problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
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Example 8. When $\mathrm{n}=1,2,3, \cdots, 1988$, find the sum $\mathrm{s}$ of the lengths of the segments cut off by the x-axis for all functions
$$
y=n(n+1) x^{2}-(2 n+1) x+1
$$ | The function $y=n(n+1) x^{2}-(2 n+1) x+1$ intersects the axis at two points, the abscissas of which are the roots of the equation
$$
n(n-1) x^{2}-(2 n+1) x+1=0 \text { . }
$$
Transforming the above equation into
$$
(n x-1)[(n+1) x-1]=0,
$$
we solve to get $x_{1}=\frac{1}{n+1}, x_{2}=\frac{1}{n}$.
Thus, $\mathrm{d}=\l... | \frac{1988}{1989} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,972 |
Example 1. There is a four-digit number. It is known that its tens digit minus 1 equals the units digit, and its units digit plus 2 equals the hundreds digit. The sum of this four-digit number and the number formed by reversing the order of its four digits equals 9878. Try to find this four-digit number. | Solution: Let the required four-digit number be $a b c d$. According to the problem, we have $\overline{a b c \bar{d}}+\overline{d c b a}=9878$,
which means $\left(10^{3} \cdot a+10^{2} \cdot b+10 \cdot c+d\right)+\left(10^{3} \cdot d\right.$ $\left.+10^{2} \cdot c+10 \cdot b+a\right)=9878$, or $10^{3} \cdot(a+d)+10^{... | 1987 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,975 |
Example 2. Rearrange the digits of a three-digit number to form the largest possible three-digit number, and subtract the smallest digit from it, which is exactly equal to the original number. Find these three digits. (Hua Luo Geng Math Contest 1988 Junior High School Level) | Let the three-digit number be $a, b, c$. If $abc$ is the largest, and $cb_{2}$ is the smallest, and
$$
\begin{array}{l}
\overline{a b c}-cba \\
=\left(10^{2} a+10 b+c\right)-\left(10^{2} \cdot c+10 b+a\right) \\
=99(a-c),
\end{array}
$$
i.e., the required three-digit number is a multiple of 99. Among such three-digit ... | 495 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,976 |
Example 1., Prove that $f(x)=\sin x^{2}$ is not a periodic function. | If $T$, then $\sin (x+T)^{2}=\sin x^{2}$. Therefore, $(x+T)^{2}$ $=x^{2}+2 n \pi$ (or $\left.(x+T)^{2}=\pi-x^{2}+2 n \pi\right)$, $n \in Z$, thus $T^{2}+2 x T-2 n \pi=0$. It is evident that, for different $x$, $T$ does not correspond to a constant, i.e., $T=-x \pm \sqrt{x^{2}+2 n} \pi$. Hence, $T$ is not a constant. Th... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 703,977 |
Example 2. Try to prove that $f(x)=\cos \sqrt{x}$ is a non-periodic function.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Example 2. Try to prove that $f(x)=\cos \sqrt{x}$ is a non-periodic function. | Assume $T>0$ is the period of $\cos \sqrt{x}$, then for all $x \geqslant 0$, we have $\cos \sqrt{x+T}=\cos x, \cos \sqrt{2 T+x} = \cos \sqrt{x+x}=\sqrt{x}$. Let $x=0$, then $\cos \sqrt{T} = \cos 0=1 \quad \cos \sqrt{2}=\cos 0=1$, $\sqrt{T}=2 n \pi, \sqrt{2 T}=2 k \pi, \quad(k, n \in \mathbb{N})$. Therefore, $\sqrt{2 T}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 703,978 |
Slope is 45, find the equation of the line 7.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | To solve $\beta=90^{\circ}-45^{\circ}=45, a=-7$.
$$
x=y \operatorname{tg} 45^{\circ}+(-7), \text { then } x-y+7=0 \text {. }
$$ | x-y+7=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,981 |
Example 4. Determine the positional relationship of the following pairs of lines:
(1)
$$
\begin{array}{l}
l_{1}: y=\frac{4}{3} x+8, \\
l_{2}: x=\frac{3}{4} y+6
\end{array}
$$
$$
\text { (2) } \begin{aligned}
l_{3}: x & =\frac{3}{4} y+6, \\
l_{4}: y & =-\frac{3}{4} x+5 .
\end{aligned}
$$ | $$
\begin{array}{l}
\text { Solution (1) } k_{x 1} \cdot k_{y 2}=\frac{4}{3} \cdot \frac{3}{4}=1, \\
\therefore l_{1} \| l_{2} \text {. }
\end{array}
$$
(2) $k_{y_{3}}=\frac{8}{4}, k_{x_{4}}=-\frac{3}{4}=-k_{y_{3}}$
$$
\therefore l_{3} \perp l_{4} .
$$ | l_{1} \| l_{2}, \, l_{3} \perp l_{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,984 |
Example 5, the line $l$ passing through the point $(1, \sqrt{3})$, its downward direction forms the smallest angle of 60 with the negative axis, find the equation of $l$.
untranslated text remains as is. | $$
\text{Solution: } \begin{array}{l}
k_{y}=\tan 120^{\circ}=-\sqrt{3}. \text{ The equation is } \\
x-1=-\sqrt{3}(y-\sqrt{3}), \\
\text{which simplifies to } x+\sqrt{3} y-2=0 .
\end{array}
$$
Examples 1 to 5, involving "intercept on the $x$-axis" and "angle formed with the $y$-axis," are directly solved using the meth... | x+\sqrt{3} y-2=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,985 |
$x^{2}+y^{2}-2 x-4 y'+4=0$ for any line intersecting the circle at points $P_{1}$ and $P_{2}$, find the locus of the midpoint $P$ of $P_{1} P_{2}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
---
$x^{2}+y^{2}-2 x-4 y'+4... | Let the coordinates of point $P$ be $(X, Y)$, then the equation of $P_{1} P_{2}$ is
$$
\begin{array}{l}
X(x-X)+Y(y-Y)-(x-X) \\
-2(y-Y)=0 .
\end{array}
$$
Substituting the coordinates of $O(0, 0)$, we get
$$
\left(X-\frac{1}{2}\right)^{2}+(Y-1)^{2}=\frac{5}{4} \text{. }
$$
The sought trajectory is a segment of the arc... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,986 |
Example 3. Suppose 1987 can be represented as a three-digit number $\overline{x y z}$ in base $b$, and $x+y+z=1+9+8+7$. Try to determine all possible values of $x$, $y$, $z$, and $b$. (Canadian 87 Competition Question) | Given the problem, we have
$$
\left\{\begin{array}{ll}
x b^{2}+y b+z=1987(x \geqslant 1), & (1) \\
x+y+z=25 & (2)
\end{array}\right.
$$
Subtracting (2) from (1) gives
$$
(b-1)[(b+1) x+y]=1962 \text {, }
$$
Thus, $(b-1) 1962=2 \times 9 \times 10$.
$$
\text { Also, } b^{3}>1987, b^{2}<1987 \text {, }
$$
Therefore, $12<... | x=5, y=9, z=11, b=19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,987 |
Example 2. A straight line intersects a hyperbola and its asymptotes, prove that the segments between the asymptotes and the hyperbola are equal.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Proof: Let the line intersect the hyperbola \( b^{2} x^{2} - a^{2} y^{2} = a^{2} b^{2} \) at points \( P \) and \( Q \), and intersect its asymptotes at points \( R \) and \( S \). The midpoint of \( P Q \) is \( M_{1}^{a}\left(x_{1}, y_{1}\right) \), and the midpoint of \( R \) and \( S \) is \( M_{2}\left(x_{2}, y_{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 703,988 |
Example 1. Find the equation of the curve symmetric to the curve $2 x^{2}-3 y^{2}-5 x=0$ with respect to the line $x+1=0$. | Solve: In the equation, substituting $-x-4$ for $x$ yields $2 x^{2}-3 y^{2}+21 x+52=0$. | 2 x^{2}-3 y^{2}+21 x+52=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,990 |
Example 2. Find the equation of the curve symmetric to the parabola $x^{2}-2 y+3=0$ with respect to $x-y-2=0$. | To solve: In the parabolic equation, by substituting $y+2$, $x-2$ for $x, y$ respectively, we get
$$
y^{2}-2 x+4 y+11=0 \text {. }
$$
(Author's affiliation: Sanhao Middle School, Luwan District, Shanghai) | y^{2}-2 x+4 y+11=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 703,991 |
Example 2. The sequence $\left\{a_{1}\right\} a_{1}=-\frac{1}{3}$, $a_{n}=\frac{3 a_{n-1}-1}{3-a_{n-1}}$, find its general term formula. | Solve $a_{n}=\frac{-\frac{1}{3}+a_{n-1}}{1-\frac{1}{3} \cdot a_{n-1}}=\frac{a_{1}+a_{n-1}}{1+a_{1} a_{n-1}}$. By property 3, we have
$$
\begin{aligned}
F\left(a_{n}\right) & =F\left(\frac{a_{1}+a_{n-1}}{1+a_{1} a_{n-1}}\right) \\
& =F\left(a_{1}\right) F\left(a_{n-1}\right) .
