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2. Let $a, b$ be real numbers, and $|a|+a=0,|a b|=a b,|c|$ $-c=0$. Then, simplify $|b|-\sqrt{(a-b)^{2}}-\sqrt{(c-b)^{2}}$ $+|a-c|$ to ( ).
(A) $2 c-b$
(B) $2 b-2 a$
(C) $-b$
(D) $b$ | 2. I))
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 | 707,748 |
3. Among the following four equations: $2 \sqrt{\frac{2}{7}}=\sqrt{\frac{4}{7}}, x \sqrt{-\frac{1}{x}}=$
$$
\sqrt{-x}, \sqrt{5}-2=\sqrt{9-4 \sqrt{5}}, \frac{2^{n+4}-2 \cdot 2^{n}}{2\left(2^{n+3}\right)}=\frac{7}{8} \text {. }
$$
The number of correct ones is ( ).
(A) 1
(B) 2
(C) 3
$(D)_{4} \hat{i}$ | 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 | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,749 |
9. In $\triangle A B C$, $D$ is the midpoint of $A C$, $C E=2 E B$, and $A E \perp B D$ at $H$. Given $B C=3, A C=4$. Then $A B=$ $\qquad$ . | $(\sqrt{5})$ | \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,750 |
6. As shown in the figure, in $\triangle A B C$, $D, E$ are on $B C$, with $B D: D E: E C=3: 2: 1$. Point $M$ is on $A C$, with $C M: M A=1: 2$. $B M$ intersects $A D, A E$ at $H, G$. Then $B H: H G: G M$ equals ( . ).
(A) $3: 2: 1$
(B) $5: 3: 1$
(C) $25: 12: 5$
(C) $51: 24: 10$ | 6. D (Hint,
Draw $M N / / B C$ intersecting $A D, A E$ at $Q, P$. Let $B H=x, H G=$ $y, G M=z$, then we have $\frac{x}{y+z}=\frac{B D}{M Q}=\frac{D C}{M Q}=\frac{3}{2}, \frac{x+y}{z}=\frac{B E}{M P}=\frac{5 E C}{M P}=\frac{5 \cdot 3}{2}$. Solving these equations simultaneously gives $x: y: z=51: 24: 10.$) | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,753 |
4. If $x, y$ are real numbers, and $-\frac{1}{2} \leqslant x^{2}+4 y^{2} \leqslant 2$, then the range of values for $x^{2}-2 x y+4 y^{2}$ is $\qquad$ | $\begin{array}{l}\text { 4. } \frac{1}{4} \leqslant x^{2}-2 x y+4 y^{2} \\ \leqslant 3\end{array}$ | \frac{1}{4} \leqslant x^{2}-2 x y+4 y^{2} \leqslant 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,757 |
Three, (12 points) Let the equation about $x$ be $x^{2}+\left(\frac{a}{x}\right)^{2}-7 x$ $-\frac{7 a}{x}+2 a+12=0$ and suppose it has equal roots. Find the value of $a$.
untranslated portion:
设关于 $x$ 的方程 $x^{2}+\left(\frac{a}{x}\right)^{2}-7 x$ $-\frac{7 a}{x}+2 a+12=0$ 有相等鳫根. 求 $a$ 的值.
translated portion:
Let t... | Three, transform the original equation into
$$
\left(x+\frac{a}{x}-4\right)\left(x+\frac{a}{x}-3\right)=0 .
$$
Then we have $x+\frac{a}{x}-4=0$ or $x+\frac{a}{x}-3=0$,
i.e., $\square$
$$
\begin{array}{l}
x^{2}-4 x+a=0, \\
x^{2}-3 x+a=0 .
\end{array}
$$
If (1) has a repeated root, we get $a=4$.
If (2) has a repeated r... | 4 \text{ or } 2 \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,758 |
In $\triangle A B C$, $A B=A C$, $D$ is any point on side $B C$, $C_{1}$ is the symmetric point of $C$ with respect to the line $A D$, and $C_{1} B$ intersects $A D$ at point $P$. Try to determine: When point $D$ moves on $B C$ (excluding the midpoint of $B C$), what happens to the value of $A D \cdot A P$? Prove your ... | Four, $\because A B=A C$,
$$
\therefore \angle A B C=\angle C \text {. }
$$
Also, $\because$ point $C_{1}$ is symmetric to point $C$ with respect to $A D$,
$$
\therefore A, C_{1}, B, D \text { are four points }
$$
concurrent.
Thus, $\angle P B C=\angle C_{1} A D$.
Similarly, points $C, A, B, P$ are concyclic.
Hence, ... | A D \cdot A P = A B^2 | Geometry | proof | Yes | Yes | cn_contest | false | 707,759 |
Five, (15 points) (1) Calculate the number of all diagonals of a convex nonagon and the number of all triangles with vertices at the vertices of the convex nonagon.
(2) Assign an arbitrary natural number to each vertex of the convex nonagon. Among the triangles with vertices at the vertices of the nonagon, if the sum o... | (1) From each vertex, 6 diagonals can be drawn, and with nine vertices, a total of 54 diagonals are emitted. Since each diagonal is counted twice, there are actually 27 diagonals.
(Combination $C_{9}^{2}=36=$ number of diagonals + number of sides)
Since the total number of sides and diagonals is 36, and each side belon... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,760 |
Example 10. Given any 7 integers. Prove: there must exist two numbers, the sum or the difference of which is divisible by 10. | Proof: Let these 7 integers be $a_{1}, a_{2}, a_{3}, \cdots, a_{7}$. Consider the residue classes modulo 10: $\{0\},\{1\},\{2\}, \cdots$, $\{9\}$.
If among these 7 numbers, there are two numbers that are congruent modulo 10, i.e., $a_{i} \equiv a_{j}(\bmod 10)$, then $a_{i}-a_{j}$ is divisible by 10. The problem is so... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,761 |
1. If real numbers $a, b, c$ satisfy $|a|+a=0,|a b|=a b,|c|$ $-c=0$, then the algebraic expression $\sqrt{b^{2}}-|a+b|-\sqrt{c^{2}-2 b c+b^{2}}$ $+|a-c|$ simplifies to ( ).
(A) $2 c-b$ or $b$
(B) $2 c-b$
(C) $b$
(D) -6 | $-1 .(A)$.
$$
\because|a|=-a, \therefore a \leqslant 0 \text {. }
$$
If $a<0$, from the problem we know $b \leqslant 0, c \geqslant 0$.
$\therefore$ the original expression $=-b+a+b-(c-b)-(a-c)=b$;
If $a=0$, from the problem we know $b \geqslant 0$ or $b<0, c \geqslant 0$.
Then when $b<0$, it is easy to know the origi... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,762 |
$+\sqrt{x-y}=a$ has ( ) solutions.
(A) 2
(B) 4
(C) 5
(D) infinitely many
The text above is translated into English, keeping the original text's line breaks and format. | 2. (C).
Since $x, y$ are integers, $x+y$ and $x-y$ have the same parity, and $\sqrt{x-y}$ can only be an integer, then $a=2$. Therefore, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+y=0, \\
x-y=4,
\end{array},\left\{\begin{array}{l}
x+y=1, \\
x-y=1,
\end{array},\left\{\begin{array}{l}
x+y=2, \\
x-y=0
\end{arr... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,763 |
3. If the real roots $x_{1}, x_{2}$ of the equation $x^{2}+p x+q=0$ satisfy $x_{1}+x_{2}+x_{1} x_{2}>3$, then $p()$.
(A) is not less than 6
(B) is greater than 6 or less than -2
(C) is greater than -2 and less than 6
(D) is less than 6 | 3. (B)
From the problem, we get $p^{2} \geqslant 4 q,-p+q>3$, which means $4 q>4 p+12$. Solving the quadratic inequality $p^{2}>4 p+12$ yields $p>6$ or $p<-2$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,764 |
4. Given that $A_{1}, B_{1}, C_{1}, D_{1}$ are points on the four sides of square $A B C D$. The following are two propositions about it:
(1) If $A_{1} B_{1} C_{1} D_{1}$ is a rectangle with unequal adjacent sides, then the rectangle $A_{1} B_{1} C_{1} D_{1}$ must have a side parallel to one of the diagonals of the squ... | 4. (D).
(1) As shown in the figure, it is easy to know that $\angle A D_{1} A_{1} + \angle D D_{1} C_{1} = 90^{\circ}, \angle D D_{1} C_{1} + \angle D C_{1} D_{1} = 90^{\circ}, \angle A D_{1} A_{1} = \angle D C_{1} D_{1}$, thus $\triangle A D_{1} A_{1} \sim \triangle D C_{1} D_{1}$. Also, Rt $\triangle A D_{1} A_{1} \c... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,765 |
5. The inequality $x^{2}-|2 x-4| \geqslant p$ holds for all real numbers $x$. Then the correct conclusion is ( ).
(A) $p$ has both a maximum value and a minimum value
(B) $p$ has a maximum value, but no minimum value
(C) $p$ has no maximum value, but has a minimum value
(D) $p$ has neither a maximum value nor a minimum... | 5. (B).
