problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
$$
\begin{array}{l}
\text { 2. Compare the sizes: } \frac{1}{2}(\sqrt{1998}-\sqrt{2000}) \\
\sqrt{1998}-\sqrt{1999} .
\end{array}
$$
3. Given that $\alpha$ is a root of the equation $x^{2}-3 x-1=0$. Then $\alpha^{6}+120 \alpha^{-2}=$ $\qquad$ . | Answer: >Answer:1309 | 1309 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,458 |
$\begin{array}{c}\text { Let } x_{1}, x_{2}, \cdots, x_{n}, y_{1}, y_{2}, \cdots, y_{n} \in[a, b] \\ (0<a<b) \text { and } \sum_{i=1}^{n} x_{i}^{2}=\sum_{i=1}^{n} y_{i}^{2} \text {. Prove: } \\ \sum_{i=1}^{n} \frac{x_{i}^{3}}{y_{i}} \leqslant \frac{a^{4}+b^{4}}{a^{3} b+a b^{3}} \sum_{i=1}^{n} x_{i}^{2} .\end{array}$ | Proof: $\because \frac{\sqrt{\frac{x_{i}^{3}}{y_{i}}}}{\sqrt{x_{i} y_{i}}}=\frac{x_{i}}{y_{i}}, x_{i}, y_{i} \in[a, b]$, $i=1,2, \cdots n$,
$$
\therefore \frac{a}{b} \leqslant \frac{\sqrt{\frac{x_{i}^{3}}{y_{i}}}}{\sqrt{x_{i} y_{i}}} \leqslant \frac{b}{a} \text {, }
$$
i.e., $\left(\frac{a}{b} \sqrt{x_{i} y_{i}}-\sqrt... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,460 |
Proposition As shown in the figure, for a point $P$ inside a regular $\triangle A B C$ with side length $a$, the distances from $P$ to the vertices are $u, v, w$, and
satisfy the relation $u^{2}+v^{2}$ $=w^{2}$, then
$$
w^{2}+\sqrt{3} u v=a^{2} .
$$ | Proof: Let $\triangle ABC$,
$\triangle PAB, \quad \triangle PBC$,
$\triangle PCA$ have areas
$S, S_{1}, S_{2}, S_{3}$, respectively.
Rotate $\triangle PCA$ around point $A$ by $60^{\circ}$ clockwise to get $\triangle P^{\prime}AB$, and connect $PP^{\prime}$. Then $\triangle APP^{\prime}$ is an equilateral triangle with... | w^{2}+\sqrt{3} u v=a^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 710,461 |
Proposition Let the circumradius of $\triangle A B C$ be $R$, and the vertices of the equilateral
$D E F$ lie on the three sides of $\triangle A B C$. Then
$\qquad$ the minimum side length of $D E F$ is
$$
\frac{2 R \sin A \sin B \sin C}{\sqrt{1+\sqrt{3} \sin A \sin B \sin C+\cos A \cos B \cos C}} .
$$ | Proof: As shown in the figure, without loss of generality, let $\angle A \leqslant \angle B \leqslant \angle C$, then $\angle A \leqslant 60^{\circ} \leqslant \angle C$. Let $BC = a$, $\angle DEC = \varphi$. The side length of $\triangle DEF$ is $x$, and $\angle AEF = 120^{\circ} - \varphi$, while $\angle AFE = 60^{\ci... | \frac{2 R \sin A \sin B \sin C}{\sqrt{1+\sqrt{3} \sin A \sin B \sin C+\cos A \cos B \cos C}} | Geometry | proof | Yes | Yes | cn_contest | false | 710,462 |
Proposition If $h_{a}, h_{b}, h_{c}, r$ are the lengths of the altitudes and the inradius of a triangle, $0<\lambda \leqslant 2$, then
$$
\frac{1}{h_{a}-\lambda r}+\frac{1}{h_{b}-\lambda r}+\frac{1}{h_{c}-\lambda r} \geqslant \frac{3}{(3-\lambda) r} \text {, }
$$
with equality if and only if the triangle is equilatera... | Proof: $\because \frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}=\frac{1}{r}$,
$$
\therefore \frac{h_{a}-\lambda r}{h_{a}}+\frac{h_{b}-\lambda r}{h_{b}}+\frac{h_{c}-\lambda r}{h_{c}}=3-\lambda \text {. }
$$
Thus, $\frac{h_{a}}{h_{a}-\lambda r}+\frac{h_{b}}{h_{b}-\lambda r}+\frac{h_{c}}{h_{c}-\lambda r} \geqslant \frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,463 |
Given
\[
\begin{array}{l}
a_{11} x_{1}^{a_{1}}+a_{12} x_{2}^{a_{1}}+\cdots+a_{1 n} x_{n}^{a_{n}}=y_{1}^{a_{1}}, \\
a_{21} x_{1}^{a_{2}}+a_{22} x_{2}^{a_{2}}+\cdots+a_{2 n} a_{n}^{a_{n}}=y_{2}^{a_{2}}, \\
\cdots \cdots \\
a_{n i} x_{1}^{a_{n}}+a_{n 2} x_{2}^{a_{2}}+\cdots+a_{n n} x_{n}^{a_{n}}=y_{n}^{a_{n}},
\end{array}... | Proof: By the weighted power mean inequality
$$
\begin{array}{l}
\frac{q_{1} a_{1}+q_{2} a_{2}+\cdots+q_{n} a_{n}}{q_{1}+q_{2}+\cdots+q_{n}} \\
\geqslant\left(a_{1}^{q_{1}} a_{2}^{q_{2}} \cdots a_{n}^{q_{n}}\right)^{\frac{1}{q_{1}+q_{2}+\cdots+q_{n}}},
\end{array}
$$
Let $q_{i}=a_{1 i}, a_{i}=x_{i}^{a_{1}}(i=1,2, \cdo... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,464 |
Example 7 Calculate $\frac{1}{2 \sqrt{1}+\sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}$
$$
+\frac{1}{4 \sqrt{3}+3 \sqrt{4}}+\cdots+\frac{1}{100 \sqrt{99}+99 \sqrt{100}} .
