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1. If $f(x) (x \in R)$ is an even function with a period of 2, and when $x \in [0,1]$, $f(x) = x^{1998}$, then the order from smallest to largest of $f\left(\frac{98}{19}\right), f\left(\frac{101}{17}\right), f\left(\frac{104}{15}\right)$ is | $$
\begin{array}{l}
\text { 2.1.f }\left(\frac{101}{17}\right), f\left(\frac{98}{19}\right), f\left(\frac{104}{15}\right) . \\
f\left(\frac{98}{19}\right)=f\left(6-\frac{16}{19}\right)=f\left(-\frac{16}{19}\right)=f\left(\frac{16}{19}\right), \\
f\left(\frac{101}{17}\right)=f\left(6-\frac{1}{17}\right)=f\left(-\frac{1}... | f\left(\frac{101}{17}\right)<f\left(\frac{98}{19}\right)<f\left(\frac{104}{15}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,345 |
2. Let the complex number $z=\cos \theta+i \sin \theta\left(0^{\circ} \leqslant \theta \leqslant\right.$ $\left.180^{\circ}\right)$, and the complex numbers $z, (1+i)z, 2\bar{z}$ correspond to three points $P, Q, R$ in the complex plane. When $P, Q, R$ are not collinear, the fourth vertex of the parallelogram formed by... | 2.3.
Let the complex number $w$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have
$$
w+z=2 \bar{z}+(1+i) z \text{, i.e., } w=2 \bar{z}+i z \text{. }
$$
Therefore, $|w|^{2}=(2 \bar{z}+i z)(2 z-i \bar{z})$
$$
\begin{array}{l}
=4+1+2 i\left(z^{2}-\bar{z}^{2}\right) \\
=5-4 \sin 2 A<5 \div 4=9 .
\end{a... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,346 |
3. From the 10 numbers $0,1,2,3,4,5,6,7,8,9$, choose 3 numbers such that their sum is an even number not less than 10.
The number of different ways to choose them is $\qquad$ . | $3.5 i$
From these 10 numbers, the number of ways to choose 3 different even numbers is $C_{5}^{3}$; the number of ways to choose 1 even number and 2 different odd numbers is $C_{5}^{1} C_{5}^{2}$.
From these 10 numbers, the number of ways to choose 3 numbers such that their sum is an even number less than 10, there a... | 51 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,347 |
4. In an arithmetic sequence with real number terms, the common difference is 4, and the square of the first term plus the sum of the remaining terms does not exceed 100. Such a sequence can have at most terms. | 4.8 .
Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence with a common difference of 4, then
$$
\begin{aligned}
& a_{1}^{2}+a_{2}+a_{3}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow & a_{1}^{2}+\frac{\left(a_{1}+4\right)+\left[a_{1}+4(n-1)\right]}{2} \cdot(n-1) \\
& \leqslant 100 \\
\Leftrightarrow & a_{1}^{2... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,348 |
5. If the ellipse $x^{2}+4(y-a)^{2}=4$ intersects the parabola $x^{2}=2 y$, then the range of the real number $a$ is $\qquad$ | 5. $-1 \leqslant a \leqslant \frac{17}{8}$.
From $x^{2}+4(y-a)^{2}=4$, we can set $x=2 \cos \theta, y=a+ \sin \theta$, substituting into $x^{2}=2 y$ gives $4 \cos ^{2} \theta=2(a+\sin \theta)$.
$$
\begin{aligned}
\therefore a & =2 \cos ^{2} \theta-\sin \theta=2-2 \sin ^{2} \theta-\sin \theta \\
& =-2\left(\sin \theta+... | -1 \leqslant a \leqslant \frac{17}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,349 |
Three. (Full marks 20 points) Given the complex number $z=1-\sin \theta + i \cos \theta \left(\frac{\pi}{2}<\theta<\pi\right)$. Find the principal value of the argument of the conjugate complex number $\bar{z}$. | $$
\begin{array}{l}
\text { Three, } \bar{z}=1-\sin \theta-i \cos \theta \\
=1-\cos \left(\frac{\pi}{2}-\theta\right)-i \sin \left(\frac{\pi}{2}-\theta\right) \\
=2 \sin ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)-2 i \sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \cos \left(\frac{\pi}{4}-\frac{\theta}{2}\right) ... | \frac{3 \pi}{4}-\frac{\theta}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,351 |
Four. (Full marks 20 points) Let the function $f(x)=a x^{2}+8 x+3(a<0)$. For a given negative number $a$, there is a largest positive number $l(a)$, such that the inequality $|f(x)| \leqslant 5$ holds for the entire interval $[0, l(a)]$.
Question: For what value of $a$ is $l(a)$ the largest? Find this largest $l(a)$. ... | Given, $f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}$, so,
$$
\max _{x \in R} f(x)=3-\frac{16}{a} \text {. }
$$
We discuss in two cases:
(i) $3-\frac{16}{a}>5$, i.e., $-8-\frac{4}{a} \text {. }
$$
Thus, $l(a)$ is the larger root of the quadratic equation $a x^{2}+8 x+3=-5$,
$$
\begin{aligned}
l(a) & =\frac{-8-... | \frac{\sqrt{5}+1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,352 |
Five. (Full marks 20 points) Given the parabola $y^{2}=2 p x$ and fixed points $A(a, b) 、 B(-a, 0)\left(a b \neq 0, b^{2} \neq\right.$ $2 \mathrm{pa}$ ). $M$ is a point on the parabola, and the other intersection points of the lines $A M 、 B M$ with the parabola are $M_{1} 、 M_{2}$, respectively.
Prove that when point... | Let the coordinates of $M$, $M_{1}$, and $M_{2}$ be $\left(\frac{y_{0}^{2}}{2 p}, y_{0}\right)$, $\left(\frac{y_{1}^{2}}{2 p}, y_{1}\right)$, and $\left(\frac{y_{2}^{2}}{2 p}, y_{2}\right)$, respectively.
From the collinearity of $A$, $M$, and $M_{1}$, we have
$$
\frac{\frac{y_{1}^{2}}{2 p}-\frac{y_{0}^{2}}{2 p}}{y_{1... | \left(a, \frac{2 p a}{b}\right) | Algebra | proof | Yes | Yes | cn_contest | false | 710,353 |
One, (Full marks 50 points) As shown in the figure, $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$ respectively, $AD$ is the altitude from $A$ to side $BC$, and $I$ lies on segment $OD$. Prove that the circumradius of $\triangle ABC$ is equal to the exradius of the excircle opposite to side $BC$.
Note... | Let $A B=c, B C=a, C A=b$. Suppose the extension of $A I$ intersects the circumcircle $O$ of $\triangle A B C$ at point $K$, then $O K$ is the radius of $\odot O$, denoted as $R$. Since $O K \perp B C$, we have $O K \parallel A D$. Therefore,
$$
\begin{array}{l}
\frac{A I}{I K}=\frac{A D}{O K}=\frac{c \sin B}{R} \\
=2 ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,354 |
Example 7 When $x$ varies, the minimum value of the fraction $\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$
(1993, National Junior High School Competition) | Let $y=\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$. Then,
$$
\left(3-\frac{1}{2} y\right) x^{2}+(6-y) x+(5-y)=0 \text {. }
$$
Since $x$ is a real number, $\Delta \geqslant 0$, so,
$$
y^{2}-10 y+24 \leqslant 0 \text {. }
$$
Thus, $4 \leqslant y \leqslant 6$.
When $y=4$, $x=1$.