When $x \geqslant 2$, $x^{2}-$
$$
|2 x-4|=x^{2}-2 x+
$$
$4=(x-1)^{2}+3 \geqslant 4$; when $x<2$, $x^{2}-|2 x-4|=x^{2}+$ $2 x-4=(x+1)^{2}-5 \geqslant-5$. Therefore, $p$ has a maximum value of -5, and no minimum value. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 707,766 |
6. Let the diagonals of convex quadrilateral $A B C D$ intersect at $O$, and $A D / / B C$. The following are four statements about it:
(1) Given $A B+B C=A D+D C$. Then $A B C D$ is a parallelogram,
(2) Given $D C+D O=A O+A B$. Then $A B C D$ is a parallelogram;
(3) Given $B C+B O+A O=A D+D O+C O$. Then $A B C D$ is n... | 6. (B).
(1) As shown in the figure, extend $CB$ to point $E$ such that $BE=AB$, and extend $AD$ to point $F$ such that $DF=DC$. Then $EC \parallel AF$, so $AECF$ is a parallelogram, $\angle E=\angle F$, $\angle ABE=\angle CDF$, $\angle ABC=\angle ADC$.
Since $AD \parallel BC$, $\angle ABC + \angle DAB = 180^{\circ}$.
$... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,767 |
7. As shown in the figure, ABCD is a cyclic quadrilateral, point $C$ is the midpoint of $\overparen{B D}$, the tangent $C E$ intersects the extension of $A D$ at $E$, and $A C$ intersects $B D$ at $F$. Then the number of pairs of line segments whose ratio is equal to $\frac{A E}{C E}$ is ( ).
(A)5
(B) 6
(C) 7
(D) 8 | 7. (C).
As shown in the figure, it is easy to know that
$$
\frac{A E}{C E}=\frac{C E}{D E}=\frac{A D}{F D}=\frac{A B}{B F}=\frac{B C}{C F}=\frac{D C}{C F}=\frac{A C}{D C}=\frac{A C}{B C} .
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,768 |
1. In the table below, the average of any three adjacent small squares in the upper row is $1(x \neq 0)$, and the square of any four adjacent small squares in the lower row is also 1. Then the value of $\frac{x^{2}+y^{2}+z^{2}-32}{y z+16}$ is $ـ$. $\qquad$ | ii. 3 .
From the problem, we know that in the above row, the numbers in every two small squares separated by one are equal, thus we can deduce
$$
\frac{x+7-y}{3}=1 \text {. }
$$
Similarly, for the row below, we get
$$
\begin{array}{l}
\frac{z-x+9.5-1.5}{4}=1 . \\
\begin{array}{l}
\therefore y=x+4, z=x-4, \text { then... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,770 |
Example 11. Let $n>3$, and $n \in \mathbb{N}, a_{0}, a_{1}, u_{2}, \cdots, a_{n}$ be $n+1$ integers, and satisfy
$$
1 \leqslant a_{0}<a_{1}<a_{i}<\cdots<a_{n} \leqslant 2 n-3 .
$$
Prove that there exist different integers $i, j, k, l, m$, such that
$$
a_{i}+a_{j}=a_{k}+a_{l}=a_{m} .
$$ | Proof: Let $b_{i}=a_{n}-a_{i}(i=0,1, \cdots, n-1)$. Then $b_{i}$ are $n$ distinct positive integers and satisfy $b_{n-1} \leq 2n-3$. If there are two numbers equal, let $b_{j}=a_{i}$, at this point there are $2n-1$ numbers no greater than $2n-3$, and there are still two numbers equal, let $b_{l}=a_{k}$. Thus,
$$
\begin... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,772 |
4. As shown in the figure, in $\triangle A B C$, $\angle A=60^{\circ}$, points $D, E$ are on side $A B$, and $F$, $G$ are on side $C A$. Connecting $B F, F E, E G, G D$ divides $\triangle A B C$ into five smaller triangles of equal area, i.e., $S_{\triangle C B F}=S_{\triangle F B E}=S_{\triangle F E C}=S_{\triangle G ... | 4. $\frac{\sqrt{7}}{2}$.
It is known that $S_{\triangle F E A}=3 S_{\triangle F B E}$, so $A E=3 B E$.
Similarly, we get $A F=4 C F$.
And $B E=2 C$, so we can set $C F=x$, then $A F=4 x, A C=$ $5 x, A E=6 x, A B=8 x$.
In $\triangle F E A$, using the cosine rule we get
$$
E F^{2}=36 x^{2}+4^{2} x^{2}-2 \cdot 6 x \cdot ... | \frac{\sqrt{7}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,774 |
One, (20 points) Given the quadratic equation in $x$, $a x^{2}+$ $2 a x+2^{2 b}-2^{b+3}+12=0$, cannot have equal real roots, and $\frac{x^{2}}{4}-\sqrt{-a} x+4+c^{2}=0$ has real roots. Find the values of $a, b, c$. | $$
\begin{array}{l}
\left\{\begin{array}{l}
4 a^{2}-4 a\left(2^{3 b}-2^{b+3}+12\right) \leqslant 0, \\
a<0, \\
-a-\left(4+c^{2}\right) \geqslant 0,
\end{array}\right. \\
\text { i.e. }\left\{\begin{array}{l}
a \geqslant 2^{3 b}-2^{b+3}+12, \\
a<0, \\
-4-c^{2} \geqslant a .
\end{array}\right. \\
\therefore-4 \geqslant-4... | a=-4, b=2, c=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,775 |
II. (20 points) As shown in the figure, in trapezoid $ABCD$, $AD = DC = CB = \frac{1}{2} AB$, $M$ is the midpoint of $AB$, $\overparen{DC}$ is part of the circumcircle of $\triangle DCM$, and point $P$ is any point on $\overparen{DC}$. Connecting $PA$ and $PB$, they intersect $DC$ at points $E$ and $F$ respectively. Pr... | II. As shown in the figure, connect $PD, PC$ and extend them to intersect line $AB$ at points $H, N$ respectively.
Given $\angle DAB = \angle CBA = 60^{\circ}, \angle DAH = \angle C1$ $\angle ADC = \angle DCB = 120^{\circ}$. $\triangle DCM$ is an equilateral triangle.
$\therefore \angle DPC = 120^{\circ}, \angle PDC +... | DE \cdot FC = \left(\frac{1}{2} EF\right)^2 | Geometry | proof | Yes | Yes | cn_contest | false | 707,776 |
Three. (20 points) There are 1995 points on the circumference of a circle. Among the 1995 small arcs formed by these points, 665 arcs have a length of 1, another 665 arcs have a length of 2, and the remaining arcs have a length of 3. Prove that among these 1995 points, there must be two points that are the endpoints of... | Three, assuming that among these 1995 points, no two points are the endpoints of a certain diameter of the circle. Then the endpoints of an arc of length 1, when connected to the center of the circle, must pass through one of the trisection points of an arc of length 3. This diameter divides the circle into two semicir... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,777 |
1. On the surface of a cube, there exist ascending lines on the diagonals of the square faces. If the distance between two of these lines is 1, then the volume of the cube is ( ).
(A) 1
(B) $3 \sqrt{3}$
(C) 1 or $3 \sqrt{3}$
(D) $3 \sqrt{3}$ or $3 \sqrt{2}$ | $-1 .(\mathrm{C})$.
Let the edge length of the cube be $x$. If the distance between the face diagonal $A C$ and $B^{\prime} D^{\prime}$ is 1, then $x=1$, and thus the volume $V$ $=x^{3}=1$.
If the distance between the face diagonal $A C$ and $B C^{\prime}$ is 1, then $\frac{\sqrt{3}}{3} x=$ $1, x=\sqrt{3}, V=x^{3}=3 \... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,778 |
3. In the acute triangle $\triangle A B C$, $\angle A: \angle B, \angle C$ are its interior angles. Let $T=\operatorname{ctg} 2 A+\operatorname{ctg} 2 B+\operatorname{ctg} 2 C$. Then it must be ( ).
(A.) T $>0$
(B) $\cdots \geqslant 0$
(c) $T<0$
(D) $T \leqslant 0$ | 3. (C).
From $\angle A + \angle B > \frac{\pi}{2}$, we get $\angle A > \frac{\pi}{2} - \angle B$, which means $\operatorname{tg} A > \operatorname{tg}\left(\frac{\pi}{2} - \angle B\right) = \operatorname{ctg} B$.
Similarly, $\operatorname{tg} B > \operatorname{ctg} C, \operatorname{tg} C > \operatorname{ctg} A$. There... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,780 |
+. $C$ is the set of complex numbers, let the set $A=\left\{z \mid z^{18}=1, z \in C\right\}, B$ $=\left\{\omega^{2 x}=1, \omega \in C\right\}, D=\{z \omega \mid z \in A, \omega \in B\}$. Then the number of elements in $D$ is ( ) .
