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4. Given positive integers $a$ and $b$ differ by 120, their least common multiple is 105 times their greatest common divisor. The larger of $a$ and $b$ is $\qquad$ | 4.225.
Let $(a, b)=d$, and $a=m d, b=n d$, where $m>n, m$ and $n$ are coprime. Thus, the least common multiple of $a$ and $b$ is $m n d$. According to the problem, we have
$$
\left\{\begin{array}{l}
m d-n d=120, \\
\frac{m n d}{d}=105,
\end{array}\right.
$$
which simplifies to
$$
\left\{\begin{array}{l}(m-n) d=2^{3} ... | 225 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,765 |
One. (20 points) Try to find a four-digit number, the square of the sum of the two-digit numbers formed by its first two digits and the last two digits is exactly equal to this four-digit number.
保留源文本的换行和格式，所以翻译结果如下：
One. (20 points) Try to find a four-digit number, the square of the sum of the two-digit numbers f... | One, Solution 1: Let the two two-digit numbers be $x$ and $y$, $10 \leqslant x, y \leqslant 99$.
We have $(x+y)^{2}=100 x+y$, which means
$$
\begin{array}{l}
x^{2}+2(y-50) x+\left(y^{2}-y\right)=0 \\
\text { When } \Delta=4(y-50)^{2}-4\left(y^{2}-y\right) \\
=4(2500-99 y) \geqslant 0,
\end{array}
$$
i.e., $2500-99 y ... | 2025 \text{ or } 3025 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,766 |
In $\triangle A B C$, $D$ is the midpoint of $A B$. Extend $C A$ and $C B$ to points $E$ and $F$ respectively, such that $D E = D F$. Draw perpendiculars from $E$ and $F$ to $C A$ and $C B$ respectively, intersecting at point $P$. Let the midpoints of segments $P A$ and $P B$ be $M$ and $N$ respectively. Prove:
(1) $\t... | (1) As shown in Figure 5, according to the given conditions,
$$
\begin{array}{l}
D M \text { II } B N, \\
D N \cong A M . \\
\text { Therefore, } \angle A M D=\angle B N D .
\end{array}
$$
Since $M$ and $N$ are the midpoints of the hypotenuses of right triangles $\triangle A E P$ and $\triangle B F P$, respectively, w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,767 |
II. (25 points) In $\triangle ABC$, $D$ is the midpoint of $AB$. Extend $CA$ and $CB$ to points $E$ and $F$ respectively, such that $DE = DF$. Draw perpendiculars from $E$ and $F$ to $CA$ and $CB$ respectively, intersecting at point $P$. Prove that $\angle PAE = \angle PBF$.
保留源文本的换行和格式，直接输出翻译结果如下：
II. (25 points) In... | As shown in Figure 6, take the midpoints $M$ and $N$ of $AP$ and $BP$ respectively. Connect $EM$, $DM$, $FN$, and $DN$. Since $D$ is the midpoint of $AB$, then $DM \cong BN$, $DN \stackrel{\|}{=} AM$. Therefore, $\angle AMD = \angle BND$.
Since $M$ and $N$ are the midpoints of the hypotenuses of right triangles $\trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,768 |
Three, (25 points) Given that the area of quadrilateral $ABCD$ is $32$, the lengths of $AB$, $CD$, and $AC$ are all integers, and their sum is $16$.
(1) How many such quadrilaterals are there?
(2) Find the minimum value of the sum of the squares of the side lengths of such quadrilaterals. | (1) As shown in Figure 7, let $AB = a$, $CD = b$, $AC = l$, and let the height from $A$ to $BC$ in $\triangle ABC$ be $h_1$, and the height from $D$ to $AC$ in $\triangle ADC$ be $h_2$. Then,
$$
\begin{array}{l}
S_{\text{quadrilateral } BCD} = S_{\triangle ABC} + S_{\triangle ADC} \\
= \frac{1}{2} (h_1 a + h_2 b) \leq ... | 192 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,769 |
Example 1 Given the parabola $y^{2}=2 p x$ and fixed points $A(a, b) 、 B(-a, 0)\left(a b \neq 0, b^{2} \neq 2 p a\right) . M$ is a point on the parabola, and the lines $A M 、 B M$ intersect the parabola at another point, denoted as $M_{1} 、 M_{2}$. Prove that when $M$ varies on the parabola (as long as $M_{1} 、 M_{2}$ ... | Analysis: Let $M\left(\frac{y_{0}^{2}}{2 p}, y_{0}\right), M_{i}\left(\frac{y_{i}^{2}}{2 p}, y_{i}\right)(i=1,2)$. It is easy to obtain the equation of the line $M M_{1}$ as
$$
y_{0} y_{1}=y\left(y_{0}+y_{1}\right)-2 p x.
$$
Similarly, the equation of the line $M M_{2}$ is
$$
y_{0} y_{2}=y\left(y_{0}+y_{2}\right)-2 p ... | Q\left(a, \frac{2 p a}{b}\right) | Geometry | proof | Yes | Yes | cn_contest | false | 713,770 |
Three. (25 points) Given real numbers $a, b, c, d$ are all distinct, and
$$
a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{d}=d+\frac{1}{a}=x \text {. }
$$
Find the value of $x$. | Three, Solution 1: From the given, we have
$$
\begin{array}{l}
a+\frac{1}{b}=x, \\
b+\frac{1}{c}=x, \\
c+\frac{1}{d}=x, \\
d+\frac{1}{a}=x .
\end{array}
$$
From equation (1), we solve for
$$
b=\frac{1}{x-a} \text {. }
$$
Substituting equation (5) into equation (2) gives
$$
c=\frac{x-a}{x^{2}-a x-1} \text {. }
$$
Sub... | x = \pm \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,772 |
2. Among all the interior angles of a convex octagon, the number of obtuse angles is at least ( ) .
(A) 3
(B) 5
(C) 7
(D) 8 | 2. (B).
Since the exterior angles of a convex polygon can have at most 3 obtuse angles, its interior angles can have at most 3 acute angles. Therefore, in a convex octagon, there are at least 5 obtuse angles among the interior angles. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,774 |
3. As shown in Figure 1, a symmetrical shape is formed by 3 squares with side lengths of 1. The minimum radius of a circle that can completely cover this shape is ( ).
(A) $\sqrt{2}$
(B) $\frac{\sqrt{5}}{2}$
(C) $\frac{5}{4}$
(D) $\frac{5 \sqrt{17}}{16}$ | 3. (D).
As shown in Figure 3, we have
$$
\left\{\begin{array}{l}
a^{2}+1=r^{2}, \\
(2-a)^{2}+\left(\frac{1}{2}\right)^{2}=r^{2}
\end{array}\right.
$$
Solving, we get $a=\frac{13}{16}, r=\frac{5 \sqrt{17}}{16}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,775 |
4. Given that $A$ and $B$ are two fixed points on a plane, and $AB=5$. If the distances from points $A$ and $B$ to line $l$ are $2$ and $3$ respectively, then the number of lines $l$ that satisfy the condition is ( ).
(A) 2
(B) 3
(C) 4
(D) infinitely many | 4. (B).
The number of common tangents to two circles with centers at points $A$ and $B$ and radii 2 and 3, respectively, is the answer. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,776 |
5. Given that $x, y, z$ are 3 non-negative real numbers, and satisfy $3x + 2y + z = 5, x + y - z = 2$. If $s = 2x + y - z$, then the sum of the maximum and minimum values of $s$ is ( ).
(A) 5
(B) $\frac{23}{4}$
(C) $\frac{27}{4}$
(D) $\frac{35}{4}$ | 5.(A).
From $\left\{\begin{array}{l}3 x+2 y+z=5, \\ x+y-z=2, \\ 2 x+y-z=s\end{array}\right.$ we get $\left\{\begin{array}{l}x=s-2 \geqslant 0, \\ y=5-\frac{4}{3} s \geqslant 0, \\ z=1-\frac{1}{3} s \geqslant 0 .\end{array}\right.$
Then $2 \leqslant s \leqslant 3, s_{\text {max }}+s_{\text {min }}=5$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,777 |
6. For the expression $2 n(n+1)(n+2)(n+3)+12$ to be representable as the sum of squares of two positive integers, the natural number $n(\quad)$.
(A) does not exist
(B) has 1
(C) has 2
(D) has infinitely many
Translate the above text into English, please retain the original text's line breaks and format, and output the... | 6. (A).
Original expression $=2\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+12$.
Let $n^{2}+3 n+1=t$, then $t$ is an odd number.
Thus, let $t=2 k+1$, the original expression $=4\left(2 k^{2}+2 k+3\right)$.