\end{aligned}
$$
$F\left(a_{1}\right)=\frac... | a_{n}=\frac{1-2^{n}}{1+2^{n}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 703,993 |
Example 5, find a four-digit number that can be divided by 11, and the quotient obtained is 10 times the sum of the digits of this four-digit number. | Let $\overline{a b c d}=110(a+b+c+d)$, then we have
$$
890 a=10 b+100 c+109 d \text {. }
$$
Also, since 11 divides $\overline{a b c d}$, we have $a-b+c-d= \pm 11$ or $a-b+c-d=0$.
If $a-b+c-d=0$,
then from $\left\{\begin{array}{l}a+c=b+d, \\ 890 a=10 b+100 c+109 d\end{array}\right.$
(2) $-10 \times(1)$ we get
$880 a=1... | 1980 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,996 |
Example 6. In the six-digit number $31 x y 13$ !, $x 、 y$ are both greater than 5, and this number is divisible by 13, find the four-digit number $\overline{1 x y 9}$. | $$
\begin{array}{l}
\text { Solution: Let } n=31 x y 13 \\
=3 \cdot 10^{5}+10^{4}+x \cdot 10^{3}+y \cdot 10^{2} \\
+13 \\
=310013+10^{3} x+10^{2} y, \\
\end{array}
$$
where $14,6 \leqslant x, y \leqslant 9$.
Since the remainders of $310013,10^{3}, 10^{2}$ when divided by 13 are $2,-1,-4$ respectively, we can set $3100... | 1989 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 703,997 |
Example 1. The nine-digit number in the form of $1234 \times \times \times \times \times$ composed of $1,2,3,4,5,6,7,8,9$, where the ten-thousands place is not 5, the thousands place is not 6, the hundreds place is not 7, the tens place is not 8, and the units place is not 9. How many such nine-digit numbers are there? | Solution: Obviously, this is a problem of five letters being placed in the wrong envelopes. And
$$
\begin{aligned}
\overline{5} & =5!-C_{5}^{1} 4!+C_{5}^{2} 3!-C_{5}^{3} 2! \\
& +C_{5}^{4} 1!-C_{5}^{5} 0! \\
& =120-120+60-20+5-1 \\
& =44 .
\end{aligned}
$$
The number of derangements it is asking for is 44. | 44 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 703,998 |
For example, the lateral edges of the pyramid $P-ABC$ form a $75^{\circ}$ angle with the base, the internal angles of $\triangle ABC$ are in the ratio $\angle A: \angle B: \angle C=1: 2: 9$, and $AC=3$. Find the height of the pyramid. | Let $P O \perp$ base $A B C$, with $O$ as the foot of the perpendicular, then $O$ is the circumcenter of $\triangle A B C$. Given that
$\angle A: \angle B: \angle C=1: 2: 9, \angle B=30^{\circ}$, $\angle C=135^{\circ}$, it is evident that $O$ is outside $\triangle A B C$, on $O C$, then $O C=\frac{3}{2 \sin 30^{\circ}}... | 3(2+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,000 |
In $\triangle A B C$, the three interior angles $A, B, C$ form a geometric sequence with a common ratio of $\frac{1}{2}$. Prove: $\frac{a}{a+b+c}=2 \sin \frac{\pi}{14}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | ```
Given $B=2 C, A=4 C$,
Prove that $A+B+C+2 C+C=7 C=$
$$
\begin{array}{l}
=\frac{\sin \frac{3 \pi}{7}}{2 \sin ^{3} \frac{\pi}{7}\left(\cos \frac{\pi}{7}+\cos \frac{3 \pi}{7}\right)} \\
=\frac{1}{4 \cos \frac{2 \pi}{7} \cos \frac{\pi}{7}} \\
=\frac{1}{\sin ^{4} \frac{\pi}{7}} \frac{\sin \frac{2 \pi}{7}}{2 \sin \frac{2... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,001 |
Example 4. Prove the Angle Bisector Theorem: The angle bisector of an interior angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides.
As shown in Figure 5, in
$$
\begin{array}{l}
\triangle A B C \text {, } \angle 1 \\
=\angle 2 .
\end{array}
$$
Prove:
$$
\frac{B D}{D C... | Prove $\because S_{S \triangle \triangle B D}=\frac{B D}{D C}$, (two triangles have the same height)
$S_{S \triangle \triangle B D}=\frac{\frac{1}{2} A B \cdot A D \sin \angle 1}{\frac{1}{2} A C \cdot A D \sin \angle 2}=\frac{A B}{A C}$.
$\therefore \frac{B D}{D C}=\frac{A B}{A C}$. (substitution of equal quantities)
T... | \frac{B D}{D C}=\frac{A B}{A C} | Geometry | proof | Yes | Yes | cn_contest | false | 704,002 |
Example 9. (i) Let three positive real numbers $a, b, c$ satisfy
$$
\left(a^{2}+b^{2}+c^{2}\right)^{2}>2\left(a^{2}+b^{4}+c^{4}\right) \text {. }
$$
Prove: $a, b, c$ must be the three side lengths of some triangle.
(ii) Let $n$ positive numbers $a_{1}, a_{2}, \cdots ; a_{n}$ satisfy
$$
\begin{array}{l}
\left.\therefor... | Proof one: Expanding and organizing (1) by $a_{\mathrm{n}}^{2}$, we obtain the equivalent inequality
$$
\begin{array}{l}
(n-2) a_{n}^{4}-2\left(a_{1}^{2}+\cdots+a_{\mathrm{n}-1}^{2}\right) a_{\mathrm{n}}^{2} \\
+ {\left[(n-1)\left(a_{1}^{4}+\cdots+a_{\mathrm{n}-1}^{4}\right)-\left(a_{1}^{2}+\cdots\right.\right.} \\
\le... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,004 |
Theorem 1 If $a>0$ and $m>0$, then the polynomial $F$ takes its minimum value when and only when $y=\sqrt{\frac{m}{3 a}}$, and the minimum value is
$$
F \text{ minimum }=f\left(\sqrt{\frac{m}{3 a}}\right)=n-\frac{2 m \sqrt{3 a m}}{9 a},
$$
and the polynomial $F$ takes its maximum value when and only when $y=-\sqrt{\fr... | It is easy to verify that (II) can be rewritten as
$$
\begin{aligned}
F=f(y)=a\left(y-\sqrt{\frac{m}{3 a}}\right)^{2}\left(y+2 \cdot \sqrt{\frac{m}{3 a}}\right) \\
+n-\frac{2 m \sqrt{3 a m}}{9 a}
\end{aligned}
$$
From this, it can be concluded that when \( y \) is slightly greater than or slightly less than \( \sqrt{\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,005 |
Theorem 1 The function $y=f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is defined when $x_{1} \in(a, b), i=1,2, \cdots ; n$ and
$x_{i}+x_{2}+\cdots+x_{n}=n c$, for any $x_{1}, x_{j}(i \neq j)$, fixing the others $x_{1}(k \neq i, j), y$ decreases (increases) as $\left|x_{1}-x_{j}\right|$ decreases. Then, the function $y=f... | Obviously, \( c \in (a, b) \). If \( x_{1}^{n} ; x_{2}^{-}, \cdots \), \( x_{\mathrm{n}} \) are not all equal. Without loss of generality, assume \( x_{1} f\left(c, x_{2}^{\prime} ; x_{3}, \cdots ; x_{n}\right) \).
If \( x_{2}^{\prime}, x_{3}^{-}, \cdots, x_{0} \) are not all equal, for example, \( x_{2}^{\prime}, x_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,006 |
Example 1. Let $F=x^{3}-3 x^{2}-9 x+5$, find the extremum of $F$. | Here $a=1, b=-3$.
Let $x=y-(b / 3a)=y+1$, then
$$
F=f(y)=y^{3}-12 y-6 .
$$
Here $a=1, m=12$. Thus, by Theorem 1, we have
$$
\begin{array}{l}
F_{\text {min }}=f(\sqrt{m} / \sqrt{3a})=f(2)=-22 ; \\
F_{\text {max }}=f(-\sqrt{m / 3a})=f(-2)=10 .
