqslant 27$. Thus, $t \geqslant 7, t-1 \geqslant 6$, i.e., $t-1 \in \mathbf{N}$.
Therefore, when $n=k+1$, equation (1) holds.
Thus, equation (1) holds, and we have
$$
a_{n}=3^{a_{n-1}}=3^{4 m+3}=(81)^{m} \cdot 27 \text {, }
$$
and its last digit is 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given an integer $n > 3$, let real numbers $x_{1}, x_{2}, \cdots, x_{n}$, $x_{n+1}, x_{n+2}$ satisfy the condition
$$
0 < x_{1} < x_{2} < \cdots < x_{n} < x_{n+1} < x_{n+2}.
$$
Find the minimum value of
$$
\frac{\left(\sum_{i=1}^{n} \frac{x_{i+1}}{x_{i}}\right)\left(\sum_{j=1}^{n} \frac{x_{j+2}}{x_{j+1}}\right)}{\left(\sum_{k=1}^{n} \frac{x_{k+1} x_{k+2}}{x_{k+1}^{2} + x_{k} x_{k+2}}\right)\left(\sum_{l=1}^{n} \frac{x_{l+1}^{2} + x_{l} x_{l+2}}{x_{l} x_{l+1}}\right)}
$$
and determine all sets of real numbers $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}, x_{n+2}$ that achieve this minimum value.
(Zhang Zhusheng provided)
|
Solution: (I) Let $t_{i}=\frac{x_{i+1}}{x_{i}}(>1), 1 \leqslant i \leqslant n+1$. The expression in the problem can be written as
$$
\frac{\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} i_{i+1}\right)}{\left(\sum_{i=1}^{n} \frac{t_{i=1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right)} \text {. }
$$
We observe that
$$
\begin{aligned}
( & \left(\sum_{i=1}^{n} \frac{t_{i} t_{i+1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
= & \left(\sum_{i=1}^{n} t_{i}-\sum_{i=1}^{n} \frac{t_{i,}^{2}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
= & \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
& -\left(\sum_{i=1}^{n} \frac{t_{i}^{2}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
\leqslant & \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
& \quad-\left(\sum_{i=1}^{n} \frac{t_{i}}{\sqrt{t_{i}+t_{i+1}}} \sqrt{t_{i}+t_{i+1}}\right)^{2} \\
= & \left(\sum_{i=1}^{n} t_{i}\right)^{2}+\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right)-\left(\sum_{i=1}^{n} t_{i}\right)^{2} \\
= & \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right) .
\end{aligned}
$$
Therefore, for the real number sequence $00$, we have
$$
x_{k}=t_{k-1} t_{k-2} \cdots t_{1} a=a b^{k-1} c^{\frac{(k-1)(k-2)}{2}}, 2 \leqslant k \leqslant n
$$
+2 .
$$
\begin{array}{l}
\because x_{2}>x_{1}, \therefore b=\frac{x_{2}}{x_{1}}>1 . \\
\text{Also, since } t_{j}=b c^{j-1}>1,1 \leqslant j \leqslant n+1, \\
\therefore c>\sqrt[n]{\frac{1}{b}}\left(\geqslant \sqrt[j]{\frac{1}{b}}, 1 \leqslant j \leqslant n+1\right) .
\end{array}
$$
(III) Conclusion:
(i) For the real number sequence $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}$, $x_{n+2}$, the minimum value of the expression in the problem is 1.
(ii) The conditions for the expression to achieve its minimum value are $00, b>1, c>\sqrt[\pi]{\frac{1}{b}}$.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $16^{9 m}=a, 4^{37 n}=\frac{1}{a}$, then $(36 m+74 n-1)^{2000}$
|
Ni.1.1.
Since $a=16^{9 m}=\left(2^{4}\right)^{9 m}=2^{36 m}$, and $\frac{1}{a}=4^{37 n}=2^{74 n}$, then $1=a \times \frac{1}{a}=2^{36 m} \cdot 2^{74 n}=2^{36 m+74 n}$,
we have $36 m+74 n=0$.
Therefore, $(36 m+74 n-1)^{2000}=(-1)^{20000}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 2, in
Rt $\triangle A B C$,
$\angle C=90^{\circ}$, point $M$
is the intersection of the three
medians of the triangle. Perpendiculars are drawn from $M$
to $A B$, $B C$, and $A C$,
with the feet of the perpendiculars being $D$, $E$, and $F$, respectively. If $A C=3$, $B C=12$, then the area of $\triangle D E F$ is $\qquad$
|
4.4.
$$
\begin{array}{l}
AB = \sqrt{3^2 + 12^2} = \sqrt{153} = 3 \sqrt{17}, \\
S_{\triangle ABC} = \frac{1}{2} \times 3 \times 12 = 18.
\end{array}
$$
Draw a perpendicular from $C$ to $AB$, with the foot of the perpendicular at $H$, and let it intersect $EF$ at $G$. We have
$$
\begin{array}{l}
S_{\triangle ABC} = \frac{1}{2} \cdot AB \cdot CH = 18, \\
CH = \frac{18 \times 2}{3 \sqrt{17}} = \frac{12}{\sqrt{17}}. \\
\because ME = FC = \frac{1}{3} AC = 1, CE = \frac{1}{3} BC = 4. \\
\therefore EF = \sqrt{CF^2 + CE^2} = \sqrt{1 + 4^2} = \sqrt{17}. \\
\because EF \cdot CG = CE \cdot CF,
\end{array}
$$
i.e., $CG = \frac{CE \cdot CF}{EF} = \frac{4 \times 1}{\sqrt{17}} = \frac{4}{\sqrt{17}}$.
$$
\begin{array}{l}
\therefore GH = CH - CG = \frac{12}{\sqrt{17}} - \frac{4}{\sqrt{17}} = \frac{8}{\sqrt{17}}. \\
\text{Therefore, } S_{\triangle DEF} = \frac{1}{2} \cdot EF \cdot GH = \frac{1}{2} \times \sqrt{17} \times \frac{8}{\sqrt{17}} = 4.
\end{array}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. A piece of lead wire of length $2 n$ (where $n$ is a natural number and $n \geqslant 4$) is folded into a triangle with integer side lengths. Let $(a, b, c)$ represent a triangle with side lengths $a, b, c$ such that $a \leqslant b \leqslant c$.
(1) For the cases $n=4, 5, 6$, write down all the $(a, b, c)$ that satisfy the conditions;
(2) Someone, based on the conclusions in (1), conjectured that when the length of the lead wire is $2 n$ ($n$ is a natural number and $n \geqslant 4$), the number of corresponding $(a, b, c)$ is always $n-3$. In fact, this is an incorrect conjecture; write down all the $(a, b, c)$ for $n=12$, and state the number of $(a, b, c)$;
(3) Try to classify all the $(a, b, c)$ that satisfy the conditions for $n=12$ according to at least two different criteria.
|
15. (1) When $n=4$, the length of the lead wire is 8. Then the only group of $(a, b, c)$ that satisfies the condition is $(2,3,3)$;
When $n=5$, the length of the lead wire is 10. Then the groups of $(a, b, c)$ that satisfy the condition are $(2,4,4),(3,3,4)$;
When $n=6$, the length of the lead wire is 12. Then the groups of $(a, b, c)$ that satisfy the condition are $(2,5,5),(3,4,5),(4,4,4)$.
(2) When $n=12$, the length of the lead wire is 24. According to the problem,
$$
a+b+c=24 \text {, and }\left\{\begin{array}{l}
a+b>c \\
a \leqslant b \leqslant c
\end{array}\right. \text {. }
$$
From this, we get $8 \leqslant c \leqslant 11$, i.e., $c=8,9,10,11$.
Therefore, the groups of $(a, b, c)$ that satisfy the condition are 12:
$$
\begin{array}{l}
A(2,11,11) ; B(3,10,11) ; C(4,9,11) ; \\
D(5,8,11) ; E(6,7,11) ; F(4,10,10) ; \\
G(5,9,10) ; H(6,8,10) ; I(7,7,10) ; \\
J(6,9,9) ; K(7,8,9) ; L(8,8,8) .
\end{array}
$$
(3) Different classification criteria determine different classifications. Here are some examples:
(1) Classify by the value of the largest side $c$, there are four categories:
i) $c=11$, there are five: $A, B, C, D, E$;
ii) $c=10$, there are four: $F, G, H, I$;
iii) $c=9$, there are two: $J, K$;
iv) $c=8$, there is only one: $L$.
(2) Classify by whether they are equilateral or isosceles triangles, there are three categories:
i) Equilateral triangles: only $L$;
ii) Isosceles but not equilateral triangles: $A, F, I, J$ four;
iii) Neither equilateral nor isosceles triangles: $B, C, D, E, G, H, K$ seven.
(3) Classify by the relationship between the largest angle and the right angle, there are three categories:
i) Acute triangles: $A, F, G, I, J, K, L$ seven;
ii) Right triangles: only $H$;
iii) Obtuse triangles: $B, C, D, E$ four.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The square number $y^{2}$ is
the sum of the squares of 11 consecutive integers. Then the smallest value of the natural number $y$ is
$\qquad$
|
3.11.
Let the middle number of these 11 consecutive integers be \( a \). Then
\[
\begin{aligned}
y^{2}= & (a-5)^{2}+(a-4)^{2}+(a-3)^{2}+(a-2)^{2} \\
& +(a-1)^{2}+a^{2}+(a+1)^{2}+(a+2)^{2} \\
& +(a+3)^{2}+(a+4)^{2}+(a+5)^{2} \\
= & 11 a^{2}+2\left(5^{2}+4^{2}+3^{2}+2^{2}+1^{2}\right) \\
= & 11\left(a^{2}+10\right) .
\end{aligned}
\]
Thus, when \( a^{2}+10=11 \), i.e., \( a= \pm 1 \), \( y^{2} \) is minimized. At this point, \( v_{\text {min }}=11 \).
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $a_{n}=6^{n}+8^{n}$. Then $a_{84} \equiv$ $\qquad$ $(\bmod 49)$
|
5.2.
$$
\begin{aligned}
a_{84}= & (7-1)^{84}+(7 \\
& +1)^{84} \\
= & 2\left(\mathrm{C}_{84}^{0} \cdot 7^{84}+\right. \\
& \mathrm{C}_{84}^{2} \cdot 7^{82}+\cdots \\
& \left.+\mathrm{C}_{84}^{82} \cdot 7^{2}+\mathrm{C}_{84}^{84}\right) \\
= & 45 \times M+2 . \\
a_{84}= & 2(\text { modulo } 49) .
\end{aligned}
$$
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$=6$ integer solutions.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Solution: Let the two integer roots of the equation be $x_{1}$ and $x_{2}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-(10 a+b), \\
x_{1} x_{2}=10 b+a .
\end{array}\right.
$$
We have $10 x_{1}+10 x_{2}+x_{1} x_{2}=-99 a(1 \leqslant a \leqslant 9)$.
Thus $\left(x_{1}+10\right)\left(x_{2}+10\right)=100-99 a$.
By the problem, $x_{1}<0, x_{2}<0$, and
$$
\overline{a b}=-\left(x_{1}+x_{2}\right), \overline{b a}=x_{1} x_{2}=99 \text {. }
$$
Let $\left|x_{1}\right| \leqslant\left|x_{2}\right|$, by $\left|x_{1}\right| \cdot\left|x_{2}\right| \leqslant 99$ we know $\left|x_{1}\right| \leqslant 9$.
Obviously, $x_{1} \neq 0$, then
$$
9 \geqslant x_{1}+10 \geqslant 1 \text {. }
$$
Discussion: (1) When $a=1$.
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=1 \text {. }
$$
By (2), only
$$
\left\{\begin{array}{l}
x_{1}+10=1, \\
x_{2}+10=1 .
\end{array}\right.
$$
That is, $x_{1}=x_{2}=-9$.
Substituting into equation (1) gives $\overline{a b}=18, \overline{b a}=81$.
(2) When $a=2$,
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=-98 \text {. }
$$
By (2) we have,
$$
x_{2}+10<0 \text {. }
$$
Thus $-98=k \cdot\left(-\frac{98}{k}\right) \cdot\left(k, \frac{98}{k}\right.$ are integers, $\left.1 \leqslant k \leqslant 9\right)$
$$
\begin{array}{l}
\therefore\left(x_{1}+10\right)\left(x_{2}+10\right) \\
=1 \times(-98)=2 \times(-49)=7 \times(-14) .
\end{array}
$$
Solving gives $\left\{\begin{array}{l}x_{1}=-9, \\ x_{2}=-108 ;\end{array}\left\{\begin{array}{l}x_{1}=-8, \\ x_{2}=-59 ;\end{array}\left\{\left\{\begin{array}{l}x_{1}=-3, \\ x_{2}=-24\end{array}\right.\right.\right.\right.$.
Substituting into equation (1) gives $\overline{a b}=27, \overline{b a}=72$.
At this time, $x_{1}=-3, x_{2}=-24$.
(3) When $a=3$,
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=-197=1 \times(-197) \text {. }
$$
By (2) we know there is no solution.
(4) When $a=4$,
$$
\begin{array}{l}
\left(x_{1}+10\right)\left(x_{2}+10\right) \\
=-296=1 \times(-296)=2 \times(-148) \\
=4 \times(-74)=8 \times(-37) .
\end{array}
$$
By (2) we know $k=1,2,4$, there is no solution.
Solving gives $x_{1}=-2, x_{2}=-47$.
Substituting into equation (1) gives $\overline{a b}=49, \overline{b a}=94$.
(5) When $a=5$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-395=5 \times(-75)$, there is no solution.
(6) When $a=6$,
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=-494=2 \times(-247) \text {, no solution. }
$$
(7) When $a=7$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-593=1 \times(-593)$, there is no solution.
(8) When $a=8$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-692=4 \times(-173)$, there is no solution.
(9) When $a=9$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-791=7 \times(-113)$, there is no solution.
In summary, when $\overline{a b}=18,27,49$, the integer solutions of the equation are: $\left\{\begin{array}{l}x_{1}=-9, \\ x_{2}=-9 ;\end{array}\left\{\begin{array}{l}x_{1}=-3, \\ x_{2}=-24 ;\end{array}\left\{\begin{array}{l}x_{1}=-2, \\ x_{2}=-47 .\end{array}\right.\right.\right.$
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Through the vertex of isosceles $\triangle A B C$, draw a line intersecting the extension of the opposite side at $D$. If the resulting triangles are all isosceles, then $\triangle A B C$ has $\qquad$ such configurations.
|
4.5.
Draw a straight line from the base angle to intersect with the extension of the opposite side, there are 5 $\triangle A B C$ that meet the conditions, with base angles of $45^{\circ} 、 72^{\circ} 、 36^{\circ} 、 \frac{180^{\circ}}{7}$ 、 $\frac{2 \times 180^{\circ}}{7}$.
Draw a straight line from the vertex angle to intersect with the extension of the opposite side, there is 1 $\triangle A B C^{\prime}$ that meets the conditions, with a base angle of $72^{\circ}$, which is repeated above.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The number of triangles with unequal integer sides and a perimeter less than 13 is $\qquad$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
(3)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 On a street $AB$, person A walks from $A$ to $B$, and person B rides a bike from $B$ to $A$. Person B's speed is 3 times that of person A. At this moment, a public bus departs from the starting station $A$ and heads towards $B$, and a bus is dispatched every $x$ minutes. After some time, person A notices that a bus catches up to him every 10 minutes, while person B feels that he encounters a bus every 5 minutes. Therefore, the interval time $x=$ $\qquad$ for the dispatch of buses at the starting station.
(9th Hope Cup National Mathematics Invitational for Junior High School)
|
Solution: According to the problem, draw the graph as shown in Figure 4.
$A C, A_{1} C_{1}, A_{2} C_{2}, A_{3} C_{3}$ represent the motion graphs of buses departing every $x$ minutes, $A D$ and $B . V$ are the motion graphs of person A and person B, respectively. $E_{1}, E_{2}$ are the points where a bus catches up with person A every 10 minutes. $F_{1}, F_{2}, F_{3}, F_{4}$ are the points where person B encounters a bus every 5 minutes.
Draw $E_{1} G \perp t$-axis through $E_{1}$, draw $F_{2} H / / t$-axis intersecting $O C$ at $H$, and draw $F_{1} M \perp H F_{2}$, then $A G=$ 10. $M F_{2}=5, H F_{2}=A A_{1}=x$.
Let $\angle E_{1} A G=\alpha, \angle F_{1} H M=\angle E_{1} A_{1} G=$ $\beta, \angle F_{1} F_{2} M=\gamma$. Also,
$\tan \alpha=\frac{E_{1} G}{A G}=\frac{E_{1} G}{10}$.
