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2. A class takes a test, which consists of 20 multiple-choice questions. Each correct answer earns 5 points, unanswered questions earn 1 point, and wrong answers earn 0 points. Therefore, which of the following scores ( ) is impossible.
(A) 91
(B) 92
(C) 95
(D) 97 | 2.D.
To score above 90 points, a student must answer 18 or 19 questions correctly.
(1) If 18 questions are answered correctly, the scoring for the remaining two questions is $0+0, 0+1, 1+0, 1+1$. Therefore, the possible scores are $90, 91, 92$.
(2) If 19 questions are answered correctly, the score is $95+0$ or 95 +1, ... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,453 |
$3.0 、 I$ are the circumcenter and incenter of acute $\triangle A B C$, respectively. $O^{\prime} 、 I^{\prime}$ are the reflections of $O 、 I$ over $B C$, respectively. It is known that $A 、 B 、 O^{\prime} 、 C$ are concyclic. Then, the position of point $I^{\prime}$ is ( ).
(A) outside $\odot O$
(B) inside $\odot O$
(C... | 3.C.
As shown in Figure 4, since points $A$, $B$, $O'$, and $C$ are concyclic, we have
$$
\begin{array}{l}
\angle A + \angle B O' C = 180^{\circ}. \\
\text{Therefore, } \angle A + \angle B O C \\
= 180^{\circ}. \\
\text{Also, } \angle B O C = 2 \angle A, \text{ so }
\end{array}
$$
$\angle A = 60^{\circ}$.
Thus, $\ang... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,454 |
4. 10 people sit around a round table, each choosing a number and telling it to their two neighbors, then each person reports the average of the two numbers they heard. Figure 1 shows the numbers everyone reported. Then the person who reported the number 6, the number he originally chose is ( ).
(A) 1
(B) 3
(C) 2
(D) 6 | 4.A.
Let the number originally chosen by the person who reported the number 6 be $x$. According to the problem, the numbers originally chosen by the people who reported the numbers $4, 2, 10, 8$ are $10-x, x-4, 6-x, x+12$ respectively. Therefore, we have
$$
x+x+12=2 \times 7 \text {. }
$$
Solving for $x$ gives $x=1$. | A | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 714,455 |
5. As shown in Figure 2, in $\triangle A B C$, $\angle C=$ $90^{\circ}, B C=a, C A=b$, $A B=c$, the two angle bisectors $A E$ and $B F$ intersect at point 0. Then $\frac{S_{\triangle O E F}}{S_{\triangle O A B}}=(\quad)$.
(A) $\frac{a+b-c}{a+b+c}$
(B) $\frac{a-b+c}{a+b+c}$
(C) $\frac{-a+b+c}{a+b+c}$
(D) $\frac{2 c-a-b... | 5.A.
From the properties of the incenter, we have
$$
\frac{O E}{O A}=\frac{a}{b+c}, \frac{O F}{O B}=\frac{b}{a+c} \text {. }
$$
Since $c^{2}=a^{2}+b^{2}$, we have
$$
\begin{array}{l}
\frac{S_{\triangle O E F}}{S_{\triangle O A B}}=\frac{O E \cdot O F}{O A \cdot O B}=\frac{a b}{(b+c)(a+c)} \\
=\frac{2 a b}{2\left(a b+... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,456 |
6. Set $R$ represents the set of points $(x, y)$ in the plane, where $x^{2}+6 x+1+y^{2}+6 y+1 \leqslant 0$, and $x^{2}+6 x+1-\left(y^{2}+\right.$ $6 y+1) \leqslant 0$. Then, the area of the plane represented by set $R$ is closest to ( ).
(A) 25
(B) 24
(C) 23
(D) 22 | 6.A.
From the given, we have $(x+3)^{2}+(y+3)^{2} \leqslant 16$,
$$
(x-y)(x+y+6) \leqslant 0 \text {. }
$$
The first inequality represents the interior and boundary of a circle with center $(-3,-3)$ and radius 4; the second inequality can be written as $x-y \geqslant 0$ and $x+y+6 \leqslant 0$, or $x-y \leqslant 0$ a... | 25 | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,457 |
$$
\begin{array}{l}
\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{49}-\frac{1}{50}\right) \div\left(\frac{1}{20}+\frac{1}{77}+\cdots+\frac{1}{50}\right) \\
= \\
\end{array}
$$ | $=、 1.1$
$$
\begin{array}{l}
\text { Original expression }=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{49}+\frac{1}{50}\right)-\right. \\
\left.\quad\left(1+\frac{1}{2}+\cdots+\frac{1}{25}\right)\right] \div\left(\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\right) \\
=\left(\frac{1}{26}+\frac{1}{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,458 |
2. Let the quadratic function $f(x)$ satisfy $f(x+2)=f(-x+2)$, and its graph intersects the $y$-axis at the point $(0,1)$, and the segment it intercepts on the $x$-axis has a length of $2 \sqrt{2}$. Find the analytical expression of $f(x)$.
(2002, Anhui Province High School Mathematics Competition) | (Given that $f(x)$ is symmetric about $x=2$, we can set $f(x)=a(x-$ $2)^{2}+b$. From $4 a+b=1,\left|x_{1}-x_{2}\right|=2 \sqrt{2}$, i.e., $2 \sqrt{-\frac{b}{a}}$ $=2 \sqrt{2}$, we know $\left.f(x)=\frac{1}{2}(x-2)^{2}-1.\right)$ | f(x)=\frac{1}{2}(x-2)^{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,459 |
2. Fold a square piece of paper to form a triangle and a quadrilateral. It is known that the ratio of their areas is $3: 5$. Then the ratio of their perimeters is | 2. $\frac{6}{7}$.
As shown in Figure 5, $ABCD$ is a square, and point $M$ is on side $AB$. Let the side length of the square be $1$, $BM=x$, then $AM=1-x$. Therefore,
$$
\frac{3}{5}=\frac{S_{\triangle ADM}}{S_{\text{quadrilateral } BCDM}}=\frac{\frac{1}{2} AM \cdot AD}{\frac{1}{2}(BM+DC) \cdot BC}=\frac{1-x}{1+x} \tex... | \frac{6}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,460 |
3. The number of real roots of the equation $x^{2}|x|-5 x|x|+2 x=0$ is $\qquad$.
The equation $x^{2}|x|-5 x|x|+2 x=0$ has $\qquad$ real roots. | 3.4 .
From the given, we have $x(x|x|-5|x|+2)=0$.
Thus, $x=0$ or $x|x|-5|x|+2=0$.
(1) When $x>0$, $x^{2}-5 x+2=0$, which has two distinct positive real roots;
(2) When $x<0$, $-x^{2}+5 x+2=0$, which has one negative real root.
Therefore, the original equation has 4 real roots. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,461 |
One. (20 points) As shown in Figure 3, the diameter $AB = m$ (fixed length) of $\odot O$. Two moving circles are internally tangent to $\odot O$ and tangent to $AB$. A tangent line through $B$ to $\odot O$ intersects the rays $AO_{1}$ and $AO_{2}$ at $E_{1}$ and $E_{2}$, respectively; a tangent line through $A$ to $\od... | Let $\odot O_{1}$ be tangent to $AB$ at point $D$, and connect $O_{1}D, OO_{1}$.
Let $AD=a, DB=b$, then
$AB=a+b$,
$OA=OB=\frac{1}{2}(a+b), OD=\frac{1}{4}(a-b)^{2}$.
Let the radius of $\odot O_{1}$ be $x$, then
$$
OO_{1}=\frac{1}{2}(a+b)-x \text{. }
$$
In the right triangle $\triangle OO_{1}D$, we have $OO_{1}^{2}=OD^{... | m^2 | Geometry | proof | Yes | Yes | cn_contest | false | 714,463 |
For each positive integer $n$, let $f(n)$ represent the last digit of $1+2+\cdots+n$. For example, $f(1)=$
$$
\begin{aligned}
1, f(2)=3, f(5) & =5 \text {. Find } \\
f(1)+f(2) & +f(3)+\cdots+f(2004) .
\end{aligned}
$$ | $$
\begin{array}{l}
\text { Given } f(1)=1, f(2)=3, f(3)=6, f(4)=0, \\
f(5)=5, f(6)=1, f(7)=8, f(8)=6, f(9)=5, \\
f(10)=5, f(11)=6, f(12)=8, f(13)=1, f(14)=5, \\
f(15)=0, f(16)=6, f(17)=3, f(18)=1, f(19)=0, \\
f(20)=0, f(21)=1, f(22)=3, f(23)=6, \cdots
\end{array}
$$
From the above, we can see that the period of $f(n)... | 7010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,464 |
Three, (25 points) The integers $x_{0}, x_{1}, \cdots, x_{2000}$ satisfy the conditions: $x_{0}=0,\left|x_{1}\right|=\left|x_{0}+1\right|,\left|x_{2}\right|=\left|x_{1}+1\right|$, $\cdots,\left|x_{2004}\right|=\left|x_{2003}+1\right|$. Find the minimum value of $\mid x_{1}+x_{2}+\cdots+$ $x_{2004} \mid$. | Three, from the known we can get
$$
\left\{\begin{array}{l}
x_{1}^{2}=x_{0}^{2}+2 x_{0}+1, \\
x_{2}^{2}=x_{1}^{2}+2 x_{1}+1, \\
\cdots \cdots . \\
x_{2004}^{2}=x_{2003}^{2}+2 x_{2003}+1 .
