problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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$17^{\circ}$. In the spatial tetrahedron $ABCD$, the lengths of $AB$, $AC$, and $AD$ are $a_{1}$, $a_{2}$, and $a_{3}$, and their opposite edge lengths are $b_{1}$, $b_{2}$, and $b_{3}$. The angles between these three pairs of edges are $\theta_{1}$, $\theta_{2}$, and $\theta_{3}$. If $a_{1}^{2}+b_{1}^{2} \geqslant a_{... | Proof: Let the projection of $A$ on plane $BCD$ be $O$. Connect $DO$ and extend it to intersect $BC$ at $E$. Denote:
$$
\angle ADO=\alpha, \angle DEC=\beta,
$$
Then $\cos \theta_{3}=|\cos \alpha \cdot \cos \beta|$ (this conclusion has been proven in high school textbooks).
$$
\begin{aligned}
\because & a_{1}^{2}=AO^{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,095 |
High $18^{\circ}$, no real roots for the $n$-th degree equation
$$
b_{n} x^{n}+b_{n-1} x^{n-1}+\cdots+b_{1} x+b_{0}=0
$$
all roots are imaginary, $b_{k}=a_{k}+a_{n-k} \cdot i(k=0,1, \cdots, n)$ and satisfy
$$
0<a_{0}<a_{1}<a_{2}<\cdots<a_{n-1}<a_{n} .
$$
If $z$ is a complex root of the equation, find $|z|$. | Let $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$, and $z_{0}$ is a root of the equation $f(x)=0$. Then $\left|z_{0}\right|<1$.
In fact, by the condition, we know that $z_{0} \neq 0,1$. Because
$$
\begin{array}{l}
0=\left(z_{0}-1\right)\left(a_{n} z_{0}^{n}+a_{n-1} z_{0}^{n-1}+\cdots+a_{1} z_{0}+a_{0}\right) ... | |z|=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,096 |
Example 1. $P$ is a point inside an equilateral $\triangle A B C$, and the ratios of the sizes of $\angle A P B, \angle B P C, \angle C P A$ are 5:6:7. Then, the ratio of the sizes of the three interior angles of the triangle with sides $P A, P B, P C$ is () (from smallest to largest).
(A) $2: 3: 4$
(B) $3: 4: 5$
(C) $... | From the given, we know that $\angle A P B=100^{\circ}, \angle B P C=$ $120^{\circ}, \angle C P A=140^{\circ}$.
From the proof of the previous problem, we know that the three interior angles of the triangle with sides of lengths $P A, P B, P C$ are (note the comparison and observation of the angles at one point):
$$
1... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,097 |
Example 3. Find the basic Pythagorean triple, where one of the numbers is
$$
16 .
$$ | Let's set the Pythagorean triplet as $x, y, z$, where $x=16$. From
$$
x=2 \times 4 \times 2=2 \times 8 \times 1
$$
and when $m=4, n=2$, $(m, n)=2 \neq 1$, so only $m=8, n=1$. At this point, $(m, n)=1$ and $m$ and $n$ are one odd and one even. Thus,
$$
\begin{array}{l}
y=m^{2}-n^{2}=8^{2}-1^{2}=63, \\
z=m^{2}+n^{2}=8^{... | 16, 12, 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,098 |
Example 2. In the equilateral $\triangle A B C$, $P$ is a point on side $B C$. Then, in the new triangle formed by $A P, B P, C P$ as sides, the largest interior angle is $\theta_{1}$, then ( ).
(A) $\theta \geqslant 90^{\circ}$
(B) $\theta \leqslant 120^{\circ}$
(C) $\theta=120^{\circ}$
(D) $\theta=135^{\circ}$ | Solution: Obviously, $P$ on side $BC$ can be regarded as an extreme case or example of $P$ inside $\triangle ABC$.
Thus, from the conclusion of the previous problem, we know that the angles formed by the lines connecting $P$ with the three vertices of the triangle are $\alpha$,
$$
\beta, 180^{\circ} \text {, }
$$
Ther... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,099 |
Example 3. In an equilateral $\triangle ABC$, a point $P$ inside the triangle is such that the segments $PA=6, PB=8, PC=10$. The integer closest to the area of $\triangle ABC$ is ().
(A) 159
(B) 131
(C) 95
(D) 79
(E) 50
(1967, 18th American High School Mathematics Examination) | From the previous example, we know that $a=6, b=8, c=10$, and the key to the problem is to find $u, v, z$.
And $S_{\triangle A B C}=2\left(S_{\triangle 1}\right.$
$$
\begin{array}{l}
\left.+S_{\triangle 1}+S_{\triangle 1}\right)+\left(S_{\triangle T}\right. \\
\left.+S_{\triangle 2}+S_{\triangle G}\right)= \\
2 S_{\tri... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,100 |
Example 4. Positive numbers $x, y, z$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{2}+x y+\frac{y^{2}}{3}=25, \\
\frac{y^{2}}{3}+z^{2}=9, \\
x^{2}+x y+z^{2}=16 .
\end{array}\right.
$$
Find: $x y+2 y z+3 z x$.
(1984, 18th All-Union Mathematical Olympiad) | Notice that $9=3^{2}, 16=4^{2}, 25=5^{2}$, then a right-angled $\triangle A B C$ with side lengths $3,4,5$ can be formed. Inside the triangle, find a point such that the angles formed by connecting this point to the three vertices are $90^{\circ}, 120^{\circ}$, $150^{\circ}$ (as shown in the figure). Let
$$
P B=x, P C=... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,101 |
Example 1. Given a cyclic quadrilateral $A B C D$ with side lengths $a, b, c, d$, and the longest side length is $a$. Under what conditions on $a$, $b, c, d$ does the perimeter of the cyclic quadrilateral $A B C D$ have a minimum value? And find this minimum value. | Let the area of quadrilateral $ABCD$ be $S$, the radius of the circumcircle be $R$, and $AB=a, BC=b, CD=c, DA=d$ (as shown in the figure). Then,
$$
\begin{aligned}
S & =\sqrt{p(p-a)(p-b)(p-c)} \\
(p= & \left.\frac{1}{2}(a+b+c)\right),
\end{aligned}
$$
$$
\begin{array}{l}
\sin A=\frac{2 S}{a d+b c}, \sin B=\frac{2 S}{a ... | 4 S \sqrt{\frac{a c+b d}{(a b+c d)(a d+b c)}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,102 |
Example 3. In the spatial quadrilateral $ABCD$, it is known that $AB = CD$, $AD = BC$, and $AC = BD = a$. Prove: The perimeter of the inscribed quadrilateral in $ABCD$ has a minimum value, and find this minimum value.
---
The translation maintains the original text's formatting and line breaks. | Prove $\because A B=C D, A D=B C$,
$$
\begin{array}{c}
A C=B D=a, \\
\therefore \triangle A B C \cong \triangle B A D \cong \triangle C D A \cong \triangle D C B, \\
\therefore \angle A B C=\angle A D C=\angle B A D \\
=\angle B C D, \\
\angle B A D+\angle D A C+\angle C A B=180^{\circ}, \\
\angle D C B+\angle B C A+\a... | 2a | Geometry | proof | Yes | Yes | cn_contest | false | 707,104 |
The 24th All-Union Mathematical Olympiad has a problem:
There are 1990 piles of stones, with the number of stones being $1, 2, \cdots$, 1990. The operation is as follows: each time, you can choose any number of piles and take the same number of stones from each of them. How many operations are needed at least to take ... | Solve the confusion of
$$
\begin{array}{l}
1990=: 2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+0 \cdot 2^{5} \\
+0 \cdot 2^{4}+0 \cdot 2^{3}+2^{2}+2^{1}+0 \cdot 2^{0},
\end{array}
$$
and write $1,2, \cdots, 1989$ in binary form. The operation is as follows:
First, take away $2^{10}=1024$ stones from each pile that has enough; seco... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,105 |
Are there infinitely many integers $n(\geqslant 2)$, such that the equation
$$
x_{1}^{2}+x_{2}^{3}+x_{3}^{4}+\cdots+x_{n}^{n+1}=x_{n+1}^{n+2} \quad \text { (*) }
$$
has infinitely many positive integer solutions $\left(x_{1}, x_{2}, \cdots, x_{n+1}\right)$? Prove your conclusion. | The answer is affirmative. First, if $\left(x_{1}, \cdots, x_{n+1}\right)$ is a solution, then for any positive integer $b$, taking $x^{\prime}_{i}=x_{i} \cdot b^{\frac{(n+2)'}{i+1}}, i=1,2, \cdots, n+1$, then $\left(x^{\prime}_{1}, \cdots, x^{\prime}_{n+1}\right)$ is also a solution.
