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5. Let $x_{1}, x_{2}, \cdots, x_{51}$ be natural numbers, $x_{1}<x_{2}$ $<\cdots<x_{51}$, and $x_{1}+x_{2}+\cdots+x_{51}=1995$. When $x_{26}$ reaches its maximum value, the maximum value that $x_{51}$ can take is | 5. To maximize $x_{26}$, then $x_{1}, x_{2}, \cdots, x_{25}$ must be minimized, thus we have
$$
\begin{array}{l}
x_{26}+x_{27}+\cdots+x_{51} \\
=1995-(1+2+\cdots+25)=1670 .
\end{array}
$$
To maximize $x_{51}$, then $x_{26}, x_{27}, \cdots, x_{50}$ must be minimized, thus we have
$$
\begin{array}{l}
x_{26}+\left(x_{26}... | 95 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,592 |
Three, (Full marks 30 points) In $\triangle A B C$, $A B=A C$, the altitude $A D=5$ on $B C$, $M$ is a point on $A D$, $M D=1$, and $\angle B M C=3 \angle B A C$. Try to find the perimeter of $\triangle A B C$. | Three, draw $M E$ $/ / A B$ intersecting $B C$ at $E$, and draw $M F / / A C$ intersecting $B C$ at $F$. Let $E D=t, M E=$ $s$.
$$
\because \triangle B D A \backsim
$$
$\triangle E D M, \therefore \frac{B D}{E D}=\frac{A D}{M D}=\frac{5}{1}, \therefore B D=5 t$.
Also, $M E$ is the angle bisector of $\angle B M F$,
$$
\... | \frac{10 \sqrt{7}}{7}(1+2 \sqrt{2}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,594 |
Four, (Full marks 30 points) Find the natural number $n$, such that $S_{n}=9$ $+17+25+\cdots+(8 n+1)=4 n^{2}+5 n$ is a perfect square.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text { Four, } \because S_{n}=4 n^{2}+5 n \\
=(2 n+1)^{2}+(n-1), \\
\therefore S_{n} \geqslant(2 n+1)^{2} .
\end{array}
$$
Also, $(2 n+2)^{2}=4 n^{2}+8 n+4>S_{n}$,
From $(2 n+1)^{2} \leqslant S_{n}<(2 n+2)^{2}$, we know $S_{n}=(2 n+1)^{2}$.
$\therefore n-1=0$, i.e., $n=1$. | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,595 |
Example 9. If the inequality $x^{2}-2 a x+a^{2}+b \geqslant 0$ holds for any real number, then the range of values for $b$ is $\qquad$
(1987, Baoji City Junior High School Mathematics Competition) | Solve: Since the coefficient of the quadratic term is greater than zero, to make the solution set of the inequality the set of all real numbers, it is only necessary that
$$
\Delta=4 a^{2}-4\left(a^{2}+b\right) \leqslant 0 .
$$
Solving for the range of $b$ yields $b \geqslant 0$. | b \geqslant 0 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 708,597 |
3. If in $\triangle A B C$, the angle bisector of $\angle B A C$ intersects $B C$ at $D, A C=A B+B D, \angle C=30^{\circ}$, then the degree measure of $\angle B$ is ( ).
(A) $45^{\circ}$
(B) $60^{\circ}$
(C) $75^{\circ}$
(D) $90^{\circ}$ | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,599 |
4. Among the following four propositions
(1) If the lengths of two sides of a right triangle are 3 and 4, then the length of the third side is 5;
(2) $(\sqrt{a})^{2}=a$;
(3) If point $P(a, b)$ is in the third quadrant, then point $P^{\prime}(-a,-b+1)$ is in the first quadrant;
(4) A quadrilateral with equal and perpend... | 4. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,600 |
5. As shown in the figure, 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 value of $B E$ is ( ).
(A) 10
(B) 11
(C) 12
(D) 15 | 5. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,601 |
6. Let $x_{1}, x_{2}$ be the two real roots of the equation $x^{2}-2(k+1) x+\left(k^{2}+2\right)=0$, and $\left(x_{1}+1\right)\left(x_{2}+1\right)=8$. Then the value of $k$ is ( ).
(A) -3 or 1
(B) -3
(C) 1
(D) all real numbers not less than $\frac{1}{2}$ | 6. C.
The above text has been translated into English, preserving the original text's line breaks and formatting. Here is the direct output of the translation result. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,602 |
8. Let the three equations $x^{2}+4 m x+4 m^{2}+2 m+3=0, x^{2}+$ $(2 m+1) x+m^{2}=0,(m-1) x^{2}+2 m x+m-1=0$ have at least one real root. Then the range of values for $m$ is ( ).
(A) $-\frac{3}{2}<m<-\frac{1}{4}$
(B) $m \leqslant-\frac{3}{2}$ or $m \geqslant-\frac{1}{4}$
(C) $m \leqslant-\frac{3}{2}$ or $m \geqslant \f... | 8. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,604 |
3. There is a four-digit number. Dividing it
in the middle into two parts, we get the front and back two $\uparrow=$ digit numbers. By adding a 0 to the end of the front two-digit number, and then adding the product of the front and back two-digit numbers, we get exactly the original four-digit number. It is also know... | 3. 1995 | 1995 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,607 |
Example 10. Let real numbers $x, y$ satisfy the equation $x^{3}+y^{3}=a^{3}(a>0)$. Find the range of values for $x+y$:
(1992, Taiyuan City Junior High School Mathematics Competition) | Let $x+y=t$. Then
$$
\begin{aligned}
x^{3}+y^{3} & =(x+y)^{3}-3(x+y) x y \\
& =t^{3}-3 t x y=a^{3} . \\
\therefore \quad x y & =\frac{t^{3}-a^{3}}{3 t} .
\end{aligned}
$$
From this, we know that $x, y$ are the two real roots of the equation about $z$
$$
z^{2}-t z+\frac{t^{3}-a^{3}}{3 t}=0
$$
By the discriminant theor... | 0<x+y \leqslant \sqrt[3]{4} a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,608 |
4. A circle with a radius of $1 \mathrm{~cm}$, moving arbitrarily within a regular hexagon with a side length of $6 \mathrm{~cm}$ (the circle can be tangent to the sides of the hexagon), the area that the circle cannot reach within the hexagon is $\qquad$ $\mathrm{cm}^{2}$. | 4. $2 \sqrt{3}-\pi\left(\mathrm{cm}^{2}\right)$ | 2 \sqrt{3}-\pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,609 |
Three, (Full marks 20 points) As shown in the figure, $A B, B C, C D$ are tangent to the circle at $E, F, G$ respectively, and $A B = B C = C D$. Connect $A C$ and $B D$ intersecting at point $P$, and connect $P F$. Prove: $P F \perp B C$.
---
The translation maintains the original text's formatting and line breaks. | Three, let the center of the circle be $O$, and connect $B O, C O$. According to the tangent length theorem, $B O, C O$ are the angle bisectors of the vertex angles of isosceles $\triangle A B C$ and isosceles $\triangle D C B$, respectively, so $B O \perp A C, C O \perp$ $B D$.
$\therefore O$ is the orthocenter of $\t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,610 |
Four, (Full marks 20 points, Sub-question (1) 6 points, Sub-question (2) 14 points) Let the two real roots of $x^{2}-p x+q=0$ be $\alpha, \beta$.
(1) Find the quadratic equation whose roots are $\alpha^{3}, \beta^{3}$;
(2) If the quadratic equation whose roots are $\alpha^{3}, \beta^{3}$ is still $x^{2}-p x$ $+q=0$, fi... | $\therefore$ The required equation is $x^{2}-p\left(p^{2}-3 q\right) x+q^{3}=0$.
(2) From the given and the derived results, we have $\left\{\begin{array}{l}p\left(p^{2}-3 q\right)=p, \\ q^{3}=q \Rightarrow q=0,1,-1 \text {. }\end{array}\right.$
Then
\begin{tabular}{c|c|c|c}
$p$ & $0,1,-1$ & $0,2,-2$ & 0 \\
\hline$q$ ... | x^{2}=0, x^{2}-x=0, x^{2}+x=0, x^{2}-2 x+1=0, x^{2}+2 x+1=0, x^{2}-1=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,611 |
Five. (Full marks 20 points, Question (1) 8 points, Question (2) 12 points) As shown in the figure, there is a cube-shaped wire frame, and the midpoints of its sides $I, J, K, L$ are also connected with wire.
(1) There is an ant that wants to crawl along the wire from point A to point G. How many shortest routes are th... | (1) "The shortest path" means that the ant can neither move left nor down, otherwise it would be taking a "detour" rather than the "shortest" path. The ant can crawl on face $ABCGFE$ or on $ADCGHE$. The number of "shortest paths" on these two faces is the same, and the possible paths can be represented by the following... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,612 |
$1.5^{1000}$ 's last three digits are ( ).
(A) 025
(B) 125
(C) 625
(D) 825 | -、1. B.
If $n \geqslant 3$, then when $n$ is odd, the last three digits of $5^{n}$ are 125. Hint: Discuss the cases when $n \geqslant 3$, $n$ is odd, and $n$ is even. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,613 |
3. If $x^{3}+x^{2}+x+1=0$, then the value of $y=x^{97}+x^{96}+\cdots$ $+x^{103}$ is ( ).
(A) -1
(B) 0
(C) 1
(Dic | 3. A.