If the original expression can be expressed as the sum of squares of two positive integers $x, y$, i.e., $x^{2}... | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,778 |
7. An editor uses the digits $0 \sim 9$ to number the pages of a book. If a total of 636 digits were written, then the book has $\qquad$ pages. | $$
\text { Ni, 7. } 248 \text {. }
$$
Assuming the book has $x$ pages, then we have
$$
(x-99) \times 3+90 \times 2+9=636 \text {. }
$$
Solving for $x$ gives $x=248$. | 248 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,779 |
8. Let the area of $\triangle A B C$ be $1, D$ be a point on side $B C$, and $\frac{B D}{D C}=\frac{1}{2}$. If a point $E$ is taken on side $A C$ such that the area of quadrilateral $A B D E$ is $\frac{4}{5}$, then the value of $\frac{A E}{E C}$ is $\qquad$. | 8. $\frac{7}{3}$.
Connect $A D$. According to the problem, we have
$$
\begin{array}{l}
S_{\triangle C D E}=\frac{1}{5}, S_{\triangle A D B}=\frac{1}{3}, \\
S_{\triangle A D E}=\frac{4}{5}-\frac{1}{3}=\frac{7}{15} . \\
\text { Therefore, } \frac{A E}{E C}=\frac{S_{\triangle A D E}}{S_{\triangle C D E}}=\frac{7}{3} .
\e... | \frac{7}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,780 |
Example 2 Suppose a circle and an equilateral hyperbola intersect at four points $A_{1}, A_{2}, A_{3}, A_{4}$, where $A_{1}$ and $A_{2}$ are the endpoints of a diameter of the circle. Prove:
(1) $A_{3}$ and $A_{4}$ are the endpoints of a diameter of the hyperbola;
(2) The tangents to the hyperbola at $A_{3}$ and $A_{4}... | Analysis: Let the equations of the hyperbola and the circle be \(xy = a\) and \(x^2 + y^2 + 2Dx + 2Ey + F = 0\), respectively, with the intersection points having coordinates \((x_i, y_i), i=1,2,3,4\). Eliminating \(y\) from these two equations yields
\[
x^4 + 2Dx^3 + Fx^2 + 2aEx + a^2 = 0.
\]
Then \(x_i (i=1,2,3,4)\) ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,781 |
9. Given the quadratic function $y=a x^{2}+b x+c$, and the linear function $y=$ $k(x-1)-\frac{k^{2}}{4}$. If their graphs have only one common point for any real number $k$, then the analytical expression of the quadratic function is $\qquad$ . | 9. $y=x^{2}-2 x+1$.
From the problem, we get $\left\{\begin{array}{l}y=a x^{2}+b x+c, \\ y=k(x-1)-\frac{k^{2}}{4} .\end{array}\right.$
Rearranging gives $a x^{2}+(b-k) x+\left(c+k+\frac{k^{2}}{4}\right)=0$.
Also, $\Delta=(b-k)^{2}-4 a\left(c+k+\frac{k^{2}}{4}\right)=0$,
which simplifies to $(1-a) k^{2}-2(b+2 a) k+\lef... | y=x^{2}-2 x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,782 |
12. There are 18 football teams participating in a round-robin tournament, where each team plays one match against every other team. Assuming there are no ties in the matches, if $a_{i}$ and $b_{i}$ represent the number of wins and losses, respectively, for the $i(i=1,2,3, \cdots, 18)$-th team throughout the tournament... | Three, 12. Since each team has to play $18-1=17$ matches, for the $i$-th team, we have
$$
a_{i}+b_{i}=17, i=1,2,3, \cdots, 18 .
$$
Since there are no ties in the matches, the total number of wins and losses for all participating teams is equal, i.e.,
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{18}=b_{1}+b_{2}+\cdots+b_{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,785 |
13. As shown in Figure 2, $PA$, $PB$ are tangent to $\odot O$ at points $A$, $B$, respectively, and $PC$ is any secant line intersecting $\odot O$ at points $E$, $C$, and intersecting $AB$ at point $D$. Prove that $\frac{AC^2}{BC^2} = \frac{AD}{BD}$. | 13. Connect $A E, B E$.
Since $P A$ is tangent to $\odot O$, we know $\angle P A E = \angle A C E$.
Also, $\angle A P C = \angle A P E$,
thus $\triangle P A C \backsim \triangle P E A, \frac{A C}{A E} = \frac{P C}{P A}$.
Similarly, $\frac{B C}{B E} = \frac{P C}{P B}$.
Since $P A = P B$, we have $\frac{A C}{A E} = \fra... | \frac{AC^2}{BC^2} = \frac{AD}{BD} | Geometry | proof | Yes | Yes | cn_contest | false | 713,786 |
14. Given the function $y=(a+2) x^{2}-2\left(a^{2}-1\right) x+1$, where the independent variable $x$ is a positive integer, and $a$ is also a positive integer. Find the value of $x$ for which the function value is minimum. | 14. $y=(a+2)\left(x-\frac{a^{2}-1}{a+2}\right)^{2}+1-\frac{\left(a^{2}-1\right)^{2}}{a+2}$,
Its axis of symmetry is $x=\frac{a^{2}-1}{a+2}=(a-2)+\frac{3}{a+2}$.
Since $a$ is a positive integer,
thus $0<\frac{3}{a+2}<1$ when $a>1$. Since $x$ is an integer and $\frac{a^{2}-1}{a+2}$ is a decimal, $x=\frac{a^{2}-1}{a+2}$ ... | x=a-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,787 |
Example 3 Given the family of curves $2(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) y=0$, where $\theta$ is a parameter. Try to find the maximum value of the length of the chord intercepted by the line $y=2 x$ on this family of curves.
$(1995$, National High School Mathematics Competition) | Analysis: Substituting $y=2 x$ into the given family of curve equations yields $2(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) 2 x=0$.
Solving for $x_{1}=0, x_{2}=\frac{8 \sin \theta+\cos \theta+1}{2 \sin \theta-\cos \theta+3}$.
To maximize the length of the chord intercepted, the absolute value of ... | 8 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,792 |
7. If the three sides of $\triangle A B C$ are all unequal, the area is $\frac{\sqrt{15}}{3}$, and the lengths of the medians $A D$ and $B E$ are 1 and 2, respectively, then the length of the median $C F$ is $\qquad$. | 7. $\sqrt{6}$ | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,795 |
Given $p$ is a prime number, such that the quadratic equation
$$
x^{2}-2 p x+p^{2}-5 p-1=0
$$
has two integer roots. Find all possible values of $p$.
| Because the quadratic equation with integer coefficients has integer roots, so,
$$
\Delta=4 p^{2}-4\left(p^{2}-5 p-1\right)=4(5 p+1)
$$
is a perfect square, thus, $5 p+1$ is a perfect square.
Let $5 p+1=n^{2}$. Note that $p \geqslant 2$, hence $n \geqslant 4$, and $n$ is an integer. Therefore,
$$
5 p=(n+1)(n-1) \text ... | p=3 \text{ or } 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,799 |
Three, (16 points) Given that $\triangle X Y Z$ is an isosceles right triangle with legs of length 1 $\left(\angle Z=90^{\circ}\right)$, its three vertices lie on the three sides of isosceles right $\triangle A B C\left(\angle C=90^{\circ}\right)$. Find the maximum possible length of the legs of $\triangle A B C$. | (1) As shown in Figure 4, vertex $Z$ is on the hypotenuse $A B$.
Take the midpoint $M$ of $X Y$, and connect $C M$, $Z M$, and $C Z$, and draw the perpendicular $C N$ from $C$ to side $A B$, then
$$
\begin{array}{l}
C Z \leqslant C M + M Z \\
= \frac{1}{2} X Y + \frac{1}{2} X Y \\
= X Y = \sqrt{2} .
\end{array}
$$
Als... | \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,800 |
Four, (18 points) On a plane, there are 7 points, and some line segments can be connected between them, so that any 3 points among the 7 points must have 2 points connected by a line segment. How many line segments are needed at least? Prove your conclusion. | (1) If one of the 7 points is isolated (i.e., it is not connected to any other points), then the remaining 6 points must be connected in pairs, which requires at least $\frac{6 \times 5}{2}=15$ lines.
(2) If one of the 7 points is connected to only one other point, then the remaining 5 points must be connected in pairs... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,801 |
Example 4 Given $A(-2,2)$, and $B$ is a moving point on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, $F$ is the left focus. When $\mid A B \mid +\frac{5}{3}|B F|$ takes the minimum value, find the coordinates of point $B$.