Second, let $x_{1}=2, x_{2}=3, x_{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,106 |
In 1992, the second question of the second round of the National Junior High School Mathematics League is:
In $\triangle A B C$, $A B=A C$, and $D$ is a point on the base $B C$.
A point $E$ is on the line segment $A D$, and $\angle B E D$
$=2 \angle C E D=\angle A$. Prove: $B D=2 C D$. | As shown in the figure, let $F$ be
the midpoint of $B D$. Draw $F G \perp$
$B D$ intersecting $A B$ at $G$. Draw
$G H / / B C$, intersecting $A C$ at
$H$. Connect $G D, H D$. Then
$\triangle G B D$ and $\triangle A G H$ are
both isosceles triangles and similar. Therefore, $\angle B G D=\angle A$
$=\angle B E D$. Hence,... | B D=2 C D | Geometry | proof | Yes | Yes | cn_contest | false | 707,107 |
Example 4. Let $p, m, r$ be a set of Pythagorean numbers, where $p$ is an odd prime, and $n>p, n>m$.
Prove: $2 n-1$ must be a perfect square. | Proof that since $p, m, n$ are Pythagorean numbers, and $n>m$, $n>p$, then
$$
\begin{array}{l}
m^{2}+p^{2}=n^{2} . \\
m^{2}=n^{2}-p^{2}=(n+p)(n-p) .
\end{array}
$$
Thus, $m>n-p$.
Let $m=n-r(1 \leqslant r<p)$.
Then we have
$$
\begin{aligned}
p^{2} & =n^{2}-m^{2}-n^{2}-(n-r)^{2} \\
& =r(2 n-r) .
\end{aligned}
$$
Since ... | 2 n-1 \text{ is a perfect square} | Number Theory | proof | Yes | Yes | cn_contest | false | 707,109 |
Example 2. As shown in Figure 2. $A D: D C=2: 1$, $\angle A D B=60^{\circ}, \angle C=$ $45^{\circ}$. Prove: $A B$ is the tangent line of the circumcircle of $\triangle B C D$ (1987, National Junior High School Mathematics League). | Prove: Draw $A E \perp B D$ at $E$, let $A D=2 k$, then $D C=k, D E=k$. Connect $E C$, it is easy to know $\angle E A C=30^{\circ}$, $\angle D C E=30^{\circ}$, so $A E=E C$. Also, $\angle E C B=45^{\circ}-30^{\circ}=15^{\circ}$, $\angle D B C=60^{\circ}-45^{\circ}=15^{\circ}$, hence $E B=E C$. Therefore, $A E=E B$,
kno... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,110 |
Example 3. As shown in Figure 3, the equilateral triangles $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ have their sides $A C$ and $A_{1} C_{1}$ bisected at $O$. Then $A A_{1}: B B_{1}=$ $\qquad$ | This problem can also be solved using the "specialization" method (let $A_{1}$ be on $AO$). However, if $B O, B_{1} O$ are considered, it is immediately known that $\triangle A A_{1} O$ ○ $\triangle B B_{1} O$, thus $A A_{1}: B O_{1}=A O: B O=\sqrt{3}$ : 3. | \sqrt{3} : 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,111 |
$B$ is a point on $A C$,
circles $\odot O_{1}$, $\odot O_{2}$, $\odot O$ are constructed with diameters $A B$, $B C$, $A C$ respectively. A line through $B$ intersects $\odot O_{1}$ at $R$, $\odot O_{2}$ at $S$, and $\odot O$ at $P, Q$. Prove: $\bar{F} R=Q S$. | If $AB = BC$, then it is obvious. Suppose $AB \neq BC$, draw $OP \perp PQ$ at $D$, then $PD = DQ$, connect $AR, SC$, then $AR \parallel OD \parallel SC$, from $AD = OC$ we know $RD = DS$, thus $PR = QS$. | PR = QS | Geometry | proof | Yes | Yes | cn_contest | false | 707,112 |
Theorem The tangents drawn from the vertices of a scalene triangle to its circumcircle intersect the opposite sides at collinear points.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
However, since the last part is an instruction and not part of the theorem, it should not be translated. He... | Proof As shown in the figure, given an unequal-sided $\triangle ABC$ with the tangents forming $\triangle A^{\prime} B^{\prime} C^{\prime}$. Let the line $AB$ intersect $A^{\prime} B^{\prime}$ at $P$, $BC$ intersect $B^{\prime} C^{\prime}$ at $Q$, and $CA$ intersect $C^{\prime} A^{\prime}$ at $R$. Connect $A^{\prime} A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,113 |
Example 1. Find the value of $\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}$. | Considering $\sin \alpha \cos \alpha=\frac{1}{2} \sin 2 \alpha$, we can use its symmetric (dual) form to assist in solving: Let $x=\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}, y=\cos 10^{\circ} \cos 30^{\circ} \cos 50^{\circ} \cos 70^{\circ}$, then
$$
\begin{aligned}
x y & =\frac{1}{16} \sin 20^{\ci... | \frac{1}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,114 |
1. Find all arrays composed of four natural numbers $a, b, c, d$, such that the product of any three numbers divided by the remaining one leaves a remainder of 1. (Supplied by Li Shangzhi, University of Science and Technology of China) | Given that $bcd$ divided by $a$ leaves a remainder of 1, then $a \geqslant 2$. Otherwise, $bcd$ divided by 1 would leave a remainder of 0. Therefore, there exists an integer $k$ such that $bcd = k a - 1$, which means $a$ is coprime with $b, c, d$. Thus, $a, b, c, a'$ are all distinct and pairwise coprime.
Without loss... | \{2, 3, 7, 41\} \text{ or } \{2, 3, 11, 13\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,115 |
2. Fill in each cell of an $n \times n$ grid paper with a number, such that each row and each column forms an arithmetic sequence. Such a filled grid paper is called an arithmetic cipher table. If knowing the numbers in certain cells of this arithmetic cipher table can decipher the entire table, then the set of these c... | (1) When $s \leqslant 2 n-1$, take $s$ squares from the first row and the first column. Clearly, these $s$ squares do not form a key.
Below, we prove that when $s=2 n(n \geqslant 4)$, any $2 n$ squares taken from an $n \times n$ grid form a key. In $n$ columns, there are $2 n$ squares $(n \geqslant 4)$, so there must ... | n+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,116 |
3. Find the smallest natural number $n$ with the following property: when any five vertices of a regular $n$-gon $S$ are colored red, there is always a line of symmetry $l$ of $S$ such that the reflection of each red point across $l$ is not a red point. (Supplied by Hu Chengzhang, Nankai University) | When $n \leqslant 9$, a regular $n$-sided polygon clearly does not possess the property mentioned in the problem.
For a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, when $n=2 k (k \in N)$, there are $2 k$ axes of symmetry: the lines $A_{i} A_{k+i} (i=1,2, \cdots, k)$ and the perpendicular bisectors of the seg... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,117 |
4. Given $5 n$ real numbers $r_{i}, s_{i}, t_{i}, u_{i}, v_{i}$ all greater than $1(1 \leqslant i \leqslant n)$, let $R=\frac{1}{n} \sum_{i=1}^{n} r_{i}, S=\frac{1}{n} \sum_{i=1}^{n} s_{i}, T$ $=\frac{1}{n} \sum_{i=1}^{n} t_{i}, U=\frac{1}{n} \sum_{i=1}^{n} u_{i}, V=\frac{1}{n} \sum_{i=1}^{n} u_{i}$. Prove that: $\prod... | To prove, first establish a lemma:
Lemma: Let $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ real numbers greater than 1, and $A = \sqrt[n]{x_{1} x_{2} \cdots x_{n}}$, then
$$
\prod_{i=1}^{n}\left(\frac{x_{i}+1}{x_{i}-1}\right) \geqslant\left(\frac{A+1}{A-1}\right)^{n}.
$$
Proof of the lemma: Let $x_{i} = \max \left(x_{1}, \cdo... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,118 |
5. $p, q$ are two distinct prime numbers, and the natural number $n \geqslant$ 3. Find all integers $a$, such that the polynomial $f(x)=x^{n}+$ $a x^{n-1}+p q$ can be factored into the product of two integer-coefficient polynomials, each of degree at least one. (Supplied by Huang Qiguo, Fudan University) | Let $f(x)=g(x) h(x)$.
Here
$$
\begin{array}{l}
g(x)=a_{1} x^{t}+\cdots+a_{1} x+a_{0}, \\
h(x)=b_{m} x^{m}+\cdots+b_{1} x+b_{0} .