From the given, we have $x^{4}=1$. Therefore, $y=x^{97}+x^{98}+\cdots+x^{103}$
$$
\begin{array}{l}
=x^{97}\left(1+x+x^{2}+x^{3}\right)+x^{100}\left(x+x^{2}+x^{3}\right)=x^{97} \times 0+ \\
\left(x^{4}\right)^{25}(-1)=-1
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,615 |
4. In $\triangle A B C$, among the following conditions: (1) two medians are equal, (2) two altitudes are equal; (3) $\cos A=\cos B$; (4) $\operatorname{tg} A=\operatorname{tg} B$. The number of conditions that can lead to $\triangle A B C$ being an isosceles triangle is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 4. D.
(1) If $A D, C E$
are two medians intersecting at $M$, and
$A D=C E$, then $A M$
$$
\begin{array}{l}
=\frac{2}{3} A D=\frac{2}{3} C E \\
=C M, M D=\frac{1}{3} A D \\
=\frac{1}{3} C E=M E, \angle A M E=\angle C M D,
\end{array}
$$
thus $\triangle A M E \cong \triangle C M D \Rightarrow A E=C D \Rightarrow A B=C ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,616 |
5. In trapezoid $ABCD$, $AB \parallel CD, AB=3CD, E$ is the midpoint of diagonal $AC$, and line $BE$ intersects $AD$ at $F$. Then the value of $AF: FD$ is ( ).
(A) 2
(B) $\frac{5}{2}$
(C) $\frac{3}{2}$
(D) 1 | 5. C.
Draw $C G / / B F$ intersecting the extension of $A D$ at $G$ (as shown in the figure). From $A B$ $/ / C D, C G / / B F$, we get $\triangle A B F \backsim$ $\triangle D C G$, thus $A F: D G=A B$ ～ $D C=3: 1$. Also, by $C G / / B F$, $F D=\varepsilon D C$, hence $A F: F l=-32=\frac{3}{2}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,617 |
6. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$, respectively. Given that $a^{2}=b(b+c)$ and $\angle C$ is an obtuse angle, the size relationship of $a, 2 b, c$ is ( ).
(A) $a<2 b<c$
(B) $a<c<2 b$
(C) $2 b<a<c$
(D) $a=2 b<c$ | 6. A.
$$
\because a+b>c, \therefore a^{2}=b(b+c)<b(2b+a),(a-2b) \text {. }
$$
$(a+b)<0, \therefore a<2b$. Also $\because \angle C$ is an obtuse angle, $\therefore a^{2}+b^{2}<c^{2}$, i.e., $b(b+c)+b^{2}<c^{2},(2b-c)(b+c)<0, \therefore 2b<c$, then $a$ $<2b<c$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,618 |
Example 1. Given in a convex pentagon $A B C D E$,
$$
\begin{array}{l}
\angle B A E=3 \alpha, B C=C D \\
=D E, \text { and } \angle B C D= \\
\angle C D E=180^{\circ}-2 \alpha .
\end{array}
$$
Prove: $\angle B A C=\angle C A D=\angle D A E$. (1990, National Junior High School Competition) | Prove connect $B D$,
$CE$.
$$
\begin{array}{c}
\because B C=C D=D E, \\
\angle B C D=\angle C D E \\
=180^{\circ}-2 \alpha,
\end{array}
$$
$\therefore \angle E C D \cong \triangle \triangle C D E$.
$$
\text { Hence } \begin{aligned}
\angle C B D & =\angle C D B=\angle D C E \\
& =\angle D E C=\alpha .
\end{aligned}
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,619 |
Example 2. As shown in the figure, $O$ is a point inside the convex pentagon $A B C D E$, and $\angle 1=\angle 2$, $\angle 3=\angle 4$, $\angle 5=\angle 6$, $\angle 7=\angle 8$. Prove that $\angle 9$ and $\angle 10$ are either equal or supplementary. (1985, National Junior High School Competition) | Prove that the locus of the vertex of an angle that subtends a fixed line segment at a fixed angle is the arc of a segment with the fixed line segment as the chord and the angle equal to the fixed angle. Therefore, from $\angle 1=\angle 2$, the circumcircles of $\triangle O A B$ and $\triangle O B C$ are equal.
Simila... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,620 |
Example 11. As shown in the figure, $\triangle P Q R$ and $\triangle P^{\prime} Q^{\prime} R^{\prime}$ are two congruent equilateral triangles. Let
the side lengths of hexagon $A B C D E F$ be
$A B=a_{1}, B C$
$$
\begin{array}{l}
=b_{1}, C D=a_{2}, D E=b_{2}, \\
E F=a_{3}, F A=b_{3} . \text { Prove that: } \\
\quad a_{... | Prove that $\triangle P A B \backsim \triangle Q^{\prime} C B \sim \triangle Q C D$ $\sim \triangle R^{\prime} E D \backsim \triangle R E F \sim \triangle P^{\prime} A F$.
Sequentially denote the areas of the above six triangles as $S_{1}$, $S^{\prime}{ }_{1}$, $S_{2}$, $S^{\prime}{ }_{2}$, $S_{3}$, $S^{\prime}{ }_{3}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,621 |
Example 12. Polygon $A_{1} A_{2} \cdots A_{n}$ contains a polygon $B_{1} B_{2} \cdots B_{n}$, with corresponding sides of these two polygons being parallel, and the distance between each pair of parallel sides being 1. The perimeter of $B_{1} B_{2} \cdots B_{n}$ is $p$, and its area is $S_{b}$, while the area of $A_{1}... | Prove that by drawing arcs with $B_{1}, B_{2}, \cdots, B_{n}$ as centers and 1 as the radius, which are tangent to the polygon $A_{1} A_{2} \cdots A_{n}$, the points of tangency are denoted as $M_{i}, N_{i}(i=1,2, \cdots, n)$. Connect $B_{i} M_{i}$, $B_{i} N_{i}$.
Then $\angle M_{1} B_{1} N_{1}$ is supplementary to $\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,622 |
Example 13. If the diagonals $A D$, $B E$, $C F$ of a convex hexagon $A B C D E F$ all bisect the area of the hexagon. Prove: $A D$, $B E$, $C F$ are concurrent.
$(1965$, Polish Mathematical Olympiad) | Prove that $S_{A B C D}=\frac{1}{2} S_{A B C D E F}=S_{B C D E}$,
and $S_{A B C D}=S_{\triangle A B D}+S_{\triangle D B C}$,
and $S_{B C D E}=S_{\triangle E B D}+S_{\triangle D B C}$,
thus $S_{\triangle A B D}=S_{\triangle E B D}$.
Since they share the same base, then $A E / / B D$.
Similarly, we can prove that $A C / ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,623 |
Example 14. If $A C, C E$ are two diagonals of a regular hexagon $A B C D E F$, and points $M, N$ internally divide $A C, C E$ such that $A M: A C=C N: C E=r$. If $B, M, N$ are collinear, find $r$. (23rd IMO Problem) | Let $A C, B E$ intersect at $K$. By the collinearity of $B, M, N$ and Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{C N}{N E} \cdot \frac{E B}{B K} \cdot \frac{K M}{M C} \\
=1 .
\end{array}
$$
Assume the side length of the regular hexagon is 1. Then
$$
A C=C E=\sqrt{3} \text {. }
$$
And $\frac{C N}{N E}=\frac{... | r=\frac{1}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,624 |
Example 15. Each diagonal of a convex pentagon is parallel to one of its sides. Prove that the ratio of each diagonal to the corresponding side is $\frac{\sqrt{5}+1}{2}$.
(45th Moscow Mathematical Olympiad) | Prove that from the given, quadrilaterals $E D C D^{\prime}$ and $D C B C^{\prime}$ are parallelograms (see figure).
$$
\begin{array}{l}
\therefore E C^{\prime}=E B-C^{\prime} B=E B-D C \\
=E B-E D^{\prime}=D^{\prime} B,
\end{array}
$$
Also, from $\triangle E C^{\prime} B^{\prime} \backsim \triangle A C^{\prime} B, \t... | \frac{\sqrt{5}+1}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 708,625 |
Example 1. Let $a, b, c \in R^{+}$. Prove:
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} \text {. }
$$
(1963, Moscow Mathematical Olympiad) | Proof: Let $x=b+c, y=c+a, z=a+b$.
