problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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Question: Let $a, b, c$ represent the side lengths opposite to $\angle A$, $\angle B$, and $\angle C$ in $\triangle ABC$, respectively. For any two points $P_{1}, P_{2}$ inside the triangle, the distances from $P_{1}, P_{2}$ to the vertices $A, B, C$ are $a_{1}, a_{2}$, $b_{1}, b_{2}$, $c_{1}, c_{2}$, respectively. Pro... | Proof 1: As shown in the figure, with $BC$, $CA$, and $AB$ as reflection axes, reflect $\triangle P_{1}BC$, $\triangle P_{1}CA$, and $\triangle P_{1}AB$ into $\triangle DBC$, $\triangle ECA$, and $\triangle FAB$. Connect $P_{2}E$, $P_{2}F$, and $P_{2}D$, then
$$
\begin{array}{l}
S_{\triangle A F P_{2}}=\frac{1}{2} A F ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,222 |
Example 1 Find the range of $y=\sqrt{x^{2}-12 x+45}-$ $\sqrt{x^{2}-2 x+5}$. | Solution: The original expression is transformed into
$$
y=\sqrt{(x-6)^{2}+3^{2}}-\sqrt{(x-1)^{2}+2^{2}},
$$
where the fixed points are $A(6,3), B(1,2)$, and point $A$ is higher than point $B$.
$$
\begin{aligned}
\because & |A B|=: \sqrt{(6-1)^{2}+(3-1)^{2}}=\sqrt{26}, \\
& \left|x_{2}-x_{1}\right|=5,
\end{aligned}
$$... | (-5, \sqrt{26}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,223 |
Example 2 Find the range of $y=\sqrt{x^{2}-2 x+5}-$ $\sqrt{x^{2}-6 x+13}$. | Solution: The original expression is transformed into
$$
y=\sqrt{(x-1)^{2}+2^{2}}-\sqrt{(x-3)^{2}+2^{2}},
$$
where the fixed points are $A(1,2), B(3,2)$.
Clearly, the line $AB \parallel x$-axis, and $|AB|=|3-1|=2$,
$\therefore$ the range is $(-2,2)$.
$2 y=\sqrt{\left(x-x_{1}\right)^{2}+y_{1}^{2}}+\sqrt{\left(x-x_{2}\r... | (-2,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,224 |
1. If points $A$ and $B$ are on the same side of the $x$-axis, you can first find the symmetric point $B^{\prime}\left(x_{2}, -y_{2}\right)$ (or $A^{\prime}\left(x_{1}, -y_{1}\right)$) of point $B$ (or $A$) with respect to the $x$-axis, then calculate $\left|A B^{\prime}\right|$. The value of $\left|A B^{\prime}\right|... | Solution 1: The original expression is transformed into
$$
\begin{aligned}
y & =\sqrt{(x-1)^{2}+(0-2)^{2}} \\
& +\sqrt{(x+2)^{2}+(0-3)^{2}},
\end{aligned}
$$
which means the moving point is $P(x, 0)$, and the fixed points are $A(1,2), B(2,3)$. The point $B$'s reflection over the $x$-axis is $B^{\prime}(2,-3)$.
The li... | \sqrt{26} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,225 |
Example 3 Find the range of $y=\sqrt{x^{2}-2 x+5}+\sqrt{x^{2}-6 x+13}$.
Analysis: If points $A$ and $B$ are placed above the $x$-axis, then $AB$ is parallel to the $x$-axis, so we can solve the problem by placing the points below the $x$-axis. | Solution: The original expression is $y=$ $\sqrt{(x-1)^{2}+(0-2)^{2}}+$ $\sqrt{(x-3)^{2}+(0+2)^{2}}$, with moving point $P(x, 0)$, and fixed points $A(1,2), B(3,-2)$. Connecting $A B$ intersects the $x$-axis at $P(2,0)$, as shown in Figure 7.
When $x=2$, $y_{\text {min }}=|A B|=2 \sqrt{5}$,
$\therefore$ the range is $... | [2 \sqrt{5},+\infty) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,227 |
Example 4 Find the range of $y=\sqrt{x^{2}-2 x+5}+$ $\sqrt{x^{2}-2 x+17}$. | Solution: The original expression can be rewritten as
$$
\begin{aligned}
y= & \sqrt{(x-1)^{2}+2^{2}}+ \\
& \sqrt{(x-1)^{2}+4^{2}},
\end{aligned}
$$
with a moving point $P(x, 0)$, and fixed points $A(1,2)$, $B(1,4)$. Clearly, $AB \parallel y$-axis, and the line segment $AB$ intersects the $x$-axis at $P(1,0)$, as shown... | [6,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,228 |
Proposition: Let $n$ be a natural number greater than 1, then the sum of all irreducible proper fractions with denominators not exceeding $n$ is
$$
\frac{1}{2} \sum_{k=2}^{n} \varphi(k),
$$
that is, for $p, q \in N, 1 \leqslant p < q \leqslant N$, we have
$$
\sum_{\substack{1 \leq p \leq q \leq n \\(p, q)=1}} \frac{p}... | Proof: According to the definition of Euler's function, for a natural number $k \geqslant 2$, the number of irreducible proper fractions with denominator $k$ is $\varphi(k)$ (thus, the number of irreducible proper fractions with denominators not exceeding $n$ is $\sum_{k=2}^{n} \varphi(k)$). Let these $\varphi(k)$ irre... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,229 |
Example 5 In the border desert area, patrol vehicles travel 200 kilometers per day, and each patrol vehicle can carry enough gasoline to travel for 14 days. There are 5 patrol vehicles that set out from base $A$ simultaneously, complete their mission, and then return along the original route to the base. To allow three... | Analysis: The key is where point $B$ is most appropriate. Suppose car A and car B travel for $x$ days to reach point $B$. After reaching point $B$, considering the fuel capacity of the patrol car, we should have
$$
\frac{2(14-2 x)}{3}+(14-x) \leqslant 14 \text {. }
$$
Thus, $x \geqslant 4$.
For the other three cars, t... | 1800 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 710,230 |
The radius of the excircle and the length of the internal angle bisector. Then
$$
\frac{r_{a}}{t_{a}}+\frac{r_{b}}{t_{b}}+\frac{r_{c}}{t_{c}} \geqslant 3 .
$$
This paper strengthens (1) to
$$
r_{a} r_{b} r_{r} \geqslant t_{a} t_{b} t_{c} .
$$ | Let $\triangle A B C$ have side lengths $a, b, c$, and semi-perimeter $p$. From $r_{a}=\sqrt{\frac{p(p-b)(p-c)}{p-a}} x_{a} \leqslant$ $\sqrt{p(p-a)}$ and similar inequalities for $r_b$ and $r_c$:
$$
\begin{array}{l}
r_{a} r_{b} r_{c}= \sqrt{\frac{p(p-b)(p-c)}{p-a}} \\
\cdot \sqrt{\frac{p(p-c)(p-a)}{p-b}} \\
\cdot \sq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,231 |
Proposition In $\triangle A B C$, $t_{a} 、 t_{b} 、 t_{c}$ are the lengths of the internal angle bisectors. Then we have
$$
\sum \frac{a^{2}}{t_{b}^{2}+t_{c}^{2}} \geqslant 2 \text {. }
$$
Equality holds if and only if $\triangle A B C$ is an equilateral triangle. | Proof: By the angle bisector length formula, we have
$$
t_{a}=\frac{2 \sqrt{b c}}{b+c} \sqrt{p(p-a)} \leqslant \sqrt{p(p-a)} .
$$
Similarly, \( t_{b} \leqslant \sqrt{p(p-b)} \),
$$
t_{c} \leqslant \sqrt{p(p-c)} .
$$
Thus, \( t_{b}^{2}+t_{c}^{2} \leqslant p(p-b)+p(p-c) \)
$$
\begin{array}{c}
\quad=a p . \\
\therefore ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,232 |
Proposition Let $F$ be the Fermat point inside $\triangle ABC$, and extend $AF, BF, CF$ to intersect the opposite sides at $A', B', C'$, respectively. Let $AA'=x, BB'=y, CC'=z$. Then
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geqslant \sqrt{\frac{3 \sqrt{3}}{\triangle}},
$$
where $\triangle$ is the area of $\triangle ABC... | - Ming: As shown in the figure, let
$F A=u, F B=v$,
$F C=w$. Then, by reference [1],
we know
$$
\begin{array}{l}
u v+v w+w u \\
=\frac{4 \triangle}{\sqrt{3}},
\end{array}
$$
where $a, b, c$ are the three side lengths of $\triangle A B C$.
$\because F$ is the Fermat point,
$$
\therefore \angle A^{\prime} F B=\angle A^{... | \sqrt{\frac{3 \sqrt{3}}{\triangle}} | Inequalities | proof | Yes | Yes | cn_contest | false | 710,233 |
Question: $P, Q, R$ are points on the sides $AB, BC$, and $CA$ of $\triangle ABC$, respectively, and $AP + AR = BP + BQ = CQ + CR = \frac{1}{3}$.
Prove: $PR + RQ + QP \geqslant \frac{1}{2}$.