\end{array}
$$
where $l, m$ are natural numbers, and the coefficients $a_{i}(0 \leqslant i \leqslant l), b_{0}(0 \leqslant \alpha \leqslant m)$ are all integers, and $a_{l} \neq 0, b_{m} \ne... | a=-(1+p q), a=1+(-1)^{n-2} p q | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,119 |
$\begin{array}{l}-0.25^{2} \div\left(-\frac{1}{2}\right)^{4} \times(-2)^{3}+\left(1 \frac{3}{8}+\right. \\ \left.2 \frac{1}{3}-3.75\right) \times 24\end{array}$ | One, 7.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,122 |
II. (10 points) Factorize $(a-b)^{2}+4 a b-c^{2}$. | $=、(a+b+c)(a+b-c)$
$= (a+b+c)(a+b-c)$ | (a+b+c)(a+b-c) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,123 |
$$
\frac{\sqrt{5}+3 \sqrt{3}+4 \sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{2}+\sqrt{3})} \text {. }
$$
Simplify the expression above. | Four, $\sqrt{3}+\sqrt{5}-2 \sqrt{2}$ (Hint: Original expression $=\frac{1}{\sqrt{3}+\sqrt{2}}$ $\left.+\frac{3}{\sqrt{5}+\sqrt{2}}\right)$ | \sqrt{3}+\sqrt{5}-2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,124 |
Five, (10 points) As shown in the figure, find all corresponding angles, alternate interior angles, and consecutive interior angles.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Five, Corresponding Angles: $\angle 1$ and $\angle 3 ; \angle 1$ and $\angle 5 ; \angle 1$ and $\angle 8 ; \angle 1$ and $\angle 7+\angle 8 ; \angle 2$ and $\angle 4 ; \angle 3$ and $\angle 5$. Alternate Interior Angles: $\angle 3$ and $\angle 6 ; \angle 2$ and $\angle 7 ; \angle 4$ and $\angle 8 ; \angle 5$ and $\angl... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,125 |
$\therefore$ (10 points) Given $a, b$ are integers, and $a+9b$ is divisible by 5, prove that $8a+7b$ is also divisible by 5. | Six, prompt:
$$
8 a+7 b=8(a+9 b)-65 b .
$$ | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,126 |
Seven, (10 points) A, B, and C each have several pieces of candy. They are required to give each other candies, starting with A giving B and C, the number of candies given being equal to the number of candies B and C originally had; then, in the same manner, B gives A and C the number of candies they currently have, an... | Seven, let the original number of sugar blocks that A, B, and C have be $x, y, z$. According to the problem, after the first distribution, they each have sugar blocks: $x-y-z, 2 y, 2 z$.
After the second distribution, they each have sugar blocks: $2(x-y-z), 2 y-(x-y-z)$ $-2 z, 4 z$.
After the third distribution, they e... | x=104, y=56, z=32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,127 |
Ten, (10 points) If $a>0, b>0$, and $\sqrt{a}(\sqrt{a} + \sqrt{b}) = 3 \sqrt{b}(\sqrt{a} + 5 \sqrt{b})$. Find the value of $\frac{2a + 3b + \sqrt{ab}}{a - b + \sqrt{ab}}$. | $$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0 . \\
\because a>0, b>0, \therefore \sqrt{a}+3 \sqrt{b}>0 . \\
\therefore \sqrt{a}-5 \sqrt{b}=0 \Rightarrow \frac{a}{b}=25 .
\end{array}
$$
Therefore, the value of the original expression is 2.
$$
\begin{array}{l}
(\sqrt{a}-5 \sqrt{b})(\sqrt{a}+3 \sqrt{b}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,130 |
Example 5. In $\triangle A B C$, if there exists a point $P$ such that $\angle P A C=\angle P B A=\angle P C B=\theta$. Find the range of values for $\theta$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
a^{2}+b^{2}+c^{2}=4 S_{\triangle A B C} \operatorname{ctg} \theta, \\
\therefore \operatorname{ctg} \theta=\frac{a^{2}+b^{2}+c^{2}}{4 S_{\triangle A B C}} . \text { Let } S_{\triangle A B C}=S, \\
S=\sqrt{p(p-a)(p-b)(p-c)} \\
\leqslant \sqrt{p} \sqrt{\left[\frac{(p-a)+(p-b)+(p-c)}{3}\right]^{3}} \\
... | \theta \in\left[0, \frac{\pi}{6}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,131 |
One, (10 points) Solve the equation
$$
5 x^{2}+10 y^{2}-12 x y-6 x-4 y+13=0 .
$$ | I. The original equation is transformed into $(2 x-3 y)^{2}+(x-3)^{2}+(y-$ $2)^{2}=0$, solving it yields $x=3, y=2$. | x=3, y=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,132 |
II. (10 points) Simplify $\frac{\sqrt{5-\sqrt{24}}}{\sqrt{10}+3-\sqrt{6}-\sqrt{15}}$. | $=,-\frac{1}{2}(\sqrt{5}+\sqrt{3})$. | -\frac{1}{2}(\sqrt{5}+\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,133 |
Three, (10 points) Given that the fractional parts of $9+\sqrt{3}$ and $9-\sqrt{13}$ are $a$ and $b$ respectively. Find the value of $a b-3 a+4 b+8$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, due to $9+\sqrt{13}=12+a, 9-\sqrt{13}=5+b$. Therefore, $a=\sqrt{13}-3, b=4-\sqrt{13}$. Then the original expression value is 8. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,134 |
Four, (10 points) Person A and Person B process a batch of parts. If A and B work together for 6 days, and then A continues alone for 5 more days, the task can be completed; if A and B work together for 7 days, and then B continues alone for 5 more days, the task can also be completed. Now, if A first processes 300 par... | Four, suppose there are $s$ parts in total, person A completes $x$ parts per day, and person B completes $y$ parts per day. According to the problem, we have
$$
\left(\begin{array}{l}
6 x+6 y+5 x=s, \\
7 x+7 y+5 y=s, \\
300+8(x+y)=s .
\end{array}\right.
$$
Solving this, we get $s=2700$ ( $\uparrow$ ). | 2700 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,135 |
Five. (10 points) Given as shown, in $\triangle A B C$, $\angle A=2 \angle B$, and $C D$ is the bisector of $\angle A C B$. Prove: $B C=A C+A D$.
保留源文本的换行和格式，直接输出翻译结果如下：
Five. (10 points) Given as shown, in $\triangle A B C$, $\angle A=2 \angle B$, and $C D$ is the bisector of $\angle A C B$. Prove: $B C=A C+A D$.
-... | Five, on $CB$ take $CE=CA$, connect $DE$, then $\triangle ACD \cong \triangle ECD$. Also, $\angle A=2 \angle B$, so $\angle B=\angle BDE$. Therefore, $BE=DE=AD$. Hence, $BC=AC+AD$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,136 |
Seven, (10 points) When $a>0$, and $b>a+c$, prove that the equation $a x^{2}+b x+c=0$ must have two distinct real roots. | (1) Assume $c>0$. Since $a>0$, we have $a+c>0$. Also, $b>a+c$, then $b>0$, so $b^{2}>(a+c)^{2}$.
Thus, $\Delta=b^{2}-4ac>(a+c)^{2}-4ac=(a-c)^{2} \geqslant 0$.
That is, $\Delta>0$.
(2) Assume $c<0$. Since $a>0$, we have $ac<0, -4ac>0$, hence $\Delta>0$.
From (1) and (2), we can conclude that the conclusion holds. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,138 |
Eight, (10 points) Given as shown, equilateral $\triangle A B C$, take a point $E$ on the extension of side $A C$, construct an equilateral $\triangle C D E$ with $C E$ as a side, and it is on the same side of line $A E$ as $\triangle A B C$. Point $M$ is the midpoint of segment $A D$, and point $N$ is the midpoint of ... | VIII. It is easy to prove that $\triangle A C D \cong \triangle B C E$.
Also, $\triangle A C M \cong \triangle B C N, \angle M C N=\angle B C N+ \angle M C B=\angle A C M+\angle M C B=60^{\circ}$, then $\triangle C M N$ is an equilateral triangle. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,139 |
Nine, (10 points) Given as shown, the angle bisector of $\angle A$ in $\triangle A B C$ is $A D, M$ is the midpoint of $B C$, and $A D \parallel M E$. Prove:
$$
B E=C F=\frac{1}{2}(A B+A C)
$$ | $\triangle B M G C \triangle C M F$.
Also, $\angle B A D=\angle D A C$, since $A D / / M E$, thus, $\angle B A D=\angle A E M, \angle D A C=\angle C F M$. Therefore, $\angle B E G=\angle B G E$.