(1999, National High School Mathematics Competition) | Analysis: Since the eccentricity of the ellipse $e=\frac{3}{5}$, we have $|A B|+\frac{5}{3}|B F|=|A B|+\frac{1}{e}|B F|$.
And $\frac{|B F|}{e}$ is the distance from the moving point $B$ to the left directrix. Therefore, this problem is transformed into:
Finding a point $B$ on the ellipse such that the sum of its dista... | \left(-\frac{5}{2} \sqrt{3}, 2\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,803 |
3. The equation $x^{2}+a x+b=0$ and $x^{2}+c x+d=0(a \neq c)$ have the same root $\alpha$. Then $\alpha=$ | 3. $\frac{d-b}{a-c}$ | \frac{d-b}{a-c} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,805 |
7. As shown in Figure 2, a right-angle ruler $A B C$ is tangent to $\odot O$ at point $D$, and $A$ touches $\odot O$ at point $A$. It is measured that $A B=a, B D=b$. Then the radius of $\odot O$ is $\qquad$. | 7. $\frac{a^{2}+b^{2}}{2 a}$ | \frac{a^{2}+b^{2}}{2 a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,809 |
2. When $50$, $72$, and $157$ are divided by the positive integer $a$, the sum of their remainders is 27. Then $a$ is ( ).
(A) 9
(B) 15
(C) 21
(D) 35 | 2. (C)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,811 |
4. Let $[x]$ denote the greatest integer not exceeding $x$. If $[x]=5$, $[y]=-3$, $[z]=-2$, then the number of possible values for $[x-y+z]$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 4. (C)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,813 |
Example 5 As shown in Figure 1, it is known that point $P$ moves on the circle $x^{2}+(y-4)^{2}=1$, and point $Q$ moves on the ellipse $\frac{x^{2}}{9}+y^{2}=1$. Try to find the maximum value of $|PQ|$.
(1994, Sichuan Province High School Mathematics Competition) | Analysis: First, fix point $Q$ on the ellipse. It is evident that $|P Q|$ is maximized when $P Q$ passes through the center $O_{1}$. Therefore, to find the maximum value of $|P Q|$, we only need to find the maximum value of $\left|O_{1} Q\right|$.
Let $Q(x, y)$, then
$$
\left|O_{1} Q\right|^{2}=x^{2}+(y-4)^{2} \text {.... | 3 \sqrt{3}+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,814 |
5. Given $0 \leqslant a-b \leqslant 1$ and $1 \leqslant a+b \leqslant 4$. Then the range of values for $a$ is ( ).
(A) $1 \leqslant a \leqslant 2$
(B) $2 \leqslant a \leqslant 3$
(C) $\frac{1}{2} \leqslant a \leqslant \frac{5}{2}$
(D) $\frac{3}{2} \leqslant a \leqslant \frac{5}{2}$ | 5. (C)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | cn_contest | false | 713,815 |
6. From 1 cent, 2 cents, and 5 cents 3 types of coins, take out 100 coins, totaling 3 yuan. The possible number of 2 cent coins is ( ) types.
(A) 13
(B) 16
(C) 17
(D) 19 | 6.
(C)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,816 |
7. As shown in Figure 3, in square $A B C D$, $A E=E F=F B, B G=2 G C$, $D E$ and $D F$ intersect $A G$ at $P$ and $Q$. Among the following statements, the incorrect one is ( ).
(B) $A Q: Q G=6: 7$
(C) $E P: P D=2: 11$
(D) $S_{\text {quadrilateral } C C D Q}: S_{\text {quadrilateral } B C O F}=17: 9$ | 7. (C)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,817 |
1. (16 points) Suppose the equation $\left|x^{2}+a x\right|=4$ has only 3 distinct real roots. Find the value of $a$ and the corresponding 3 roots. | Three, 1. The equation $\left|x^{2}+a x\right|=4$ is equivalent to the following two equations:
$x^{2}+a x-4=0$,
$x^{2}+a x+4=0$.
Let $x_{0}$ be a root of equation (1), then $x_{0}^{2}+a x_{0}-4=0$.
Also, $x_{0}^{2}+a x_{0}+4=x_{0}^{2}+a x_{0}-4+8=8 \neq 0$,
Therefore, $x_{0}$ is definitely not a root of equation (2), ... | a=4, \text{ roots: } -2, -2 \pm 2\sqrt{2}; \text{ or } a=-4, \text{ roots: } 2, 2 \pm 2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,818 |
2. (16 points) Two vegetable bases, A and B, supply the same type of vegetables to three farmers' markets, $A$, $B$, and $C$. According to the signed contract, 45 t should be supplied to $A$, 75 t to $B$, and 40 t to $C$. Base A can arrange for 60 t, and Base B can arrange for 100 t. The distance (in km) between A and ... | 2. Let base B supply $x \text{t}$ to $A$, $y \text{t}$ to $B$, and $[100-(x+y)] \text{t}$ to $C$. Then base A supplies $(45-x) \text{t}$ to $A$, $(75-y) \text{t}$ to $B$, and $[40-(100-x-y)] \text{t} = [(x+y)-60] \text{t}$ to $C$. According to the problem, the total transportation cost is
$$
\begin{aligned}
W= & 10(45-... | 960 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,819 |
3. (18 points) As shown in Figure 4, in $\triangle ABC$, altitudes $BE$ and $CF$ intersect at $H$, and $\angle BHC=135^{\circ}$. $G$ is a point inside $\triangle ABC$ such that $GB=GC$ and $\angle BGC=3\angle A$. Connect $HG$. Prove that $HG$ bisects $\angle BHF$. | 3. From $B E \perp A C, C F \perp A B$, we know that $A, F, H, E$ are concyclic. Thus, $\angle A=180^{\circ}-135^{\circ}=45^{\circ}$.
Therefore, $\angle A B E=\angle B H F=45^{\circ}$.
Also, $\angle B G C=3 \angle A=135^{\circ}, \angle B H C=135^{\circ}$,
Thus, $B, G, H, C$ are concyclic.
Since $G B=G C$, then $\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,820 |
1. Given $a+\frac{1}{b}=\frac{2}{a}+2 b \neq 0$. Then $\frac{a}{b}$ is ( ).
(A) -1
(B) 1
(C) 2
(D) cannot be determined | $-、 1 .(\mathrm{C})$.
From the condition, we get $a-2 b=\frac{2}{a}-\frac{1}{b}=\frac{2 b-a}{a b}$, $(a-2 b)\left(1+\frac{1}{a b}\right)=0$.
Since $\frac{1}{a}+b \neq 0$, i.e., $a b+1 \neq 0$,
and obviously $a b \neq 0$, hence $1+\frac{1}{a b} \neq 0$.
Therefore, $a-2 b=0$, i.e., $\frac{a}{b}=2$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,821 |
2. Given $\frac{3 x+4}{x^{2}-x-2}=\frac{A}{x-2}-\frac{B}{x+1}$, where $A$ and $B$ are constants. Then the value of $4 A-B$ is ( ).
(A) 7
(B) 9
(C) 13
(D) 5 | 2. (C).
From $\frac{A}{x-2}-\frac{B}{x+1}=\frac{(A-B) x+A+2 B}{x^{2}-x-2}$, we get $A-B=3, A+2 B=4$.
Thus, $4 A-B=3(A-B)+(A+2 B)=13$. | 13 | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,822 |
3. In a polygon, except for 2 interior angles, the sum of the remaining interior angles is $2002^{\circ}$. Then the number of sides of this polygon is ( . ).
(A) 12
(B) 12 or 13
(C) 14
(D) 14 or 15 | 3. (D).
Let this polygon be an $n$-sided polygon (where $n$ is a positive integer). According to the problem,
$$
2002^{\circ}<(n-2) \times 180^{\circ}<2002^{\circ}+360^{\circ} \text {, }
$$
which simplifies to $13 \frac{11}{90}<n<15 \frac{11}{90}$.
Therefore, $n=14$ or 15. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,823 |
4. Given a linear function $y=k x-k$. If $y$ decreases as $x$ decreases, then the graph of the function passes through ().
(A) the first, second, and third quadrants
(B) the first, second, and fourth quadrants
(C) the first, third, and fourth quadrants
(D) the second, third, and fourth quadrants | 4. (C)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,824 |
Example 6 In quadrilateral $ABCD$, diagonal $AC$ bisects $\angle BAD$. Take a point $E$ on $CD$, and let $BE$ intersect $AC$ at point $F$. Extend $DF$ to intersect $BC$ at $G$. Prove: $\angle GAC = \angle EAC$.