This inequality is known as the trisecting perimeter type inequality. A proof of this inequality is provided below. | Prove: As shown in the figure, construct
$A D \perp B C, P F \perp$
$A D, R E \perp A D$. Then
$$
\begin{array}{l}
\angle A P F=\angle B, \\
\angle A R E=\angle C .
\end{array}
$$
Let $B C=a, A C$
$=h, A B=c$. Then
$$
\begin{aligned}
A H^{\circ} & =c-B P, A R=b-C R . \\
\because P F & =A P \cos B=(c-B P) \cos B \\
& =... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,234 |
1. In a convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ are perpendicular to each other, and the opposite sides $AB$ and $DC$ are not parallel. Point $P$ is the intersection of the perpendicular bisectors of segments $AB$ and $CD$, and $P$ is inside the quadrilateral $ABCD$. Prove that $ABCD$ is a cyclic quadr... | Proof: First, prove the necessity: That is, when points $A$, $B$, $C$, and $D$ are concyclic, we have $S_{\triangle A B P}=S_{\triangle C P P}$.
Let the two perpendicular diagonals $A C$ and $B D$ intersect at point $K$. Thus,
$$
\begin{array}{l}
90^{\circ}=\angle A K B \\
= \angle D B C + \angle A C B \\
= \frac{1}{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,235 |
3. For any positive integer $n$, let $d(n)$ denote the number of positive divisors of $n$ (including 1 and $n$ itself). Determine all possible positive integers $k$ such that there exists a positive integer $n$ satisfying $\frac{d\left(n^{2}\right)}{d(n)}=k$.
| Solution: First, if $k$ is a possible positive integer, i.e., there exists $n \in N$ such that
$$
\frac{d\left(n^{2}\right)}{d(n)}=k,
$$
we can assume the prime factorization of $n$ is $n=p_{1}^{a_{1}^{1}} p_{2}^{\alpha} \cdots p_{s}^{o}$, then we have
$$
d(n)=\prod_{i=1}^{3}\left(\alpha_{i}+1\right).
$$
And $d\left(... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,237 |
5. Let $I$ be the incenter of $\triangle ABC$, and let the incircle of $\triangle ABC$ touch the sides $BC$, $CA$, and $AB$ at points $K$, $L$, and $M$, respectively. The line through $B$ parallel to $MK$ intersects lines $LM$ and $LK$ at points $R$ and $S$, respectively. Prove that $\angle RIS$ is acute. | Proof: Given $R S // M K$,
we have
$$
\begin{array}{l}
\angle M B R=\angle K M B, \\
\angle K B S=\angle M K B .
\end{array}
$$
Since $B M$ and $B K$ are two
tangents, then
$$
\begin{aligned}
& B M=B K, \angle K M B \\
= & \angle M K B . \\
& \therefore \angle M B R=\angle K B S .
\end{aligned}
$$
Connecting $I B$, t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,238 |
A function in itself, and for any $s$ and $t$ in $N$, it satisfies $f\left(t^{2} f(s)\right)=s(f(t))^{2}$. Determine the minimum value that $f(1998)$ can achieve among all functions $f$.
untranslated part:
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。
(As requested, the untranslated part is not included in the translation.) | Solution: For every function $f$ that satisfies the conditions of the problem,
it is known that $f\left(t^{2} f(s)\right)=s(f(t))^{2}$.
Substituting $t=1$ yields $f(f(s))=s(f(1))^{2}$,
and substituting $s=1$ into (1) gives
$f\left(t^{2} f(1)\right)=(f(t))^{2}$.
Now, let $f(1)=k$, then the above two equations can be rew... | 120 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,239 |
Example 6 In a mountain bike race held in a city, two cyclists, A and B, start from point A to point B at the same time. Cyclist A runs $\frac{1}{3}$ of the time at speeds $v_{1}, v_{2}, v_{3}$ respectively, while Cyclist B runs $\frac{1}{3}$ of the distance at speeds $v_{1}, v_{2}, v_{3}$ respectively. Which cyclist, ... | Solution: Let the distance between the two places be $s$, and the time taken by A and B to complete the entire journey be $t_{1}$ and $t_{2}$, respectively. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{1}{3} t_{1}\left(v_{1}+v_{2}+v_{3}\right)=s, \\
\frac{s}{3}\left(\frac{1}{v_{1}}+\frac{1}{v_{2}}+\frac{1}{v_... | t_{1}<t_{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,240 |
Let $R_{1}, R_{2}, \cdots$, be a family of finite sequences of positive integers defined by the following rules:
$$
R_{1}=\text { (1); }
$$
If $R_{4} =\left(x_{1}, \cdots, x_{s}\right)$, then
$$
R_{n}=\left(1,2, \cdots, x_{1}, 1,2, \cdots, x_{2}, \cdots, 1,2, \cdots, x_{s}\right. \text {, }
$$
$n)$.
For example: $R_{2... | Solution: (a) The easiest way to see that the sum in (a) is $2^{n}-1$ using the knowledge of the pseudo-Pascal triangle is as follows: write out the pseudo-Pascal triangle, and in the sequence, replace the sequence itself with the sum of the numbers, which is represented as follows
By adding a string of 1s (as shown a... | 2^{n}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,241 |
5. Let $A B C D$ be a regular tetrahedron, and let $M, N$ be distinct points on the planes $A B C, A D C$ respectively. Prove that the segments $M N, B N, M D$ are the sides of a triangle.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result d... | Proof: We start by proving the following lemma.
Lemma: Let $\triangle BDE$ be an isosceles triangle with side $BD = l$, $BE = DE = \frac{1}{2} \sqrt{3} l$, and $M_1, N_1$ be different points on segments $BE$ and $DE$, respectively. Then $M_1N_1$, $BN_1$, and $M_1D$ are the three sides of a triangle.
Proof of the lemm... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,243 |
6. (a) Let $n$ be a positive integer. Prove: there exist distinct positive integers $x, y, z$ such that $x^{n-1} + y^n = z^{n+1}$.
(b) Let $a, b, c$ be positive integers, and $a$ and $b$ are coprime, $c$ is coprime with either $a$ or $b$. Prove: there exist infinitely many distinct positive integer triples $(x, y, z)$ ... | Proof: (a) For example,
$$
\begin{array}{l}
x=2^{n^{2}} \cdot 3^{n+1}, y=2^{(n-1) n} \cdot 3^{n}, \\
z=2^{n^{2}-2 n+2} \cdot 3^{n-1} .
\end{array}
$$
The idea behind this solution is simple: $1+3=2^{2}$.
(b) Let $P (P \geqslant 3)$ be a positive integer, then $Q=P-1>1$. We seek solutions of the form.
$$
x=Q^{m}, y=Q^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,244 |
7. Let $A B C D E F$ be a convex hexagon, and $A B=B C, C D=$ $D E, E F=F A$. Prove:
$$
\frac{B C}{B E}+\frac{D E}{D A}+\frac{F A}{F C} \geqslant \frac{3}{2},
$$
and determine when equality holds. | Proof: As shown in the figure, let \( AC = a, CE = b, AE = c \). Applying Ptolemy's inequality to quadrilateral \( ACEF \), we get
\[
\begin{array}{l}
AC \cdot EF + CE \cdot AF \\
\geqslant AE \cdot CE.
\end{array}
\]
Since \( EF = AF \), this implies \( \frac{N}{F} \geqslant \frac{c}{i} \).
Similarly, we have
\[
\beg... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,245 |
Three, let $M=\{2,3,4, \cdots, 1000\}$. Find the smallest natural number $n$ such that in any $n$-element subset of $M$, there exist 3 pairwise disjoint 4-element subsets $S, T, U$ satisfying the following three conditions:
(1) For any two elements in $S$, the larger number is a multiple of the smaller number, and the ... | Solution: Note that $999=37 \times 27$, let $A=\{3,5, \cdots, 37\}$, $B=M-A$, then $,|A|=18,|B|=981$.
Below we prove that the subset $B$ of $M$ cannot simultaneously satisfy conditions (1) $\sim$ (3). If not, let $S=\left\{s_{1}, s_{2} ; s_{3}, s_{4}\right\}, T=\left\{t_{1}, t_{2}, t_{3}, t_{4}\right\}$ and there is $... | 982 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,248 |
Four, $A, B, C=$ Three countries are holding a Go broadcast tournament, each team with 9 members. The rules are as follows: Each match involves one person from each of two teams competing, the winner stays to defend, the loser is eliminated, and the other team sends one person to challenge. The first match starts with ... | Solution: (1) When the champion team finally wins, the other two teams' 18 people have all been eliminated. Since team $C$ plays later, when $C$ team wins the championship, it can win one less game. To minimize the number of wins for team $C$, teams $A$ and $B$ need to eliminate as many of their own members as possible... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 710,249 |
Example 7 A shipping company has a ship leaving Harvard for New York every noon, and at the same time every day, a ship also leaves New York for Harvard. It takes seven days and seven nights for the ships to complete their journeys in both directions, and they all sail on the same route. How many ships of the same comp... | Solution 1: Establishing Equal Relationships of Distance and Time.