\end{array}\right.
$$
Thus, $x_{2004}^{2}=x_{0}^{2}+2\left(x_{0}+x_{1}+\cdots+x_{2003}\right)+2004$.
Given $x_{0}=0$, then
$$
\begi... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,465 |
1. If $x, y>0, \log _{x} y+\log , x=\frac{10}{3}, x y=$ 144, then the value of $x+y$ is ( ).
(A) $20 \sqrt{3}$
(B) $26 \sqrt{3}$
(C) $24 \sqrt{3}$
(D) $10 \sqrt{3}$ | $-、 1 . B$.
Since $\log _{x} y=\frac{1}{\log _{y} x}$, and $\log _{x} y+\log x=\frac{10}{3}$, thus $\log _{x} y=3$ or $\frac{1}{3}$.
Assume $\log _{x} y=3$, then $y=x^{3}$.
Substitute into $x y=144$, solve to get $x=2 \sqrt{3}, y=24 \sqrt{3}$.
Therefore, $x+y=26 \sqrt{3}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,466 |
2. Let $z=m+n \mathrm{i}$ be a complex root of the equation
$$
a z^{4}+\mathrm{i} b z^{3}+c z^{2}+\mathrm{i} d z+e=0
$$
where $a, b, c, d, e, m, n \in \mathbf{R}$. Then which of the following numbers must also be a root of the equation? ( ).
(A) $m-n$ i
(B) $-m+n$ i
(C) $n+m$ i
(D) $1-m+n \mathrm{i}$ | 2.B.
Since $z=m+n$ i is a root of the known equation, we have
$$
a(\mathrm{i} z)^{4}-b(\mathrm{i} z)^{3}-c(\mathrm{i} z)^{2}+d(\mathrm{i} z)+e=0 \text {. }
$$
This means that $z^{\prime}:(m+n \mathrm{i}) \mathrm{i}=-n+m \mathrm{i}$ is a root of the equation
$$
a x^{4}-b x^{3}-c x^{2}+d x+e=0
$$
Since the complex roo... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,467 |
3. Given that the terms of the arithmetic sequence $\left\{a_{n}\right\}$ and the geometric sequence $\left\{b_{n}\right\}$ are all positive, and $a_{1}=b_{1}, a_{2 n+1}=b_{2 n+1}$. Then, it must be true that ( ).
(A) $a_{n+1}b_{n+1}$
(D) $a_{n+1} \geqslant b_{n+1}$ | 3.D.
$$
a_{n+1}=\frac{a_{1}+a_{2 n+1}}{2}=\frac{b_{1}+b_{2 n+1}}{2} \geqslant \sqrt{b_{1} b_{2 n+1}}=b_{n+1} \text{. }
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,468 |
4. In a Cartesian coordinate system, points $A$, $B$, and $C$ are not collinear. Now, with points $B$ and $C$ as foci, a hyperbola is drawn passing through $A$. With points $A$ and $B$ as foci, a hyperbola is drawn passing through $C$. With points $A$ and $C$ as foci, a hyperbola is drawn passing through $B$. Then, the... | 4.D.
Let the first two hyperbolas intersect at point $E$, then
$$
E B - E C = A B - A C, \quad E B - E A = B C - A C.
$$
Subtracting the two equations gives $E C - E A = B C - A B$.
Thus, $E$ lies on the third hyperbola. Therefore, the three hyperbolas intersect at one or two points (since point $E$ may have two posi... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,469 |
3. If on the interval $[1,2]$, the function $f(x)=x^{2}+p x+q$ and $g(x)=x+\frac{1}{x^{2}}$ take the same minimum value at the same point, then the maximum value of $f(x)$ on this interval is $(\quad)$. | (Hint: $g(x)=x+\frac{1}{x^{2}}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^{2}} \geqslant \frac{3}{2} \sqrt[3]{2}$. Also, $f(x)$ and $g(x)$ achieve the same minimum value at the same point, so $-\frac{p}{2}=$ $\sqrt[3]{2}, \frac{4 q-p^{2}}{4}=\frac{3}{2} \sqrt[3]{2}$. Solving these, we get $p=-2 \sqrt[3]{2}, q=\frac{3}{2} \sqrt... | 4-\frac{5}{2} \sqrt[3]{2}+\sqrt[3]{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,470 |
5. Given $f(x)=x^{2}+p x+q, p^{2}+1 \leqslant 4 q+$ $2 p$ holds. Let the set of real solutions to the equation $f(x)=x$ be $P$, and the set of real solutions to the equation $f(f(x))=x$ be $Q$. Then ( ).
(A) $P=Q$
(B) $P \subset Q$
(C) $Q \subset P$
(D) $P \not \subset Q, Q \not \subset P$ | 5.A.
When $p=1, q=0$, $f(x)=x^{2}+x$.
Obviously, we can get option (A). Here, we also provide a theoretical basis.
From the given, we have $(p-1)^{2}-4 q \leqslant 0$, thus,
$$
f(x)-x=x^{2}+(p-1) x+q \geqslant 0
$$
holds for any $x$. For any $x_{0} \in P, f\left(x_{0}\right)=x_{0}$.
Therefore, $f\left(f\left(x_{0}\ri... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,471 |
6. In space, a finite set of points $M$ not all lying on the same plane has the property:
For any $A, B \in M$, there exist $C, D \in M$ such that $A B / / C D$.
Then a possible value for the number of points in the set $M$ is ( ).
(A)6
(B) 8
(C) 10
(D) 5 | 6.C.
As shown in Figure 2, as long as the plane $\alpha$ of the regular hexagon is rotated around one of its diagonals that can serve as an axis of symmetry by an angle $\theta$, four points of the hexagon will be rotated out of plane $\alpha$. These 10 points clearly meet the requirements. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,472 |
3. In $\triangle A B C$,
$$
\cos A+\sin A-\frac{2}{\cos B+\sin B}=0 \text {. }
$$
Then the value of $\frac{a}{c}$ is $\qquad$ | 3. $\frac{\sqrt{2}}{2}$.
From the given equation, we have
$$
\sqrt{2} \sin \left(A+\frac{\pi}{4}\right)-\frac{\sqrt{2}}{\sin \left(B+\frac{\pi}{4}\right)}=0,
$$
which means $\sin \left(A+\frac{\pi}{4}\right) \cdot \sin \left(B+\frac{\pi}{4}\right)=1$.
Since $0 \leqslant \sin \left(A+\frac{\pi}{4}\right), \sin \left(B... | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,475 |
4. In the Cartesian coordinate system, color the point set
$$
\{(m, n) \mid m, n \in \mathbf{N}, 1 \leqslant m, n \leqslant 6\}
$$
with red or blue. Then, the number of different coloring schemes where each unit square has exactly two red vertices is | 4.126.
First, color the first row (points with a y-coordinate of 6), which has $2^{6}$ ways of coloring. We can divide these into two cases:
(1) No two adjacent points are the same color (i.e., red and blue alternate), which gives 2 ways of coloring. In this case, there are only 2 ways to color the second row. Similar... | 126 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,476 |
5. The four sides of the quadrilateral pyramid are isosceles triangles with a leg length of $\sqrt{7}$ and a base length of 2. Then the maximum possible volume of this quadrilateral pyramid is | 5. 3 .
There are three scenarios for the quadrilateral pyramid that meet the requirements.
(1) As shown in Figure 4, all four lateral edges are $\sqrt{7}$, and the volume is calculated as $V_{1}=\frac{4}{3} \sqrt{5}$.
(2) As shown in Figure 5, two lateral edges are $\sqrt{7}$.
Construct $O E \perp$ plane $A B C D$, a... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,477 |
6. For any $\theta \in\left[0, \frac{\pi}{2}\right]$, such that
$$
|\sin \theta-p \cos \theta-q| \leqslant \frac{\sqrt{2}-1}{2}
$$
the real number pairs $(p, q)$ that always satisfy the condition are | $6 .\left(-1, \frac{1+\sqrt{2}}{2}\right)$.
The problem is equivalent to: For any $x \in [0,1]$, we have
$\left|\sqrt{1-x^{2}}-(p x+q)\right| \leqslant \frac{\sqrt{2}-1}{2}$.
Let $y_{1}=p x+q, y_{2}=\sqrt{1-x^{2}}$. Thus,
$y_{2}-\frac{\sqrt{2}-1}{2} \leqslant y_{1} \leqslant y_{2}+\frac{\sqrt{2}-1}{2}$.
Therefore, the ... | \left(-1, \frac{1+\sqrt{2}}{2}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,478 |
Three. (20 points) Determine all real numbers $\theta$ such that the complex sequence $\{\cos n \theta+\mathrm{i} \sin n \theta\}$ forms an arithmetic sequence.
| Three, let $a_{n}=\cos n \theta+\mathrm{i} \sin n \theta$. Then $\left\{a_{n}\right\}$ is an arithmetic sequence, and $a_{n}-a_{n-1}$ is a constant value independent of $n$. Also,
$$
\begin{array}{l}
a_{n}-a_{n-1} \\
=(\cos n \theta-\cos (n-1) \theta)+\mathrm{i}(\sin n \theta-\sin (n-1) \theta) \\
=2 \sin \frac{\theta}... | \theta=2 k \pi, k \in \mathbf{Z} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,479 |
Four, (20 points) Given the parabola $C: y=x^{2}-2$. Is there any three points $M, N, P$ on $C$ such that the incircle of $\triangle M N P$ is exactly $x^{2}+y^{2}=1$? If they exist, find all such points; if not, explain the reason.