$$ | $\begin{array}{l}\text { Solution: } \because \frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}} \\ =\frac{1}{\sqrt{n} \cdot \sqrt{n+1}(\sqrt{n}+\sqrt{n+1})} \\ =\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n} \cdot \sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}, \\ \therefore \text { the original expression }=\left(\frac{1}{\sqrt{1}}-\fr... | \frac{9}{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,465 |
Find a positive integer $k$ such that
(a) For any positive integer $n$, there does not exist $j$ satisfying $0 \leqslant j \leqslant n - k + 1$, and $C_{n}^{j}, C_{n}^{j+1}, \cdots, C_{n}^{j+k-1}$ form an arithmetic sequence;
(b) There exists a positive integer $n$ such that there is $j$ satisfying $0 \leqslant j \le... | Proof: Since any two numbers must form an arithmetic sequence, let $k=7=$ $1, k \neq 2$. Now consider three numbers:
$$
C_{n}^{j-1}, C_{n}^{j}, C_{0}^{j+i}(1 \leqslant j \leqslant n-1) .
$$
If they form an arithmetic sequence, then
$$
2 C_{n}^{j}=C_{n}^{j-1}+C_{n}^{j+1} \text {. }
$$
This implies $n+2=(n-2 j)^{2}$,
i... | k=4, n=m^{2}-2(m \in \mathbf{N}, m \geqslant 3) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,467 |
II. $n(\geqslant 5)$ football teams participate in a round-robin tournament. Each pair of teams plays one match, with the winning team getting 3 points, the losing team getting 0 points, and each team getting 1 point in the case of a draw. The team that finishes in 3rd to last place scores fewer points than all the tea... | Solution: Let the $n$ teams be $A_{1}, A_{2}, \cdots, A_{n-3}, B, C_{1}$, $C_{2}$.
Since the $A$ group teams score more but have fewer wins, and the $C$ group teams score less but have more wins, each team in the $A$ group must have at least 8 draws, team $B$ must have at least 1 win, 3 losses, and 4 draws, and each t... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,468 |
For a fixed $\theta \in\left(0, \frac{\pi}{2}\right)$, find the smallest positive number $a$ that satisfies the following two conditions:
(i) $\frac{\sqrt{a}}{\cos \theta}+\frac{\sqrt{a}}{\sin \theta}>1$;
(ii) there exists $x \in\left(1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right)$ such that
$$
\be... | $$
\begin{array}{l}
\left.x \sqrt{\frac{a}{\sin ^{2} \theta}-(1-x)^{2}}\right) \geqslant a \text { . } \\
\text { First, prove a lemma: Let } 0 \\
1, p^{2}+q^{2} \leqslant 1, f(x)=(1-x) \sqrt{p^{2} \cdots x^{2}}+ \\
x \sqrt{q^{2}-(1-x)^{2}} \quad(1-q \leqslant x \leqslant p) \text {. } \\
\text { When } \sqrt{p^{2}-x^{... | a=\frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 710,469 |
Four, As shown in Figure 1, in acute triangle $\triangle ABC$, $H$ is the orthocenter, $O$ is the circumcenter, and $I$ is the incenter. Given that $\angle C>\angle B>\angle A$. Prove: $I$ is inside $\triangle BOH$. | Proof: Let the angle bisector of $\angle B$ intersect $O H$ at $P$, then $B P$ is also the angle bisector of $\angle O B H$.
$$
\therefore \frac{B H}{B O}=\frac{H P}{O P} \text {. }
$$
Let the angle bisector of $\angle A$ intersect $O H$ at $Q$, then $A Q$ is also the angle bisector of $\angle O A H$,
$$
\therefore \f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,470 |
Five, natural number $n \geqslant 3$. In the plane, a line $l$ is given, on which there are $n$ distinct points $P_{1}, P_{2}, \cdots, P_{n}$ in sequence. The product of the distances from point $P_{i}$ to the other $n-1$ points is denoted as $d_{i}(i=1,2, \cdots, n)$. There is also a point $Q$ in the plane not on $l$,... | Let's assume these $n$ points are on the real axis, with coordinates $P_{i}\left(x_{i}, 0\right) (i=1,2, \cdots, n), x_{1}3 \text { when. }
\end{array}\right.
$$
Substituting into (1) we get
$$
\begin{aligned}
S_{n} & =\left(a^{2}+\beta^{2}\right) T_{0}-2 \alpha T_{1}+T_{2}=T_{2} \\
& =\left(\begin{array}{l}
1, n=3 \t... | 1, n=3 \text { when, } \\
0, n>3 \text { when. } | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,471 |
Six, for any given $h=2^{r} (r$ is a non-negative integer). Find all natural numbers $k$ that satisfy the following conditions: for each such $k$, there exist an odd natural number $m>1$ and a natural number $n$, such that
$$
k\left|m^{h}-1, m\right| n^{\frac{m^{h}-1}{k}}+1 \text {. }
$$ | For $h=2^{r}$, we agree to denote the set of all $k$ that satisfy the problem conditions as $k(h)$. We will prove:
$$
k(h)=\left\{\left.2^{r+s} t\right|_{s, t} \in \mathbf{N}, 2 \nmid t\right\} \text {. }
$$
We will use the following fact:
$$
m \equiv 1(\bmod 4) \Rightarrow 2^{r} \| \frac{m^{2^{r}}-1}{m-1} \text {. }
... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,472 |
15.- An infinite arithmetic sequence of positive integers contains one term that is a perfect square and another term that is a perfect cube. Prove: This sequence contains a term that is a perfect sixth power. | Proof: Let the sequence $\{a+i h: i=0,1,2, \cdots\}$ contain $x^{2}$ and $y^{3}$ terms, where $x$ and $y$ are integers. Using mathematical induction on the common difference $h$, for $h=1$, it is obviously true. For some fixed $h>1$, assume that all arithmetic sequences with a common difference less than $h$ and satisf... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,473 |
18. The altitudes of acute $\triangle ABC$ from vertices $A, B, C$ intersect the opposite sides at points $D, E, F$ respectively. A line through $D$ parallel to $EF$ intersects $AC$ and $AB$ at points $Q$ and $R$ respectively, and $EF$ intersects $BC$ at point $P$. Prove that the circumcircle of $\triangle P Q K$ passe... | Proof: The existence of point $P$ implies $AB \neq AC$. By symmetry, assume $AB > AC$. Then $P, D$ lie on the line $MC_1$, and points $B, C, E, F$ are concyclic, so $PH \cdot PC = PC \cdot PF$. The circumcircle of $\triangle IEF$ is also the nine-point circle of $\triangle ABC$, which must pass through the midpoint $M$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,475 |
Example 8 Simplify $\sqrt{11+2(1+5)(1+\sqrt{7})}$ | $\begin{array}{l}\text { Solution: The radicand }=1^{2}+(\sqrt{5})^{2}+(\sqrt{7})^{2} \\ \quad+2 \sqrt{5}+2 \sqrt{7}+2 \sqrt{5} \cdot \sqrt{7} \\ \quad=(1+\sqrt{5}+\sqrt{7})^{2} . \\ \therefore \text { the expression }=1+\sqrt{5}+\sqrt{7} .\end{array}$ | 1+\sqrt{5}+\sqrt{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,476 |
10. Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant a_{, \cdots}=0$ be a sequence of real numbers. Prove:
$$
\sqrt{\sum_{k=1}^{1} u_{k}} \leqslant \sum_{k=1}^{n} \sqrt{k}\left(\sqrt{a_{k}}-\sqrt{a_{k+1}}\right) .