Analysis: If this ship meets another ship opposite to it at point $A$, then the next ship opposite to it is at point $B$. By the equal relationship of distance, it is easy to know that they meet at the midpoint of $A B$, and the interval is exactly half... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 710,251 |
Six, does there exist non-zero complex numbers $a$, $b$, $c$ and a natural number $h$, such that for any integers $k$, $l$, $m$ satisfying $|k|+|l|+|m| \geqslant 1996$, it is always true that $|k a+l b+m c|>\frac{1}{h}$? | Solution: It does not exist. Otherwise, suppose there are non-zero complex numbers $a, b, c$ and a natural number $h$ that satisfy the requirements of the problem.
Consider the complex plane. Without loss of generality, assume that the angle between the vectors $\vec{a}$ and $\vec{b}$ corresponding to the complex numb... | proof | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 710,252 |
4. A person takes a boat from place $A$ downstream to place $B$, and then upstream to place $C$, spending a total of 4 hours on the boat. It is known that the speed of the boat in still water is 7.5 kilometers per hour, and the speed of the current is 2.5 kilometers per hour. If the distance between $A$ and $C$ is 10 k... | $\begin{array}{l}4.20 \\ \text { or } \frac{20}{3}\end{array}$ | \frac{20}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,256 |
一、Multiple Choice Questions (7 points each, total 42 points)
1. The number of integer solutions to the equation $\left(x^{2}-x-1\right)^{x+10}=1$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | From $x+10=0$, we get $x=-10$. When $x+10$ is odd, from $x^{2}-x-1=1$, we get $x=-1$; when $x+10$ is even, from $x^{2}-x-1= \pm 1$, we get $x=0$ or $x=2$.
In summary, the integer solutions of the original equation are $x_{1}=-10, x_{2}=-1$, $x_{3}=0, x_{4}=2$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,259 |
2. If the equation $\frac{x+7}{x-1}+\frac{1}{x}-\frac{4 x+k}{x^{2}-x}=0$ has only one real root (counting repeated roots as one). Then the value of $k$ is ( ).
(A) -5
(B) -1 or 4
(C) -5 or -1
(D) -5 or -1 or 4 | 2. (I)).
The original equation can be transformed into $x^{2}+4 x-(k+1)=0$.
By $\Delta=16+4(k+1)=0$, we get $k=-5$. Therefore, when $k=-5$, the roots of the original equation are $x=-2$.
When $k>-5$, equation (1) has two equal real roots. According to the problem, one of the roots must be an extraneous root of the or... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,260 |
3. The typhoon center moves northeast at a speed of 20 kilometers per hour, and the area within 30 kilometers of the typhoon center is the danger zone. City $B$ is 40 kilometers due east of city $A$. The time that city $B$ is in the danger zone is ( ).
(A) 0.5 hours
(B) 1 hour
(C) 1.5 hours
(D) 2 hours | 3. (A).
As shown in the figure, let the line through which the center of the typhoon passes intersect the circle with center $B$ and radius 30 kilometers at points $C$ and $D$. Clearly, finding the length of $C D$ will allow us to calculate the required time.
Draw $B E \perp C D$ at $E$, then $B E=A B \sin 45^{\circ}=... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,261 |
Example 8 A truck arrives at the edge of a 563-mile-wide desert. The truck consumes on average one gallon of gasoline per mile. The truck, in addition to its own fuel tank, has a spare fuel tank that can hold a maximum of 315 gallons of gasoline, and there is no other way to carry fuel (there is an ample supply of gaso... | Analysis: Operation method. To consume the least amount of gasoline, one should try to avoid retracing steps as much as possible. Therefore, with one full tank, 315 gallons of gasoline can cover a maximum of 315 miles; with two full tanks, a fuel storage point can be established 315 miles from the opposite side, allowi... | B | Other | MCQ | Yes | Yes | cn_contest | false | 710,262 |
4. In a convex pentagon $A B C D E$, $\angle A=\angle E$ $=120^{\circ}, E A=A B=B C$ $=2, C D=D E=4$. Then its area is ( ).
(A) $6 \sqrt{3}$
(B) $7 \sqrt{3}$
(C) $8 \sqrt{3}$
(D) $9 \sqrt{3}$ | 4. (B).
Extend $E A$ and $C B$ to intersect at point $F$.
$$
\because \angle E A B=\angle C B A=
$$
$120^{\circ}$,
$$
\therefore \angle F A B=\angle F B A=60^{\circ} \text {, }
$$
Therefore, $\triangle A B F$ is an equilateral triangle.
$$
\begin{array}{c}
A F=F B=A B=2 . \\
\therefore C D=D E=E F=F C=4 .
\end{array}... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,263 |
5. Given that the graph of the parabola $y=a x^{2}+b x+c$ passes through the points $(1,4)$ and $(2,7)$, and the axis of symmetry is the line $x=k$, with $|k| \leqslant 1$. Then the range of values for $a$ is ( ).
(A) $\frac{3}{5} \leqslant a \leqslant 3$
(B) $a \geqslant 3$
(C) $a \leqslant \frac{3}{5}$
(D) $a<0$ | 5. (A).
According to the problem, we have
$$
\left\{\begin{array}{l}
a+b+c=4, \\
4 a+2 b+c=17 .
\end{array}\right.
$$
(2) - (1), we get $3 a+b=3$, then $b=3-3 a$.
The axis of symmetry of the parabola is given by $x=-\frac{b}{2 a}$.
From $\left|-\frac{b}{2 a}\right| \leqslant 1$, we know $-1 \leqslant-\frac{b}{2 a} \l... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,264 |
6. As shown in the figure, $A B C D$ is a square inscribed in $\odot O$, and $P$ is any point on $\overparen{A D}$. Then the value of $\frac{P A+P C}{P B}$ is ).
(A) 1
(B) $\sqrt{2}$
(C) $\sqrt{3}$
(D) 2 | 6. (B).
Draw a perpendicular from point $B$ to $BP$ intersecting the extension of $PC$ at point $E$.
In $\triangle B A P$ and $\triangle B C E$,
$$
\begin{array}{l}
\angle B C E=180^{\circ}-\angle B C P \\
=\angle B A P, \\
\angle C B E=90^{\circ}-\angle P B C \\
=\angle A B P, \\
B C=A B, \\
\therefore \triangle B C ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,265 |
1. Given that $p$ and $q$ are both natural numbers, and satisfy $19 p + 37 q = 1997$. Then $p q =$ $\qquad$ | $=、 1.100$ or 60 .
From $19 p+97 q=1997$, we get
$$
p=\frac{1997-97 q}{19}=100+\frac{97(1-q)}{19} \text {. }
$$
Since $97$ and $19$ are coprime,
Therefore, $1-q$ is a multiple of 19.
$$
\begin{array}{l}
\text { Also } \frac{1997-97 q}{19} \geqslant 1, \\
\therefore q \leqslant 20 .
\end{array}
$$
Therefore, $1-q=0$ or... | 100 \text{ or } 60 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,266 |
2. Given $a^{n}=x, a^{-n}=\sqrt{2}+1$. Then $a^{3 n}+$ $a^{-3 n}=$
保留了源文本的换行和格式。 | $2.10 \sqrt{2}$.
$$
\begin{aligned}
\because & a^{n} \cdot a^{-n}=1, \text{ then } x=\sqrt{2}-1 . \\
\therefore & a^{n}+a^{-n}=2 \sqrt{2} . \\
& a^{2 n}+a^{-2 n}=\left(a^{n}+a^{-n}\right)^{2}-2 \\
& =(2 \sqrt{2})^{2}-2=6 . \\
\therefore & a^{3 n}+a^{-3 n}=\left(a^{n}+a^{-n}\right)\left(a^{2 n}-1+a^{-2 n}\right) \\
& =2... | 10 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,267 |
3. Given that $\alpha, \beta$ are the roots of the equation $x^{2}-3 x+1=0$, and $\alpha>\beta$. Then $\alpha^{4}+\frac{3}{\beta}=$ $\qquad$ . | $$
3.28+12 \sqrt{5}
$$
By Vieta's formulas, we know $\alpha+\beta=3, \alpha \beta=1$.
$$
\begin{aligned}
\therefore & \alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=7, \\
& \alpha^{4}+\beta^{4}=\left(\alpha^{2}+\beta^{2}\right)^{2}-2 \alpha^{2} \beta^{2}=47, \\
& (\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha ... | 28+12 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,268 |
4. In $\triangle A B C$, $A B=15, A C=13$, and the altitude $A D=12$. Let the radius of the circle that can completely cover $\triangle A B C$ be $R$. Then the minimum value of $R$ is $\qquad$ | 4. $\frac{65}{8}$ or $\frac{15}{2}$.
As shown in figure (a), if $D$ is on $BC$, by the Pythagorean theorem, we get
$$
\begin{array}{l}
B D=\sqrt{A B^{2}-A D^{2}}=\sqrt{15^{2}-12^{2}}=9, \\
D C=\sqrt{A C^{2}-A D^{2}}=\sqrt{13^{2}-12^{2}}=5 . \\
\therefore B C=14 . \\
\because B C^{2}+A C^{2}=14^{2}+13^{2}>15^{2}=A B^{2... | \frac{65}{8} \text{ or } \frac{15}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,269 |
One, (Full marks 20 points) Given that $x$, $y$, $z$ are positive integers, and satisfy $x^{3}-y^{3}-z^{3}=3 x y z$, $x^{2}=2(y+z)$. Find the value of $x y+y z+z x$.