(1999, National High School Mathematics Competition) | Analysis: There are many ways to establish a coordinate system, but using the line $AC$ as the $x$-axis and point $A$ as the origin to establish a Cartesian coordinate system can reduce the computational load.
Proof: Establish a Cartesian coordinate system as shown in Figure 2. Let $A(0,0)$, $C(c, 0)$, $D(d, k d)$, $B(... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,825 |
6. As shown in Figure 2, in $\triangle A B C$, $\angle A B C=90^{\circ}$, $D$ is the midpoint of $A C$, and $B E \perp B D$ intersects the extension of $C A$ at $E$. Which of the following conclusions is correct? ( ).
(A) $\triangle B E D \backsim \triangle B C A$
(B) $\triangle B E A \backsim \triangle B C D$
(C) $\tr... | 6. (C).
From the problem, we have $\angle A B E=90^{\circ}-\angle A B D=\angle D B C$. Since $D$ is the midpoint of the hypotenuse $A C$ of the right triangle $\mathrm{Rt} \triangle A B C$, it follows that $\angle D B C=\angle D C B$. Therefore, $\angle A B E=\angle D B C$, so $\triangle A B E \backsim \triangle B C E... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,827 |
8. If 4 lines in a plane intersect each other pairwise and no three lines are concurrent, then there are $\qquad$ pairs of consecutive interior angles. | 8.24.
Each line intersects with 3 other lines, forming 3 intersection points. Each 2 intersection points determine a line segment, making a total of 3 line segments. Since each line has a pair of consecutive interior angles on both sides, there are $3 \times 4=12$ line segments, and a total of 24 pairs of consecutive ... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,829 |
9. The solution to the equation $\frac{1}{x^{2}+x-2}+\frac{1}{x^{2}+7 x+10}=2$ is | 9. $-2 \pm \sqrt{10}$.
The original equation is transformed into
$$
\frac{1}{(x+2)(x-1)}+\frac{1}{(x+2)(x+5)}=2,
$$
which simplifies to $x^{2}+4 x-6=0$, solving for $x$ yields $x=-2 \pm \sqrt{10}$.
Upon verification, the two roots of the original equation are $x=-2 \pm \sqrt{10}$. | x=-2 \pm \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,830 |
11. Given
$$
\left(x+\sqrt{x^{2}+2002}\right)\left(y+\sqrt{y^{2}+2002}\right)=2002 .
$$
Then $x^{2}-3 x y-4 y^{2}-6 x-6 y+58=$ $\qquad$ . | 11.58
$$
\begin{array}{l}
\text { Given }\left(x+\sqrt{x^{2}+2002}\right)\left(\sqrt{x^{2}+2002}-x\right) \\
=2002,
\end{array}
$$
we get $\sqrt{x^{2}+2002}-x=\sqrt{y^{2}+2002}+y$.
Similarly, $\sqrt{y^{2}+2002}-y=\sqrt{x^{2}+2002}+x$.
Adding these two equations gives $x+y=0$.
Therefore, $x^{2}-3 x y-4 y^{2}-6 x-6 y+58... | 58 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,832 |
12. As shown in Figure 4, line $AB$ intersects $\odot O$ at points $A$ and $B$, point $O$ is on $AB$, and point $C$ is on $\odot O$, with $\angle AOC=40^{\circ}$. Point $E$ is a moving point on line $AB$ (not coinciding with point $O$), and line $EC$ intersects $\odot O$ at another point $D$. The number of points $E$ t... | 12.3.
Consider the position of point $E$: Point $E$ can be on the extension of line segment $O A$; Point $E$ can be on line segment $O A$ (excluding point $O$); Point $E$ can be on the extension of line segment $O B$; but point $E$ cannot be on line segment $O B$ (excluding point $O$). Therefore, point $E$ has a total... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,833 |
13. There are two equations:
Good + Good $=$ Wonderful, Wonderful $\times$ GoodGood $\times$ ReallyGood $=$ WonderfulProblemProblemWonderful, where each Chinese character represents a digit from $0 \sim 9$, the same character represents the same digit, and different characters represent different digits. Therefore, th... | 13.16.
From the addition formula, we get “好” $<5$, “妙” $\neq 0$, so “好” $=1$, “妙” $=2$ or “好” $=2$, “妙” $=4$ or “好” $=3$, “妙” $=6$ or “好” $=4$, “妙” $=8$. Clearly, the middle two scenarios do not satisfy the multiplication formula, so it can only be:
(1) “好” $=1$, “妙” $=2$, thus the multiplication formula becomes
$2 \t... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,834 |
14. Given real numbers $a, b, c$, satisfying $a+b+c=0, a^{2}+b^{2}+c^{2}=6$. Then the maximum value of $a$ is | 14.2.
From the problem, we get $c=-(a+b)$, thus,
$$
a^{2}+b^{2}+[-(a+b)]^{2}=6,
$$
which simplifies to $b^{2}+a b+a^{2}-3=0$.
Since $b$ is a real number, the above equation, which is a quadratic equation in $b$, must have real roots, hence,
$$
\Delta=a^{2}-4\left(a^{2}-3\right) \geqslant 0, \quad -2 \leqslant a \leqs... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,835 |
Example 1: Does there exist an isosceles triangle with integer sides, an integer area, and a perimeter of 2003? Prove your conclusion.
If there exists such a triangle, let the two equal sides be \(a\) and the base be \(b\). The perimeter condition gives us:
\[ 2a + b = 2003 \]
The area \(A\) of the isosceles trian... | Explanation: First, express the area of a triangle using its three side lengths, then analyze whether the area is an integer given that the perimeter is 2003 and the side lengths are integers.
Assume such a triangle exists, with the base and the equal sides being \(a\) and \(b\) (where \(a\) and \(b\) are integers), t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,836 |
Example 2: Prove that in a right-angled triangle with integer sides, one of the legs must be a multiple of 3. | Explanation: We can classify and discuss according to the remainder when divided by 3.
Let the three sides of a right triangle be $a, b, c$, and $a^{2}+b^{2}=c^{2}$, where $a, b, c$ are all integers.
(1) If $a, b, c$ are not multiples of 3, then they can be expressed in the form $3n+1$ or $3n-1$ ($n$ is a positive inte... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,837 |
4. In Rt $\triangle A B C$, $\angle B C A=90^{\circ}$, $C D$ is the altitude. It is known that the three sides of $\mathrm{Rt} \triangle A B C$ are all integers, and $B D=11^{3}$. Find the ratio of the perimeters of $\mathrm{Rt} \triangle B C D$ and $\mathrm{Rt} \triangle A C D$. | (Given: Let the three sides of $\triangle A B C$ be $a, b, c$. From $11^{3} c = a^{2}$, set $a = 11^{2} p, c = 11 p^{2}$, then $b^{2} = c^{2} - a^{2} = (11 p)^{2} (p^{2} - 11^{2})$. Set $b = 11 p q$, then $p^{2} - q^{2} = 11^{2}$. Therefore, $p = 61, q = 60$. Answer: $\frac{11}{60}$.) | \frac{11}{60} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,838 |
2. $\frac{\left(2^{3}-1\right)\left(3^{3}-1\right)\left(4^{3}-1\right) \cdots\left(100^{3}-1\right)}{\left(2^{3}+1\right)\left(3^{3}+1\right)\left(4^{3}+1\right) \cdots\left(100^{3}+1\right)}$ is closest to ( ).
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{5}$
(D) $\frac{5}{8}$ | 2.(B).
$$
\begin{array}{l}
\text { Since } \frac{a^{3}-1}{(a+1)^{3}+1}=\frac{(a-1)\left(a^{2}+a+1\right)}{(a+2)\left(a^{2}+a+1\right)}=\frac{a-1}{a+2}, \\
\text { the original expression }=\frac{(2-1)(3-1) \cdots(99-1)\left(100^{3}-1\right)}{\left(2^{3}+1\right)(3+1) \cdots(99+1)(100+1)} \\
=\frac{1 \times 2 \times 3 \... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,839 |
3. As shown in Figure 1, in $\triangle A B C$, $A B=A C, \angle A=40^{\circ}$, extend $A C$ to $D$ such that $C D=B C$, point $P$ is the incenter of $\triangle A B D$. Then $\angle B P C=(\quad)$.
(A) $145^{\circ}$
(B) $135^{\circ}$
(C) $120^{\circ}$
(D) $105^{\circ}$ | 3. (A).
Connect $P D$, by the property of the incenter we know $\angle B P D=90^{\circ}+\frac{1}{2} \angle A$.
Also, $\angle B C D=90^{\circ}+\frac{1}{2} \angle A$, thus
$\angle B P D=\angle B C D$.