In $\triangle E_{1} A_{1} G$,
$\tan \beta=\frac{E_{1} G}{A_{1} G}=\frac{E_{1} G}{A G-A A_{1}}=\frac{E_{1} G}{10-x}$.
In $\triangle F_{1} H M$,
$\tan \beta=\frac{F_{1} M}{H M}=\frac{F_{1} M}{H F_{2}-M F_{2}}=\frac{F_{1} M}{x-5}$.
$\tan \gamma=\frac{F_{1} M}{M F_{2}}=\frac{F_{1} M}{5}$.
(1) $\div$ (2) gives $\frac{\tan \alpha}{\tan \beta}=\frac{10-x}{10}$.
(3) $\div$ (4) gives $\frac{\tan \beta}{\tan \gamma}=\frac{5}{x-5}$.
Since person B's speed is 3 times that of person A, then $\tan \gamma=3 \tan \alpha$.
(3) $\times$ (6) gives $\frac{1}{3}=\frac{10-x}{2(x-5)}$. Solving for $x$ yields $x=8$.
Therefore, the interval between departures at the starting station is 8 minutes.
The key here is equations (3) and (6). Their practical significance is the same as in Example 2.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example $8 A$ and $B$ are two fixed points on a plane. Find a point $C$ on the plane such that $\triangle A B C$ forms an isosceles right triangle. There are $\qquad$ such points $C$.
|
Solution: As shown in Figure 5, the vertices of the two isosceles right triangles with $AB$ as the hypotenuse are $C_{1}$ and $C_{2}$; the vertices of the four isosceles right triangles with $AB$ as the legs are $C_{3}, C_{4}, C_{5}, C_{6}$.
Therefore, there are a total of 6 points $C$ that meet the conditions.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. As shown in Figure 5, in $\square A B C D$, $P_{1}$, $P_{2}, \cdots$, $P_{n-1}$ are the $n$ equal division points on $B D$. Connect $A P_{2}$ and extend it to intersect $B C$ at point $E$, and connect $A P_{n-2}$ and extend it to intersect $C D$ at point $F$.
(1) Prove that $E F \parallel B D$;
(2) Let the area of $\square A B C D$ be $S$. If $S_{\triangle A E F} = \frac{3}{8} S$, find the value of $n$.
|
14. (1) Since $A D / / B C, A B / / D C$. Therefore, $\left.\triangle P_{n-2} F I\right) \triangle \triangle P_{n-2} A B, \triangle P_{2} B E \subset \triangle P_{2} D . A$.
Thus, we know
$$
\frac{A P_{n-2}}{P_{n+2} F}=\frac{B P_{n}}{P_{n-2} D}=\frac{n-2}{2}, \frac{A P_{2}}{P_{2} E}=\frac{D P_{2}}{P_{2} B}=\frac{n-2}{2},
$$
which means $\frac{A P_{n-2}}{P_{n} \frac{{ }_{2}}{} F}=\frac{A P_{2}}{P_{2} E}$; hence $E F / / B D$.
(2) From (1), we know $\frac{A P_{2}}{A E}=\frac{n-2}{n}, \frac{S_{\triangle A P_{2} P_{n-2}}}{S_{\text {ZAFF }}}=$ $\left(\frac{A P_{2}}{A E}\right)^{2}$. Therefore,
$$
\begin{aligned}
& S_{\triangle_{2} P_{n-2}}=\left(\frac{n-2}{n}\right)^{2} \cdot S_{\text {IAE }} \\
= & \left(\frac{n-2}{n}\right)^{2} \cdot \frac{3}{8} S .
\end{aligned}
$$
Also, since $S_{\triangle A A S}=\frac{1}{2} S_{\triangle A / \times D}=\frac{1}{2} S$, and $P_{1}, P_{2}, \cdots$ 。 $P_{n}$ 。are the $n$ equal division points of $B D$. Therefore,
$$
\begin{array}{l}
S_{-N_{2} p_{n}}=\left(\frac{n-4}{n}\right) \cdot S_{\text {U(S) }} \\
=\left(\frac{n-4}{n}\right) \cdot \frac{1}{2} S .
\end{array}
$$
From (1) and (2), we get $\left(\frac{n-2}{n}\right)^{2} \cdot \frac{3}{8} S=\left(\frac{n-4}{n}\right) \cdot \frac{1}{2} S$,
which simplifies to $n^{2}-4 n-12=0$.
Solving, we get $n=6, n=-2$ (discard). Therefore, $n=6$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) Several containers are unloaded from a cargo ship, with a total weight of 10 tons, and the weight of each container does not exceed 1 ton. To ensure that these containers can be transported in one go, how many trucks with a carrying capacity of 3 tons are needed at least?
|
First, note that the weight of each container does not exceed 1 ton, so the weight of containers that each vehicle can carry at one time will not be less than 2 tons; otherwise, another container can be added.
Let $n$ be the number of vehicles, and the weights of the containers they carry be $a_{1}, a_{2}, \cdots, a_{n}$, then $2 \leqslant a_{1} \leqslant 3(i=1,2, \cdots, n)$.
Let the total weight of all the containers carried be $s$, then
$$
2 n \leqslant s=a_{1}+a_{2}+\cdots+a_{n} \leqslant 3 n \text {. }
$$
That is, $2 n \leqslant 10 \leqslant 3 n$.
Thus, $\frac{10}{3} \leqslant n \leqslant 5$, which means $n=4$ or 5.
In fact, 4 vehicles are not enough.
Suppose there are 13 containers, each weighing $\frac{10}{13}$ tons. Since $\frac{10}{13} \times 3 < 3$, each vehicle can only carry 3 containers, so 4 vehicles cannot carry all of them.
Therefore, at least 5 trucks are needed.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 In Figure 1, there are 8 vertices, each with a real number. The real number at each vertex is exactly the average of the numbers at the 3 adjacent vertices (two vertices connected by a line segment are called adjacent vertices). Find
$$
a+b+c+d-(e+f+g+h)
$$
|
The following solution has appeared in a journal:
Given
$$
\begin{array}{l}
a=\frac{b+e+d}{3}, \\
b=\frac{a+f+c}{3}, \\
c=\frac{b+g+d}{3}, \\
d=\frac{c+h+a}{3}.
\end{array}
$$
Adding the four equations yields
$$
\begin{array}{l}
a+b+c+d \\
=\frac{1}{3}(2 a+2 b+2 c+2 d+e+f+g+h),
\end{array}
$$
which simplifies to $a+b+c+d-(e+f+g+h)=0$.
The above solution, though clever, has two drawbacks.
First, it is too coincidental. Adding the equations to get $a+b+c+d$ is expected, but the fact that it simplifies to $a+b+c+d-(e+f+g+h)=0$ is purely coincidental. If the problem were to find $2a+b+c+d-(2e+f+g+h)$, the above method would no longer work.
Second, the solution fails to reveal the essence of the problem, which is the relationship among the 8 numbers $a, b, c, d, e, f, g, h$.
To solve this problem, one should first consider a specific example. To make each vertex number equal to the average of the adjacent vertex numbers, the simplest example is to take
$$
a=b=c=d=e=f=g=h.
$$
In fact, this is the only possibility. Without loss of generality, assume $a$ is the largest of the 8 numbers. Since $a$ is the average of $b, e, d$, $b, e, d$ must also be the largest, i.e., equal to $a$. By the same reasoning, all 8 numbers must be equal. Thus, whether it is $a+b+c+d-(e+f+g+h)$ or $2a+b+c+d-(2e+f+g+h)$, the result is clearly 0.
Therefore, equations (1), (2), etc., are just superficial, while the equality of the 8 numbers is the true essence.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: When A was B's current age, B was 10 years old; when B was A's current age, A was 25 years old. Who is older, A or B? How many years older?
|
Solution: Let the age difference between A and B be $k$ years, which is an undetermined constant. When $k>0$, A is older than B; when $k<0$, A is younger than B. Then, A's age $y$ and B's age $x$ have a linear relationship:
$$
y=x+k \text {. }
$$
After designing this dynamic process, the given conditions become 3 "moments" of this process:
(1) The current ages of B and A are $(x, y)$;
(2) "When A was B's current age, B was 10 years old", which is equivalent to taking $(10, x)$, i.e.,
$$
x=10+k \text {; }
$$
(3) "When B was A's current age, A was 25 years old", which is equivalent to taking $(y, 25)$, i.e.,
$$
25=y+k .
$$
By combining (1)+(2)+(3) and eliminating $x, y$, we get
$$
k=5 \text {. }
$$
Answer: A is 5 years older than B.
2. Solve local problems in a global context
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 If $\left(1+x+x^{2}+x^{3}\right)^{5}\left(1-x+x^{2}-\right.$ $\left.x^{3}\right)^{5}=a_{30}+a_{29} x+\cdots+a_{1} x^{29}+a_{0} x^{30}$, find $a_{15}$.
|
Let $f(x)=\left(1+x+x^{2}+x^{3}\right)^{5}$, then the original expression is
$$
F(x)=f(x) f(-x)
$$
which is an even function. Therefore, we have
$$
\begin{array}{l}
\boldsymbol{F}(x)=\frac{1}{2}[\boldsymbol{F}(x)+\boldsymbol{F}(-x)] \\
=a_{30}+a_{28} x^{2}+\cdots+a_{2} x^{28}+a_{0} x^{30},
\end{array}
$$
where all the coefficients of the odd powers of $x$ are zero, hence $a_{15}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 There are 5 medicine boxes, every 2 boxes contain one same medicine, each medicine appears in exactly 2 boxes, how many kinds of medicines are there?
|
Solution: Represent the medicine boxes as 5 points. When a medicine box contains the same medicine, draw a line segment between the corresponding points.
Since every 2 medicine boxes have one kind of the same medicine, a line should be drawn between every two points. Also, because each kind of medicine appears in exactly 2 medicine boxes, there is exactly one line between every two points. This results in Figure 2. The number of lines in Figure 2 is the number of different medicines, so there are 10 kinds of medicines.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a=1999 x+2000, b=1999 x+2001, c$ $=1999 x+2002$. Then the value of the polynomial $a^{2}+b^{2}+c^{2}-a b-b c-$ $c a$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
|
2. (D).
$$
\begin{array}{l}
\because a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] . \\
\text { Also } a-b=-1, b-c=-1, c-a=2, \\
\therefore \text { the original expression }=\frac{1}{2}\left[(-1)^{2}+(-1)^{2}+2^{2}\right]=3 .
\end{array}
$$
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
11. The number of integers $n$ that satisfy $\left(n^{2}-n-1\right)^{n+2}=1$ is.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if the task is to translate the problem statement itself, the translation would be:
11. The number of integers $n$ that satisfy $\left(n^{2}-n-1\right)^{n+2}=1$ is.
|
11.4 .
From $n+2=0, n^{2}-n-1 \neq 0$, we get $n=-2$;
From $n^{2}-n-1=1$, we get $n=-1, n=2$;
From $n^{2}-n-1=-1$ and $n+2$ is even, we get $n=0$.
Therefore, $n=-1,-2,0,2$ for a total of 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, 18 football teams are participating in a single round-robin tournament, meaning each round the 18 teams are divided into 9 groups, with each group's two teams playing one match. In the next round, the teams are regrouped to play, for a total of 17 rounds, ensuring that each team plays one match against each of the other 17 teams. After $n$ rounds of the tournament according to any feasible schedule, there always exist 4 teams that have played a total of only 1 match among themselves. Find the maximum possible value of $n$.
(Li Chengzhang, contributor)
|
```
3. Consider the following competition program:
1.(1.2)(3.4)(5.6)(7.8)(9,18)
(10,11)(12,13)(14,15)(16,17)
2.(1,3)(2.4)(5,7)(6.9)(8,17)
(10,12)(11,13)(14,16)(15.18)
3.(1.4)(2.5)(3.6)(8.9)(7.16)
(10,13)(11,14)(12,15)(17,18)
4.(1,5)(2.7)(3.8)(4,9)(6.15)
(10.14)(11,16)(12.17)(13.18)
5.(1,6)(2,8)(3,9)(4,7)(5,14)
(10,15)(11,17)(12,18)(13.16)
6.(1,7)(2.9)(3.5)(6,8)(4,13)
(10,16)(11,18)(12,14)(15,17)
7.(1.8)(2.6)(4.5)(7.9)(3.12)
(10.17)(11,15)(13,14)(16,18)
8. (1,9)(3,7)(4,6)(5,8)(2,11)
(10,18)(12,16)(13.15)(14.17)
9.(2.3)(4.8)(5.9)(6.7)(1.10)
(11,12)(13,17)(14,18)(15,16)
10.(1,11)(2.12)(3,13)(4,14)(5,15)
(6,16)(7,17)(8,18)(9,10)
11.(1.12)(2,13)(3,14)(4,15)(5,16)
(6,17)(7,18)(8,10)(9,11)
12.(1,13)(2.14)(3,15)(4,16)(5,17)
(6,18)(7.10)(8.11)(9.12)
.....
17.(1,18)(2,10)(3,11)(4,12)(5,13)
(6,14)(7,15)(8,16)(9,17)
The first 9 teams are called Group A, and the last 9 teams are called Group B. It is easy to see that after 9 rounds, any two teams in the same group have already played against each other. Therefore, any 4 teams have already played at least two matches, which of course does not meet the requirements of the problem.
If the above program is reversed and the matches are played in the new order, then after 8 rounds, any two teams in the same group have not played against each other. Each team has played 1 match against 8 of the 9 teams in the other group. At this point, any 4 teams in the same group have not played against each other, and any 4 teams not all in the same group have played at least two matches. Of course, none of these meet the requirements of the problem.
The above discussion shows that the maximum possible value of \( n \) is no more than 7.
Assume that 7 rounds of matches have been played and no 4 teams meet the requirements of the problem.
Select two teams \( A_1 \) and \( A_2 \) that have played one match. Thus, each team has played against 6 other teams. The two teams have played against at most 12 other teams. Therefore, there are at least 4 teams \( B_1, B_2, B_3, B_4 \) that have not played against \( A_1 \) and \( A_2 \). By the assumption of contradiction, \( B_1, B_2, B_3, \) and \( B_4 \) have all played against each other.
Teams \( B_1 \) and \( B_2 \) have each played 3 matches among the 4 teams \( \{B_1, B_2, B_3, B_4\} \). Thus, each team has played 1 match against 4 of the other 14 teams. This leads to at least 6 teams \( C_1, C_2, \cdots, C_6 \) that have not played against \( B_1 \) and \( B_2 \). By the assumption of contradiction, \( C_1, C_2, \cdots, C_6 \) have all played against each other.
Teams \( C_1 \) and \( C_2 \) have each played 5 matches among the 6 teams \( \{C_1, C_2, \cdots, C_6\} \). Thus, each team has played 1 match against 2 of the other 12 teams. Therefore, there are at least 8 teams \( D_1, D_2, \cdots, D_8 \) that have not played against \( C_1 \) and \( C_2 \). By the assumption of contradiction, these 8 teams have all played against each other. This way, \( D_1 \) and \( D_2 \) have not played against 10 other teams. Since only 7 rounds have been played, there must be two teams \( E_1 \) and \( E_2 \) among the 10 other teams that have not played against each other. Thus, \( D_1, D_2, E_1, \) and \( E_2 \) have played a total of 1 match, which contradicts the assumption of contradiction.
In summary, the maximum number of rounds that can be played is 7.
```
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. Given positive integers $x, y$, then $\frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5}$ has ( ) solutions of the form $(x, 1)$.
(A) 0
(B) 1
(C) 2
(D) More than 2, but finite
(E) Infinite
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
16. (C).
$$
\text { Given } \frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5} \Rightarrow 50 y=x^{2}(y+5) \Rightarrow x^{2}=\frac{50 y}{y+5} \text {. }
$$
If $(y, 5)=1$, then $(y, y+5)=1$, so $(y+5) \mid 50=5^{2} \times 2$. Since $y+5>2$, then $(y+5) \mid 5^{2}$, which contradicts $(y+5,5)=1$. Therefore, $5 \mid y$.
Let $y=5 y_{0}$, then $x^{2}=\frac{50 y_{0}}{y_{0}+1}=\frac{5^{2} \cdot 2 y_{0}}{y_{0}+1} \geqslant 5^{2}$.
In (I), from $y+5>y$ we have $x^{2}<50$.