(A) 864
(B) 432
(C) 288
(D) 144 | 4. (D).
The elements of sets $A, B, D$ are
$$
\begin{aligned}
z & =\cos \frac{2 k \pi}{18}+i \sin \frac{2 k \pi}{18} \\
\omega & =\cos \frac{2 t \pi}{48}+i \sin \frac{2 t \pi}{48} \\
z \omega & =\cos \left(\frac{2 k \pi}{18}+\frac{2 t \pi}{48}\right)+i \sin \left(\frac{2 k \pi}{18}+\frac{2 t \pi}{48}\right) \\
& =\cos... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,781 |
5. Given positive numbers $a, b, c$ satisfy $a b+b c+c a=1995$. Then the minimum value of $\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b}$ is ( ).
(A) 1995
(B) $3 \sqrt{665}$
(C) $2 \sqrt{665}$
(D) $\sqrt{665}$ | 5. (B).
$$
\begin{array}{l}
\text { Given } a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a=1995, \text { we have } \\
\frac{a^{2} b^{2}}{c^{2}}+\frac{c^{2} a^{2}}{b^{2}}+\frac{b^{2} c^{2}}{a^{2}} \\
\geqslant \frac{a b}{c} \cdot \frac{c a}{b}+\frac{c a}{b} \cdot \frac{b c}{a}+\frac{b c}{a} \cdot \frac{a b}{c} \\
=a^{2}+b^{2}+... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,782 |
Example 12. Is there a north-facing 100 petals arrangement, such that they exactly have 1985 intersections. | Solution 1: Since $x$ parallel lines and another set of $100 - x$ parallel lines can yield $x(100-x)$ intersection points, and
$$
x(100-x)=1985
$$
has no integer solutions, we can consider a grid formed by 99 lines. Since
$$
x(99-x)<1955
$$
has solutions $x \leqslant 26$ or $x \geqslant 73, x \in \mathbb{N}$, and
$$
... | 1985 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,783 |
6. Given that the function $f(x)$ is defined and increasing on $(0,+\infty)$, and satisfies $f(x) \cdot f\left(f(x)+\frac{1}{x}\right)=1$. Then $f(1)=$ ).
(A) 1
(B) 0
(C) $\frac{1+\sqrt{5}}{2}$
(D) $\frac{1-\sqrt{5}}{2}$ | 6. (D).
Let $f(1)=a \cdot x=1$. From the given functional equation, we have
$$
\begin{array}{l}
f(1) f(f(1)+1)=1, \\
\quad a \int(a+1)=1, f(a+1)=\frac{1}{a} .
\end{array}
$$
Let $x=a+1$. Then
$$
\begin{array}{l}
f(a+1) f\left(f(a+1)+\frac{1}{a+1}\right)=1, \\
\frac{1}{a} f\left(\frac{1}{a}+\frac{1}{a+1}\right)=1, \\
... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,784 |
1. Given the parabola equation $y=-\frac{x^{2}}{2}+h(h>0)$, and point $P(0, h)$, and the parabola intersects at points $A, B$. The inclination angles of lines $PA$ and $PB$ are complementary. Then the maximum area of $\triangle PAD$ is
Translate the above text into English, please retain the original text's line break... | $$
=1 . \frac{64 \sqrt{3}}{9} \text {. }
$$
Given $P(2,4)$ lies on $y=-\frac{x^{2}}{2}+h$, we get $h=6$.
The equations of lines $P A$ and $P B$ are $y-4=-k(x-2)$ and $y-4=k(x-2)$, respectively.
Substituting $y-4=k(x-2)$ into $y=-\frac{x^{2}}{2}+6$, we get $x^{2}+2kx-4k-1=0$.
Thus, $x_{1}=2k-2, x=-2k-2$,
and $y_{B}=k(... | \frac{64 \sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,785 |
3. The equation $a \cos (x+1)+b \cos (x+2)+c \cos (x+3)$ $=0$ has at least two roots in the open interval $(0, \pi)$. Then all the roots of this equation are $\qquad$ . | 3. All real numbers.
Given that the left side of the equation can be transformed into
$$
\begin{array}{l}
(a \cos 1+b \cos 2+c \cos 3) \cos x- \\
\quad(a \sin 1+b \sin 2+c \sin 3) \sin x \\
=A \cos x+b \sin x
\end{array}
$$
where \( A=a \cos 1+b \cos 2+c \cos 3 \),
$$
B=a \sin 1+b \sin 2+c \sin 3 .
$$
Thus, the equa... | All real numbers | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,787 |
5. Let the polynomial $p(x)$ have a degree not exceeding 3, and $p(0)=$ $1, p(3)=0,|p(2+x)|=|p(2-x)|$. If the leading coefficient of $p(x)$ is negative, then $p(x)=$ $\qquad$ $ـ$ | 5. $p(x)=-\frac{1}{6}(x-1)(x-2)(x-3)$.
From $p(3)=0$ we know that $(x-3)$ is a factor of $p(x)$, and from $|p(2+x)|=|p(2-x)|$ we know that $|p(3)|=|p(1)|=0$, so $(x-1)$ is also a factor of $p(x)$.
Let $p(x)=(x-1)(x-3)(a x+b)$, and from $p(0)=1$ we get $f(0)=(-1)(-3) b=1, b=\frac{1}{3}$, and from $|p(2+x)|=|p(2-x)|$ w... | p(x)=-\frac{1}{6}(x-1)(x-2)(x-3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,789 |
6. In a tennis tournament, $n$ women and $2 n$ men participate, and each player plays against all other players exactly once. If there are no ties, the ratio of the number of games won by women to the number of games won by men is 7:5. Then $n=$
| 6. $n=3$.
Let $k$ represent the number of times the woman wins over Luo Ziyue.
Here, we set $\frac{k+\mathrm{C}_{0}^{2}}{\mathrm{C}_{\mathrm{jn}}^{2}}=\frac{7}{12}, k=\frac{7}{12} \mathrm{C}_{3 n}-\mathrm{C}_{n}^{2}$. From $k \leq 2 n^{2}$, we can solve to get $n \leq 3$.
When $n=1$, we get $k=\frac{7}{4}$; when $n=... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,790 |
One, (25 points) Find all values of $\alpha$, $\alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that the system of equations
$$
\left\{\begin{array}{l}
y=\operatorname{arasin}\left(\sin \frac{1}{x}\right), \\
y=\operatorname{tg} \alpha\left(x-\frac{1}{10 \pi}\right)
\end{array}\right.
$$
has exactly 10 solu... | $-, y=\arcsin \left(\sin \frac{1}{x}\right)$
$$
=\left\{\begin{array}{c}
\frac{1}{x}, x \geqslant \frac{2}{\pi} \\
\pi-\frac{1}{x}, \frac{2}{3 \pi} \leqslant x < \frac{2}{\pi} \\
-\pi-\frac{1}{x}, x < \frac{2}{3 \pi}
\end{array}\right.
$$
When $x > 0$, the above curve intersects with the line at most 9 points in $\lef... | \left\{\operatorname{arctg}\left[(190-60 \sqrt{10}) \pi^{2}\right]\right\} U\left(\operatorname{arctg}\left(-\frac{15}{17} \pi^{2}\right), 0\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,791 |
29. $P$ is on the side $AC$ or its extension of $\triangle ABC$, and
$\frac{AP}{AC}=k, D, E$ are on $BC$, and $BD=DE=EC, AE, AD$ intersect $BP$ at $F, G$. Prove that $\frac{FG}{BP}=\frac{3k}{(2k+1)(k+2)}$. | Prove $\frac{S_{\triangle A B D}}{S_{\triangle A D C}}=\frac{B D}{D C}=\frac{1}{2}=\frac{A B \cdot \sin \angle B A D}{A C \cdot \sin \angle D A C}$
$$
\frac{S_{\triangle A B F}}{S_{\triangle A F P}}=\frac{B F}{F P}=\frac{A B \cdot \sin \angle B A F}{A P \cdot \sin \angle F A P} \text {. }
$$
Divide the two equations a... | \frac{3k}{(2k+1)(k+2)} | Geometry | proof | Yes | Yes | cn_contest | false | 707,796 |
Given 30. Let $H$ be a point inside the acute $\triangle ABC$, and $AH, BH$, $CH$ intersect $BC, CA, AB$ at $D, E, F$, respectively, and $\angle EDH = \angle FDH$. Prove: $AD \perp BC$.
---
The translation maintains the original format and line breaks as requested. | Draw a line through point $A$ parallel to $BC$ intersecting the extensions of $DE$ and $DF$ at points $M$ and $N$ (see figure).