Therefore, when $x=1$, the minimum value... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,355 |
$$
\begin{array}{l}
a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, \\
b_{n} \in[1,2] \text { and } \sum_{i=1}^{n} a_{i}^{2}= \\
\sum_{i=1}^{n} b_{i}^{2} \text {. Prove: } \sum_{i=1}^{n} \frac{a_{i}^{3}}{b_{i}} \leqslant \frac{17}{10}
\end{array}
$$
- $\sum_{i=1}^{n} a_{i}^{2}$. And ask: When does the equality hold? | Given $a_{i}, b_{i} \in [1,2], i=1,2, \cdots, n$, therefore,
$$
\frac{1}{2} \leqslant \frac{\sqrt{\frac{a_{i}^{3}}{b_{i}}}}{\sqrt{a_{i} b_{i}}}=\frac{a_{i}}{b_{i}} \leqslant 2 \text {. }
$$
From this, $\left(\frac{1}{2} \sqrt{a_{i} b_{i}}-\sqrt{\frac{a_{i}^{3}}{b_{i}}}\right)\left(2 \sqrt{a_{i} b_{i}}-\sqrt{\frac{a_{i... | \sum_{i=1}^{n} \frac{a_{i}^{3}}{b_{i}} \leqslant \frac{17}{10} \sum_{i=1}^{n} a_{i}^{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 710,356 |
Three, 50 points 5) For positive integers $a, n$, define $F_{a}(a)=q+r$, where $q, r$ are non-negative integers, $a=$ $q n+r$, and $0 \leqslant r<n$. Find the largest positive integer $A$ such that there exist positive integers $n_{1}, n_{2}, n_{3}, n_{4}, n_{5}, n_{6}$, for any positive integer $a \leqslant A$, we hav... | Three, the maximum positive integer $B$ that satisfies the condition "there exist positive integers $n_{1}, n_{2}, \ldots, n_{k}$, such that for any positive integer $a \leqslant B$,
$$
F_{n_{k}}\left(F_{\pi_{k-1}}\left(\cdots\left(F_{n_{1}}(a)\right) \cdots\right)\right) \doteq 1^{\prime}
$$
is denoted as $x_{k}$. Cle... | 53590 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,357 |
8. Take four distinct points $A, B, C, D$ on the circle $\Gamma$, such that $\angle BCD$ is not a right angle. Prove:
(a) The perpendicular bisectors of $AB$ and $AC$ intersect the line $AD$ at points $W$ and $V$, respectively, and the lines $CV$ and $BW$ intersect at a point $T$;
(b) The length of one of the segments ... | Wei Ming: (a) Let the perpendicular bisectors of $AB$ and $AC$ be $b$ and $c$, respectively. Assume that line $b$ and $AD$ do not intersect, then they are parallel and $\angle DAB = 90^{\circ}$. Points $D$ and $B$ are endpoints of a certain diameter, and $\angle DCB = 50^{\circ}$, which contradicts the known informatio... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,358 |
$(i=1,2,3)$ are the indices of the small circles (tangent) passing through $I$ and tangent to $A_{i} A_{i+1}$ and $A_{i} A_{i+2}$ (indices taken modulo 3), $B_{i}(i=1,2,3)$ are the other intersection points of circles $C_{i+1}$ and $C_{i+2}$. Prove: The circumcenters of $\triangle A_{1} B_{1} I$, $\triangle A_{2} B_{2}... | Proof 1: Since $\triangle A_{1} A_{2} A_{3}$ is a non-equilateral triangle, it is easy to see that the circumcenters of $\triangle A_{1} B_{1} I$, $\triangle A_{2} B_{2} I$, and $\triangle A_{3} B_{3} I$ can all be defined.
We first look at the following lemma.
Lemma: Let the incenter of $\triangle ABC$ be $I$, and $T$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,359 |
11. Let $P(x)$ be a polynomial with real coefficients, and for all $x \geqslant 0$, $P(x)>0$. Prove that there exists a positive integer $n$, such that $(1+x)^{n} P(x)$ is a polynomial with non-negative coefficients. | Proof: Since $P(x)$ cannot have positive roots, all its real roots (if any) must be negative. Let them be denoted as $-a_{1}, -a_{2}, \cdots, -a_{k}$. It follows that $P(x)$ has a factorization of the form
\[
\begin{aligned}
P(x)= & c\left(x+a_{1}\right) \cdots\left(x+a_{k}\right)\left(x^{2}-p_{1} x+a_{1}\right) \\
& \... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,361 |
12. Let $p$ be a prime, $f(x)$ be a $d$-degree polynomial with integer coefficients and satisfy:
(i) $f(0)=0, f(1)=1$;
(ii) for any positive integer $n$, the remainder when $f(n)$ is divided by $p$ is either 0 or 1.
Prove: $d \geqslant p-1$. | Proof 1: Using proof by contradiction:
Assume $d \leqslant p-2$, then the polynomial $f(x)$ is completely determined by its values at $0,1, \cdots, p-2$. By the Lagrange interpolation formula, for any $x$, we have
$$
\begin{array}{l}
f(x)= \\
\sum_{k=1}^{p-2} f(k) \frac{x(x-1) \cdots(x-k+1)(x-k-1) \cdots(x-p+2)}{k!(-1)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,362 |
13. City $A$ has $n$ girls and $n$ boys, and each girl knows all the boys. City $B$ has $n$ girls $g_{1}, g_{2}, \cdots, g_{n}$ and $2 n-1$ boys $b_{1}, b_{2}, \cdots, b_{2 n-1}$. Girl $g_{i}$ (where $i=1,2, \cdots, n$) knows boys $b_{1}, b_{2}, \cdots, b_{2 n-1}$ but does not know any other boys. For any $r=1,2, \cdot... | Prove: Let $A(r)$ and $B(r)$ be denoted as $A(n, r)$ and $B(n, r)$, respectively. The sequence $A(n, r)$ can be directly found. We have
$$
A(n, r)=C_{n}^{r} \frac{n!}{(n-r)!}, \quad r=1,2, \cdots, n.
$$
In fact, selecting $r$ girls from $n$ girls in city $A$ can be done in $C_{n}^{r}$ ways, and selecting $r$ boys from... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,363 |
14. Let $b, m, n$ be positive integers, satisfying $b>1$ and $m \neq n$. Prove: if $b^{m}-1$ and $b^{n}-1$ have the same set of divisors, then $b+1$ is a power of 2. | Proof: For any positive integers $x$ and $y$, we use $x-y$ to denote that $x$ and $y$ have the same prime divisors.
Since $b^{m}-1 \sim b^{n}-1$, any prime that divides $b^{m}-1$ also divides $b^{n}-1$, and thus also divides $\left(b^{m}-1, b^{n}-1\right)$, where $(x, y)$ denotes the greatest common divisor of $x$ and... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,364 |
1. Let $0<a<b<c$, real numbers $x, y$ satisfy $2 x+$ $2 y=a+b+c, 2 x y=a c$. Then the range of $x, y$ is ( ).
(A) $a<x<b, 0<y<a$
(B) $b<x<c, 0<y<a$
(C) $0<x<a, a<y<c$
(D) $a<x<c, b<y<c$ | ,$- 1 . C$.
It can be known that $x, y$ are the roots of the equation $f(t)=t^{2}-\frac{1}{2}(a+b+c) t+$ $\frac{1}{2} a c=0$, and $f(0)=\frac{1}{2} a c>0, f(a)=\frac{1}{2} a(a$ $-b)0$, so the two roots $x, y$ of the equation $f(t)=0$ satisfy $0<x<a, a<y<c$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,365 |
2. In acute $\triangle A B C$, $B E$ and $C F$ are altitudes on sides $A C$ and $A B$, respectively. Then $S_{\triangle A E F}: S_{\triangle A B C}=(\quad$.
(A) $\sin ^{2} A$
(B) $\cos ^{2} A$
(C) $\operatorname{tg}^{2} \hat{A}$
(D) ctg $A$ | 2. B.
In the right triangle $\triangle A B E$,
$$
\begin{aligned}
\cos A & =\frac{A E}{A B}, \text { in the right triangle } \triangle A C F, \\
\cos A & =\frac{A F}{A C}, \\
& \therefore \frac{S_{\triangle A E F}}{S_{\triangle A B C}} \\
& =\frac{\frac{1}{2} A E \cdot A F \sin A}{\frac{1}{2} A B \cdot A C \sin A} \\
... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,367 |
3. Among the following four propositions:
(1) Two triangles are congruent if they have two sides and one angle correspondingly equal;
(2) Two triangles are congruent if they have two angles and one side correspondingly equal;
(3) Two triangles are congruent if they have the same perimeter and area;
(4) Two triangles ar... | 3. A.
Obviously, (1) is not true, (2) is true. (3) is not true, for example, in $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$, $A B=A C=29, B C=40 ; A^{\prime} B^{\prime}=A^{\prime} C^{\prime}=37, B^{\prime} C^{\prime}=24$, the height $A D=21, A^{\prime} D^{\prime}=35$, the perimeters of both tria... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,368 |
4. In $\triangle A B C$, $A D$ and $C F$ intersect at $E$, $D$ is on $B C$, $F$ is on $A B$, and $A E \cdot B F=2 A F \cdot D E$. Then $A D$ must be ( ).
(A) the median on side $B C$
(B) the altitude on side $B C$
(C) the angle bisector of $\angle B A C$
(D) the diameter of the circumcircle of $\triangle A C F$ | 4. A.
As shown in Figure 2, from the given conditions, we have $\frac{A E}{E D} = \frac{2 A F}{F B}$. And $\frac{A E}{E D} = \frac{A F}{D G}$, so $\frac{D G}{F B} = \frac{1}{2}$. Also, $\frac{C D}{B C} = \frac{D G}{F B}$, which means $C D = \frac{1}{2} B C$. Therefore, $D$ is the midpoint of $B C$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,369 |
5. Given that $n$ is the product of two distinct prime numbers. Then the number of positive integer solutions $(x, y)$ to the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$ is ( ).
(A) 1
(B) 4
(C) 9
(D) infinitely many | 5.C.
It is known that $x>n, y>n$. Let $x=n+t_{1}, y=n+t_{2}\left(t_{1}\right.$, $t_{2}$ be natural numbers). The original equation becomes $\frac{1}{n+t_{1}}+\frac{1}{n+t_{2}}=\frac{1}{n}$.
Simplifying, we get $t_{1} t_{2}=n^{2}$.
Let $n=a b(a, b$ both be prime numbers, and $a \neq b)$, then
$$
\begin{array}{l}
t_{1}... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 710,370 |
6. The lengths of the three altitudes of a triangle are $3, 4, 5$. When all three sides are the smallest integers, the length of the shortest side is ( ).
(A) 60
(B) 12
(C) 15
(D) 20 | 6. B.