---
The translation is provided as requested, maintaining the original format and line breaks. | $$
\begin{aligned}
- & \because x^{3}-y^{3}-z^{3}-3 x y z \\
= & x^{3}-(y+z)^{3}-3 x y z+3 y^{2} z+3 y z^{2} \\
= & (x-y-z)\left(x^{2}+x y+x z+y^{2}+2 y z+z^{2}\right) \\
& -3 y z(x-y-z) \\
= & (x-y-z)\left(x^{2}+y^{2}+z^{2}-x y-y z+x z\right),
\end{aligned}
$$
X. $x^{3}-y^{3}-z^{3}=3 x y z$,
$$
\begin{array}{l}
\there... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,270 |
II. (Full marks 25 points) As shown in the figure, in $\triangle ABC$, $AB=AC$, $AD$ is the altitude, $E$ is a point on $AB$, $CF \perp BC$ intersects the extension of $ED$ at $F$, $M$ and $N$ are the midpoints of $DE$ and $DF$ respectively. Prove: $\angle MAD = \angle NAD$. | $$
\begin{array}{l}
\quad \because AB=AC, AD \perp \\
BC, CF \perp BC, \\
\therefore BD=DC, AD \parallel \\
CF. \\
\text { Hence } SD=DT .
\end{array}
$$
$\because M, N$ are the midpoints of $DE, DF$ respectively,
$$
\begin{array}{l}
\therefore SM \parallel AB, NT \parallel CF . \\
\therefore AP=PD, NQ \parallel PD . \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,271 |
Three. (Full marks 25 points) On the blackboard, all natural numbers from 1 to 1997 are written. Students $A$ and $B$ take turns to perform the following operations: Student $A$ subtracts the same natural number from each number on the blackboard (the number subtracted can be different in different operations); Student... | Three, because after each operation by student $A$ and student $B$, the number of numbers written on the blackboard decreases by 1. Since students $A$ and $B$ take turns operating, when $B$ completes the last operation, only one number remains on the blackboard, and both have performed 1996 operations. Let $d_{k} (k=1,... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,272 |
For example, $1 H$ is the orthocenter of acute $\triangle A \overline{S C}$, and the reflections of $H$ over the sides $B C$, $C A$, and $A B$ are $H_{1}$, $H_{2}$, and $H_{3}$, respectively. If $P$ is any point on the circumcircle of $\triangle A B C$, and $P H_{1}$, $P H_{2}$, and $P H_{3}$ intersect the sides $B C$,... | Prove: Connect $C H_{1}$. Since $H$ and $H_{3}$ are symmetric with respect to $A B$, we have
$$
\begin{array}{l}
\angle H_{3} H F= \\
\angle H H_{3} F .
\end{array}
$$
Similarly, $\angle C H_{1} D=\angle C H D$.
By the property of the orthocenter, the symmetric points of $H$ with respect to the three sides of $\triang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,273 |
1. For all real numbers $x$, the value of all quadratic functions $f(x)=a x^{2}+b x+c(a<b)$ is always non-negative. Then the minimum value of $M=\frac{a+b+c}{b-a}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) 2
(D) 3 | -1. (D).
From the given conditions, we know $a>0$, and $b^{2}-4 a c \leqslant 0$, which means $c \geqslant \frac{b^{2}}{4 a}$. Let $b-a=t(t>0)$, then
$$
\begin{aligned}
M & \geqslant \frac{a+(t+a)+\frac{(a+t)^{2}}{4 a}}{t} \\
& =\frac{(3 a+t)^{2}}{4 a t} \geqslant \frac{(2 \sqrt{3 a t})^{2}}{4 a t}=3 .
\end{aligned}
$$... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,274 |
2. Given the curve $y^{2}=a x$ and its curve symmetric about the point $(1,1)$ intersect at two different points. If the inclination angle of the line passing through these two points is $45^{\circ}$, then the value of the real number $a$ is $(\quad)$.
(A)2
(B) 4
(C) $\frac{1}{2}$
(D) $\frac{1}{4}$ | 2. (A).
The curve $y^{2}=a x$ symmetric to the point $(1,1)$ is $(2-y)^{2}=a(2-x)$.
From $\left\{\begin{array}{l}y^{2}=a x, \\ (2-y)^{2}=a(2-x)\end{array}\right.$ we get $y^{2}-2 y+2-a=0$.
$\therefore y_{1}+y_{2}=2$.
Thus, $k=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{a\left(y_{1}-y_{2}\right)}{y_{1}^{2}-y_{2}^{2}}=\frac{a... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,275 |
3. Given the sides of $\triangle A B C$ are $a, b, c$ and satisfy the equations $a^{2}-a-2 b-2 c=0, a+2 b-2 c+3=0$. Then the largest interior angle of $\triangle A B C$ is ( ).
(A) $150^{\circ}$
(B) $120^{\circ}$
(C) $90^{\circ}$
(D) $60^{\circ}$ | 3. (B).
Given $\left\{\begin{array}{l}(a+2 b)-2 c=a^{2}, \\ (a+2 b)+2 c=-3 .\end{array}\right.$
Multiplying the two equations and simplifying, we get
$$
a^{2}+b^{2}-c^{2}=-a b<0 .
$$
Therefore, $\angle C$ is an obtuse angle, and $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=-\frac{1}{2}$. Hence, the largest interior angle ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,276 |
4. Let $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants. If $f(1)=1, f(2)$ $=2, f(3)=3$, then the value of $\frac{1}{4}[f(4)+f(0)]$ is ( ).
(A) 1
(B) 4
(C) 7
(D) 8 | 4. (C).
Let $g(x)=f(x)-x$, then
$$
g(1)=g(2)=g(3)=0 \text {. }
$$
Thus, $g(x)=(x-1)(x-2)(x-3)(x-r)$ (where $r$ is a constant).
$$
\begin{array}{l}
\therefore \frac{1}{4}[f(4)+f(0)]=\frac{1}{4}[g(4)+4+g(0)] \\
=\frac{1}{4}[6(4-r)+4+6 r]=7 .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,277 |
5. Let the function $y=f(x)=\frac{x^{2}-x+n}{x^{2}+x+1}(x \in R$ and $\left.x \neq \frac{n-1}{2}, n \in N\right)$ have the minimum value $\Omega_{n}$, the maximum value $b_{n}$, and let $c_{n}=\left(1-a_{n}\right)\left(1-b_{n}\right)$. Then the sequence $\left\{c_{n}\right\}$ ).
(A) is an arithmetic sequence with a non... | 5. (C).
From $y=\frac{x^{2}-x+n}{x^{2}+x+1}$, we get
$$
\begin{array}{l}
(y-1) x^{2}+(y+1) x+(y-n)=0 . \\
\because x \neq \frac{n-1}{2}, \therefore y \neq 1 .
\end{array}
$$
$\because$ Equation (1) has real roots,
$$
\therefore \Delta=(y+1)^{2}-4(y-1)(y-n) \geqslant 0 \text {, }
$$
which means $3 y^{2}-(4 n+6) y+(4 n... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,278 |
6. Let $M$ be a subset of the set $S=\{1,2,3, \cdots, 1998\}$, and each natural number (element) in $M$ contains exactly one 0. Then the maximum number of elements in the set $M$ is ( ).
(A) 324
(B) 243
(C) 495
(D) 414 | 6. (D).
Divide the set $S$ into four subsets
$$
\begin{array}{l}
S_{1}=\{1,2, \cdots, 9\}, \\
S_{2}=\{10,11, \cdots, 99\}, \\
S_{3}=\{100,101, \cdots, 999\}, \\
S_{4}=\{1000,1001, \cdots, 1998\} .
\end{array}
$$
Let $M_{i}^{r}$ represent the set formed by all elements in $S_{i}(i=1,2,3,4)$ that contain exactly one di... | 414 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 710,279 |
1. Given the real-coefficient equations about $x$: $x^{2}-2 x+2=0$ and $x^{2}+2 m x+1=0$, the four distinct roots of these equations correspond to points in the complex plane that lie on a circle. Then the set of values for $m$ is | II. 1. $|m|-10$, i.e., $m1$ when the equation $x^{2}+2 m x+1=0$ has two distinct real roots $x_{3}, x_{4}$, then the points corresponding to $x_{1}, x_{2}$ lie on the circle with the diameter endpoints corresponding to $x_{3}, x_{4}$. By the equation of the circle with diameter, the equation of the circle is
$$
\left(x... | \left\{m \mid-1<m<1 \text { or } m=-\frac{3}{2}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,280 |
2. Given $\frac{\sin ^{3} \alpha}{\sin \left(\frac{\pi}{6}-3 \alpha\right)}=\frac{\cos ^{3} \alpha}{\cos \left(\frac{\pi}{6}-3 \alpha\right)}$ $=\sqrt{3}\left(\alpha \neq \frac{k \pi}{4}, k \in Z\right)$. Then the value of $\operatorname{tg} 2 \alpha$ is | 2. $\frac{\sqrt{3}}{4}$.