That is, $B E=B G=C F$.
Furthermore, since $B G=A B+A E, A E=A F$, hence $2 B G=A B+A E+C F$,
which means $B G=\frac{1}{2}(... | B E=C F=\frac{1}{2}(A B+A C) | Geometry | proof | Yes | Yes | cn_contest | false | 707,140 |
Ten, (10 points) Given $n$ lines in a plane that intersect each other pairwise. Prove: their intersection angles include at least one angle that is not greater than $\frac{180^{\circ}}{n}$. | Ten, on a plane, $n$ lines intersecting each other at most have
$$
4[(n-1)+(n-2)+\cdots+2+1]=2 n(n-1)
$$
angles.
Take any point $C$ on the plane, and translate the $n$ lines so that they all pass through point $O$, becoming $n$ lines intersecting at $O$. Thus, these $n$ lines divide the circle's angles with $O$ as the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,141 |
Example 1. In a Cartesian coordinate system, points with both coordinates as integers are called integer points. Please design a method to color all integer points, with each integer point being colored white, red, or black, such that
(1) every integer point appears on infinitely many parallel lines to the x-axis:
(2) ... | Analyzing, the construction of this problem is quite strong. Here, a coloring method is given and intuitively explained to meet the requirements.
As shown in the figure, color the lattice points on the line $y=x+2m$ (where $m$ is an integer) red; color the lattice points on the line $y=x+2m+1$ (where $m$ is an integer... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,142 |
3. If the length of a rectangle is increased by $p \%$, to keep the area of the rectangle unchanged, the width of the rectangle should be reduced by ( ).
(A) $p \%$
(B) $(1-p) \%$
(C) $\frac{p}{1+p} \%$
(D) $\frac{100 p}{100+p} \%$ | 3. D Hint: Let the length of the rectangle be $a$, and the width be $b$. Then according to the selection, the length increases to $a(1+p) \%$. And the width decreases by $x \%$. That is, the width is $b(1-$ $x \%)$, and the area remains unchanged, so $a(1+p \%) b(1-x \%)=a b$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,145 |
4. $p$ is a prime number, and the sum of all positive divisors of $P^{4}$ is exactly a perfect square. Then the number of prime numbers $p$ that satisfy the above condition is ( ).
(A) 3
(B) 2
(C) 1
(D) 0 | 4. C Since $p$ is a prime number, $p^{+}$ has 5 positive divisors $1 \cdot p$, $p^{2}, p^{3}, p^{4}$. According to the problem, we have
$1+p+p^{2}+p^{3}+p^{4}=n^{2}$ ( $n$ is a positive integer).
Then $(2 n)^{2}>4 p^{4}+4 p^{3}+p^{2}=\left(2 p^{2}+p\right)^{2}$.
Also, $(2 n)^{2}<4 p^{2}+p^{2}+4+4 p^{3}+8 p^{2}+4 p$
$$... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,146 |
5. The number of integer solutions $(x, y)$ for the equation $\sqrt{x}+\sqrt{y}=\sqrt{336}$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 5. D $x, y$ are both positive integers, and $\sqrt{x}, \sqrt{y}$, when simplified, should be like radicals with $\sqrt{336}$ simplified to $4 \sqrt{21}$.
Let $\sqrt{x}=m \sqrt{21}, \sqrt{y}=n \sqrt{21}$. Then $m+n$ $=4$ (where $m, n \geqslant 0$ and are positive integers). The values are
$$
(0,4),(1,3),(2,2),(3,1),(4,... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,147 |
1. If natural numbers $m, n$ satisfy $m+n>m n$, then the value of $m$ $+n-m n$ is $\qquad$ | 2. 1. From the given, we have $\frac{1}{n}+\frac{1}{m}>1$. If $n, m \geqslant 2$, then $\frac{1}{n}$ $+\frac{1}{m} \leqslant \frac{1}{2}$, which is a contradiction. Therefore, at least one of $m, n$ must be 1. Without loss of generality, let $m=1$, then $m+n-m n=1+n-1 \cdot n=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,149 |
2. In $\triangle A B C$, $A B \leqslant A C \leqslant B C$, and the smallest interior angle is not less than $59^{\circ}$. Then the maximum value of the largest interior angle is $\qquad$ degrees. | 2. According to the problem, we have $\angle C \leqslant \angle B \leqslant \angle A$. From $59^{\circ} \leqslant \angle C \Rightarrow$ $59^{\circ} \leqslant \angle B$, then $\angle B+\angle C \geqslant 118^{\circ}$. Therefore, $\angle A \leqslant 62^{\circ}$. | 62 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,150 |
3. Let $x, y, z$ be non-negative real numbers not less than 1. If $k=x+y(1-x)+z(1-x)(1-y)$, then the value range of $k$ is $\qquad$ . | 3. Prompt: From the given, we have
$$
k=1-(1-x)(1-y)(1-z) \text { . }
$$
And $0 \leqslant 1-x \leqslant 1,0 \leqslant 1-y \leqslant 1,0 \leqslant 1-z \leqslant 1$.
Thus $0 \leqslant 1-(1-x)(1-y)(1-z) \leqslant 1$. That is, $0 \leqslant k \leqslant 1$. | 0 \leqslant k \leqslant 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,151 |
4. Given the inequality $a x+3 \geqslant 0$ has positive integer solutions of $1,2,3$. Then the range of values for $a$ is $\qquad$ | 4. Obviously $a<0$, so $x \leqslant-\frac{3}{a}$. According to the value of $x$, we get 3 $\leqslant-\frac{3}{a}<4$. Solving this, we get $-1 \leqslant a<-\frac{3}{4}$. | -1 \leqslant a<-\frac{3}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,152 |
Example 2. In three-dimensional space, the set of all lattice points is denoted as $T$. Two lattice points $(x, y, z)$ and $(u, v, w)$ are said to be adjacent if and only if $|x-u|+|y-v|+|z-w|=1$. Prove: There exists a subset $S$ of $T$ such that for every $P \in T$, $P$ and its adjacent points contain exactly one poin... | Solve the similar problem only in two-dimensional space. In the Cartesian coordinate system, two lattice points $(x, y)$ and $(u, v)$ are adjacent if $|x-u|+|y-v|=1$. There are four points adjacent to the lattice point $(x, y)$: $(x+1, y)$, $(x, y+1)$, $(x-1, y)$, and $(x, y-1)$. The subset $S$ chosen should contain an... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,153 |
5. In $\triangle A B C$, one side is 5, and the other two sides are exactly the two roots of the equation $2 x^{2}-12 x+m=0$. Then the range of values for $m$ is | 5. Let the two roots of $2 x^{2}-12 x+m=0$ be $x_{1}, x_{2}$. Clearly, $x_{1}>0, x_{2}>0$. Therefore, $x_{1} x_{2}=\frac{m}{2}>0, x_{1}+x_{2}=6>5$, and $\left|x_{1}-x_{2}\right|0$, so from $\left|x_{1}-x_{2}\right|^{2}\frac{11}{2}$. That is, $\frac{11}{2}<m \leqslant 18$. | \frac{11}{2}<m \leqslant 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,154 |
6. $a, b, c$ are all positive integers, and satisfy $a b+b c=$ $3984, a c+b c=1993$. Then the maximum value of $a b c$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 6. Since 1993 is a prime number, from $c(a+b)=1993$, we get $c=1, a+b=1993$. Substituting $b=1993-a$ into the other equation, we get $a^{2}-1992 a+1991=0$. Solving this, we get $a_{1}=1, a_{2}=1991$, then $b_{1}=1992, b_{2}=2$. Therefore, $(a, b, c)$ has two sets $(1,1992,1),(1991,2,1)$. Thus, the maximum value of $abc... | 3982 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,155 |
7. In quadrilateral $A B C D$, $A B=B C, \angle A=$ $\angle C=90^{\circ}, \angle B=13 E^{\prime}, K$ is a point on $A B$, $N$ is a point on $B C$. If the perimeter of $\triangle B K N$ is equal to 2 times $A B$, then the degree measure of $\angle K D N$ is | 7. Connect $B D$, obviously $\mathrm{Rt} \triangle A B D \cong \mathrm{Rt} \triangle G B D$. Extend $B C$ to $A_{1}$, such that $C A_{1}=A K$, it is easy to prove $\mathrm{Rt} \triangle D C A_{1} \cong \mathrm{Rt} \triangle D A K$.
Therefore, $\angle A_{1} D K=\angle C D A=\angle 45^{\circ}$.