Then
$$
\begin{aligned}
a= & \frac{1}{2}(-x+y+z), b=\frac{1}{2}(x-y+z), \\
c= & \frac{1}{2}(x+y-z) . \\
\therefore & \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \\
= & \frac{1}{2}\left(\frac{-x+y+z}{x}+\frac{x-y+z}{y}\right. \\
& \left.+\frac{x+y-z}{z}\right) \\
= & \frac{1}{2}\left(\fra... | \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 708,626 |
Example 2. Given $a, b, c \in R^{+}$, and $\frac{a^{2}}{1+a^{2}}+\frac{b^{2}}{1+b^{2}}$ $+\frac{c^{2}}{1+c^{2}}=1$. Prove: $a b c \leqslant \frac{\sqrt{2}}{4}$. | Proof: Let $x=\frac{a^{2}}{1+a^{2}}, y=\frac{b^{2}}{1+b^{2}}$,
$$
z=\frac{c^{2}}{1+c^{2}} \text {, then } 0<x, y, z<1, x+y+z=
$$
1, and
$$
a=\sqrt{\frac{x}{1-x}}, b=\sqrt{\frac{y}{1-y}}, c=\sqrt{\frac{z}{1-z}} .
$$
Thus, $a b c \leqslant \frac{\sqrt{2}}{4}$
$$
\begin{array}{l}
\Leftrightarrow \sqrt{\frac{x y z}{(1-x)(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,627 |
Example 3. In $\triangle A B C$, prove that: $\cos A+\cos B+\cos C \leqslant \frac{3}{2}$. (1979, Shandong Province Mathematics Competition) | $$
\begin{array}{l}
\text { Prove that } t=\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\
=\frac{1}{2}\left(\cos \frac{B-C}{2}-\cos \frac{B+C}{2}\right) \sin \frac{A}{2} \\
=\frac{1}{2}\left(\cos \frac{B-C}{2}-\sin \frac{A}{2}\right) \sin \frac{A}{2},
\end{array}
$$
we get a quadratic equation in $\sin \frac{A}... | \cos A+\cos B+\cos C \leqslant \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 708,628 |
Example 4. Given $x^{2}+y^{2}=1$. Prove: $\left|x^{2}+2 x y-y^{2}\right| \leqslant \sqrt{2}$.
(1979, Beijing Mathematical Competition) | Proof: Let $x=r \cos \theta, y=r \sin \theta$, where $0 \leq r \leq 1$. Then we have
$$
\begin{array}{l}
\left|x^{2}+2 x y-y^{2}\right| \\
=\left|r^{2} \cos ^{2} \theta+2 r^{2} \sin \theta \cos \theta-r^{2} \sin ^{2} \theta\right| \\
=r^{2}|\cos 2 \theta+\sin 2 \theta| \\
=\sqrt{2} r^{2}\left|\sin \left(2 \theta+\frac{... | \sqrt{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 708,629 |
Example 5. Given $x, y, z \in R^{+}$, and $x+y+z=1$. Prove: $\frac{1}{x}+\frac{4}{y}+\frac{9}{z} \geqslant 36$.
(1990, Japan IMO Team Selection Test) | Prove that for $x=\sin ^{2} \alpha \cos ^{2} \beta, y=\cos ^{2} \alpha \cos ^{2} \beta$, $z=\sin ^{2} \beta$, where $\alpha, \beta$ are acute angles, we have
$$
\begin{array}{l}
\frac{1}{x}+\frac{4}{y}+\frac{9}{z} \\
=\left(1+\operatorname{ctg}^{2} \alpha\right)\left(1+\operatorname{tg}^{2} \beta\right)+4\left(1+\opera... | 36 | Inequalities | proof | Yes | Yes | cn_contest | false | 708,630 |
Example 3. In the equilateral convex hexagon $A B C D E F$, $\angle A+\angle C+\angle E=\angle B+\angle D+\angle F$. Prove: $\angle A=\angle D, \angle B=\angle E, \angle C=\angle F$. (1953, Hungarian Mathematical Olympiad) | Connect $B D$, $D F, F B$. According to the problem, we have
$$
\angle A+\angle C+\angle E=360^{\circ} \text {. }
$$
Also, $A B=B C=\cdots=F A$, thus
$\triangle A B F, \triangle B C D, \triangle D E F$ can form a hexagon.
And it is congruent to $\triangle S D F$ (proof omitted).
From this, we can get
$C D / / B O / / ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,631 |
Example 7. Given positive numbers $m_{i} \in \mathbb{R}^{+}(i=1,2, \cdots, n)$, $p \geqslant 2$, and $p \in \mathbb{N}$, and satisfying
$$
\frac{1}{1+m_{1}^{p}}+\frac{1}{m_{2}^{\rho}}+\cdots+\frac{1}{m_{n}^{\rho}}=1 \text {. }
$$
Prove: $m_{1} m_{2} \cdots m_{n} \geqslant(n-1) \frac{n}{p}$.
(First National Mathematica... | Proof: Let $\frac{1}{1+m_{i}^{p}}=\frac{\alpha_{i}}{\sum_{i=1}^{n} \alpha_{i}}, a_{i} \in R^{+}$. Then
$$
m_{i}^{p}=\frac{\sum_{i=1}^{n} \alpha_{i}-\alpha_{i}}{\alpha_{i}} \geqslant(n-1) \sqrt[n-1]{\frac{\prod_{i=1}^{n} \alpha_{i}}{\alpha_{i}}} \cdot \frac{1}{\alpha_{i}} .
$$
Thus, $\left(\prod_{i=1}^{n} m_{i}\right)^... | m_{1} m_{2} \cdots m_{n} \geqslant(n-1)^{\frac{n}{p}} | Inequalities | proof | Yes | Yes | cn_contest | false | 708,633 |
Example 9. Let $a, b, c$ be positive real numbers, and satisfy $abc = 1$. Prove:
$$
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} \text {. }
$$
(36th IMO Problem) | Proof $\quad$ Let $2 a^{-1}=b^{-1}+c^{-1}+\alpha, 2 b^{-1}=c^{-1}+a^{-1}+\beta, 2 c^{-1}=a^{-1}+b^{-1}+\gamma$, then $a+\beta+\gamma=0$. Therefore,
$$
\begin{array}{l}
\frac{4}{a^{3}(b+c)}+\frac{4}{b^{3}(c+a)}+\frac{4}{c^{3}(a+b)} \\
=\frac{4 a b c}{a^{3}(b+c)}+\frac{4 a b c}{b^{3}(c+a)}+\frac{4 a b c}{c^{3}(a+b)} \\
=... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,635 |
Example 10. Given that $x, y, z$ are non-negative real numbers, and $x+y+z=1$. Prove:
$$
0 \leqslant x y+y z+z x-2 x y z \leqslant \frac{7}{27} \text {. }
$$
(25th IMO Problem) | Assume without loss of generality that $x \geqslant y \geqslant z \geqslant 0$. From $x+y+z=1$, it is easy to see that $z \leqslant \frac{1}{3}, x+y \geqslant \frac{2}{3}$. Therefore,
$$
\begin{array}{l}
2 x y z \leqslant \frac{2}{3} x y \leqslant x y . \\
\therefore \quad y z+z x+x y-2 x y z \geqslant 0 .
\end{array}
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,636 |
Example 11. Given $a+b+c+d+e=8, a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16$. Prove that $0 \leqslant e \leqslant \frac{16}{5}$. | Proof: Let $a=2-\frac{e}{4}+t_{1}, b=2-\frac{e}{4}+t_{2}$, $c=2-\frac{e}{4}+t_{3}, d=2-\frac{e}{4}+t_{4}$, then $\sum_{i=1}^{4} t_{i}=0$.
Thus, we have
$$
\begin{aligned}
16= & \sum_{i=1}^{4}\left[\left(2-\frac{e}{4}\right)+t_{i}\right]^{2}+e^{2} \\
= & 4\left(2-\frac{e}{4}\right)^{2}+2\left(2-\frac{e}{4}\right) \sum_{... | 0 \leqslant e \leqslant \frac{16}{5} | Algebra | proof | Yes | Yes | cn_contest | false | 708,637 |
Example 12. Given $x^{2}+y^{2}=1$. Prove:
$$
\sqrt{a^{2} x^{2}+b^{2} y^{2}}+\sqrt{a^{2} y^{2}+b^{2} x^{2}} \geqslant a+b .
$$ | Proof: Let $z_{1}=a x+b y i, z_{2}=b x+a y i$. Then
$$
\begin{array}{l}
\sqrt{a^{2} x^{2}+b^{2} y^{2}}+\sqrt{a^{2} y^{2}+b^{2} x^{2}} \\
=\left|z_{1}\right|+\left|z_{2}\right| \geqslant\left|z_{1}+z_{2}\right| \\
=|(a+b)(x+y i)|=|a+b| \geqslant a+b .
\end{array}
$$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,638 |
Example 13. Let $x, y, z$ be any real numbers. Prove:
$$
\begin{array}{l}
\sqrt{x^{2}+x y+y^{2}}+\sqrt{x^{2}+x z+z^{2}} \\
\geqslant \sqrt{y^{2}+y z+z^{2}} .
\end{array}
$$ | Prove that in the plane, establish a coordinate system $x o y$, and take $\exists$ point $A(x, 0), B\left(-\frac{y}{2},-\frac{\sqrt{3}}{2} y\right) \cdot C\left(-\frac{z}{2}\right.$, $\left.\frac{\sqrt{3}}{2} z\right)$, then
$$
\begin{array}{l}
|A B|=\sqrt{\left(x+\frac{y}{2}\right)^{2}+\left(0+\frac{\sqrt{3}}{2} y\rig... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,639 |
Example 14. Let positive real numbers $a, b, c$ satisfy $a+b+c=1$. Prove that: $3-\sqrt{3}<\sqrt{1-3 a^{2}}+\sqrt{1-3 b^{2}}+$ $\sqrt{1-3 x^{2}} \leqslant \sqrt{6}$. | Proof of the figure, let
$$
\begin{array}{l}
A F=\sqrt{3} a, E D= \\
\sqrt{3} b, G F=\sqrt{3} c, \text { and } \\
A D=D F=B F=1 . \text { Then } \\
\quad D H=\sqrt{1-3 a^{2}}, \\
\therefore B^{2}=\sqrt{1-3 b^{2}}, B G= \\
\sqrt{1-3 c^{2}} .