From the given, we have
(1) + (2), we get
$$
\begin{array}{l}
\frac{\sqrt{3} \sin \left(\frac{\pi}{6}-3 \alpha\right)}{\sin \alpha}+\frac{\sqrt{3} \cos \left(\frac{\pi}{6}-3 \alpha\right)}{\cos \alpha}=1 . \\
\therefore \sqrt{3} \sin \left(\frac{\pi}{6}-3 \alpha\right) \cos \alpha+\sqrt{3} \co... | \frac{\sqrt{3}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,281 |
3. From the center of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, two perpendicular chords $A C$ and $B D$ are drawn. Connecting $A, B, C, D$ in sequence forms a quadrilateral. Then, the maximum value of the area $S$ of quadrilateral $A B C D$ is | 3. 12.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Given $O A \perp O B$, we have
$$
\begin{array}{l}
x_{1} x_{2}+y_{1} y_{2}=0 . \\
\begin{aligned}
\therefore S & =4 S_{\triangle A O B}=2|O A| \cdot|O B| \\
& =2 \sqrt{\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)} \\
& =2 \sqrt{\l... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,282 |
4. Let the sequence of positive numbers be $\left\{a_{n}\right\}$, with the sum of the first $n$ terms being $S_{n}$, and there exists a positive number $t$, such that for all natural numbers $n$, $\sqrt{t S_{n}}=\frac{t+a_{n}}{2}$ holds. If $\lim _{n \rightarrow \infty} \frac{\sqrt{S_{n}}}{a_{n}}<t$, then the range of... | 4. $t>\frac{\sqrt[3]{2}}{2}$.
From $\sqrt{t S_{n}}=\frac{t+a_{n}}{2}$, we get $S_{n}=\frac{1}{4 t}\left(t+a_{n}\right)^{2}$.
When $n=1$, $a_{1}=S_{1}=\frac{1}{4 t}\left(t+a_{1}\right)^{2}$, we get $a_{1}=t$;
When $n \geqslant 2$,
$$
a_{n}=S_{n}-S_{n-1}=\frac{1}{4 t}\left[\left(t+a_{n}\right)^{2}-\left(t+a_{n-1}\right)... | t>\frac{\sqrt[3]{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,283 |
Example 2 (Simson's Theorem) Take any three points $A$, $B$, $C$ on a circle. Prove that the feet of the perpendiculars from any point $M$ on the circle to the lines $AB$, $BC$, and $CA$ lie on a straight line.
(1940-1950, Polish Mathematical Olympiad) | Proof: As shown in the figure, let $M$ be any point on the circumcircle of $\triangle ABC$, and the feet of the perpendiculars from $M$ to the sides or their extensions are $L$, $P$, and $N$. We need to prove that $L$, $P$, and $N$ are collinear.
In fact, since $L$, $P$ lie on the circle with diameter $BM$, and $P$, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,284 |
5. Given a pyramid $S-ABCD$ inscribed in a sphere with the base being a rectangle $ABCD$, and $SA=4, SB=8, SD=7$, $\angle SAC=\angle SBC=\angle SDC$. Then the length of $BD$ is | 5. $B D=9$.
As shown in the figure, $\because \angle S A C=\angle S B C=\angle S D C$,
$$
\therefore \frac{S C}{\sin \angle S A C}=\frac{S C}{\sin \angle S B C}=\frac{S C}{\sin \angle S D C} \text {. }
$$
Therefore, the circumcircles of $\triangle S A C$,
$\triangle S B C, \triangle S D C$ have equal diameters, meani... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,285 |
6. Given positive integers $m, n$ satisfying $m+n=19$. Then the maximum number of solutions to the equation $\cos m x=\cos n x$ in the interval $[0, \pi]$ is | 6.18.
Since $m+n=19$ is odd, then $m \neq n$. Without loss of generality, assume $m>n$.
The solutions to the equation $\cos m x=\cos n x$ are
$x=\frac{2 k_{1} \pi}{m+n}, x=\frac{2 k_{2} \pi}{m-n}\left(k_{1}, k_{2} \in \mathbb{Z}\right)$.
From $0 \leqslant \frac{2 k_{1} \pi}{m+n} \leqslant \pi, 0 \leqslant \frac{2 k_{2... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,286 |
Three, (Full marks 20 points) The complex numbers $z_{1}, z_{2}, z_{3}, z_{4}, z_{5}$ satisfy
$$
\left\{\begin{array}{l}
\left|z_{1}\right| \leqslant 1, \\
\left|z_{2}\right| \leqslant 1, \\
\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right|, \\
\left|2 z_{4}-\left(z_{1}+z_{2}\right)\right|... | $$
\begin{array}{l}
\text { 3, } \because\left|z_{1}-z_{2}\right| \geqslant\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \\
\quad \geqslant|2| z_{3}|-| z_{1}+z_{2}||, \\
\therefore\left|z_{1}+z_{2}\right|-\left|z_{1}-z_{2}\right| \leqslant 2\left|z_{3}\right| \\
\quad \leqslant\left|z_{1}+z_{2}\right|+\left|z_{1}-z_{2}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,287 |
Four. (Full marks 20 points) Let $t_{1}, t_{2}$ be the two distinct real roots of the equation $t^{2}-$ $(5 a-2) t-3 a^{2}-7 a+1=0$. Find the value of the real number $a$ such that for any non-zero real number $m$, the function $f(x)=\cos (m \pi x) \cos \left[\left(t_{1}^{3}+t_{2}^{3}\right) \cdot \pi x\right]$ is a pe... | Let $T$ be the smallest positive period of the function $f(x)$, then we have $f(T) = f(0)$, i.e.,
$$
\cos (m \pi T) \cos \left(\left(t_{1}^{3} + t_{2}^{3}\right) \pi T\right) = 1.
$$
By the boundedness of the cosine function, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
|\cos (m \pi T)| = 1, \\
\left|\cos \left[\... | \frac{2}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,288 |
Five. (Full marks 20 points) As shown in the figure, the base $ABC$ of the tetrahedron $P-ABC$ and the base $\odot O$ of the cone $SO$ are both on the plane $\alpha$, and $\odot O$ passes through point $A$. Furthermore, the diameter $AD \perp BC$ of $\odot O$, with the foot of the perpendicular being $E$. Given that al... | Five, first find the distance from $S$ to the side $P B C$.
Let $E A$ intersect $B C$ at $E$, and connect $P E$, then $P E \perp B C$. Draw $P H \perp A E$, with the foot of the perpendicular at $H$, then $P H \perp \alpha$. It is known that $S O, A E, P E$ are in the same plane $\beta$. Let the intersection point of $... | \frac{1}{6}(2 \sqrt{2} + 2 \sqrt{6} - \sqrt{3}), \frac{1}{6}(\sqrt{3} + \sqrt{2}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,289 |
One, (Full marks 50 points) Let $0<x_{1}<x_{2}<\cdots<x_{n}<1$. Prove:
$$
\begin{array}{l}
\left(1-x_{n}\right)^{2}\left(\frac{x_{1}}{\left(1-x_{1}^{2}\right)^{2}}+\frac{x_{2}^{2}}{\left(1-x_{2}^{3}\right)^{2}}+\cdots+ \\
\frac{x_{n}^{n}}{\left(1-x_{n}^{n+1}\right)^{2}}\right)<1 .
\end{array}
$$ | \begin{array}{l}-, \because 0<x_{k} \leqslant x_{n}<1(k=1,2, \cdots, n), \\ \therefore 0<\frac{1-x_{n}}{1-x_{k}} \leqslant 1 \text { (equality holds only when } k=n \text { ). } \\ \therefore\left(1-x_{n}\right)^{2} \cdot \frac{x_{k}^{k}}{\left(1-x_{k}^{k+1}\right)^{2}} \\ =\frac{\left(1-x_{n}\right)^{2}}{\left(1-x_{k}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,290 |
Three. (Full marks 50 points) Given $n(n \geqslant 2)$ distinct points in the plane. Prove: the number of point pairs with a distance of 1 does not exceed $\frac{1}{4} n+\frac{1}{\sqrt{2}} n^{\frac{3}{2}}$.
---
Given $n(n \geqslant 2)$ distinct points in the plane. Prove: the number of point pairs with a distance of ... | Let $n$ points be $P_{i}(i=1,2, \cdots, n)$.
Let $S_{i}=\left\{P_{i}|| P_{i} P_{j} \mid=1\right\}, x_{i}=\left|S_{i}\right|$, where $\left|S_{i}\right|$ represents the number of elements in the set $S_{i}$. Then the number of point pairs at a distance of 1 is
$$
x=\frac{1}{2}\left(x_{1}+x_{2}+\cdots+x_{n}\right) .
$$
... | \frac{1}{4} n+\frac{\sqrt{2}}{2} n^{\frac{3}{2}} | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,292 |
Example 12 Let real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{2}-y z-8 x+7=0, \\
y^{2}+z^{2}+y z-6 x+6=0 .
\end{array}\right.
$$
Find the range of values for $x$.
(1980, National High School Mathematics Competition) | Solution: From (1), we get $y z=x^{2}-8 x+7$.
From (2), we get
$$
\begin{array}{l}
(y+z)^{2}=y z+6 x-6=(x-1)^{2} . \\
\therefore y+z= \pm(x-1) . \\
\because y 、 \pm \frac{x-1}{2}, z \text { form an arithmetic sequence, so let } \\
y= \pm \frac{x-1}{2}+d, z= \pm \frac{x-1}{2}-d . \text { Substituting into }
\end{array}
... | 1 \leqslant x \leqslant 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,293 |
Example 13 Determine the largest real number $z$, such that $x+y+z=5, xy+yz+zx=3$, and $x, y$ are also real numbers.