Also, $K N+B K+B N=2 A B=A... | 22.5^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,156 |
1. The roots $x_{1}, x_{2}$ of the equation $x^{2}-a x-a=0$ satisfy the relation $x_{1}{ }^{3}+x_{2}{ }^{3}+x_{1}{ }^{3} x_{2}{ }^{3}=75$. Then $1993+5 a^{2}$ $+9 a^{4}=$ $\qquad$ | 1. According to Vieta's formulas, we have $x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3}=75$. Thus, $a^{2}=25, a^{4}=625$. Therefore, $1993+5 a^{2}+9 a^{4}=7743$. | 7743 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,158 |
2. Given $a-b=5+\sqrt{6}, b-c=5-$ $\sqrt{6}$. Then $a^{2}+b^{2}+c^{2}-a b-b c-c a=$. | 2. Given, we have $a-c=10$. Then
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=\underline{81} .
\end{array}
$$ | 81 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,159 |
3. In $\triangle A B C$, $\angle B=100^{\circ}$, the angle bisector of $\angle C$ intersects side $A B$ at $E$, take point $D$ on side $A C$ such that $\angle C B D=20^{\circ}$, and connect $D, E$. Then the degree measure of $\angle C E D$ is | 3. Let the reverse extension line of $CB$ be $BF$, then $\angle ABF=80^{\circ}$, $\angle ABD=80^{\circ}$. Therefore, $BA$ is the bisector of $\angle DBF$. Since $E$ is on $BA$, the distances from $E$ to $BF$ and $BD$ are equal.
Since $E$ is on the bisector $CE$ of $\angle ACB$, the distances from $E$ to $CF$ and $CA$ ... | 10^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,160 |
4. $a, b, c$ are all natural numbers greater than 20, one of them has an odd number of positive divisors, the other two each have exactly three positive divisors, and $a+b=c$. The smallest value of $c$ that satisfies the above conditions is . $\qquad$ | 4. Note first that a natural number has an odd number of positive divisors if and only if it is a perfect square. In particular, if a natural number has exactly 3 positive divisors, this number must be the square of a prime.
Let \(a=p^{2}, b=q^{2}, c=r^{2}\). From \(a+b=c\), we get
\[
p^{2}+q^{2}=r^{2} \text {. }
\]
w... | 169 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,161 |
5. Person A and Person B start walking towards each other at a constant speed from points $A$ and $B$ respectively, and they meet for the first time at a point 700 meters from $A$; then they continue to walk, with A reaching $B$ and B reaching $A$, and both immediately turning back, meeting for the second time at a poi... | 5. Let the distance between places $A$ and $B$ be $x$ meters, the time for the first meeting be $t$, and the time for the second meeting be $2t$. According to the problem,
$$
\frac{700}{t}=\frac{x-700+400}{2 t} \text{, solving for } x=1700 \text{. }
$$ | 1700 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,162 |
$$
\begin{array}{l}
\left(\frac{x^{3}+x y^{2}+1}{x^{3}+x y^{2}-x^{2} y-y^{3}}\right)\left(\frac{x^{2} y-x y^{2}}{x^{3}+x y^{2}-x^{2} y-y^{3}}\right) \\
+\left(\frac{x^{2} y+x y^{2}}{x^{3}+x^{2}}\right)\left(\frac{x^{2} y-y^{2}+y^{3}+1}{x^{2} y+y^{3}-x^{3}-x y^{2}}\right) \\
+\left(\frac{x^{3}+y^{3}}{x^{3}+x^{2} y+x y^{... | \begin{array}{l}\text { Combined ratio } \frac{x^{2} y+x y^{2}}{x^{3}+x y^{2}+x^{2} y+y^{3}}=C \text {. } \\ \text { I) }=\frac{x^{2} y+y^{3}+1}{x^{2} y+y^{3}-x^{3}-x y^{2}}, E=\frac{x^{3}+y^{3}}{x^{4}+x^{2} y+x y^{2}+y^{3}} \text {. } \\ \text { Then, } A+D=1 \cdots+E=1 \text {. } \\ \text { Therefore, the original ex... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,163 |
Example 3. Draw the largest circle that contains exactly $n$ lattice points internally, and calculate the diameter of the circle.
Make the circle with the largest possible diameter that contains exactly $n$ lattice points inside it, and compute the diameter of this circle. | This is a rather difficult question. When $n$ is small, the problem can be illustrated intuitively with diagrams. The following five figures depict the largest circles that contain exactly 0, 1, 2, 3,
and 4 lattice points inside, with diameters of $\sqrt{2}$, 2, $\sqrt{5}$, $\frac{5 \sqrt{5}}{3}$, and $\sqrt{10}$, re... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,164 |
Three, (15 points) Let non-zero real numbers $p_{1}, p_{2}, q_{1}, q_{2}$ satisfy the relation $p_{1} p_{2}=4\left(q_{1}+q_{2}\right)$. Prove: The equations $x^{2}+$ $p_{1} x+q_{1}=0$ and $x^{2}+p_{2} x+q_{2}=0$ have at least one with unequal real roots. | Three, using proof by contradiction
Suppose the equations $x^{2}+p_{1} x+q_{1}=0$ and $x^{2}+p_{2} x+q_{2}=0$ both do not have unequal real roots, then their discriminants should satisfy $p_{1}^{2}-4 q_{1} \leqslant 0$ and $p_{2}^{2}-4 q_{2} \leqslant 0$. Adding them yields $p_{1}^{2}+p_{2}^{2}-4\left(q_{1}+q_{2}\right... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,165 |
Four, (15 points) As shown in the figure, in $\triangle A B C$, $\angle B=90^{\circ}, M$ is a point on $A B$ such that $A M = B C, N$ is a point on $B C$ such that $C N = B M$. Connect $A N, C M$ intersecting at point $P$. Try to find the degree of $\angle A P M$ and write out your reasoning and proof process. | IV. $\angle A l^{\prime} M=45^{\circ}$. Hint:
Solution 1 Draw a perpendicular to $BC$ through $C$, and intercept $CD=AM=BC$ on it, then connect $DN, DA, DM$. It can be proven that $\triangle DCN \cong \triangle CBM$. Thus, $\triangle ADN$ is an isosceles right triangle.
Solution 2 Outside $\triangle ABC$, draw a perpe... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,166 |
Five. (15 points) Please find 6 distinct natural numbers that satisfy the following conditions:
(1) Any two of the 6 numbers are coprime;
(2) The sum of any 2, 3, 4, 5, or 6 of these numbers is a composite number.
Briefly explain the reason for your choice of numbers and how they meet the conditions. | Five, choose 6 distinct natural numbers as
$$
a_{i}=i \times 1 \times 2 \times 3 \times 4 \times 5 \times 6+1(i=1,2 \cdots, 6) .
$$
Only prove that any two are coprime. Without loss of generality, prove that $a_{2}$ and $a_{5}$ are coprime. Let $\left(a_{6}, a_{5}\right)=d \neq 1$. That is, $d\left|a_{2}, d\right| a_{... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,167 |
5. Given that the area of $\triangle A B C$ is $1, D$ is the midpoint of $B C, E, F$ are on $A C, A B$ respectively, and $S_{\triangle B D F}=$ $\frac{1}{5}, S_{\triangle C D E}=\frac{1}{3}\left(S_{\triangle C D E}\right.$ represents the area of $\triangle C D E$, the same applies below). Then $S_{\triangle D E F}=$ $\... | 5. $\frac{4}{15}$ | \frac{4}{15} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,172 |
Example 4. In the Cartesian coordinate system, points whose both coordinates are integers are called integer points. Try to prove: for any positive integer $n$, there exists a circle such that its interior contains exactly $n$ integer points. | Prove that on the circumference of each circle centered at the point $(\sqrt{2}, \sqrt{3})$, there is at most one integer point. Otherwise, let $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$ be two integer points on the circumference, and at least one of $x_{1} \neq x_{2}$ and $y_{1} \neq y_{2}$ holds. T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,175 |
$二 、$ (16 points) $1,2,3,4,5,6$ each is used once to form a six-digit number $\overline{a b c d e f}$, such that the three-digit numbers $\overline{a b c}$, $\tilde{b} c \bar{d}, \bar{c} d e, \overline{d e f}$ can be successively divisible by $4,5,3,11$. Find this six-digit number. | II. Since $5 \mid \overline{b c d}$, therefore, $d=5$.
Also, since $11 \mid \overline{d e f}$, therefore, $d+f-e$ is a multiple of 11.
But $3 \leqslant d+f \leqslant 5+6=11, 1 \leqslant e \leqslant 6$,
so $-3 \leqslant d+f-e \leqslant 10$.
Therefore, it can only be that $d+f-e=0$, i.e., $5+f=e$.