\end{array}
$$ | F_{\text {min }} = -22, F_{\text {max }} = 10 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 704,007 |
Example 2. Let $F=-2 x^{3}+3 x^{2}+12 x-1$, find the extremum of $F$.
| Solve: Here $a=-2, b=3$.
Let $x=y-(b / 3a)=y+(1 / 2)$, then
$$
F=f(y)=-2 y^{3}+\frac{27}{2} y+\frac{11}{2} \text{. }
$$
Here $a=-2, m=-27 / 2$. Thus, by Theorem 2, we know
$F$ maximum $=f(\sqrt{m} / 3a)=f(3 / 2)=19$;
$F_{\text{minimum}}=f(-\sqrt{m / 3a})=f(-3 / 2)=-8$. | F_{\text{maximum}}=19, F_{\text{minimum}}=-8 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 704,008 |
Example 1. Using the three vertices of a triangle and several points inside it as vertices, the number of small triangles into which the original triangle can be divided is ( .
(A)15; (B)19; (C) 22; (D) cannot be determined・ (88 Jiangsu Province Junior High School Mathematics Competition Question) | Solution 1: To find the number of small triangles after the division, we first calculate the sum of all the interior angles of these small triangles. Since each of the $n$ interior points of the original triangle is a common vertex of several of the divided small triangles, the sum of the interior angles of these small... | 2n + 1 | Geometry | MCQ | Yes | Yes | cn_contest | false | 704,011 |
Inversion 6 ・Butterfly Theorem・As shown in Figure 7; if $EF$ is a chord of $\odot O$, $M$ is the midpoint of $EF$, $AB, CD$ are other chords passing through $M$, $AC$ and $BD$ intersect $EF$ at $P, Q$ respectively. Prove: $M$ is the midpoint of $PQ$. | Prove as shown in the figure:
$$
\angle 1 = \angle 3,
$$
$$
\begin{array}{l}
\angle 2 = \angle 4, \\
\angle 5 = \angle 6, \\
\angle 7 = \angle 8 .
\end{array}
$$
Let the areas of $\triangle C M P$, $\triangle A M P$, $\triangle D M O$, and $\triangle B M Q$ be denoted as $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$, respect... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,014 |
For example. Given that all edges of a parallelepiped are equal, and its four diagonals are $a$, $b$, $c$, and $d$, find the edge length of this parallelepiped. | Let the edge length of the parallelepiped be \( x \). According to the above theorem, we have:
\[
\begin{aligned}
& 4 \cdot 3 x^{2}=a^{2}+b^{2}+c^{2}+d^{2} \\
\therefore x & =\sqrt{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{12}} \\
& =\frac{\sqrt{3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)},}{6}
\end{aligned}
\]
Thus, the edge length o... | \frac{\sqrt{3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,015 |
Example 2. Given $x \in(3 \pi ; 3.5 \pi)$, find the angle $x$ in the following expressions:
(1) $\operatorname{tg} x=4$,
(2) $\operatorname{ctg} x=\sqrt{5}$. | (1) $\because \operatorname{arctg} 4$ is an angle in the I quadrant, and $x$ is an angle in the III quadrant, so the terminal sides of angle $x$ and $\operatorname{arctg} 4$ are symmetric with respect to the origin. Thus,
$$
\begin{array}{c}
x-\operatorname{arctg} 4=3 \pi, \\
\therefore \quad x=3 \pi+\operatorname{arct... | x=3 \pi+\operatorname{arctg} 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,017 |
Example 1. If for $m, n \in N, f(m+n)=\boldsymbol{f}(m) +f(n)+m n$, and $f(1)=1$, find the expression for $f(n)$.
| Solve $f(m+1)=f(m)+m+1$.
Let $m=1,2, \cdots, n-1$, we have
$f(2)=f(1)+2$,
$$
f(3)=f(2)+3, \cdots, f(n)=f(n-1)+n .
$$
Adding both sides respectively, simplifying yields
$$
f(n)=f(1)+2+3+\cdots+n=\frac{1}{2} n(n+1) \text {. }
$$ | \frac{1}{2} n(n+1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,018 |
Example 3. The ends of a line segment $AB$ are dyed in red and blue, respectively. Insert $n$ points into the line segment, and randomly mark each point with red or blue, thus dividing the original line segment $AB$ into non-overlapping smaller segments. Prove that the number of segments with different colors at their ... | Solve the problem by assigning values to points: assign +1 to red points, and -1 to blue points. Establish a coordinate system with $AB$ as the $x$-axis. Then, consider the assigned values as the $y$-coordinates of the points and connect adjacent points with line segments (readers are encouraged to draw the figure them... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,020 |
On a circle with radius 1, given two point sets $A$ and $B$, both consisting of a finite number of disjoint arcs, where each arc in $B$ has a length of $\frac{\pi}{m}$, and $m$ is a natural number. Let $A^{\mathrm{J}}$ denote the set obtained by rotating set $A$ counterclockwise on the circle by $\frac{j \pi}{m}$ radia... | Proof: Let $B=\bigcup_{\mathrm{j}=1}^{0} B_{1}$,
where $\bar{i}_{1}$ and $B_{1}=\varnothing(i \neq j)$ and $i\left(B_{1}\right)=\frac{\pi}{m}$, $j=1, \cdots$, ... Denote the set obtained by rotating $B$ clockwise by $\frac{k \pi}{m}$ radians as $B_{1}^{-k}$, and denote
$$
B^{-k}=\bigcup_{j=1}^{\mathrm{P}} B_{j}^{-\mat... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,021 |
Four, Let points $D, E, F$ be on the sides $BC, CA, AB$ of $\triangle ABC$, respectively, and the incircles of $\triangle AEF, \triangle BFD, \triangle CDE$ have equal radii $r$. Let $r_{0}$ and $R$ be the radii of the incircles of $\triangle DEF$ and $\triangle ABC$, respectively. Prove that $r + r_{0} = R$.
---
Thi... | Prove: Draw auxiliary lines as shown in the figure. Then we have
$$
S_{\mathrm{ABC}}-S_{\mathrm{BDF}}-S_{\mathrm{CED}}-S_{\mathrm{A}^{P} E}=S_{\mathrm{DE}} \text{. (1)}
$$
Let the perimeters of $\triangle A B C$ and $\triangle D E F$ be denoted by $l$ and $l^{\prime}$, respectively. Then (1) can be rewritten as
$$
\be... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,024 |
In space, there are 1989 points, where no three points are collinear. Divide them into 30 groups with different numbers of points. For any three different groups, take one point from each to form a triangle. To maximize the total number of such triangles, how many points should each group have? | When dividing these 1989 points into 30 groups, the number of points in each group is denoted as $n_{1}n_{1}^{n} n_{2}$. When $n_{1}^{\prime}, n_{2}^{\prime}$ replace $n_{1} n_{2}$ while $n_{3}, \cdots, n_{3}$ remain unchanged, the value of $S$ increases, leading to a contradiction. $2^{\circ}$ The number of $i$ values... | 51,52, \cdots, 56,58,59, \cdots, 82 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,026 |
1. Prove: $\quad(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$ has a minimum value of 8 on $0<u<\sqrt{2}, v>0$. | $(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$ can be seen as the square of the distance between point $P_{1}(u, \sqrt{2-u^{2}})$ and $P_{2}\left(v, \frac{9}{v}\right)$.
$P_{1}$ satisfies the equation of the circle:
$$
x^{2}+y^{2}=2 \text{. }
$$
$P_{2}$ satisfies the equation of the hyperbola:
$$
x y=9, x>0 \t... | 8 | Algebra | proof | Yes | Yes | cn_contest | false | 704,028 |
1. In triangles with three sides of unequal natural numbers, the number of triangles where the longest side is exactly $n$ is 600. What is the value of $n$? $\qquad$ . | Let the lengths of the three sides of a triangle be $x, r, n$, and $x < y < n$, where $x, y, n$ are all natural numbers.
Obviously, the length of the shortest side $x$ satisfies the relation: $2 \leqslant x \leqslant n-2$. Now, fix $x$ to find the number of triangles formed (assuming $n$ is odd).
\begin{tabular}{llc}
... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,031 |
2. Let $\alpha, \beta$ be acute angles. When
$$
-\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta}
$$
takes the minimum value, the value of $\operatorname{tg}^{2} \alpha+\operatorname{tg}^{2} \beta$ is | To find the minimum value of the original expression, we need
$$
\sin ^{2} \beta \cos ^{2} \beta=\frac{1}{4} \sin ^{2} 2 \beta
$$
to take the maximum value. Since $\beta$ is an acute angle, we know $\beta=\frac{\pi}{4}$.