Let the heights from $A$, $B$, and $C$ to the opposite sides $BC$, $CA$, and $AB$ of $\triangle ABC$ be $3$, $4$, and $5$, respectively. Then,
$$
2 S_{\triangle A I C}=3 BC=4 AC=5 AB \text{. }
$$
It is known that $AB$ is the shortest side.
Given $BC=\frac{5}{3} AB$, $AC=\frac{5}{4} AB$, $AB$ should be the least... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,371 |
1. For the polynomial in $x$
$$
x^{3}-g x^{2}+(g+m) x+6 m^{2}+6 m-8
$$
to be divisible by $x-2$, and when $m$ is a certain real number, $g$ has a minimum value. Then, this polynomial should be $\qquad$ . | $=, x^{3}-\frac{4}{3} x^{2}+\frac{2}{3} x-\frac{28}{3}$.
The remainder when the given polynomial is divided by $x-2$ is
$$
2(4+m-g)+\left(6 m^{2}+6 m-8\right) \text {. }
$$
Since it can be divided exactly, the above expression is complex, i.e., $g=3 m^{2}+4 m$. When $m$ $=-\frac{2}{3}$, $g_{\text {min }}=\frac{4}{3}$,... | x^{3}-\frac{4}{3} x^{2}+\frac{2}{3} x-\frac{28}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,372 |
2. In $\triangle A B C$, segment $A D$ intersects $B C$ at $D$, and $A D$ divides $\triangle A B C$ into two similar triangles, with a similarity ratio of $\sqrt{3}: 3$. Then the interior angles of $\triangle A B C$ are $\qquad$ . | 2. $\angle B A C=90^{\circ}, \angle B=30^{\circ}, \angle C=60^{\circ}$.
As shown in Figure 3, since $\angle A D C \neq$
$$
\begin{array}{l}
\angle B, \angle A D C \neq \angle D A B, \\
\angle A D B \neq \angle C, \angle A D B \neq \\
\angle D A C \text { (these four inequalities can only be } \\
\text { greater), and ... | \angle B A C=90^{\circ}, \angle B=30^{\circ}, \angle C=60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,373 |
3. Given $\sqrt{x}+\sqrt{y}=35, \sqrt[3]{x}+\sqrt[3]{y}=13$. Then $x+y=$ $\qquad$ | 3.793 .
$$
\begin{aligned}
x+y= & (\sqrt[3]{x}+\sqrt[3]{y})\left[(\sqrt[3]{x}+\sqrt[3]{y})^{2}-3 \sqrt[3]{x y}\right] \\
= & 13\left[13^{2}-3 \sqrt[3]{x y}\right], \\
& (\sqrt{x}+\sqrt{y})^{2}=x+y+2 \sqrt{x y}=35^{2}, \\
& \therefore 35^{2}-2 \sqrt{x y}=13^{3}-39 \sqrt[3]{x y},
\end{aligned}
$$
i.e., $2 \sqrt{x y}-39 ... | 793 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,374 |
4. In $\triangle A B C$, $\angle A B C=90^{\circ}, A C=\sqrt[3]{2}, D$ is a point on the extension of $A C$, $C D=A B=1$. Then the degree measure of $\angle C B D$ is $\qquad$ | $4.30^{\circ}$.
As shown in Figure 4, draw $D E / / B C$ intersecting the extension of $A B$ at $E$, then $\angle C B D=\angle B D E$.
From $\frac{A C}{C D}=\frac{A B}{B E}$ we get
$$
\begin{array}{l}
B E=\frac{1}{\sqrt[3]{2}}=\frac{\sqrt{4}}{2} . \\
\begin{array}{l}
\left.\therefore D E^{2}=A i\right)-A E^{2} \\
=(\sq... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,375 |
One. (20 points) The two roots of the equation $x^{2}-2 \sqrt{2} x+1=0$ are $\alpha$ and $\beta$. Find the quadratic function $f(x)$ that satisfies $f(\alpha)=$ $\beta, f(\beta)=\alpha, f(1)=1$. | Given $\alpha+\beta=2 \sqrt{2}, \alpha \beta=1(\alpha \neq \beta)$.
Let $f(x)=a x^{2}+b x+c$. From the conditions, we have
$$
\begin{array}{l}
a \alpha^{2}+b \alpha+c=\beta, \\
a \beta^{2}+b \beta+c=\alpha, \\
a+b+c=1 .
\end{array}
$$
Adding (1) and (2) and applying $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \be... | f(x)=x^{2}-(2 \sqrt{2}+1) x+2 \sqrt{2}+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,376 |
For example, the minimum value of $9 \sqrt{x^{2}-4 x+8}+\sqrt{x^{2}+2 x+2}$ is $\qquad$ | Solution: The original expression $=\sqrt{(x-2)^{2}+2^{2}}+$ $\sqrt{(x+1)^{2}+(-1)^{2}}$, which represents the sum of the distances from point $(x, 0)$ to points $A(2,2)$ and $B(-1,1)$. From the previous problem, we know that the required minimum value is
$$
\sqrt{(2+1)^{2}+(2+1)^{2}}=3 \sqrt{2} \text {. }
$$ | 3 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,377 |
II. (25 points) In $\triangle A B C$, $A D$ is the angle bisector, and $\frac{1}{A D}=\frac{1}{A B}+\frac{1}{A C}$. Find the measure of $\angle B A C$. | As shown in Figure 5, draw $B E \parallel A D$ intersecting the extension of $C A$ at $E$.
From $\frac{1}{A D}=\frac{1}{A B}+\frac{1}{A C}$
$=\frac{A B+A C}{A B \cdot A C}$, we get
$A B+A C=\frac{A B \cdot A C}{A D}$.
(1)
From $\frac{B D}{D C}=\frac{A B}{A C}$, we get
$\frac{B C}{D C}=\frac{A B + A C}{A C}$.
From (1) a... | 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,378 |
Three. (25 points) Does there exist a four-digit number $\overline{a b c d}$, such that the last four digits of its square are also $\overline{a b c d}$? Does there exist a five-digit number $\overline{a b c d e}$, such that the last five digits of its square are also $\overline{a b c d e}$? If they exist, find all of ... | Let $\overline{a b c d}=x$. Since $x^{2}$ and $x$ have the same last four digits, the last four digits of $x^{2}-x$ are 0000, i.e., $10000 \mid x^{2}-x$, or equivalently, $2^{4} \cdot 5^{4} \mid x(x-1)$.
(i) If $2^{4} \mid x$ and $5^{4} \mid x-1$, then $x-1$ is a four-digit odd number and a multiple of 625. Therefore, ... | 9376 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,379 |
1. $|b| \leqslant 1$ is the ( ) for the inequality $a \cos x + b \cos 3 x > 1$ to have no solution with respect to $x$.
(A) necessary but not sufficient condition
(B) sufficient but not necessary condition
(C) necessary and sufficient condition
(D) neither sufficient nor necessary condition | ,- 1 (A). $+b \alpha \cos 3 x \leqslant 1$ holds for all real numbers $x$.
By substituting $x=0, x$ into $\left\{\begin{array}{l}a+b \leqslant 1, \\ -a-b \leqslant 1\end{array}\right.$, we get $-1 \leqslant a+b \leqslant 1$.
By substituting $x=\frac{\pi}{3}, \frac{2 \pi}{3}$ into $\left\{\begin{array}{l}\frac{a}{2}-b \... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 710,380 |
2. In the triangular pyramid $S$ $A B C$, $S A \perp$ plane $A B C, A B \perp A C, S A=$ $A B, S B=$ i.,$E$ is the midpoint of $S C$, $D \perp \perp S C$ intersects $A C$ at $D$. Then the degree of the dihedral angle $E-D B-C$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$ | 2. (C).
Let $S A=A B=1$, then $S B=B C=\sqrt{2}$.
Since $S B \perp B C, S C=\sqrt{2} \cdot \sqrt{2}=2, \angle S C A=30^{\circ}$, $\angle E D C=60^{\circ}$, it is easy to prove that $\angle E D C$ is the dihedral angle of $E-D B-C$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,381 |
4. $a, b, c \in R^{+}$. Then the minimum value of $f(x)=\sqrt{x^{2}+a}+$ $\sqrt{(c-x)^{2}+b}$ is ( ).
(A) $\sqrt{a}+\sqrt{c^{2}+b}$
(B) $\sqrt{c^{2}+a}+\sqrt{b}$
(C) $\frac{\sqrt{2}}{2} c+\sqrt{a}+\sqrt{b}$
(D) $\sqrt{c^{2}+(\sqrt{a}+\sqrt{b})^{2}}$ | 4. (D).
$$
\begin{aligned}
f(x) & =\sqrt{(x-0)^{2}+(0+\sqrt{a})^{2}} \\
& +\sqrt{(x-c)^{2}+(0-\sqrt{b})^{2}},
\end{aligned}
$$
This can be seen as the sum of the distances from a moving point $(x, 0)$ on the $x$-axis to two fixed points $A(0,-\sqrt{a})$ and $B(c, \sqrt{b})$. Its minimum value is
$$
\begin{aligned}
|A ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,383 |
$6 .(a, b)$ represents the greatest common divisor of the natural numbers $a, b$. Let $(a, b)=1$, then $\left(a^{2}+b^{2}, a^{3}+b^{3}\right)$ is ( ).
(A) 1
(b, 2
(C) only 1 or 2
(D) possibly greater than 2 | 6. (C).