(7th Canadian Intermediate Mathematics Competition) | Solution: $\because x+y=2 \cdot \frac{5-z}{2}$,
$\therefore x, \frac{5-z}{2}, y$ form an arithmetic sequence.
Thus, let $x=\frac{5-z}{2}+d, y=\frac{5-z}{2}-d$. Substituting into $x y+y z+z x=3$ and simplifying, we get
$$
\begin{array}{l}
3 z^{2}-10 z-13=-d^{2} . \\
\therefore 3 z^{2}-10 z-13 \leqslant 0 .
\end{array}
$... | \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,294 |
Example 2 Polynomial $M=2 a^{2}-8 a b+17 b^{2}-$ $16 a-4 b+1998$. Then, what is the minimum value of $M$? | $$
\text { Sol: } \begin{aligned}
M= & 2\left(a^{2}+4 b^{2}+16-4 a b-8 a+\right. \\
& 16 b)+9\left(b^{2}-4 b+4\right)+1930 \\
= & 2(a-2 b-4)^{2}+9(b-2)^{2}+ \\
& 1930 .
\end{aligned}
$$
It is evident that for all real numbers $a, b$, we always have $M \geqslant 1930$. The minimum value of $M$ is 1930, which occurs onl... | 1930 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,296 |
Example 11 In a scalene $\triangle A B C$, two of the altitudes are 4 and 12. If the altitude to the third side is also an integer, then the longest possible value for this altitude is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | Solution: Let the area of $\triangle A B C$ be $S$. Then the sides corresponding to the two given heights are $\frac{2 S}{4}$ and $\frac{2 S}{12}$. Let the third height be $h$, then the third side is $\frac{2 S}{h}$. Clearly, $\frac{2 S}{12}<\frac{2 S}{4}, \\
\frac{2 S}{12}+\frac{2 S}{4}>\frac{2 S}{h}
\end{array} \Righ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,297 |
Initial 69. Given that $m$ is an odd number. Prove that:
$$
k^{m}+(2 k)^{m}+\cdots+(1998 k)^{m}
$$
is a multiple of $k+2 k+\cdots+1998 k$. | Proof: Let $M=1^{m}+2^{m}+\cdots+1998^{m}$. Then
$$
\begin{aligned}
2 M= & \left(1^{m}+1998^{m}\right)+\left(2^{m}+1997^{m}\right)+\cdots \\
& +\left(1998^{m}+1^{m}\right)
\end{aligned}
$$
Since $m$ is odd, $2 M$ is a multiple of 1999.
$$
\begin{array}{l}
\text { Also, } M=0^{m}+1^{m}+2^{m}+\cdots+1998^{m}, \text { t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,298 |
69. Given the three sides of $\triangle A B C$ are $a, b, c$, and $2 \leqslant n \in N$. Prove:
$$
\frac{\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}}{a+b+c}<1+\frac{\sqrt[n]{2}}{2} .
$$ | Proof: Without loss of generality, let $a \leqslant b \leqslant c$.
$$
\begin{array}{l}
\because a+b>c, \\
\therefore \frac{\sqrt[n]{2}}{2}(a+b+c)>\frac{\sqrt[n]{2}}{2}(c+c) \\
\quad=\sqrt[n]{2 c^{n}} \geqslant \sqrt[n]{b^{n}+c^{n}},
\end{array}
$$
That is,
$$
\begin{array}{l}
\sqrt[n]{b^{n}+c^{n}}<\frac{\sqrt[n]{2}}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,300 |
$\begin{array}{l}\cos ^{k} \frac{\pi}{2 n+1}+\cos ^{k} \frac{3 \pi}{2 n+1}+\cdots+\cos ^{k} \frac{(2 n-1) \pi}{2 n+1} \\ =\frac{1}{2} .\end{array}$ | Proof: First, we prove a lemma:
If $\sin \frac{d}{2} \neq 0$, then $\sum_{i=1}^{n} \cos [x+(i-1) d]=$
$$
\frac{\cos \left(x+\frac{n-1}{2} d\right) \sin \frac{n}{2} d}{\sin \frac{d}{2}} \text {. }
$$
In fact, $\sin \frac{d}{2} \sum_{i=1}^{n} \cos [x+(i-1) d]$
$$
\begin{array}{l}
=\sum_{i=1}^{n} \cos [x+(i-1) d] \sin \f... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,301 |
Example 12 Let $a, b, c, d, e$ be real numbers, and $a+b+c+d+e=8, a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16$. Then the maximum value of $e$ is . $\qquad$ | Solution: From the given, we have
$$
a+b+c+d=8-e
$$
and $a^{2}+b^{2}+c^{2}+d^{2}=16-e^{2}$.
Let $f(x)=4 x^{2}-2(a+b+c+d) x$
$+\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$
$=(x-a)^{2}+(x-b)^{2}+(x-$
$c)^{2}+(x-d)^{2} \geqslant 0$.
On the other hand, the coefficient of the quadratic term is 4, so $\Delta \leqslant 0$,
i.e., $... | \frac{16}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,302 |
Example 13 Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ all be natural numbers, and $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5}$. Try to find the maximum value of $x_{5}$. | Solution: Without loss of generality, let $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}$.
$$
\because x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} \text {. }
$$
$x_{4} x_{5}$,
$$
\begin{aligned}
\therefore 1= & \frac{1}{x_{2} x_{3} x_{4} x_{5}}+\frac{1}{x_{1} x_{3} x_{4} x_{5}} \\
& +\frac{1... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,303 |
Example 14 Try to find the maximum value of the following expression:
$$
P=|\cdots|\left|x_{1}-x_{2}\right|-x_{3}|-\cdots|-x_{1995} \mid,
$$
where $x_{1}, x_{2}, \cdots, x_{1995}$ are different natural numbers from 1 to 1995. | Solution: It is evident that when $x_{1} \geqslant 0, x_{2} \geqslant 0$, $\left|x_{1}-x_{2}\right|$ does not exceed the largest of $x_{1}$ and $x_{2}$; for $x_{1} \geqslant 0, x_{2} \geqslant$ $0, x_{3} \geqslant 0$, $\left|\left|x_{1}-x_{2}\right|-x_{3}\right|$ does not exceed the largest of $x_{1}$, $x_{2}$, and $x_... | 1994 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,304 |
3. If $x \neq 0$, then the maximum value of $\frac{\sqrt{1+x^{2}+x^{4}}-\sqrt{1+x^{4}}}{x}$ is $\qquad$ (1992, National Junior High School Competition) | 3. $\sqrt{3}-\sqrt{2}$ | \sqrt{3}-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,307 |
4. If real numbers $x, y$ satisfy the condition $2 x^{2}-6 x+y=0$, then the maximum value of $x^{2}+y^{2}+2 x$ is ( ).
(A) 14
(B) 15
(C) 16
(D) Cannot be determined | 4.B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,308 |
5. Given $|y| \leqslant 1$, and $2 x+y=1$. Then the minimum value of $2 x^{2}+16 x+$ $3 y^{2}$ is ( ).
(A) $\frac{19}{7}$
(B) 3
(C) $\frac{27}{7}$
(D) 13
(1997, Taiyuan City Junior High School Mathematics Competition) | 5.B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,309 |
6. If $a, b, c, d$ are four positive numbers whose product is 1, then the minimum value of the algebraic expression $a^{2}+b^{2}+c^{2}+d^{2}+a b+a c+a d+b c+b d+c d$ is ( ).
(A) 0
(B) 4
(C) 8
(D) 10
(1997, Tianjin City Junior High School Mathematics Competition) | 6. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,310 |
Example 3 If $x y=1$, then the minimum value of the algebraic expression $\frac{1}{4}+\frac{1}{4 y^{4}}$ is $\qquad$ | Solution: From $a+b \geqslant 2 \sqrt{a b}$ we know $\frac{1}{x^{4}}+\frac{1}{4 y^{4}} \geqslant 2 \sqrt{\frac{1}{x^{4} \cdot 4 y^{4}}}=1$. Therefore, the minimum value of $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,311 |
Example 1: From 10 English letters $A, B, C, D, E$, $F, G, X, Y, Z$, select 5 letters (letters can be repeated) to form a word. Arrange all possible words in "dictionary order" (i.e., the order in which English words are arranged in an English-Chinese dictionary), resulting in a word list $A 1 A A A A, A A A A E, A A A... | Analysis: 10 letters, allowing repetition, are used to form 5-letter words, resulting in a total of 100,000 possible words. Solving this problem directly through permutations would be extremely difficult. However, if we assign values to the letters $A, B, \cdots, Y, Z$, the problem can be transformed into a numerical p... | EFFGX | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,312 |
Example 2 On a circle, there are $n$ (a fixed value) points, some of which are colored red, and the remaining points are colored white. They divide the circumference into non-overlapping arcs. We define: arcs with both endpoints red are labeled with the number 2, arcs with both endpoints white are labeled with the numb... | Analysis: From the problem, we know that the value of an arc segment is related to the colors of its endpoints. If we can assign values to the endpoints such that the product of the values of the two endpoints equals the value of the arc segment, the problem can be transformed into a numerical operation, making it easi... | 2^{2m-n} | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,313 |
Example: $3 A, B, C, D, E$ five people participate in an exam, with 7 questions, all of which are true or false questions. The scoring rule is: for each question, a correct answer earns 1 point, a wrong answer deducts 1 point, and no answer neither earns nor deducts points. Figure 1 records the answers of $A, B, C, D, ... | Let: Assign $k=1,2, \cdots, 7$. When the conclusion of the $k$-th question is correct, i.e., $x_{k}:=1$, if it is judged as correct (i.e., marked with the symbol “$\checkmark$”), then $x_{k}$ points are scored; if it is judged as incorrect (i.e., marked with the symbol “$X$”), then $-x_{k}$ points are scored. When the ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 710,314 |
Example 4: Now there are $2n$ men and $2n$ women, forming two concentric circles for dancing, with $2n$ people in each circle, including both men and women. The outer circle faces inward, and the inner circle faces outward. The dancing rules are as follows: whenever the music starts, if the people facing each other are... | Proof: Let men be denoted as +1 and women as -1, then in the sequence of $2n$ numbers $a_{1}, a_{2}, \cdots, a_{2 n}$ and the sequence of $2n$ numbers $b_{1}, b_{2}, \cdots, b_{2 n}$, there are $2n$ +1s and $2n$ -1s. Therefore,
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2 n}+b_{1}+b_{2}+\cdots+b_{2 n} \\
=0 .