---
Will translate the provided text into English, preserving the original formatting... | Let $M\left(x_{1}, x_{1}^{2}-2\right), N\left(x_{2}, x_{2}^{2}-2\right), P\left(x_{3}, x_{3}^{2}-2\right)$ be three points on a parabola.
The equations of the lines $MN$, $NP$, and $MP$ are:
$$
\begin{array}{l}
y-\left(x_{1}+x_{2}\right) x+x_{1} x_{2}+2=0, \\
y-\left(x_{2}+x_{3}\right) x+x_{2} x_{3}+2=0, \\
y-\left(x_{... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,480 |
4. Find the range of real numbers $a$ such that the inequality
$$
\begin{array}{l}
\sin 2 \theta-(2 \sqrt{2}+a \sqrt{2}) \sin \left(\theta+\frac{\pi}{4}\right)-2 \sqrt{2} \sec \left(\theta-\frac{\pi}{4}\right) \\
>-3-2 a
\end{array}
$$
holds for all $\theta \in\left[0, \frac{\pi}{2}\right]$.
(2000, Tianjin High School... | (Let $x=\sin \theta+\cos \theta$. Since $\theta \in\left[0, \frac{\pi}{2}\right]$, then $x \in[1, \sqrt{2}]$. Thus, $\sin 2 \theta=x^{2}-1, \sin \left(\theta+\frac{\pi}{4}\right)=$ $\cos \left(\theta-\frac{\pi}{4}\right)=\frac{x}{\sqrt{2}}$. The original inequality can be transformed into $(x-2)$. $\left(x+\frac{2}{x}-... | a>3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,481 |
Five. (20 points) Let $a$ be a given positive constant, and $n$ be a given natural number. If there exists a sequence of positive real numbers $\left\{x_{i}\right\}(i=1$, $2, \cdots, n)$ satisfying
(1) $\sum_{i=1}^{n} x_{i}=a$;
(2) $c \geqslant \frac{x_{j}}{1+x_{1}+x_{2}+\cdots+x_{j}}, 1 \leqslant j \leqslant n$,
find... | For $\left\{x_{i}\right\}$ that satisfy the requirements, we have
$$
1-c \leqslant \frac{1+x_{1}+\cdots+x_{j-1}}{1+x_{1}+\cdots+x_{j}}, 1 \leqslant j \leqslant n \text {. }
$$
Multiplying the above $n$ inequalities, we get
$(1-c)^{n}$
$\leqslant \frac{1}{1+x_{1}} \cdot \frac{1+x_{1}}{1+x_{1}+x_{2}} \cdots \cdot \frac{... | c = 1-(1+a)^{-\frac{1}{n}}, \quad x_{j}=(1+a)^{\frac{j}{n}}-(1+a)^{\frac{j-1}{n}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,482 |
One, (50 points) As shown in Figure 1, in the convex quadrilateral $ABCD$, points $P$ and $Q$ are located on sides $CD$ and $AB$ respectively. It is known that
\[
\begin{array}{l}
\angle CDQ = \angle CQB, \\
\angle BAP = \angle CPB. \text{ If } \\
\angle APB = \angle DQC, \text{ prove: } AD \parallel BC.
\end{array}
\] | Given $\angle C D Q=\angle C Q B, \angle B A P=\angle C P B$, we know that the necessary and sufficient condition for $\angle A P B=\angle D Q C$ is
$$
\angle C P B=\angle C Q B \text { and } \angle D P A=\angle A Q D \text {. }
$$
In fact, $\angle A P B=\angle D Q C$
$$
\begin{aligned}
\Leftrightarrow & \angle B A P+... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,483 |
For the given natural number $n(n>$ 5), find all real numbers $a$ such that there exist non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying
$$
\sum_{k=1}^{n} k x_{k}=a, \sum_{k=1}^{n} k^{3} x_{k}=a^{2}, \sum_{k=1}^{n} k^{5} x_{k}=a^{3} .
$$ | Assume $\left\{x_{i}\right\}$ satisfies the conditions, then,
$$
\sum_{k=1}^{n} k x_{k}=a, \sum_{k=1}^{n} k^{3} x_{k}=a^{2}, \sum_{k=1}^{n} k^{5} x_{k}=a^{3} .
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
a a^{3} & =\left(\sum_{k=1}^{n} k x_{k}\right)\left(\sum_{k=1}^{n} k^{5} x_{k}\right) \\
& \geq... | 1, 2^2, 3^2, \cdots, n^2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,484 |
Three, (50 points) Let $s$ be a natural number greater than 1. Find all natural numbers $n$ such that for $n$ there exist coprime natural numbers $x$ and $y$ satisfying $3^{n}=x^{s}+y^{s}$. | Three, first prove a lemma.
Lemma Let $x, y, p, n, k \in \mathbf{N}$, and satisfy $x^{n}+y^{n}=p^{k}$. If $n$ is an odd number greater than 1, and $p$ is an odd prime, then $n$ can be expressed as a power of $p$ with a natural number as the exponent.
Proof of the lemma: Let $m$ be the greatest common divisor of $x, y$... | n=2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,485 |
Let $E, F, G, H$ be points on the sides $AB, BC, CD, DA$ of square $ABCD$, respectively, and the areas of $\triangle AHE, \triangle BEF, \triangle CFG, \triangle DGH$ are $2, 28, t, 7$, respectively, where $t$ is a given positive real number. If the lines $EG, FH, BD$ intersect at a single point, find the area of quadr... | Solution: As shown in Figure 1, let the side length of the square $ABCD$ be $a$, and $AH = x, DG = y, CF = z, BE = w$, then $HD = a - x, GC = a - y, FB = a - z, EA = a - w$.
Thus, from the given conditions, we have
$$
\left\{\begin{array}{l}
x(a - w) = 4, \\
y(a - x) = 14, \\
z(a - y) = 2t, \\
w(a - z) = 56.
\end{array... | 3\sqrt{t^2 + 10t + 81} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,486 |
As 138 As shown in Figure 2, point $G$ is the centroid of $\triangle A B C$, and ray $A G$ intersects the circumcircle of $\triangle A B C$ at point $P$. Prove that: $A G \cdot G P \geqslant 2 R r$, where $R, r$ represent the circumradius and inradius of $\triangle A B C$, respectively. | Proof: Let $BC = a$, $CA = b$, $AB = c$, and $AP$ intersects $BC$ at point $D$. By the properties of the centroid, we know $BD = DC$.
Since $AG = \frac{2}{3} AD$,
$$
\begin{array}{l}
GD = \frac{1}{3} AD, \\
DP = \frac{BD \cdot DC}{AD} = \frac{a^2}{4 AD}.
\end{array}
$$
Also, $AD^2 = \frac{1}{2}\left(b^2 + c^2 - \frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,487 |
Given a height of 137, find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in $S$ is a positive integer not exceeding 100;
(2) For any two distinct elements $a$ and $b$ in $S$, there exists another element $c$ in $S$ such that the greatest common divisor (gcd) of $... | Solution: Construct 50 arrays:
$$
(1,100),(2,99) \cdots,(50,51) \text {, }
$$
The sum of the two numbers in each array is 101.
Since 101 is a prime number, there does not exist an element $c$ in $S$ such that the greatest common divisor of 101 and $c$ is greater than 1. Therefore, it is impossible for $S$ to contain b... | 50 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,488 |
For $a 、 b>0, a+b=1 . \alpha \in \mathbf{R}, \alpha \geqslant 2.0<\lambda$ $<\min \left\{\frac{2^{a}}{3}, 2\right\}$. Then
$$
\frac{2+\lambda}{1+\lambda}<\frac{1}{\lambda a^{a}+1}+\frac{1}{\lambda b^{a}+1} \leqslant \frac{2^{\alpha+1}}{2^{\alpha}+\lambda} .
$$ | Prove: The right-hand side of inequality (1) is equivalent to
$$
\begin{array}{l}
\left(2^{a}+\lambda\right)\left[\lambda\left(a^{a}+b^{a}\right)+2\right] \\
\leqslant 2^{a+1}\left(\lambda a^{a}+1\right)\left(\lambda b^{a}+1\right) \\
\Leftrightarrow \lambda\left[2^{a+1} \lambda a^{a} b^{a}+\left(2^{a}-\lambda\right)\l... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,489 |
Given in $\triangle A B C$, $\angle A C B=$ $90^{\circ}$, as shown in Figure 1. When point $D$ is on the hypotenuse $A B$ (excluding the endpoints), prove:
$$
\frac{C D^{2}-B D^{2}}{B C^{2}}=\frac{A D-B D}{A B} \text {. }
$$
(2003, National Junior High School Mathematics League) | Prove: Draw $C E \perp B A$ at point $E$. Let $B C=a, A B=c, C E=h, B D=m, A D=n, C D=t, B E=p, E D=k$. Clearly, $p+k=m$. Then
$$
\begin{array}{l}
\frac{C D^{2}-B D^{2}}{B C^{2}}=\frac{t^{2}-m^{2}}{a^{2}} \\
=\frac{h^{2}+k^{2}-m^{2}}{a^{2}}=\frac{h^{2}+(k+m)(k-m)}{a^{2}} .
\end{array}
$$
Substitute $h^{2}=p(k+n), k-m=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,490 |
5. If the function $f(x)=-\frac{1}{2} x^{2}+\frac{13}{2}$ has a minimum value of $2a$ and a maximum value of $2b$ on the interval $[a, b]$, find $[a, b]$.