According to the Euclidean algorithm: If $a=b q+r$, then $(a, b)=(b; r)$. Since $a^{3}+b^{3}=\left(a^{2}+b^{2}\right)(a b)-a b(a+b)$, we have $\left(a^{3}+b^{3}, a^{2}+b^{2}\right)=\left(a^{2}+b^{2}, a b\left(a+b\right)\right)$; given $(a, b)=1$. Therefore,
$$
\begin{array}{l}
\left(a^{2}+b^{2}, a b(a+b)\right... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 710,385 |
Sure, here is the translated text:
```
II. Fill-in-the-Blanks (Total 54 points, 9 points per question)
1. If $x^{2}-x+2$ is a factor of $a x^{9}+b x^{8}+1$, then $a=$ $\qquad$ .
``` | Ni. $1 . a=-\frac{3}{256}$.
Let the roots of $x^{2}-x+2=0$ be $x_{1} 、 x_{2}\left(x_{1}+x_{2}=1\right.$. $x_{1} x_{2}=2$ )
From $\left\{\begin{array}{l}a x_{1}^{9}+b x_{1}^{8}+1=0, \\ a x_{2}^{9}+b x_{2}^{8}+1=0\end{array}\right.$
Eliminating $b$ gives
$$
\begin{array}{l}
a x_{1}^{8} x_{2}^{8}\left(x_{1}-x_{2}\right)=... | -\frac{3}{256} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,386 |
2. Four cities each send 3 political advisors to participate in $k$ group inspection activities (each advisor can participate in several groups), with the rules: (1) advisors from the same city are not in the same group; (2) any two advisors from different cities exactly participate in one activity together. Then the m... | 2.9 Group.
First, consider the CPPCC members from two cities, Jia and Yi. Let the members from Jia city be $A_{1}, A_{2}, A_{3}$, and the members from Yi city be $B_{1}, B_{2}, B_{3}$. They can form 9 pairs: $A_{1} B_{1}, A_{1} B_{2}, A_{1} B_{3}, A_{2} B_{1}, A_{2} B_{2}, A_{2} B_{3}$, $A_{3} B_{1}, A_{3} B_{2}, A_{3... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,387 |
Example 10 When $a$ takes all real values from 0 to 5, the number of integer $b$ that satisfies $3 b=a(3 a-8)$ is $\qquad$
(1997, National Junior High School Competition) | Analysis: From $3 b=a(3 a-8)$, we have $b=a^{2}-$ $\frac{8}{3} a$. This is a quadratic function, and its graph is a parabola. When $a$ takes all real numbers from 0 to 5, finding the number of integer $b$ is equivalent to finding the number of integers between the maximum and minimum values of $b$.
Solution: First, dr... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,388 |
$\begin{array}{r}\text { 4. Let }\left(x^{1997}+x^{1999}+3\right)^{1998}=a_{0}+a_{1} x+ \\ a_{2} x^{2}+\cdots+a_{n} x^{n} \text {. Then } a_{0}-\frac{a_{1}}{2}-\frac{a_{2}}{2}+a_{3}-\frac{a_{4}}{2} \\ -\frac{a_{5}}{2}+\cdots+a_{3 k}-\frac{a_{3 k+1}}{2}-\frac{a_{3 k+2}}{2}+\cdots+a_{n}=\end{array}$ | $4.2^{1998}$
Let $x=\frac{-1+\sqrt{3} i}{2}=w$, then $x^{3 k}=1, x^{3 k+1}=w$, $x^{3 k+2}=w^{2}, 1+w+w^{2}=0$. Substitute $x=w$ into the original expression,
$$
\begin{aligned}
\text { Left side } & =\left(w^{2}+w+3\right)^{1998}=\left(w^{2}+w+1+2\right)^{1998} \\
& =2^{1998},
\end{aligned}
$$
$$
\begin{aligned}
\text ... | 2^{1998} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,390 |
5. For a real number $a>1$, written as a reduced fraction $a=\frac{q}{p}$, $(p, q)=1$. The number of values of $a$ that satisfy $p q=30!$ is $\qquad$.
| 5.512.
30! has 10 prime factors, $\left(30!=2^{26} \times 3^{14} \times 5^{7} \times 7^{7} \times 11^{2} \times 13^{2} \times 17 \times 19 \times 23 \times 29\right),(p, q)=$ 1. Each prime factor $p_{i}^{\circ}$ is either all in the numerator or all in the denominator, which gives $2^{10}$ cases in total, with those gr... | 512 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,391 |
6. In $\triangle A B C$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given $a^{2}+b^{2}-7 c^{2}=0$, then $\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=$ | 6.3 .
$$
\begin{array}{l}
\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}=\frac{\cos C \sin A \sin B}{\sin (A+B) \sin C} \\
=\cos C \cdot \frac{\sin A \sin B}{\sin ^{2} C}=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \cdot \frac{a b}{c^... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,392 |
Three. (Full marks 20 points) Let real numbers $x^{2}+y^{2} \leqslant 5$. Find the maximum and minimum values of $f(x, y)=3|x+y|+|4 y+9|+|7 y-3 x-18|$. | Three, let $x=r \cos \theta, y=r \sin \theta(r>0,0 \leqslant \theta < 2\pi)$.
$$
\begin{array}{l}
7 y-3 x-18=7 r \sin \theta-3 r \cos \theta-18 \\
=r \sqrt{7^{2}+3^{2}} \sin (\theta-\alpha)-18 \leqslant \sqrt{5} \times \sqrt{58}-18 \\
=\sqrt{290}-18<0 . \\
\begin{aligned}
\therefore f(x, y)= & 3|x+y|+4 y+9-7 y+3 x+18 \... | f(x, y)_{\text {max }}=27+6 \sqrt{5}, \quad f(x, y)_{\text {min }}=27-3 \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,393 |
Four. (Full marks 20 points) Given point $A(1,1)$ and line $l: x=3$ in the coordinate plane, let the moving point $P$ have a distance $m$ to point $A$ and a distance $n$ to line $l$. If $m+n=4$,
(1) Find the equation of the trajectory of point $P$ and draw the graph;
(2) Draw a line $s$ through $A$ with an inclination ... | (1) Let $P(x, y)$, then
$$
m=\sqrt{(x-1)^{2}+(y-1)^{2}}, n=|x-3| \text {. }
$$
From the condition $m+n=4$ we have
$$
\sqrt{(x-1)^{2}+(y-1)^{2}}+|x-3|=4 \text {. }
$$
It is easy to see that $0 \leqslant x \leqslant 4$ (otherwise $m+n>4$ ).
When $0 \leqslant x \leqslant 3$, the equation is
$$
(x-1)^{2}+(y-1)^{2}=(x+1)^... | \frac{16}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,394 |
Five. (Full marks 20 points) $n$ is a natural number, $r>0$ is a real number. Prove: The equation $x^{n+1}+r x^{n}-r^{n+1}=0$ has no complex root with modulus $r$.
---
The translation maintains the original format and line breaks as requested. | Five, as shown in Figure 5, assuming $|z|=r$ is a root of the equation, then we have
$$
z^{n}(z+r)=r^{n+1} \text {. }
$$
Taking the modulus, $r^{n}|z+r|=r^{n+1} \Rightarrow|z+r|=r$.
It is known that $z$ is the intersection point of the two circles $|z|=r$ and $|z+r|=r$. It is easy to see that
$$
\begin{array}{l}
z_{1}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,395 |
一、(Full marks 50 points) (1) Find natural numbers $x, y$ such that $\frac{6}{1997}=\frac{1}{x}+\frac{1}{y}$.
(2) How many natural numbers $n(n<1997)$ can make $\frac{n}{1097}=\frac{1}{x}+\frac{1}{y} ?(x, y$ are natural numbers $)$ | (1) From $\frac{6}{1997}=\frac{x+y}{x y}$, we have $6 x y=1997(x+y)$.
Since $1997$ is a prime number,
$\therefore 6 \mid x+y$.
Let $\left\{\begin{array}{l}x+y=6 k^{2}, \\ x y=1997 k^{2}\end{array}\right.$.
To find $k$, we can take $x=k, y=1997 k$
Substituting into (1) gives $k+1997 k=6 k^{2}, 1998=6 k$.
Thus, $k=333$.
... | 333, 1997 \times 333 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,396 |
II. (Full marks 50 points) In Figure 3,
$\odot O$ has a radius of $R, \odot O_{1}$ has a radius of $r(R > r)$, the two circles are externally tangent at $A, O D$ is tangent to $\odot O_{1}$ at $D, O_{1} E$ is tangent to $\odot O$ at $E, B$ and $C$ are the midpoints of $O E$ and $O_{1} D$ respectively.
(1) Prove: $B, A,... | $\therefore$ (i) Let $\angle O A B=\angle 1, \angle O_{1} A C=\angle 2$. The necessary and sufficient condition for points $B$, $A$, and $C$ to be collinear is $\angle 1=\angle 2$, i.e., $\sin \angle 1=\sin \angle 2$. We will now prove that this is not true.