Therefore, $B, D, C, P$ are concyclic.
It is easy to know $\angle D=35^{\circ}$, so, $\angle B P C=180^{\circ}-35^{\circ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,840 |
$4 . a 、 b 、 c 、 d$ are pairwise distinct positive integers, and $a+b=c d, a b=c+d$. Then the number of quadruples $a 、 b 、 c 、 d$ that satisfy the above requirements is $(\quad)$.
(A) 4
(B) 6
(C) 8
(D) 10 | 4. (C).
(1) If $a, b, c, d$ are all not equal to 1, and $a \neq b, c \neq d$, then $(a-1)(b-1) \geqslant 2$, i.e., $a b > a + b$, $(c-1)(d-1) \geqslant 2$, i.e., $c d > c + d$.
But $a + b = c d > c + d, c + d = a b > a + b$, which is a contradiction. Therefore, there is no solution in this case.
(2) If one of $a, b, c,... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,841 |
6. In $\triangle A B C$, $\angle A$ and $\angle B$ are both acute angles, and $C D$ is the altitude. Given that $\frac{A D}{D B}=\left(\frac{A C}{B C}\right)^{2}$. Then $\triangle A B C$ is ( ).
(A) a right triangle
(B) an isosceles triangle
(C) an isosceles right triangle
(D) an isosceles triangle or a right triangle | 6. (D).
As shown in Figure 5, let $A D=a, D B=b$. According to the given, let $A C=\sqrt{a x}$, then $B C=\sqrt{b x}$.
Since $A C^{2}-A D^{2}=C D^{2}=B C^{2}-D B^{2}$, we have
$$
a x-a^{2}=b x-b^{2},
$$
which simplifies to
$$
(a-b)[x-(a+b)]=0.
$$
When $a=b$, it is easy to see that
$$
\triangle A D C \cong \triangle ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,843 |
1. All real numbers $c$ that make the equation $|x-1|-|x-2|+2|x-3|=$ $c$ have exactly two solutions are $\qquad$ . | 二、1. $c>3$ or $13$;
(2) When $1 \leqslant x3$, the original equation has exactly two solutions $\frac{5-c}{2}, \frac{5+c}{2}$;
When $1<c<3$, the original equation has exactly two solutions $\frac{7-c}{2}, \frac{5+c}{2}$. | c>3 \text{ or } c=13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,844 |
3. As shown in Figure 3, $\triangle A B C$ is inscribed in $\odot O, B C=a, C A=$ $b, \angle A-\angle B=90^{\circ}$. Then the area of $\odot O$ is | 3. $\frac{1}{4} \pi\left(a^{2}+b^{2}\right)$
The auxiliary lines are shown in Figure 6.
It is easy to see that $\angle A=\angle A B D$,
i.e., $\overparen{C D B}=\overparen{A C D} \Rightarrow \overparen{A C}=\overparen{B D}$
$$
\Rightarrow B D=A C=b \text {. }
$$
At this point, the radius $R$ of $\odot O$ is
$$
\frac{... | \frac{1}{4} \pi\left(a^{2}+b^{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,846 |
4. A group of 17 middle school students went to several places for a summer social survey, with a budget for accommodation not exceeding $x$ yuan per person per day. One day, they arrived at a place with two hostels, $A$ and $B$. $A$ has 8 first-class beds and 11 second-class beds; $B$ has 10 first-class beds, 4 second... | 4. $x=10$.
If staying at location $A$, even choosing the most economical beds, the average accommodation cost per person will exceed 10 yuan (since $8 \times 11 + 14 \times 6 = 172$ (yuan), $172 \div 17 \approx 10.12$ (yuan)).
If staying at location $B$, with a reasonable choice of beds, the budget can be met, and th... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,847 |
One. (20 points) A batch of goods is prepared to be transported to a certain place, and there are three trucks, A, B, and C, available for hire. It is known that the cargo capacity of trucks A, B, and C remains constant each time, and trucks A and B would need $2a$ and $a$ trips, respectively, to transport this batch o... | Obviously, the load capacity of vehicle B is twice that of vehicle A.
Let the total weight of this batch of goods be $M \mathrm{t}$, and vehicles A and C together transport the goods in $b$ trips, while vehicles B and C together transport the goods in $c$ trips. Since the load capacity of vehicle C remains unchanged wh... | 2160 \text{ yuan, } 4320 \text{ yuan, } 4320 \text{ yuan} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,848 |
5. Given that the lengths of the four sides of a rectangle are all integers less than 10, these four length numbers can form a four-digit number, where the thousand's digit and the hundred's digit of this four-digit number are the same, and this four-digit number is a perfect square. Find the area of this rectangle. | (Tip: Let the two adjacent sides of the rectangle be $a$ and $b$, then $\overline{a a b b}=$ $11(100 a+b)$ is a perfect square, which implies 11 divides $(a+b)$. Thus, $a + b = 11$. Upon verification, only $a=7, b=4$. Answer: 28.) | 28 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,849 |
II. (25 points) As shown in Figure 4, in the circumscribed convex hexagon $A B C D E F$, $A B / / D E$, $B C / / E F$, $C D$ $/ / F A$. Prove that the convex hexagon is a centrally symmetric figure. | As shown in Figure 7, since $C D / / A F$ and $A B$ intersects $A F$, extending $A B$ and $D C$ will definitely intersect at point $Q$; similarly, we get intersection points $P$ and $S$. Let $X$, $Y$, and $Z$ be the three points of tangency. It is easy to see that
$\triangle P A F \backsim \triangle B Q C$
$\sim \trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,850 |
Three. (25 points) Find all positive integers $a$ such that the quadratic equation
$$
a x^{2}+2(2 a-1) x+4(a-3)=0
$$
has at least one integer root. | Three, the original equation can be transformed into $(x+2)^{2} a=2 x+12$.
It is easy to see that $x \neq-2$, at this time $a=\frac{2 x+12}{(x+2)^{2}}$.
Since $a$ is a positive integer, i.e., $\frac{2 x+12}{(x+2)^{2}} \geqslant 1$ is a positive integer.
Also, $(x+2)^{2}>0$, then
$$
(x+2)^{2} \leqslant 2 x+12 \text {, }... | a=1, 6, 10, 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,851 |
1. If the set $S=\{n \mid n$ is an integer, and $22 n+2$ divides $2003 n+2004\}$, then $S$ is ( ).
(A) empty set $\varnothing$
(B) singleton set
(C) set with two elements
(D) infinite set | -、1.(C).
Obviously, $2 \mid n$. Let $n=2k$, then $(22k+1) \mid (2003k+1002)=91(22k+1)+k+911$.
Thus, $(22k+1) \mid (k+911)$, which is equivalent to $(22k+1) \mid (22k+911 \times 22)=(22k+1)+20041$, that is,
$$
(22k+1) \mid 20041=7^2 \times 409 \text{, }
$$
where 409 is a prime number.
Therefore, $22k+1$ can take $\pm 1... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,852 |
2. If the polynomial $x^{2}-x+1$ can divide another polynomial $x^{3}+x^{2}+a x+b(a, b$ are constants). Then $a+b$ equals ( ).
(A) 0
(B) -1
(C) 1
(D) 2 | 2. (C).
By synthetic division, it is easy to obtain the remainder $r(x)=(a+1) x+b-2$. Since it can be divided exactly, we have $a+1=0$ and $b-2=0$, i.e., $a=-1, b=2$. Therefore, $a+b=1$. | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,853 |
3. Let $a$ be an integer, and the two real roots of the equation $x^{2}+(a-3) x+a^{2}=0$ be $x_{1}$ and $x_{2}$. If $\tan \left(\arctan x_{1}+\arctan x_{2}\right)$ is also an integer, then the number of such $a$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 4 | 3. (B).
By Vieta's formulas, we have $x_{1}+x_{2}=3-a, x_{1} x_{2}=a^{2}$. Using the tangent addition formula, we get
$$
\begin{array}{l}
\tan \left(\arctan x_{1}+\arctan x_{2}\right) \\
=\frac{\tan \left(\arctan x_{1}\right)+\tan \left(\arctan x_{2}\right)}{1-\tan \left(\arctan x_{1}\right) \tan \left(\arctan x_{2}\r... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,854 |
4. Let the volume of a tetrahedron be $V_{1}$, and the midpoints of its edges form a convex polyhedron with volume $V_{2}$. Then $\frac{V_{2}}{V_{1}}$ is $(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) a constant, but not equal to $\frac{1}{2}$ and $\frac{2}{3}$
(D) uncertain, its value depends on the specific shape... | 4. (A).