\end{array... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,315 |
Example 6 Given two points 1 and $\sqrt{2}$ on the number line, take $n$ points arbitrarily in the interval $(1, \sqrt{2})$. Among these $n+2$ points, connect each pair of adjacent points with a line segment, resulting in $n+1$ line segments. Prove: Among these $n+1$ line segments, the number of segments with one ratio... | Proof: Arrange the $n+2$ points in ascending order and denote them as $A_{1}, A_{2}, \cdots, A_{n+2}$. Assign integer values to each point as follows:
$$
a_{i}=\left\{\begin{array}{l}
1, \text { when } A_{i} \text { is a rational point, } \\
-1, \text { when } A_{i} \text { is an irrational point. }
\end{array}\right.
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,317 |
Prove that 15 rectangular tiles of size $1 \times 4$ and 1 rectangular tile of size $2 \times 2$ cannot exactly cover an $8 \times 8$ rectangular floor. | Prove: First, assign each small square in an $8 \times 8$ grid the values as shown in Figure 2(a). It is evident that for each $1 \times 4$ tile, no matter how it is placed, the four small squares it covers must have the values $1, 2, 3, 4$. For a $2 \times 2$ tile, no matter how it is placed, the four small squares it... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,318 |
Example 8 If $1,2,3, \cdots, 14$ are taken in increasing order as $a_{1}, a_{2}, a_{3}$, such that both $a_{2}-$ $a_{1} \geqslant 3$ and $a_{3}-a_{2} \geqslant 3$ are satisfied. Find the total number of all different ways of selection.
(1989, National High School Competition) | Solution: Assignment
$i=1,2, \cdots, 14$. Then any selection of three numbers from the 14 numbers corresponds to a permutation $\left(x_{1}, x_{2}, \cdots, x_{14}\right)$, and conversely, any permutation $\left(x_{1}, x_{2}, \cdots, x_{14}\right)$ must correspond to a selection. Therefore, a selection $\rightarrow\left... | C_{10}^{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,319 |
Example 9 Use $n$ numbers (repetition allowed) to form a sequence of length $N$, and $N \geqslant 2^{n}$. Prove: It is possible to find several consecutive terms in this sequence, whose product is a perfect square. | Proof: Let $n$ numbers $a_{1}, a_{2}, \cdots, a_{n}$ form a sequence of length $N$: $b_{1}, b_{2}, \cdots, b_{n}$, where $b_{i} \in \{a_{1}, a_{2}, \cdots, a_{n}\}, i=1,2, \cdots, N$.
Establish the mapping
$$
B=\{b_{1}, b_{2}, \cdots, b_{n}\} \rightarrow \{v_{1}, v_{2}, \cdots, v_{n}\},
$$
where $v_{j}=(c_{1}, c_{2}, ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,320 |
Example 4 Given that $x, y, z$ are all positive numbers, and $x y z \cdot (x+y+z)=1$. Then, the minimum value of $(x+y)(y+z)$ is . $\qquad$ | Analysis: To find the minimum value of the original expression, that is, to find the minimum value of $x y+y^{2}+$ $x z+y z$, we can achieve this by appropriately combining terms so that the sum becomes the sum of two terms, and the product of these two terms should be a constant.
$$
\text { Solution: } \begin{array}{l... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,322 |
Proposition $1 \quad A_{n} \geqslant \frac{1}{12}\left(n^{2}-4 n-5\right), n \geqslant 6$. | Proof: By drawing a diameter through one of the fixed points, and then drawing two more diameters, the unit circle is divided into six congruent closed sectors. In each closed sector, the distance between any two points does not exceed 1. Points on these three diameters can be considered as belonging to two sectors, wi... | A_{n} \geqslant \frac{1}{12}\left(n^{2}-4 n-5\right) | Inequalities | proof | Yes | Yes | cn_contest | false | 710,323 |
Proposition: Among the 100 natural numbers from 1 to 100, if any 16 numbers are chosen, there must exist four numbers such that the difference between two of them equals the difference between the other two (the "difference" refers to the difference between the larger and smaller numbers).
Analysis: To prove this prop... | Proof: In the 100 natural numbers from 1 to 100, if we arbitrarily select 16 numbers, denoted as
$$
a_{1}s$, subtracting the two equations, we get
$$
a_{k}-a_{s}=a_{t}-a_{j} \text {. }
$$
Since $s>i>t$, it follows that $a_{k}>a_{s}>a_{t}>a_{j}$, so $a_{k}-a_{s}$ and $a_{t}-a_{j}$ form a pair of equal non-adjacent diff... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,326 |
Proposition 1 In the tetrahedron $A_{1} A_{2} A_{3} A_{4}$, the distances between the three pairs of opposite edges $A_{1} A_{2} 、 A_{3} A_{4}, A_{1} A_{3} 、 A_{2} A_{4}, A_{1} A_{4} 、 A_{2} A_{3}$ are $d_{1} 、 d_{2} 、 d_{3}$, respectively. Then for the Fermat point and $L$, we have
$$
L \geqslant \frac{2}{3} \sqrt{3}\... | Proof: In the tetrahedron \( A_{1} A_{2} A_{3} A_{4} \), take the Fermat point \( M \), and construct planes \( S_{i}^{\prime} \perp M A_{i} \) at \( A_{i} (1 \leqslant i \leqslant 4) \). Then \( S_{i}^{\prime} \) intersect to form a tetrahedron \( A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime} A_{4}^{\prime} \) (where \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,327 |
In the tetrahedron $A_{1} A_{2} A_{3} A_{4}$, the centroid of the face opposite $A_{i}$ is $G_{i}(1 \leqslant i \leqslant 4)$, then we have
$$
\sum A_{1} G_{1} \geqslant \frac{8}{9} \sqrt{3} \sum d_{1} \text {. }
$$ | Proof: Let the centroid of the tetrahedron \( A_{1} A_{2} A_{3} A_{4} \) be \( G \). Then, by the minimality of \( L \) and (7), we have
\[
\sum A_{1} G \geqslant L \geqslant \frac{2}{3} \sqrt{3} \sum d_{1} .
\]
By the properties of the centroid of a tetrahedron and \( A_{1} G: G G_{1} = 3: 1 \), i.e., \( A_{1} G = \f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,328 |
In the tetrahedron \(A_{1} A_{2} A_{3} A_{4}\), the distance from \(A_{i}\) to the face \(S_{i}\) is \(h_{i}(1 \leqslant i \leqslant 4)\). For the Fermat point \(M\), we have
\[
\begin{array}{l}
\frac{2}{27} \sqrt{3} \sum\left(M A_{1} \cdot M A_{2} \cdot M A_{3}\right) \geqslant V \\
\geqslant \frac{\sqrt{3}}{32} \sum ... | Prove: In the tetrahedron \( A_{1} A_{2} A_{3} A_{4} \), there is the inequality \({ }^{[3]}\)
\[
\sum \lambda_{1} S_{1}^{2} \geqslant \frac{9}{2} \sqrt[3]{6}\left(\sum_{2} \lambda_{1} \lambda_{2} \lambda_{3}\right)^{\frac{1}{3}} V^{\frac{4}{3}}.
\]
If we let \(\lambda_{i}=\dot{h}_{i}^{2} (1 \leqslant i \leqslant 4)\)... | V \geqslant \frac{\sqrt{3}}{32} \sum h_{1} h_{2} h_{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 710,330 |
Proposition Let $a_{i} \in R(1 \leqslant i \leqslant n), n \in N$, and denote $y=a\left|\prod_{i=1}^{n} \cos \alpha_{i}\right|+b\left|\prod_{i=1}^{n} \sin \alpha_{i}\right|$. Then
(I) When $a b0, b>0$, if $n=1$, then $\min (a, b) \leqslant y \leqslant \sqrt{a^{2}+b^{2}}$; if $n \geqslant 2$, then $0 \leqslant y$ $\leqs... | Proof: (I) Without loss of generality, let $a<0<b$, then $a\left|\prod_{i=1}^{n} \cos \alpha_{i}\right| \leqslant 0 \leqslant b\left|\prod_{i=1}^{n} \sin \alpha_{i}\right|$.