\end{array}
$$
$$
\text { From } D H>A D-A H, E F>D F-D E, B G
$$
$>B F-K G$, a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,640 |
Example 15. Let $a, b, c$ be the lengths of the sides of a triangle. Prove that: $a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \geqslant 0$.
(24th IMO Problem) | Proof: Let $a=y+z, b=z+x, c=x+y$ $\left(x, y, z \in R^{+}\right)$. Then the original inequality is equivalent to
$$
\begin{array}{l}
(y+z)^{2}(z+x)(y-x) \\
+(z+x)^{2}(x+y)(z-y) \\
+(x+y)^{2}(y+z)(x-z) \geqslant 0 \\
\Leftrightarrow x y^{3}+y z^{3}+z x^{3}-x y z(x+y+z) \geqslant 0 \\
\Leftrightarrow x z(x-y)^{2}+x y(y-z... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,641 |
Example 4. In a convex pentagon with equal sides, there exists a point on the longest diagonal such that the angles subtended by the other vertices do not exceed $90^{\circ}$.
（3rd All-Russian Mathematical Olympiad） | Proof Let $K$ be the midpoint of the longest diagonal $AD$ of pentagon $ABCDE$.
From $AE=ED$, we know
$EK \perp AD$.
Also, since $AC \leqslant AD$, we have $\angle BAC > \angle DAE$, and thus $\angle BAK > \angle KAE$.
This implies that $A, B$ are on the same side of line $EK$.
$\therefore$ Points $C, D$ are also on t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,642 |
Theorem The necessary and sufficient condition for the area of a right triangle $ABC$ to be equal to its perimeter is that there exists a positive number $m$, such that
$$
\left\{\begin{array}{l}
a=4+m, \\
b=4+\frac{8}{m}, \\
c=4+m+\frac{8}{m} .
\end{array}\right.
$$ | Prove that for a right-angled triangle, the inradius \( r = \frac{ab}{a + b + c} \), the necessary and sufficient condition for the area to equal the semiperimeter is that the inradius is 2.
$$
\left\{\begin{array}{l}
a = 2\left(\operatorname{ctg} \frac{B}{2} + \operatorname{ctg} \frac{C}{2}\right), \\
b = 2\left(\oper... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,644 |
3. Let $S=\{0,1,2,3, \cdots\}$ be the set of all non-negative integers. Find all functions $f$ defined on $S$ and taking values in $S$ that satisfy the following condition: $f(m+f(n))=f(f(m))+f(n)$ for all $m$, $n \in S$. | Let $f: S \rightarrow S$ satisfy
$f(m+f(n))=f(f(m))+f(n)$ (for any $m, n \in S$).
Taking $m=n=0$, we get $f(f(0))=f(f(0))+f(0)$ (1).
Therefore, $f(0)=0$.
Also, taking $m=0$ in (1), we have
$f(f(n))=f(n)$ (for any $n \in S$).
Thus, (1) becomes $f(m+f(n))=f(m)+f(n)$ (for any $m, n \in S$).
Let the range of $f$ be $T(T=... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,646 |
4. Let positive integers $a, b$ make $15a + 16b$ and $16a - 15b$ both squares of positive integers. Find the smallest value that the smaller of these two squares can take. | Let positive integers $a, b$ be such that $15a + 16b$ and $16a - 15b$ are both squares of positive integers, i.e.,
$$
15a + 16b = r^2, \quad 16a - 15b = s^2 \quad (r, s \in \mathbb{N}).
$$
$$
\text{Thus, } 15^2 a + 16^2 a = 15r^2 + 16s^2,
$$
i.e., $481a = 15r^2 + 16s^2$;
$$
16^2 b + 15^2 b = 16r^2 - 15s^2,
$$
i.e., $... | 481^2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,647 |
6. Let $n, p, q$ be positive integers such that $n > p + q$. If $x_{0}, x_{1}, \cdots, x_{n}$ are integers satisfying the following conditions:
(a) $x_{0} = x_{n} = 0$;
(b) For each integer $i (1 \leqslant i \leqslant n)$, either $x_{i} - x_{i-1} = p$ or $x_{i} - x_{i-1} = -q$.
Prove: There exists a pair of indices $(... | Proof: First, without loss of generality, let $(p, q)=1$. (When $(p, q)=d>1$, let $p=d p_{1}, q=d q_{1}$, then $\left(p_{1}, q_{1}\right)=1$. We only need to consider $\frac{x_{0}}{d}$, $\frac{x_{1}}{d}, \cdots, \frac{x_{n}}{d}$. The difference between any two consecutive terms is $p_{1}$ or $-q_{1}$, and $n>p_{1} + q_... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,649 |
2. (Zh-En) Let $a, b$ be non-negative integers, and satisfy $a b \geqslant c^{2}$, where $c$ is an integer. Prove: There exists a number $n$, and integers $x_{1}, x_{2}, \cdots, x_{n}$; $y_{1}, y_{2}, \cdots, y_{n}$, such that
$$
\sum_{i=1}^{n} x_{i}^{2}=a, \quad \sum_{i=1}^{n} y_{i}^{2}=b, \quad \sum_{i=1}^{n} x_{i} y... | Proof Let the above question be denoted as $(a, b, c)$. It is easy to see that the problem holds for $(a, b, c)$ if and only if it holds for $(a, b, -c)$, so we can assume $c \geqslant 0$. Since the problem is symmetric with respect to $a, b$, we can also assume $a \geqslant b$. Therefore, from $a b \geqslant c^{2}$, w... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,650 |
4. (USA) Let $a, b, c$ be given positive constants.
Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+z=a+b+c, \\
4 x y z-\left(a^{2} x+b^{2} y+c^{2} z\right)=a b c
\end{array}\right.
$$
for all positive real numbers $x, y, z$. | The second equation is equivalent to
$$
\begin{array}{l}
4=\frac{a^{2}}{y z}+\frac{b^{2}}{z x}+\frac{c^{2}}{x y}+\frac{a b c}{x y z} . \\
\text { Let } x_{1}=\frac{a}{\sqrt{y z}}, y_{1}=\frac{b}{\sqrt{z x}}, z_{1}= \\
4=x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+x_{1} y_{1} z_{1} .
\end{array}
$$
where, $0<x_{1}<2,0<y_{1}<2,0<z_{1... | (x, y, z)=\left(\frac{1}{2}(b+c), \frac{1}{2}(c+a), \frac{1}{2}(a+b)\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,652 |
Example 5. In a convex pentagon with equal sides, each interior angle is less than $120^{\circ}$. Prove that all its interior angles are obtuse.
(35th Moscow Mathematical Olympiad) | Proof by contradiction. Suppose $\angle B C D$ is not an obtuse angle. Connect $A C$, $C E$. From the isosceles $\triangle B C A$ and the isosceles $\triangle C D E$ with the vertex angles less than $120^{\circ}$, we know that their base angles are no less than $30^{\circ}$. Therefore, $\angle A C E<90^{\circ}-2 \cdot ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,653 |
5. (Ukraine) Let $R$ be the set of real numbers. Does there exist a function $f: R \rightarrow R$ that satisfies the following three conditions simultaneously?
(a) There exists a positive number $M$ such that for all $x$,
$$
-M \leqslant f(x) \leqslant M \text {; }
$$
(b) The value of $f(1)$ is 1,
(c) If $x \neq 0$, th... | The function $f$ that satisfies all the above conditions does not exist. Proof by contradiction. Otherwise, let $f: R \rightarrow R$ satisfy all the conditions, and let $c$ be a real number greater than any $f(x)$, and $c$ is the smallest integer multiple of $\frac{1}{4}$. It can be concluded that $c \geqslant 2$, beca... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,654 |
2. (Germany) Let points $A, B, C$ be non-collinear. Prove: there exists a unique point $X$ in the plane $ABC$ such that
$$
\begin{array}{l}
X A^{2}+X B^{2}+A B^{2}=X B^{2}+X C^{2}+B C^{2} \\
=X C^{2}+X A^{2}+C A^{2} .
\end{array}
$$ | Prove the construction of $\triangle A^{\prime} B^{\prime} C^{\prime}$, such that points $A, B, C$ are the midpoints of sides $B^{\prime} C^{\prime}, C^{\prime} A^{\prime}, A^{\prime} B^{\prime}$, respectively. From the conditions satisfied by $\triangle X A B$ and $\triangle X A C$, we have
$$
B X^{2}-C X^{2}=A C^{2}-... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,656 |
3. (Turkey)
In $\triangle A B C$, the incircle touches the sides
$B C, C A, A B$ at
points $D, E, F$, respectively. Point $X$
is a point on $\triangle A B C$, and the incircle of $\triangle X B C$
touches the side $B C$ and is tangent to $C X, X B$ at points $Y, Z$, respectively.
Prove: $E F Z Y$ is a cyclic quadrilate... | Prove that if $E F$ is parallel to $B C$, then $A B=A C, A D$ is the axis of symmetry of $E F Z Y$, and thus the quadrilateral is a cyclic quadrilateral.