By the Law of Sines, $\frac{\sin \angle 1}{O B}=\frac{\sin \a... | \frac{24 \sqrt{10}-15}{41} | Geometry | proof | Yes | Yes | cn_contest | false | 710,397 |
Three. (Full marks 50 points) (1) Towns $A$ and $B$ on a highway are 5 kilometers apart. Each town has two side roads forming a $>$ < shape. It is planned to build one gas station on each side road. The requirement is that the distance from each station to towns $A$ and $B$, and to other stations (via the highway throu... | Three, (1) Two pairs
of road conditions can be realized.
As shown in Figure 6:
There are $C_{6}^{?}=15$ different distances,
all of which are distinct.
They are $1,2,3, \cdots, 15$ kilometers.
(2) The three-way road situation
cannot be realized. Assume
in Figure 7, the six stations together need
to meet the requiremen... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,398 |
Example 1 If the simplest radical expressions $\sqrt[3 x+y]{2 x-y}$ and $\sqrt[y+6]{4 x+y-2}$ are of the same degree, and $y$ is an even number, then the sum of all possible values of $y$ is ( ).
(A) 0
(B) 2
(C) 4
(D) 6 | Solution: By the definition of like radicals, we have
$3 x+y=y+6$, which gives $x=2$.
$\because$ The root index $6+y$ is even,
$\therefore$ the radicand $4-y$ and $6+y$ are both non-negative, i.e., $4-y \geqslant 0$, and $6+y \geqslant 0$.
Solving this, we get $-6 \leqslant y \leqslant 4$.
Thus, the even number $y=-6,-... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,399 |
Example 2: Given $x=\sqrt[3]{a+\sqrt{a^{2}+b^{3}}}-$ $\sqrt[y]{\sqrt{a^{2}}+b^{3}}-a$. Prove: $x^{3}+3 b x=2 a$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 2: Given $x=\sqrt[3]{a+\sqrt{a^{2}+b^{3}}}-$ $\sqrt[y]{\sqrt{a^{2}}+b^{3}}-a$. Prove: $x^{3}+3 b x=2 a$. | $$
\begin{array}{c}
x^{3}=a+\sqrt{a^{2}+b^{3}}-\left(\sqrt{a^{2}+b^{3}}-a\right)- \\
3 b \cdot\left(\sqrt[3]{a+\sqrt{a^{2}+b^{3}}}-\sqrt[3]{\sqrt{a^{2}+b^{3}}-a}\right) .
\end{array}
$$
Prove: Cubing both sides of the known condition, we get
$$
\begin{array}{c}
x^{3}=a+\sqrt{a^{2}+b^{3}}-\left(\sqrt{a^{2}+b^{3}}-a\rig... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,400 |
Example 11 Let $x>0, y>0, \sqrt{x}(\sqrt{x}+2 \sqrt{y})$ $=\sqrt{y}(6 \sqrt{x}+5 \sqrt{y})$. Find the value of $\frac{x+\sqrt{x y}-y}{2 x+\sqrt{x y}+3 y}$. | Solution: From the given, we have
$$
(\sqrt{x})^{2}-4 \sqrt{x} \sqrt{y}-5(\sqrt{y})^{2}=0,
$$
which is $(\sqrt{x}-5 \sqrt{y})(\sqrt{x}+\sqrt{y})=0$.
$$
\begin{array}{l}
\because \sqrt{x}+\sqrt{y}>0, \\
\therefore \sqrt{x}-5 \sqrt{y}=0,
\end{array}
$$
which means $x=25 y$.
Substituting $x=25 y$ into the original fract... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,401 |
6. On the front and back of four cards, 0 and 1, 0 and 2, 3 and 4, 5 and 6 are written respectively. By placing any three of them side by side to form a three-digit number, a total of $\qquad$ different three-digit numbers can be obtained. | 6.124.
When the hundreds digit is 1 (or 2): $P_{3}^{2} \times 2 \times 2 \times 2=48$.
When the hundreds digit is 3 (or $4,5,6$), and the tens or units digit uses 0 pieces, there are $(05,06,50,60,00)$ 5 repetitions, so
$$
\left(P_{3}^{2} \times 2 \times 2-5\right) \times 4=76 .
$$
Therefore, the total number of diff... | 124 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,402 |
Three. (20 points) Prove: (1) For any $x$, at least one of the numbers $|\sin x|$ and $|\sin (x+1)|$ is greater than $\frac{1}{3}$;
(2) $\frac{|\sin 10|}{10}+\frac{|\sin 11|}{11}+\frac{|\sin 12|}{12}+\cdots+$ $\frac{|\sin 29|}{29}>\frac{1}{6}$. | Three, (1) Consider the unit circle and the lines $y=\frac{1}{3}$ and $y=-\frac{1}{3}$. Let $A$ and $B$ be the intersection points of the unit circle and the lines, as shown in Figure 7. It can be proven that $\angle AOB$ is less than 1 radian.
In fact,
$$
\begin{array}{l}
\sin \angle AOB = \sin \left(2 \arcsin \frac{1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,403 |
Four. (20 points) Through the edges $AI$ and $BC$ of the tetrahedron $ABCD$, a plane is made passing through the midpoints $K$ and $N$ of these edges, intersecting the edge $CD$ at point $M$ and the edge $AB$ at point $L$. Prove:
(1) $|DM|:|MC|=|AL|:|LB|$;
(2) Area $S_{\triangle KIN}=S_{\triangle KMN}$. | Four, using a simple lemma: If line segment $F G$ intersects plane $\alpha$ at point $H$, and $h_{F}$ and $h_{G}$ are the distances from points $F$ and $G$ to plane $\alpha$, respectively, then, $\frac{|F H|}{|H G|} = \frac{h_{F}}{h_{G}}$.
Let $h_{\mathrm{A}}, h_{B}, h_{\mathrm{C}}, h_{\mathrm{D}}$ be the distances fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,404 |
Five. (20 points) In the complex plane, there are three points: $c_{1}=a + b i, c_{2}=m + b i, c_{3}=a + n i$, where $a > m$, $n > b$, and $C_{1} C_{2} C_{3}$ (here $C_{i}$ represents the point corresponding to the complex number $c_{i}$) form a triangle. Prove: The complex number $z$ representing the point $Z$ that sa... | Five, by using translation, point $C_{1}$ can be moved to the origin $O$, and the entire triangle can be moved to the first quadrant (with $O S_{2}$ on the $x$-axis), becoming $\triangle O S_{2} S_{3}$. Clearly, this does not affect the essence of the problem.
Thus, $O=0, s_{2}=m-a=t, s_{3}=(n-b) i=f i$, $t, f \in \ma... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,405 |
In $\triangle C D F$, $\angle D=90^{\circ}$, $D O \perp C F$, $O$ is the foot of the perpendicular. A circle is drawn with $C$ as the center and $C D$ as the radius. $A A^{\prime}$ is a moving chord of circle $C$ passing through point $O$. $E$ is a point on the line $A^{\prime} A \perp$, and $E F \perp C F$. Prove: The... | As shown in Figure 10, draw perpendiculars $A B$ and $A^{\prime} B^{\prime}$ from $A$ and $A^{\prime}$ to $E F$, and let $A A^{\prime}$ intersect $E F$ at $E$. Then,
$$
\begin{array}{c}
A B=\frac{O F}{O E} \cdot O E, \\
A^{\prime} B^{\prime}=\frac{O F}{O E} \cdot A^{\prime} E .
\end{array}
$$
Thus, $A B \cdot A^{\prim... | \frac{2}{O F} | Geometry | proof | Yes | Yes | cn_contest | false | 710,406 |
$$
\begin{array}{c}
\text { II. (50 points) }(1) \text { For } 0 \leqslant x \leqslant 1, \text { find the range of the function } \\
h(x)=(\sqrt{1+x}+\sqrt{1-x}+2) . \\
\left(\sqrt{1-x^{2}}+1\right)
\end{array}
$$
(2) Prove: For $0 \leqslant x \leqslant 1$, there exists a positive number $\beta$ such that the inequal... | $$
\begin{array}{l}
=(1) 00$, the inequality $\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{a}}{\beta}(x \in[0,1])$ does not hold. Conversely, i.e., $-\frac{2 x^{2}}{h(x)} \leqslant-\frac{x^{0}}{\beta}$, which means $x^{2 \cdots} \geqslant \frac{h(x)}{2 \beta}$ holds.
Since $2-\alpha>0$, let $x \rightarrow 0$, we get
$$
0... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 710,407 |
Three. (50 points) (1) For three points $A_{1}\left(x_{1}, y_{1}\right)$, $A_{2}\left(x_{2}, y_{2}\right)$, $A_{3}\left(x_{3}, y_{3}\right)$ forming a triangle, with $x_{1}<x_{2}<x_{3}$. Prove: when $d$ is sufficiently small, the points $\left(x_{2}, y_{2}-d\right)$ and $\left(x_{2}, y_{2}+d\right)$, one is inside the ... | (1) Let $d\left(A_{i}, A_{j} A_{k}\right)$ denote the distance from point $A_{i}$ to the line $A_{j} A_{k}$. Take $d = \frac{1}{2} d\left(A_{2}, A_{1} A_{3}\right)$.