$$ | Proof 1 Restate the conclusion to be proved as follows:
For each non-increasing, non-negative real number sequence $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, the inequality
$$
\sqrt{\sum_{k=1}^{n} a_{k}} \leqslant \sum_{k=1}^{n-1} \sqrt{k}\left(\sqrt{a_{k}}-\sqrt{a_{k+1}}\right)+\sqrt{n a_{n}}
$$
holds.... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,477 |
20. Let $D$ be an interior point on side $B C$ of $\triangle A B C$. $A D$ intersects the circumcircle of $\triangle A B C$ at $X$. $P$ and $Q$ are the feet of the perpendiculars from $X$ to $A B$ and $A C$, respectively. $\Gamma$ is the circle with diameter $X D$. Prove that $P Q$ is tangent to $\Gamma$ if and only if... | Proof: Let $A^{\prime}$ be the point on the circumcircle's diameter opposite to $A$. Due to the symmetry of $B$ and $C$, without loss of generality, we can assume that $X$ lies on $\overparen{B A}^{\prime}$. Then $Q$ lies on the ray $C A$, and $P$ lies on the ray $A B$, and is outside the circumcircle (note: this arran... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,478 |
22. (a) Do there exist functions $f: \mathbf{R} \rightarrow \mathbf{R}, g: \mathbf{R} \rightarrow \mathbf{R}$, such that for all $x \in \mathbf{R}$, we have $f(g(x))=x^{2}, g(f(x))=x^{3}$?
(b) Do there exist functions $f: \mathbf{R} \rightarrow \mathbf{R}, g: \mathbf{R} \rightarrow \mathbf{R}$, such that for all $x \in... | Solution: (a) If such functions $f, g$ exist, from $g(f(x))=x^{3}$, we know that when $x_{1} \neq x_{2}$, $f\left(x_{1}\right) \neq f\left(x_{2}\right)$, especially $f(0), f(1), f(-1)$ are three different real numbers. On the other hand, according to the problem's requirements, we can get $(f(x))^{2}=f(g(f(x)))=f\left(... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,479 |
23. Let $A B C D$ be a convex quadrilateral, and $O$ the intersection of diagonals $A C$ and $B D$. If $O A \sin A + O C \sin C = O B \sin B + O D \sin D$, prove that $A B C D$ is a cyclic quadrilateral. | Proof: First, consider the following fact. Let $O$ be a point inside $\angle U A V$, as shown in Figure 9 11. Let $\angle O A U=\varphi, \angle O A V=\psi$. If $X$ and $Y$ are the orthogonal projections of $O$ onto $A U$ and $A V$, respectively, then by the Law of Sines and the fact that $X$ and $Y$ lie on the circle w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,480 |
25. In $\triangle ABC$, the angle bisectors of $\angle A$, $\angle B$, and $\angle C$ intersect the circumcircle at points $K$, $L$, and $M$ respectively. $R$ is an inner point on side $AB$. Points $P$ and $Q$ are defined by the conditions: $RP \parallel AK$, $BP \perp BL$; $RQ \parallel BL$, $AQ \parallel AK$. Prove: ... | Proof: As shown in Figure 13, let $M R$ intersect the circumcircle $\Gamma$ of $\triangle A B C$ at $X$, then $X$ is the concurrency point of $K P$, $L Q$, and $M R$.
Let $I_a$, $I_b$, $I_c$, and $I$ be the excenters and incenter of $\triangle A B C$. The line $I_a I$ is the external angle bisector of $\angle B$, perp... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,481 |
26. For each integer $n \geqslant 2$, determine the minimum value of
$$
a_{0}+a_{1}+\cdots+a_{n}
$$
satisfying the conditions
$$
a_{0}=1, a_{i} \leqslant a_{i+1}+a_{i+2}, i=0,1, \cdots, n-2
$$
where $a_{0}, a_{1}, \cdots, a_{n}$ are non-negative numbers. | Consider the special case where $a_{i}=a_{i+1}+a_{i+2}$.
Let $u=a_{n-1}, v=a_{n}$, then we have $a_{k}=F_{n-k} u + F_{n-k-1} v, k=0,1, \cdots, n-1$. Here, $F_{i}$ is the $i$-th Fibonacci number $\left(F_{0}=0, F_{1}=1, F_{i+2}=F_{i}+F_{i+1}\right)$. The sum is
$$
a_{0}+a_{1}+\cdots+a_{n}=\left(F_{n+2}-1\right) u+F_{n+1... | \frac{F_{n+2}-1}{F_{n}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,482 |
1. Let $a, b, c, d$ be integers, and $a<2b, b<3c, c<4d$. Given $d<100$, then the maximum possible value of $a$ is ( ).
(A) 2367
(B) 2375
(C) 2391
(D) 2399 | $-.1 .(\mathrm{A})$.
Then, if $d=99$, we have $c<4 \times 99=395$. Taking $c=395$, we get $b<3 \times 395=1185$. Taking $b=1184$, we get $a<2368$. Therefore, the greatest integer for $a$ is 2367. | 2367 | Inequalities | MCQ | Yes | Yes | cn_contest | false | 710,483 |
2. Given the three sides of an obtuse triangle are 3, $4$, and $x$. Then the range of $x$ is ( ).
(A) $1<x<7$
(B) $5<x<$ ?
(C) $1<x<\sqrt{7}$
(I)) $5<x<7$ or $1<x<\sqrt{7}$ | 2. (i)).