Notice that $\left|\cos \alpha_{i}\right| \leqslant 1,\left|\sin \alpha_{i}\right| \leqslant 1$, we get
$$
\begin{array}{l}
\min (a, b)=a \leqslan... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,331 |
Proposition: Let the three sides of $\triangle A B C$ be $a, b, c$, and the radii of the excircles be $r_{a}, r_{b}, r_{c}$. Then
$$
\left(\frac{a}{r_{a}}\right)^{n}+\left(\frac{b}{r_{b}}\right)^{n}+\left(\frac{c}{r_{c}}\right)^{n} \geqslant 2^{n} \cdot 3^{1-\frac{n}{2}}(n>0) .
$$ | $$
\begin{array}{l}
\left(\frac{a}{r_{a}}\right)^{n}+\left(\frac{b}{r_{b}}\right)^{n}+\left(\frac{c}{r_{c}}\right)^{n} \geqslant 3\left(\frac{a b c}{r_{a} r_{b} r_{c}}\right)^{\frac{n}{3}} . \\
\because a+b+c \leqslant 3 \sqrt{3} R, \\
\text { and } p=\frac{1}{2}(a+b+c), \\
\therefore 2 p \leqslant 3 \sqrt{3} R .
\end{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,332 |
Example 5 If $a, c, d$ are integers, $b$ is a positive integer, and they satisfy $a+b=c, b+c=d, c+d=a$, then the maximum value of $a+b+c+d$ is ( ).
(A) -1
(B) -5
(C) 0
(D) 1
(1991, National Junior High School Competition) | Analysis: Since $b$ is a positive integer, we can consider $b$ as the main variable and try to eliminate $a, c, d$.
Solution: From the given, we have $c-a=b, d-c=b, c+d-a=0$. Solving the system of equations for $a, c, d$ yields
$$
a=-3b, c=-2b, d=-b \text{. }
$$
Thus, $a+b+c+d=-5b \leqslant-5$. (Since $b$ is a positi... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,333 |
Proposition Let $m=\min \left(a_{1}, a_{2}, \cdots, a_{n}\right), M$
$$
\begin{array}{l}
=\max \left(a_{1}, a_{2}, \cdots, a_{n}\right), m>0 . \text { Then } \\
\quad\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) \\
\geqslant n^{2}+\left(\sqrt{\frac{M}{m}}-\sqrt... | Proof: Without loss of generality, let $a_{1}=m, a_{n}=M$, and $\frac{a_{2}+\cdots+a_{n-1}}{n-2}=a$. Then
$$
\begin{aligned}
\frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n-1}} \\
\geqslant \frac{(n-2)^{2}}{a_{2}+a_{3}+\cdots+a_{n-1}}=\frac{n-2}{a} .
\end{aligned}
$$
The equality holds if and only if $a_{2}=a_{3}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,334 |
In $\triangle A B C$, the three sides $a, b, c$ satisfy $b+c \leqslant 2 a, a+c \leqslant 2 b$. Then the range of $\frac{b}{a}$ is ( ).
(A) $(0, \infty)$
(B) $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$
(C) $\left(\frac{1}{2}, 2\right)$
(D) $(0,2)$ | Given pipe:
From the conditions, $c$ is the smallest (otherwise, if $b < 2b$, it leads to a contradiction; if $a < 2a$, it contradicts the given information).
From $\frac{b}{a}+\frac{c}{a} \leqslant 2$ and $1+\frac{c}{a} \leqslant 2 \cdot \frac{b}{a}$, we have $\frac{1}{2}+\frac{c}{2a} \leqslant \frac{b}{a} \leqslant 2... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 710,336 |
This article presents a method of solving the Pell equation
$x^{2}-A y^{2}=1$ (where $A$ is a non-square integer) through a specific formulation.
Proposition For the equation (1), express $A$ in the form $A=\left(\frac{p}{q}\right)^{2}+B$ (where $p$ and $q$ are positive integers, $p>q, (p, q)=1, B \neq 0$), such that ... | Solution: Let $A=\left(\frac{p}{q}\right)^{2}+B$, then $p=2 h m^{2} \pm 2$, $q=m$. Therefore,
$$
B=A-\left(\frac{p}{q}\right)^{2}=-\frac{4}{m^{2}}<0
$$
and $\frac{2 p}{A q^{2}-p^{2}}=-h m^{2} \mp 1$,
which means $\left(A q^{2}-p^{2}\right) \mid 2 p$.
Thus, the solution to (1) is
$$
\left\{\begin{array}{l}
y=h m^{3} \p... | \left\{\begin{array}{l}
y=h m^{3} \pm m, \\
x=2\left(h m^{2} \pm 1\right)^{2}-1
\end{array}\right.
} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,337 |
1. If $a>1, b>1$, and $\lg (a+b)=\lg a+$ $\lg b$, then the value of $\lg (a-1)+\lg (b-1)$ is ( ).
(A) equal to $\lg 2$
(B) equal to 1
(C) equal to 0
(D) not a constant independent of $a$ and $b$ | $$
\begin{array}{l}
-1 . C . \\
\because \lg (a+b)=\lg a+\lg b, \\
\therefore a+b=a b, \text { i.e., }(a-1)(b-1)=1 .
\end{array}
$$
Therefore, $\lg (a-1)+\lg (b-1)=0$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,338 |
3. For a geometric sequence $\left\{a_{n}\right\}$ with all terms being real numbers, the sum of the first $n$ terms is denoted as $S_{n}$. If $S_{10}=10, S_{30}=70$, then $S_{40}$ equals ( ).
(A)150
(B) -200
(C)150 or -200
(D) 400 or -50 | 3. A.
Let $b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, b_{3}=S_{30}-S_{20}, b_{4}=$ $S_{\infty 0}-S_{30}$. Suppose $q$ is the common ratio of $\left\{a_{n}\right\}$, then $b_{1}, b_{2}, b_{3}, b_{4}$ form a geometric sequence with the common ratio $r=q^{10}$. Therefore, $70=S_{30}=b_{1}+$
$$
\begin{array}{l}
b_{2}+b_{3}=b_{1}\... | 150 | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,340 |
4. Let proposition $P$ : The solution sets of the inequalities $a_{1} x^{2}+$ $b_{1} x+c_{1}>0$ and $a_{2} x^{2}+b_{2} x+c_{2}>0$ with respect to $x$ are the same; Proposition $Q: \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$. Then proposition $Q$ ( ).
(A) is a sufficient and necessary condition for prop... | 4.D.
For example, the solution sets of $x^{2}-3 x+2>0$ and $-x^{2}+3 x-2>0$ are different, thus eliminating (A) and (B); also, the solution sets of $x^{2}+x+1>0$ and $x^{2}+x+3>0$ are the same, thus eliminating (C). | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 710,341 |
5. Let $E$, $F$, and $C$ be the midpoints of the edges $\overline{AB}$, $BC$, and $CD$ of a regular tetrahedron $ABCD$, respectively. Then the size of the dihedral angle $C$ $F G-E$ is ( ).
(A) $\arcsin \frac{\sqrt{6}}{3}$
(B) $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$
(C) $\frac{\pi}{2}-\operatorname{arctg} \sqrt{2}$
... | 5.D.
As shown in the figure, construct $E E_{1} \perp$
plane $B C D$, then $E E_{1} \perp$
$B G$.
$\because A B C D$ is a regular
tetrahedron,
$$
\begin{array}{l}
\therefore A C \perp B D . \\
\text { Also, since } E F / / A C, \\
F G / / B D, \\
\therefore E F \perp F G .
\end{array}
$$
Thus, $E_{1} F \perp F G$. T... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,342 |
6. Among the 8 vertices, 12 midpoints of edges, 6 centers of faces, and the center of the cube, a total of 27 points, the number of groups of three collinear points is ( ).
(A) 57
(B) 49
(C) 43
(D) 37 | 6. B.
The number of collinear triplets with both endpoints being vertices is $\frac{8 \times 7}{2}=28$ (groups); the number of collinear triplets with both endpoints being the centers of faces is $\frac{6 \times 1}{2}$ $=3$ (groups); the number of collinear triplets with both endpoints being the midpoints of edges is ... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 710,343 |
Example 6 Given that $a, b, c$ are three non-negative real numbers, and satisfy $3a+2b+c=5, 2a+b-3c=1$. If $S=$ $3a+b-7c$, then the sum of the maximum and minimum values of $S$ is $\qquad$ | Analysis: There are three variables in $S$, and we should try to eliminate two variables through the method of elimination to facilitate finding the extremum.
Solution: From $3a + 2b = 5 - c$ and $2a + b = 3c + 1$, we get $a = 7c - 3, b = 7 - 11c$.
Thus, $S = 3(7c - 3) + 7 - 11c - 7c = 3c - 2$.
Since $a, b, c$ are non... | -\frac{62}{77} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,344 |
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