If $E F$ is not parallel to $B C$, assume that the extension of $B C$ intersects the extension of $E F$ at $P$. By Menelaus' theorem, we have
$$
\frac{\overrightarro... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,657 |
4. (Ukraine) Given an acute triangle $\triangle A B C$, take points $A_{1}, A_{2}\left(A_{2}\right.$ is between $A_{1}$ and $C$) on side $B C$, points $B_{1}$, $B_{2}\left(B_{2}\right.$ is between $B_{1}$ and $A$) on side $A C$, and points $C_{1}, C_{2}$ ( $C_{2}$ is between $C_{1}$ and $B$) on side $A B$, such that $\... | Proof: Let the two triangles be $\triangle U V W$ and $\triangle X Y Z$ as shown in the figure. Since $\angle A B_{2} X = \angle A C_{1} U$, $\triangle A B_{2} B$ and $\triangle A C_{1} C$ are similar, thus $\frac{A C_{1}}{A C} = \frac{A B_{2}}{A B}$, and $\angle A B B_{2} = \angle A C C_{1}$. Similarly, we can get $\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,658 |
8. (Bolivia) Let $ABC$ be a triangle, and a circle passing through points $B, C$ intersects sides $AB, AC$ at $C', B'$ respectively. Prove: $BB', CC', HH' \equiv$ are concurrent, where $H$ and $H'$ are the orthocenters of $\triangle ABC$ and $\triangle AB'C'$ respectively. | Prove that from
$\angle A B^{\prime} C^{\prime}=\angle A B C$, we know $\triangle A B^{\prime} C^{\prime}$ and $\triangle A B C$ are similar triangles. Similarly, $\triangle H^{\prime} B^{\prime} C^{\prime}$ and $\triangle H B C$ are also similar (using the properties of the orthocenter and $\triangle A B C \sim \trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,660 |
2. Find all real solutions of the system
$$
\left\{\begin{array}{l}
\frac{4 x^{2}}{1+4 x^{2}}=y, \\
\frac{4 y^{2}}{1+4 y^{2}}=z, \\
\frac{4 z^{2}}{1+4 z^{2}}=x
\end{array}\right.
$$
and prove that your solution is correct. | \%. Let $x=y$. Then $x=y=z$.
From $4 x^{2}=x+4 x^{3}$, we get $x=0$ or $\frac{1}{2}$.
Let $x>y$. Then
$1-\frac{1}{1+4 x^{2}}>1-\frac{1}{1+4 x^{2}}$ or $z>x$,
and $1-\frac{1}{1+4 y^{2}}>1-\frac{1}{1+4 z^{2}}$ or $y>z$.
But $x>y>z>x$ is impossible.
When $x<y$, a similar contradictory statement is obtained.
Therefore, the... | (x, y, z)=(0,0,0) \text{ and } (x, y, z)=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,662 |
3. Let $a_{1}, a_{2}, \cdots, a_{n}$ represent any permutation of the integers $1,2, \cdots, n$. Let $f(n)$ be the number of such permutations such that
(i) $a_{1}=1$;
(ii) $\left|a_{i}-a_{i+1}\right| \leqslant 2 . i=1,2, \cdots, n-1$.
Determine whether $f(1996)$ is divisible by 3. | 3. Verify that $f(1)=f(2)=1$ and $f(3)=2$.
Let $n \geqslant 4$. Then it must be that $a_{1}=1, a_{2}=2$ or 3.
For $a_{2}=2$, the number of permutations is $f(n-1)$, because by deleting the first term and reducing all subsequent terms by 1, we can establish a one-to-one correspondence of sequences.
If $a_{2}=3$, then ... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,663 |
$1.1996^{1996}$ The tens digit is ( ).
$\begin{array}{llll}\text { (A) } 1 & \text { (B) } 3 & \text { (C) } 5 & \text { (D) } 9\end{array}$ | -1. (D).
Obviously, we only need to find the tens digit of $96^{1096}$. In the following discussion, $m \equiv n(\bmod k)$ means that $m$ and $n$ have the same remainder when divided by $k$ (referred to as “$m$ is congruent to $n$ modulo $k$”). It is easy to see that,
$$
\begin{array}{l}
96^{2} \equiv 16(\bmod 100) . \... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,666 |
2. The three altitudes of a triangle are $2.4, 3, 4$. This triangle is ( ).
(A) acute triangle
(1)）right triangle
(C) obtuse triangle
(i) cannot be determined | 2. (B).
Let the area of this triangle be $S$. Then its three sides are $\frac{2 S}{2.4}$, $\frac{2 S}{3}$, $\frac{2 S}{4}$, where $\frac{2 S}{2.4}$ is the longest side. Let the largest angle opposite to this side be $A$. By the cosine rule, we get
$$
\begin{array}{l}
\cos A=\frac{\left(\frac{2 S}{3}\right)^{2}+\left(\... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,667 |
3. Given that $a, \dot{c}, c$ are not all equal real numbers. Then, the equation about $x$: $x^{2}+(a+b+c) x+\left(a^{2}+b^{2}+c^{2}\right)=0(\quad)$.
(A) has two negative roots
(B) has two positive roots
(C) has two real roots with the same sign
(D) has no real roots. | 3. (D).
$$
\begin{aligned}
\Delta= & (a+b+c)^{2}-4\left(a^{2}+b^{2}+c^{2}\right) \\
= & a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c \\
& -4\left(a^{2}+b^{2}+c^{2}\right) \\
= & -(a-b)^{2}-(b-c)^{2}-(c-a)^{2} \\
& -\left(a^{2}+b^{2}+c^{2}\right) \\
& <0,
\end{aligned}
$$
$\therefore$ the original equation has no real roots. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,668 |
4. As shown in Figure 1, in quadrilateral $A B C D$, $\angle A C D=$ $20^{\circ}, \angle B D C=25^{\circ}, A C$ $=10, B D=8$. Then, the area of quadrilateral $A B C D$ is
(A) $40 \sqrt{2}$
(B) $20 \sqrt{2}$
(C) $18 \sqrt{2}$
(D) Uncertain | 4. (B).
$$
\begin{aligned}
S= & \frac{1}{2} O C \cdot O D \cdot \sin \angle C O D+\frac{1}{2} O D \cdot O A \\
& \cdot \sin \angle D O A+\frac{1}{2} O A \cdot O B \cdot \sin \angle C O D \\
& +\frac{1}{2} O B \cdot O C \cdot \sin \angle D O A \\
= & \frac{1}{2} \sin 45^{\circ} \cdot(O C+O A)(O B+O D) \\
= & \frac{1}{4}... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,669 |
5. Among the 100 integers $\{1,2, \cdots, 100\}$, if $k$ numbers are chosen, such that among these $k$ numbers, there are always two numbers whose sum equals the sum of two other different numbers. Then, the smallest value of $k$ that satisfies this condition is ( ).
(A) 21
(B) 24
(C) 27
(D) 30 | 5. (A).
Among the first 100 natural numbers, the smallest sum of any two different numbers is 3, and the largest is $19 \%$, making 197 different values.
If we take $k$ numbers, each number can form a different sum with the other $k-1$ numbers. $k$ numbers can form $\frac{1}{2} k(k-1)$ sums (these sums are not necess... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,670 |
1. If $360^{x}=3, 360^{y}=5$, then $72^{\frac{1-2 x-y}{3(1-y)}}=$ | \begin{array}{l}=1.2. \\ \because 72=\frac{360}{5}=\frac{360}{360^{y}}=360^{1-y}, \\ \therefore \text { original expression }=\left(360^{1-y}\right)^{\frac{1}{3(2 x-y)}} \\ =(360)^{\frac{1}{3}(1-2 x-y)}-\left(360^{1-2 x-y}\right)^{\frac{1}{3}} \\ =\left(\frac{360}{\left(360^{x}\right)^{2} \cdot 360^{5}}\right)^{\frac{1... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,672 |
2. If $x=\frac{1}{2}-\frac{1}{4 x}$, then $1-2 x+2^{2} x^{2}-2^{3} x^{3}+2^{4} x^{4}$ $-\cdots-2^{1995} x^{1995}$ is. $\qquad$. | 2. 1 .
From $x=\frac{1}{2}-\frac{1}{4 x}$ we can get $1-2 x+(2 x)^{2}=0$.
$$
\begin{array}{c}
\therefore \text { the original expression }=1-2 x\left[1-2 x+(2 x)^{2}\right]+(2 x)^{4}[1-2 x \\
\left.+(2 x)^{2}\right]-\cdots-(2 x)^{1993}\left[1-2 x+(2 x)^{2}\right]=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,673 |
3. If $B$ is a point on the diameter $AC$ of a circle with radius 3 and $\angle ABC = 90^\circ$, $BC = 2$, and circles $\odot A, \odot C, \odot P$ are constructed with $AB, BC$ as radii, respectively, such that $\odot P$ is tangent to $\odot A, \odot C, \odot O$, then the radius $r$ of $\odot P$ is $\qquad$ | 3. $\frac{2}{3}$.
As shown in Figure 6, connect $P A$, $P O$, $P B$, we get $P A=4+r$. $O P=3-r$, $P B=2+r$. Let $\angle A O P=\alpha, \angle B O P=\beta$, then
$$
\begin{array}{l}
\alpha+\beta=180^{\circ}, \\
\cos \alpha=-\cos \beta .