Since $AN \parallel BD$, we have
$$
\frac{AN}{BD} = \frac{AF}{BF},
$$
Since $AM \parallel CD$, we have
$$
\frac{AM}{CD} = \frac{AE}{CE},
$$
From (1) and (2), we get
$$
\frac{AN}{AM} = \frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,797 |
30. Given a monotonic real sequence $\left\{x_{n}\right\}$. And $x_{i} \in(0$,
$$
\begin{array}{l}
\left.\frac{\pi}{2}\right), i=1,2, \cdots, n \\
\text { Prove: } \sum_{i=1}^{n} \sin 2 x_{i}<\frac{\pi}{2}+2 \sum_{i=1}^{n-1} \sin x_{i} \cos x_{i+1} .
\end{array}
$$ | Proof. Chen's Inequality is equivalent to
$$
\sum_{i=1}^{n-1} \sin x_{i}\left(\cos x_{i}-\cos x_{i+1}\right)+\sin x_{n} \cos x_{n}<\frac{\pi}{4} \text {. }
$$
Taking $n$ points on a unit circle, with each point's coordinates being $Z_{1}\left(\cos x_{1}, \sin x_{i}\right)$. The area of the $n$-sided polygon formed by ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,799 |
Create 3. In the right-angled triangle $\triangle A B C$, $A D$ is the median on the right side $B C$, $B E \perp A D$ at $G$, and intersects $A C$ at $E, E F \perp B C$ at $F$. Given $A B=B C=a$, find the length of $E F$.
(A) $\frac{1}{3} a$
(E) $\frac{1}{2} a$
(C) $\frac{2}{3} a$
(D) $\frac{2}{5} a$ | Analysis: From the right triangle $\mathrm{Rt} \triangle A B C$ and the median $A D \perp B E$, we can derive that $A G \cdot G D=4 \cdot 1$. Since $D$ is the midpoint of $B C$, and $\triangle A B C$ is the original triangle, from (*) we have $(1+1) \cdot \frac{A E}{E C}=4$, so $\frac{A E}{E C}=2$. Therefore, $\frac{E ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,802 |
Example 16. Does a right-angled triangle with a perimeter of 6 and an integer area exist? If not, please provide a proof; if it does exist, please prove how many there are. | Solve: Such a right-angled triangle exists, and there is exactly one.
Let the hypotenuse of this right-angled triangle be $c$, and the two legs be $a, b$, with the area being $S$. Then we have
$$
\left\{\begin{array}{l}
a \leqslant b<c<a+b, \\
a+b+c=6, \\
a^{2}+b^{2}=c^{2}, \\
S=\frac{1}{2} a b \text { is an integer. ... | a=\frac{5-\sqrt{7}}{3}, b=\frac{5+\sqrt{7}}{3}, c=\frac{8}{3} | Geometry | proof | Yes | Yes | cn_contest | false | 707,803 |
Example 18. On the coordinate plane, the vertical coordinates are all integers, which are referred to as true tests: there exists a set of concentric circles, such that
(2) on this set, there is exactly one integer on each circumference.
保留源文本的换行和格式，翻译结果如下：
Example 18. On the coordinate plane, the vertical coordinate... | Let the center of the circle be $(m, n)$, and assume that there are two integer points $(a, b), (c, d)$ on the circle with center $(m, n)$, then
\[
\begin{array}{l}
(a-m)^{2}+(b-n)^{2} \\
=(c-m)^{2}+(d-n)^{2},
\end{array}
\]
which simplifies to
\[
2(c-a) m=c^{2}+d^{2}-a^{2}-b^{2}+2(b-d) n.
\]
Clearly, when $m$ is an i... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,805 |
Example 1. Prove: $\left(x^{209}+x^{188}+\cdots+x^{11!}+1\right)$ is divisible by $\left(x^{9}+x^{8}+\cdots+x-1\right)$. | Proof Let $\omega$ be any 10th root of unity not equal to 1, then
$$
\begin{array}{l}
x^{999}+x^{888}+\cdots+x^{111}+1 \\
=\left(\omega^{10}\right)^{99} \omega^{9}+\left(\omega^{10}\right)^{88}\left(\omega^{8}+\cdots\right. \\
+\left(\omega^{10}\right)^{11} \omega^{1}+1 \\
=\omega^{9}+\omega^{8}+\cdots+\omega+1 \\
=\fr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,807 |
Example 2. Let $P(x), Q(x), R(x)$, and $S(x)$ be polynomials in $x$ that satisfy the condition
$$
\begin{array}{l}
P\left(x^{5}\right)+x \cdot Q\left(x^{5}\right)+x^{2} \cdot R\left(x^{5}\right) \\
=\left(x^{4}+x^{3}+x^{2}+x+1\right) \cdot S(x) .
\end{array}
$$
Prove that $(x-1)$ is a factor of $P(x)$. | Proof: Let $x_{k}=\cos \frac{2 k \pi}{5}+i \cdot \sin \frac{2 k \pi}{5}(k=$ $1,2,3,4)$. Then $x_{k}$ is a fifth root of unity, hence
$$
\begin{array}{l}
x_{1}=\frac{\sqrt{5}-1}{4}+\frac{1}{4} \sqrt{10+2 \sqrt{5}} i, \\
x_{2}=x_{1}^{2}=-\frac{\sqrt{5}+1}{4} \\
+\frac{1}{4} \sqrt{10-2 \sqrt{5}} i, \\
x_{3}=x_{1} x_{2}=-\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,808 |
Example 3. Factorize: $x^{12}+x^{9}+x^{6}+x^{3}+1$. | Let $\omega$ be a fifth root of unity other than 1. Substituting $x = \omega$ into the polynomial, we get
$$
\begin{array}{l}
\omega^{12}+\omega^{9}+\omega^{6}+\omega^{3}+1 \\
=\left(\omega^{5}\right)^{2} \omega^{2}+\omega^{5} \omega^{4}+\omega^{5} \omega^{1}+\omega^{3}+1 \\
=\omega^{2}+\omega^{4}+\omega^{1}+\omega^{3}... | x^{12}+x^{9}+x^{6}+x^{3}+1 = \left(x^{4}+x^{3}+x^{2}+x+1\right)\left(x^{8}-x^{7}+x^{5}-x^{4}+x^{3}-x+1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,809 |
Example 4. Factorize
\[
\begin{aligned}
1+2 x+3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+5 x^{6}+4 x^{7} \\
+3 x^{8}+2 x^{9}+x^{10} .
\end{aligned}
\] | Solve for the non-1 roots of the polynomial $x^{5}=1$. Then its value is 0, similar to Example 2, the polynomial must be the same as the polynomial $x^{5}$
$$
\begin{array}{l}
+x^{4}+x^{3}+x^{2}+x+1 \text {, so } \\
\quad \text { the original expression }=\left(x^{2}+x^{4}+x^{3}+x^{2}+x+1\right)^{2} \\
=\left[x^{3}\lef... | (x+1)^{2}\left(1-x+x^{2}\right)^{2}\left(1+x+x^{2}\right)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,810 |
Example 5. Given
hexagon inscribed in a circle
$A B C D E F$ with sides
satisfying the relation $A B$
$$
=C D=E F=r \text {, }
$$
$r$ being the target radius,
and $G, H, K$ are
the midpoints of $B C, D E, F A$ respectively. Prove: $\triangle G H K$ is an equilateral triangle. | To simplify, let $r=1$. Therefore, we can consider the known circle as the unit circle in the complex plane, and let point $\Lambda$ correspond to the number 1. From $A B=C D=E F=r$, we get $\angle A O B = \angle C O D = \angle E Q F = \frac{\pi}{3}$. Without loss of generality, assume points $A, B, C, D, E, F$ are arr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,811 |
Example 6. Solve the equation: $1+2 x+3 x^{2}+4 x^{3}+5 x^{4}+$ $6 x^{5}+5 x^{6}+4 x^{7}+3 x^{8}+2 x^{9}+x^{10}=0$. | From Example 4, we know that the equation is equivalent to:
$$
\begin{array}{l}
(x+1)^{2}\left(1-x+x^{2}\right)^{2}\left(1+x+x^{2}\right)^{2}=0 \\
\Leftrightarrow(x+1)^{2}=0 \text { or }\left(1-x+x^{2}\right)^{2}=0 \text { or } \\
\left(1+x+x^{2}\right)^{2}=0 .