Also, since $e \leqslan... | 324561 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,179 |
Three, (16 points) In acute $\triangle ABC$, it is known that $AB=$ 4, $AC=5, BC=6, AD, BE, CF$ are the altitudes to sides $BC, CA, AB$ respectively, with $D, E, F$ being the feet of the perpendiculars. Find $\frac{S_{\triangle DEF}}{S_{\triangle ABC}}$ | Three, let $B D=x, C E=y$, $A F=z$, then by the Pythagorean theorem, we get
$$
\left\{\begin{array}{l}
4^{2}-x^{2}=5^{2}-(6-x)^{2}, \\
6^{2}-y^{2}=4^{2}-(5-y)^{2}, \\
5^{2}-z^{2}=6^{2}-(4-z)^{2} .
\end{array}\right.
$$
Solving, we get $x=\frac{9}{4}, y=\frac{9}{2}, z=\frac{5}{8}$.
$$
\begin{aligned}
\frac{S_{\triangle... | \frac{27}{256} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,180 |
Four, (18 points) Suppose that among $r$ people, the transmission of messages is carried out through telephone calls. When two people $A$ and $B$ are talking on the phone, $A$ tells $B$ everything he knows at that moment, and $B$ also tells $A$ everything he knows at that moment. Let $a_{r}$ denote the minimum number o... | Four, let $r$ people be denoted by $A_{1}, A_{2}, A_{3}, \cdots, A_{r}$.
(1) When $r=3$, first $A_{1}$ and $A_{2}$ communicate, then $A_{2}$ and $A_{3}$ communicate. At this point, $A_{1}$ only knows the information of $A_{1}, A_{2}$, and $A_{3}$ (or $A_{2}$) must communicate with $A_{1}$. Therefore, $a_{3}=3$.
(2) Whe... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,181 |
3. As shown in the figure, in $\triangle A B C$
, $\angle A C B=90^{\circ}$,
$\angle C D A=120^{\circ}$, $C E$ is
the median on side $A B$, and $D E$
$=D C$, then the ratio of the degree measures of $\angle B$ to $\angle A$ is ( ).
(A) 6
(B) 5
(C) 4
(D) 3 | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,184 |
4. Given the quadratic equation in $x$, $x^{2}-2(p+1)$ - $x+p^{2}+8=0$, the absolute value of the difference of its roots is 2. Then the value of $p$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | 4. $\mathrm{B}$ (Hint: $\left(x_{1}-x_{2}\right)^{2}$ $=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}$, then apply Vieta's formulas. ) | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,185 |
Example 1. In $\triangle A B C$, $O$ is the circumcenter of $\triangle A B C$, $I$ is the incenter, and $\angle B O C=$ $\angle B I C$. Find $\angle A$.
保留源文本的换行和格式，直接输出翻译结果。 | Since $I$ is the incenter of $\triangle A B C$,
$$
\begin{array}{l}
\therefore \angle I B C+\angle I C B \\
=\frac{1}{2}(\angle B+\angle C) \\
=\frac{1}{2}\left(180^{\circ}-\angle A\right),
\end{array}
$$
Therefore, $\angle B I C=180^{\circ}-(\angle I B C+\angle I C B)$
$$
\begin{array}{l}
=180^{\circ}-\frac{1}{2}(180... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,186 |
Example 2. In Rt $\triangle A B C$, $\angle C=90^{\circ}, \angle A B C$ $=66^{\circ}, \triangle A B C$ is rotated around $C$ to the position of $\triangle A^{\prime} B^{\prime} C^{\prime}$, with vertex $B$ on the hypotenuse $A^{\prime} B^{\prime}$, and $A^{\prime} C$ intersects $A B$ at $D$. Find $\angle B D C$. (1993,... | Solve $\because \triangle A B C$ is a right triangle, $\angle A B C=66^{\circ}$,
$$
\therefore \angle A=90^{\circ}-66^{\circ}=24^{\circ} \text {. }
$$
Also, $\triangle A B C$ is rotated around $C$ to $\triangle A^{\prime} B^{\prime} C^{\prime}$, so
$$
\begin{array}{l}
\angle A C D=\angle A B A^{\prime} \\
=\angle B C ... | 72^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,187 |
2. In $\triangle A B C$, $\angle A$ is an obtuse angle, $O$ is the orthocenter, and $A O=B C$. Find the value of $\cos (\angle O B C+\angle O C B)$. (1993, National Competition) | (Answer: $-\frac{\sqrt{2}}{2}$ ) | -\frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,188 |
1. Given $1 \leqslant a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5} \leqslant a_{6} \leqslant$ 64. Then the minimum value of $Q=\frac{a_{1}}{a_{2}}+\frac{a_{3}}{a_{4}}+\frac{a_{5}}{a_{6}}$ is $\qquad$ . | $\begin{array}{l}\text { II. 1. } \frac{3}{4} \\ \because Q \geqslant \frac{1}{a_{2}}+\frac{a_{3}}{a_{4}}+\frac{a_{4}}{64} \geqslant 3 \sqrt[3]{\frac{1}{a_{3}} \cdot \frac{a_{3}}{a_{4}} \cdot \frac{a_{4}}{64}}=\frac{3}{4}\end{array}$ | \frac{3}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,189 |
2. Divide the sequence $2,6,10,14, \cdots$ into groups in order, the first group has 2 terms $(2,6)$, the second group has 6 terms $(10,14, \cdots, 30)$, $\cdots$ the $k$-th group has $4 k-2$ terms. Then 1994 belongs to the $\qquad$ group. | 2.16
1994 is the 499th term. The first $k$ groups have a total of $2+6+\cdots+(4 k-2)=2 k^{2}$ terms. From $2 k^{2}<499$, we have $k \leqslant 15$. Therefore, 1994 is in the 16th group. | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,190 |
3. The area of $\triangle A B C$ is $S, \angle A=45^{\circ}$, line $M N$ divides the area of $\triangle A B C$ into two equal parts, and $M$ is on $A B$, $N$ is on $A C$. Then the shortest length of $M N$ is $\qquad$ | 3. $\sqrt{2(\sqrt{2}-1) S}$
Let $A M=x, A N=y$, then $x y=\sqrt{2} S$,
$$
\begin{aligned}
M N^{2} & =x^{2}+y^{2}-2 x y \cos 45^{\circ} \\
& =x^{2}+\frac{(\sqrt{2} S)^{2}}{x^{2}}-2 \cdot \sqrt{2} S \cdot \frac{\sqrt{2}}{2} \\
& \geqslant 2 \sqrt{2} S-2 S=2 S(\sqrt{2}-1) .
\end{aligned}
$$
Thus, $M N \geqslant \sqrt{2(... | \sqrt{2(\sqrt{2}-1) S} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,191 |
4. Let $0 \leqslant \theta \leqslant \frac{\pi}{2}$, the range of real numbers $m$ that satisfy the inequality $\sin ^{2} \theta+3 m$ - $\cos \theta-6 m-4<0$ is $\qquad$ . | 4. $m>-\frac{1}{2}$.
The inequality becomes $3 m>\frac{3+\cos ^{2} \theta}{\cos \theta-2}$ for all $0 \leqslant \theta \leqslant \frac{\pi}{2}$, so $3 m$ should be greater than the maximum value of $f(\theta)=-\frac{3+\cos ^{2} \theta}{2-\cos \theta}$. It is easy to see that $-\frac{3+\cos ^{2} \theta}{2-\cos \theta} ... | m>-\frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,192 |
5. For different real numbers $m$, the equation $y^{2}-6 m y-4 x+$ $9 m^{2}+4 m=0$ represents different parabolas. A line intersects all these parabolas, and the length of the chord intercepted by each parabola is $\frac{8 \sqrt{5}}{9}$. Then the equation of this line is $\qquad$ . | 5. $y=3 x-\frac{1}{3}$.