Therefore, $x$ can only be 5, 6, or 7. Upon verification, $(x, y)$ has only two positive integer solutions: $(5,5),(7,245)$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13 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 League)
|
Solution: Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and $x_{1} < 0 < x_{2}$,
then $x_{1} < 0, \\
\therefore b > 2 \sqrt{a c} . \\
\text{Also, } \because |O A| = |x_{1}| > 1$. Therefore, the parabola opens upwards, and when $x = -1$, $y > 0$, so $a(-1)^{2} + b(-1) + c > 0$, which means $b < a + c + 1$. Since $b > 2 \sqrt{a c}$, we have $2 \sqrt{a c} + 1 < a + c$,
i.e., $(\sqrt{a} - \sqrt{c})^{2} > 1$.
From equation (2), we get $\sqrt{a} - \sqrt{c} > 1$,
i.e., $\sqrt{a} > \sqrt{c} + 1$.
Thus, $a > (\sqrt{c} + 1)^{2} \geq (\sqrt{1} + 1)^{2} = 4$.
Solving, we get $a \geq 5$.
$$
\begin{array}{l}
\text{Also, } \because b > 2 \sqrt{a c} \geq 2 \sqrt{5 \times 1} > 4, \\
\therefore b \geq 5 .
\end{array}
$$
When $a = 5, b = 5, c = 1$, the parabola $y = 5 x^{2} + 5 x + 1$ satisfies the given conditions.
Therefore, the minimum value of $a + b + c$ is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $[A]$ denote the greatest integer less than or equal to $A$, and set $A=38+$ $17 \sqrt{5}$. Then $A^{2}-A[A]=$ $\qquad$ .
|
$\begin{array}{l}\text { II.1.1. } \\ \because A=38+17 \sqrt{5}=(\sqrt{5}+2)^{3} \text {, let } B=(\sqrt{5}-2)^{3} \text {, } \\ \therefore A-B=76 . \\ \text { Also } \because 0<(\sqrt{5}-2)^{3}<1 \text {, } \\ \therefore[A]=76 \text {, then } A-[A]=B . \\ \text { Therefore } A^{2}-A[A]=A(A-[A])=A B=1 .\end{array}$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the function $y=\frac{a-x}{x-a-1}$, the graph of its inverse function is symmetric about the point $(-1,4)$. Then the value of the real number $a$ is $\qquad$ .
|
2.3.-
From the problem, we know that the graph of the function $y=\frac{a-x}{x-a-1}$ is centrally symmetric about the point $(4,-1)$.
$\because y=\frac{a-x}{x-a-1}=-1-\frac{1}{x-(a+1)}$, we have $(y+1)[x-(a+1)]=-1$,
$\therefore$ the graph of the function is a hyperbola with its center at $(a+1,-1)$.
Also, $\because$ the graph of the hyperbola is centrally symmetric about its center,
$\therefore a+1=4$, which means $a=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Find the largest constant $k$, such that for all real numbers $a, b, c, d$ in $[0,1]$, the inequality
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant k\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
holds.
|
Solution: First, estimate the upper bound of $k$.
When $a=b=c=d=1$, we have $4 k \leqslant 4+4, k \leqslant 2$.
Next, we prove that for $a, b, c, d \in[0,1]$, it always holds that
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
First, we prove a lemma.
Lemma If $x, y \in[0,1]$, then
$$
x^{2} y+1 \geqslant x^{2}+y^{2} \text {. }
$$
This inequality is equivalent to $(y-1)\left(x^{2}-y-1\right) \geqslant 0$.
Thus, inequality (2) holds. By substituting $a, b$ and $b, c$ and $c, d$ and $d, a$ for $x, y$ in inequality (1), we get
$$
\begin{array}{l}
a^{2} b+1 \geqslant a^{2}+b^{2}, b^{2} c+1 \geqslant b^{2}+c^{2}, \\
c^{2} d+1 \geqslant c^{2}+d^{2}, d^{2} a+1 \geqslant d^{2}+a^{2} .
\end{array}
$$
Adding the above four inequalities, we obtain
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
In conclusion, the maximum value of $k$ is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Find the smallest positive integer $k$, such that for all $a$ satisfying $0 \leqslant a \leqslant 1$ and all positive integers $n$, we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$
|
Solution: First, we aim to eliminate the parameter $a$, and then it will be easier to find the minimum value of $k$. Using the arithmetic-geometric mean inequality, we get
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{k+n}=\frac{k}{k+n} .
\end{array}
$$
Therefore, $a^{k}(1-a)^{n} \leqslant \frac{k^{k} n^{n}}{(n+k)^{n+k}}$.
Equality holds if and only if $a=\frac{k(1-a)}{n}$, i.e., $a=\frac{k}{n+k}$.
Thus, we need to find the smallest positive integer $k$ such that for any positive integer $n$, we have
$$
\frac{k^{k} n^{n}}{(n+k)^{n+k}}<\frac{1}{(n+1)^{3}}.
$$
When $k=1$, taking $n=1$ leads to a contradiction in equation (4);
When $k=2$, taking $n=1$ leads to a contradiction in equation (4);
When $k=3$, taking $n=3$ leads to a contradiction in equation (4).
Therefore, $k \geqslant 4$.
Next, we prove that when $k=4$, equation (4) holds, i.e.,
$$
4^{4} n^{n}(n+1)^{3}<(n+4)^{n+4}.
$$
When $n=1,2,3$, it is easy to verify that equation (5) holds.
When $n \geqslant 4$, using the arithmetic-geometric mean inequality, we get
$$
\begin{array}{l}
\sqrt[n+4]{4^{4} n^{n}(n+1)^{3}} \\
=\sqrt[n+4]{16(2 n)(2 n)(2 n)(2 n) n^{n-4}(n+1)^{3}} \\
\leqslant \frac{16+4 \times 2 n+n(n-4)+3(n+1)}{n+4} \\
=\frac{n^{2}+7 n+19}{n+4}<n+4 .
\end{array}
$$
In conclusion, the minimum value of $k$ is 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 If $xy=1$, then the minimum value of the algebraic expression $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is $\qquad$ .
(1996, Huanggang City, Hubei Province, Junior High School Mathematics Competition)
|
$$
\text { Sol: } \begin{aligned}
\because & \frac{1}{x^{4}}+\frac{1}{4 y^{4}}=\left(\frac{1}{x^{2}}\right)^{2}+\left(\frac{1}{2 y^{2}}\right)^{2} \\
& \geqslant 2 \cdot \frac{1}{x^{2}} \cdot \frac{1}{2 y^{2}}=1,
\end{aligned}
$$
$\therefore \frac{1}{x^{4}}+\frac{1}{4 y^{4}}$'s minimum value is 1.
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Person A and Person B go to a discount store to buy goods. It is known that both bought the same number of items, and the unit price of each item is only 8 yuan and 9 yuan. If the total amount spent by both on the goods is 172 yuan, then the number of items with a unit price of 9 yuan is $\qquad$ pieces.
Person A and Person B go to a discount store to buy goods,
it is known that both bought
the same number of items, and the unit price of each
item is only 8 yuan and 9 yuan. If the total amount spent by both on the goods is
172 yuan, then the number of items with a unit price of 9 yuan is $\qquad$ pieces.
|
3.12.
Suppose each person bought $n$ items, among which $x$ items cost 8 yuan each, and $y$ items cost 9 yuan each. Then we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ x + y = 2 n , } \\
{ 8 x + 9 y = 172 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=18 n-172, \\
y=172-16 n .
\end{array}\right.\right. \\
\because x \geqslant 0, y \geqslant 0, \therefore\left\{\begin{array}{l}
18 n-172 \geqslant 0, \\
172-16 n \geqslant 0 .
\end{array}\right.
\end{array}
$$
Solving, we get $9 \frac{5}{9} \leqslant n \leqslant 10 \frac{3}{4}$.
Thus, $n=10$. Therefore, $y=172-160=12$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, (15 points) 1 and 0 alternate to form the following sequence of numbers:
$$
101,10101,1010101,101010101, \cdots
$$
Please answer, how many prime numbers are there in this sequence? And please prove your conclusion.
|
Obviously, 101 is a prime number.
Below is the proof that $N=\underbrace{101010 \cdots 01}_{k \uparrow 1}(k \geqslant 3)$ are all composite numbers (with $k-1$ zeros in between).
$$
\begin{aligned}
11 N= & 11 \times \underbrace{10101 \cdots 01}_{k \uparrow 1} \\
& =\underbrace{1111 \cdots 11}_{2 k \uparrow 1}=\underbrace{11 \cdots 1}_{k \uparrow 1} \times\left(10^{k}+1\right) .
\end{aligned}
$$
(1) When $k$ is an odd number not less than 3, according to the divisibility rule for 11, 11 does not divide $\underbrace{11 \cdots 11}_{k \uparrow 1}$, so, $111\left(10^{k}+1\right)$, i.e.,
$$
\frac{10^{k}+1}{11}=M_{1}>1 \text {. }
$$
Thus, $N=\underbrace{11 \cdots 11}_{k \uparrow 1} \times\left(\frac{10^{k}+1}{11}\right)=\underbrace{11 \cdots 11}_{k \uparrow 1} \times M_{1}$. Therefore, $N$ is a composite number.
(2) When $k$ is an even number not less than 3, it is easy to see that $11 \mid \underbrace{11 \cdots 11}_{k \uparrow 1}$.
i.e., $\frac{11 \cdots 11}{11}=M_{2}>1$.
Thus, $N=\frac{\overbrace{11 \cdots 11}^{k \uparrow 1}}{11} \times\left(10^{k}+1\right)=M_{2}\left(10^{k}+1\right)$.
Therefore, $N$ is a composite number.
In summary, when $k \geqslant 3$, $N=\underbrace{10101 \cdots 01}_{k \uparrow 1}$ must be a composite number.
Therefore, in the sequence $101,10101,1010101,101010101, \cdots \cdots$, only 101 is a prime number.
(Provided by Zhou Chunluo, Capital Normal University)
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Under the conditions $x+2 y \leqslant 3, x \geqslant 0, y \geqslant 0$, the maximum value that $2 x+y$ can reach is $\qquad$
(2000, Hope Cup Junior High School Mathematics Competition Second Trial)
|
Solution: As shown in Figure 1, draw the line $x + 2y = 3$. The set of points satisfying the inequalities $x \geqslant 0, y \geqslant 0, x + 2y \leqslant 3$ is the region $\triangle ABO$ (including the boundaries) enclosed by the line and the $x$ and $y$ axes. To find the maximum value of $s = 2x + y$, we transform $s = 2x + y$ into $y = -2x + s$, whose corresponding image is a bundle of parallel lines with a slope of -2. To find the maximum value of $s$, it is converted into finding the maximum intercept of the parallel lines when they pass through $\triangle ABO$. Clearly, when the line $y = -2x + s$ passes through $A(3,0)$, the intercept $s$ is maximized, at which point $s = 6$.
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given that a square has three vertices $A, B, C$ on the parabola $y=x^{2}$. Find the minimum value of the area of such a square.
(1998, Shanghai High School Mathematics Competition)
|
Solution: As shown in Figure 3, without loss of generality, assume that two of the three vertices are on the right side of the $y$-axis (including the $y$-axis). Let the coordinates of points $A$, $B$, and $C$ be $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$, respectively, and the slope of $BC$ be $k$ $(k>0)$. Then we have
$$
\begin{aligned}
y_{3}-y_{2} & =k\left(x_{3}-x_{2}\right), \\
y_{1}-y_{2} & =-\frac{1}{k}\left(x_{1}-x_{2}\right).
\end{aligned}
$$
Since points $A$, $B$, and $C$ lie on the parabola, we have $y_{1}=x_{1}^{2}$, $y_{2}=x_{2}^{2}$, and $y_{3}=x_{3}^{2}$. Substituting these into the above equations, we get
$$
x_{3}=k-x_{2}, \quad x_{1}=-\frac{1}{k}-x_{2}.
$$
Since $|AB|=|BC|$,
$$
\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}} = \sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}},
$$
we have $\sqrt{1+\frac{1}{k^{2}}}\left(x_{2}-x_{1}\right)=\sqrt{1+k^{2}}\left(x_{3}-x_{2}\right)$. Therefore, $x_{2}-x_{1}=k\left(x_{3}-x_{2}\right)$. From equation (1), we get
$$
\frac{1}{k}+2 x_{2}=k\left(k-2 x_{2}\right).
$$
Thus, $k^{2}-\frac{1}{k}=(2 k+2) x_{2} \geqslant 0$. Solving this, we get $k \geqslant 1$, and $x_{2}=\frac{k^{3}-1}{2 k(k+1)}$. Therefore, the side length of the square is
$$
\begin{array}{l}
\sqrt{1+k^{2}}\left(x_{3}-x_{2}\right)=\sqrt{1+k^{2}}\left(k-2 x_{2}\right) \\
=\sqrt{1+k^{2}}\left[k-\frac{k^{3}-1}{k(k+1)}\right] \\
=\frac{k^{2}+1}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \geqslant \frac{2 k}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \\
\geqslant \frac{2 k}{k} \cdot \frac{\sqrt{\frac{1}{2}(k+1)^{2}}}{k+1}=\sqrt{2}.
\end{array}
$$
The equality holds if and only if $k=1$, i.e., when point $B$ is at the origin. Therefore, the minimum area of the square is 2.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. When $s$ and $t$ take all real values, the minimum value that can be reached by $(s+5-3|\cos t|)^{2}$ $+(s-2|\sin t|)^{2}$ is $\qquad$
(1989, National High School Mathematics Competition)
|
(The original expression can be regarded as the square of the distance between any point on the line $\left\{\begin{array}{l}x=s+5, \\ y=s\end{array}\right.$ and any point on the ellipse arc $\left\{\begin{array}{l}x=3|\cos t| \\ y=2|\sin t|\end{array}\right.$. It is known that the square of the distance from the point $(3,0)$ to the line is the smallest, with a value of 2.)
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given that $p$, $q$, $\frac{2p-1}{q}$, $\frac{2q-1}{p}$ are all integers, and $p>1$, $q>1$. Try to find the value of $p+q$.
|
Analysis: The conditions of this problem do not provide specific numbers. Yet, it requires finding the value of $p+q$, which poses a certain difficulty. If we analyze the given conditions one by one, we almost get no useful information. However, if we look at the conditions as a whole, i.e., consider $\frac{2 p-1}{q}$ and $\frac{2 q-1}{p}$ together, we will find that at least one of $\frac{2 p-1}{q}$ and $\frac{2 q-1}{p}$ is less than 2. Otherwise, if $\frac{2 p-1}{q} \geqslant 2$ and $\frac{2 q-1}{p} \geqslant 2$ were both true, then we would have
$2 p-1 \geqslant 2 q, 2 q-1 \geqslant 2 p$.
Adding these two inequalities gives
$$
2 p+2 q-2 \geqslant 2 p+2 q \text {. }
$$
which simplifies to
$$
-2 \geqslant 0 \text{. }
$$
This is clearly impossible.
By observing the given conditions as a whole, we have found a way to solve the problem.
Since at least one of $\frac{2 p-1}{q}$ and $\frac{2 q-1}{p}$ is less than 2, let's assume $\frac{2 p-1}{q} < 2$. From $\frac{2 p-1}{q} < 2$, we get $2 p - 1 < 2 q$, or $2 p - 2 q < 1$. Since $p$ and $q$ are integers, the only possible solution is $2 p - 2 q = 0$, which implies $p = q$. However, this contradicts the condition that $\frac{2 p-1}{q} < 2$. Therefore, we need to consider the next possible integer values. From $\frac{2 p-1}{q} < 2$, we can try $p = 3$ and $q = 5$. Substituting these values, we get $\frac{2 \cdot 3 - 1}{5} = \frac{5}{5} = 1 < 2$, which satisfies the condition. Thus, $p = 3$ and $q = 5$. Therefore, $p+q=8$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.4 In Greek mythology, the "many-headed serpent" god is composed of some heads and necks, with each neck connecting two heads. With each strike of a sword, one can sever all the necks connected to a certain head $A$. However, head $A$ immediately grows new necks connecting to all the heads it was not previously connected to (each head is connected by only one neck). Only by severing the "many-headed serpent" into two mutually disconnected parts can it be defeated. Try to find the smallest natural number $N$ such that for any "many-headed serpent" god with 100 necks, it can be defeated with no more than $N$ strikes.
|
9.410 .
We will reformulate the problem using graph theory terminology, with heads as vertices, necks as edges, and a strike that cuts the necks connected to head $A$ as a "reversal" of vertex $A$. It is easy to see that if a vertex $X$ has a degree no greater than 10, then it is sufficient to perform a "reversal" on all vertices adjacent to $X$ to make vertex $X$ "isolated". If a vertex $X$ is not adjacent to at most $n (n \leqslant 9)$ vertices, then it is sufficient to first perform a "reversal" on $X$, and then perform a "reversal" on each of these $n$ vertices to make vertex $X$ "isolated".