\end{array}
$$
Therefore, the roots of the original equat... | x=-1 \text{ or } x=\frac{1 \pm \sqrt{3} i}{2} \text{ or } x=\frac{-1 \pm \sqrt{3} i}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,812 |
Make the opposite vertices $A, C$ coincide, and fold the paper. Find the length of the crease.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Given that the crease $EF$ is the perpendicular bisector of the diagonal $AC$, $O$ is the midpoint of $AC$ and also the midpoint of $EF$. Let $OE = OF = x \, \text{cm}$. Then
\[
\begin{aligned}
AF & = \sqrt{AO^2 + OF^2} = \sqrt{\left(\frac{5}{2}\right)^2 + x^2} \\
& = \frac{1}{2} \sqrt{25 + 4x^2} \, (\text{cm}), \\
FD ... | 3.75 \, \text{cm} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,814 |
Example 2. In parallelogram $A B C D$, it is known that $B C=$ $\sqrt{6} \mathrm{~cm}, A C=(3+$ $\sqrt{3}) \mathrm{cm}, A B=2 \sqrt{3}$ $\mathrm{cm}$. If the paper is folded so that point $A$ coincides with point $C$. Find the length of the fold. | Fold line $E S$ is the perpendicular bisector of $A C$.
$$
\begin{aligned}
\because \cos \angle C A B & =\frac{A B^{2}+A C^{2}-B C^{2}}{2 A B \cdot A C} \\
& =\frac{\sqrt{3}}{2},
\end{aligned}
$$
$$
\therefore \angle C A B=30^{\circ} \text {. }
$$
$\because A B C D$ is a $\square$,
$$
\therefore \angle D C A=\angle C A... | 1+\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,815 |
Example 3. The side length of square $ABCD$ is $a$. Fold the square so that $C$ falls on the midpoint $G$ of $AD$. Find the length of the crease $EF$.
Fold the square so that $C$ falls on the midpoint $G$ of $AD$. Find the length of the crease $EF$. | Draw $E H \perp$ $C D$, with the foot of the perpendicular being $H$. Given $E F$ $\perp C G$, then $\triangle E F H \cong$ $\triangle C G D$.
Thus, $E F=C G$
$$
=\frac{\sqrt{5}}{2} a \text {. }
$$ | \frac{\sqrt{5}}{2} a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,816 |
Example 4. Fold a square $ABCD$ with side length 3, with the crease being $EF$ (as shown in the figure), so that point $B$ lands on point $B'$ on $CD$, the extensions of $BA$ and $FE$ intersect at point $A'$, and $\angle B'BC=30^{\circ}$. Try to calculate the area of $\triangle A'EG$. | Let $B B^{\prime}$ intersect $A^{\prime} F$ at point $O$. It is easy to prove that $\triangle B C B^{\prime} \cong \triangle A^{\prime} O B$, and $B B^{\prime} = 2 \sqrt{3}$.
$$
\begin{array}{l}
\quad \therefore A A^{\prime} = A^{\prime} B - A B = 2 \sqrt{3} - 3, \\
\angle A G A^{\prime} = 30^{\circ}. \\
\therefore A G... | 7 \sqrt{3} - 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,817 |
Example 5. In rectangle $ABCD$, $AB=a, BC=b(a>b)$, $\triangle ADC$ is folded along the diagonal $AC$ as the axis of symmetry, so that point $D$ moves to point $E$, and $CE$ intersects $AB$ at $F$. Then the area of the shaded part in the figure is
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
This part is a... | Obviously, $\triangle C B F \cong \triangle A E F$, so $C F=A F$.
Let $B F=x$, then
$$
C F=\sqrt{b^{2}+x^{2}}, A F=a-x .
$$
From $a-x=\sqrt{b^{2}+x^{2}}$, we get $x=\frac{a^{2}-b^{2}}{2 a}$.
Therefore, $S_{\text {by|E }}=S_{\triangle A B C}-S_{\triangle C B F}$
$$
=\frac{1}{2}-\frac{1}{2} \times \frac{a^{2}-b^{2}}{2 a... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,818 |
Example 2. Non-negative real numbers $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ satisfy $a_{1}+$ $a_{2}+\cdots+a_{n}=1$. Find the minimum value of $\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+$ $\frac{a_{2}}{1+a_{3}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{1+a_{2}+\cdots+a_{n-1}}$. | Let $S=\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}$
$$
\begin{array}{l}
+\frac{a_{2}}{1+a_{3}+\cdots+a_{n}}+\cdots \\
+\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n-1}},
\end{array}
$$
then $S:=\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}=2 \sum_{i=1}^{n} \frac{1}{2-a_{i}}-n$.
Since $2-a_{i}>0$, by (*) we get
$\frac{1}{2-a_{i}} \geqslant 2 \la... | \frac{n}{2 n-1} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,821 |
Example 3. Let $a, b, c, d$ be positive real numbers satisfying $a b+b c+c d+$ $d a=1$. Prove that: $\frac{a^{3}}{b+c+d}+\frac{b^{3}}{a+c+d}+$ $\frac{c^{3}}{a+b+d}+\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3}$. | Prove that for $a, b, c, d \in R^{+}$ and (*), we have
$$
\frac{a^{2}}{b+c+d} \geqslant 2 \lambda a-(b+c+d) \lambda^{2},
$$
i.e., $\frac{a^{3}}{b+c+d} \geqslant 2 \lambda a^{2}-a(b+c+d) \lambda^{2}$. Equality holds if and only if $\lambda=\frac{a}{b+c+d}$.
Similarly, $\frac{b^{3}}{a+c+d} \geqslant 2 \lambda b^{2}-b(a+... | \frac{a^{3}}{b+c+d}+\frac{b^{3}}{a+c+d}+\frac{c^{3}}{a+b+d}+\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 707,822 |
Example 5. As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at $D$, $P$ is the midpoint of $AD$, $BP$ intersects $AC$ at $E$, $EF \perp BC$ at $F$. Given $AE=3$, $EC=$ 12. Find the length of $EF$. | Analysis: Consider $\triangle A B C$ as its machine triangle, given that $E$, $P$ are the fixed ratio points of $A C, A D$ respectively, hence from (*) we get $\left(1+\frac{C D}{D B}\right) \cdot \frac{3}{12}=1$, i.e., $\frac{C D}{D B}=3$.
From the projection theorem, we can find $B D=\frac{5}{2} \sqrt{3}, D C$ $=\fr... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,824 |
Example 2. If $a, b$ are both positive real numbers, and $\frac{1}{a}-\frac{1}{b}-$ $\frac{1}{a+b}:=0$, then $\left(\frac{b}{a}\right)^{3}+\left(\frac{a}{b}\right)^{3}=$ $\qquad$ | Given that both sides are multiplied by $a+b$, we know $\frac{b}{a}-\frac{a}{b}=1$. Considering its dual form $\frac{b}{a}+\frac{a}{b}=A$, adding the two equations gives $\frac{2 b}{a}$ $=A+1$, and subtracting them gives $\frac{2 a}{b}=A-1$. Multiplying the last two equations yields $A^{2}$ $-1=4, A=\sqrt{5}(\because A... | 2 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,825 |
Example 4. $\cos ^{2} 10^{\circ}+\cos ^{2} 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}=$ | Consider the dual form $B=\sin ^{2} 10^{\circ}+$ $\sin ^{2} 50^{\circ}-\cos 40^{\circ} \cos 80^{\circ}$ of the original expression $A=$, then $A+B=2-$ $\cos 40^{\circ}, A-B=\frac{1}{2}+\cos 40^{\circ}$.
Thus, $A=\frac{3}{4}$. | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,827 |
1. Given that the lengths of the two diagonals of a trapezoid are $m$ and $n$, and the angle between the two diagonals is $60^{\circ}$. Then the area of the trapezoid is ( ).
(A) $\frac{\sqrt{3}}{8} m n$
(B) $\frac{\sqrt{3}}{4} m n$
(C) $\frac{\sqrt{3}}{2} m n$
(D) $\sqrt{3} m n$ | 1. B As shown in the figure, let $A C=m, B D=n$. Extend $A B$ to $E$ such that $B E=C D$. Then we have
$$
C E=B I=n \cdot \angle A C E=
$$
$\angle A O B$ is $60^{\circ}$ or $120^{\circ}$,
Also, $S_{\triangle A C D}=S_{\triangle H C E}$, thus
$$
S_{\triangle M E E}=\frac{1}{2} A C \cdot C E \sin \angle A C E=\frac{\sqrt... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,828 |
3. Given $a=3^{55}, b=4^{44}, c=5^{33}$. Then ( $\quad$ ).
(A) $a<b<c$
(B) $c<b<a$
(C) $c<a<b$
(D) $a<c<b$ | 3. C: From the problem, we have $a=243^{11}, b=256^{11}, c=125^{11}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,830 |
4. The number of positive integer solutions to the system of equations $\left\{\begin{array}{l}x y+y z=63, \\ x z+y z=23\end{array}\right.$ is ).
(A) 1
(B) 2
(C) 3
(D) 4 | 4. April From the second equation, we have $(x+y) z=23$. Since $x+y \geqslant 2$ and 23 is a prime number, it follows that $z=1, x+y=23$. Substituting $z=1, y=23-x$ into the first equation, we get $x^{2}-22 x+40=0$. The two roots of the equation are $x_{1}=2, x_{2}=20$. Solving this, we find that the system of equatio... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,831 |
5. If the three roots of the equation $(x-1)\left(x^{2}-2 x+m\right)=0$ can serve as the lengths of the three sides of a triangle, then the range of the real number $m$ is ( ).