If $x$ is the longest side, then by the cosine rule we get $5<x<7$; if $x$ is the shortest side, then by the cosine rule we get $1<x<\sqrt{7}$. | 5<x<7 \text{ or } 1<x<\sqrt{7} | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,484 |
4. If $x+\frac{1}{y}=1, y+\frac{1}{z}=1$, then the value of $x y z$ is ( ).
(A) 1
(B) -1
(C) $\pm 1$
(J) $\pm 2$ | 4. (13).
From $1 \frac{1}{y}=1$ we get $x y=y-1$.
From $y \frac{1}{z}=1$ we get $\quad y z=z-1$.
$$
\begin{array}{l}
\quad \mathrm{L}, x y=\frac{(y-1)(z-1)}{y} . \\
x \frac{1}{y}=1-x, y-1=-\frac{1}{z} ; \text { substituting into the above, we get } \\
x y z=-1 .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,486 |
2. The set of natural number solutions for the equation $x^{2}-12 x+y^{2}+2=0$ is $\qquad$ | 2. $\left\{\begin{array}{l}x_{1}=1, \\ y_{1}=3 ;\end{array}\left\{\begin{array}{l}x_{2}=11, \\ y_{2}=3 ;\end{array}\left\{\begin{array}{l}x_{3}=3, \\ y_{3}=5 ;\end{array}\left\{\begin{array}{l}x_{4}=9, \\ y_{4}=5 .\end{array}\right.\right.\right.\right.$
If we consider the equation as a quadratic equation in $x$, then... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,491 |
4. For a $4 n+2$-sided polygon $A_{1} A_{2} \cdots A_{4 n-2}$ (where $n$ is a natural number), each interior angle is an integer multiple of $30^{\circ}$, and $\angle A_{1}=\angle A_{2}=\angle A_{3}=90^{\circ}$, then the possible values of $n$ are $\qquad$ | 4. 1 .
From the problem, we get
$$
\begin{array}{l}
90^{\circ} \times 3+(4 n+2-3) \times 30^{\circ} \mathrm{k} \\
=(4 n+2-2) \times 180^{\circ},
\end{array}
$$
where $k \geqslant 4 n-1$.
Simplifying, we get $(4 n-1) k=24 n-9$,
which means $24 n-9 \geqslant(4 n-1)^{2}$.
Thus, $8 n^{2}-16 n+5 \leqslant 0$.
Clearly, $0... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,493 |
One, (20 points) As shown in Figure 3, goods are to be transported from a riverside city $A$ to a location $B$ 30 kilometers away from the riverbank. According to the distance along the river, the distance from $C$ to $A$, $AC$, is 40 kilometers.
If the waterway transportation cost is half of the highway transportation... | Let the waterway transportation cost per kilometer be 1, and let $AD$ be $x$ kilometers, then $DC$ is $40-\pi$ kilometers. Since the highway transportation cost per kilometer is twice that of the waterway per kilometer, according to the problem, the total cost is
$$
y=x+2 \sqrt{(40-x)^{2}+30^{2}} \text {. }
$$
Simplif... | x=40-10 \sqrt{3} \approx 23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,494 |
1. Given three quadratic trinomials: $:_{1}(x)=x^{2}+$ $b_{1} x+c_{1}, P_{2}(x)=x^{2}+b_{2} x+c_{2}, P_{3}(x)=$ $x^{2}+b_{3} x+c_{3}$. Then the equation $\left|P_{1}(x)\right|+P_{2}(x)=$ $\left|P_{3}(x)\right|$ has at most $(\quad)$ roots.
(A) 4
(B) 6
(C) 8
(D) None of the above | $-1 .(\mathrm{C})$
Each root of the original equation should be a root of a quadratic trinomial of the form $\pm P_{1}(x)+P_{2}(x) \pm P_{3}(x)$, and there are 4 such quadratic trinomials. Since the coefficient of the $x^{2}$ term has the form $\pm 1+1 \pm 1$, no matter how the signs are chosen, the coefficient of the ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,497 |
Example 10 Given that $n$ is a natural number, and
$m=\left[\frac{(\sqrt{2}+1)^{n}+(\sqrt{2}-1)^{n}}{2}\right]^{2}$. Prove:
$$
(\sqrt{2}-1)^{n}=\sqrt{m}-\sqrt{m-1} \text {. }
$$ | Proof: Let $a=(\sqrt{2}+1)^{n}, b=(\sqrt{2}-1)^{n}$. Then $a b=1(a>b)$, and
$$
\begin{array}{l}
\sqrt{m}=\frac{a+b}{2}, \\
\sqrt{m-1}=\sqrt{\left(\frac{a+b}{2}\right)^{2}-1}=\frac{a-b}{2} .
\end{array}
$$
Therefore, $\sqrt{m}-\sqrt{m-1}=b=(\sqrt{2}-1)^{n}$. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,498 |
2. The function $f(x)=-\log _{\frac{1}{2}}\left(x^{2}-a x-a\right)$ is decreasing on the interval $(-\infty, 1-\sqrt{3})$. Then the range of values for $a$ is ( ).
(A) $0 \leqslant a \leqslant 2$
(B) $2(1-\sqrt{3}) \leqslant a \leqslant 2$
(C) $0 \leqslant a \leqslant 2$ or $a \leqslant-4$
(D) $2(1-\sqrt{3}) \leqslant ... | 2. (B).
$$
y=-\log _{\frac{1}{2}}\left(x^{2}-a x-a\right)=\log _{2}\left(x^{2}-a x-a\right) .
$$
To make $y=\log _{2}\left(x^{2}-a x-a\right)$ a decreasing function on the interval $(-\infty, 1-\sqrt{3})$, it is sufficient that the domain of the function includes $(-\infty, 1-\sqrt{3})$, and that $y^{*}=x^{2}-a x-a$ i... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,499 |
3. In space, there are nine points, among which no four points are coplanar. Connect several line segments between these nine points so that there is no tetrahedron in the graph. Then the graph can have at most ( ) triangles.