When $\beta=\frac{\pi}{4}$,
$$
\begin{aligned}
\text { the original expression } & \geqslant \frac... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,032 |
3. Let [x] denote the greatest integer not exceeding $x$, then the sum of the squares of all real roots of the equation $x^{2}-5[x]+4=0$ is
保留源文本的换行和格式，直接输出翻译结果。 | Solve: The intersection points of the graphs of the functions $y=x^{2}$ and $y=5[x]-4$ (see the figure below) are the solutions to the original equation.
From $5[x]=x^{2}+4>0$, we know that $[x]>0$.
Therefore, we only
need to consider the case where $x \geqslant 1$.
Since $x \geqslant[x]$,
substituting into t... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,033 |
4. In space, there are 5 points, the distance between any two points is different, and any 4 points can form three closed broken lines. For each closed broken line formed, the shortest segment is painted red. $A$ is one of the 5 points.
Let the number of red segments starting from point $A$ be $m$;
When no segments sta... | Solve step by step for the maximum value of \( m \):
Consider 5 points \( A, B, C, D, E \). The line segments \( AB, AC, AD, AE \) can all be colored red. Without loss of generality, assume \( AB < AC < AD < AE \). If \( AE \) is also colored red, then it must be the shortest side of some closed broken line, but any cl... | 18 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,034 |
5. A regular triangle with a side length of 2 is divided into four smaller triangles by its median lines. Except for the central small triangle, three other small triangles are used as bases to construct three regular tetrahedrons of height 2 on the same side. There is a sphere that is tangent to the plane of the origi... | Let the center of the sphere be point $O$, and the radius be $R$. By symmetry, the point where the sphere $O$ touches the base $\triangle ABC$ is the centroid of $\triangle ABC$.
For the regular tetrahedron $O_{1}-AEF$, the foot of the perpendicular from the apex to the base, $M$, is the centroid of $\triangle AEF$. P... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,035 |
Example 7. In $\triangle A B C$, $2 b=a+c$.
Prove: $a c=6 R r$.
The text has been translated while preserving the original line breaks and format. | Prove that from $\triangle=\frac{a b c}{4 R}$, we get
$$
\begin{aligned}
\therefore=r s & =r \cdot \frac{1}{2}(a+b+c)=r \cdot \frac{3 b}{2}, \\
\therefore \frac{a b c}{4 R} & =\frac{3 b r}{2} . \\
a c & =6 Rr,
\end{aligned}
$$
From the above examples, we can see that appropriately using the area transformation of a tr... | a c = 6 R r | Geometry | proof | Yes | Yes | cn_contest | false | 704,036 |
6. A bag of peanuts has a total of 1988 peanuts. On the first day, a monkey takes one peanut. Starting from the second day, the number of peanuts taken each day is the sum of all the peanuts taken in previous days. If on a certain day the number of peanuts left in the bag is less than the total number of peanuts alread... | Let's assume the monkey used $k$ days in the first round, meaning that on the $k+1$-th day, it started again by taking one peanut. Therefore, the total number of peanuts taken in the first $k$ days is
$$
1+1+2+2^{2}+2^{3}+\cdots+2^{k-1}=?^{k-1}
$$
peanuts, and $1988-2^{k-1}<2-1$. Also, $1988-2^{k-1}-2^{k-1}<2^{t-1}, t... | 48 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,037 |
7. The general term of the sequence is $a_{\mathrm{n}}=b[\sqrt{n+c}]+d$, and the terms are calculated successively as
$$
1,3,3,3,5,5,5,5,5, \cdots \text {. }
$$
where each positive odd number $m$ appears exactly $m$ times consecutively. The above $b, c, d$ are undetermined integers. Then, the value of $b+c+d$ is $\qqu... | Solve: First determine $b$.
Since $a_{n}$ is odd, and $a_{n}+1 \geqslant a_{n}$, we know that $a_{n}+1-a_{n}$ $\in\{0,2\}$,
i.e., $b[\sqrt{n+1+c}]-b[\sqrt{n+c}]$ is 0 or 2.
For any natural number $n$, it always holds that
$[\sqrt{n+1+c}]-[\sqrt{n+c}] \in\{0,1\}$.
Clearly, $b \neq 0$.
When $-[\sqrt{n+c}])=2$, by (1), it... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,038 |
8. For a square field with side length 1 (including all four sides), if 3 points are placed arbitrarily, the minimum value of $a$ such that the distance between at least two points does not exceed $a$ is
| Solve: According to the problem, the minimum value of $a$ is actually the maximum of the shortest sides of triangles within a unit square. For this, we only need to examine the maximum triangle with the shortest side in the square $ABCD$.
Assume $\triangle AEF$ is the triangle that meets the requirement, with $A$ bein... | \sqrt{6}-\sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,039 |
9. $f(n)$ is a function defined on the set of natural numbers, when $p$ is a prime number, $f(p)=1$, and for any natural numbers $r$, $s$, we have
$$
f(r s)=r f(s)+s f(r) .
$$
Then, the sum of all $n$ that satisfy the condition
$$
f(n)=n, 1 \leqslant n \leqslant 10^{4}
$$
is. | Given the problem, for any natural numbers $r, s$, we have
$$
\frac{f(r s)}{r s}=\frac{f(r)}{r}+\frac{f(s)}{s} \text {. }
$$
Thus,
$$
\begin{array}{l}
\frac{f\left(p^{m}\right)}{p^{m}}=\frac{f(p)}{p}+\frac{f(p)}{p}+\cdots+\frac{f(p)}{p} \\
\quad=m \cdot \frac{f(p)}{p} .
\end{array}
$$
When $p$ is a prime number, we g... | 3156 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,040 |
10. Tank A contains 4 liters of liquid A, Tank B contains 2 kilograms of liquid B, and Tank C contains 2 kilograms of liquid C. These liquids can all be mixed. First, 1 liter of liquid from Tank A is poured into Tank B, and then 1 liter of the mixed liquid from Tank B is poured into Tank C. Finally, 1 liter of the mixe... | Let $a_{n}, b_{n}, c_{n}$ be the amounts of liquid A in containers 甲, 乙, and 丙, respectively, after $n$ mixings. Then,
$$
a_{n}+b_{n}+c_{n}=4 \text{. }
$$
By symmetry, we have
$$
b_{n}=c_{n} \text{. }
$$
Now, let's examine the amount of liquid A in container 甲 after the $(n+1)$-th mixing. According to the problem,
$$... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,041 |
Example 1. Let positive numbers $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ satisfy $a_{1}+a_{2}+\cdots+a_{\mathrm{n}}=1$. Find
$$
\begin{array}{l}
\frac{a_{1}}{1+a_{2}+\cdots+a_{\mathrm{n}}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}} \\
+\cdots+\frac{a_{1}}{1+a_{1}+\cdots+a_{\mathrm{n}-1}}
\end{array}
$$
the minimum value. ... | Let the sum of the original expression be $S$, then
$$
S=\sum_{k=1}^{n} \frac{a_{k}}{2-a_{k}}=2 \sum_{k=1}^{n} \frac{1}{2-a_{k}}-n .(1)
$$
Since $2-a_{k}>0(k=1,2 ; \cdots ; n)$, by the Cauchy-Schwarz inequality (corollary), we have
$$
\sum_{k=1}^{n} \frac{1}{2-a_{k}} \geqslant \frac{n^{2}}{\sum_{k=1}^{n}\left(2-a_{k}\... | \frac{n}{2 n-1} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 704,042 |
Example 3. On the lower base $AB$ of trapezoid $ABCD$, there are two fixed points $M, N$, and on the upper base $CD$, there is a moving point $P$. Let $E = DN \cap AP, F = DN \cap MC, G = MC \cap PB$, and $DP = \lambda DC$. For what value of $\lambda$ is the area of quadrilateral $PEFG$ maximized? (88 China Training Te... | Solve: Because
$$
S_{\mathrm{PE}}{ }^{\mathbb{F G}}=S_{\mathrm{ABP}}-S_{\mathrm{ANE}}-S_{\mathrm{MBG}}+S_{\mathrm{MNE}},
$$
and among them, $S_{A B P}$ and $S_{\mathrm{ANF}}$ are constants, so $S_{\mathrm{PEFG}}$ is maximized if and only if $S_{\mathrm{ANE}}+S_{\mathrm{MBG}}$ takes the minimum value.