(A) $0<m \leqslant 1$
(B) $m \geqslant \frac{3}{4}$
(C) $\frac{3}{4}<m \leqslant 1$
(D) $m \geqslant 1$ | 5. C. Let $\alpha, \beta$ be the roots of the equation $x^{2}-2 x+m=0$. From the discriminant, we get $m \leqslant 1$. It is evident that $\alpha+\beta=2>1$. Also, by the triangle inequality, we have $|\alpha-\beta|<\frac{3}{4}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,832 |
1. The minimum value of the function $y=2 x^{2}-x+1$ is | 二、1. $\frac{7}{8} \quad y=2\left(x-\frac{1}{4}\right)^{2}+\frac{7}{8}$.
---
Second, 1. $\frac{7}{8} \quad y=2\left(x-\frac{1}{4}\right)^{2}+\frac{7}{8}$. | \frac{7}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,834 |
3. Among the 35 numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 35^{2}$, the numbers with an odd digit in the tens place are $\qquad$ in total. | 3. 7. Among $1^{2}, 2^{2}, \cdots, 9^{2}$, those with an odd tens digit are only $4^{2}=16, 6^{2}=36$.
The square of a two-digit number can be expressed as
$$
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2} \text {. }
$$
It is evident that, when the unit digit is $A$ and $\hat{C}$, the tens digit of the square is odd. That is, on... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,837 |
4. Take line segment $AB$ as the diameter to draw a semicircle, with the center at $O$. $C$ is a point on the semicircle, and $OC^2 = AC \cdot BC$. Then $\angle CAB=$ $\qquad$ | 4. $15^{\circ}$ or $75^{\circ}$. As
shown in the figure, construct $C D \perp A B$, with
D being the foot of the perpendicular. By the area formula and the given
information, we have $C D=\frac{1}{2} O C$,
$$
\angle C O D=30^{\circ} \text {. }
$$
If point $D$ is on line segment $O B$, since $\triangle A O C$ is an i... | 15^{\circ} \text{ or } 75^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,838 |
One, (20 points) Given $C A=C B=C D$, the circle passing through points $A, C, D$ intersects $A B$ at point $F$ (see figure). Prove: $C F$ is the angle bisector of $\angle D C B$.
untranslated text remains the same as requested. | Connect $A D, D F$. Since $A$, $C, F, D$ are concyclic, then $\angle C F A=$ $\angle C D A$. And $\angle C D A=\angle C A D$, then $\angle C F A=\angle C A D$.
Thus, $\angle C F B=180^{\circ}-$ $\angle C F A=\angle C F D$.
Also, $\angle C B F=\angle C A F=\angle C D F$,
Therefore, $\angle B C F=\angle D C F$,
Hence, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,839 |
二、(25 points) Given positive integers $a, b, c$ satisfying the following conditions: $a>b>c,(a-b)(b$ $-c)(a-c c)=72$ and $a b c<100$. Find $a, b, c$. | II. Since $a-b, b-c, a-c$ are all positive integers, and $a-c=(a-b)+(b-c)$, when decomposing 72 into three factors, the only possibilities that satisfy the above conditions are $1,8,9$ or $8,1,9$. Therefore, we have
(1) $\left\{\begin{array}{l}a-b=1, \\ b-c=8, \\ a-c=9,\end{array}\right.$
(2) $\left\{\begin{array}{l}a-... | 10,9,1 \text{ or } 10,2,1 \text{ or } 11,3,2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,840 |
Three, (25 points) Given $y=\frac{x^{2}}{10}-\frac{x}{10}+\frac{9}{5}$, and $y \leqslant|x|$. Find the range of values for $x$.
| Three, when $x \geqslant 0$, the $x$ that meets the condition should satisfy $y-x \leqslant 0$, i.e., $\frac{1}{10}\left(x^{2}-11 x+18\right) \leqslant 0$. Since
$$
x^{2}-11x+18=(x-2)(x-3) \text {, }
$$
it is solved that $2 \leqslant x \leqslant 9$. Print $1+2 \leqslant x \leqslant 9$. Also, $\frac{1}{10}\left(x^{2}+9... | 2 \leqslant x \leqslant 9 \text { or }-6 \leqslant x \leqslant-3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,841 |
$$
\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)\left(1-\frac{1}{4^{2}}\right) \cdots \cdots \cdot\left(1-\frac{1}{10^{2}}\right) .
$$
Calculate:
$$
\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)\left(1-\frac{1}{4^{2}}\right) \cdots \cdots \cdot\left(1-\frac{1}{10^{2}}\right) .
$$ | $\begin{array}{l}\text { One Guest Formula }=\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right) \\ \cdots \cdots\left(1-\frac{1}{10}\right)\left(1+\frac{1}{10}\right) \\ =\frac{1}{8} \cdot \frac{8}{8} \cdot \cdots \cdot \frac{8}{10} \cdot \frac{8}{2} \cdot \frac{1}{... | \frac{11}{20} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,842 |
$$
\begin{array}{l}
\frac{a}{a^{3}+a^{2} b+6 b^{2}+b^{3}}+\frac{b}{a^{3}-a^{2} b+a b^{2}-b^{3}} \\
+\frac{1}{a^{2}-b^{2}}-\frac{1}{b^{2}+a^{2}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} .
\end{array}
$$ | \[
\begin{aligned}
=、 \text { Original expression } & =\frac{a}{(a+b)\left(a^{2}+b^{2}\right)} \\
& +\frac{b}{(a-b)\left(a^{2}+b^{2}\right)}+\frac{2 b^{2}}{a^{4}-b^{4}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} \\
= & \frac{a^{2}+b^{2} .}{\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)}-\frac{a^{2}+b^{2}}{a^{4}-b^{4}} \\
&... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,843 |
Three, given the system of equations $\left\{\begin{array}{l}x-y=2, \\ m x+y=6 .\end{array}\right.$ If the system has non-negative integer solutions, find the value of the positive integer $m$, and solve the system of equations. | Three, adding the two equations, we get $(m+1) x=8$.
Since the system of equations has non-negative integer solutions, and $m$ is a positive integer, the possible values of $m$ are $1,3,7$.
When $m=1$, we have $x=4$, solving for $y=2$.
When $m=3$, we have $x=2$, solving for $y=0$. When $m=7$, we have $x=1$. Solving for... | m=1, (x,y)=(4,2); m=3, (x,y)=(2,0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,844 |
Given, as shown in the figure, in $\triangle ABC$, $AB=AC$, and points $D$, $E$, $F$ are on $AB$, $BC$, $CA$ respectively, and $DE=EF=FD$.
Prove: $\angle DEB = \frac{1}{2}(\angle ADF + \angle CFE)$. | Obviously, $\triangle D E F$ is an equilateral triangle.
Since $\angle D E B=\angle A D F+60^{\circ}-\angle B$.
And $\angle D E B+60^{\circ}=\angle C F E+\angle C$,
which means $\angle D E B=\angle C F E-60^{\circ}+\angle C$.
(1) + (2) gives
$2 \angle D E B=\angle A E C+C E E-\angle B+\angle C$.
Since $\angle S=\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,845 |
Example 7. As shown in the figure, from the right-angle vertex $A$ of a right triangle, draw a perpendicular to the median $BP$ of the waist $AC$ (the foot of the perpendicular is $E$) and intersect the hypotenuse at $D$. Prove: $BD=2CD$.
保留源文本的换行和格式，直接输出翻译结果。 | Analysis: This problem can be transformed into proving that $C D: D B = \frac{1}{2}$. From (*) we know that it is only necessary to find the value of $A E * E D$, and the proof is complete. Note that $\mathrm{Rt} \triangle B A E \subset \mathrm{R} t \triangle A=E$, then $\frac{A E}{E P}=\frac{A B}{A P}=2$, i.e., $A E=2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,846 |
Six, a boat sails between
$A, B$ two docks. When sailing downstream, it takes 40 minutes to be 4 kilometers away from the destination, and when sailing upstream, it needs $1 \frac{1}{3}$ hours to arrive. It is known that the upstream speed is 12 kilometers per hour. Find the speed of the boat in still water. | Six, let the speed of the current be $x$ kilometers/hour. Then
$$
\frac{2}{3} x+4=1 \frac{1}{3} \times 12 \text {. }
$$
Solving for $x$ gives $x=18$.
Therefore, the speed of the boat in the stream is
$$
(18+12) \div 2=15 \text { (kilometers/hour). }
$$
Answer: The speed of the boat in still water is 15 kilometers/hou... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,847 |
Eight, given the figure, in $\triangle A B C$, $\angle E A C=90^{\circ}, A B=A C, B E$ bisects $\angle A B C, C E \perp B E$. Prove:
$$
C E=\frac{1}{2} B D .
$$ | Eight, Hint: Extend $C E$ and $B A$ to intersect at point $F$. Try to prove $\triangle A B D \cong \triangle A C F$.