(2000, National High School Mathematics Competition) | (Discuss in three cases. (1) If $0 \leqslant a<b$, $f(x)$ is decreasing on $[a, b]$, so $f(a)=2 b, f(b)=2 a$. Solving gives $[a, b]=[1,3]$. (2) If $a<0<b$, by monotonicity $(f(x))_{\max }=f(0)=2 b$, which gives $b=\frac{13}{4}, a=-2-\sqrt{17}$. Thus, $[a, b]=\left[-2-\sqrt{17}, \frac{13}{4}\right]$. (3) When $a<b \leqs... | [1,3] \text{ or } \left[-2-\sqrt{17}, \frac{13}{4}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,491 |
Example 2 Given positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying $x_{1} x_{2} \cdots x_{n}=1$. Prove:
$$
\frac{1}{n-1+x_{1}}+\frac{1}{n-1+x_{2}}+\cdots+\frac{1}{n-1+x_{n}} \leqslant 1 \text {. }
$$
Analysis: Start with a special case, when $x_{1}=x_{2}=\cdots=x_{n}=1$, the inequality holds. Then study t... | Prove: When $x_{1}=x_{2}=\cdots=x_{n}=1$, the inequality holds.
When $x_{1}, x_{2}, \cdots, x_{n}$ are not all 1, there must be one in $(0,1)$ and one in $(1,+\infty)$. By symmetry, without loss of generality, assume
$0 < x_{1} < 1 < x_{n}$. Then
$$
x_{1} x_{n} < (n-1)^2.
$$
First adjustment: Let $x_{1}^{\prime}=1, x... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,493 |
Example 3: In front of each number in $1, 2, 3, \cdots, 1989$, add a “+” or “-” sign to make their algebraic sum the smallest non-negative number, and write out the equation.
(1989, All-Russian Mathematical Olympiad) | Proof: First, we prove that the algebraic sum is an odd number.
Consider the simplest case:
If all are filled with “+”, then at this moment
$$
1+2+\cdots+1989=995 \times 1989
$$
is an odd number.
For the general case, it only requires adjusting some “+” to “-”.
Since $a+b$ and $a-b$ have the same parity, the parity o... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,494 |
Example 3 Given that $\alpha, \beta$ are the roots of the equation $x^{2}-x-1=0$. Then the value of $\alpha^{4}+3 \beta$ is $\qquad$
(2003, National Junior High School Mathematics Competition, Tianjin Preliminary Contest) | Explanation: $\alpha^{4}+3 \beta$ is not a symmetric expression of the two roots of the equation, and it is obviously impossible to directly substitute it using Vieta's formulas. We can construct the dual expression $\beta^{4}+3 \alpha$ of $\alpha^{4}+3 \beta$, and calculate the sum and difference of the two expression... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,495 |
Example 4 In space, there are 2003 points, where no three points are collinear. Divide them into 30 groups with different numbers of points. From any three different groups, take one point each as the vertex to form a triangle. To maximize the total number of such triangles, how many points should each group have? | Solution: Let the number of points in the 30 groups be $n_{1}$, $n_{2}, \cdots, n_{30}$, where $n_{i}(i=1,2, \cdots, 30)$ are all distinct, then the total number of triangles satisfying the conditions is
$$
s=\sum_{1 \leqslant i < j < k \leqslant 30} \left\{
\begin{array}{l}
1, \text{ if } n_{i} + n_{j} > n_{k}, \\
0, ... | 52, 53, ..., 73, 75, ..., 82 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,496 |
Example 8 As shown in Figure 10, in trapezoid $ABCD$, $AD // BC$. Two squares $ABCE$ and $DCHF$ are constructed on the legs $AB$ and $CD$ respectively. Let the perpendicular bisector $l$ of line segment $AD$ intersect line segment $EF$ at point $M$, $EP \perp l$ at $P$, and $FQ \perp l$ at $Q$. Prove: $EP = FQ$.
保留源文本... | Prove: Establish a complex plane with $BC$ as the real axis and $l$ as the imaginary axis, and use the letter of the point to represent the complex number of that point. Let
$$
\begin{array}{l}
A=-a+h \mathrm{i}, B=-b, \\
C=c, D=a+h \mathrm{i},
\end{array}
$$
where $a, b, c, h$ are all positive real numbers. According... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,498 |
Example 9 As shown in Figure 11, construct rhombuses $A B D E, C A F G, B C H K$ outward from the three sides of $\triangle A B C$, such that $\angle A B D=\angle C A F=\angle B C H$. Connect $E F, G H$, and $K D$. Prove: A triangle can be formed with $E F, G H$, and $K D$ as sides, and the area of the triangle formed ... | Proof: (1) It is easy to see that
$$
\begin{array}{l}
B D \perp A E, B K \perp C H, A F \perp C G ; \\
\angle A B D + \angle C B K = 180^{\circ} \text{. } \\
\end{array}
$$
Thus, $\angle D B K = 180^{\circ} - \angle A B C$.
Similarly, $\angle E A F = 180^{\circ} - \angle B A C$,
$$
\angle G C H = 180^{\circ} - \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,499 |
Example 10 In $\triangle A B C$, the opposite sides of the internal angles $A, B, C$ are $a, b, c$ respectively. Prove: that with
$$
\begin{array}{l}
a_{1}=\sqrt{a^{2}+b^{2}+2 a b \cos C}, \\
b_{1}=\sqrt{b^{2}+c^{2}+2 b c \cos A}, \\
c_{1}=\sqrt{c^{2}+a^{2}+2 c a \cos B}
\end{array}
$$
as sides, a triangle can be form... | Proof: Without loss of generality, let $a_{1} \leqslant b_{1} \leqslant c_{1}$,
$$
\begin{aligned}
( & \left.a_{1}+b_{1}+c_{1}\right)\left(a_{1}+b_{1}-c_{1}\right) . \\
& \left(a_{1}+c_{1}-b_{1}\right)\left(b_{1}+c_{1}-a_{1}\right) \\
= & {\left[\left(a_{1}+b_{1}\right)^{2}-c_{1}^{2}\right]\left[c_{1}^{2}-\left(a_{1}-b... | 3 S_{\triangle A B C} | Geometry | proof | Yes | Yes | cn_contest | false | 714,500 |
In the acute triangle $\triangle ABC$, $AD$ is the internal angle bisector of $\angle BAC$, and point $D$ is on side $BC$. Through point $D$, draw $DE \perp AC$ and $DF \perp AB$, with the feet of the perpendiculars being $E$ and $F$ respectively. Connect $BE$ and $CF$, and let them intersect at point $H$. The circumci... | Proof: As shown in Figure 1, construct $D G^{\prime} \perp B E$ at $G^{\prime}$, $A M \perp B C$ at $M$, and connect $F G^{\prime}$.
Let $\angle A B C = \alpha$, $\angle A C B = \beta$, then
$B M = A M \cot \alpha$,
$C M = A M \cot \beta$.
From the given information,
$$
\begin{array}{l}
B F = D F \cot \alpha, C E = D E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,501 |
1. Given $a b c \neq 0$, and $a+b+c=0$. Then the value of the algebraic expression $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}$ is ( ).
(A) 3
(B) 2
(C) 1
(D) 0 | $\begin{array}{l}\text { I. 1.A. } \\ \text { Original expression }=\frac{-(b+c) a}{b c}+\frac{-(a+c) b}{a c}+\frac{-(a+b) c}{a b} \\ =-\left(\frac{a}{b}+\frac{a}{c}\right)-\left(\frac{b}{a}+\frac{b}{c}\right)-\left(\frac{c}{a}+\frac{c}{b}\right) \\ =\frac{a}{a}+\frac{b}{b}+\frac{c}{c}=3 .\end{array}$ | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,502 |
2. Given that $p$ and $q$ are both prime numbers, and satisfy $5 p^{2} + 3 q = 59$. Then the triangle with side lengths $p+3, 1-p+q, 2 p+q-4$ is ( ).
(A) acute triangle
(B) right triangle
(C) obtuse triangle
(D) isosceles triangle | 2.B.
Since $5 p^{2}+3 q$ is odd, $p$ and $q$ must be one odd and one even. Since $p$ and $q$ are both prime numbers, one of $p$ and $q$ must be 2.
If $q=2$, then $p^{2}=\frac{53}{5}$, which does not satisfy the condition, so we discard it;
If $p=2$, then $q=13$. In this case,
$$
p+3=5,1-p+q=12,2 p+q-4=13 \text {. }
$$... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,503 |
3. The three sides of one triangle are $a$, $a$, and $b$, and the three sides of another triangle are $a$, $b$, and $b$, where $a > b$. If the smallest interior angles of the two triangles are equal, then $\frac{a}{b}$ equals ( ).
(A) $\frac{\sqrt{3}+1}{2}$
(B) $\frac{\sqrt{5}+1}{2}$
(C) $\frac{\sqrt{3}+2}{2}$
(D) $\fr... | 3.B.
As shown in Figure 7, in $\triangle ABC$, $AB = AC = a, BC = b$. $D$ is a point on $AB$ such that $AD = b$. Since $a > b$, $\angle A$ is the smallest angle in $\triangle ABC$, denoted as $\angle A = \theta$. Therefore, the smallest angle of the triangle with sides $b, b, a$ is also $\theta$. Thus, this triangle i... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,504 |
4. Draw a line through point $P(-1,3)$, so that the triangle formed with the two coordinate axes has an area of 5. How many such lines can be drawn?
(A) 4
(B) 3
(C) 2
(D) 1 | 4.C.