\end{array}
$$
By the Law of Cosines, we have
$$
\begin{array}{l}
\frac{3^{2}+(3-r... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,674 |
Example 7. (1) In a convex hexagon $A B C D E F$, all its interior angles are equal. Prove:
$$
A B-D E=E F-B C=C D-F A .
$$
(2) Conversely, if the lengths of the segments $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ satisfy $a_{1}-a_{4}=a_{5}-a_{2}=a_{3}-a_{6}$, then these segments can form a convex hexagon with all inte... | Proof (1) From the six interior angles of the hexagon all being $120^{\circ}$, we know that their opposite sides are respectively parallel. Let $A B \geqslant D E$. Construct $\square A B C K, \square C D E L$, $\square A F E M$. If $K, M, L$ do not coincide, then $\angle M K L=\angle K L M=\angle L M K=60^{\circ}$. Th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,675 |
4. As shown in Figure 4, in $\triangle A B C$, $A B=A C$, $\angle B=40^{\circ}$, $B D$ is the angle bisector of $\angle B$, and $B D$ is extended to $E$ such that $D E=A D$. Then the degree measure of $\angle E C A$ is | 4. $40^{\circ}$.
As shown in Figure 7, on $BC$, take $BF=AB$, connect $DF$, then $\triangle ABD \cong \triangle FBD$.
$$
\therefore DF=DA=DE.
$$
From $AC=AB$, we know $\angle ACB=40^{\circ}$.
$$
\begin{aligned}
\angle DFC & =180^{\circ}-\angle DFB \\
& =180^{\circ}-80^{\circ}=100^{\circ},
\end{aligned}
$$
$$
\therefo... | 40^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,676 |
One, (20 points) As shown in Figure $5, O$ is any point inside $\triangle A B C$, and lines $A O, B O, C O$ intersect the three sides at $P, Q, R$ respectively. If $a > b > c$, prove: $O P + O Q + O R < a$.
---
Note: The translation maintains the original text's line breaks and formatting. | In $\triangle B S M$,
$$
\begin{array}{l}
\angle B S M=\angle B R C>\angle A \\
>\angle B . \\
\quad \therefore B M>S M=O R .
\end{array}
$$
Similarly, $N C>O Q$.
Also, from $\angle A P C>\angle B>\angle C$, we know $A C>A P$,
$$
\therefore B C>A C>A P \text {. }
$$
But $\triangle A B C \sim \triangle O M N$, so $M N... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,677 |
II. (25 points) $\left\{F_{4} \mid k\right.$ $=1,2, \cdots\}$ is a sequence of positive integers, $F_{1}=F_{2}=1, F_{t+1}=$ $F_{4}+F_{k-1}^{*}, k=2,3, \cdots$, prove that among $5 F_{k}^{2}+4$ and $5 F_{k}-4$, at least one is a perfect square, $k=2,3, \cdots$. | It is easy to know that $F_{2}^{2}-F_{2} F_{1}-F_{1}^{2}=1-1-1=-1$,
$$
\begin{array}{l}
F_{3}^{2}-F_{3} F_{2}-F_{2}^{2}=4-2-1=1, \\
F_{k+1}^{2}-F_{k+1} F_{4}-F_{k}^{2} \\
=\left(F_{k}+F_{k-1}\right)^{2}-\left(F_{t}+F_{k-1}\right) F_{4}-F_{k}^{2} \\
=-\left(F_{k}^{2}-F_{k} F_{k-1}-F_{k-1}^{2}\right) .
\end{array}
$$
Th... | 5 F_{k}^{2}+4(-1)^{k}=\left(2 F_{k-1}+F_{k}\right)^{2} | Number Theory | proof | Yes | Yes | cn_contest | false | 708,678 |
1. The number of proper subsets of the set $\left\{n \left\lvert\,-\frac{1}{2}<\log _{\frac{1}{n}} 2<-\frac{1}{3}\right., n \in N\right\}$ is ().
(A) 7
(B) 8
(C) 31
() 32 | $$
\begin{array}{l}
-\sqrt{1}(\mathrm{~A}) \\
-\frac{1}{2}<\log _{\frac{1}{n}} 2<-\frac{1}{3} \\
\Leftrightarrow \log _{\frac{1}{n}}\left(\frac{1}{n}\right)^{-\frac{1}{2}}<\log _{\frac{1}{n}} 2<\log _{\frac{1}{n}} \cdot\left(\frac{1}{n}\right)^{-\frac{1}{3}}, \\
\log _{\frac{1}{n}} \sqrt{n}<\log _{\frac{1}{2}} 2<\log _... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,680 |
2. From .1 to Э these nine natural numbers, select two, separately as the logarithm's number and base, to get different logarithmic values ( ).
(A) 52
(B) 53
(C) 57
(D) 72 | 2. (B).
Since the base is not 1, the logarithm and the base cannot take the same value, so only 64 different logarithmic forms can be taken. Among them, there are 8 cases where the true value is 1, all taking the value 0. In addition,
$$
\begin{array}{ll}
\log _{2} 3=\log _{4} 9 & \log _{2} 4=\log _{3} 9, \\
\log _{3}... | 53 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,681 |
3. In space, there are four different planes. The set of possible numbers of intersection lines formed by these four planes is ( ).
(A) $\{1,2,3,4,5,6\}$
(B) $\{0,1,2,3,4,5,6\}$
(C) $\{0,1,3,4,5,6\}$
(D) $\{0,1,2,3,5,6\}$ | 3. (C).
If four planes are parallel to each other, the number of intersection lines is 0;
If these four planes are like an open book with only four pages, the number of intersection lines is 1;
If three planes are parallel to each other, and the fourth plane intersects with them, the number of intersection lines is 3... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,682 |
4. The domain and range of functions $y=f(x), y=g(x)$ are both $R$, and they both have inverse functions, then the inverse function of $y=f^{-1}\left(g^{-1}(f(x))\right)$ is ( ).
(A) $y=f\left(g\left(f^{-1}(x)\right)\right)$
(B) $y=f^{-1}(g(f(x)))$
(C) $y=\int^{-1}\left(g^{-1}(f(x))\right)$
(D) $y=f\left(g^{-1}\left(f^... | 4. (B).
From $y=f^{-1}\left(g^{-1}(f(x))\right)$ we get $g^{-1}(f(x))=f(y)$, and thus, $f(x)=g(f(x))$. Therefore, $x=f^{-1}(g(f(y)))$. By swapping the letters $x$ and $y$, the inverse function of $y=f^{-1}\left(g^{-1}(f(x))\right)$ is $y=f^{-1}(g(f(x)))$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,683 |
5. If $\omega=\cos 40^{\circ}+i \sin 40^{\circ}$, then $\mid \omega+2 \omega^{2}+3 \omega^{3}+\cdots$ $+\left.9 \omega^{9}\right|^{-1}$ equals ( ).
(A) $\frac{1}{18} \cos 20^{\circ}$
(B) $\frac{1}{9} \sin 40^{\circ}$
(C) $\frac{1}{9} \cos 40^{\circ}$
(D) $\frac{2}{9} \sin 20^{\circ}$ | 5. (D).
Let $s=\omega+2 \omega^{2}+3 \omega^{3}+\cdots+9 \omega^{9}$, where $\omega=e^{i \frac{2 \pi}{9}}$.
$$
\begin{array}{l}
\omega s=\omega^{2}+2 \omega^{3}+3 \omega^{4}+\cdots+9 \omega^{10}, \\
s(1-\omega)=\omega+\omega^{2}+\omega^{3}+\cdots+\omega^{9}-9 \omega^{10} . \\
\because \omega \neq 1, \\
\therefore s(1-... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,684 |
6. When $0<x<1$, the size relationship of $\frac{\sin x}{x},\left(\frac{\sin x}{x}\right)^{2}, \frac{\sin x^{2}}{x^{2}}$ is ( ).
(A) $\frac{\sin x}{x}<\left(\frac{\sin x}{x}\right)^{2}<\frac{\sin x^{2}}{x^{2}}$
(B) $\left(\frac{\sin x}{x}\right)^{2}<\frac{\sin x}{x}<\frac{\sin x^{2}}{x^{2}}$
(C) $\frac{\sin x^{2}}{x^{2... | 6. (B).
$0B D=\operatorname{tg} x^{2}>x^{2}, \\
\therefore \frac{\sin x}{\sin x^{2}}<\frac{x-x^{2}}{x^{2}}+1=\frac{x}{x^{2}}, \\
\text { hence } \frac{\sin x}{x}<\frac{\sin x^{2}}{x^{2}} .
\end{array}
$$
It is clear that $\left(\frac{\sin x}{x}\right)=\frac{\sin x}{x}<\frac{\sin x^{2}}{x^{2}}$ holds. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 708,685 |
1. Given $f(x)=x^{2}, g(x)=-\frac{1}{2} x+5, g^{-1}(x)$ represents the inverse function of $g(x)$. Let
$$
F(x)=f\left(g^{-1}(x)\right)-g^{-1}(f(x)) .
$$
Then the minimum value of $F(x)$ is $\qquad$ . | $$
\text { II.1. } \frac{70}{3} \text {. }
$$
$$
\begin{array}{l}
\text { Given } f(x)=x^{2}, g(x)=-\frac{1}{2} x+5 \\
\begin{array}{l}
\Rightarrow g^{-1}(x)=10-2 x . \\
F(x)=f\left(g^{-1}(x)\right)-g^{-1}(f(x)) \\
=(10-2 x)^{2}-\left(10-2 x^{2}\right) \\
=6\left(x-\frac{10}{3}\right)^{2}+\frac{70}{3} .