If each vertex has at least 11 adjacent vertices and at least 10 non-adjacent vertices, then there must be at least 22 vertices. Thus, the number of edges (necks) is no less than $22 \times 11 > 100$, which is not within the scope of consideration.
We provide an example to show that 9 strikes may not be sufficient.
Assume there are two groups of 10 heads each, with each head in one group connected to each head in the other group, resulting in exactly 100 necks.
If 9 strikes are made, then there is at least one head in each group that is not struck, denoted as $A$ and $B$. For the remaining 18 heads, by the problem's conditions, any head $C$ is connected to exactly one of $A$ or $B$ at any time (during the 9 strikes), and not to the other. Since $A$ and $B$ are connected, the "many-headed snake" remains connected.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of intersections of the function $y=x \cdot|x|-\left(4 \cos 30^{\circ}\right) x+2$ with the $x$-axis is $\qquad$
|
3.3 .
$$
y=\left\{\begin{array}{ll}
x^{2}-2 \sqrt{3} x+2, & x>0, \\
2, & x=0 . \\
-x^{2}-2 \sqrt{3} x+2, & x<0
\end{array}\right.
$$
When $x>0$, $y=x^{2}-2 \sqrt{3} x+2$ intersects the x-axis at 2 points;
When $x=0$, $y=2$ does not intersect the x-axis;
When $x<0$, $y=-x^{2}-2 \sqrt{3} x+2$ intersects the x-axis at 1 point.
Therefore, the graph of this function intersects the x-axis at 3 points.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there are always 3 numbers, where the sum of any two of them is still irrational.
|
Three, take 4 irrational numbers $\{\sqrt{2}, \sqrt{3},-\sqrt{2},-\sqrt{3}\}$, clearly they do not satisfy the condition, hence $n \geqslant 5$.
Consider 5 irrational numbers $a, b, c, d, e$. View them as 5 points. If the sum of two numbers is a rational number, then connect the corresponding two points with a red line; otherwise, connect them with a blue line.
(1) There is no red triangle. Otherwise, assume without loss of generality that $a+b, b+c, c+a$ are all rational numbers. Since $(a+b)+(c+a)-(b+c)=2a$, this contradicts the fact that $a$ is irrational.
(2) There must be a monochromatic triangle. Otherwise, the graph must contain a red cycle, with the 5 numbers at the vertices, where the sum of any two is a rational number. Suppose $a+b, b+c, c+d, d+e, e+a$ are rational numbers, then by $(a+b)-(b+c)+(c+d)-(d+e)+(e+a)=2a$ we derive a contradiction. Thus, the monochromatic triangle must be a blue triangle. Therefore, the minimum value of $n$ is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Determine the smallest natural number $k$, such that for any $a \in [0,1]$ and any $n \in \mathbf{N}$ we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$
|
Solution: By the arithmetic-geometric mean inequality, we have
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{n+k}=\frac{k}{n+k} .
\end{array}
$$
Thus, $a^{k}(1-a)^{n} \leqslant \frac{k^{k} n^{n}}{(n+k)^{n+k}}$.
Equality holds if and only if $a=\frac{k(1-a)}{n}\left(\right.$ i.e., $\left.a=\frac{k}{n+k}\right)$.
This means that when $a=\frac{k}{n+k}(\in[0,1])$, $a^{k}(1-a)^{n}$ attains its maximum value. Therefore, the problem can be reformulated as: determine the smallest natural number $k$ such that for any $n \in \mathbf{N}$,
$$
\frac{k^{k} n^{n}}{(n+k)^{n+k}}<\frac{1}{(n+1)^{3}} .
$$
It is easy to verify that for the pairs $(k, n)=(1,1),(2,1)$, $(3,3)$, inequality (1) does not hold. Therefore, for it to hold for all $n \in \mathbf{N}$, we must have $k \geqslant 4$.
Next, we prove that when $k=4$, inequality (1) holds, i.e., for all $n \in \mathbf{N}$, we have $4^{4} n^{n}(n+1)^{3}<(n+4)^{n+4}$.
Indeed, for $n=1,2,3$, this can be directly verified.
For $n \geqslant 4$, we have
$$
\begin{array}{l}
\sqrt[n+1]{4^{4} n^{n}(n+1)^{3}} \\
=\sqrt[n+4]{16 \times(2 n)(2 n)(2 n)(2 n) n^{n-4}(n+1)^{3}} \\
\leqslant \frac{16+8 n+(n-4) n+3(n+1)}{n+4} \\
=\frac{n^{2}+7 n+19}{n+4}<\frac{n^{2}+8 n+16}{n+4}=n+4,
\end{array}
$$
i.e., $4^{4} n^{n}(n+1)^{3}<(n+4)^{n+4}$.
In summary, the smallest value of $k$ is 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given that the three altitudes of $\triangle A B C$ are $A D=3, B E=4, C F=5$, and the lengths of the three sides of this triangle are all integers. Then the minimum value of the length of the shortest side is ( ).
(A) 10
(B) 12
(C) 14
(D) 16
|
5. (B).
From $S_{\triangle A B C}=\frac{1}{2} B C \cdot A D=\frac{1}{2} C A \cdot B E=\frac{1}{2} A B \cdot C F$,
we get $3 B C=4 C A=5 A B$.
It is clear that $A B$ is the shortest side.
From $B C=\frac{5}{3} A B, C A=\frac{5}{4} A B$ and the lengths of $B C, C A, A B$ are all
integers, we know that $3 \mid A B$ and $4 \mid A B$.
$$
\because(3,4)=1, \therefore 12 \mid A B \text {. }
$$
Therefore, the minimum possible value of the shortest side $A B$ is 12.
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) Let $a$, $b$, and $c$ be three distinct real numbers, and $c \neq 1$. It is known that the equations $x^{2} + a x + 1 = 0$ and $x^{2} + b x + c = 0$ have a common root, and the equations $x^{2} + x + a = 0$ and $x^{2} + c x + b = 0$ also have a common root. Find the value of $a + b + c$.
|
Let the common root of the first two equations be $x_{1}$, then
$$
\begin{array}{l}
x_{1}^{2}+a x_{1}+1=0, \\
x_{1}^{2}+b x_{1}+c=0 .
\end{array}
$$
(2)
(1) - (2) gives $(a-b) x_{1}+(1-c)=0$.
$$
\because a \neq b, \quad \therefore x_{1}=\frac{c-1}{a-b} \text {. }
$$
Similarly, the common root of the last two equations is $x_{2}=\frac{a-b}{c-1}$.
$$
\therefore x_{2}=\frac{1}{x_{1}} \text {. }
$$
By Vieta's formulas, the other root of the equation $x^{2}+a x+1=0$ is $\frac{1}{x_{1}}$. Therefore, $x_{2}$ is the common root of the equations $x^{2}+a x+1=0$ and $x^{2}+x+a=0$. Thus, we have
$$
x_{2}^{2}+a x_{2}+1=0, \quad x_{2}^{2}+x_{2}+a=0 .
$$
Subtracting these gives $(a-1)\left(x_{2}-1\right)=0$.
When $a=1$, these two equations have no real roots, so $a \neq 1$. Hence, we have $x_{2}=1$, and thus $x_{1}=1$. Therefore, $a=-2, b+c=-1$.
$$
\therefore a+b+c=-3 .
$$
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. For any natural numbers $m, n$ satisfying $\frac{m}{n}<\sqrt{7}$, the inequality $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\lambda}{n^{2}}$ always holds. Find the maximum value of $\lambda$.
|
(Let $G=|(m, n)| m<\sqrt{7} n, m, n \in \mathbf{N}$. $\lambda_{\text {max }}=\min _{(m, n \in 6} 17 n^{2}-m^{2}$, then perform $\bmod 7$ analysis on $7 n^{2}-m^{2}$, obtaining $\lambda_{\text {max }}=3$. )
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. For the polynomial $\left(\sqrt{x}+\frac{1}{2 \sqrt[4]{x}}\right)^{n}$ expanded in descending powers of $x$, if the coefficients of the first three terms form an arithmetic sequence, then the number of terms in the expansion where the exponent of $x$ is an integer is $\qquad$ .
|
8.3.
It is easy to find that the coefficients of the first three terms are $1, \frac{1}{2} n, \frac{1}{8} n(n-1)$. Since these three numbers form an arithmetic sequence, we have $2 \times \frac{1}{2} n=1+\frac{1}{8} n(n-1)$. Solving this, we get $n=8$ and $n=1$ (the latter is discarded).
When $n=8$, $T_{r+1}=\mathrm{C}_{8}^{r}\left(\frac{1}{2}\right)^{r} x^{\frac{(16-3 r)}{4}}$, where $r=0$, $1, \cdots, 8$. $r$ must satisfy $4 \mid(16-3 r)$, so $r$ can only be $0,4,8$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given that $f(x)$ is a function defined on
$\mathbf{R}$, $f(1)=1$ and for any $x \in \mathbf{R}$ we have
$$
f(x+5) \geqslant f(x)+5, f(x+1) \leqslant f(x)+1 \text {. }
$$
If $g(x)=f(x)+1-x$, then $g(2002)=$
|
10.1.
$$
\begin{array}{l}
\text { From } g(x)=f(x)+1-x \text { we get } f(x)=g(x)+x-1 . \\
\text { Then } g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5 \text {, } \\
g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1 . \\
\text { Therefore, } g(x+5) \geqslant g(x), g(x+1) \leqslant g(x) . \\
\therefore g(x) \leqslant g(x+5) \leqslant g(x+4) \leqslant g(x+3) \\
\quad \leqslant g(x+2) \leqslant g(x+1) \leqslant g(x) . \\
\therefore g(x+1)=g(x) .
\end{array}
$$
Thus, $g(x)$ is a periodic function with a period of 1.
Given $g(1)=1$, hence $g(2002)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 In the school football championship, it is required that each team must play a match against all the other teams. Each winning team gets 2 points, a draw gives each team 1 point, and a losing team gets 0 points. It is known that one team scored the most points, but it played fewer matches than any other team. How many teams participated at least?
|
Explanation: We call the team $A$ with the highest score the winning team.
Suppose team $A$ wins $n$ matches and draws $m$ matches, then the total score of team $A$ is $2n + m$ points.
From the given conditions, every other team must play at least $n+1$ matches, meaning they score no less than $2(n+1)$ points. Therefore,
$2n + m > 2(n+1)$. This implies $m \geqslant 3$.
Thus, there must be a team that draws with the winning team $A$, and this team's score should be no less than $2(n+1) + 1$ points, i.e.,
$$
2n + m > 2(n+1) + 1, \quad m \geqslant 4.
$$
Let there be $x$ teams in total, then the winner must win at least one match. Otherwise, its score would not exceed $x-1$ points. In this case, any other team's score would be strictly less than $x-1$ points, and the total score of all participating teams would be less than $x(x-1)$ points, but the total score of $x$ participating teams is $x(x-1)$ points, leading to a contradiction.
Thus, $m \geqslant 4, n \geqslant 1$, meaning the winning team $A$ must play at least 5 matches, and therefore, there must be at least 6 teams participating.
From this, we can estimate: at least 6 teams are participating in the competition.
We can construct a match points table: Table 1 shows 6 teams (let's call them $A, B, C, D, E, F$) participating, with the winning team $A$ having the highest score and the fewest draws.
Table 1 Match Points Table
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & Score \\
\hline 4 & lle & 1 & 1 & 1 & 1 & 2 & 6 \\
\hline$B$ & 1 & lle & 2 & 0 & 0 & 2 & 5 \\
\hline i & 1 & 0 & Illo & 0 & 2 & 2 & 5 \\
\hline$D$ & 1 & 2 & 2 & & 0 & 0 & 5 \\
\hline$E$ & 1 & 2 & 0 & 2 & & 0 & 5 \\
\hline $\boldsymbol{r}$ & 0 & 0 & 0 & 2 & 2 & 16 & 4 \\
\hline
\end{tabular}
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}$, $a \neq 0$ ) satisfy the following conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$;
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.
Find the largest $m(m>1)$. Such that there exists $t \in \mathbf{R}$, for any $x \in[1, m]$, we have $f(x+t) \leqslant x$.
|
15. Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about $x=-1$. Therefore, $-\frac{b}{2a}=-1, b=2a$.
From (3), when $x=-1$, $y=0$, i.e., $a-b+c=0$.
From (1), $f(1) \geqslant 1$, and from (2), $f(1) \leqslant 1$,
thus $f(1)=1$, i.e., $a+b+c=1$.
Also, $a-b+c=0$, so $b=\frac{1}{2}, a=\frac{1}{4}, c=\frac{1}{4}$.
Hence, $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x+\frac{1}{4}$.
Assume there exists $t \in \mathbf{R}$, such that for any $x \in[1, m]$, $f(x+t) \leqslant x$.
Taking $x=1$, we have $f(t+1) \leqslant 1$, i.e.,
$\frac{1}{4}(t+1)^{2}+\frac{1}{2}(t+1)+\frac{1}{4} \leqslant 1$. Solving this, we get $-4 \leqslant t \leqslant 0$.
For a fixed $t \in[-4,0]$, taking $x=m$, we have $f(t+m) \leqslant m$, i.e., $\frac{1}{4}(t+m)^{2}+\frac{1}{2}(t+m)+\frac{1}{4} \leqslant m$. Simplifying, we get $m^{2}-2(1-t) m+\left(t^{2}+2 t+1\right) \leqslant 0$.
Solving this, we get $1-t-\sqrt{-4 t} \leqslant m \leqslant 1-t+\sqrt{-4 t}$.
Thus, $m \leqslant 1-t+\sqrt{-4 t} \leqslant 1-(-4)+\sqrt{-4(-4)}=9$.
When $t=-4$, for any $x \in[1,9]$, we always have
$$
\begin{array}{l}
f(x-4)-x=\frac{1}{4}\left(x^{2}-10 x+9\right) \\
=\frac{1}{4}(x-1)(x-9) \leqslant 0 .
\end{array}
$$
Therefore, the maximum value of $m$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $A B C D$ be a rectangle with an area of 2, and let $P$ be a point on side $C D$. Let $Q$ be the point where the incircle of $\triangle P A B$ touches side $A B$. The product $P A \cdot P B$ varies with the changes in rectangle $A B C D$ and point $P$. When $P A \cdot P B$ is minimized,
(1) Prove: $A B \geqslant 2 B C$;
(2) Find the value of $A Q \cdot B Q$.
(Ruo Zengru)
|
Thus, $\frac{1}{2} P A \cdot P B \sin \angle A P B=1$,
which means $P A \cdot P B=\frac{2}{\sin \angle A P B} \geqslant 2$.
Equality holds only when $\angle A P B=90^{\circ}$. This indicates that point $P$ lies on the circle with $A B$ as its diameter, and this circle should intersect with $C D$,
Therefore, when $P A \cdot P B$ takes its minimum value, we should have $B C \leqslant \frac{A B}{2}$, i.e., $A B \geqslant 2 B C$.
(2) Let the inradius of $\triangle A P B$ be $r$, then
$$
\begin{array}{l}
P A \cdot P B=(r+A Q)(r+B Q) \\
=r(r+A Q+B Q)+A Q \cdot B Q .
\end{array}
$$
And $P A \cdot P B=2 S_{\triangle A P B}, r(r+A Q+B Q)=S_{\triangle A P B}$,
Thus, $A Q \cdot B Q=S_{\triangle A P B}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. Given $a+b+c=0$, and $a, b, c$ are all non-zero. Then simplify $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)$ to
|
16. -3 .
$$
\begin{aligned}
\text { Original expression }= & \left(\frac{a}{a}+\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{b}+\frac{c}{b}\right) \\
& +\left(\frac{a}{c}+\frac{b}{c}+\frac{c}{c}\right)-\frac{a}{a}-\frac{b}{b}-\frac{c}{c} \\
= & \frac{0}{a}+\frac{0}{b}+\frac{0}{c}-3=-3 .
\end{aligned}
$$
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Question: How many real roots does the equation $x^{2}|x|-5 x|x|+2 x=0$ have (where $|x|$ represents the absolute value of $x$)?
|
1.4 .