Let the line $l: y=kx+b$. Since $P(-1, 3)$ lies on the line $l$, we have $3=-k+b$, hence $b=k+3$.
Therefore, $l$ is $y=kx+k+3$.
The intersection points of $l$ with the two coordinate axes are $A\left(-\frac{k+3}{k}, 0\right)$, $B(0, k+3)$, so the area of the triangle formed by $l$ and the two coordinate axes is
$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,505 |
Example 4 Real numbers $a, b, x, y$ satisfy $a x + b y = 3$, $a x^2 + b y^2 = 7$, $a x^3 + b y^3 = 16$, $a x^4 + b y^4 = 42$. Find the value of $a x^5 + b y^5$.
(8th American High School Mathematics Examination) | Explanation: Let $S_{n}=a x^{n}+b y^{n}$. Then
$$
S_{n}=(x+y) S_{n-1}-x y S_{n-2}, n=3,4, \cdots \text {. }
$$
Substituting the known conditions into the above equation, we get
$$
\begin{array}{l}
7(x+y)-3 x y=16,16(x+y)-7 x y=42 . \\
\text { Solving, we get } x+y=-14, x y=-38 . \\
\text { Therefore, } S_{n}=-14 S_{n-... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,506 |
5. Given that $b^{2}-4 a c$ is a real root of the quadratic equation $a x^{2}+b x+c=0(a \neq 0)$. Then $(\quad)$.
(A) $a b \geqslant \frac{1}{8}$
(B) $a b \leqslant \frac{1}{8}$
(C) $a b \geqslant \frac{1}{4}$
(D) $a b \leqslant \frac{1}{4}$ | 5.B.
Since the equation has real solutions, we have $b^{2}-4 a c \geqslant 0$. According to the problem, we have
$$
\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}=b^{2}-4 a c
$$
or $\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}=b^{2}-4 a c$.
Let $u=\sqrt{b^{2}-4 a c}$, we get the equation
$2 a u^{2}-u+b=0$ or $2 a u^{2}+u+b=0$.
Since $u=\sqr... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,507 |
6. As shown in Figure 1, on a $2 \times 3$ rectangular grid paper, the vertices of each small square are called lattice points. Then the number of isosceles right triangles with lattice points as vertices is ( ) .
(A) 24
(B) 38
(C) 46
(D) 50 | 6.D.
The length of line segments with grid points as vertices can take 8 values: $1, \sqrt{2}$, $2, \sqrt{5}, 2 \sqrt{2}, 3, \sqrt{10}, \sqrt{13}$. The isosceles right triangles formed by these line segments can be classified into 4 cases based on their side lengths:
$$
1,1, \sqrt{2} ; \sqrt{2}, \sqrt{2}, 2 ; 2,2,2 \s... | 50 | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,508 |
1. Calculate $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots+$ $\frac{1}{\sqrt{2003}+\sqrt{2004}}=$ $\qquad$ | $\begin{array}{l}\text { II. 1.2 } \sqrt{501}-1 . \\ \text { Original expression }=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+ \\ \quad(\sqrt{2004}-\sqrt{2003}) \\ =\sqrt{2004}-1=2 \sqrt{501}-1 .\end{array}$ | 2\sqrt{501}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,509 |
2. Given real numbers $a, b$ satisfy $a^{3}+b^{3}+3 a b=1$. Then $a+b$ $=$ . $\qquad$ | 2.1 or -2.
Let $a+b=x$. From $a^{3}+b^{3}+3 a b=1$, we easily get $x^{3}-3 a b x+3 a b-1=0$.
Factoring, we get $(x-1)\left(x^{2}+x+1-3 a b\right)=0$.
Therefore, $x=1$ or $x^{2}+x+1-3 a b=0$.
If $x^{2}+x+1-3 a b=0$, then
$$
a^{2}+b^{2}-a b+a+b+1=0 \text {. }
$$
Completing the square, we get $\frac{1}{2}\left[(a-b)^{2}... | 1 \text{ or } -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,510 |
3. Let $m$ be the largest integer that cannot be expressed as the sum of three distinct composite numbers. Then $m=$ $\qquad$
Set $m$ is the largest integer that cannot be represented as the sum of three distinct composite numbers. Then $m=$ $\qquad$ | 3.17 .
The smallest three composite numbers are 4, 6, and 8, and $4+6+8=18$. Therefore, 17 is an integer that cannot be expressed as the sum of three distinct composite numbers.
When $m>18$, if $m=2k>18$, then $m=4+6+2 \times (k-5)$; if $m=2k-1>18$, then $m=4+9+2(k-7)$.
Therefore, any integer greater than 18 can be ... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,511 |
4. As shown in Figure 2, $ABCD$ is a square with side length $a$. A circular arc centered at $D$ with radius $DA$ intersects a semicircle with diameter $BC$ at another point $P$. Extending $AP$ intersects $BC$ at point $N$. Then $\frac{BN}{NC}=$ $\qquad$ . | 4. $\frac{1}{2}$.
As shown in Figure 8, connect $B P$ and extend it to intersect $A D$ at $E$, connect $C P$ and extend it to intersect $A B$ at $M$, and extend it to intersect the extension of $D A$ at $F$. Clearly, $B E \perp C F$.
Since $M A^{2}=M P \cdot M C$, $M B^{2}=M P \cdot M C$, we have
$M A=M B$.
Also, $\a... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,512 |
One, (20 points) Given that the roots of the equation $x^{2}-6 x-4 n^{2}-32 n=0$ are integers. Find the integer value of $n$.
---
The above text has been translated into English, preserving the original text's line breaks and format. | I. Solving for $x=3 \pm \sqrt{4 n^{2}+32 n+9}$.
Since the roots of the equation are integers, $4 n^{2}+32 n+9$ must be a perfect square. Let $4 n^{2}+32 n+9=m^{2}, m>0$. Then we have $(2 n+8+m)(2 n+8-m)=55$.
Since $55=1 \times 55=5 \times 11$ $=(-1) \times(-55)=(-5) \times(-11)$, solving these gives $n=10, n=0, n=-18, ... | -18, -8, 0, 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,513 |
II. (25 points) As shown in Figure 3, in trapezoid $ABCD$, $AD \parallel BC$. Squares $ABCE$ and $DCHF$ are constructed on the legs $AB$ and $CD$ respectively. Let the perpendicular bisector $l$ of line segment $AD$ intersect line segment $EF$ at point $M$, and $EP \perp l$ at $P$, $FQ \perp l$ at $Q$. Prove: $EP = FQ$... | In the right triangle $\triangle P^{\prime} P N$ and right triangle $\triangle L N R$, it is easy to see that $\angle P P^{\prime} N = \angle L N R$.
Furthermore, since $R N = A B = A E = P^{\prime} N$, we know that right triangle $\triangle N P^{\prime} P \cong$ right triangle $\triangle R N L$.
Therefore, $P P^{\prim... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,514 |
Three. (25 points) As shown in Figure 4, given points $A(0,3), B(-2, -1), C(2,-1) \cdot P\left(t, t^{2}\right)$ is a moving point on the parabola $y=x^{2}$ located within $\triangle A B C$ (including the boundary), the line $B P$ intersects $A C$ at point $E$, and the line $C P$ intersects $A B$ at point $F$. Express $... | Three, suppose the line $AB$ is the graph of the linear function $y=ax+b$. Substituting $A(0,3)$ and $B(-2,-1)$, we get
$$
3=b, -1=-2a+b \text{. }
$$
Solving these, we get $a=2, b=3$.
Therefore, the line $AB$ is the graph of the linear function $y=2x+3$. The x-coordinates of the intersection points of this line with t... | \frac{t^2+2t+5}{t^2-2t+5}(-1 \leqslant t \leqslant 1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,515 |
II. (25 points) As shown in Figure 5, in trapezoid $ABCD$, $AD \parallel BC$. Squares $ABGE$ and $DCHF$ are constructed on the legs $AB$ and $CD$ respectively. Let the perpendicular bisector $l$ of line segment $AD$ intersect line segment $EF$ at point $M$. Prove that point $M$ is the midpoint of $EF$.
untranslated p... | II. Draw perpendiculars $EP$ and $FQ$ from $E$ and $F$ to $l$, intersecting $l$ at $P$ and $Q$, respectively, as shown in Figure 9. From the solution to the second problem in the A paper, we have $EP = FQ$.
Since $EP \parallel FQ$, quadrilateral $EQFP$ is a parallelogram, and its diagonals bisect each other. Therefore... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,516 |
Example 5 Given that $a$ and $b$ are positive numbers, and $a+b=2$. Find the minimum value of $u=\sqrt{a^{2}+1}+\sqrt{b^{2}+4}$.
(2003, Beijing Middle School Mathematics Competition (Grade 8)) | Explanation: If we consider this problem from an algebraic perspective, it would be quite challenging. By observation, it is not difficult to find that the conditions in the problem have a clear geometric background.
Let's consider $\sqrt{a^{2}+1}$ and $\sqrt{b^{2}+4}$ as the hypotenuses of right triangles with legs $... | \sqrt{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,517 |
II. (25 points) As shown in Figure 6, in trapezoid $ABCD$, $AD \parallel BC$. Squares $ABGE$ and $DCHF$ are constructed on the legs $AB$ and $CD$ respectively. Connect $EF$, and let the midpoint of segment $EF$ be $M$. Prove: $MA = MD$.