(A) 21
(B) 24
(C) 25
(D) 27 | 3. (D)
Divide nine points $A_{1}, A_{2}, \cdots, A_{9}$ into three groups $\mid A_{1}, A_{2}$, $\left.A_{3}\right\},\left\{A_{4}, A_{5}, A_{6}\right\},\left\{A_{7}, A_{8}, A_{4}\right\}$, such that no two points within the same group are connected, while any two points from different groups are connected. For any poin... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 710,500 |
4. Let $A=\{(x, y) \mid 0 \leqslant x \leqslant 2,0 \leqslant y \leqslant 2\}$, $B=\{(x, y) \mid x \leqslant 10, y \geqslant 2, y \leqslant x-4\}$ be point sets on the Cartesian plane $x O y$. Then $C=$ $\left\{\left.\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \right\rvert\,\left(x_{1}, y_{1}\right) \in A... | 4. (D).
Set $A$ is
square $O A B C$,
Set $B$ is
Rt $\triangle D E F$, as
shown in Figure 1.
The midpoint of $O D$
is $M(3,1)$,
The midpoint of $A E$ is
$N(6,1)$, the midpoint of $B F$ is $P(6,4)$, the midpoint of $C F$ is $Q(5,4)$, and the midpoint of $C D$ is $R(3,2)$.
The area of the formed figure is
$S_{\text {MNTQ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,501 |
5. $a_{1}, a_{2}, \cdots, a_{6}$ are six distinct natural numbers whose sum is 23. Then, $a_{1} a_{2}+a_{2} a_{3}+\cdots+$ $a_{5} a_{6}+a_{6} a_{1}$ has the minimum value of ( ).
(A) 62
(B) 64
(C) 65
(D) 67 | 5. (B).
For 6 numbers whose sum is 21, arranged as shown in Figure 2, i.e., $a_{1}=1, a_{2}=6$, $\cdots, a_{6}=5$, the required "sum value" $a_{1} a_{2} +\cdots+a_{6} a_{1}$ is the smallest. At this time, the largest number 6 is surrounded by 1,2; the second largest number 5 is surrounded by 1,3; $\cdots$. Any adjustm... | 64 | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,502 |
$$
\begin{array}{l}
\text { 6. Let } a \in \mathbf{R}^{+}, \\
A=\left\{(x, y) \left\lvert\,(x-1)^{2}+(y-2)^{2} \leqslant \frac{4}{5}\right.\right\} \\
\text { and } B=\{(x, y)|| x-1|+2| y-2 \mid \leqslant a\}
\end{array}
$$
are point sets in the plane $x O y$. Then $A \subseteq B$ if
(A) $a \geqslant 2$
(B) $a \geqsla... | 6. (A).
Set $A$ is a circular region with center at $(1,2)$ and radius $\frac{2}{\sqrt{5}}$. Set $B$ is a rhombus with the diagonal intersection at $(1,2)$, and the two diagonals are parallel to the coordinate axes, with the diagonal parallel to the $x$-axis being twice the length of the other diagonal.
$A \subseteq B... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,503 |
1. On a flat desktop, there are three wooden balls with radii of 1, 2, and 2, each touching the desktop and externally tangent to the other two balls. Additionally, there is a smaller wooden ball with a radius less than 1 between the three balls, touching the desktop and externally tangent to all three balls. What is t... | Two, $1.4-2 \sqrt{3}$.
For two tangent spheres with radii $r_{1}$ and $r_{2}$, the distance between the two points of tangency on the table is $2 \sqrt{r_{1} r_{4}}$.
Let the radius of the smaller sphere be $x$. The problem simplifies to: In $\triangle ABC$, $AB = AC = 2 \sqrt{2}, BC = 4$. $D$ is a point inside the tr... | 4 - 2 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,504 |
2. Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$ be $n$ real numbers satisfying the following condition: for any integer $k>0$, we have $a_{1}^{k}+a_{2}^{k}+$ $\cdots+a_{n}^{k} \geqslant 0$. Then, $p=\max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\right.$ $\left.\cdots,\left|a_{n}\right|\right\}=$ $\q... | 2. $a_{1}$.
Obviously, $p$ is either equal to $a_{1}$, or equal to $\left|a_{n}\right|$.
Assume $p \neq a_{1}$, then $p=\left|a_{n}\right|$, and $p>a_{1}, a_{n}a_{n-k+1}=$ $a_{n-k+2}=\cdots=a_{n}$, then $\left|\frac{a_{1}}{a_{n}}\right|0 .
\end{array}
$$
Thus, $a_{1}^{2 l+1}+\cdots+a_{n}^{2 l+1}$
$$
\begin{array}{l}
... | a_{1} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 710,505 |
3. The sum of $m$ distinct positive even numbers and $n$ distinct positive odd numbers is 117. For all such $m$ and $n$, the maximum value of $3m + 2n$ is $\qquad$ . | 3.37.
Let the $m$ positive even numbers be $a_{i}$, then
$$
\sum_{i=1}^{m} a_{i} \geqslant 2+4+\cdots+2 m=m(m+1) \text {. }
$$
Let the $n$ positive odd numbers be $b_{j}$, then
$$
\sum_{j=1}^{n} b_{j} \geqslant 1+3+\cdots+(2 n-1)=n^{2} .
$$
Thus, the problem reduces to finding the maximum value of $y=3 m+2 n$ under ... | 37 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,506 |
5. The solution set of the inequality $\frac{1}{x-1}+\frac{2}{x-2} \geqslant \frac{3}{2}$, is the union of some non-overlapping intervals with a total length of $\qquad$. | 5.2.
For $g(x)=(x-1)(x-2)>($ or $\text { (or } \leqslant \text { ) } 0 \text {. }$
The graph of $y=f(x)$ is a parabola opening upwards, and the graph of $y=$ $g(x)$ is a parabola opening upwards.
For $f(x):$ when $x=1$, $y0$.
Therefore, the graph intersects the
$x$-axis at two points
$$
\begin{array}{l}
\left(x_{1}, 0... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 710,508 |
Example 1 As shown in Figure 1, in $\triangle A B C$, $A B=A C, D$ is a point on the base $B C$, $E$ is a point on the line segment $A D$, and $\angle B E D=2 \angle C E D=\angle A$. Prove: $B D=2 C D$.