The original equation simplifies to $x(x|x|-5|x|+2)=0$. After removing the absolute value signs and discussing, we get
$$
x_{1}=0, x_{2}=\frac{5+\sqrt{17}}{2}, x_{3}=\frac{5-\sqrt{17}}{2}, x_{4}=\frac{5-\sqrt{33}}{2}
$$
as the four real roots of the original equation.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $x_{1}=\sqrt[3]{3}, x_{2}=\left(x_{1}\right)^{\sqrt[3]{3}}$, for $n>1$ define $x_{n+1}$ $=\left(x_{n}\right)^{\sqrt[3]{3}}$. Find the smallest positive integer $n$ such that $x_{n}=27$.
|
4.7.
It is known that $x_{n}=x_{1}^{x_{1}^{n-1}}(n>1)$, and $27=(\sqrt[3]{3})^{(\sqrt[3]{3})^{6}}$. Therefore, $n-1=6$. Hence, $n=7$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $\triangle A B C$ be an acute triangle, and construct isosceles triangles $\triangle D A C$, $\triangle E A B$, and $\triangle F B C$ outside $\triangle A B C$ such that $D A = D C$, $E A = E B$, and $F B = F C$, with $\angle A D C = 2 \angle B A C$, $\angle B E A = 2 \angle A B C$, and $\angle C F B = 2 \angle A C B$. Let $D^{\prime}$ be the intersection of line $D B$ and $E F$, $E^{\prime}$ be the intersection of $E C$ and $D F$, and $F^{\prime}$ be the intersection of $F A$ and $D E$. Find the value of $\frac{D B}{D D^{\prime}} + \frac{E C}{E E^{\prime}} + \frac{F A}{F F^{\prime}}$.
|
Solution: Since $\angle ABC$ is an acute triangle, then $\angle ADC$, $\angle BEA$, $\angle CFB < \pi$. Therefore,
$$
\begin{array}{l}
\angle DAC = \frac{\pi}{2} - \frac{1}{2} \angle ADC = \frac{\pi}{2} - \angle BAC. \\
\angle BAE = \frac{\pi}{2} - \frac{1}{2} \angle BEA = \frac{\pi}{2} - \angle ABC. \\
\text{Thus, } \angle DAE = \angle DAC + \angle BAC + \angle BAE \\
= \pi - \angle ABC < \pi. \\
\end{array}
$$
Similarly, $\angle EBF < \pi$, $\angle FCD < \pi$.
Therefore, the hexagon $DAEBCF$ is a convex hexagon, and
$$
\begin{array}{l}
\angle ADC + \angle BEA + \angle CFB \\
= 2(\angle BAC + \angle ABC + \angle ACB) = 2\pi.
\end{array}
$$
Let $w_1$, $w_2$, $w_3$ be the circles with centers at $D$, $E$, $F$ and radii $DA$, $EB$, $FC$ respectively. Since $\angle ADC + \angle BEA + \angle CFB = 2\pi$, these three circles intersect at a common point. Let this point be $O$, as shown in Figure 7. Then $O$ is the reflection of $C$ over $DF$.
Similarly, $O$ is also the reflection of $A$ over $DE$ and $B$ over $EF$. Therefore,
$$
\begin{array}{l}
\frac{DB}{DD'} = \frac{DD' + D'B}{DD'} = 1 + \frac{D'B}{DD'} \\
= 1 + \frac{S_{\triangle AKB}}{S_{\triangle FDF}} = 1 + \frac{S_{\triangle EOF}}{S_{\triangle DHF}}. \\
\end{array}
$$
Similarly,
$$
\begin{array}{l}
\frac{EC}{EE'} = 1 + \frac{S_{\triangle DFO}}{S_{\triangle DEF}} \cdot \frac{FA}{FF'} = 1 + \frac{S_{\triangle LDE}}{S_{\triangle UEF}}. \\
\text{Thus, } \frac{DB}{DD'} + \frac{EC}{EE'} + \frac{FA}{FF'} \\
= 3 + \frac{S_{\triangle OEF} + S_{\triangle ODF} + S_{\triangle UOF}}{S_{\triangle OHF}} = 4. \\
\end{array}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a_{1}=11^{11}, a_{2}=12^{12}, a_{3}=13^{13}$, and
$$
a_{n}=\left|a_{n-1}-a_{n-2}\right|+\left|a_{n-2}-a_{n-3}\right|, n \geqslant 4 \text {. }
$$
Find $a_{4^{4}}$.
|
For $n \geqslant 2$, define $s_{n}=\left|a_{n}-a_{n-1}\right|$. Then for $n \geqslant 5, a_{n}=s_{n-1}+s_{n-2}, a_{n-1}=s_{n-2}+s_{n-3}$. Thus, $s_{n}=$ $\left|s_{n-1}-s_{n-3}\right|$. Since $s_{n} \geqslant 0$, if $\max \left\{s_{n}, s_{n+1}, s_{n+2}\right\} \leqslant T$, then for all $m \geqslant n$, we have $s_{m} \leqslant T$. In particular, the sequence $\left\{s_{n}\right\}$ is bounded. We will prove the following proposition:
If for some $i, \max \left|s_{i}, s_{i+1}, s_{i+2}\right|=T \geqslant 2$, then $\max \left\{s_{1+6}, s_{1+7}, s_{1+8}\right\} \leqslant T-1$.
Proof by contradiction. If the above conclusion does not hold, then for $j=i, i+1, \cdots, i+6$, we have $\max \left\{s_{j}, s_{j+1}, s_{j+2}\right\}=T \geqslant 2$. For $s_{t}, s_{t+1}$, or $s_{t+2}=T$, take $j=i, i+1$, or $i+2$ accordingly, so the sequence $s_{j}, s_{j+1}, s_{j+2}, \cdots$ has the form $T, x, y, T-y, \cdots$, where $0 \leqslant x, y \leqslant T, \max \{x, y, T-y\}=T$. Therefore, either $x=T$, or $y=T$, or $y=0$.
(1) If $x=T$, then the sequence has the form $T, T, y, T-y, y, \cdots$. Thus $\max \mid y, T-y, y\}=T$. So $y=T$ or $y=0$.
(2) If $y=T$, then the sequence has the form $T, x, T, 0, x, T-x, \cdots$. Since $\max \mid 0, x, T-x=T$, so $x=0$ or $x=T$.
(3) If $y=0$, then the sequence has the form $T, x, 0, T, T-x, T-x, x, \cdots$. Since $\max \{T-x, T-x, x\}=T$, so $x=0$ or $x=T$.
In each of the above cases, $x$ and $y$ are either 0 or $T$. In particular, $T$ must divide every term in $s_{j}, s_{j+1}$, and $s_{j+2}$. Since $\max \left\{s_{2}, s_{3}, s_{4}\right\}=s_{3}=T$, for $n \geqslant 4$, $T$ divides $s_{n}$. However, when $s_{4}=11^{11}6(N-1)+4$, $a_{n}=s_{n-1}+s_{n-2}=0,1$ or 2. In particular, $a_{M}=0,1$ or 2. Since $M$ is a multiple of 7, $a_{M} \equiv a_{7} \equiv 1(\bmod 2)$, hence $a_{M}=1$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 The three edge lengths of an isosceles tetrahedron are 3, $\sqrt{10}$, and $\sqrt{13}$. Then the radius of the circumscribed sphere of this tetrahedron is
|
Analysis: According to Basic Conclusion 10, the triangular pyramid can be expanded into a rectangular parallelepiped, making the known three edges the diagonals of the faces of the rectangular parallelepiped. At this point, the original triangular pyramid and the rectangular parallelepiped have the same circumscribed sphere. Let the three edges at the same vertex of the rectangular parallelepiped be $x$, $y$, and $z$, respectively. Then, from the conditions, we have $y^{2}+z^{2}=9$, $z^{2}+x^{2}=10$, and $x^{2}+y^{2}=13$. According to this, the radius of the required circumscribed sphere is
$$
\frac{1}{2} \sqrt{x^{2}+y^{2}+z^{2}}=2 \text {. }
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, try to find all positive integers $k$, such that for any positive numbers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a, b, c$.
|
Three, due to $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0$,
thus $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
It can be known that $k>5$. Noting that $k$ is a positive integer, therefore, $k \geqslant 6$.
Since there does not exist a triangle with side lengths $1,1,2$, according to the problem, we have
$$
k(1 \times 1+1 \times 2+1 \times 2) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right),
$$
which means $k \leqslant 6$.
The following proves that $k=6$ satisfies the problem's requirements.
Assume without loss of generality that $a \leqslant b \leqslant c$, then
$$
6(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right),
$$
that is
$$
\begin{array}{l}
5 c^{2}-6(a+b) c+5 a^{2}+5 b^{2}-6 a b5\left(a^{2}+b^{2}+c^{2}\right)
\end{array}
$$
a contradiction. Hence $c<a+b$.
Method two: Construct a function
$$
f(x)=5 x^{2}-6(a+b) x+5 a^{2}+5 b^{2}-6 a b \text {. }
$$
Then $f(c)<0$.
Since $f(x)$ is increasing in the interval $\left[\frac{3}{5}(a+b),+\infty\right)$, and
$$
\begin{aligned}
f(a+b)= & 5(a+b)^{2}-6(a+b)(a+b)+5 a^{2}+ \\
& 5 b^{2}-6 a b=4(a-b)^{2} \geqslant 0,
\end{aligned}
$$
thus $c<a+b$.
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Lift Your Veil
A 101-digit natural number $A=\underbrace{88 \cdots 8}_{\text {S0 digits }} \square \underbrace{99 \cdots 9}_{\text {S0 digits }}$ is divisible by 7. What is the digit covered by $\square$?
|
Explanation: “ $\square$ ” is a veil, covering the number to be found. To lift the veil, whether approaching from the front or the back, there are 50 digits in between, which is too many. The key is to find a way to reduce the number of digits.
The difference between two multiples of 7 is still a multiple of 7.
We already know that $A$ is a multiple of 7.
Find another multiple of 7, $B$, such that all its digits are 1. Through trial division, we find
$$
B=111111=7 \times 15873 .
$$
From this, we see that $8 B=888888$ and $9 B=999999$ are also multiples of 7.
Removing 888888 from the front of the number $A$ is equivalent to subtracting $88888800 \cdots 0$ (95 zeros) from $A$. Therefore, the resulting number is still a multiple of 7.
Continuing this process, we can remove 6 eights from the front each time, until a total of 48 eights have been removed.
Similarly, we can remove 6 nines from the back of $A$ each time, until a total of 48 nines have been removed.
Finally, a 5-digit number 88 $\square$ 99 remains, which is a multiple of 7.
The problem is simplified to finding a 5-digit number $C$, which starts with 88 and ends with 99, and is a multiple of 7. The result is
$$
\begin{aligned}
C & =(49+350)+(84000+4200) \\
& =88599 .
\end{aligned}
$$
Therefore, the digit covered by $\square$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure $1, \angle A O B=30^{\circ}$, within $\angle A O B$ there is a fixed point $P$, and $O P$ $=10$, on $O A$ there is a point $Q$, and on $O B$ there is a fixed point $R$. To make the perimeter of $\_P Q R$ the smallest, the minimum perimeter is
|
3.10 .
As shown in Figure 5, construct the symmetric points $P_{1}, P_{2}$ of $P$ with respect to the sides $O A, O B$ of $\angle A O B$. Connect $P_{1} P_{2}$, intersecting $O A, O B$ at $Q, R$. Connect $P Q, P R$. It is easy to see that
$$
P Q=P_{1} Q, P R=P_{2} R,
$$
which implies
$$
\begin{aligned}
l_{\triangle P Q N} & =P_{1} Q+Q R+P_{2} R \\
& =P_{1} P_{2} .
\end{aligned}
$$
Thus, $\triangle P Q R$ is the triangle with the minimum perimeter.
Given that $O P=O P_{1}=O P_{2}=10$,
$$
\angle P O Q=\angle P_{1} O Q, \angle P O R=\angle P_{2} O R,
$$
we have
$$
\begin{aligned}
\angle P_{1} O P_{2} & =2 \angle P O Q+2 \angle P O R \\
& =2(\angle P O Q+\angle P O R)=2 \angle A O B=60^{\circ} .
\end{aligned}
$$
Therefore, $\triangle P_{1} O P_{2}$ is an equilateral triangle,
Thus, $P_{1} P_{2}=P_{1} O=10$.
Hence, the minimum perimeter of $\triangle P Q R$ is 10.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Given theorem: “For any prime number $n$ greater than 3, $b$ and $c$ satisfy the equation $2a + 5b = c$. Then $a + b + c$ is how many times the integer $n$? Prove your conclusion.”
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Three, the maximum possible value of $n$ is 9.
First, we prove that $a+b+c$ is divisible by 3.
In fact, $a+b+c=a+b+2a+5b=3(a+2b)$.
Thus, $a+b+c$ is a multiple of 3.
Let the remainders when $a$ and $b$ are divided by 3 be $r_{a}$ and $r_{b}$, respectively, with $r_{a} \neq 0$ and $r_{b} \neq 0$.
If $r_{a} \neq r_{b}$, then $r_{a}=1, r_{b}=2$ or $r_{a}=2, r_{b}=1$. In this case, $2a+5b$ must be a multiple of 3, meaning $c$ is a composite number. This is a contradiction.
Therefore, $r_{a}=r_{b}$, so $r_{a}=r_{b}=1$ or $r_{a}=r_{b}=2$. In this case, $a+2b$ must be a multiple of 3, and thus $a+b+c$ is a multiple of 9.
Next, we prove that 9 is the largest.
Since $2 \times 11 + 5 \times 5 = 47$ and $11 + 5 + 47 = 63$,
and $2 \times 13 + 5 \times 7 = 61$ and $13 + 7 + 61 = 81$,
and $(63, 81) = 9$,
therefore, 9 is the maximum possible value.
|
9
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Multiplicative Magic Square
Figure 1 shows a partially filled magic square. Fill in the following nine numbers: $\frac{1}{4}, \frac{1}{2}, 1,2,4,8,16,32,64$ in the grid so that the product of the numbers in each row, column, and diagonal is the same. The number that should be filled in the “ $x$ ” cell is $\qquad$ .
|
Explanation: Label the unfilled cells with letters, as shown in Figure 2.
The product of the nine known numbers is
$$
\frac{1}{4} \times \frac{1}{2} \times 8 \times 1 \times 2 \times 32 \times 4 \times 16 \times 64=64^{3} \text {. }
$$
Therefore, the fixed product of the three numbers in each row, each column, and each diagonal is 64.
Since the fixed product is 64, we get from the second column and the third row respectively
$$
a c=1, e f=1 .
$$
Thus, $a, c, e, f$ are each one of the numbers $\frac{1}{4}, \frac{1}{2}, 2, 4$.
Considering the product of the first row, we get $a x=2$.
This means $x$ can only be 1 or 8.
Considering the product of the diagonal, we get $64=c e x$.
If $x=1$, then $c e=64$, which is impossible.
The only possibility is $x=8$. The filling method is shown in Figure 3.
The answer to this problem is: The number to be filled in the “$x$” cell should be 8.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. $x, y$ are real numbers, and $\left(x+\sqrt{x^{2}+1}\right)(y+$ $\left.\sqrt{y^{2}+1}\right)=1$. Then $x+y=$ $\qquad$ .
|
二、1.0.
Multiply both sides of the original equation by $\sqrt{x^{2}+1}-x$, we get $y+\sqrt{y^{2}+1}=\sqrt{x^{2}+1}-x$; multiply both sides of the original equation by $\sqrt{y^{2}+1}-y$, we get $x+\sqrt{x^{2}+1}=\sqrt{y^{2}+1}-y$. Adding the two equations immediately yields $x+y=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If $\left|\log \frac{a}{\pi}\right|<2$, then the number of $a$ for which the function $y=\sin (x+a)+\cos (x-a)(x \in \mathbf{R})$ is an even function is $\qquad$.
|
2.10.
$$
\begin{aligned}
\because-2 & <\log \frac{\alpha}{\pi}<2, \therefore \frac{1}{\pi}<\alpha<\pi^{3} . \\
f(x)= & \sin (x+\alpha)+\cos (x+\alpha) \\
& =\sqrt{2} \sin \left(x+\alpha+\frac{\pi}{4}\right) .
\end{aligned}
$$
Since $f(x)$ is an even function,
$$
\therefore \sin \left(\frac{\pi}{4}+\alpha+x\right)=\sin \left(-x+\alpha+\frac{\pi}{4}\right) \text {, }
$$
i.e., $\alpha=k \pi+\frac{\pi}{4}, k \in \mathbf{Z}$. Therefore, $\frac{1}{\pi}<k \pi+\frac{\pi}{4}<\pi^{3}$. Thus, $\frac{1}{\pi^{2}}-\frac{1}{4}<k<\pi^{2}-\frac{1}{4}$,
Hence $0 \leqslant k \leqslant 9, k \in \mathbf{Z}$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Total 20 points) The sequence $\left\{x_{n}\right\}$ satisfies
$$
\begin{array}{l}
x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, n \in \mathbf{N}, y_{n}=\frac{1}{1+x_{n}}, \\
S_{n}=y_{1}+y_{2}+\cdots+y_{n}, P_{n}=y_{1} y_{2} \cdots y_{n} .