\end{array}
\e... | \frac{70}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,687 |
3. In the tetrahedron $P-ABC$, $PC \perp$ plane $ABC$, $AB=8$, $BC=6$, $PC=9$, $\angle ABC=120^{\circ}$. Then the cosine value of the dihedral angle $B-AP-C$ is $\qquad$ | 3. $1 \frac{\sqrt{111}}{148}$.
Draw $C K \perp A B$ extended at $K$, and connect $P K$. We have
$$
\begin{array}{l}
P K \perp A B, C K=6 \sin 60^{\circ}=3 \sqrt{3} . \\
P K=\sqrt{9^{2}+(3 \sqrt{3})^{2}}=6 \sqrt{3}, \\
B K=3 .
\end{array}
$$
Let $\angle P K C=\alpha$. Then
$$
\begin{array}{c}
\cos \alpha=\frac{3 \sqrt... | \frac{11 \sqrt{111}}{148} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,689 |
4. Let $P=\{$ natural numbers no less than 3 $\}$. Define the function $f$ on $P$ as follows: if $n \in P, f(n)$ represents the smallest natural number that is not a divisor of $n$, then $f(360360)=$ $\qquad$ . | 4. 16 .
Since $360360=2^{3} \times 3^{2} \times 5 \times 7 \times 11 \times 13$, we know that the divisors of 360360 in ascending order are $1,2,3,4,5,6,7,8,9,10$, $11,12,13,14,15,18,20, \cdots$, the smallest natural number that is not a divisor of 360360 is 16. Therefore, $f(360360)=16$. | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,690 |
$5 . n$ is a positive integer not exceeding 1996. If there is a 0 such that $(\sin \theta+i \cos \theta)^{n}=\sin n \theta+i \cos n 0$ holds, then the number of $n$ values satisfying the above condition is $\qquad$. | 5. 498 .
$$
\begin{array}{l}
\because(\sin \theta+i \cos \theta)^{n}=[i(\cos \theta-i \sin \theta)]^{n} \\
\quad=i^{n}(\cos n \theta-\sin n \theta)=i^{n-1}(\sin n \theta+i \cos n \theta), \\
\text { and }(\sin \theta+i \cos \theta)^{n}=\sin n \theta+i \cos n \theta, \\
\therefore i^{n-1}(\sin n \theta+i \cos n \theta)=... | 498 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,691 |
6. In the sequence of natural numbers starting from 1, certain numbers are colored red according to the following rules. First, color 1; then color two even numbers 2, 4; then color the three consecutive odd numbers closest to 4, which are $5, 7, 9$; then color the four consecutive even numbers closest to 9, which are ... | 6. 3929 .
The first time, color one number red: $1,1=1^{2}$;
The second time, color 2 numbers red: $2,4,4=2^{2}$;
The third time, color 3 numbers red: $5,7,9,9=3^{2}$;
Guessing, the last number colored red in the $k$-th time is $k^{2}$. Then the $k+1$ numbers colored red in the $(k+1)$-th time are:
$$
k^{2}+1, k^{2}+3... | 3929 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,692 |
One, (25 points) Point $M$ is a point inside an equilateral triangle - Prove: The area of the triangle formed by the segments $M A, M B$, and $M C$ does not exceed $\frac{1}{3}$ of the area of the original equilateral triangle. | Through point $M$, draw lines parallel to the sides of the equilateral triangle, labeled as shown in the figure.
$$
\triangle M A_{1} A_{2}, \triangle M F_{1} B_{2} \text {, }
$$
$\triangle M C_{1} C_{2}$ are all equilateral triangles, with side lengths $a$, $b$, and $c$ respectively, and the side length of the equilat... | S^{\prime} \leqslant \frac{1}{3} S | Geometry | proof | Yes | Yes | cn_contest | false | 708,693 |
II. (25 points) If $2 x+y \geqslant 1$. Try to find the minimum value of the function $u=y^{2}-2 y$ $+x^{2}+4 x$.
| $\therefore$ From $u=y^{2}-2 y+x^{2}+4 x$ completing the square, we get
$$
(x+2)^{2}+(y-1)^{2}=u+5 \text {. }
$$
Since $(x+2)^{2}+(y-1)^{2}$ can be regarded as the square of the distance from point $P(x, y)$ to the fixed point $(-2,1)$. The constraint $2 x+y \geqslant 1$ indicates that point $P$ is in the region $G$ w... | -\frac{9}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,694 |
Three, (35 points) Prove: From any four positive integers, one can always select two numbers $x$ and $y$, such that the following inequality holds
$$
0 \leqslant \frac{x-y}{1+x+y+2xy}<2-\sqrt{3} .
$$ | Three, if $x_{1}, x_{2}, x_{3}, x_{4}$ have two that are equal, then the conclusion is obviously true. Now assume four positive numbers $x_{1}<x_{2}<x_{3}<x_{4}$. Since
$$
\begin{array}{l}
\frac{x-y}{1+x+y+2 x y}=\frac{(x y+x)-(x y+y)}{(1+x)(1+y)+x y} \\
=\frac{\left(1+\frac{1}{y}\right)-\left(1+\frac{1}{x}\right)}{\le... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,695 |
Four. (35 points) Connecting nine different points on a circle with 36 chords, these chords are either painted red or blue, referred to as "red edges" or "blue edges". Suppose that every triangle formed by any three of these nine points contains a "red edge". Prove: There exist four points among these nine points such ... | (1) If there exists a point $y_{1}$ that draws at least four blue edges to other points, let's assume these four blue edges are $y_{1} y_{2}, y_{1} y_{3}, y_{1} y_{4}, y_{1} y_{5}$. Then $y_{2} y_{3}, y_{2} y_{4}, y_{2} y_{5}, y_{3} y_{4}, y_{3} y_{5}, y_{4} y_{5}$ are all red edges. That is, there exist four points $y... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,696 |
Example 9. A cyclic polygon with an odd number of sides, if all its interior angles are equal, is necessarily a regular polygon. (1966, Polish Mathematical Olympiad) | Prove that the conclusion is obvious when $n=3$. Now, we prove the case when $n \geqslant 5$. Let the polygon be $A_{1} A_{2} \cdots$
$A_{n}$. For convenience, let $A_{0} = A_{n}, A_{1} = A_{n+1}$.
Let $A_{i-1}, A_{i}, A_{i+1}$ be three consecutive vertices of the polygon, and let $\angle A_{i-1} A_{i} A_{i+1} = \alph... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,697 |
For what real values of $k$ are the solutions of the equation in $x$
$$
(6-k)(9-k) x^{2}-(117-15 k) x+54=0
$$
all integers? | It is known that when $k=6$ or 9, the solutions of the original equation are all integers.
When $k \neq 6$ and $k \neq 9$, the original equation can be transformed into
$$
[(6-k) x-9][(9-k) x-6]=0 \text {. }
$$
Solving, we get $x_{1}=\frac{9}{6-k}=\frac{3}{2-\frac{k}{3}}, x_{2}=\frac{6}{9-k}=\frac{2}{3-\frac{k}{3}}$.
... | k=3, 6, 7, \frac{15}{2}, \frac{39}{5}, \frac{33}{4}, 9, \frac{21}{2}, 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,698 |
44. On the outside of $\triangle ABC$, construct $\triangle BCP$, $\triangle CAQ$, and $\triangle ABR$ such that $\angle PBC = \angle QAC = 30^{\circ}$, $\angle PCB = \angle QCA = 45^{\circ}$, and $\angle RAB = \angle RBA = 15^{\circ}$. Prove that $\triangle PQR$ is an equilateral triangle. | Proof As shown in the figure, construct an equilateral $\triangle R B G$ with $R B$ as one side inside the shape, and connect $A G, Q G$.
$$
\because R A=R B=R G \text {, }
$$
i.e., $R$ is the circumcenter of $\triangle A B G$,
$$
\begin{array}{l}
\therefore \angle G A B=\frac{1}{2} \angle G R B=30^{\circ} . \\
\text ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,699 |
43. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{n+1}=\frac{1}{2} a_{n}+\frac{1}{a_{n}}(n \geqslant 1)$. And $a_{1}=1$. Prove that for any $n>1$, the number $\frac{2}{\sqrt{a_{n}^{2}-2}}$ is always a natural number. | Prove that for any $n>1$,
$$
a_{n} \geqslant 2 \sqrt{\frac{1}{2} a_{n-1} \cdot \frac{1}{a_{n-1}}}=\sqrt{2} \text {, }
$$
with equality holding only when $\frac{1}{2} a_{n-1}=\frac{1}{a_{n-1}}$, i.e., $a_{n-1}=\sqrt{2}$. Given $a_{1}=1$, we know $a_{n}>\sqrt{2}(n>1)$, i.e., $a_{n}^{2}>2$.