As shown in Figure 2, $E, F, G, H, P$, and $Q$ are the midpoints of the respective edges, forming a convex polyhedron $P-EFGH-Q$. It is easy to see that the volumes of $A-EPH$, $B-EFQ$, $C-FGP$, and $D-GHQ$ are all $\left(\frac{1}{2}\right)^{3}$ of $V_{1}$, so,
$$
\begin{aligned}
V_{2} & =V_{1}-4\left(\frac{1}... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,855 |
5. In decimal, if a positive integer with at least two digits has all its digits, except the leftmost one, smaller than the digit to their left, then it is called a descending positive integer. The total number of such descending positive integers is ( ).
(A) 1001
(B) 1010
(C) 1011
(D) 1013 | 5.(D).
Let the leftmost digit be $a$, then the digits $a-1$, $a-2, \cdots, 2,1,0$ that are smaller than $a$ can each appear at most once to the right of $a$. Therefore, the number of decreasing positive integers starting with $a$ is equal to the number of non-empty subsets of $\{a-1, a-2, \cdots, 2,1,0\}$, which is $2... | 1013 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 713,856 |
6. Among the 8 vertices of a cube, the number of right-angled triangle groups formed by 3 vertices is ( ).
(A) 36
(B) 37
(C) 48
(D) 49 | 6. (C).
As shown in Figure 3, note that each square or rectangle has 4 "right-angle triplets." There are a total of 6 side faces and 6 diagonal sections, so there are $4 \times(6+6)=48$ "right-angle triplets." | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 713,857 |
1. If the line $x \cos \theta+y \sin \theta=\cos ^{2} \theta-\sin ^{2} \theta$ $(0<\theta<\pi)$ intersects the circle $x^{2}+y^{2}=\frac{1}{4}$, then the range of values for $\theta$ is $\qquad$ | 二、1. $\frac{\pi}{6} \leqslant \theta \leqslant \frac{\pi}{3}$ or $\frac{2 \pi}{3} \leqslant \theta \leqslant \frac{5 \pi}{6}$.
Let the distance from the center of the circle to the line be
$$
d=\frac{\left|\cos ^{2} \theta-\sin ^{2} \theta\right|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=|\cos 2 \theta| \leqslant r=\f... | \frac{\pi}{6} \leqslant \theta \leqslant \frac{\pi}{3} \text{ or } \frac{2 \pi}{3} \leqslant \theta \leqslant \frac{5 \pi}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,858 |
2. In the Cartesian coordinate system $x O y$, a circle passes through $(0,2)$ and $(3,1)$, and is tangent to the $x$-axis. Then the radius of this circle is $\qquad$ . | 2. $15 \pm 6 \sqrt{5}$.
Let the equation of the circle be $(x-a)^{2}+(y-r)^{2}=r^{2} (r>0$ being the radius), then we have
$$
\left\{\begin{array}{l}
(0-a)^{2}+(2-r)^{2}=r^{2}, \\
(3-a)^{2}+(1-r)^{2}=r^{2},
\end{array}\right.
$$
which simplifies to
$$
\left\{\begin{array}{l}a^{2}=4 r-4, \\ a^{2}-6 a=2 r-10 .\end{arra... | 15 \pm 6 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,859 |
7. In $\triangle A B C$, $\angle B$ is an acute angle, $A D$ is the altitude to side $B C$, and $C E$ is the altitude to side $A B$. When $\frac{2 B D}{B C}$ and $\frac{2 B E}{A B}$ are both integers, determine the shape of $\triangle A B C$ and prove your conclusion. | (Let $\frac{2 B D}{B C}=a, \frac{2 B E}{A B}=b$. From $B D<A B, B E<B C$, we get $0<a b<4$. When $a=b=1$, $\triangle A B C$ is an equilateral triangle; when $a=1, b=2$ or $a=2, b=1$, $\triangle A B C$ is an isosceles right triangle; when $a=1, b=3$ or $a=3, b=1$, $\triangle A B C$ is an isosceles obtuse triangle.) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,860 |
3. If the constant $a$ makes the equation about $x$
$$
\lg \left(x^{2}+20 x\right)-\lg (8 x-6 a-3)=0
$$
have a unique solution. Then the range of values for $a$ is $\qquad$ | 3. $-\frac{163}{6}0$, that is
$$
\left\{\begin{array}{l}
x(x+20)>0, \\
x^{2}+12 x+6 a+3=0 .
\end{array}\right.
$$
From (2), solve for $x_{1}=-6-\sqrt{33-6 a}, x_{2}=-6+\sqrt{33-6 a}$. At this point, there are two possible scenarios:
(1) $\left\{\begin{array}{l}-6-\sqrt{33-6 a}0 .\end{array}\right.$
(1) has no solution... | -\frac{163}{6}<a<-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,861 |
4. $f(x)=\frac{x^{2}}{8}+x \cos x+\cos (2 x)(x \in \mathbf{R})$'s minimum value is $\qquad$ . | 4. -1 .
$$
\begin{array}{l}
f(x)=\frac{x^{2}}{8}+x \cos x+2 \cos ^{2} x-1 \\
=\frac{1}{8}(x+4 \cos x)^{2}-1 \geqslant-1 .
\end{array}
$$
Since the equation $\cos x=-\frac{x}{4}$ (as can be seen from the graph) has a solution, we have
$$
f(x)_{\min }=-1
$$ | -1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 713,862 |
6. Let $A B C D$ be a convex quadrilateral, $A B=7, B C=4$, $C D=5, D A=6$, and its area $S$ has a range of $(a, b]$. Then $a+b=$ $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $6.2 \sqrt{210}$.
As shown in Figure 4, connect $B D$, and let
$\angle B A D=a, \angle B C D=\beta$, then
$S=21 \sin a+10 \sin \beta$.
By the cosine rule, we have
$$
\begin{array}{l}
6^{2}+7^{2}-2 \times 6 \times 7 \cos a \\
\quad=B D^{2}=4^{2}+5^{2}-2 \times 4 \times 5 \cos \beta,
\end{array}
$$
which means $21 \cos ... | 2 \sqrt{210} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,864 |
Three. (20 points) Let the left and right foci of an ellipse be $F_{1}$ and $F_{2}$, respectively, and the left directrix be $l$. Point $P$ is on the ellipse. Draw $P Q \perp$ $l$, with $Q$ being the foot of the perpendicular. Try to determine: For what kind of ellipse does there exist such a point $P$ that $P Q F_{1} ... | Three, as shown in Figure 5, let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and point $P$ is above the $x$-axis, with coordinates $\left(x_{P}, y_{P}\right)$, $c^{2}=a^{2}-b^{2}(c>0)$, then $F_{1}(-c, 0)$, $F_{2}(c, 0)$, and the left directrix $l: x=-\frac{a^{2}}{c}$.
Since $P Q... | \frac{1}{2} \leqslant e < 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,865 |
Four. (20 points) Let $a_{0}=1, a_{1}=2, a_{n+1}=$ $2 a_{n-1}+n, n=1,2,3, \cdots$. Try to find the expression for $a_{n}$ (the answer should be in the form of a finite sum of expressions involving $n$, and the number of terms should be independent of $n$). | First, we have $a_{2}=2 a_{0}+1=3$. From $a_{n+1}=2 a_{n-1}+n$ and $a_{n+2}=2 a_{n}+n+1$, subtracting these gives
$$
a_{n+2}-a_{n+1}=2\left(a_{n}-a_{n-1}\right)+1.
$$
Let $b_{n}=a_{n+1}-a_{n}$, then from (1) we get
$$
b_{n+1}=2 b_{n-1}+1,
$$
where $b_{0}=a_{1}-a_{0}=1, b_{1}=a_{2}-a_{1}=3-2=1, b_{2}=2 b_{0}+1=3$.
Fro... | a_{2 n}=2^{n+2}-2 n-3, \quad a_{2 n+1}=3 \times 2^{n+1}-2 n-4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,866 |
Five. (20 points) Try to find all ordered integer pairs $(a, b)$, such that the roots of the equation $x^{4}+\left(2 b-a^{2}\right) x^{2}$ $-2 a x+b^{2}-1=0$ are all integers.
Find all ordered integer pairs $(a, b)$, such that the roots of the equation $x^{4}+\left(2 b-a^{2}\right) x^{2} -2 a x+b^{2}-1=0$ are all inte... | Transform the original equation into
$$
\left(x^{2}+b\right)^{2}-(a x+1)^{2}=0 \text {, }
$$
which is $\left(x^{2}+a x+b+1\right)\left(x^{2}-a x+b-1\right)=0$.