Extend $C E$ and $B A$ to intersect at point $F$. Try to prove $\triangle A B D \cong \triangle A C F$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,849 |
Nine, the circumference of a circular track is 400 meters. Person A and Person B start running clockwise along the circular track at the same time and place. A runs 52 meters per minute, and B runs 46 meters per minute. Both A and B rest for 1 minute after every 100 meters they run. When will A catch up with B? | Nine, let $x$ meters be the distance at which Yi is caught by Jia. Then
$$
\frac{x+400}{52}+\left(\frac{x+400}{100}\right)=\frac{x}{46}+\left[\frac{x}{100}\right) .
$$
where $[x]$ denotes the integer part of $x$.
Solving for $x$ gives $x=\frac{13984}{3}$.
The time when Yi is caught is:
$$
\frac{13984}{3 \times 46}+\le... | 136 \frac{5}{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,850 |
Ten, given that $a, b, c$ are all positive integers, and satisfy $a^{2}+b^{2}=c^{2}$, and $a$ is a prime number. Prove:
(1) $b$ and $c$ must be one odd and one even,
(2) $2(a+b+1)$ is a perfect square. | (1) From the given,
$$
a^{2}=c^{2}-b^{2}=(c+b)(c-b) \text {. }
$$
If $a=2$, then $4=(c+b)(c-b)$.
In this case, $c+b=2, c-b=2$, then $b=0$.
Or $c+b=4, c-b=1$, then $c=\frac{5}{2}$.
Both lead to contradictions with the assumption. Hence, $a \neq 2$.
Therefore, $a$ is an odd prime number.
$$
\begin{array}{l}
\text { Also... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,851 |
3. If $D$ is a point on side $B C$ of $\triangle A B C$, and the centroids of $\triangle A B D$ and $\triangle A C D$ are $E, F$ respectively, then $S_{\triangle A E F}=S_{\triangle A B C}$ equals ).
(A) $4 \div 9$
(B) $1: 3$
(C) $2 \div 9$
(D) $1: 9$ | 3. $\mathrm{C}$ (Hint: Extend $A E, A F$ to intersect $B C$ at $M, N$. Then $M, N$ are the midpoints of $B D, D C$ respectively.) | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,854 |
5. Let $\frac{x}{x^{2}-m x+1}=1$. Then the value of $\frac{x^{3}}{x^{6}-m^{2} x^{3}+1}$ is ( ).
(A) 1
(B) $\frac{1}{m^{3}+3}$
(C) $\frac{1}{3 m^{2}-2}$
(D) $\frac{1}{3 m^{2}+1}$ | 5. C (Hint: From the condition, we know $x \neq 0$, thus $\frac{x^{2}-m x+1}{x}$ $=1$, which means $x+\frac{1}{x}=m+1$, then transform the required expression accordingly.) | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,856 |
3. $\frac{\sqrt{3}+2 \sqrt{5}+\sqrt{7}}{\sqrt{15}+\sqrt{21}+\sqrt{35}+5}$ simplifies to | 3. $\frac{1}{2}(\sqrt{7}-\sqrt{3})$ | \frac{1}{2}(\sqrt{7}-\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,861 |
4. In trapezoid $A B C D$, $A D / / B C, E, F$ are points on $A B, C D$ respectively, $E F / / B C$ and $E F$ bisects the area of the trapezoid. If $A D=a, B C=b$, then the length of $E F$ is $\qquad$ | 4. $\sqrt{\frac{a^{2}+b^{2}}{2}}$ | \sqrt{\frac{a^{2}+b^{2}}{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,862 |
Three, (20 points) Solve the equation
$$
\left(\frac{x}{x-1}\right)^{2}+\left(\frac{x}{x+1}\right)^{2}=\frac{40}{9} .
$$ | Three, removing the denominators, we get
$$
x^{2}(x+1)^{2}+x^{2}(x-1)^{2}=\frac{40}{9}\left(x^{2}-1\right)^{2},
$$
After simplification, we have $x^{2}\left(x^{2}+1\right)=\frac{20}{9}\left(x^{2}-1\right)^{2}$. Let $y=x^{2}$, the original equation becomes $y(y+1)=\frac{20}{9}(y-1)^{2}$, which simplifies to $11 y^{2}+4... | x= \pm 2, x= \pm \frac{\sqrt{55}}{11} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,863 |
1. Given the equation $2 x^{2}+2 k x-13 k+1=0$ has two real roots whose squares sum to 13. Then $k=$ $\qquad$ . | 1.1. By the relationship between roots and coefficients, we solve to get $k=1$ or $k=$ -14. Substituting these values back into the original equation, we find that $k=-14$ does not satisfy the conditions, so it is discarded; $k=1$ satisfies the conditions. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,866 |
$\begin{array}{l}\text { 3. } \frac{1}{2 \sqrt{1}+\sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}+\cdots+ \\ \frac{1}{25 \sqrt{24}+24 \sqrt{25}}=\end{array}$ | 3. $\frac{4}{5}$ Hint: Apply $\frac{1}{(k+1) \sqrt{k}+k \sqrt{k+1}}=$ $\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$, let $k=1,2, \cdots, 24$ be substituted respectively, and the result is obtained. | \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,869 |
4. In $\triangle A B C$, $A C=2, D$ is the midpoint of $A B, E$ is a point on $C D$, $E D=\frac{1}{3} C D$. If $C E=\frac{1}{3} A B$, and $C E \perp$ $A E$, then $B C=$ $\qquad$. | 4. $2 \sqrt{2}$ From the given, $E$ is the centroid of $\triangle ABC$, and $CE=\frac{2}{3} CD$. Also, $CE=\frac{1}{3} AB$, so $CD=\frac{1}{2} AB$, thus $\angle ACB=90^{\circ}$. Extending $AE$ to intersect $BC$ at $F$, then $AE \cdot AF = AC^2 = 4$, hence $CF^2 = EF \cdot AF = \frac{1}{2} AE \cdot AF = 2$, so $BC = 2 C... | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,870 |
5. A four-digit number has the following property: dividing this four-digit number by its last two digits yields a perfect square (if the tens digit is zero, then divide by the units digit), and this perfect square is exactly the square of the first two digits plus 1. For example, $4802 \div 2$ $=2401=49^{2}=(48+1)^{2}... | 5. 1085 Let this four-digit number be $100 c_{1}+c_{2}$, where $10 \leqslant c_{1} \leqslant 99, 1 \leqslant c_{2} \leqslant 99$. According to the problem, we have
$$
\begin{array}{l}
100 c_{1}+c_{2}=\left(c_{1}+1\right)^{2} c_{2}=c_{1}^{2} c_{2}+2 c_{1} c_{2}+c_{2}, \\
\therefore 100 c_{1}=c_{1} c_{2}\left(c_{2}+2\ri... | 1085 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,871 |
II. (30 points) Given positive numbers $m, n$ are the roots of the quadratic equation $x^{2}+$ $p x+q=0$, and $m^{2}+n^{2}=3, m n=1$. Find the value of the polynomial $x^{3}-(\sqrt{5}-1) x^{2}-(\sqrt{5}-1) x+1994$. | $$
=, \because m, n>0, \therefore m+n=5 .
$$
Thus, $p=-(m+n)=-\sqrt{5}, q=mn=1$,
$$
\therefore x^{2}-\sqrt{5} x+1=0 \text {. }
$$
$$
\text { Also } \begin{array}{l}
x^{3}-(\sqrt{5}-1) x^{2}-(\sqrt{5}-1) x+1994 \\
=\left(x^{3}+1\right)-(\sqrt{5}-1) x(x+1)+1993 \\
=(x+1)\left[\left(x^{2}-x+1\right)-(\sqrt{5}-1) x\right... | 1993 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,872 |
Three. (30 points) As shown in the figure, it is known that $\angle x O y=90^{\circ}$, points $A$, $B$ move on rays $O x$, $O y$ respectively, the internal angle bisector of $\angle O A B$ intersects with the external angle bisector of $\angle O B A$ at $C$. Is the size of $\angle A C B$ variable? Prove your conclusion... | Three, the size of $\angle A C B$ remains constant.
$$
\begin{aligned}
\because \angle O B C & =\frac{1}{2}\left(180^{\circ}-\angle O B A\right) \\
& =90^{\circ}-\frac{1}{2} \angle O B A, \\
\therefore \angle A B C & =\angle O B C+\angle O B A \\
& =90^{\circ}-\frac{1}{2} \angle O B A+\angle O B A \\
& =90^{\circ}+\fra... | 45^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 707,873 |
Four. (30 points) As shown in the figure, $AC$ is a vertical pole standing on a flat ground, and $AB$ is a taut steel cable used to secure the pole. It is known that the length of the steel cable is $l$, and the point $D$ on the steel cable is at a distance $d$ from both the pole and the ground. Try to express the dist... | Four, $\because \triangle A D E \sim \triangle A B C$,
$$
\therefore \frac{D E}{B C}=\frac{A E}{A C}, \frac{d}{h_{2}}=\frac{h_{1}-d}{h_{1}},
$$
It is known that $h_{1}+h_{2}=\frac{h_{1} h_{2}}{d}$. Let the value of the above expression be $m$. By Vieta's formulas, $h_{1}, h_{2}$ are the roots of the equation $x^{2}-m ... | x=\frac{1}{2}\left(d+\sqrt{d^{2}+l^{2}} \pm \sqrt{l^{2}-2 d^{2}-2 d \sqrt{d^{2}+l^{2}}}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,874 |
1. Given
$$
\left\{\begin{array}{l}
-a+b+c+d=x, \\
a-b+c+d=y, \\
a+b-c+d=z, \\
a+b+c-d=w,
\end{array}\right.