Let $A B=a, C D=... | \lambda=\mu | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,043 |
Example 7. Find all real numbers $a$ such that there exist non-negative real numbers $x_{\mathfrak{k}}, k=1,2,3,4,5$, satisfying the relations
$$
\begin{array}{c}
\sum_{\mathrm{k}=1}^{5} k x_{\mathrm{k}}=a, \quad \sum_{\mathrm{k}=1}^{5} k^{3} x_{k}=a^{2}, \\
\sum_{\mathrm{k}=1}^{5} k^{5} x_{\mathrm{k}}=a^{3} .
\end{arr... | If there exists a number $a$ such that the above three equations hold for the corresponding $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, then we have
$$
\left(\sum_{k=1}^{5} k^{3} x_{k}\right)^{2}=\left(\sum_{k=1}^{5} k x_{k}\right)\left(\sum_{k=1}^{5} k^{5} x_{k}\right) \text {. }
$$
On the other hand, by the Cauchy-Schwarz ... | 0,1,4,9,16,25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,046 |
Example 9. Ptolemy's Theorem: Prove that in a cyclic quadrilateral, the sum of the products of the two pairs of opposite sides is equal to the product of the diagonals. | Proof As shown in Figure 9, let $A B=a, B C=b$, $C D=c, D A=d, B D=e, A C=f$.
According to the cosine rule:
In $\triangle D A B$, we have
$$
e^{2}=a^{2}+d^{2}-2 a d \cos \angle D A B ;
$$
In $\triangle D C B$, we have
$$
\begin{array}{r}
c^{2}=c^{2}+c^{2} \\
-2 b c \cos \angle D C B \\
\text { Also, } \angle D A B \\
... | e f=a c+b d | Geometry | proof | Yes | Yes | cn_contest | false | 704,048 |
Example 2. Write three integers on the blackboard, then erase one of them and replace it with the difference of 1 and the sum of the other two numbers. After repeating this process several times, the result is
(17, 1967, 1983).
Could the three numbers initially written on the blackboard have been
(1.) $(2,2,2)_{3}$
(2... | Given the problem, if in a certain operation the set of numbers is
$$
(a, b, c), 0<a<b<c \text{, }
$$
then, $c=a+b-1$.
After removing $c$, we have
$$
b=a+d-1 \text{. }
$$
Substituting (2) into (1) gives
$$
c=d+2a-2=d+2(a-1) \text{. }
$$
After some operations, it must be that
$$
\begin{array}{l}
c=p+k(a-1) . \quad(p<... | (3,3,3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,049 |
Example 3. Write the numbers $1, 2, 3, \cdots$, 1986, 1987 on the blackboard. At each step, determine some numbers from those written and erase them, replacing them with the remainder of their sum divided by 7. After several steps, two numbers remain on the blackboard, one of which is 987. What is the second remaining ... | Solution: Clearly, at each step, the sum of all the numbers written down modulo 7 is preserved.
Let the remaining number be $x$, then $x+987$ is congruent to $1+2+\cdots+1987$ modulo 7.
Since $1+2+\cdots+1987=1987 \times 7 \times 142$ is divisible by 7, the remainder is 0, so $x+987$ is also divisible by 7. Since 987... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,050 |
Casually write a decimal natural number (such as 2583), then find the sum of the squares of each digit of this number $\left(2^{2}+\right.$ $\left.5^{2}+8^{2}+3^{2}=102\right)$, for the resulting number (102), apply this method again $\left(1^{2}+0^{2}+2^{2}=5\right)$, and continue to do so $\left(5^{2}=25,2^{2}+5^{2}=... | $$
\begin{array}{c}
\text { Proof: Let } A=\overline{a_{n} a_{n-1} \cdots a_{2} a_{1}}, \text { then } \\
A_{1}=a_{n}^{2}+a_{n-1}^{2}+\cdots+a_{2}^{2}+a_{1}^{2}, \\
A-A_{1}=\left(10^{n-1}-a_{n}\right) a_{n}+\left(10^{n-2}-a_{n-1}\right) \\
\cdot a_{n-1}+\cdots+\left(10-a_{2}\right) a_{2}+\left(1-a_{1}\right) a_{1} \\
\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,051 |
Example 1. Prove by analytical method that the sum of the distances from a point to the three sides of a triangle equals the height of the triangle. | Proof: Let the side length of a regular triangle be $a (a>0)$, and establish a Cartesian coordinate system (as shown in Figure 1) such that the coordinates of the vertices are $O(0,0)$, $A\left(\frac{a}{2}, \frac{\sqrt{3}}{2} a\right)$, and $B\left(-\frac{a}{2}\right.$, $\left.\frac{\sqrt{3}}{2} a\right)$. The equation... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,052 |
Example 2. On a line segment $AB$ of length 1, take two points $C, D$ in the order of $A, B, C, D$. Let $AC=x$, $CD=y$.
(1) Draw a diagram to represent the range of point $P(x, y)$.
(2) If the segments $AC, CD, DB$ can form the sides of a triangle, draw a diagram to represent the range of point $P(x, y)$. | $$
\begin{array}{l}
\text { (1) A, B, C, and D are four distinct points arranged in sequence, } x>0, y>1, x+y 1 - ( x + y ) , } \\
{ x + 1 - ( x + y ) > y , } \\
{ y + 1 - ( x + y ) > x , }
\end{array} \text { that is } \left\{\begin{array}{l}
x+y>\frac{1}{2}, \\
0<x<\frac{1}{2} \\
0<y<\frac{1}{2} .
\end{array}\right.\... | 0<x<\frac{1}{2}, 0<y<\frac{1}{2}, x+y>\frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,053 |
Example 3. Find the maximum and minimum values of $x+3y$ under the conditions $x \leqslant 2, y \leqslant 2, x+y>2$.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given the solution region of the system of inequalities is the triangle $\triangle A B^{B} C$ with vertices $A(0,2), B(2,0), C(2,2)$ (Figure 3). Let $x+3 y=k$, the graph of which is a set of parallel lines with a slope of $-\frac{1}{3}$ and a y-intercept of $\frac{k}{3}$. It is evident that $k$ takes its maximum or min... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,054 |
Example 5 ・At
7:00 AM, depart from port $A$,
take a motorboat at a constant speed of $v_{1}$ kilometers/hour (4
$\leqslant v_{1} \leqslant 20$) to
head to port $B$ 50 kilometers
away from port $A$, then immediately switch to a steamship at a constant speed of $v_{2}$ kilometers/hour $\left(30 \leqslant v_{2} \leqslant ... | (1) From the given conditions, we have
\[
\left\{
\begin{array}{l}
x=\frac{300}{v_{2}}, \quad 30 \leqslant v_{2} \leqslant 100, \\
y=\frac{50}{v_{1}}, \quad 4 \leqslant v_{1} \leqslant 20,
\end{array}
\right.
\]
and
\[
9 \leqslant x+y \leqslant 14,
\]
which implies
\[
\left\{
\begin{array}{l}
3 \leqslant x \leqslant 10... | 930 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,056 |
Example 1. The radii of two concentric circles are $R, r$. Prove: The sum of the squares of the distances from any point on one circumference to the vertices of a regular $n$-sided polygon inscribed in the other circle is a constant. | To prove that in a regular polygon $A_{1} A_{2} \cdots A_{n}$ inscribed with its center at the origin, without loss of generality, let $A_{n}$ be on the real axis, and the vertices $A_{1}, A_{2}, \cdots, A_{n}$ are represented by the complex numbers $z_{1}=r e^{\frac{13 \pi}{n}}$, $z_{2}^{-}=r e^{\frac{i \frac{x}{n}}{n... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,057 |
Example 2. Prove that on $\odot O$ there exists a point whose distance squared to a fixed point on the plane is equal to the arithmetic mean of the distance squared from that fixed point to the vertices of a regular $n$-sided polygon inscribed in the circle.
Translate the above text into English, please keep the origi... | Proof: Let the distance from point $P$ to the center $O$ of the circle be $a$, i.e., $|PO| = a$, and the radius of $\odot O$ be $R$. From the result of Example 1, the sum of the squares of the distances from $P$ to the vertices of the inscribed regular $n$-sided polygon in $\odot O$ is $n\left(R^{2}+a^{2}\right)$. It i... | null | Geometry | proof | Yes | Yes | cn_contest | false | 704,058 |
Example 10. As shown in Figure 10, given that $P$ is any point on the arc $\widehat{B C}$ of the circumcircle of equilateral triangle $A B C$. Prove: $P A = P B + P C$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 10. As shown in Figure 10, given that $P$ is any point on the arc $\widehat{B C}$ of the circumcircle of equilateral ... | ```
Prove that according to Ptolemy's theorem:
\[
\begin{array}{c}
P A \cdot B C \\
=A C \cdot B P + A B \cdot P C . \\
\text { But } B C = A C = A B, \therefore P A = P B + P C .