We will prove by mathematical... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,700 |
44. Let $E$ be the set of all even numbers. For any real number $x$, define $d(x, E)$ as the distance from $x$ to the nearest even number, such as $d(1.2, E) = 2 - 1.2 = 0.8$, $d(0.7, E) = 0.7$, etc. Given the sequence $\{x_n\}$ satisfying $x_{n+1} = d(2x_n, E)$, $x_1 = \frac{1}{p}$, where $p$ is an odd prime. Find the... | Let $\mathbb{R}$ and $\mathbb{Z}$ denote the sets of all real numbers and all integers, respectively. Any real number $x$ can be expressed as $x = 2k \pm b$, where $k \in \mathbb{Z}$ and $0 \leq b \leq 1$. By the problem statement, we should have $d(x, E) = b$.
First, we prove the three properties of $d(x, E)$:
(1) $... | p - 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,701 |
Example 10. A given convex pentagon $A B C D E$ has the following property: each of the five triangles $\triangle A B C, \triangle B C D$, $\triangle C D E, \triangle D E A, \triangle E A B$ has an area of 1. Prove: every pentagon with this property is equiareal, and there are infinitely many such non-congruent pentago... | Prove that since $S_{\triangle B C D}=S_{\triangle E C D}=1$, and the two triangles share the same base, hence
$B E / / C D$.
Similarly, $B D / / A E, E C / /$
$A B$.
E.
Let the intersection point of $E C$ and $B D$ be $P$, then $A B P E$ is a parallelogram.
Thus, $S_{\triangle E A B}=S_{\triangle E P B}=1$.
Let $S_{\t... | \frac{5+\sqrt{5}}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 708,702 |
Example 1. If $x=\sqrt{19-8 \sqrt{3}}$, then the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15}=$ $\qquad$ | Solve: From $x=\sqrt{19-2 \sqrt{48}}=4-\sqrt{3}$, we get $x-4=-\sqrt{3}$. Squaring both sides and rearranging, we obtain
$$
x^{2}-8 x+13=0 \text {. }
$$
Therefore,
$$
\begin{aligned}
\text { Original expression } & =\frac{\left(x^{2}-8 x+13\right)\left(x^{2}+2 x+1\right)+10}{\left(x^{2}-8 x+13\right)+2} . \\
& =5 .
\e... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,703 |
Example 2. If $x+\frac{1}{x}=3$, then $\frac{x^{2}}{x^{4}+x^{2}+1}$ $=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve: From $x+\frac{1}{x}=3$, squaring both sides, we get $x^{2}+\frac{1}{x^{2}}$ $=7$.
$$
\begin{array}{l}
\because \frac{x^{4}+x^{2}+1}{x^{2}}=x^{2}+\frac{1}{x^{2}}+1=7+1=8, \\
\therefore \frac{x^{2}}{x^{4}+x^{2}+1}=\frac{1}{8} .
\end{array}
$$ | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,704 |
Example 11. Given $x=\frac{1}{2}\left(1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}\right)$ ( $n$ is a natural number). Then, the value of $\left(x-\sqrt{1+x^{2}}\right)^{n}$ is ( ).
(A) 1991
(B) $-1991^{1}$
(C) $(-1)^{\star} 1991$
(D) $(-1)^{\star} 1991^{-1}$ | Let $a=x+\sqrt{1+x^{2}}, b=x-\sqrt{1+x^{2}}$. Then $a+b=2 x, a b=-1$ or $a=-b^{-1}$. Therefore, $b-b^{-1}=2 x$.
And $2 x=1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}$,
so $b-b^{-1}=1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}$.
Solving, we get $b=1991^{\frac{1}{n}}$ or $b=-1991^{-\frac{1}{n}}$.
Since $b=x-\sqrt{1+x^{2}}<0$, which ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,705 |
4. If $0<x<\frac{\pi}{2}$, then the range of $\operatorname{tg} x+\operatorname{ctg} x+\frac{1}{\sin x}-$ $\frac{1}{\cos x}$ is ().
(A) $(-\infty,+\infty)$
(B) $(0,+\infty)$
(C) $\left(\frac{1}{2},+\infty\right)$
(D) $(1,+\infty)$ | 4. (D) .
$$
\begin{array}{l}
f(x)=\frac{1+\cos x}{\sin x}+\frac{\sin x-1}{\cos x} \\
\frac{\cos x+\cos ^{2} x+\sin ^{2} x-\sin x}{\sin x \cos x} \\
=\frac{1+\cos x-\sin x}{\sin x \cos x}=\frac{2 \cos ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2} \cos x} \\
=-\frac{\cos \frac{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,706 |
5. There are 7 boys and 3 girls standing in a row for a photo, and any two girls want to be adjacent. Then the number of possible arrangements is ().
(A) $\frac{8!}{5!}$
(B) $\frac{7!6!}{4!}$
(C) $\frac{10!3!}{7!}$
(D) $\frac{10!7!}{3!}$ | 5. (A).
If we think about it using the insertion method, we can easily get the answer as $C_{8}^{3} \cdot P_{7}^{7} \cdot P_{3}^{3}$ $=\frac{8!}{5!}$.
First, let the 7 boys stand in a row. At this point, the number of arrangements is $P_{7}^{7}$. Then, from the 8 possible positions (the spaces between the boys or the... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,707 |
6. The smallest positive integer $n$ that satisfies $n \sin 1 > 5 \cos 1 + 1$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 6. (B).
From $n \sin 1>1+5 \cos 1$ we get $n \sin \frac{\pi}{3}>n \sin 1>1+5 \cos 1>1+5 \cos \frac{\pi}{3}$, which means $n>\frac{2+5}{\sqrt{3}}=\frac{7}{3} \sqrt{3}>4$. Therefore, $n \geqslant 5$.
Next, we prove that $5 \sin 1>5 \cos 1+1$, which is equivalent to proving $\sin 1 - \cos 1 > \frac{1}{5}$, or equivalent... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 708,708 |
1. Let $a \in R$. If the function $y=f(x)$ is symmetric to $y=10^{x}+3$ with respect to the line $y=x$, and $y=f(x)$ intersects with $y=$ $\lg \left(x^{2}-x+a\right)$, then the range of values for $a$ is $\qquad$ | $$
=, 1 . a3)$, and the equation $x^{2}-x+a=x-3$ has a root $x$ greater than 3. Therefore, the range of values for $a$ is the range of the function $a=-x^{2}+2x-3=-(x-1)^{2}-2$ $(x>3)$. Hence, $a<-6$ is the solution. (Please note the equivalent transformation)
$$ | a<-6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,709 |
2. Let $a, b \in \mathbb{R}^{+}, i^{2}=-1$, and there exists $z \in \mathbb{C}$, such that
$$
\left\{\begin{array}{l}
z+\bar{z}|z|=a+b i, \\
|z| \leqslant 1 .
\end{array}\right.
$$
Then the maximum value of $a b$ is | 2. $\frac{1}{8}$.
Let $z=x+y i\left(x, y \in R, i^{2}=-1\right)$, substitute into the given equation and by the definition of equality of complex numbers we get
$$
\left\{\begin{array}{l}
a=x \cdot\left(1+\sqrt{x^{2}+y^{2}}\right), \\
b=y \cdot\left(1-\sqrt{x^{2}+y^{2}}\right) .
\end{array}\right.
$$
From $|z| \leqsl... | \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,710 |
3. Let $0^{\circ}<\alpha<90^{\circ}$. If $1+\sqrt{3} \operatorname{tg}(60-\alpha)$ $=\frac{1}{\sin \alpha}$, then $\alpha$ equals $\qquad$. | 3. $30^{\circ}$ or $50^{\circ}$.
The original equation is equivalent to the following three equations:
$$
\begin{array}{l}
\sin \alpha\left(1+\sqrt{3} \cdot \frac{\sqrt{3}-\tan \alpha}{1+\sqrt{3} \tan \alpha}\right)=1, \\
\sin \alpha(1+\sqrt{3} \tan \alpha+3-\sqrt{3} \tan \alpha) \\
=1+\sqrt{3} \tan \alpha, \\
4 \sin ... | 30^{\circ} \text{ or } 50^{\circ} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,711 |
4. Let $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be a cube with edge length 1. Then the minimum distance $d=$ $\qquad$ between a point $P$ on the incircle of the top face $A B C D$ and a point $Q$ on the circle passing through the vertices $A, B, C^{\prime}, D^{\prime}$. | 4. $\frac{\sqrt{3}-\sqrt{2}}{2}$.
As shown in the figure, let $\angle P O Q = 0$, where point $O$ is the center of $A B C^{\prime} D^{\prime}$ (also the center of the cube). Clearly, $O Q = \frac{\sqrt{3}}{2}$, $O P = \frac{\sqrt{2}}{2}$. By the triangle inequality, we get $P Q \geqslant O Q - O P = \frac{\sqrt{3} - \... | \frac{\sqrt{3} - \sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,712 |
6. Let $a, b$ be positive integers, and $a+b \sqrt{2}$ $=(1+\sqrt{2})^{100}$. Then the units digit of $a b$ is $\qquad$ | 6.4.
By the binomial theorem, we have
$$
a-b \sqrt{2}=(1-\sqrt{2})^{100} \text {. }
$$
Therefore, $a=\frac{1}{2}\left((1+\sqrt{2})^{100}+(1-\sqrt{2})^{100}\right)$,
$$
\begin{array}{l}
b=\frac{1}{2 \sqrt{2}}\left((1+\sqrt{2})^{100}-(1-\sqrt{2})^{100}\right) . \\
\text { Hence } \left.a b=\frac{1}{4 \sqrt{2}}(1+\sqrt{... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,714 |
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