The parabola $(y-3 m)^{2}=4(x-m)$. The vertex $(m, 3 m)$ lies on the line $y=3 x$, so the required line must be parallel to it, and can be set as $y=3 x+b$. Take any parabola $y^{2}=4 x$ (when $m=0$), substitute $y=3 x+b$ into it to get $x^{2}+\frac{6 b-4}{9} x+\frac{b^{2}}{9}=0$. We can find t... | y=3 x-\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,193 |
6. If the complex number $z$ satisfies $3 z^{6}+2 i \cdot z^{5}-2 z-3 i=$ 0 . Then $|z|=$ $\qquad$ . | 6. $|z|=1$.
The equation is $z^{5}=\frac{2 z+3 i}{3 z+2 i}$. Let $z=a+b i$.
$$
\left|z^{5}\right|=\frac{|2 z+3 i|}{|3 z+2 i|}=\sqrt{\frac{4\left(a^{2}+b^{2}\right)+12 b+9}{9\left(a^{2}+b^{2}\right)+12 b+4}} \text {. }
$$
If $a^{2}+b^{2}>1$, then the left side of (1) $=|z|^{5}=\left(\sqrt{a^{2}+b^{2}}\right)^{5}>$ 1. B... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,194 |
One, (20 points) Let $a_{n}=(2+\sqrt{3})^{n}$. Prove that for all $n \in N, [a_{n}]$ is an odd number $([x]$ denotes the greatest integer not exceeding $x$). | Given $a_{1}=2+\sqrt{3}$, and $2^{2}-(\sqrt{3})^{2}=1$, $a_{2}=(2+\sqrt{3})^{2}=7+4 \sqrt{3}$, and $7^{2}-(4 \sqrt{3})^{2}=1$,
$a_{3}=(2+\sqrt{3})^{3}=26+15 \sqrt{3}$, and $26^{2}-(15 \sqrt{3})^{2}=1$.
Conjecture $a_{n}=x_{n}+y_{n} \cdot \sqrt{3}$, and $x_{n}^{2}-\left(y_{n} \cdot \sqrt{3}\right)^{2}=1$ $\left(x_{n}, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,195 |
II. (20 points) In the sequence of natural numbers $1,2,3, \cdots, n, \cdots$, remove all natural numbers containing the digits $0,7,8,9$, resulting in the sequence $\left\{a_{n}\right\}: 1,2,3,4,5,6,11,12, \cdots, 16,21$, $22, \cdots$.
Prove: $\sum_{n=1}^{\infty} \frac{1}{a_{n}}<\frac{49}{8}$. | For the sequence $\left\{a_{n}\right\}$:
When $a_{n}$ is a single-digit number,
$$
\sum \frac{1}{a_{n}}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{49}{20} \text {. }
$$
When $a_{n}$ is a two-digit number,
$$
\begin{array}{l}
\sum \frac{1}{a_{n}}=\left(\frac{1}{11}+\frac{1}{12}+\cdots+\frac{1}{1... | \frac{49}{8} | Number Theory | proof | Yes | Yes | cn_contest | false | 707,196 |
Three, (30 points) In $\triangle A B C$, $A B=A C$, point $M$ is on $A B$ and $M A=M C$, point $N$ is on $A C$ and $C N=C B, \angle A: \angle N B A=2: 3$. Find the degree measure of $\angle N M C$. | Let $\angle A=\alpha$, then $\angle N B A=\frac{3}{2} \alpha, \angle B N C=\frac{5}{2} \alpha$, $\angle C B i V=\frac{5}{2} \alpha, \angle C B A=\frac{5}{2} \alpha+\frac{3}{2} \alpha=4 \alpha$.
$\angle M C A=20^{\circ}, \angle M C B=60^{\circ}$.
Draw $M K / / B C$. Then quadrilateral $B M K C$ is an isosceles trapezoid... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,197 |
Four. (30 points) Given 1994 points in the plane, no three of which are collinear, label each line segment with endpoints among these points as +1 or -1. If the product of the numbers labeled on the three sides of a triangle formed by these points is -1, the triangle is called negative. Prove that the number of negativ... | $$
\begin{array}{l}
\because AB=AC, \\
\therefore \frac{\alpha}{2}+4 \alpha=90^{\circ}, \\
\alpha=20^{\circ} .
\end{array}
$$
Thus, in $\triangle ABC$
$$
\begin{array}{l}
\angle A=20^{\circ}, \\
\angle ACB=80^{\circ},
\end{array}
$$
Four, the number of triangles with 1994 points as vertices is $C_{1994}^{3}$. Suppose... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,198 |
3. $\triangle A B C$ is an isosceles triangle, with vertex angle $A$ being $20^{\circ}$. On $A B$, take $A D = B C$, and connect $D C$. Find $\angle B D C$. | (Answer: $30^{\circ}$. Hint: Inside $\triangle ABC$, construct an equilateral triangle $BCN$ with $BC$ as a side, and connect $AN$) | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,199 |
19. As shown in the figure, $P B A$ is a secant of $\odot O$, $P C$ is a tangent, $C D$ is a diameter of $\odot O$, and $D B, D P$ intersect at $E$. Prove: $A C \perp C E$.
保留源文本的换行和格式，直接输出翻译结果如下：
19. As shown in the figure, $P B A$ is a secant of $\odot O$, $P C$ is a tangent, $C D$ is a diameter of $\odot O$, and $... | Prove as follows: Draw another tangent line through $P$ to $\odot O$, touching $\odot O$ at $F$. Connect $D F, F E, C E, A C, B F$. Then,
$$
\begin{array}{l}
\angle C D F=\frac{1}{2} \overparen{F Q C}=\overparen{C G}=\angle C O P \\
\Rightarrow D F / / O P \Rightarrow \angle 7=\angle 8 . \\
\because \angle 6=\angle 8, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,200 |
Initially $20^{\circ}$. There are 800 points on the circumference, labeled in a clockwise direction as $1,2, \cdots, 800$. They divide the circumference into 800 gaps. Now, choose one point and color it red, then follow the rule to color other points red one by one: if the $k$-th point has been colored red, then move $... | Proof Consider a circle with $2n$ points in general.
(1) On a circle with $2n$ points, if the first red point is an even-numbered point, for example, the $2k$-th point, then according to the coloring rule, each red point dyed afterward will also be an even-numbered point. At this time, if the points numbered $2, 4, 6, ... | 25 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,201 |
19. Prove: The lines connecting the midpoints of the three pairs of opposite edges of a tetrahedron, together with the three edges of any face of the tetrahedron, can form a new tetrahedron, and its volume is equal to half the volume of the original tetrahedron. | Proof As shown in the figure, let $c$
$B$
$GH, JK, LM$ be the midline connections of the three pairs of opposite edges of the tetrahedron $ABCD$.
Construct the parallelogram $ABCE$, and let $F$ be the midpoint of $DE$. Consider the edges of the tetrahedron $FACE$.
$\because F G \underline{\mathbb{L}} \frac{1}{2} E C \u... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,202 |
20. Let the three sides, semi-perimeter, and area of $\triangle A_{i} B_{i} C_{i}$ be $a_{i}, b_{i}, c_{i}, p_{i}, \triangle_{i} (i=1,2)$, respectively, then
$$
a_{1}\left(p_{1}-a_{1}\right)\left(p_{2}-b_{2}\right)\left(p_{2}-c_{2}\right)+b_{1}\left(p_{1}-b_{1}\right)\left(p_{2}-c_{2}\right)\left(p_{2}-a_{2}\right)+c_{... | Prove that according to the half-angle formula $\operatorname{tg} \frac{A_{1}}{2}=\frac{1}{\triangle_{1}}\left(p_{1}-b_{1}\right)\left(p_{1}-c_{1}\right)$ and five similar formulas, it is easy to see that the inequality to be proved is equivalent to
$$
\begin{array}{l}
Q=\operatorname{tg} \frac{A_{1}}{2}\left(\operator... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,203 |
Example 1. Let $(a x+b y)^{1993}(a b \neq 0,|a|+|b|=$ 1 ) be expanded, and the absolute value of the coefficient of the $(k+1)$-th term is the largest. Prove: $1994|b|-1 \leqslant k \leqslant 1994|b|$. | $$
\begin{array}{l}
\left\{\begin{array}{l}
C_{1993}^{k}\left|a^{1993-k} b^{k}\right| \geqslant C_{1993}^{k-1}\left|a^{1993-k+1} b^{k-1}\right| \\
C_{1993}^{k}\left|a^{1993-k} b^{k}\right| \geqslant C_{1993}^{k+1}\left|a^{1993-k-1} b^{k+1}\right|
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
\frac{k}{1994-k}... | 1994|b|-1 \leqslant k \leqslant 1994|b| | Algebra | proof | Yes | Yes | cn_contest | false | 707,205 |
Example 2. Prove: The coefficients of the expansion of $(a+b)^{2^{1994}-1}$ have the same parity.