(1992, National Junior High School Mathematics League) | Analysis: The key is to find the connection between $\angle B E D=2 \angle C E D$ and the conclusion. It is easy to think of constructing the angle bisector of $\angle B E D$, but since $B E \neq E D$, it cannot directly prove that $B D=2 C D$. If $A D$ is extended to intersect the circumcircle of $\triangle A B C$ at ... | B D = 2 C D | Geometry | proof | Yes | Yes | cn_contest | false | 710,509 |
Example 2 In a convex quadrilateral $ABCD$, $\angle ABC=60^{\circ}, \angle BAD=$ $\angle BCD=90^{\circ}, AB=2, CD=$ 1 , the diagonals $AC$ and $BD$ intersect at point $O$, as shown in Figure 2. Then $\sin \angle AOB=$ $\qquad$
(1996, Beijing Middle School Mathematics Competition) | Analysis: From $\angle B A D=$ $\angle B C D=90^{\circ}$, we know that points $A, B, C, D$ are concyclic. To find $\sin \angle A O B$, we think of Ptolemy's theorem, which requires us to find $B C$ and $A D$.
Solution: Since $\angle B A D=\angle B C D=90^{\circ}$, points $A, B, C, D$ are concyclic. Extend $B A$ and $C... | \frac{15+6 \sqrt{3}}{26} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,510 |
Example 3 Set $A=\{0,1,2, \cdots, 9\},\left\{B_{1}\right.$, $\left.B_{2}, \cdots, B_{k}\right\}$ is a family of non-empty subsets of $A$, when $i \neq j$, $B_{i} \cap B_{j}$ has at most two elements. Find the maximum value of $k$. | Solution: There are 175 subsets of one, two, and three elements, which clearly meet the requirements, so $k \geqslant 175$.
Assume $k>175$, then among $\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$, there is at least one subset with more than 3 elements. Without loss of generality, let $B_{i}$ have more than 3 elements,... | 175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,511 |
Three, (Full marks 25 points) $a, b, c$ are distinct numbers. If any two of the following three equations hold, prove: the third equation also holds.
$$
\begin{array}{l}
\quad\left(b^{2}+b c+c^{2}\right) x^{2}-b c(b+c) x+b^{2} c^{2}=0 ; \\
\left(c^{2}+c a+a^{2}\right) x^{2}-c a(c+a) x+c^{2} a^{2}=0 ; \\
\left(a^{2}+a b... | Three, due to the symmetry of $a$, $b$, and $c$ in the three equations, without loss of generality, we can consider only the first two equations to prove that the third equation also holds.
The first two equations, after rearrangement, are
$$
\begin{array}{l}
\left(c^{2}-c x+x^{2}\right) b^{2}-c x(c-x) b+c^{2} x^{2}=0,... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,512 |
1. $a$ is a 1998-digit number consisting of 1998 nines, $b$ is a 1998-digit number consisting of 1998 eights, then the sum of the digits of $a \cdot b$ is ( ).
(A) 19980
(B) 19971
(C) 17982
(D) 17991 | $$
\begin{array}{l}
=\underbrace{88 \cdots 87}_{1987 \uparrow 8} \underbrace{11 \cdots 12}_{11 \cdots 71 \cdot 1} . \\
\end{array}
$$
Therefore, the sum of the digits of $a b$ is
$$
8 \times 1997+7+1 \times 1997+2=17982 \text {. }
$$ | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 710,513 |
2. Given $x \in(0,2 \pi)$, then the sum of all roots of the equation $3 \operatorname{ctg}^{2} x+$ $8 \operatorname{ctg} x+3=0$ is ( ).
(A) $3 \pi$
(B) $4 \pi$
(C) $5 \pi$
(D) $6 \pi$ | 2. (C).
Let $\operatorname{ctg} x_{1}, \operatorname{ctg} x_{2}$ be the two roots of the equation in terms of $\operatorname{ctg} x$.
Then $\operatorname{ctg} x_{1}+\operatorname{ctg} x_{2}=-\frac{8}{3}$,
$$
\begin{array}{l}
\operatorname{ctg} x_{1} \cdot \operatorname{ctg} x_{2}=1 . \\
\therefore \operatorname{ctg} x... | 5 \pi | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,514 |
4. Given $x_{n}=\frac{1}{2}\left[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right]$ $(n \in \mathbf{N}), x_{n}$ is a positive integer. Then the units digit of $x_{19981999}$ is ( ).
(A) 1
(B) 2
(C) 6
(D) 7 | From equation (1), the unit digits of $x_{n} (n=1,2, \cdots)$ are $2,7,6,7,2,1,2,7,6, \cdots$.
Let $m\left(x_{n}\right)$ denote the unit digit of $x_{n}$, clearly $m\left(x_{n}\right)=$ $m\left(x_{n}+6 k\right)$, where $1 \leqslant n \leqslant 6, n k \in \mathbf{N}$,
$$
\begin{array}{l}
\therefore m\left(x_{19981999}\... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 710,516 |
5. In $\triangle A B C$, it is known that $\lg \operatorname{tg} \frac{A}{2}, \lg \operatorname{tg} \frac{B}{2}$, $\operatorname{lgtg} \frac{C}{2}$ form an arithmetic sequence. Then the range of $\angle B$ is
(A) $0<\angle B \leq \frac{\pi}{6}$
(B) $0<\angle B \leqslant \frac{\pi}{3}$
(C) $\frac{\pi}{3} \leqslant \angl... | 5. (B).
$$
\begin{array}{l}
\because \operatorname{lgtg} \frac{A}{2}+\operatorname{lgtg} \frac{C}{2}=2 \operatorname{lgt} \frac{B}{2}, \\
\therefore \operatorname{sg} \cdot \operatorname{tg} \frac{C}{2}=\operatorname{tg}^{2} \frac{B}{2}
\end{array}
$$
In $\triangle A \bar{x}$, we always have
$$
\begin{array}{l}
\opera... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,517 |
6. A small ball is placed inside a rectangular container and is in contact with three faces that meet at a point. A point on the ball is $3 \mathrm{~cm}$, $3 \mathrm{~cm}$, and $6 \mathrm{~cm}$ away from these three faces, respectively. Then the radius of this ball is ( ).
(A) only $3 \mathrm{~cm}$
(B) only $6 \mathrm{... | 6. (D).
For a small sphere with center $O$, a point $P$ on the sphere projects onto three planes $\alpha, \beta, \gamma$ intersecting at point $M$ as points $A, B, C$ respectively. The projection of point $O$ onto plane $\gamma$ is $O_{1}$, and $P Q \perp O O_{1}$ at $Q$.