\end{array}
$$
Find $P_{n}+\frac{1}{2} S_{n}$.
|
$$
\begin{array}{l}
\text { Three, } \because x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, \\
\therefore x_{n+1}>x_{n}>0, x_{n+1}=x_{n}\left(1+x_{n}\right) . \\
\therefore y_{n}=\frac{1}{1+x_{n}}=\frac{x_{n}^{2}}{x_{n} x_{n+1}}=\frac{x_{n+1}-x_{n}}{x_{n} x_{n+1}}=\frac{1}{x_{n}}-\frac{1}{x_{n+1}} . \\
\therefore P_{n}=y_{1} y_{2} \cdots y_{n}=\frac{x_{1}}{x_{2}} \cdot \frac{x_{2}}{x_{3}} \cdots \cdots \frac{x_{n}}{x_{n+1}}=\frac{1}{2 x_{n+1}}, \\
S_{n}=\frac{1}{x_{1}}-\frac{1}{x_{2}}+\frac{1}{x_{2}}-\frac{1}{x_{3}}+\cdots+\frac{1}{x_{n}}-\frac{1}{x_{n+1}} \\
\quad=2-\frac{1}{x_{n+1}} .
\end{array}
$$
Therefore, $P_{n}+\frac{1}{2} S_{n}=\frac{1}{2 x_{n+1}}+1-\frac{1}{2 x_{n+1}}=1$ :
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The number of sets $X$ that satisfy the condition $\{1,2,3\} \subseteq X \subseteq\{1,2,3,4,5,6\}$ is $\qquad$ .
|
ii. 7.8 items.
$X$ must include the 3 elements $1,2,3$, while the numbers 4,5,6 may or may not belong to $X$. Each number has 2 possibilities, so the total number of different $X$ is $2^{3}=8$.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. The function $y=f(x)$ defined on $\mathbf{R}$ has the following properties:
（1）For any $x \in \mathbf{R}$, $f\left(x^{3}\right)=f^{3}(x)$;
(2) For any $x_{1} 、 x_{2} \in \mathbf{R}, x_{1} \neq x_{2}$, $f\left(x_{1}\right)$ $\neq f\left(x_{2}\right)$.
Then the value of $f(0)+f(1)+f(-1)$ is $\qquad$
|
10.0 .
From $f(0)=f^{3}(0)$, we know $f(0)[1-f(0)][1+f(0)]=0$,
thus, $f(0)=0$ or $f(0)=1$, or $f(0)=-1$;
from $f(1)=f^{3}(1)$, similarly $f(1)=0$ or 1 or 1;
from $f(-1)=f^{3}(-1)$, similarly $f(-1)=0$ or 1 or -1.
However, $f(0)$, $f(1)$, and $f(-1)$ are pairwise distinct,
so $\{f(0), f(1), f(-1)\}=\{0,1,-1\}$.
Therefore, $f(0)+f(1)+f(-1)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. (13 points) Let the quadratic equation in $x$, $2 x^{2}-t x-2=0$, have two roots $\alpha, \beta(\alpha<\beta)$.
(1) If $x_{1} 、 x_{2}$ are two different points in the interval $[\alpha, \beta]$, prove that $4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-4<0$;
(2) Let $f(x)=\frac{4 x-t}{x^{2}+1}$, and the maximum and minimum values of $f(x)$ on the interval $[\alpha, \beta]$ be $f_{\max }$ and $f_{\min }$, respectively. Let $g(t)=f_{\max }-f_{\min }$. Find the minimum value of $g(t)$.
|
$$
\begin{aligned}
14. (1) & \text{ From the conditions, we have } \alpha+\beta=\frac{t}{2}, \alpha \beta=-1. \\
& \text{ Without loss of generality, assume } \alpha \leqslant x_{1}4 x_{1} x_{2}-2(\alpha+\beta)\left(x_{1}+x_{2}\right)+4 \alpha \beta=4 x_{1} x_{2} \\
& -t\left(x_{1}+x_{2}\right)-4 .
\end{aligned}
$$
Therefore,
$$
4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-40, f_{\min }=f(\alpha)<0 . \\
g(t) & =f(\beta)-f(\alpha) \\
& =|f(\beta)|+|f(\alpha)| \\
& \geqslant 2 \sqrt{|f(\beta)| \cdot|f(\alpha)|}=4 .
$$
Equality holds if and only if $f(\beta)=-f(\alpha)=2$,
i.e., $\frac{8}{\sqrt{t^{2}+16}+t}=2$, which implies $t=0$.
Thus, the minimum value of $g(t)$ is 4.
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. (13 points) Given $a_{1}=1, a_{2}=3, a_{n+2}=(n+3) a_{n+1}$ $-(n+2) a_{n}$. If for $m \geqslant n$, the value of $a_{m}$ can always be divided by 9, find the minimum value of $n$.
|
15. From $a_{n+2}-a_{n+1}=(n+3) a_{n+1}-(n+2) a_{n}$
$$
\begin{array}{l}
-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)=(n+2)(n+1) \cdot \\
\left(a_{n}-a_{n-1}\right)=\cdots=(n+2) \cdot(n+1) \cdot n \cdots+4 \cdot 3 \cdot\left(a_{2}-\right. \\
\left.a_{1}\right)=(n+2)!, \\
\text { Therefore, } a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+\left(a_{n}-a_{n-1}\right) \\
\quad=1+2!+3!+\cdots+n!(n \geqslant 1) .
\end{array}
$$
Therefore, $a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+\left(a_{n}-a_{n-1}\right)$
Given $a_{1}=1, a_{2}=3, a_{3}=9, a_{4}=33, a_{5}=153$, at this point 153 is divisible by 9.
When $m \geqslant 5$, $a_{n}=a_{5}+\sum_{k=6}^{m} k!$, and when $k \geqslant 6$, $k!$ is divisible by 9.
Thus, when $m \geqslant 5$, $a_{n}$ is divisible by 9. Therefore, the smallest value of $n$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. As shown in Figure 1, the area of square $A B C D$ is 256, point $F$ is on $A D$, and point $E$ is on the extension of $A B$. The area of right triangle $\triangle C E F$ is 200. Then the length of $B E$ is $(\quad)$.
(A) 10
(B) 11
(C) 12
(D) 15
|
2. (C).
It is easy to prove that $\mathrm{Rt} \triangle C D F \cong \mathrm{Rt} \triangle C B E$. Therefore, $C F=C E$.
Since the area of Rt $\triangle C E F$ is 200, i.e., $\frac{1}{2} \cdot C F \cdot C E=200$, hence $C E=20$.
And $S_{\text {square } B C D}=B C^{2}=256$, so $B C=16$.
By the Pythagorean theorem, $B E=\sqrt{C E^{2}-B C^{2}}=12$.
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure $11, \angle A O B=$ $30^{\circ}, \angle A O B$ contains a fixed point $P$, and $O P=10, O A$ has a point $Q, O B$ has a fixed point $R$. If the perimeter of $\triangle P Q R$ is minimized, find its minimum value.
|
(Tip: Draw auxiliary lines with $O A$ and $O B$ as axes of symmetry. Answer:
10.)
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Let the function $f_{0}(x)=|x|, f_{1}(x)=$ $\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed figure formed by the graph of $y$ $=f_{2}(x)$ and the $x$-axis is $\qquad$
(1989, National High School Mathematics Competition)
|
Analysis: If it is not easy to directly draw the graph of $y=f_{2}(x)$, but we know the relationship between the graphs of $y=f(x)$ and $y=|f(x)|$. If we follow the sequence
$$
\begin{array}{l}
f_{0}(x)=|x| \rightarrow y=f_{0}(x)-1 \\
\rightarrow f_{1}(x)=\left|f_{0}(x)-1\right| \rightarrow y=f_{1}(x)-2 \\
\rightarrow f_{2}(x)=\left|f_{1}(x)-2\right|
\end{array}
$$
to perform graphical transformations, it is easier to draw the graph of $y=f_{2}(x)$.
It is easy to get the answer as 7.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 The sequence $a_{1}, a_{2}, a_{3}, \cdots, a_{2 n}, a_{2 n+1}$ forms an arithmetic sequence, and the sum of the terms with odd indices is 60, while the sum of the terms with even indices is 45. Then the number of terms $n=$ $\qquad$
|
Analysis: From $\left\{\begin{array}{l}a_{1}+a_{3}+\cdots+a_{2 n+1}=60, \\ a_{2}+a_{4}+\cdots+a_{2 n}=45\end{array}\right.$ $\Rightarrow\left\{\begin{array}{l}(n+1) a_{n+1}=60, \\ n a_{n+1}=45\end{array} \Rightarrow \frac{n+1}{n}=\frac{4}{3}\right.$, solving gives $n=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12 Given the three side lengths $a$, $b$, and $c$ of $\triangle ABC$ satisfy: (1) $a>b>c$, (2) $2b=a+c$, (3) $b$ is an integer, (4) $a^{2}+b^{2}+c^{2}=84$. Then the value of $b$ is $\qquad$
|
Analysis: Starting from $2 b=a+c$, let $a=b+d$, $c=b-d(d>0)$, then
$$
(b+d)^{2}+b^{2}+(b-d)^{2}=84 .
$$
That is, $3 b^{2}+2 d^{2}=84$.
Obviously, $2 d^{2}$ is a multiple of 3. Also, by $b+c>a \Rightarrow$ $b>2 d$, substituting into (4) gives $2 d^{2}<12$. Therefore, $2 d^{2}=3,6,9$. Verification shows $b=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 19 Let $f(x)=|1-2 x|, x \in[0,1]$. Then, the number of solutions to the equation $f\{f[f(x)]\}=\frac{1}{2} x$ is $\qquad$ .
|
Analysis: Let $y=f(x), z=f(y), w=f(z)$, use “ $\rightarrow$ ” to indicate the change, we have
$$
\begin{array}{l}
x: 0 \rightarrow 1, y: 1 \rightarrow 0 \rightarrow 1, \\
z: 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1, \\
w: 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1 .
\end{array}
$$
Since the “ $\rightarrow$ ” change is linear and occurs within the non-negative real number range, the number of intersections between the graph of $w=f\{f[f(x)]\}$ and $w=\frac{x}{2}$ is 8, which means the number of solutions is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 If $x \in \mathbf{R}$, find the maximum value of $F(x)=\min \{2 x+1$, $x+2,-x+6\}$.
(38th AHSME)
|
Solution: As shown in Figure 1, draw the graph of $F(x)$ (the solid part). From the graph, we can see that the maximum value of $F(x)$ is equal to the y-coordinate of the intersection point of $y=x+2$ and $y=-x+6$.
Therefore, the maximum value of $F(x)$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}$, $\cdots, b_{n}$ satisfying
$$
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1 .
$$
Find the maximum value of $F=\min \left\{\frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \cdots, \frac{a_{n}}{b_{n}}\right\}$.
(1979, Guangdong Province High School Mathematics Competition)
|
Solution: It is easy to see that when all the letters are equal, the value of $F$ is 1.
Below we prove that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$; $b_{1}, b_{2}, \cdots, b_{n}$, we have $F \leqslant 1$.
If not, then $F>1$.
$$
\text { Hence } \frac{a_{1}}{b_{1}}>1, \frac{a_{2}}{b_{2}}>1, \cdots, \frac{a_{n}}{b_{n}}>1 \text {, }
$$
which implies $a_{1}^{2}>b_{1}^{2}, a_{2}^{2}>b_{2}^{2}, \cdots, a_{n}^{2}>b_{n}^{2}$.
Thus, $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}>b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}$, contradicting the given condition.
Therefore, the maximum value of $F$ is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If real numbers $x, y, z$ satisfy $x+\frac{1}{y}=4, y+\frac{1}{z}=1, z+$ $\frac{1}{x}=\frac{7}{3}$, then the value of $x y z$ is $\qquad$.
|
7.1 .
Since $4=x+\frac{1}{y}=x+\frac{1}{1-\frac{1}{z}}=x+\frac{z}{z-1}$
$=x+\frac{\frac{7}{3}-\frac{1}{x}}{\frac{7}{3}-\frac{1}{x}-1}=x+\frac{7 x-3}{4 x-3}$, then
$4(4 x-3)=x(4 x-3)+7 x-3$,
which simplifies to $(2 x-3)^{2}=0$.
Thus, $x=\frac{3}{2}$.
Therefore, $z=\frac{7}{3}-\frac{1}{x}=\frac{5}{3}, y=1-\frac{1}{z}=\frac{2}{5}$.
Hence, $x y z=\frac{3}{2} \times \frac{2}{5} \times \frac{5}{3}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given the quadratic function $y=a x^{2}+b x+c$ (where $a$ is a positive integer) whose graph passes through the points $A(-1,4)$ and $B(2,1)$, and intersects the $x$-axis at two distinct points. Then the maximum value of $b+c$ is $\qquad$ .
|
10. -4 .
From the given, we have $\left\{\begin{array}{l}a-b+c=4, \\ 4 a+2 b+c=1,\end{array}\right.$ solving this yields $\left\{\begin{array}{l}b=-a-1, \\ c=3-2 a .\end{array}\right.$
Since the graph of the quadratic function intersects the $x$-axis at two distinct points, we have,
$$
\begin{array}{l}
\Delta=b^{2}-4 a c>0, \\
(-a-1)^{2}-4 a(3-2 a)>0,
\end{array}
$$
which simplifies to $(9 a-1)(a-1)>0$.
Since $a$ is a positive integer, we have $a>1$, so $a \geqslant 2$.
Since $b+c=-3 a+2 \leqslant-4$, and when $a=2, b=$ $-3, c=-1$, the conditions are satisfied, hence the maximum value of $b+c$ is -4.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$.
|
14. (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$. From the problem, we know $a>0$, and $b+c=2-a, bc=\frac{4}{a}$. Therefore, $b, c$ are the two real roots of the quadratic equation $x^{2}-(2-a)x+\frac{4}{a}=0$, then
$$
\begin{array}{l}
\Delta=(2-a)^{2}-4 \times \frac{4}{a} \geqslant 0, \\
a^{3}-4 a^{2}+4 a-16 \geqslant 0, \\
\left(a^{2}+4\right)(a-4) \geqslant 0 .
\end{array}
$$
Thus, $a \geqslant 4$.
When $a=4, b=c=-1$, the conditions are satisfied.
Therefore, the minimum value of the maximum among $a, b, c$ is 4.
(2) Since $abc>0$, $a, b, c$ are either all positive or one positive and two negative.
(i) If $a, b, c$ are all positive, then from (1), the maximum of $a, b, c$ is not less than 4, which contradicts $a+b+c=2$.
(ii) If $a, b, c$ are one positive and two negative, let $a>0, b<0, c<0$, then
$$
\begin{array}{l}
|a|+|b|+|c|=a-b-c \\
=a-(2-a)=2a-2 .
\end{array}
$$
From (1), we know $a \geqslant 4$, hence $2a-2 \geqslant 6$.
When $a=4, b=c=-1$, the conditions are satisfied and the equality holds. Therefore, the minimum value of $|a|+|b|+|c|$ is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $0 \leqslant a-b \leqslant 1,1 \leqslant a+b \leqslant 4$. Then, when $a-2 b$ reaches its maximum value, the value of $8 a+2002 b$ is $\qquad$ .
|
3.8.
$$
\begin{array}{l}
\text { Let } 0 \leqslant a-b \leqslant 1, \\
1 \leqslant a+b \leqslant 4, \\
\text { and } m(a-b)+n(a+b)=a-2 b .
\end{array}
$$
By comparing the coefficients of $a$ and $b$ on both sides, we get the system of equations and solve to find
$$
m=\frac{3}{2}, n=-\frac{1}{2} \text {. }
$$
Thus, $a-2 b=\frac{3}{2}(a-b)-\frac{1}{2}(a+b)$.
From (1) and (2), we have $-2 \leqslant a-2 b \leqslant 1$.
Therefore, the maximum value of $a-2 b$ is 1, at which point $b=\frac{a-1}{2}$.
Substituting into (1) and (2) gives $0 \leqslant a \leqslant 1, 1 \leqslant a \leqslant 3$.