By contradiction. As shown in Figure 11, if points $Q_{1}$ and $Q_{2}$ are both outside the shape, then $\left|A_{2} Q_{i}\right| > \left|A_{2} E\right| ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,408 |
Given 71. $\odot O_{1} 、 \odot O_{2}$ have radii both $r, \odot O_{1}$ passes through two vertices $A 、 B$ of $\triangle A B C D$, $\odot O_{2}$ passes through vertices $B 、 C$, $M$ is another intersection point of $\odot O_{1} 、 \odot O_{2}$. Prove: The circumradius of $\triangle A M D$ is also $r$. | Proof: Construct $\square A B M N$, connect $N D, M C$, then quadrilateral $D C M N$ is also a parallelogram.
Since the radii of $\odot O_{1}$ and $\odot O_{2}$ are both $r$,
let $\angle B A M=\alpha, \angle B C M=\beta$.
Then $\alpha=\beta$.
By the construction,
$$
\begin{array}{l}
\angle A D N=\beta=\alpha \\
=\angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,409 |
For a constant $p \in N$, if the indeterminate equation $x^{2}+y^{2}=$ $p(x y-1)$ has positive integer solutions, prove that $p=5$ must hold. | Proof: Let $x_{0}, y_{0}$ be positive integer solutions of the equation. If $x_{0}=y_{0}$, substituting gives $p=(p-2) x_{0}^{2}$.
$$
\therefore x_{0}^{2}=\frac{p}{p-2}=1+\frac{2}{p-2} \in \mathbb{N} \text { : }
$$
Then $x_{0}^{2}=2$ or 3, which contradicts $x_{0} \in \mathbb{N}$.
Assume $x_{0}>y_{0} \geqslant 2$. Con... | 5 | Number Theory | proof | Yes | Yes | cn_contest | false | 710,410 |
71. Find all triples $(p, q, r)$ that satisfy the following conditions:
1) $p, q, r \in \mathbf{N}, p \geqslant q \geqslant r$;
2) At least two of $p, q, r$ are prime numbers;
3) $\frac{(p+q+r)^{2}}{p q r}$ is a positive integer. | Let $(a, b, c)$ be a solution that meets the problem's conditions, where $a$ and $b$ are prime numbers, and $a \geqslant b$. Using $\frac{(a+b+c)^{2}}{a b c} \in \mathbf{N}$ and the fact that $a$ and $b$ are prime, it is easy to deduce that $a|b+c, b| c+a, c \mid(a+b)^{2}$.
$1^{\circ}$ Suppose $a=b$.
In this case, from... | (3,3,3), (4,2,2), (12,3,3), (3,2,1), (25,3,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,411 |
Example 12 Let $m>0, n>0$, and real numbers $a, b, c, d$ satisfy $a+b+c+d=m, ac=bd=n^{2}$. Try to find the value of $\sqrt{(a+b)(b+c)(c+d)(d+a)}$. (Express the result in terms of $m, n$) | Solution: $\because a c=b d$,
$\therefore \frac{a}{b}=\frac{d}{c}=\frac{a+d}{b+c} \cdot(b+c \neq 0$, otherwise it leads to $m=0$ )
Thus, $\frac{a+b}{b}=\frac{m}{b+c}$,
which means $(a+b)(b+c)=b m$.
Similarly, $(c+d)(d+a)=d m$.
Therefore, the original expression $=\sqrt{b d m^{2}}=\sqrt{n^{2} m^{2}}=m n$. | mn | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,412 |
Let $P$ be any point inside $\triangle A B C$, and let $B C$, $C A$, and $A B$ be denoted as $a$, $b$, and $c$ respectively. Let $P A=x$, $B P=y$, and $P C=z$. Then
$$
\begin{array}{l}
a \cos A+b \cos B+c \cos C \\
\leqslant x \sin A+y \sin B+z \sin C .
\end{array}
$$ | Prove: Draw perpendiculars from point $P$ to the three sides, with the feet of the perpendiculars being $D$, $E$, and $F$. Let $\angle C A P = \alpha$. Then
$$
\begin{array}{l}
\cos B \cos \alpha + \cos C \cos (A - \alpha) \\
= -\cos (A + C) \cos \alpha \\
+ \cos C (\cos A \cos \alpha + \\
\sin A \sin \alpha) \\
= \sin... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,413 |
Example 13 Suppose the equation $\sqrt{a(x-a)}+\sqrt{a(y-a)}$ $=\sqrt{x-a}-\sqrt{a-y}$ holds in the real number range, where $a, x, y$ are distinct real numbers. Then the value of $\frac{3 x^{2}+x y-y^{2}}{x^{2}-x y+y^{2}}$ is ( ).
(A) 3
(B) $\frac{1}{3}$
(C) 2
(I) $\frac{5}{3}$. | Solution: From $x-a \geqslant 0$ and $a(x-a) \geqslant 0$ we know $a \geqslant 0$; from $a-y \geqslant 0$ and $a(y-a) \geqslant 0$ we know $a \leqslant 0$. Therefore, $a=0$,
and $\sqrt{x}-\sqrt{-y}=0$,
which means $x=-y \neq 0$.
Substituting the above equation into the original fraction, we get
the original expression ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,414 |
Example 14 Let positive integers $a, m, n$ satisfy $\sqrt{a^{2}-4 \sqrt{2}}=\sqrt{m}-\sqrt{n}$. Then the number of such values of $a, m, n$ is ( ).
(A) one set
(B) two sets
(C) more than two sets
(D) does not exist | Solving: Squaring both sides of the original equation, we get
$$
a^{2}-4 \sqrt{2}=m+n-2 \sqrt{m n} \text {. }
$$
Since $a$, $m$, and $n$ are positive integers, $a^{2}-4 \sqrt{2}$ and $\sqrt{m n}$ are irrational numbers. Therefore,
$$
\left\{\begin{array}{l}
\sqrt{m n}=\sqrt{8}, \\
m+n=a^{2},
\end{array}\right.
$$
Thu... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,415 |
Example 15 Find the minimum value of $\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Construct right triangles $\triangle P A C$ and $\triangle P B I$ as shown in Figure 1, such that
$$
\begin{array}{l}
A C=1, B D=2, P C \\
=x, C I=4, \text { and }
\end{array}
$$
$P C$ and $P D$ lie on line $l$. Then the problem of finding the minimum value is converted to "finding a point $I$ on line $l$ suc... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,416 |
Example 16 The greatest integer not exceeding $(\sqrt{7}+\sqrt{5})^{6}$ is $\qquad$ . | Solution: Let $a=\sqrt{7}+\sqrt{5}, b=\sqrt{7}-\sqrt{5}$. Then
$$
\begin{array}{l}
a+b=2 \sqrt{7}, a b=2 . \\
a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b) \\
=44 \sqrt{7}, \\
a^{6}+b^{6}=\left(a^{3}+b^{3}\right)^{2}-2(a b)^{3}=13536 . \\
\because 0<b<1, \\
\therefore 0<b^{6}<1 .
\end{array}
$$
Therefore, the integer part of $a^{6... | 13535 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,417 |
1. Given the simplest radical form $a \sqrt{2 a+b}$ and $^{a} \sqrt[b]{7}$ is a similar radical, then the values of $a$ and $b$ that satisfy the condition ( ).
(A) do not exist
(B) there is one set
(C) there are two sets
(D) more than two sets | Answer: B | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,418 |
3. When $x=\frac{1+\sqrt{1994}}{2}$, the value of the polynomial $\left(4 x^{3}-\right.$ $1997 x-1994)^{2001}$ is ( ).
(A) 1
(B) -1
(C) $2^{2001}$
(D) $-2^{2001}$ | Answer: B | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,419 |
Example 3 Solve the equation
$$
\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}(x+y+z) .
$$ | Solution: Completing the square for the original equation, we get
$$
\begin{array}{l}
(\sqrt{x}-1)^{2}+(\sqrt{y-1}-1)^{2} \\
+(\sqrt{z-2}-1)^{2}=0 .
\end{array}
$$
Thus, $\sqrt{x}-1=\sqrt{y-1}-1=\sqrt{z-2}-1$
$$
=0 \text {, }
$$
which means $x=1, y=2, z=3$. | x=1, y=2, z=3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,421 |
9. In the range of real numbers, let
$$
\begin{aligned}
x= & \left(\frac{\sqrt{(a-2)(|a|-1)}+\sqrt{(a-2)(1-|a|)}}{1+\frac{1}{1-a}}\right. \\
& \left.+\frac{5 a+1}{1-a}\right)^{1988} .
\end{aligned}
$$
Then the unit digit of $x$ is ( ).
(A) 1
(B) 2
(C) 4
(D) 6 | Answer: D | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,426 |
Example 1 Proof: 1992 divides
$$
1 \cdot 2 \cdot 3 \cdots \cdots \cdot 82 \cdot\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{82}\right) .
$$ | Proof: First note that $1992=3 \cdot 2^{3} \cdot 83$. Let the original expression be $A$, denoted as $A=82!\cdot \sum_{i=1}^{82} \frac{1}{i}$.