保留源文本的换行和格式，直接输出翻译结果。 | As shown in Figure 10, perpendiculars are drawn from points $B, C, E, F$ to line $AD$, with the feet of the perpendiculars marked as $B_{0}, C_{0}, E_{0}, F_{0}$. The midpoint of segment $AD$ is taken as $N$, and $MN$ is connected. It is easy to see that $AB = EA$.
$$
\begin{array}{c}
\text{Also, } \angle B A B_{0} = 9... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,518 |
2. As shown in Figure 1, $M$ is the midpoint of side $BC$ of $\triangle ABC$, $AN$ bisects $\angle BAC$, $BN \perp AN$ at point $N$, and $AB=10, BC=15, MN=3$. Then the perimeter of $\triangle ABC$ is ( ).
(A) 38
(B) 39
(C) 40
(D) 41 | 2.D.
As shown in Figure 4, extend $B N$ to intersect $A C$ at $D$.
Since $A N$ bisects $\angle B A C$, and $B N \perp A N$, we have $\triangle A N B \cong \triangle A N D$. Therefore, $A D=A B=10$, and $N$ is the midpoint of $B D$. Also, since $M$ is the midpoint of $B C$, we have $C D=2 M N=6$. Thus, the perimeter o... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,520 |
3. On Tree Planting Day, the average number of trees planted per person in a class is 6. If only the female students were to complete the task, each should plant 15 trees; if only the male students were to complete the task, each should plant ( ) trees.
(A)9
(B) 10
(C) 12
(D) 14 | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,521 |
$6 . a 、 b$ are integers, it is known that the equation about $x$
$$
\frac{1}{4} x^{2}-a x+a^{2}+a b-a-b-1=0
$$
has two identical real roots. Then $a-b$ equals ( ).
(A) 1
(B) 2
(C) $\pm 1$
(D) $\pm 2$ | 6.C.
According to the problem,
$$
\begin{aligned}
\Delta & =a^{2}-4 \times \frac{1}{4}\left(a^{2}+a b-a-b-1\right) \\
& =-a b+a+b+1=0, \\
& (a-1)(b-1)=2=2 \times 1=1 \times 2 \\
& =(-1) \times(-2)=(-2) \times(-1) .
\end{aligned}
$$
Solving, we get $a-b=1$ or $a-b=-1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,524 |
1. In $\triangle A B C$, $A B=3, A C=4, B C=5$. Now fold $\triangle A B C$ so that point $B$ coincides with point $C$, then the length of the fold line is $\qquad$ . | II. $1 . \frac{15}{8}$.
As shown in Figure 5, it is easy to know that $\angle A=90^{\circ}$. Let $DE$ be the fold line, then $DE \perp BC$. Therefore, $\triangle CDE \backsim \triangle CAB$.
We have $\frac{CD}{AC}=\frac{DE}{AB}$,
which is $\frac{\frac{5}{2}}{4}=\frac{DE}{3}$.
Solving for $DE$ gives $DE=\frac{3 \times ... | \frac{15}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,525 |
2. The minimum value of the function $f(x)=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$ is $\qquad$ . | 2. 5 .
Obviously, if $xf(-x)$.
Therefore, when $f(x)$ takes its minimum value, there must be $x \geqslant 0$.
As shown in Figure 6, draw a line segment $A B=4$, $A C \perp A B, D B \perp A B$, and $A C=$ $1, B D=2$. For any point $O$ on $A B$, let $O A=x$, then
$$
\begin{array}{l}
O C=\sqrt{x^{2}+1}, \\
O D=\sqrt{(4-x... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,526 |
3. As shown in Figure 2, there is a point $O$ inside $\triangle A B C$. Lines parallel to each side are drawn through $O$, dividing $\triangle A B C$ into three triangles and three parallelograms. If the areas of the three triangles are $1,1,2$, then the area of $\triangle A B C$ is $\qquad$ | 3. $6+4 \sqrt{2}$.
As shown in Figure 7, it is easy to see that the three triangles are similar to $\triangle A B C$. Let the area of $\triangle A B C$ be $S$, then
$$
\begin{aligned}
& \frac{\sqrt{S_{1}}}{\sqrt{S}}+\frac{\sqrt{S_{2}}}{\sqrt{S}}+\frac{\sqrt{S_{3}}}{\sqrt{S}} \\
& =\frac{O R}{B C}+\frac{O T}{B C}+\frac... | 6+4 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,527 |
Example 6 Given $a>0, b>0$. Find the area of the triangle with side lengths $\sqrt{a^{2}+b^{2}}$, $\sqrt{a^{2}+4 b^{2}}$, and $\sqrt{4 a^{2}+b^{2}}$.
(The 12th "Hope Cup" National Mathematics Invitational (Junior $\Rightarrow)$ | Explanation: Directly using the formula for the area of a triangle is quite complicated, but by constructing the rectangular figure as shown in Figure 2, the area of the desired triangle can be easily obtained through the relationship between the whole and the parts. Let the area of the desired triangle be \( S \), the... | \frac{3}{2} a b | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,528 |
4. For the equation $x^{3}-a x^{2}-2 a x+a^{2}-1=0$ with respect to $x$, if it has only one real root, then the range of values for $a$ is $\qquad$ | 4. $a<\frac{3}{4}$.
The original equation can be factored as
$$
[a-(x-1)]\left[a-\left(x^{2}+x+1\right)\right]=0 \text {. }
$$
Solving gives $a=x-1$ or $a=x^{2}+x+1$.
Thus, $x=a+1$ or $x^{2}+x+(1-a)=0$.
Since the original equation has only one real root, we have
$$
\Delta=1-4(1-a)<0 \text {. }
$$
Therefore, $a<\frac... | a<\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,529 |
Three. (20 points) As shown in Figure 3, the scalene $\triangle ABC$ is inscribed in $\odot O, I$ is its incenter, and $AI \perp OI$. Prove:
$$
AB + AC = 2BC \text{.}
$$ | Three, as shown in Figure 8, extend \( A I \) to intersect \(\odot O\) at \( D \), and connect \( O A \), \( O D \), \( B D \), and \( B I \).
Since \( O A = O D \),
\( O I \perp A D \),
therefore, \( A I = I D \).
Also, \(\angle D B I\)
\( = \angle D B C + \angle C B I \)
\( = \angle D A C + \angle C B I \)
\( = \frac... | AB + AC = 2BC | Geometry | proof | Yes | Yes | cn_contest | false | 714,530 |
Four. (25 points) Given the quadratic function $f(x)=x^{2}+p x+q$, and the equation $f(x)=0$ has the same non-zero real roots as $f(2 x)=0$.
(1) Find the value of $\frac{q}{p^{2}}$;
(2) If $f(1)=28$, solve the equation $f(x)=0$. | (1) Let the two roots of $f(x)=0$ be $x_{1}$ and $x_{2}$, and $x_{1} \leqslant x_{2}$,
then
$$
\begin{array}{l}
f(x)=\left(x-x_{1}\right)\left(x-x_{2}\right), \\
f(2 x)=\left(2 x-x_{1}\right)\left(2 x-x_{2}\right) \\
=4\left(x-\frac{x_{1}}{2}\right)\left(x-\frac{x_{2}}{2}\right) .
\end{array}
$$
Thus, the two roots of... | x_{1}=-3, x_{2}=-6 \text{ or } x_{1}=\frac{9}{2}, x_{2}=9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,531 |
Five. (25 points) Given three integers $a, b, c$ whose sum is 13, and $\frac{b}{a} = \frac{c}{b}$. Find the maximum and minimum values of $a$, and the corresponding values of $b$ and $c$ at these times. | Five, let $\frac{b}{a}=\frac{c}{b}=x$, then $b=a x, c=a x^{2}$.
Thus, $a+b+c=13$ becomes $a\left(x^{2}+x+1\right)=13$.
Since $a \neq 0$, we have
$$
x^{2}+x+1-\frac{13}{a}=0 \text {. }
$$
Also, since $a, b, c$ are integers, the solutions to equation (1) must be rational numbers, i.e., $\Delta=1-4\left(1-\frac{13}{a}\ri... | a_{\min }=1, b=-4, c=16 ; a_{\min }=1, b=3, c=9; a_{\max }=16, b=-12, c=9 ; a_{\max }=16, b=-4, c=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,532 |
1. Given $m^{2}+n^{2}+m n+m-n+1=0$. Then the value of $\frac{1}{m}+\frac{1}{n}$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2 | $-1 . B$.
Completing the square gives $-\frac{1}{2}\left[(m+n)^{2}+(m+1)^{2}+(n-1)^{2}\right]=0$.
Therefore, $m=-1, n=1$. Hence $-\frac{1}{m}+\frac{1}{n}=0$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,533 |
$2 . a 、 b 、 c$ are non-zero real numbers, and $a+b+c \neq 0$. If $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}$, then $\frac{(a+b)(b+c)(c+a)}{a b c}$ equals ( ).
(A) 8
(B) 4
(C) 2
(D) 1 | 2.A.
Let $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=k$, then $a+b+c=k(a+b+c)$.
Since $a+b+c \neq 0$, we have $k=1$.
Therefore, the original expression $=\frac{2 c \cdot 2 a \cdot 2 b}{a b c}=8$. | 8 | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,534 |
3. The integer solutions of the equation $\frac{x+3}{x+1}-y=0$ are $(\quad)$ groups.
(A) 1
(B) 2
(C) 3
(D) 4 | 3.D.