Given that $P A = P B = 3 \text{ cm}, P C = 6 ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,518 |
1. Given $1998^{n} \mid 1999$!, then the maximum value of the positive integer $n$ is | $=1.55$.
$$
\because 1998=2 \times 3^{3} \times 37,3^{3}<37 .
$$
$\therefore 1999$ ! contains the factor of the form $37^{m}$, where the maximum value of $m$ is the maximum value of $n$.
$$
\therefore n \leqslant \sum_{i=1}^{\infty}\left[\frac{1999}{37^{i}}\right]=55 \text {. }
$$
Therefore, the maximum value of $n$ i... | 55 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,519 |
2. Given that $\odot O_{0}$ is the incircle of the equilateral $\triangle A B C$, $\odot O_{1}$ is externally tangent to $\odot O_{0}$ and tangent to two sides of $\triangle A B C$, $\cdots, \odot O_{n+1}$ is externally tangent to $\odot O_{n}$ and tangent to two sides of $\triangle A B C$ $(n \in \mathbf{N})$. Then, t... | 2. $11 \pi: 24 \sqrt{3}$.
Let the radius of $\odot O_{n}$ be $R_{n}(n=0,1,2, \cdots)$.
Since the angle between the two external common tangents of $\odot O_{n}$ and $\odot O_{n+1}$ is $60^{\circ}$,
we have $\sin 30^{\circ}=\frac{R_{n}-R_{n+1}}{R_{n}+R_{n+1}}$,
which means $R_{n}=3 R_{n+1}, \frac{R_{n+1}}{R_{n}}=\frac{... | 11 \pi: 24 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,520 |
3. $P$ is a moving point on the plane of the equilateral $\triangle A B C$ with side length 2, and $P A^{2}+P B^{2}+P C^{2}=16$. Then the trajectory of the moving point $P$ is $\qquad$ . | 3. A circle with the center of $\triangle A B C$ and a radius of 2.
Let $O$ be the center of the equilateral $\triangle A B C$, $\angle P O C=\alpha$, then
$$
\angle A O C=\angle B O C=
$$
$120^{\circ}$.
$$
\begin{array}{l}
\quad O A=O B=O C=R= \\
\frac{2}{3} \sqrt{3} .
\end{array}
$$
In $\triangle O A P, \triangle O ... | a circle with the center of \triangle ABC and a radius of 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,521 |
Example 4 The first 100 natural numbers are arranged in a certain order, then the sum of every three consecutive numbers is calculated, resulting in 98 sums, of which the maximum number of sums that can be odd is how many?
(21st Russian Mathematical Olympiad) | Solution: First, it is impossible for all 98 sums to be odd; otherwise, the arrangement of these 100 natural numbers could only be one of the following:
(1) odd odd odd odd....$;$
(2) odd even even odd even even
(3) even odd even even odd even
(4) even even odd even even odd
These four cases contradict the sequence of ... | 97 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,522 |
4. Given the equation $x+8 y+8 z=n(n \in \mathbf{N})$ has 666 sets of positive integer solutions $(x, y, z)$. Then, the maximum value of $n$ is . $\qquad$ | 4.304.
When $m>1, m \in \mathbf{N}$,
$y+z=m$ has $m-1$ sets of positive integer solutions.
The original equation also has $m-1$ sets of positive integer solutions.
$$
\begin{array}{l}
\because 1+2+\cdots+(m-1)=666, \\
\therefore \frac{1}{2} m(m-1)=666 .
\end{array}
$$
Solving gives $m=37$ or -36 (discard).
$$
\theref... | 304 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,523 |
5. Given a tetrahedron $ABCD$ with the lengths of its six edges being $4 \mathrm{~cm}$, $7 \mathrm{~cm}$, $20 \mathrm{~cm}$, $22 \mathrm{~cm}$, $28 \mathrm{~cm}$, and $x \mathrm{~cm}$. Then the minimum value of $[x]$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and f... | 5.8
When $x \leqslant 4$, the triangular side with a side length of $28 \mathrm{~cm}$ must satisfy the condition that the sum of any two sides is greater than the third side. Here, $22+7>28$, $22+20>28$, so the edge with a length of $22 \mathrm{~cm}$ cannot appear on both sides that have edges of length $28 \mathrm{~c... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,524 |
6. Given that for every real number $x$ and $y$, the function $f(x)$ satisfies $f(x)+f(y)=f(x+y)+x y$. If $f(1)=m$, then the number of positive integer pairs $(m, n)$ that satisfy $f(n)=1998$ is $\qquad$. | 6.16.
Let $y=1$, we get
$$
\begin{array}{l}
f(x)+f(1)=f(x+1)+x, \\
\therefore f(x+1)-f(x)=f(1)-x, \\
\therefore f(x)-f(x-1)=f(1)-(x-1), \\
\quad f(x-1)-f(x-2)=f(1)-(x-2), \\
\quad \cdots \cdots . \\
\quad f(2)-f(1)=f(1)-1 .
\end{array}
$$
Adding the above $x-1$ equations, we get
$$
\begin{array}{l}
f(x)-f(1)=(x-1) f(... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,525 |
Four, for what real number $x$ does $y=x^{2}-x+1+$ $\sqrt{2(x+3)^{2}+2\left(x^{2}-5\right)^{2}}$ have a minimum value? What is the minimum value? | As shown in Figure 5, let \( P\left(x, x^{2}\right) \) be a point on the parabola \( y=x^{2} \). The distance from \( P \) to the line \( y=x-1 \) is
\[
\begin{aligned}
|P Q| & =\frac{\left|x^{2}-x+1\right|}{\sqrt{(-1)^{2}+1^{2}}} \\
& =\frac{\sqrt{2}}{2} \left| x^{2}-x+1 \right| \\
& =\frac{\sqrt{2}}{2}\left(x^{2}-x+1... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,527 |
Five, the function $f(x)$ is defined on $\mathbf{R}^{+}$ and satisfies the following conditions:
(1) $f(x)$ is strictly decreasing on $\mathbf{R}^{+}$, and $f(x) > \frac{1}{x^{2}}$.
(2) On $\mathbf{R}^{+}$, it always holds that
$$
f^{2}(x) f\left(f(x)-\frac{1}{x^{2}}\right)=f^{3}(1) \text {. }
$$
(1) Find the function ... | (1) When $x=1$, we get
$$
\left.f^{2}(1) f^{f} f(1)-1\right)=f^{3}(1) \text {. }
$$
Given (1) $f(1)>1$.
$$
\therefore f(f(1)-1)=f(1) \text {. }
$$
Since $f(x)$ is strictly decreasing on $\mathbf{R}^{+}$,
$$
\begin{array}{l}
\therefore f(1)-1=1 . \\
\therefore f(1)=2 .