\end{array}
\]
``` | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,059 |
Example 18. Find the sum: $1 \times 2+2 \times 3+3 \times 4$ $+\cdots+n(n+1)$ | \[
\begin{array}{l}
\text { Solution } \because n(n+1)=2 \times \frac{n(n+1)}{1 \cdot 2} \\
=2 C_{n+1}^{2}=2\left[C_{n+2}^{2}-C_{n+1}^{3}\right], \\
\begin{aligned}
\therefore \text { Original expression } & =2\left[C_{2}^{2}+C_{3}^{2}+C_{1}^{2}+\cdots+C_{1+1}^{2}\right] \\
= & 2\left[1+C_{4}^{3}-C_{3}^{3}+C_{5}^{3}-C_... | \frac{n(n+1)(n+2)}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,064 |
Example 20. Find the sum: $1^{2}+2^{2}+3^{2}+\cdots+n^{2}$. | \begin{aligned} \text { Sol } & \because n^{2}=2 \cdot \frac{n(n-1)}{2}+n \\ & =2 C_{2}^{2}+n=2\left(C_{n+1}^{3}-C_{n}^{3}\right)+n, \\ \therefore & \text { Original expression }=1^{3}+2^{2}+2\left[\left(C_{4}^{3}-C_{3}^{3}\right)\right. \\ + & \left.\left(C_{5}^{3}-C_{4}^{3}\right)+\cdots+\left(C_{n+1}^{3}-C_{n}^{3}\r... | \frac{n(n+1)(2n+1)}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,066 |
則22. Prove: $1^{2}+2^{2}+3^{2}+\cdots+n^{2}$
$$
=\frac{n(n+1)(2 n+1)}{6} \text {. }
$$ | $\begin{array}{l}\text { Proof: Let } a_{1}^{*}=1^{2}, a_{2}=2^{2}, a_{3}=3^{2}, \cdots \\ a_{n}=n^{2}, \frac{n(n+1)(2 n+1)}{6}=S_{n} \\ \because S_{n}-S_{n-1}=\frac{n(n+1)(2 n+1)}{6} \\ -\frac{(n-1) n(2 n-1)}{6} \\ \quad=n^{2}=a_{n}\end{array}$
$\begin{array}{l}\text { Therefore, } S_{2}-S_{1}=a_{2}, S_{3}-S_{2}=a_{3}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,068 |
Example: Given the radii of the upper and lower bases of a frustum are 3 and 6, respectively, and the height is $3 \sqrt{3}$, the radii $O A$ and $O B$ of the lower base are perpendicular, and $C$ is a point on the generatrix $B B^{\prime}$ such that $B^{\prime} C: C B$ $=1: 2$. Find the shortest distance between point... | Given $B^{\prime} D \perp O B$, then $B^{\prime} D=3 \sqrt{3}$.
$$
\begin{array}{l}
\because O^{\prime} B^{\prime}=3, O B=6, \\
\therefore B^{\prime} B=\sqrt{(3 \sqrt{3})^{2}+(6-3)^{2}}=6,
\end{array}
$$
and $O^{\prime} B^{\prime}: O B=3: 6=1: 2$,
$\therefore$ In the development diagram, $B^{\prime}$ is the midpoint o... | 4 \sqrt{13-6 \sqrt{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,069 |
Example 11. As shown in Figure 11, $ABCD$ is a square inscribed in a circle. Take any point $P$ on $\widehat{AB}$, prove:
$$
\begin{aligned}
& P D^{2}-P B^{2} \\
= & 2 P A \cdot P C .
\end{aligned}
$$ | Prove that by connecting $A C$, if the side length of the square is 1, then
$$
A C=\sqrt{2} \text {. }
$$
In the cyclic quadrilateral $A P B C$,
$$
\begin{array}{l}
P C \cdot A B=A P=B C+P B \cdot A C, \\
\text { i.e., } P C=4 D+v^{\prime} D B \text {, } \\
\sqrt{2} F B=P C-A P \text {. } \\
\text { In the cyclic quad... | P D^{2}-P B^{2}=2 P A \cdot P C | Geometry | proof | Yes | Yes | cn_contest | false | 704,070 |
Example 1. What day of the week was October 1, 1949? | $$
\begin{array}{l}
\text { Solve } C=19, Y=49, m=8, d=1 \\
W \equiv 1+20+49+12+4-38(\bmod 7) \\
\equiv 6(\bmod 7) .
\end{array}
$$
That is, October 1, 1949 is a Saturday. | Saturday | Other | math-word-problem | Yes | Yes | cn_contest | false | 704,071 |
Example 2. What day of the week was January 1, 2000? | $$
\begin{array}{c}
\text { Sol } C=19, Y=99 . m=11 . d=1 \text {. } \\
W \equiv 1+28+1+3+4-38(\bmod 7) \\
\equiv 6(\bmod 7) .
\end{array}
$$
So the first day of the next century is Saturday.
$$ | Saturday | Other | math-word-problem | Yes | Yes | cn_contest | false | 704,072 |
5. Use toothpicks of equal length to form the rectangular pattern shown in the figure. If the length of the rectangle is 20 toothpicks long and the width is 10 toothpicks long, then the number of toothpicks used is $\qquad$
(A) 30,
(B) 200,
(C) 410,
(D) 420,
(E) 430. | E
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 | 704,077 |
Example 12. If $a, b, c$ are all real numbers greater than zero, prove that: $\sqrt{a^{2}+b^{2}}+\sqrt{b^{2}+c^{2}}$ $+\sqrt{c^{2}+a^{2}} \geqslant \sqrt{2}(a+b+c)$, and discuss the conditions under which equality holds. | Prove as shown in Figure 12, construct a square with side length $a+b+c$.
On $AB$, take $a, b, c$ in sequence; on $AD$, take $b, c, a$ in sequence.
Draw perpendiculars to $AB$ and $AD$ at the division points, resulting in nine rectangles. The positions of $E, F$ are as shown in the figure.
Connect $AE, EF, FC$, then ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,081 |
9. Mr. and Mrs. Zeta want to name their child, Little Zeta. If the first letters of Little Zeta's given name, middle name, and last name must be in alphabetical order and the letters cannot be repeated, how many possibilities are there? $\qquad$
(A)276, (B) 300, (C)552, (D) 600,
(E) 15600 | $B$ | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 704,082 |
10. For the sequence $\left\{u_{n}\right\}, \boldsymbol{v}_{1}=a$ (where $a$ is any positive number), $u_{n+1}=-\frac{1}{u_{n}+1}, n=1,2,3, \cdots$, which of the following values of $n$ can make $u_{n}=a$?
(A) 14, (B)15, (C)16, (D) 17
(E) 18 | C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 704,083 |
17. The perimeter of an equilateral triangle is $1989 \mathrm{~cm}$ longer than the perimeter of a square. The side length of the triangle is $d \mathrm{~cm}$ longer than the side length of the square, and the perimeter of the square is greater than 0. The number of positive integer values that $d$ cannot take is $\qqu... | $D$ | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 704,090 |
Example 13. A regular octagon inscribed in a circle has four adjacent sides of length 2, and the other four sides of length 3. Find the area of this octagon.
| Solve as shown in Figure 13,
Let $AB = BC = CD = DE = 3$,
$EF = FG = GH = HA = 2$. Connect
$O$ to each point. Clearly, the area of the octagon (denoted as $S$) is the sum of the areas of the 8 triangles, i.e., four times the sum of the areas of $\triangle OHA$ and $\triangle OAB$. That is,
$$
\begin{array}{l}
S = 4\lef... | 13 + 12 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,092 |
25. In a cross-country competition, each side has 5 members. If a member finishes in the $n$th place, they score $n$ points for their team, and the team with fewer points wins. If there is no relationship between the members, then the number of possible winning scores is $\qquad$ (A) 10, , (B) $13,(C) 27,(D) 120,(E) 12... | $B$ | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 704,099 |
27. $n$ is a positive integer. The equation $2 x+2 y+z=n$ has 28 sets of positive integer solutions for $x, y, z$, then $n$ is $\qquad$
(A)14 or 15,
(B)15 or 16,
(C) 16 or 17,
(D)17 or 18,
(E) 18 or 19. | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 704,101 |
Example 14. As shown in Figure 14; $P$ is a point inside an equilateral triangle $ABC$, and $PA=3, PB=4, PC=5$. Find the side length of this equilateral triangle. | Prove that outside $\triangle ABC$ there is a point $D$, such that $AD=AP, CD=CP$.
It is easy to know that $\triangle APB \cong \triangle ADC$.