From the problem, we know that $y=\frac{x+3}{x+1}=1+\frac{2}{x+1}$ is an integer. Therefore, $x+1= \pm 1$ or $x+1= \pm 2$. Solving this, we get $x=0,-2,1,-3$. Hence, the original equation has 4 integer solutions. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,535 |
4. As shown in Figure 1, in $\triangle A B C$, $M$ is the midpoint of $A C$, and $P, Q$ are the trisection points of side $B C$. If $B M$ intersects $A P$ and $A Q$ at points $D$ and $E$ respectively, then the length ratio of the three segments $B D, D E, E M$ is $(\quad$.
(A) $3: 2: 1$
(B) $4: 2: 1$
(C) $5: 3: 2$
(D) ... | 4.C.
As shown in Figure 5, draw $M H / / B C$ intersecting $A Q$ at $F$ and $A P$ at $H$. Therefore,
$$
\begin{array}{l}
\frac{M D}{B D}=\frac{M H}{B P}=\frac{1}{2} \cdot \frac{P C}{B P}=1, \\
\frac{E M}{B E}=\frac{M F}{B Q}=\frac{1}{2} \cdot \frac{C Q}{B Q}=\frac{1}{4} .
\end{array}
$$
Solving, we get $B D=\frac{5}{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,536 |
5. In $\triangle A B C$, $\angle B A C: \angle A B C: \angle A C B=4: 2: 1$, $A D$ is the angle bisector of $\angle B A C$. There are the following three conclusions:
(1) $B C: A C: A B=4: 2: 1$;
(2) $A C=A D+A B$;
(3) $\triangle D A C \sim \triangle A B C$.
Among the correct conclusions are ( ).
(A)(1)(2)
(B)(2)(3)
(C... | 5.B.
It is clear that (1) is not correct. As shown in Figure 6, on $AC$, intercept $AE$ such that $AE = AB$. Connect $DE$, then $\triangle ABD \cong \triangle AED$. Therefore,
$$
CE = DE = AD \text{.}
$$
Thus, $AC = AD + AB$. Hence (2) is correct.
Since the ratio of the interior angles of $\triangle DAC$ and $\triang... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,537 |
6. As shown in Figure 2, in isosceles $\triangle ABC$, $O$ is the midpoint of the base $BC$. A semicircle is drawn with $O$ as the center, tangent to $AB$ and $AC$ at points $D$ and $E$, respectively. A tangent line is drawn from a point $F$ on the semicircle, intersecting $AB$ and $AC$ at points $M$ and $N$, respectiv... | 6. B.
As shown in Figure 7, connect $OD$, $OM$, $OF$, $ON$, $OE$, then $\angle 1=\angle 6$, $\angle 2=\angle 3$, $\angle 4=\angle 5$.
Therefore, $\angle 1+\angle 2=90^{\circ}-\angle 5$.
Also, $\angle B O M$
$=\angle 1+\angle 2$
$=90^{\circ}-\angle 5$
$=\angle O N E$,
$\angle M B O=\angle N C O$,
Thus,
$$
\begin{arra... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,538 |
Example 7 Given that $a, b, x, y$ are positive real numbers, and $a^{2}+b^{2}=1, x^{2}+y^{2}=1$. Prove: $a x+b y \leqslant 1$.
(1996, Handan City Middle School Students Mathematics Competition) | As shown in Figure 3, construct a circle $\odot O$ with $AB=1$ as its diameter. On both sides of $AB$, construct right triangles $\triangle ABC$ and $\triangle ADB$ such that $AC=a$, $BC=b$, $BD=x$, and $AD=y$.
By the Pythagorean theorem, $a$, $b$, $x$, and $y$ satisfy the given conditions.
According to Ptolemy's theo... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,539 |
7. Given $x=\frac{\sqrt{3}-1}{\sqrt{3}+1}, y=\frac{\sqrt{3}+1}{\sqrt{3}-1}$. Then $x^{4}+y^{4}$ equals | $=.7 .194$.
Given $x=2-\sqrt{3}, y=2+\sqrt{3}$. Then
$$
\begin{array}{l}
x+y=4, x y=1, \\
x^{2}+y^{2}=(x+y)^{2}-2 x y=14, \\
x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=194 .
\end{array}
$$ | 194 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,540 |
9. On the beach, there is a pile of apples that belongs to 3 monkeys. The first monkey comes, divides the apples into 3 equal piles with 1 apple left over, then it throws the extra apple into the sea and takes one pile; the second monkey comes, divides the remaining apples into 3 equal piles, again with 1 apple left ov... | 9. 25 .
Let there be $n$ apples initially, and the remaining number of apples each time be $y_{1}, y_{2}, y_{3}$. Then
$$
\begin{array}{l}
y_{1}=\frac{2}{3}(n-1)=\frac{2}{3}(n+2)-2, \\
y_{2}=\left(\frac{2}{3}\right)^{2}(n+2)-2, \\
y_{3}=\left(\frac{2}{3}\right)^{3}(n+2)-2 .
\end{array}
$$
To make $y_{3}$ a positive i... | 25 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,542 |
10. As shown in Figure 4, $AB$ is the diameter of $\odot O_{1}$, $AO_{1}$ is the diameter of $\odot O_{2}$, chord $MN \parallel AB$, and $MN$ is tangent to $\odot O_{2}$ at point $C$. If the radius of $\odot O_{1}$ is 2, then the area of the figure enclosed by $O_{1}B$, $\overparen{BN}$, $NC$, and $\overparen{CO_{1}}$ ... | 10. $\frac{\pi}{12}+\frac{\sqrt{3}}{2}+1$.
As shown in Figure 8, draw $O_{1} D \perp M N$, with the foot of the perpendicular at $D$, and connect $C O_{2}$. Then $O_{1} D=1, O_{1} N=2$.
Therefore, $\angle O_{1} N C^{-\prime}=30^{\circ}$, $\angle D O_{1} N=60^{\circ}$, $\angle N O_{1} B=30^{\circ}$.
Thus, $S_{\text {w... | \frac{\pi}{12}+\frac{\sqrt{3}}{2}+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,543 |
11. If $1 \leqslant p \leqslant 20,1 \leqslant q \leqslant 10$, and the roots of the equation $4 x^{2}-p x+q$ $=0$ are both odd numbers. Then the roots of this equation are $\qquad$ . | 11. $x_{1}=x_{2}=1$.
Let $x_{1}$ and $x_{2}$ be the two roots of the equation, then
$$
x_{1}+x_{2}=\frac{p}{4}, x_{1} x_{2}=\frac{q}{4} \text {. }
$$
Since $x_{1}$ and $x_{2}$ are both odd numbers, $x_{1}+x_{2}$ is even, and $x_{1} x_{2}$ is odd.
$$
\begin{array}{l}
\text { Also, } 1 \leqslant p \leqslant 20,1 \leqsl... | x_{1}=x_{2}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,544 |
12. Given two quadratic functions $y_{1}$ and $y_{2}$, when $x=\alpha(\alpha>0)$, $y_{1}$ reaches its maximum value of 5, and $y_{2}=25$. Also, the minimum value of $y_{2}$ is $-2$, and $y_{1}+y_{2}=x^{2}+16 x+13$. Find the value of $\alpha$ and the analytical expressions of the quadratic functions $y_{1}, y_{2}$. | Three, 12. Let $y_{1}=m(x-\alpha)^{2}+5$.
Then $y_{2}=x^{2}+16 x+13-m(x-\alpha)^{2}-5$.
When $x=\alpha$, $y_{2}=25$, i.e., $\alpha^{2}+16 \alpha+8=25$.
Solving, we get $\alpha_{1}=1, \alpha_{2}=-17$ (discard).
Thus, $y_{2}=x^{2}+16 x+13-m(x-1)^{2}-5$
$$
=(1-m) x^{2}+(16+2 m) x+(8-m) \text {. }
$$
Also, the minimum val... | y_{1}=-2 x^{2}+4 x+3, y_{2}=3 x^{2}+12 x+10, \alpha=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,545 |
13. In a regular 2004-gon $A_{1} A_{2} \cdots A_{2001}$, each vertex is randomly assigned a number from $1,2, \cdots, 501$. Prove that there must exist four vertices satisfying the following conditions:
(1) These four vertices form a rectangle;
(2) The sum of the numbers assigned to the opposite vertices of this rectan... | 13. From the problem, we know that vertices $A_{i}$ and $A_{i+1002} \infty 0$ form a pair of points symmetric about the center, where $i=1,2, \cdots, 1002$.
Thus, the 2004 vertices can be divided into 1002 pairs.
Connecting the vertices of each pair in sequence, we can obtain a quadrilateral. Since the diagonals of the... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,546 |
14. Given that line $l$ intersects circle $\odot O$ at two distinct points $E$ and $F$, $CD$ is a diameter of $\odot O$, $CA \perp l$, $DB \perp l$, with the feet of the perpendiculars being $A$ and $B$ respectively. If $AB=7$, $BD-AC=1$, and $AE=1$, does there exist a point $P$ on segment $AB$ such that the triangle w... | 14. (1) If $l$ does not intersect with the diameter $C D$, as shown in Figure 9.
(i) Draw $O H \perp A B$ at $H$, it is easy to see that $A E=B F$. At this point, $\triangle A C E \backsim \triangle B E D, \triangle A F C \backsim \triangle B D F$. Therefore, $E$ and $F$ are the points that satisfy the conditions.
Thus... | 1, 6, \frac{14}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,547 |
1. If $(2 x-1)^{5}=a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+$ $a_{1} x+a_{0}$, then $a_{2}+a_{4}=$ $\qquad$ | ,$- 1 .-120$.
Let $x=0$, we get $a_{0}=-1$.