This implies $a=1, b=0$. Hence, $8 a+2002 b=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A $9 \times 9$ grid of squares is colored in two colors, black and white, such that the number of black squares adjacent to each white square is greater than the number of white squares, and the number of white squares adjacent to each black square is greater than the number of black squares (squares sharing a common edge are considered adjacent). Find the maximum difference in the number of black and white squares.
(53rd Belarusian Mathematical Olympiad (Final D Class))
|
Solution: A coloring scheme of a square matrix that satisfies the problem's conditions is called a good scheme. For a good coloring scheme, it is easy to prove the following conclusions:
(1) If two adjacent cells have the same color and are in different rows (or columns), then any two adjacent cells in these two rows (columns) have the same color, and these colors alternate between black and white (as shown in Figure 2).
(2) Three consecutive cells cannot have the same color.
(3) If a corner cell is white (black), then the adjacent cell is black (white).
(4) If there are two adjacent cells of the same color in a certain column (row), then it is impossible to have two adjacent cells of the same color in any other row (column).
From these conclusions, we can deduce that if there are no two adjacent cells of the same color in a certain column (row), then the colors of the other cells in the matrix depend solely on the coloring of this column (row). Therefore, we only need to check the color distribution of the first column (row). It is easy to see that if the matrix is colored like a chessboard, the maximum difference between the number of black and white cells is 1.
Consider the case where there are two adjacent cells of the same color in the first row. From (1) to (4), we can deduce that the number of black and white cells in any two adjacent rows is the same, so the difference between the number of black and white cells is equal to the difference in the first row. It is easy to see that the difference in the number of black and white cells in the first row is 3 (as shown in Figure 3).
When there are two adjacent cells of the same color in the first column, we can draw the same conclusion. From (4), we know that any good coloring scheme cannot have two adjacent cells of the same color in both the first row and the first column simultaneously.
In summary, the maximum difference between the number of black and white cells is 3.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. Given a wire of length $150 \mathrm{~cm}$, it is to be cut into $n(n>2)$ smaller segments, each of which has an integer length of no less than $1(\mathrm{~cm})$. If no three segments can form a triangle, find the maximum value of $n$, and how many ways there are to cut the wire into $n$ segments that satisfy the condition.
|
17. Since the sum of $n$ segments is a fixed value of $150(\mathrm{~cm})$, to make $n$ as large as possible, the length of each segment must be as small as possible. Given that the length of each segment is no less than $1(\mathrm{~cm})$, and no three segments can form a triangle, the lengths of these segments can only be $1,1,2,3,5,8$, $13,21,34,55,89, \cdots$.
But $1+1+2+\cdots+34+55=143<150$,
hence the maximum value of $n$ is 10.
There are 7 ways to divide a wire of length $150(\mathrm{~cm})$ into 10 segments that meet the conditions:
$$
\begin{array}{l}
1,1,2,3,5,8,13,21,34,62 ; \\
1,1,2,3,5,8,13,21,35,61 ; \\
1,1,2,3,5,8,13,21,36,60 ; \\
1,1,2,3,5,8,13,21,37,59 ; \\
1,1,2,3,5,8,13,22,35,60 ; \\
1,1,2,3,5,8,13,22,36,59 ; \\
1,1,2,5,8,14,22,36,58
\end{array}
$$
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.
保留源文本的换行和格式，翻译结果如下：
```
Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.
```
|
Three, let $P=\frac{a+b+c}{a b c}$.
In equation (1), let $a$ and $b$ remain unchanged, 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, so $c=1$.
Thus, $P=\frac{a+b+1}{a b}=\frac{1}{a}+\frac{a+1}{a b}$.
In equation (2), let $a$ remain unchanged, and only let $b$ vary, where $b$ can take any integer from 2 to 9.
Then, from equation (2), we know that when $b=2$, $P$ reaches its maximum value, so $b=2$.
At this point, $P=\frac{1}{2}+\frac{3}{2 a}$.
In equation (3), $P$ reaches its maximum value when $a$ takes its minimum value, and $a$ can take any integer from 3 to 9, so $a=3$.
Therefore, when $a=3, b=2, c=1$ (the letters can be interchanged), $P$ reaches its maximum value of 1.
(Shen Xueming, Changqiao Middle School, Wuzhong District, Suzhou City, Jiangsu Province, 215128)
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) In a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)(n$ $\geqslant 2)$ grid, each cell is filled with 1 or -1. If the number in any cell is equal to the product of the numbers in the cells that share an edge with it, then this filling method is called "successful". Find the total number of "successful" fillings.
In a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)(n$ $\geqslant 2)$ grid, each cell is filled with 1 or -1. If the number in any cell is equal to the product of the numbers in the cells that share an edge with it, then this filling method is called "successful". Find the total number of "successful" fillings.
|
Three, assuming there exists some successful filling method that contains -1. First, prove: if this successful filling method is symmetric about the middle column (row), then the middle column (row) is entirely 1.
Let $a_{0}=1, a_{2}^{n}=1$.
If $a_{1}=1$, by $a_{1}=a_{0} \times a_{2} \times 1$, we get $a_{2}=1$.
Similarly, $a_{3}=1, \cdots$, $a_{2^{n}-1}=1$.
If $a_{2}^{n}-1=1$, similarly, we get $a_{1}=a_{2}=\cdots=a_{2}-2=1$.
Therefore, if $a_{1}=-1$, then $a_{2}^{n}-1=-1$. This way,
$$
\begin{array}{l}
a_{1}=-1, a_{2}=-1, a_{3}=1, a_{4}=-1, a_{5}=-1, a_{6} \\
=1, \cdots, a_{2}^{n}-2=-1, a_{2}^{n}-1=-1, a_{2}^{n}=1 \text {. Thus } 312^{2},
\end{array}
$$
a contradiction.
Therefore, the middle column (row) is entirely 1.
Secondly, if this successful filling method is not symmetric about the middle column, first rotate this filling method 180° along the middle column, to get another filling method, then multiply the numbers in the same positions of the two filling methods, to get a successful filling method symmetric about the middle column, which contains -1. This successful filling method can be further transformed into a successful filling method that is symmetric about both the middle row and the middle column, which contains -1.
For such a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)$ successful filling method, removing all 1s in the middle row and the middle column, we can get 4 $\left(2^{n-1}-1\right) \times\left(2^{n-1}-1\right)$ successful filling methods, at least one of which contains -1. For this successful filling method, repeat the above operation, and eventually, we can get a $3 \times 3$ successful filling method, which contains -1, but the middle row and the middle column are all 1, which contradicts the definition of a successful filling method.
Therefore, for any successful filling method, it must not contain -1. Hence, there is only 1 successful filling method, which is to fill each cell with 1.
(Anqing No.1 High School, Anhui Province, 246004)
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c$ $=$ $\qquad$ .
|
-1 .
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$. Since $\triangle A B C$ is a right triangle, it follows that $x_{1} 、 x_{2}$ must have opposite signs. Thus, $x_{1} x_{2}=\frac{c}{a}<0$.
By the projection theorem, we know $|O C|^{2}=|A O| \cdot|B O|$, i.e., $c^{2}=\left|x_{1}\right| \cdot\left|x_{2}\right|=\left|\frac{c}{a}\right|$. Therefore, $|a c|=1, a c=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $m$ be an integer, and the two roots of the equation $3 x^{2}+m x-2=0$ are both greater than $-\frac{9}{5}$ and less than $\frac{3}{7}$. Then $m=$ $\qquad$ .
|
2.4 .
From the problem, we have
$$
\left\{\begin{array}{l}
3 \times\left(-\frac{9}{5}\right)^{2}+m \times\left(-\frac{9}{5}\right)-2>0, \\
3 \times\left(\frac{3}{7}\right)^{2}+m \times\left(\frac{3}{7}\right)-2>0 .
\end{array}\right.
$$
Solving this, we get \(3 \frac{8}{21}<m<4 \frac{13}{45}\). Therefore, \(m=4\).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (18 points) On a plane, there are 7 points, and some line segments can be connected between them, so that any 3 points among the 7 points must have 2 points connected by a line segment. How many line segments are needed at least? Prove your conclusion.
|
(1) If one of the 7 points is isolated (i.e., it is not connected to any other points), then the remaining 6 points must be connected in pairs, which requires at least $\frac{6 \times 5}{2}=15$ lines.
(2) If one of the 7 points is connected to only one other point, then the remaining 5 points must be connected in pairs, which requires at least $1+\frac{5 \times 4}{2}=11$ lines.
(3) If each point is connected to at least 3 other points, then at least $\frac{7 \times 3}{2}$ lines are required. Since the number of lines must be an integer, at least 11 lines are required in this case.
(4) If each point is connected to at least 2 other points, and there is one point (denoted as $A$) that is connected to only 2 other points $A B$ and $A C$, then the 4 points not connected to $A$ must be connected in pairs, requiring $\frac{4 \times 3}{2}=6$ lines. The lines extending from $B$ must be at least 2, i.e., in addition to $B A$, there must be at least one more. Therefore, at least $6+2+1=9$ lines are required in this case.
Figure 6 shows a situation with 9 lines connected.
Combining (1) and (4), at least 9 lines are required to meet the requirements.
|
9
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
12. As shown in Figure 4, line $AB$ intersects $\odot O$ at points $A$ and $B$, point $O$ is on $AB$, and point $C$ is on $\odot O$, with $\angle AOC=40^{\circ}$. Point $E$ is a moving point on line $AB$ (not coinciding with point $O$), and line $EC$ intersects $\odot O$ at another point $D$. The number of points $E$ that satisfy $DE = DO$ is $\qquad$ .
|
12.3.
Consider the position of point $E$: Point $E$ can be on the extension of line segment $O A$; Point $E$ can be on line segment $O A$ (excluding point $O$); Point $E$ can be on the extension of line segment $O B$; but point $E$ cannot be on line segment $O B$ (excluding point $O$). Therefore, point $E$ has a total of 3 positions.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Given real numbers $a, b, c$, satisfying $a+b+c=0, a^{2}+b^{2}+c^{2}=6$. Then the maximum value of $a$ is
|
14.2.
From the problem, we get $c=-(a+b)$, thus,
$$
a^{2}+b^{2}+[-(a+b)]^{2}=6,
$$
which simplifies to $b^{2}+a b+a^{2}-3=0$.
Since $b$ is a real number, the above equation, which is a quadratic equation in $b$, must have real roots, hence,
$$
\Delta=a^{2}-4\left(a^{2}-3\right) \geqslant 0, \quad -2 \leqslant a \leqslant 2.
$$
When $a=2$, $b=c=-1$, which satisfies the problem's conditions.
Therefore, the maximum value of $a$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A group of 17 middle school students went to several places for a summer social survey, with a budget for accommodation not exceeding $x$ yuan per person per day. One day, they arrived at a place with two hostels, $A$ and $B$. $A$ has 8 first-class beds and 11 second-class beds; $B$ has 10 first-class beds, 4 second-class beds, and 6 third-class beds. It is known that the daily rates for first-class, second-class, and third-class beds are 14 yuan, 8 yuan, and 5 yuan, respectively. If the entire group stays in one hostel, they can only stay at $B$ according to the budget. Then the integer $x=$ $\qquad$.
|
4. $x=10$.
If staying at location $A$, even choosing the most economical beds, the average accommodation cost per person will exceed 10 yuan (since $8 \times 11 + 14 \times 6 = 172$ (yuan), $172 \div 17 \approx 10.12$ (yuan)).
If staying at location $B$, with a reasonable choice of beds, the budget can be met, and the most economical value is
$$
\begin{array}{l}
5 \times 6 + 8 \times 4 + 14 \times 7 = 160 \text{ (yuan) }, \\
160 \div 17 \approx 9.41 \text{ (yuan). }
\end{array}
$$
Since $9.41 \leqslant x < 10.12$, and $x$ is an integer, therefore, $x = 10$.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If the polynomial $x^{2}-x+1$ can divide another polynomial $x^{3}+x^{2}+a x+b(a, b$ are constants). Then $a+b$ equals ( ).
(A) 0
(B) -1
(C) 1
(D) 2
|
2. (C).
By synthetic division, it is easy to obtain the remainder $r(x)=(a+1) x+b-2$. Since it can be divided exactly, we have $a+1=0$ and $b-2=0$, i.e., $a=-1, b=2$. Therefore, $a+b=1$.
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. $f(x)=\frac{x^{2}}{8}+x \cos x+\cos (2 x)(x \in \mathbf{R})$'s minimum value is $\qquad$ .
|
4. -1 .
$$
\begin{array}{l}
f(x)=\frac{x^{2}}{8}+x \cos x+2 \cos ^{2} x-1 \\
=\frac{1}{8}(x+4 \cos x)^{2}-1 \geqslant-1 .
\end{array}
$$
Since the equation $\cos x=-\frac{x}{4}$ (as can be seen from the graph) has a solution, we have
$$
f(x)_{\min }=-1
$$
|
-1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Try to find the smallest possible value of the positive integer $k$, such that the following proposition holds: For any $k$ integers $a_{1}, a_{2}, \cdots, a_{k}$ (equality is allowed), there must exist corresponding $k$ integers $x_{1}, x_{2}, \cdots, x_{k}$ (equality is also allowed), and $\left|x_{i}\right| \leqslant 2(i=1,2, \cdots, k),\left|x_{1}\right|+\left|x_{2}\right|+\cdots+$ $\left|x_{k}\right| \neq 0$, such that 2003 divides $x_{1} a_{1}+x_{2} a_{2}+\cdots$ $+x_{k} a_{k}$.
|
Three, first prove that the proposition holds when $k=7$. For this, consider the sum
$$
S\left(y_{1}, y_{2}, \cdots, y_{7}\right)=y_{1} a_{1}+y_{2} a_{2}+\cdots+y_{7} a_{7} \text {, }
$$
where $y_{i} \in\{-1,0,1\}$.
There are $3^{7}=2187$ such sums, and since $2187>2003$, by the pigeonhole principle, there must be two different sums $S\left(y_{1}, y_{2}, \cdots, y_{7}\right)$ and $S\left(y_{1}^{\prime}, y_{2}^{\prime}, \cdots, y_{7}^{\prime}\right)$ that have the same remainder when divided by 2003.
Therefore, 2003 divides
$$
\begin{array}{l}
S\left(y_{1}, y_{2}, \cdots, y_{7}\right)-S\left(y_{1}^{\prime}, y_{2}^{\prime}, \cdots, y_{7}^{\prime}\right) \\
=\left(y_{1}-y_{1}^{\prime}\right) a_{1}+\left(y_{2}-y_{2}^{\prime}\right) a_{2}+\cdots+\left(y_{7}-y_{7}^{\prime}\right) a_{7} .
\end{array}
$$
where $y_{i}-y_{i}^{\prime} \in\{-2,-1,0,1,2\}, i=1,2, \cdots, 7$ (since $\left.y_{i} 、 y_{i} \in\{-1,0,1\}, i=1,2, \cdots, 7\right)$, and at least one $y_{i}-y_{i}^{\prime} \neq 0$ (because the arrays $\left(y_{1}, y_{2}, \cdots, y_{7}\right) \neq\left(y_{1}^{\prime}, y_{2}^{\prime}, \cdots\right.$, $\left.y_{7}^{\prime}\right)$ ). At this point, taking $x_{i}=y_{i}-y_{i}^{\prime}, i=1,2, \cdots, 7$, satisfies the requirement.
Thus, the proposition holds when $k=7$.
Next, prove that the proposition does not hold when $k=6$. For this, we provide a counterexample.
Take $a_{1}=3^{1}, a_{2}=3^{2}, a_{3}=3^{3}, a_{4}=3^{4}, a_{5}=3^{5}, a_{6}=$ $3^{6}$, then for any 6 integers $x_{1}, x_{2}, \cdots, x_{6},\left|x_{i}\right| \leqslant$ $2,\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{6}\right| \neq 0$, the sum
$$
S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{6} x_{6}
$$
is always a multiple of 3.
$$
\begin{array}{l}
\text { Then } S\left(x_{1}, x_{2}, \cdots, x_{6}\right) \text { | } \\
\leqslant 2\left(3^{1}+3^{2}+\cdots+3^{6}\right) \\
=3\left(3^{6}-1\right)=3\left(3^{3}-1\right)\left(3^{3}+1\right) \\
=3 \times 26 \times 28=3 \times 728 .
\end{array}
$$
Assume without loss of generality that the largest index $k_{1}$ of the non-zero elements $x_{1}, x_{2}, \cdots, x_{6}$ is $x_{k_{1}}$, i.e., $x_{k_{1}} \neq 0$, and $x_{k_{1}+1}=x_{k_{1}+2}=\cdots=x_{6}=$ 0, then
$$
\begin{array}{l}
S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{k_{1}} x_{k_{1}} \\
=3^{1} x_{1}+3^{2} x_{2}+\cdots+3^{k_{1}} x_{k_{1}} .