$\therefore \angle 1=\angle 2$.
Therefore, $\angle PAD=60^{\circ}$.
$\therefore \triangle APD$ is an equilateral triangle.
Also, $\because PD=3, DC=4, PC=5$,
$\therefore \tria... | \sqrt{25+12 \sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,103 |
30. There are 7 boys and 13 girls standing in a row. Let $S$ be the number of positions where a boy and a girl are adjacent. For example, in the arrangement $G B B G G G B G B G G G B G B G G B G G$, $S=12$. What is the average value of $S$ (considering all 20 positions)? $\qquad$
(A)9, (B)10, (C)11, (D)12, (E)13. | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 704,105 |
1. Calculate $\sqrt{(31)(30)(29)(28)+1}$. | \begin{aligned} \text { 1. } & \because(k+1) \cdot k \cdot(k-1) \cdot(k-2)+1 \\ & =[(k+1)(k-2)][k(k-1)]+1 \\ & =\left(k^{2}-k-2\right)\left(k^{2}-k\right)+1 \\ & =\left(k^{2}-k\right)^{2}-2\left(k^{2} \cdots k\right)+1 \\ & =\left[\left(k^{2}-k\right)-1\right]^{2} \\ \therefore \quad & v^{\prime}(31) \cdot(30) \cdot(29... | 869 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,106 |
2. Mark 10 points on a circle, how many different polygons can be formed using these points (all or part)? (Polygons are considered the same only if their vertices are exactly the same).
| $$
\begin{array}{l}
\left.+C_{10}^{2}+\cdots+C_{10}^{0}+C_{10}^{10}\right]-\left[C_{10}^{0}+C_{10}^{\overrightarrow{1}}\right. \\
\left.+C_{10}^{2}\right]=(1+1)^{10}-(1+10+45)=968 \text{ different convex polygons.} \\
\end{array}
$$
For $3 \leqslant k \leqslant 10$, every selection of $k$ points can form a convex $k$-... | 968 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,107 |
3 . Let $n$ be a positive integer, $d$ be a digit among the 10 decimal digits, and suppose $\frac{n}{810}=0 . d 25 d 25 d 25 \cdots$, find $n$.
| $$
\begin{array}{c}
3 . \because \frac{n}{810}=0 . d 25 d 25 d 25 \cdots, \\
\therefore 1000 \cdot \frac{n}{810}=d 25 . d 25 d 25 \cdots \\
\therefore \frac{999 n}{810}=1000 \cdot \frac{n}{810}-\frac{n}{810} \\
=d 25=100 d+25, \\
\therefore \quad 999 n=810(100 d+25),
\end{array}
$$
And $(750,37)=1$, so $37 \mid (4d+1)... | 750 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,108 |
4. If $a<b<c<d<e$ are consecutive positive integers, $b+c+d$ is a perfect square, and $a+b+c+d+e$ is a perfect cube, what is the minimum value of $c$? | $$
\begin{aligned}
4 . & b+c+d=3 c \\
& a+b+c+a+e=5 c
\end{aligned}
$$
and $b+c+d=m^{2}, a+b+c+d+e=n^{3}$
$$
\begin{aligned}
\therefore \quad 3 c & =m^{2}, \\
5 c & =n^{3} .
\end{aligned}
$$
$$
\begin{array}{l}
\therefore 3^{3} \mid c_{0} \text { and } 5\left|n_{0} \quad \therefore 5^{2}\right| c, \quad \therefore 25 ... | 675 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,109 |
6. Two Skaters
Allie and Bllie are located
at points $A$ and $B$ on
a smooth ice surface, with $A$ and $B$
being 100 meters apart. If Allie starts from $A$
and skates at 8 meters/second
along a line that forms a $60^{\circ}$
angle with $A B$, while Bllie starts from $B$ at 7 meters/second
along the shortest line to me... | 6. Let the two people meet at point $C$ after $t$ seconds, then the sides of $\triangle A B C$ are $A B=100, A C=8 t, B C=7 t$. By the cosine rule, we get
$$
\begin{aligned}
(7 t)^{2}= & 100^{2}+(8 t)^{2}-2(100)(8 t) \cos 60^{\circ} \\
\therefore \quad 0 & =3 t^{2}-160 t+2000 \\
& =(3 t-100)(t-20) .
\end{aligned}
$$
$\... | 160 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,111 |
7. If the integer $k$ is added to $36,300,596$, respectively, the results are the squares of three consecutive terms in an arithmetic sequence, find the value of $k$. | 7. Let $\left\{\begin{array}{l}36+k=(n-d)^{2} , \\ 300+k=n^{2} \\ 596+k=(n+d)^{2}\end{array}\right.$
(1)
From (3) - (1) we get $4 n d=560$,
$\therefore n d=140$.
From (2) $\times 2-(1)-$ (3) we get $2 d^{2}=32$,
$\therefore d= \pm 4$, then $n= \pm 35$.
From (2), $k=n^{2}-300=35^{2}-300=925$. | 925 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,112 |
9. Euler's conjecture was refuted by American mathematicians in 1960, who proved that there exists a positive integer $n$ such that $133^{5}+110^{5}+84^{5}+27^{5}=n^{5}$. Find $n$ when it is satisfied. | 9. Clearly $n \geqslant 134$. Now we find the upper bound of $n$.
$$
\begin{aligned}
\because \quad n^{5}= & 133^{5}+110^{5}+84^{5}+27^{5}<133^{5} \\
& +110^{5}+(27+84)^{5}<3(133)^{5} \\
& <\frac{3125}{1024}(133)^{5}=\left(\frac{5}{4}\right)^{5}(133)^{5}, \\
\therefore \quad & n<\frac{5}{4}(133), \text { i.e., } n \leq... | 144 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,115 |
11. Given an integer array consisting of 121 integers, where each number is between 1 and 1000 (repetition is allowed), this array has a unique mode (i.e., the integer that appears most frequently). Let $D$ be the difference between this mode and the arithmetic mean of the array. When $D$ reaches its maximum value, wha... | 11. Let $M$ and $m$ be the "mode" and "arithmetic mean" respectively. Without loss of generality, assume $M \geqslant m$. To maximize $D$, then $M=1000$ (otherwise, if $M=1000-k$, then increasing $M$ by $k$ while $m$ increases by no more than $k$, $D$ is obviously non-decreasing). In the case of $M=1000$, we must make ... | 947 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,117 |
12. As shown in the figure, $ABCD$ is a tetrahedron, $AB=41$, $AC=7$, $AD=18$, $BC=36$, $BD=27$, $CD=13$. Let $d$ be the distance between the midpoints of $AB$ and $CD$. Find the value of $d^{2}$. | 12. By the median formula, we get $m_{s}^{2}=\frac{1}{4}\left(2 b^{2}+2 c^{2}\right.$ $\left.-a^{2}\right)$, so $P C^{2}=\frac{1}{4}\left[2(A C)^{2}+2(B C)^{2}\right.$ $\left.-(A B)^{2}\right]=\frac{1009}{4}$.
$$
\begin{array}{l}
P D^{2}\left.=\frac{1}{4}\left[2(A D)^{2}+2(B D)^{2}\right]-(A B)^{2}\right] \\
=\frac{42... | 137 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,118 |
1. On the vertices of a regular octagon, can the numbers $1,2, \cdots, 8$ be placed such that the sum of the numbers on any three adjacent vertices is:
(1) greater than 11,
(2) greater than 13? | (1) It is possible. Figure 1 is an example that meets the conditions.
(2) Proof that the required numbering method is impossible. We make the opposite assumption: the numbers $1,2, \cdots, 8$ can be arranged in such a way that the sum of the numbers on any three consecutive vertices of the octagon is greater than 13, i... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,119 |
4. Solve the system of equations
$$
\left\{\begin{array}{l}
x y^{2}-2 y+3 x^{2}=0 \\
y^{2}+x^{2} y+2 x=0
\end{array}\right.
$$ | Solution: Obviously, the solution of the form $(0, y)$ is unique, which is $(0,0)$. Now we find the solution $(x, y)$ where $x \neq 0$. (2) Multiply by $x$ and subtract (1) to get $\left(2+x^{3}\right) y = x^{2}$. Thus, $2+x^{3} \neq 0$, meaning $y = x^{2} / \left(2+x^{3}\right)$. Substituting the obtained expression f... | (0,0), (-1,1), \left(-\frac{2}{\sqrt[3]{3}}, -2\sqrt[3]{3}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,122 |
5 . There are two piles of stones. If 100 stones are taken from the first pile and placed in the second pile, then the number of stones in the second pile will be twice that in the first pile. Conversely, if some stones are taken from the second pile and placed in the first pile, then the number of stones in the first ... | Let the number of stones in the first and second piles be denoted as $x, y$. Let $z$ be the number of stones taken from the second pile and placed into the first pile. Thus, we obtain the equations
$$
\begin{array}{l}
2(x-100)=y+100, \\
x+z=6(y-z) .