Let $x=1$, we get $a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=1$;
Let $x=-1$, we get
$-a_{5}+a_{4}-a_{3}+a_{2}-a_{1}+a_{0}=-243$.
Adding the last two equations gives $a_{4}+a_{2}+a_{0}=-121$.
Therefore, $a_{2}+a_{4}=-120$. | -120 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,548 |
Example 1 Let $a \in \mathbf{R}$, and the function
$$
f(x)=a x^{2}+x-a \quad(|x| \leqslant 1) .
$$
(1) If $|a| \leqslant 1$, prove: $|f(x)| \leqslant \frac{5}{4}$;
(2) Find the value of $a$ that makes the function $f(x)$ have a maximum value of $\frac{17}{8}$.
(2003, Anhui Province High School Mathematics Competition) | $$
\begin{array}{l}
\text { (1) Proof: }|f(x)|=\left|a x^{2}+x-a\right| \\
=\left|a\left(x^{2}-1\right)+x\right| \leqslant\left|a\left(x^{2}-1\right)\right|+|x| \\
\leqslant\left|x^{2}-1\right|+|x|=1-|x|^{2}+|x| \\
=-\left(|x|-\frac{1}{2}\right)^{2}+\frac{5}{4} \leqslant \frac{5}{4} .
\end{array}
$$
(2) Solution: When ... | a=-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,550 |
3. $a, b, c$ are non-negative real numbers, and satisfy $3a+2b+c=$ $5, 2a+b-3c=1$. Let $m=3a+b-7c$, and let $x$ be the minimum value of $m$, $y$ be the maximum value of $m$. Then $xy=$ $\qquad$ | $3 \cdot \frac{5}{77}$.
From $3a + 2b + c = 5, 2a + b - 3c = 1$, we get
\[
\left\{
\begin{array}{l}
3a + 2b = 5 - c, \\
2a + b = 1 + 3c
\end{array}
\Rightarrow
\left\{
\begin{array}{l}
3a + 2b = 5 - c, \\
4a + 2b = 2 + 6c
\end{array}
\right.
\right.
\]
Thus, $a = 7c - 3, b = 7 - 11c$.
Since $a, b, c$ are non-negative ... | \frac{5}{77} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,551 |
4. In $\triangle A B C$, $A D$ is the median on side $B C$, $A B=\sqrt{2}$, $A D=\sqrt{6}$, $A C=\sqrt{26}$. Then $\angle A B C=$ $\qquad$ | $4.60^{\circ}$.
Extend $B A$ to $E$ such that $A E=A B=\sqrt{2}$, i.e., $B E=2 \sqrt{2}$. Connect $C E$, then $C E / / A D$, and
$$
C E=2 A D=2 \sqrt{6} \text {. }
$$
In $\triangle A C E$, we have $A E^{2}+C E^{2}=2+24=26=A C^{2}$. Therefore, $\angle A E C=90^{\circ}$.
In the right triangle $\triangle B C E$, $C E=\sq... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,552 |
5. Given $x y z=1, x+y+z=2, x^{2}+y^{2}+z^{2}=16$. Then $\frac{1}{x y+2 z}+\frac{1}{y z+2 x}+\frac{1}{z x+2 y}=$ | 5. $-\frac{4}{13}$.
Since $x+y+z=2$, squaring both sides gives
$$
x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x=4 \text {. }
$$
Given $x^{2}+y^{2}+z^{2}=16$, so, $x y+y z+z x=-6$.
Also, $z=2-x-y$, so,
$$
\frac{1}{x y+2 z}=\frac{1}{x y+4-2 x-2 y}=\frac{1}{(x-2)(y-2)} \text {. }
$$
Similarly, $\frac{1}{y z+2 x}=\frac{1}{(y-2)(z... | -\frac{4}{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,553 |
II. (15 points) If positive numbers $a, b, c$ satisfy $a+c=2b$, prove:
$$
\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2}{\sqrt{c}+\sqrt{a}} \text {. }
$$ | Given that $a-b=b-c$.
$$
\begin{array}{l}
\frac{1}{\sqrt{c}+\sqrt{a}}-\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{(\sqrt{a}+\sqrt{b})-(\sqrt{c}+\sqrt{a})}{(\sqrt{c}+\sqrt{a})(\sqrt{a}+\sqrt{b})} \\
=\frac{\sqrt{b}-\sqrt{c}}{(\sqrt{c}+\sqrt{a})(\sqrt{a}+\sqrt{b})} \\
=\frac{b-c}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sq... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,554 |
Three, (15 points) The side lengths of a right-angled triangle are all integers, and its area is numerically equal to its perimeter. Determine the lengths of the three sides of this right-angled triangle.
---
Please note that the translation retains the original format and line breaks as requested. | Three, let $a$ and $b$ be the lengths of the two legs of a right triangle, then the length of the hypotenuse $c = \sqrt{a^2 + b^2}$. Since $a$, $b$, and $c$ are all positive integers, $a \neq b$. Without loss of generality, assume $a > b$. According to the problem, we have
$$
a + b + \sqrt{a^2 + b^2} = \frac{ab}{2}.
$$... | (12, 5, 13) \text{ or } (8, 6, 10) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,555 |
Four. (15 points) As shown in Figure 1, construct squares $ABDE$, $CAFG$, and $BCHK$ outwardly on the sides of $\triangle ABC$. Connect $EF$, $GH$, and $KD$. Prove that a triangle can be formed with $EF$, $GH$, and $KD$ as sides, and the area of the triangle formed is three times the area of $\triangle ABC$. | As shown in Figure 4, draw $DP \parallel KH$, then quadrilateral $DPHK$ is a parallelogram.
Therefore, $PH \parallel DK$.
Since $DP \parallel BC$, quadrilateral $DPCB$ is also a parallelogram. Thus, $PC \parallel DB$. Also, $EA \parallel DB$, so $EA \parallel PC$, which means quadrilateral $EACP$ is also a parallelogra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,556 |
Five. (15 points) 13 athletes, their sportswear numbers are 1 to 13. Ask: Can these 13 athletes stand in a circle so that the absolute difference in numbers between any two adjacent athletes is not less than 3 and not more than 5? If yes, provide an example; if not, explain the reason. | Five, it cannot be done. The reasons are as follows:
Assume it is possible to arrange the numbers in a circle such that the numbers meet the requirements of the problem. We divide the numbers into two groups $A$ and $B$:
$$
A=\{1,2,3,11,12,13\}, B=\{4,5,6,7,8,9,10\} \text {. }
$$
Clearly, the difference between any tw... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,557 |
Example 2 Given $a>0$, the function $f(x)=a x-b x^{2}$.
(1) When $b>0$, if for any $x \in \mathbf{R}$, we have $f(x) \leqslant 1$, prove: $a \leqslant 2 \sqrt{b}$;
(2) When $b>1$, prove: for any $x \in[0,1]$, $|f(x)| \leqslant 1$ if and only if $b-1 \leqslant a \leqslant 2 \sqrt{b}$;
(3) When $0<b<1$, discuss: for any ... | (1) Proof: Assume for any $x \in \mathbf{R}$, we have $f(x) \leqslant 1$.
Since $f(x)=-b\left(x-\frac{a}{2 b}\right)^{2}+\frac{a^{2}}{4 b}$, it follows that $f\left(\frac{a}{2 b}\right)=\frac{a^{2}}{4 b} \leqslant 1$.
Also, $a>0, b>0$, thus, $a \leqslant 2 \sqrt{b}$.
(2) Proof: Necessity.
For any $x \in [0,1]$, from $... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,561 |
7. The numbers $1,2,3, \cdots, k^{2}$ are arranged in the following way:
$$
\begin{array}{cccc}
1 & 2 & \cdots & k \\
k+1 & k+2 & \cdots & 2 k \\
\vdots & \vdots & \cdots & \vdots \\
(k-1) k+1 & (k-1) k+2 & \cdots & k^{2}
\end{array}
$$
Select any one of these numbers and strike out the row and column in which it lies... | 7. $\frac{1}{2} k\left(k^{2}+1\right)$ | \frac{1}{2} k\left(k^{2}+1\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,565 |
8. As shown in Figure 1, an equilateral triangle $\triangle A N B$ with side length 1 is placed inside a rectangle $M N P Q$ with side lengths $M N=3, N P=4$, and $N B$ lies on side $N P$. If the equilateral triangle rolls around the inside of the rectangle along sides $N P, P Q, Q M$, and $M N$ and returns to its orig... | $8.5 \pi$ | 8.5 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,566 |
9. As shown in Figure 2, in $\triangle A B C$, $A B=B C=10$, points $M$ and $N$ are on $B C$ such that $M N=A M=4$, and $\angle M A C = \angle B A N$. Then the area of $\triangle A B C$ is $\qquad$. | 9. $\frac{50 \sqrt{57}}{19}$ | \frac{50 \sqrt{57}}{19} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,567 |
II. (16 points) Given that $m$ and $n$ are positive integers. If the two real roots of the equation $4 x^{2}-2 m x+n=0$ are both greater than 1 and less than 2, find the values of $m$ and $n$. | Let $f(x)=4 x^{2}-2 m x+n$, then the graph of $y=f(x)$ is a parabola opening upwards, with the axis of symmetry at $x=\frac{m}{4}$.
$$
\begin{array}{l}
10, \\
f(2)=16-4 m+n>0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
m^{2} \geqslant 4 n, \\
4+n>2 m,
\end{array}\right. \\
\hline
\end{array}
$$
From (... | m=6, n=9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,569 |
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