Therefore, $x^{2}+a x+b+1=0$ and $x^{2}-a x+b-1=0$
have all integer roots. Thus,
$$
\Delta_{1}=a^{2}-4(b+1)=t_{1}^{2}, \Delta_{2}=a^{2}-4(b-1)=t_{2}^{2} \text... | (a, b) = (2k-1, k^2-k-1) \, (\forall k \in \mathbf{Z}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,867 |
One, (50 points) As shown in Figure 1, point $P$ is inside $\triangle A B C$ and $\angle B A P = \angle C A P$. Connect $B P$ and extend it to intersect $A C$ at point $Q$. Given $\angle B A C = 60^{\circ}$ and $\frac{1}{B P} + \frac{1}{P C} = \frac{1}{P Q}$. Prove: $P$ is the incenter of $\triangle A B C$. | As shown in Figure 6, let $\angle P C Q = \gamma, \angle P B A = \theta$. From the given conditions, we have
$$
\frac{P C}{B P} + 1 = \frac{P C}{P Q}, \text{ i.e., } \frac{P C}{P A} \cdot \frac{P A}{B P} + 1 = \frac{P C}{P Q}.
$$
By the Law of Sines, we get
$$
\begin{array}{l}
\frac{\sin 30^{\circ}}{\sin \gamma} \cdot... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,868 |
II. (50 points) Let positive numbers $a, b$ satisfy $a>\frac{b}{\sqrt{2}}$ and such that the inequality about $x$
$$
\sqrt{\dot{x}-1} \geqslant a \sqrt{x+1}-b
$$
always has real solutions. Try to find the range of $f(a, b)=a^{2}-3 a b+b^{2}$. | First, find the conditions that $a$ and $b$ should satisfy. From the original inequality, we get the following equivalent forms: $\sqrt{x-1}+b \geqslant a \sqrt{x+1}$. Squaring both sides and rearranging, we get
$$
\begin{array}{l}
2 b \sqrt{x-1} \geqslant\left(a^{2}-1\right) x+\left(a^{2}-b^{2}+1\right) \text {. } \\
... | (-\infty, 1) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,869 |
Three, (50 points) Try to find the smallest possible value of the positive integer $k$, such that the following proposition holds: For any $k$ integers $a_{1}, a_{2}, \cdots, a_{k}$ (equality is allowed), there must exist corresponding $k$ integers $x_{1}, x_{2}, \cdots, x_{k}$ (equality is also allowed), and $\left|x_... | Three, first prove that the proposition holds when $k=7$. For this, consider the sum
$$
S\left(y_{1}, y_{2}, \cdots, y_{7}\right)=y_{1} a_{1}+y_{2} a_{2}+\cdots+y_{7} a_{7} \text {, }
$$
where $y_{i} \in\{-1,0,1\}$.
There are $3^{7}=2187$ such sums, and since $2187>2003$, by the pigeonhole principle, there must be two... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,870 |
Example 1 A factory plans to arrange 214 workers to produce 6000 $A$ components and 2000 $B$ components. It is known that the time it takes for each worker to produce 5 $A$ components can produce 3 $B$ components. Now the workers are divided into two groups to produce these two types of components, and they start at th... | Solution: Let $x$ be the number of people producing component $A$, then the number of people producing component $B$ is $(214-x)$. Taking the time for one person to produce 5 $A$ components as the unit, the time required to produce 6000 $A$ components and 2000 $B$ components is respectively
$$
y_{1}=\frac{6000}{5 x} \t... | 137 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,871 |
Given $ABCD$ is a unit square, $O$ is its center, $P$ is a point on $CD$, line $AP$ intersects the extension of $BC$ at point $Q$, intersects $DO$ at point $E$, and $OQ$ intersects $PC$ at point $F$. If $EF \| AC$, find the length of $AP$. | Solution: As shown in Figure 1, let $DP$
$$
=x(0<x<1), CQ=y,
$$
then $AP=\sqrt{1+x^{2}}, PC=$
1 - $x$. From the similarity of Rt $\triangle PCQ \sim \text{Rt}$
$\triangle PDA$, we get $\frac{CQ}{AD}=\frac{PC}{PD}$.
Therefore, $CQ=y=$
$$
\frac{AD \cdot PC}{PD}=\frac{1-x}{x}.
$$
Since $EF \parallel AC$, we have $\frac{... | \frac{\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,872 |
Given a positive integer $n(n \geq 5)$, try to provide a set of distinct positive even numbers $p_{1}, p_{2}, \ldots, p_{n}$, such that
$$
\frac{1}{p_{1}}+\frac{1}{p_{2}}+\ldots+\frac{1}{p_{n}}=\frac{2003}{2002} .
$$ | Solution: Since $\frac{2003}{2002}=1+\frac{1}{2002}$, we can construct some $p_{i}$ such that the sum of their reciprocals is 1.
Because $\frac{1}{k}-\frac{1}{k+1}=\frac{1}{k(k+1)}$, then
$$
\frac{1}{k}=\frac{1}{k+1}+\frac{1}{k(k+1)}.
$$
Using identity 11 to construct the required expression:
$$
\begin{aligned}
1 & =\... | p_{1}, p_{2}, \ldots, p_{5} = 2, 4, 6, 12, 2002 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,873 |
127 Given $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\frac{b^{2}}{a}+\frac{c^{2}}{b}+\frac{a^{2}}{c} \geq \sqrt{3\left(a^{2}+b^{2}+c^{2}\right)} \text {. }
$$ | Proof: First, let $abc=1$. At this point, (1) becomes
$$
b^{3} c+c^{3} a+a^{3} b \geq \sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}.
$$
(1) When $a \geq b \geq c$,
$$
\begin{array}{l}
b^{3} c+c^{3} a+a^{3} b-a^{2} b^{2}-b^{2} c^{2}-c^{2} a^{2} \\
=a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \\
=a(a-b)\left(a b-c^{2}\right)+c(b-c)\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,874 |
Let $a, b$ be real numbers satisfying $a^{3}+b^{3}=a^{n}+b^{n}(n \in$ $\{0,1,2\})$. Prove: $a+b \leq 2$.
The text has been translated while preserving the original formatting and line breaks. | Proof: If $n=0$, then $a^{3}+b^{3}=2$. Therefore,
$$
\begin{array}{l}
(2-b)^{3}=8-12 b+6 b^{2}-b^{3} \\
=a^{3}+6(b-1)^{2} \geq a^{3} .
\end{array}
$$
Thus, $2 - b \geq a$, i.e., $a+b \leq 2$.
If $n=1$, then $a^{3}+b^{3}=a+b$. Therefore,
$$
\begin{array}{l}
(a+b)\left(a^{2}-a b+b^{2}-1\right)=0 . \\
a+b=0 \text {, then... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,875 |
Example 2 Let $a, b$ be positive integers, $n$ be a given positive odd number, and $a+b=n$. Find the maximum value of $a b$.
Analysis: Since under the given conditions it is impossible to achieve $a=$ $b$, the solution cannot be obtained using the arithmetic mean inequality. | Solution 1: Let $n=2k+1$, i.e., $a+b=2k+1$.
By symmetry, assume without loss of generality that $a \leqslant k, b \geqslant k+1$, then,
$$
\begin{array}{l}
(a-k)(b-k) \\
=ab-k(a+b)+k^{2} \leqslant 0 .
\end{array}
$$
Thus, $ab \leqslant k(a+b)-k^{2}$
$$
\begin{array}{l}
=kn-k^{2}=k(n-k)=k(k+1) \\
=\frac{n-1}{2} \cdot \... | \frac{n^2-1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,876 |
Example 3 Given that $a_{1}, a_{2}, \cdots, a_{n}$ are $n(n \geqslant 3)$ positive integers, and $a_{1}+a_{2}+\cdots+a_{n}=m, m(m \geqslant n)$ is a constant. Find
$$
M=\sum_{1 \leqslant i<j<k \leqslant n} a_{i} a_{j} a_{k}
$$
the maximum value. | Proof: First, we prove that when $M$ attains its maximum value $M_{1}$, for any $i, j (1 \leqslant i < j \leqslant n)$, we have $a_{i} \leqslant a_{j} + 1$. Otherwise, if there exist $i, j$ such that $a_{i} \geqslant a_{j} + 2$, then by setting $a_{i}' = a_{i} - 1$ and $a_{j}' = a_{j} + 1$, we get a new sequence $a_{1}... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,877 |
Example 4 Let $n$ be a given positive integer. Try to find non-negative integers $k, l$, satisfying $k+l \neq 0$, and $k+l \neq n$, such that
$$
s=\frac{k}{k+l}+\frac{n-k}{n-(k+l)}
$$
takes the maximum value. | If $l=0$, then $s=2$.