The power of 2 in $82!$ is $\left[\frac{82}{2}\right]+\left[\frac{82}{2^{2}}\right]+$
$$
\begin{array}{l}
{\left[\frac{82}{2^{3}}\right]+\left[\frac{82}{2^{4}}\right]+\left[\fra... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,428 |
Example 2 Proof: For any $m, n \in \mathbf{N}$,
$$
S_{m, n}=1+\sum_{k=1}^{m}(-1)^{k} \frac{(n+k+1)!}{n!(n+k)}
$$
can be divided by $m$!. But for some $m, n \in \mathbf{N}, S_{m, n}$ cannot be divided by $m!(n+1)$. | Proof: The form of the sum $S_{m, n}$ suggests that we should apply mathematical induction on the natural number variable $m$. We will prove that $S_{m, n} = (-1)^{m} \frac{(n+m)!}{n!}$.
When $m=1$,
$$
\begin{aligned}
S_{1, n} & =1-\frac{(n+2)!}{n!(n+1)}=1-(n+2) \\
& =-(n+1)=-\frac{(n+1)!}{n!} .
\end{aligned}
$$
Assu... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,429 |
Example 3: Does there exist a natural number $n$ such that the first 4 digits of $n!$ are 1993? | Solution: According to the decimal notation and the definition of factorial, we take $m=1000100000$. When $n=m, m+1, m+2, \cdots$, consider the change in the first 4 digits of $n!$.
If $m!=\overline{a b c d e \cdots}$, then $(m+1)!=m! \times 1000100001=\overline{a b c x \cdots}$, where $x=d$ or $d+1$. It is easy to se... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,430 |
Example 4 Define a positive integer $n$ to be a "tail" of a factorial if there exists a positive integer $m$ such that the decimal representation of $m$! ends with exactly $n$ zeros. How many positive integers less than 1992 are not tails of a factorial? | Solution: Clearly, the number of trailing zeros in $m!$ = the number of times 10 is a factor in $m!$ = the number of times 5 is a factor in $m!$.
Thus, we have $f(m)=\sum_{k=1}^{+\infty}\left[\frac{m}{5^{k}}\right]$.
From 1 CुS $1=\sum_{k=1}^{\infty}\left[\frac{m}{5^{k}}\right]7964$.
It is also easy to see that when $5... | 396 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,431 |
Example 4 If the fractional parts of $9+\sqrt{13}$ and $9-\sqrt{13}$ are $a$ and $b$ respectively, then $a b-4 a+3 b-2=$ $\qquad$ | Solution: $\because 3<\sqrt{13}<4$,
$\therefore 9+\sqrt{13}$ has an integer part of 12, and a decimal part
$$
\begin{array}{l}
a=\sqrt{13}-3 . \\
\because-4<-\sqrt{13}<-3,
\end{array}
$$
i.e., $0<4-\sqrt{13}<1$,
$\therefore 9-\sqrt{13}$ has an integer part of 5, and a decimal part $b$ $=4-\sqrt{13}$.
Thus, $a b-4 a+3... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,432 |
Example 5 Find all positive integers $w, x, y, z$ that satisfy $w!=x!+y!+z!$. | Solution: Without loss of generality, let $w>x \geqslant y \geqslant z$.
If $y>z$, then dividing both sides of the equation by $z!$ yields
$$
\begin{array}{l}
w(u-1) \cdots(z+1) \\
=x \cdots(z+1)+y \cdots(z+1)+1 .
\end{array}
$$
Here, $z+1>1$ can divide the left side of the above equation, but cannot divide the right ... | x=y=z=2, w=3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,433 |
Example 6 Find all positive integers $x, y'$, satisfying
$$
1!+2!+3!+\cdots+x!=y^{2} \text {. }
$$ | Solution: Since when $x \geqslant 5$, $5!+\cdots+x!=0$ $(\bmod 5)$, and $1!+2!+3!+4!=33=3$ $(\bmod 5)$, hence at this time $1!+2!+3!+\cdots+x$ ! $=3(\bmod 5)$. However, for any $y \in Z^{1}, y^{2} \equiv 0$ or $1$ or $4(\bmod 5)$. Therefore, when $x \geqslant 5$, the equation has no positive integer solutions.
If $x<5$... | x=1, y=1 \text{ and } x=3, y=3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,434 |
Example 7 Assume the sum of positive integers $a_{1}, a_{2}, \cdots, a_{n}$ is less than an integer $k$, prove that
$$
a_{1}!a_{2}!\cdots a_{n}!<k!\text {. }
$$ | Proof: Suppose we have $a_{1}$ white balls, $a_{2}$ black balls, $\cdots$,
$a_{n}$ red balls arranged in a row, then there are
$$
r=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)!}{a_{1}!a_{2}!\cdots a_{n}!}
$$
different arrangements. Since $r \geqslant 1$, we get
$$
\begin{array}{l}
a_{1}!a_{2}!\cdots a_{n}!\leqslant\le... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,435 |
Let $\mathbf{S}$ be $n \in \mathbf{N}, n>3$. Prove:
$$
\begin{aligned}
\frac{100}{120}<1 & -\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}+\cdots \\
& \quad+(-1)^{n+1} \frac{1}{(2 n-1)!}<\frac{101}{120} .
\end{aligned}
$$ | Prove: Let the middle expression be $a_{n}$. Then
$$
\begin{aligned}
a_{n}= & \left(1-\frac{1}{3!}\right)+\left(\frac{1}{5!}-\frac{1}{7!}\right) \\
& +\left(\frac{1}{9!}-\frac{1}{11!}\right)+\cdots>1-\frac{1}{3!}=\frac{100}{120} .
\end{aligned}
$$
(If $n$ is even, then $a_{n}$ can be divided into $\frac{n}{2}$ groups; ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,436 |
Example 9 Divide 1998 into 5 positive integers $a_{i}(i=1,2, \cdots, 5)$, so that $a_{1}!\cdot a_{2}!\cdot a_{3}!\cdot a_{4}!\cdot a_{5}!$ has the minimum value. | For $m, n \in \mathbf{N}$, if $m > n + 1$, then we have
$$
\begin{aligned}
m! \cdot n! & = (m-1)! \cdot m \cdot n! \\
& > (m-1)!(n+1)!
\end{aligned}
$$
This tells us that to minimize the product under discussion, the $a_i$ must be as close as possible.
Since $1998 = 399 \times 5 + 3$, the minimum value sought is $(39... | (399!)^2(400!)^3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,437 |
Example 10 We notice that $6!=8 \cdot 9 \cdot 10$. Try to find the largest positive integer $n$ such that $n!$ can be expressed as the product of $n-3$ consecutive natural numbers. | Solution: According to the requirements of the problem, write $n!$ as
$$
n!=n(n-1) \cdot \cdots \cdot 5 \cdot 24 \text {. }
$$
Therefore, $n+1 \leqslant 24, n \leqslant 23$. Hence, the maximum value of $n$ is 23. At this point, we have
$$
23!=24 \times 23 \times \cdots \times 5 .
$$
The right-hand side of the above e... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,438 |
Example 11 Write 10 consecutive natural numbers, each of which is a composite number. | Solution: Construct the following decimal natural numbers:
$$
11!+2,11!+3, \cdots, 11!+11 \text {. }
$$
Obviously, they satisfy the requirements of the problem.
In general, for any given $n \in \mathbf{N}$, we can write $n$ consecutive natural numbers, all of which are composite:
$$
\begin{array}{l}
(n+1)!+2,(n+1)!+3,... | (n+1)!+2, (n+1)!+3, \cdots, (n+1)!+(n+1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,439 |
Example 12 Given any integer $n \geqslant 2$. Prove: there exists a set of $n$ points in the coordinate plane, no three of which are collinear, and the centroid of any subset of the point set is an integer point. | Proof: Consider a set of $n$ points $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right), \cdots,\left(x_{n}, y_{n}\right)$ in the coordinate plane. For any integer $k(2 \leqslant k \leqslant n)$, the centroid coordinates of the subset of $k$ elements $\left(x_{i_{1}}, y_{i_{1}}\right)$, $\left(x_{i_{2}}, y_{i_{2}}\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,440 |
1 . Find the value of $1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\cdots+$ $n \cdot n!$. | \begin{array}{l}\text { 1. Solution: Original expression }=(2!-1!)+(3!-2!)+\cdots+ \\ {[(n+1)!-n!]=(n+1)!-1 .}\end{array} | (n+1)!-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,441 |
2. If $(n-1)$ ! is not divisible by $n^{2}$, find all possible values of $n$.
If $(n-1)$ ! cannot be divided by $n^{2}$, find all possible $n$. | 2. Solution: If $n$ is a prime number, then obviously $n^{2} \times (n-1)!$. If $n$ is not a prime number.
(1) Let $n=ab$, where $a$ and $b$ are coprime integers greater than 2, then $(n-1)!$ contains factors $a, 2a, b, 2b$, so $n^{2} \mid (n-1)!$.
(2) Let $n=p^{2}$, where $p$ is a prime number greater than 5, then $p^... | n=8,9, p, 2p \text{ (where } p \text{ is a prime number)} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,442 |
Example 5 Given $x^{2}+\sqrt{2} y=\sqrt{3}, y^{2}+\sqrt{2} x=$ $\sqrt{3}, x \neq y$. Then the value of $\frac{y}{x}+\frac{x}{y}$ is ().