$$
and $b \neq 0$. If $x-y+z+w=k b$, then the value of $k$ is $\qquad$ | $-1.4$ | -1.4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,875 |
2. After being renovated by a worker, a boiler can save $20 \%$ coal and the thermal efficiency is improved by $10 \%$. Later, the worker renovated it again, and compared to the boiler after the first renovation, it saves another $20 \%$ coal, and the thermal efficiency is improved by another $10 \%$. After the two ren... | 2. 36,21 | 36, 21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,876 |
3. Given that the area of $\triangle A B C$ is $1, D$ is the midpoint of $B C, E$, $F$ are on $A C, A B$ respectively, and $S_{\triangle B D P}=\frac{1}{5}, S_{\triangle C D E}=\frac{1}{3}$. Then $S_{\triangle D S P}=$ $\qquad$ | (7)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems there was a misunderstanding in your request. You asked me to translate a text, but the text you provided is just a number in parentheses. If you need a ... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,879 |
5. In $\triangle A B C$, $A B=A C=7, B C=4$, point $M$ is on $A B$, and $B M=\frac{1}{3} A B$. Draw $E F \perp B C$, intersecting $B C$ at $E$ and the extension of $C A$ at $F$. Then the length of $E F$ is $\qquad$ | 5. $5 \sqrt{5}$ | 5 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,880 |
6. Solve
$$
\left\{\begin{array}{l}
\sqrt{x+5}+\frac{2}{y}=3, \\
x+\frac{4}{y^{2}}=0
\end{array}\right.
$$
The solution is $\qquad$ | 6. $x_{1}=-4, y_{1}=1, x_{2}=-1, y_{2}=2$ | x_{1}=-4, y_{1}=1, x_{2}=-1, y_{2}=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,881 |
8. The square of the first and last digits of a four-digit number adds up to 13, and the square of the middle two digits adds up to 85. If 1089 is subtracted from this number, the result is a four-digit number written with the same digits but in reverse order. The original four-digit number is $\qquad$ | 8. 3762 | 3762 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,883 |
II. (16 points) As shown in the figure, extend the sides $AB, BC, CD, DA$ of quadrilateral $ABCD$ to $E, F, G, H$ respectively, such that $\frac{BE}{AB} = \frac{CF}{BC} = \frac{DG}{CD} = \frac{AH}{DA} = m$. If $S_{EFGH} = 2 S_{ABCD}$ ($S_{EFGH}$ represents the area of quadrilateral $EFGH$), find the value of $m$.
---
... | II. Connect $G A, H B, E C, F D$ and the diagonals $A C, B D$. Then, by the given conditions, we have
$S_{\triangle H A B}=(m+1) S_{\triangle H A B}=(m+1) m S_{\triangle A B D}$.
Similarly, $S_{\triangle F C C}=(m+1) m S_{\triangle B C D}$.
Therefore, $S_{\triangle H A E}+S_{\triangle F C C}=m(m+1) S_{\triangle B C D}$... | \frac{\sqrt{3}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,886 |
Three, (16 points) Does there exist a four-digit number such that when any 3 of its 4 digits are replaced by any other digits (without changing their original positions), the resulting number is not a multiple of 1994? Prove your conclusion.
| Three, Existence
There are 6 numbers that are multiples of 1994 and less than 10000: $0000, 1994, 3988, 5982, 7976, 9970$.
As long as a four-digit number $\overline{a_{1} a_{3} a_{3} a_{4}}$ (for example, 4721) is selected such that $a_{i} (i = 1,2,3,4)$ does not equal the $i$-th digit of the above 6 numbers, the cond... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,887 |
Four, (18 points) In a round-robin football tournament with $n$ teams (i.e., each team must play a match against every other team), each match awards 2 points to the winning team, 1 point to each team in the case of a draw, and 0 points to the losing team. The result is that one team has more points than any other team... | Let the team with the highest score be Team $A$, and assume Team $A$ wins $k$ games, draws $m$ games, and thus Team $A$'s total points are $2k + m$.
From the given conditions, every other team must win at least $k+1$ games, and their points must be no less than $2(k+1)$. Therefore, we have $2k + m > 2(k+1)$, which sim... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,888 |
Example 1. If $m$ satisfies the equation
$$
\begin{array}{l}
\sqrt{3 x+5 y-2-m}+\sqrt{2 x+3 y-m} \\
=\sqrt{x-199+y} \cdot \sqrt{199-x-y},
\end{array}
$$
try to determine the value of $m$. | Analyzing this problem, we have one equation with three unknowns, so we can only solve it cleverly. Observing the characteristics of each algebraic expression in the original equation, we know that
$$
x-199+y \geqslant 0,
$$
and $199-x-y \geqslant 0$.
Thus, $x+y \geqslant 199$,
and $x+y \leqslant 199$.
Therefore, $x+y... | 201 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,890 |
Simplify $\sqrt{8+2 \sqrt{10+2 \sqrt{5}}}$ $+\sqrt{8-2 \sqrt{10+2 \sqrt{5}}}$, the value equals ( ).
(A) 4
(B) $2+\sqrt{5}$
(C) $\sqrt{10}+\sqrt{2}$
(D) $2 \sqrt{5}+2$
(1994, Zu Chongzhi Cup Mathematics Competition) | $$
\begin{array}{l}
\text { Sol } \quad \text { Let the original expression }=x, \quad \text { then } x^{2}=8+ \\
2 \sqrt{10+2 \sqrt{5}}+2 \sqrt{64-4(10+2 \sqrt{5})} \\
+8-2 \sqrt{10-2 \sqrt{5}}, \\
x^{2}=16+4(\sqrt{5}-1) \\
\quad=(\sqrt{10}+\sqrt{2})^{2} . \\
\because x>0, \\
\therefore x=\sqrt{10}+\sqrt{2} .
\end{arr... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,891 |
Example 11. If the decimal part of $\sqrt{11}$ is $a$, and the decimal part of $\frac{1}{a}$ is $b$, then $b=(\quad$.
(1994, Hope Cup Mathematics Competition) | Since $9<11<16$, then $3<\sqrt{11}<4$.
Let $a=\sqrt{11}-3$,
$$
\begin{array}{l}
\frac{1}{a}=\frac{1}{\sqrt{11}-3}=\frac{\sqrt{11}+3}{2} . \\
\text { Also } b<\sqrt{11}+3<7,3<\frac{\sqrt{11}+3}{2}<3.5,
\end{array}
$$
Therefore, $b=\frac{\sqrt{11}+3}{2}-3=\frac{\sqrt{11}-3}{2}$. | \frac{\sqrt{11}-3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,892 |
4. As shown in the upper right figure, the radii of semicircles $A$ and $B$ are equal, they are tangent to each other, and both are internally tangent to the semicircle $O$ with radius 1. $\odot O_{1}$ is tangent to both of them, and $\odot O_{2}$ is tangent to $\odot O_{1}$, semicircle $O$, and semicircle $B$. Then th... | 4. $\frac{1}{6}$.
Connect $\mathrm{AO} \mathrm{O}_{1}, \mathrm{O}_{1} \mathrm{O}, \mathrm{OO}_{2}, \mathrm{O}_{1} \mathrm{O}_{2}, \mathrm{O}_{2} \mathrm{~B}$. Let the radius of $\odot \mathrm{O}_{1}$ be $r_{1}$, then $A O^{2}+O O_{1}^{2}=A O_{1}^{2}$, i.e.,
$$
\begin{array}{l}
\left(\frac{1}{2}\right)^{2}+\left(1-r_{1... | \frac{1}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,893 |
$$
\left\{\begin{array}{l}
x-y=29, \\
\sqrt[3]{x-1}-\sqrt[3]{y+2}=2 .
\end{array}\right.
$$
One, (20 points) Solve the system of equations | Let $\sqrt[3]{x-1}=a, \sqrt[3]{y+2}=b$, then the original system of equations becomes
$$
\left\{\begin{array}{l}
a^{3}-b^{3}=26, \\
a-b=2 .
\end{array}\right.
$$
From (1), we get
$$
a^{3}-b^{3}=(a-b)\left[(a-b)^{2}+3 a b\right]=26,
$$
which simplifies to $a b=3$.
Solving (2) and (3) together, we get $a=3, b=1$, or $a... | x=28, y=1 \text{ or } x=0, y=-29 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,894 |
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