\end{array}
$$
Additionally, assume without loss of generality that $x_{k_{1}}>0$ (if $x_{k_{1}}<0$, the argument is similar).
If $k_{1} \geqslant 2$, then
If $k_{1}=1$, then
$$
x_{1} \neq 0, S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}>0 \text {. }
$$
In summary,
$$
\begin{array}{l}
3 \mid S\left(x_{1}, x_{2}, \cdots, x_{6}\right) \text {, and } S\left(x_{1}, x_{2}, \cdots, x_{6}\right) \neq 0 \text {, } \\
\left|S\left(x_{1}, x_{2}, \cdots, x_{6}\right)\right| \leqslant 3 \times 728 .
\end{array}
$$
Clearly, 2003 and 3 are coprime.
If 2003 divides $S\left(x_{1}, x_{2}, \cdots, x_{6}\right)$, then $S\left(x_{1}\right.$, $\left.x_{2}, \cdots, x_{6}\right)=3 t$, where $t$ is an integer, and $1 \leqslant|t| \leqslant 728$. Thus, $2003 \mid t$, which contradicts $1 \leqslant|t| \leqslant 728$. Therefore, when taking $a_{i}=3^{i}$, $i=1,2, \cdots, 6$, it is impossible to have $x_{1}, x_{2}, \cdots, x_{6} \in \mathbf{Z},\left|x_{i}\right|$ $\leqslant 2, i=1,2, \cdots, 6,\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{6}\right| \neq 0$, such that
$2003$ divides $S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{6} x_{6}$. This counterexample shows that the proposition does not hold when $k=6$.
From the above two steps, the smallest positive integer $k$ is 7.
(Wu Weizhao, Department of Mathematics, College of Science, Guangzhou University, 510405)
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $n$ be a given positive integer. Try to find non-negative integers $k, l$, satisfying $k+l \neq 0$, and $k+l \neq n$, such that
$$
s=\frac{k}{k+l}+\frac{n-k}{n-(k+l)}
$$
takes the maximum value.
|
If $l=0$, then $s=2$.
If $l>0$, let $x=k+l$, then $00 .
\end{array}
$
Therefore, $f(1)1$ when, $s$ reaches the maximum value 2;
(2) If $n=2$, then when $l=1, k=0$ or $l=0, 0 < k \neq 2$, $s$ reaches the maximum value 2;
(3) If $n \geqslant 3$, then when $l=n-1, k=0$, $s$ reaches the maximum value $n$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Find the smallest positive integer $n$, such that
$$
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2002}
$$
has integer solutions.
(Uzbekistan provided)
|
Solution: Since $2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9), 2002$ $=667 \times 3+1$, therefore,
$$
2002^{2002} \equiv 4^{2002} \equiv 4(\bmod 9) \text {. }
$$
Also, $x^{3} \equiv 0, \pm 1(\bmod 9)$, where $x$ is an integer, thus,
$$
x_{1}^{3}, x_{1}^{3}+x_{2}^{3}, x_{1}^{3}+x_{2}^{3}+x_{3}^{3} \equiv 4(\bmod 9) \text {. }
$$
Since $2002=10^{3}+10^{3}+1^{3}+1^{3}$, then
$$
\begin{array}{l}
2002^{2002}=2002 \times\left(2002^{667}\right)^{3} \\
=\left(10 \times 2002^{667}\right)^{3}+\left(10 \times 2002^{667}\right)^{3}+ \\
\left(2002^{667}\right)^{3}+\left(2002^{667}\right)^{3} .
\end{array}
$$
Therefore, $n=4$.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. For all non-negative integers $x, y$, find all functions $f: \mathbf{N} \rightarrow \mathbf{N}$, satisfying $f(3 x+2 y)=f(x) f(y)$, where $\mathbf{N}$ is the set of non-negative integers.
(53rd Romanian Mathematical Olympiad (Final))
|
Solution: Let $x=y=0$, we get $f(0)=f(0)^{2}$. Therefore, $f(0)=0$ or $f(0)=1$.
If $f(0)=0$, for $x=0$ or $y=0$, we get $f(2 y)=$ $f(3 x)=0$ for all $x, y \in \mathrm{N}$. Let $f(1)=a$, then
$$
\begin{array}{l}
f(5)=f(3 \times 1+2 \times 1)=f(1) f(1)=a^{2}, \\
f(25)=f(3 \times 5+2 \times 5)=f(5) f(5)=a^{4} .
\end{array}
$$
Since $f(25)=f(2 \times 2+3 \times 7)=f(2) f(7)=0$, we have $a=0$.
For any odd number $k>4$, there exists $y \in \mathbb{N}$, such that $k=3+2 y$. Then $f(k)=0$, hence for all $x \in \mathbb{N}, f(x)=0$ satisfies the condition.
If $f(0)=1$, let $x=0$ or $y=0$, we get
$f(2 y)=f(y)$ or $f(3 x)=f(x)$.
Let $f(1)=a$, then
$$
\begin{array}{l}
f(2)=a, \\
f(5)=f(3 \times 1+2 \times 1)=f(1) f(1)=a^{2}, \\
f(25)=f(3 \times 5+2 \times 5)=f(5) f(5)=a^{4} .
\end{array}
$$
Since $f(25)=f(3 \times 3+2 \times 8)=f(3) f(8)=$ $f(1) f(4)=f(1) f(2)=a^{2}$, we have $a=0$ or $a=1$.
Similarly, we have
$$
f(x)=\left\{\begin{array}{ll}
1, & x=0, \\
0, & x>0
\end{array} \text { or } f(x)=1, x \in \mathbb{N} .\right.
$$
In summary, there are 3 functions that satisfy the conditions.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Consider a square on the complex plane, whose 4 vertices correspond to the 4 roots of a certain monic quartic equation with integer coefficients $x^{4}+p x^{3}+q x^{2}+r x+s=0$. Find the minimum value of the area of such a square.
|
3. According to the problem, the 4 roots of the equation can only be in two scenarios: 2 real roots and 1 pair of conjugate complex roots; 2 pairs of conjugate complex roots.
(1) If the 4 roots of the equation are 2 real roots and 1 pair of conjugate complex roots, then we can set these 4 roots as $a \pm b, a \pm b \mathrm{i}$. Thus, the original equation is
$$
(x-a)^{4}=b^{4},
$$
which is $x^{4}-4 a x^{3}+6 a^{2} x^{2}-4 a^{3} x+a^{4}-b^{4}=0$.
From $-4 a \in \mathbf{Z}$ and $4 a^{3} \in \mathbf{Z}$, we know $a \in \mathbf{Z}$. From $a^{4}-b^{4} \in \mathbf{Z}$, we know $b^{4} \in \mathbf{Z}$, so $b^{4} \geqslant 1, b^{2} \geqslant 1$. In this case, the area of the square is $2 b^{2} \geqslant 2$, with equality holding when $a \in \mathbf{Z}, b= \pm 1$.
(2) If the 4 roots of the equation are 2 pairs of conjugate complex roots, then we can set these 4 roots as $a \pm b \mathrm{i}, a+2 b \pm b \mathrm{i}$. These 4 roots are the roots of the equation $(x-(a+b))^{4}=-4 b^{4}$. Similarly, we know $4 b^{4} \in \mathbf{Z}$, thus $4 b^{4} \geqslant 1, b^{2} \geqslant \frac{1}{2}$. Therefore, the area of the square is $4 b^{2} \geqslant 2$. Equality holds when $b= \pm \frac{\sqrt{2}}{2}, a+b \in \mathbf{Z}$.
In summary, the area of such a square is greater than or equal to 2. The equation $x^{4}=1$ has 4 roots which are the 4 vertices of a square on the complex plane with an area of 2, so the minimum area of such a square is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Find a point $P$ on the plane of equilateral $\triangle A B C$ such that $\triangle P A B$, $\triangle P B C$, and $\triangle P C A$ are all isosceles triangles. How many points $P$ with this property are there?
|
Solution: Let any side of the known triangle be denoted as $a$. When $a$ is the base of the sought isosceles triangle, point $P$ lies on the perpendicular bisector of $a$; when $a$ is one of the legs of the sought isosceles triangle, point $P$ lies on the circumference of the circle with the vertex of $\triangle ABC$ as the center and $a$ as the radius. Thus, the 3 perpendicular bisectors and 3 circles yield a total of 10 intersection points (as shown in Figure 1).
Therefore, there are 10 points $P$ that satisfy the conditions. These can be categorized into 4 types: the first type is the intersection of the 3 perpendicular bisectors $\left(P_{1}\right)$. The 2nd, 3rd, and 4th types are the intersections of the circles and the perpendicular bisectors, each with 3 points $\left(P_{2}, P_{3}\right.$, $\left.P_{4}\right),\left(P_{5}, P_{6}, P_{7}\right),\left(P_{8}, P_{9}, P_{10}\right)$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given an $n \times n$ ($n$ is an odd number) chessboard where each unit square is colored in a checkerboard pattern, and the 4 corner unit squares are colored black. A figure formed by 3 connected unit squares in an L-shape is called a "domino". For what value of $n$ can all the black squares be covered by non-overlapping "dominoes"? If it can be covered, what is the minimum number of "dominoes" needed?
|
Solution: Let $n=2m+1$, consider the odd rows, then each row has $m+1$ black cells, with a total of $(m+1)^2$ black cells. Any two black cells cannot be covered by a single "domino", therefore, at least $(m+1)^2$ "dominoes" are needed to cover all the black cells on the chessboard. Since when $n=1,3,5$, we have $3(m+1)^2 > n^2$, thus, $n \geqslant 7$.
Below, we use mathematical induction to prove: when $n \geqslant 7$, $(m+1)^2$ "dominoes" can cover all the black cells on the chessboard.
When $n=7$, since two "dominoes" can form a $2 \times 3$ rectangle, and two $2 \times 3$ rectangles can form a $4 \times 3$ rectangle, then 4 such $4 \times 3$ rectangles can be placed on a $7 \times 7$ chessboard, such that all cells except the middle black cell are covered (as shown in Figure 1). Adjust the "domino" adjacent to the middle black cell, so that this "domino" covers the middle black cell and also covers the only black cell that the original "domino" covered. Thus, 16 "dominoes" cover all cells except one white cell on the chessboard (as shown in Figure 1).
Assume that when $n=2m-1$, on a $(2m-1) \times (2m-1)$ chessboard, $m^2$ "dominoes" can cover all the black cells. When $n=2m+1$, divide the $(2m+1) \times (2m+1)$ chessboard into three parts: $(2m-1) \times (2m-1)$, $(2m-1) \times 2$, and $(2m+1) \times 2$. Since the $(2m-1) \times 2$ rectangle can be divided into $m-2$ $2 \times 2$ squares and one $2 \times 3$ rectangle, the black cells in the $(2m-1) \times 2$ rectangle can be covered by $(m-2) + 2$ "dominoes". Similarly, the $(2m+1) \times 2$ rectangle can be covered by $(m-1) + 2$ "dominoes" (as shown in Figure 2). Therefore, the $(2m+1) \times (2m+1)$ chessboard can be covered by $m^2 + m + (m+1) = (m+1)^2$ "dominoes".
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $T$ be a set of ordered triples $(x, y, z)$, where $x, y, z$ are integers, and $0 \leqslant x, y, z \leqslant 9$. Two players, A and B, play the following game: A selects a triple $(x, y, z)$ from $T$, and B has to guess A's chosen triple using several "moves". One "move" consists of: B giving A a triple $(a, b, c)$ from $T$, and A responding with the number $|x+y-a-b|+|y+z-b-c|+|z+x-c-a|$. Find the minimum number of "moves" required for B to determine A's chosen triple.
(Bulgaria provided)
|
Solution: Two "movements" are not enough. Because each answer is an even number between 0 and 54, i.e., there are 28 possible values for each answer. The maximum number of possible results from two "movements" is $28^{2}$, which is less than the 1000 possible values for $(x, y, z)$.
Below, we prove that 3 "movements" are sufficient to determine the triplet chosen by A.
In the first "movement," B gives A the triplet $(0,0,0)$, and A's response is $2(x+y+z)$, so B can determine the value of $s=x+y+z$. Clearly, $0 \leqslant s \leqslant 27$. Without loss of generality, assume $s \leqslant 13$. In fact, if $s \geqslant 14$, then the $(a, b, c)$ given by B to A in the following process can be changed to $(9-a, 9-b, 9-c)$, and the triplet $(x, y, z)$ can still be determined.
If $s \leqslant 9$, in the second "movement," B gives A $(9,0,0)$, and A's response is $(9-x-y)+y+z+(9-x-z)=18-2x$, thus the value of $x$ can be determined. Similarly, in the third "movement," B gives A $(0,9,0)$, and the value of $y$ can be determined. Therefore, $z=s-x-y$ can be determined.
If $9 < s \leqslant 13$, in the second "movement," B gives A $(0, s, 0)$, and A's response is $|s-x-y|+|s-y-z|+|s-x-z|=3s-2(x+y+z)=3s-2s=s$, thus the value of $s$ can be determined. In the third "movement," B gives A $(9, s-9, 0)$, and A's response is
$$
\begin{array}{l}
|s-9-x-y|+|s-9-y-z|+|9-x-z| \\
=(9-x)+(s-9-y)+(9-x-z) \\
=18-2(x+z) .
\end{array}
$$
Thus, the value of $x+z$ can be determined. Similarly, the values of $x$ and $z$ can be determined. Therefore, the value of $y$ can be determined.
In summary, the minimum number of "movements" required is 3.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=\frac{1}{2}, x_{k+1}=x_{k}^{2}+x_{k}$. Then the integer part of the sum $\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{200 B}+1}$ is $\qquad$
|
4. 1 .
From the problem, we know that $\left\{x_{n}\right\}$ is an increasing sequence, and $x_{3}>1$. From $\frac{1}{x_{k+1}}=\frac{1}{x_{k}\left(x_{k}+1\right)}=\frac{1}{x_{k}}-\frac{1}{x_{k}+1}$, we get $\frac{1}{x_{k}+1}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}}$.
Then $S=\sum_{k=1}^{2008} \frac{1}{x_{k}+1}=\sum_{k=1}^{2008}\left(\frac{1}{x_{k}}-\frac{1}{x_{k+1}}\right)$ $=\frac{1}{x_{1}}-\frac{1}{x_{2004}}=2-\frac{1}{x_{2004}}$.
Since $x_{2} \cos >x_{3}>1$, then $1<S<2$, so $[S]=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given an infinite sequence $\left\{a_{n}\right\}$ where all terms are positive integers, and the sum of any consecutive terms is not equal to 100. Find the minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$.
|
Solution: The minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$ is 3.
First, we prove: If all terms of the sequence $\left\{a_{n}\right\}$ are 1 or 2, then there must exist a continuous subsequence whose sum equals 100.
Consider the first 100 terms of the sequence $\left\{a_{n}\right\}$, and let
$$
S_{0}=0, S_{1}=a_{1}, \cdots, S_{k}=\sum_{i=1}^{k} a_{i}, k \leqslant 100 .
$$
By the pigeonhole principle, there exist $0 \leqslant m<n \leqslant 100$ such that $S_{n}-S_{m}=100 t$, i.e.,
$$
\sum_{i=m+1}^{n} a_{i}=100 t .
$$
Since $a_{i}=1$ or 2, then $t=1$ or 2.
If $t=1$, the conclusion holds.
If $t=2$, by $a_{i} \leqslant 2, n \leqslant 100$, we get
$$
200 \geqslant \sum_{i=1}^{100} a_{i} \geqslant \sum_{i=m+1}^{n} a_{i}=200 .
$$
Then $a_{i}=2, i=1,2, \cdots, 100$. Hence $\sum_{i=1}^{50} a_{i}=100$.
In this case, the conclusion still holds. Therefore, $\max \left\{a_{n}, n \in \mathbf{N}\right\} \geqslant 3$.
Next, we construct an infinite periodic sequence such that the largest term is 3. The first 100 terms of the sequence $\left\{a_{n}\right\}$ are
$$
1,2, \cdots, 2,3,2, \cdots, 2, \cdots,
$$
where $a_{100 k+1}=1, a_{100 k+51}=3, k=0,1, \cdots$, and the other terms $a_{i}=2$.
It is easy to verify that the sum of any continuous subsequence of this sequence does not equal 100.
In summary, the minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$ is 3.
(Zhang Yanwei, Suqian Education Bureau, Jiangsu Province, 223800)
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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