(A) $2-2 \sqrt{3}$
(B) $2+2 \sqrt{3}$
(C) $2-\sqrt{3}$
(D) $2+\sqrt{3}$ | Solution: From $x^{2}+\sqrt{2} y=\sqrt{3}$,
$$
y^{2}+\sqrt{2} x=\sqrt{3} \text {, }
$$
(1) - (2) gives
$$
\begin{array}{l}
\left(x^{2}-y^{2}\right)-\sqrt{2}(x-y)=0 . \\
\because x \neq y, \\
\therefore x+y=\sqrt{2} .
\end{array}
$$
(1) + (2) gives
$$
\left(x^{2}+y^{2}\right)+\sqrt{2}(x+y)=2 \sqrt{3} \text {. }
$$
Thus... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,443 |
3. Let $m, n$ be non-negative integers. Prove that:
$$
\frac{(2 m)!(2 n)!}{m!n!(m+n)!}
$$
is an integer (with the convention $0!=1$). | 3. Proof: Let $f(m, n)=\frac{(2 m)!(2 n)!}{m!n!(m+n)!}$.
Using the method of undetermined coefficients, we easily obtain
$$
f(m+1, n)=4 f(m, n)-f(m, n+1) \text {. }
$$
We apply mathematical induction on $m$. When $m=0$, $f(0, n)=$ $C_{2 n}^{n} \in \mathbf{Z}$. Assume that when $m=k(k \geqslant 0)$, for any non-negati... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,444 |
4. Let the binary representation of $n$ be $\left(a_{m} a_{m-1} \cdots\right.$ $\left.a_{1} a_{0}\right)_{2}, S(n)=\sum_{i=0}^{m} a_{i}$. Prove that the exponent of 2 in $n!$ is $t=n-S(n)$. | \begin{array}{l}\text { 4. Proof: } \because n=\left(a_{m} a_{m-1} \cdots a_{1} a_{11}\right)_{2} \text {, } \\ \therefore t=\left[\frac{n}{2}\right]+\left[\frac{n}{2^{2}}\right]+\cdots+\left[\frac{n}{2^{m-1}}\right]+\left[\frac{n}{2^{m}}\right] \\ =\left(a_{m} a_{m}, \cdots a_{1}\right)_{2}+\left(a_{m} a_{m-1} \cdots ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,445 |
5. Write a given rational number as a reduced fraction, and calculate the product of the resulting numerator and denominator. How many rational numbers between 0 and 1, when processed this way, yield a product of 20!? | 5. Solution: There are 8 prime numbers within 20: $2,3,5,7,11$. $13,17,19$. Each prime number either appears in the numerator or the denominator, but not in both. Therefore, we can form $2^{\times}$ irreducible fractions $\frac{p}{q}$ such that $pq=20!$. Since in $\frac{p}{q}$ and $\frac{q}{p}$, only one lies in the in... | 128 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,446 |
6. Calculate the highest power of 2 contained in $\left(2^{n}\right)$ !.
保留源文本的换行和格式，翻译结果如下：
6. Calculate the highest power of 2 contained in $\left(2^{n}\right)$!. | 6. Solution: The highest power of 2 in $\left(2^{n}\right)$ ! is $\left[\frac{2^{n}}{2}\right]+$
$$
\left[\frac{2^{n}}{2^{2}}\right]+\cdots+\left[\frac{2^{n}}{2^{n}}\right]=2^{n-1}+2^{n-2}+\cdots+1=2^{n}-1 .
$$ | 2^{n}-1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,447 |
8. Find all integers \( m, n, k \) greater than 1 such that
\[
1! + 2! + \cdots + m! = n^k.
\] | 8. Solution: First, we prove that when \( m \geqslant 8 \), it must be that \( k=2 \). It is known that \( 1!+2!+\cdots+8! \) has a factor of \( 3^{2} \), but not a factor of \( 3^{3} \). Since \( 9! \), \( 10! \), etc., all contain powers of 3 greater than 3, when \( m \geqslant 8 \), \( 1!+2!+\cdots+m! \) has a facto... | m=n=3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,449 |
9. Prove: there exists a set $S_{n}$ composed of $n$ distinct positive integers, such that the geometric mean of the elements of any non-empty subset of $S_{n}$ is an integer. | 9. Proof: Construct the set of numbers
$$
S_{n}=\left\{p_{1}^{n!}, p_{2}^{n!}, \cdots, p_{n}^{n!}\right\} \text {. }
$$
where $p_{1}, p_{2}, \cdots, p_{n}$ are distinct prime numbers. Clearly, $S_{n}$ satisfies the condition. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,450 |
As shown in Figure $1, O, I$ are the circumcenter and incenter of $\triangle A B C$, respectively. $A D$ is the altitude on side $B C$. $I$ lies on the line segment $OI$.
Prove: The circumradius of $\triangle A B C$ is equal to the radius of the excircle opposite to side $B C$. | Proof 1: $\because A D 、 O K 、 O^{\prime} Y$ are all perpendicular to $B C$, $\therefore A D / / O K / / O^{\prime} Y$.
Therefore, $\triangle A I I \mathcal{\triangle} \triangle K I O$.
We get $\frac{A D}{O K}=\frac{A I}{I K}$,
which is $\frac{A D}{R}=\frac{A I}{I K}$.
Also, from $\triangle A D J \sim \triangle O^{\pri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,451 |
Example 1 Simplify $\frac{2 b-a-c}{(a-b)(b-c)}$
$$
+\frac{2 c-a-b}{(b-c)(c-a)}+\frac{2 a-b-c}{(c-a)(a-b)} .
$$ | $\begin{array}{l}\text { Solution: Original expression }=\frac{(b-c)-(a-b)}{(a-b)(b-c)} \\ \quad+\frac{(c-a)-(b-c)}{(b-c)(c-a)}+\frac{(a-b)-(c-a)}{(c-a)(a-b)} \\ =\frac{1}{a-b}-\frac{1}{b-c}+\frac{1}{b-c}-\frac{1}{c-a} \\ \quad+\frac{1}{c-a}-\frac{1}{a-b}=0 .\end{array}$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,452 |
Example 2 Given that $\alpha$ is a root of the equation $x^{2}-x-1=0$. Try to find the value of $\alpha^{18}+323 \alpha^{-6}$.
Translating the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given: $\because \alpha$ is a root of $x^{2}-x-1=0$,
$$
\therefore \alpha^{2}-\alpha-1=0 \text {. }
$$
Thus, since $\alpha \neq 0$, we have $\alpha-\alpha^{-1}=1$.
Therefore, $\alpha^{2}+\alpha^{-2}=\left(\alpha-\alpha^{-1}\right)^{2}+2=3$,
$$
\begin{array}{l}
\alpha^{4}+\alpha^{-4}=7, \alpha^{6}+\alpha^{-6}=18, \\
\a... | 5796 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,453 |
Example 6 If $x=\sqrt{19-8 \sqrt{3}}$. Then the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15}=$
Preserve the original text's line breaks and format, output the translation result directly. | Given: From the above, $x=\sqrt{(4-\sqrt{3})^{2}}=4-\sqrt{3}$, thus $x-4=-\sqrt{3}$.
Squaring and simplifying the above equation yields
$$
x^{2}-8 x+13=0 \text {. }
$$
Therefore, the denominator $=\left(x^{2}-8 x+13\right)+2=2$.
By long division, we get
the numerator $=\left(x^{2}-8 x+13\right)\left(x^{2}+2 x+1\right)... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,454 |
Example 3 Solve the equation:
$$
|2 x-3|=2 \sqrt{(x-5)(x+2)}+1 .
$$ | Solution: According to the problem,
$$
|2 x-3|-2 \sqrt{(x-5)(x+2)}=1 \text {, }
$$
which is $|2 x-3|-\sqrt{4 x^{2}-12 x-40}=1$,
$$
\sqrt{(2 x-3)^{2}}-\sqrt{(2 x-3)^{2}-49}=1 \text {. }
$$
Let $\sqrt{(2 x-3)^{2}}=m, \sqrt{(2 x-3)^{2}-49}=$ $n$, then
$$
\left\{\begin{array}{l}
m-n=1, \\
m^{2}-n^{2}=49 .
\end{array}\rig... | x_{1}=-11, x_{2}=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,455 |
Example 4 Given that the real number $x$ and the acute angle $\theta$ satisfy
$$
\begin{array}{l}
x^{2}+2 x \cos \theta=\sin \theta-\frac{5}{4} . \\
\text { Find the value of } \frac{x+\operatorname{tg} \theta}{x-\operatorname{tg} \theta} \text { . }
\end{array}
$$ | Solution: Transposing terms, we get $x^{2}+2 x \cos \theta-\sin \theta+\frac{5}{4}=0$, which is $x^{2}+2 x \cos \theta-\sin \theta+1+\frac{1}{4}=0$.
We have $x^{2}+2 x \cos \theta-\sin \theta+\sin ^{2} \theta+\cos ^{2} \theta+\frac{1}{4}=0$,
$$
(x+\cos \theta)^{2}+\left(\sin \theta-\frac{1}{2}\right)^{2}=0
$$
Thus, $x... | \frac{1}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,456 |
1. Given the equation $x^{2}+999 x-1=0$ has a root $\alpha$. Then $(1998-5 \alpha)\left(\alpha-\frac{1}{6} \alpha^{-1}\right)^{-1}=$ $\qquad$ . | answer: - 6 | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,457 |
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