If $l>0$, let $x=k+l$, then $00 .
\end{array}
$
Therefore, $f(1)1$ when, $s$ reaches the maximum value 2;
(2) If $n=2$, then when $l=1, k=0$ or $l=0, 0 < k \neq 2$, $s$ reaches the maximum value 2;
(3) If $n \geqslant 3$, then when $l=n-1, k=0$, $s$ reaches the maximum value $n$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,878 |
Example 5 There are 1988 unit cubes, and they (all or part of them) are arranged into 3 "squares" (i.e., 3 one-layer rectangular prisms with dimensions $a \times a \times 1, b \times b \times 1, c \times c \times 1$ where $a \leqslant b \leqslant c$) $A, B$, and $C$. Now, place square $C$ in the first quadrant of the $... | Solution: According to the problem, the condition for $B$ to be placed on $C$ is $b \leqslant c - 2$, and there are $(c-b-1)^{2}$ different ways to place it; the condition for $A$ to be placed on $B$ is $a \leqslant b-2$, and there are $(b-a-1)^{2}$ different ways to place it. Thus, the problem is transformed into:
Fin... | 345 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,879 |
Example 6 Let positive integers $a, b, c, d$ satisfy $\frac{a}{b}+\frac{c}{d} < 1$ and $a+c=20$. Find the maximum value of $\frac{a}{b}+\frac{c}{d}$. | Solution: By symmetry, we may assume $a \leqslant c$. From the given conditions, we know $b>a, d>c$, so we can let
$b=a+p, d=c+q$ (where $p, q$ are positive integers).
From $1-\left(\frac{a}{b}+\frac{c}{d}\right)=1-\left(\frac{a}{a+p}+\frac{c}{c+q}\right)$
$$
=\frac{p q-a c}{(a+p)(c+q)}>0,
$$
we get $p q>a c$, thus, $... | \frac{1385}{1386} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,880 |
Example 3 Given that the lengths of the three sides of an acute triangle $\triangle ABC$ and the length of one of its altitudes are four consecutive integers, and that this altitude divides $\triangle ABC$ into two right triangles with integer side lengths. Find the lengths of the three sides of this triangle.
---
Th... | Explanation: It can be considered in two steps: First, determine which of the 4 consecutive integers is the height, then discuss the base.
As shown in Figure 1, let the height be $AD$, and the 4 consecutive integers be $n, n+1, n+2, n+3$. Also, assume $AB < AC$, then $AD < AB < AC$. Therefore, $AD = n$ or $n+1$.
If $... | 13, 14, 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,882 |
1. Positive integers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1}+x_{2}+\cdots+x_{n}=$ $m$ (constant). Try to find the maximum value of $f=x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}$. | ( Hint: First prove that the necessary condition for $f$ to take the maximum value is $x_{1}=1$. Similarly, we get $x_{i}=1(i \neq n)$, so $f_{\text {max }}=(n-1)+(m-n+$ 1) ${ }^{2}$. ) | (n-1)+(m-n+1)^2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,883 |
2. Find the smallest positive integer $n$ such that the equation $\left[\frac{10^{n}}{x}\right]=1989$ has an integer solution.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | (The equation is equivalent to $10^{n} \times 0.000502512 \cdots < x \leqslant$ $10^{n} \times 0.000502765 \cdots$. When $n \leqslant 6$, there is no integer that satisfies the above equation; when $n=7$, the equation has an integer solution.) | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,884 |
3. The values of the sequence $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 2)$ can independently take $-1, 0$, and $1, S$ represents the sum of all products obtained by multiplying these numbers in pairs. Try to find the minimum value of $S$.
---
The translation maintains the original text's format and line breaks. | $$
\begin{array}{r}
\text { (Hint: } S=\sum_{1 \leqslant i<j<n} x_{1} x_{j}=\frac{1}{2}\left[\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2}\right. \\
\left.-\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)\right] \geqslant-\frac{1}{2}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) .
\end{array}
$$
If $n$ is even, $S_{\t... | S_{\text {min }} = \begin{cases} -\frac{n}{2}, & \text{if } n \text{ is even} \\ -\frac{n-1}{2}, & \text{if } n \text{ is odd} \end{cases} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,885 |
Example 1 The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=0,2 a_{n+1}$ $=3 a_{n}+\sqrt{5 a_{n}^{2}+4}(n \geqslant 1)$. Prove: There cannot be a natural number $m$ such that $a_{2 m}$ is divisible by 1992. | Proof: Transforming and rearranging the original recurrence relation, we get
$$
a_{n+1}^{2}-3 a_{n} a_{n+1}+a_{n}^{2}-1=0 \text {. }
$$
Thus, $a_{n-1}^{2}-3 a_{n} a_{n-1}+a_{n}^{2}-1=0$.
Hence, $a_{n-1}$ and $a_{n}$ are the roots of the equation $x^{2}-3 a_{n} x+a_{n}^{2}-1=0$. Therefore, we have
$$
a_{n+1}+a_{n-1}=3 ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,886 |
Example 2 Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{0}=0, x_{n+1}=$ $x_{n}+a+\sqrt{b^{2}+4 a x_{n}}, n=0,1,2, \cdots$, where $a$ and $b$ are given positive real numbers. Find the general term of this sequence. | Solution: Let $y_{n}=\sqrt{b^{2}+4 a x_{n}}$, then
$$
y_{0}=b, x_{n}=\frac{y_{n}^{2}-b^{2}}{4 a} \text {. }
$$
Thus, we have $y_{n+1}^{2}=\left(y_{n}+2 a\right)^{2}$.
Since $y_{n}>0$, then $y_{n+1}-y_{n}=2 a$, hence $\left\{y_{n}\right\}$ is an arithmetic sequence with a common difference of $2 a$.
From this, we deduc... | x_{n}=a n^{2}+b n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,887 |
5. Given the sum of several integers is 1,976. Find the maximum value of their product. | (Given: $x_{1}+x_{2}+\cdots+x_{n}=1976, u=x_{1} x_{2} \cdots$ $x_{1976}$. If $u$ takes the maximum value, then $x_{i} \leqslant 4(i=1,2, \cdots, n)$. This is because if some $x_{j}>4$, then replacing $x_{j}$ with 2 and $x_{j}-2$ will increase $u$; similarly, it can be shown that all $x_{i} \geqslant 2$. Since a $x_{i}=... | 2 \times 3^{658} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,889 |
6. There are 1989 points in space, where no three points are collinear. Divide them into 30 groups with different numbers of points. From any 3 different groups, take one point each to form a triangle. To maximize the total number of such triangles, how many points should each group have? | (Tip: Let the number of points in the $i$-th group be $x_{i}$, then the problem is transformed into: when $x_{1}$ $+x_{2}+\cdots+x_{30}=1989$, and $x_{1}, x_{2}, \cdots, x_{30}$ are all different, find the maximum value of $M=\sum_{1<i<j<k \leqslant 30} x_{i} x_{j} x_{k}$. Without loss of generality, assume $x_{1}<x_{2... | 51 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,890 |
Example 4 In the positive term sequence $\left\{a_{n}\right\}$, $a_{1}=10, a_{n+1}$ $=10 \sqrt{a_{n}}$. Find the general formula for this sequence. | Solution: Given $a_{1}>0, a_{n+1}=10 \sqrt{a_{n}}$, taking the logarithm on both sides, we get
$\lg a_{n+1}=1+\frac{1}{2} \lg a_{n}$.
Let $x_{n}=\lg a_{n}$, then we have
$$
x_{n+1}=\frac{1}{2} x_{n}+1, x_{n}=\frac{1}{2} x_{n-1}+1 \text {. }
$$
Subtracting the two equations, we get $x_{n+1}-x_{n}=\frac{1}{2}\left(x_{n}... | a_{n}=10^{2-\left(\frac{1}{2}\right)^{n-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,892 |
Example 4 In $\triangle A B C$, $A B=37, A C=58$, a circle with center $A$ and radius $A B$ intersects $B C$ at point $D$, and $D$ is between $B$ and $C$. If the lengths of $B D$ and $D C$ are both integers, find the length of $B C$.
In $\triangle A B C$, $A B=37, A C=58$, a circle with center $A$ and radius $A B$ int... | Explanation: A triangle with two known sides is indeterminate, but note that the lengths of $B D$ and $D C$ are both integers, so the length of $B C$ is also an integer. Therefore, we can start by determining the range of possible values for $B C$, and then find the value of $B C$ that meets the requirements.
As shown... | 57 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,893 |
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