\end{array}
$$
From (1), we get $y=2 x-300$, substituting this into (... | 170 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,123 |
1. Can we mark all cells of a (1) $3 \times 3$ size, (2) $198 \times 8$ size table with crosses or circles, such that in every cell with a cross, there is exactly one circle in an adjacent cell, and in every cell with a circle, there is exactly one cross in an adjacent cell (adjacent cells are those that share a common... | Solution (1) No. Suppose the center cell is marked with a cross, then among the cells $A, B, C, D$ in Figure 7, there is exactly one circle. Without loss of generality, assume the circle is in $A$. Thus, cells $C$ and $D$ must be crosses. Therefore, cell $E$ (in Figure 7) cannot be a cross, as there is no circle in any... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 704,124 |
Example 16. To completely cover a square with a side length of 1 using three circles of the same radius, what is the minimum radius of these equal circles?
To completely cover a square with a side length of 1 using three circles of the same radius, the minimum radius of these equal circles is to be determined. | Solve $A H=\sqrt{1+(2 x)^{2}}$,
$M G=\sqrt{\left(\frac{1}{2}\right)^{2}+(1-2 x)^{2}}$.
Also, $A H=M G$,
Solving for $x$ gives $x=\frac{1}{16}$.
$$
\therefore A E=\sqrt{\left(\frac{1}{2}\right)^{2}+x^{2}}=\frac{1}{16} \sqrt{65} \text {. }
$$
Thus, these three semicircles have a radius of $\frac{\sqrt{65}}{16}$, and th... | \frac{\sqrt{65}}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,125 |
4. Given a regular tetrahedron $A B C D$. Let $E$ be the intersection of the medians of the base $\triangle A B C$, connect $D E$, and take a point $M$ on the line segment $D E$ such that $\angle A M B=90^{\circ}$. Find $E M: M D$. | Since the tetrahedron $ABCD$ is a regular tetrahedron, $D$ is the apex. Let $F$ be the midpoint of edge $AC$ (Figure 10). Without loss of generality, we can assume that the edge length of the tetrahedron $ABCD$ is 1. Therefore,
$$
BF=\frac{\sqrt{3}}{2}, AE=BE=\frac{2}{3} BF=\frac{1}{\sqrt{3}},
$$
$DE=\sqrt{\frac{2}{3}}... | 1:1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,127 |
5. When $x$ takes the values $-1,0,1,2$, the polynomial $P(x)=a x^{3}+b x^{2}+c x+d$ has integer values.
Prove: This polynomial takes integer values for all integer $x$. | Prove that the original polynomial can be transformed into
$$
\begin{aligned}
P(x) & =6 a \cdot \frac{(x-1) x(x+1)}{6}+2 l \cdot \frac{x(x-1)}{2} \\
& +(a+b+c) x+d
\end{aligned}
$$
According to the conditions, \(d = P(0)\) and \(a + b + c + d = P(1)\) are integers, so \(a + b + c\) is also an integer. Additionally, si... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,128 |
1. Try to find a set of natural numbers $x, y, z$ that satisfy the equation $x^{3}+y^{4}=z^{3}$. Is the set of natural number solutions to this equation finite or infinite? | Solution: Since $2^{24}+2^{24}=2^{25}$, then $\left(2^{8}\right)^{3} +\left(2^{0}\right)^{4}=\left(2^{5}\right)^{5}$, which means the three numbers $x=2^{8}=256$, $y=2^{0}=64, z=2^{5}=32$ are solutions to the given equation. Furthermore, it is not difficult to deduce that if $\left(x_{0}, y_{0}, z_{0}\right)$ is a set ... | x=256, y=1, z=32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,129 |
3. Given a circle, there are two fixed points on this circle. Now there are two circles that are externally tangent to the given circle at points $A, B$, and these two circles are also externally tangent to each other at point $C$. Find the locus of point $C$. | Let $\mathrm{O}$ be the center of the given circle, $\mathrm{O}_{1}$ and $\mathrm{O}_{2}$ be the centers of the other two circles, and $a$ and $b$ be the tangents to circle $O$ at points $A$ and $B$ (Figure 11). Clearly, points $A, B, C$ lie on the sides of $\triangle O O_{1} O_{2}$. Let $\alpha=\angle O_{1} O O_{2}$, ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,131 |
5. Can two regular tetrahedra with edge length 1 be placed inside a cube with edge length 1 so that the two tetrahedra do not intersect? | We prove that the required placement of the tetrahedron is possible. Let \(ABCD A_1 B_1 C_1 D_1\) be the given cube with edge length 1, and let \(N, P, Q, R, S, T\) be the midpoints of the edges \(DA, AB, BB_1, B_1 C_1, C_1 D_1, D_1 D\) respectively (Figure 12). Thus, \(N P Q R S T\) is a regular hexagon with side leng... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,132 |
Example 1. The decimal representation of the natural number $A$ is $\overline{a_{n} a_{n}-\overline{1} \cdots a_{1} a_{0}}$, let
$f(A)=2^{\mathrm{n}} a_{0}+2^{\mathrm{n}-1} a_{1}+\cdots+2 a_{\mathrm{n}-1}^{\mathrm{n}}+a_{\mathrm{n}}$ denote $A_{1}=f(A), A_{\mathrm{i}+\mathrm{i}}=f\left(A_{\mathrm{i}}\right),(i=1,2, \cd... | Prove $f(A)=2^{n} a_{0}+2^{n-1} a_{1}^{\gamma}+\cdots+2 a_{n-1}$
$$
\begin{array}{l}
+a_{n} \leqslant\left(2^{n}+2^{n-1}+\cdots+2+1\right)_{9} \\
=9\left(2^{n+1}-1\right) .
\end{array}
$$
While $A \geqslant 10^{\text { }} a_{\text {n }}$
Next, we prove that when $n \geqslant 2$, $A>f(A)$.
By $a_{n} \cdot 10^{n} \geqsl... | 19 | Number Theory | proof | Yes | Yes | cn_contest | false | 704,133 |
Example 6. Given the sides of a triangle $a, b, c$, and area $S$. Prove that $a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S$, note | $$
\begin{aligned}
a^{2}+ & b^{2}+c^{2}-4 \sqrt{3} S \\
& =a^{2}+b^{2}+\left(a^{2}+b^{2}-2 a \cdot \cos C\right) \\
& -4 \sqrt{3} \cdot\left(\frac{1}{2} a b \sin C\right) \\
= & 2 a^{2}+2 b^{2}-2 a b(\cos C+\sqrt{3} \sin C) \\
= & 2\left[a^{2}+b^{2}-2 a b \sin \left(C+\frac{\pi}{6}\right)\right] \\
\geqslant & 2\left(a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,136 |
Example 7. Let $P$ be any point in $\triangle ABC$, and $M$ be the centroid.
Prove:
$$
\begin{array}{l}
P A^{2}+P B^{2}+ \\
P C^{2} \geqslant M A^{2}+ \\
M B^{2}+M C^{2} .
\end{array}
$$ | Prove: Connect $P M, P D$ (as shown in the figure).
By the median length formula, we have
$$
\begin{array}{l}
M B^{2}+M C^{2}=2 M D^{2}+\frac{1}{2} D C^{2} . \\
P B^{2}+F C^{2}=2 D^{2} D^{2}+\frac{1}{2} B C^{2} .
\end{array}
$$
By the cosine rule, we get
$$
\begin{aligned}
A P^{2} & =A M^{2}+P M^{2} \\
& -2 A M \cdot ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,137 |
Example 8. Given two circles intersecting at $M$ and $N$, how can a line segment $A B$ be constructed, passing through point $M$ and with endpoints $A, B$ on the two circles respectively, such that $A M \cdot M B$ is maximized? (75 USA Mathematical Competition)
---
Note: The translation preserves the original text's ... | Solution: Clearly, $M N$ is a fixed length while $\angle A, \angle B$ are both remote angles. Therefore,
$\alpha=\angle A N B=\pi-\angle A-\angle B$ is a fixed angle.
Let $\angle A N M=x$. Then $\angle B N M=\alpha-x$.
By the Law of Sines, we have
$$
\begin{array}{l}
A M=\frac{M N \sin x}{\sin A}, M B=\frac{M N \sin (\... | 2 x-\alpha=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,138 |
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