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | Proof According to the general term of the binomial expansion, the coefficients of each term are $C_{2}^{k}{ }^{1994}{ }_{-1}\left(0 \leqslant 1 \leqslant 2^{19 S_{4}}--1\right)$. Clearly, when $k=0$, the function is a constant,
$$
\begin{array}{l}
\text { then } C_{2^{1099}}^{k_{-1}}=\prod_{j=1}^{k} \frac{2^{i 994}-j}... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 707,206 |
Example 3. Given $x=(\sqrt{1994}+44)^{1993}$, $y=(\sqrt{1994}-44)^{1903}$. Prove: The fractional part of $x$ is $y$.
untranslated text remains unchanged. | To prove that $x-y \in N$, and $0<y<1$.
$$
\begin{array}{l}
\because x-y=\sum_{k=0}^{1993} C_{1993}^{k} \cdot 1994^{\frac{1993-k}{2}} \cdot 44^{k} \\
-\sum_{k=0}^{1993} C_{1993}^{k} \cdot 1994^{\frac{1993-k}{2}} \cdot(-44)^{k} \\
=2 \sum_{k=0}^{996} C_{1993}^{2 k+1} \cdot 1994^{996-k} \cdot 44^{2 k+1} \\
\in N, \\
\tex... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,207 |
Example 4. Let $p$ be a prime number. Can $p$ divide $1994^{p}-1994$? | Given that $2\left|\left(1994^{2}-1994\right), 3\right|\left(1994^{3}-\right.$ 1994), we conjecture that $p \mid\left(1994^{p}-1994\right)$.
To prove the above conjecture, we first prove a more general conclusion: If $p$ is a prime, then $p \mid\left(n^{p}-n\right),(n \in N)$.
(1) When $n=1$, it is obvious that $p \mi... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,208 |
Example 5. Find the value of $\sum_{k=1}^{n} k^{2} C_{n}^{k}$. (3rd Putnam Mathematical Competition, USA) | $\begin{array}{l}\text { Solve } \sum_{k=1}^{n} k^{2} C_{n}^{k}=\sum_{k=1}^{n} k \cdot k C_{n}^{k} \\ =\sum_{k=1}^{n} k \cdot n C_{n-1}^{k-1}=n \sum_{k=1}^{n}[(k-1)+1] C_{n-1}^{k-1} \\ =n \sum_{k=1}^{n} C_{n-1}^{k-1}+n \sum_{k=2}^{n}(n-1) C_{n-2}^{k-2} \\ =n \sum_{k=1}^{n} C_{n-1}^{k-1}+n(n-1) \sum_{k=2}^{n} C_{n-2}^{k... | n(n+1) \cdot 2^{n-2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,209 |
Example 6. For a given function $f(x)$, define $f_{n}(x)$ as follows:
$$
\text { follows: } f_{n}(x)=\sum_{k=0}^{n} C_{n}^{k} f\left(\frac{k}{n}\right) x^{k}(1-x)^{n-k}(n \in
$$
$N$.
(1) When $f(x)=1 \quad(0 \leqslant x \leqslant 1)$, prove that $f_{n}(x)=1(n \in N)$.
(2) When $f(x)=x$, compare the sizes of $f_{93}(94)... | (3) When $f(x)=x^{2}$, $f\left(\frac{k}{n}\right)=\frac{k^{2}}{n^{2}}$. When $n \geqslant 2$,
$$
\begin{aligned}
f_{n}(x)= & \sum_{k=0}^{n} C_{n}^{k} \cdot \frac{k^{2}}{n^{2}} \cdot x^{k} \cdot(1-x)^{n-k} \\
= & x \sum_{k=0}^{n} \frac{k}{n} C_{n-1}^{k-1} \cdot x^{k-1}(1-x)^{n-k} \\
= & x\left[\frac{1}{n} \sum_{k=1}^{n}... | \frac{1993}{1994} x^{2}+\frac{1}{1994} x | Algebra | proof | Yes | Yes | cn_contest | false | 707,210 |
Example 7. Given $a_{n}=C_{1994}^{3 n-1}$. Find $\sum_{n=1}^{655} a_{n}$. | Consider the binomial expansion
$$
(1+x)^{1994}=\sum_{k=0}^{1994} C_{1994}^{k} x^{k} \text {. }
$$
Substitute \(1, \omega, \omega^{2}\left(\omega=\frac{-1+\sqrt{3} i}{2}\right)\) into (1), we get
$$
\begin{array}{l}
2^{1994}=\sum_{k=0}^{1994} C_{1994}^{k}, \\
(1+\omega)^{1994}=\sum_{k=0}^{1994} C_{1994}^{k} \cdot \ome... | \frac{2^{1994}-1}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,211 |
Example 3. In an equilateral $\triangle ABC$, take a point $D$ inside such that $DA = DB$; also take a point $E$ outside $\triangle ABC$ such that $\angle DBE = \angle DBC$, and $BE = BA$. Find $\angle BED$. (1992, Sichuan Province Junior High School Mathematics League) | Connect $D C$.
$$
\because A D=B D, A C=
$$
$B C, D C$ is common,
$$
\therefore \triangle A D C \cong \triangle B D C \text {. }
$$
Thus $\angle B C D=30^{\circ}$.
Also, $\angle D B E=\angle D B C$, $B E=A B=B C, B D$ is common, $\therefore \triangle B D E \cong \triangle B D C$. Therefore, $\angle B E D=\angle B C D=... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,212 |
Example 8. Prove:
$$
C_{1949}^{49} C_{55}^{45}+C_{1999}^{50} C_{55}^{44}+\cdots+C_{1949}^{94} C_{55}^{0}=C_{1994}^{94} .
$$ | Prove the construction $(1+x)^{1949} \cdot(1+x)^{15}$
$$
=(1+x)^{1994} \text {. }
$$
Consider the coefficient of the $x^{94}$ term on both sides of the equation.
The coefficient on the left side is exactly equal to the sum of the numbers on the left side of the equation to be proved, and the coefficient on the right s... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,213 |
Example 9. If $a, b \in R^{+}$, prove:
$$
\frac{a^{1994}+b^{1994}}{2} \geqslant\left(\frac{a+b}{2}\right)^{1994} .
$$ | Proof: Let $a+b=2x (x>0), a-b=2y$. Then $a=x+y, b=x-y$. Thus,
$$
\begin{array}{l}
a^{1994}+b^{1994} \\
=(x+y)^{1994}+(x-y)^{1994} \\
= 2\left[x^{1994}+C_{1994}^{2} x^{1992} y^{2}+C_{1994}^{4} x^{1990} y^{4}+\cdots\right. \\
\left.+C_{1994}^{1994} y^{1994}\right] \\
\geqslant 2 x^{1994}=2 \cdot\left(\frac{a+b}{2}\right... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,214 |
Example 10. If $a, b, c, d$ are all positive numbers, and $a^{2}+b^{2}+$ $c^{2}+d^{2}=1$, prove:
$$
\begin{array}{l}
(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4} \\
+(b+d)^{4}+(c+d)^{4} \leqslant 6 .
\end{array}
$$
(28th IMO Shortlist Problem). | Proof: Let the left side of the inequality to be proved be $A$, and construct $B$ as follows:
$$
\begin{aligned}
B= & (a-b)^{4}+(a-c)^{4}+(a-d)^{4} \\
& +(b-c)^{4}+(b-d)^{4}+(c-d)^{4},
\end{aligned}
$$
Then $A+B=6\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}=6$.
Clearly, $B \geqslant 0$.
$\therefore A \leqslant 6$, the pro... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,215 |
Example 11. Let $a_{n}=6^{n}-8^{n}$. Find the remainder when $a_{94}$ is divided by 49. (Adapted from the first American Mathematical Invitational Competition) | $$
\begin{aligned}
a_{94} & =6^{94}-8^{94}=(7-1)^{94}-(7+1)^{94} \\
= & -2\left(C_{94}^{1} \cdot 7^{93}+C_{94}^{3} \cdot 7^{91}+\cdots\right. \\
& \left.+C_{94}^{91} \cdot 7^{3}+C_{94}^{93} \cdot 7\right) \\
= & 49 k-2 \cdot 94 \cdot 7 \\
= & 49(k-27)+7 .(k \in Z)
\end{aligned}
$$
$\therefore a_{94}$ when divided by 49... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,216 |
Example 12. Given $x=19^{94}-1, y=2^{m} \cdot 3^{n} \cdot 5^{l}$ $(m, n, l$ are non-negative integers, and $m+n+l \neq 0)$. Find the sum $S$ of all divisors of $x$ that are of the form $y$. | $$
\begin{array}{l}
\text { Solve } x=(20-1)^{94}-1 \\
=20^{94}-C_{94}^{1} \cdot 20^{93}+\cdots-C_{94}^{93} \cdot 20 \\
=2^{3}\left(2 n_{1}-235\right) \\
=2^{3}\left[2\left(n_{1}-118\right)+1\right] .\left(n_{1} \in \mathbb{N}\right) \\
x=(18+1)^{94}-1 \\
=18^{94}+C_{94}^{1} \cdot 18^{93}+\cdots+C_{94}^{93} \cdot 18 \\... | 1169 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,217 |
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