\end{array}
$$
(2) Let $f(x)=\frac{a}{x^{2}}$. Fro... | f(x)=\frac{2}{x^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,528 |
Given as follows, $A D$ is the angle bisector of acute $\triangle A B C$, $\angle B A C=$ $\alpha, \angle A D C=\beta$, and $\cos \alpha=\cos ^{2} \beta$. Prove: $A D^{2}=B D \cdot C D$. | Extend $A D$ to intersect $\odot O$ at $E$, connect $O B, O D, O E$. $O E$ intersects $B C$ at $F$. Let $\angle D O E=\theta$. The radius of $\odot O$ is $R$.
$$
\because A D \text{ is the angle bisector of } \angle B A C,
$$
$$
\begin{array}{l}
\therefore \overparen{B E}=\overparen{E C}, \\
\therefore O E \perp B C .
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,529 |
II. Find the last four digits of $2^{1999}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\varphi(625)=\varphi\left(5^{4}\right)=5^{4} \times\left(1-\frac{1}{5}\right)=500 \text {. }
$$
Since $(2,625)=1$, by Euler's theorem, we get $2^{500} \equiv 1(\bmod 625)$.
$$
\begin{array}{l}
\therefore 2^{2000} \equiv 1(\bmod 625) . \\
\therefore \text { Let } 2^{2000}=625 m+1(m \in \mathbf{N}) .
\end{array}
$$
... | 4688 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,530 |
Three, given that $n$ is a positive integer, $m$ is a positive odd number, $a, b$ are positive constants, and $a > b + 1$. The function $f(x, n) = \sum_{i=1}^{2 n} x^{\frac{i}{m}} + a x^{\frac{2 n+1}{m}} - b$. If real numbers $s, t$ satisfy $f(s, n) = f(t, n+1) = 0$, prove that $s < t$. | Three, obviously, for any natural number $n$, when $x \geqslant 0$, $f(x, n)$ and $f(x, n+1)$ are both strictly increasing functions, then when $x \geqslant 1$,
$$
f(x, n) \geqslant f(1, n)=2 n+a-b>0 \text {. }
$$
From $f(s, n)=0$ we know, $s0$.
$\therefore$ By the monotonicity of $f(x, n)$ on $[0,+\infty)$, $s$ is th... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,531 |
In $\triangle ABC$ inscribed in $\odot O$, $AB > AC > BC$. Point $D$ is on $\overparen{BC}$, and perpendiculars from $O$ to $AB$ and $AC$ intersect $AD$ at $E$ and $F$, respectively. The rays $BE$ and $CF$ intersect at point $P$. When $PB = PC + PO$, what is $\angle BAC$? | Solution: Connect $O B, O C$. It is easy to know that $\triangle E A B$ and $\triangle F A C$ are both isosceles triangles, and the exterior angle of the vertex angle in an isosceles triangle is equal to twice the base angle. Therefore,
$$
\begin{array}{l}
\angle B P C=\angle A E P+ \\
\angle C F D=2(\angle B A D+\angl... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,532 |
Example 5 Given $n$ points $A_{1}, A_{2}, \cdots, A_{n}(n \geqslant 3)$ in the plane, no three of which are collinear. By selecting $k$ pairs of points, determine $k$ lines (i.e., draw a line through each pair of the $k$ selected pairs), such that these $k$ lines do not form a triangle with all three vertices being giv... | Solution: Let the line passing through the point pair $A_{1}, A_{2}$ be $l$, then $A_{1}, A_{2}$ cannot both connect to any of the remaining $n-2$ points, meaning the lines passing through $A_{1}$ or $A_{2}$ are at most $n-1$ (including $l$).
Similarly, for the $n-2$ points $A_{3}, A_{4}, \cdots, A_{n}$, the lines pas... | \left\{\begin{array}{ll}
\frac{n^{2}}{4} & (n \text { is even }), \\
\frac{n^{2}-1}{4} & (n \text { is odd }) .
\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,533 |
Initial 74. Given isosceles $\triangle A B C$ with vertex angle $A$ being $108^{\circ}, D$ is a point on the extension of $A C$, and $A D=B C, M$ is the midpoint of $B D$. Find the degree measure of $\angle C M A$. | Given that $AB = AC$, $\angle ABC = \angle ACB = 36^\circ$.
Construct the angle bisector $BP$ of $\angle ABC$ intersecting $AC$ at $P$. Take a point $Q$ on $BC$ such that $BQ = AB$. Connect $PQ$. From the construction, we know that $\triangle ABP \cong \triangle QBP$.
$$
\begin{array}{l}
\text{Therefore, } \angle PQB ... | 90^\circ | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,534 |
73. Let $a, b, c$ be the sides of $\triangle ABC$, and $n$ be any real number. Prove: $\sum \frac{c}{a+b+c}\left(a^{n}-b^{n}\right)^{2} \geqslant$ $\frac{1}{4} \sum\left(a^{n}-b^{n}\right)^{2}$
where $\sum$ denotes the cyclic sum over three elements. | $$
\begin{array}{l}
\text { Prove: When } n \geqslant 0 \text {, equation (1) is equivalent to } \\
\sum(3 c-a-b)\left(a^{n}-b^{n}\right)^{2} \text { (2) } \\
\text { Due to the symmetry of equation (2) }
\end{array}
$$
Due to the symmetry of equation (2) $\sqrt{2} \cdot \mathrm{R} \sqrt{2} \sqrt{2}$, then $3 a-b-c \p... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,535 |
74. Let the set $A=\{1,2,3, \cdots, 1997\}$, for any 999-element subset $X$ of $A$, if there exist $x, y \in X$, such that $x<y$ and $x \mid y$, then $X$ is called a good set. Find the largest natural number $a(a \in A)$, such that any 999-element subset containing $a$ is a good set. | Solution: We prove that $\max a=665$.
First, we prove that $a \leqslant 665$. Clearly, the 999-element subset $X_{0}$ $=\{999,1000,1001, \cdots, 1997\}$ does not contain any $x, y \in$ $X_{0}$ such that $x1997$, i.e., larger than the maximum element of $X_{0}$, which holds. Thus, $a$ cannot be any of the numbers $999,1... | 665 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,536 |
For example, 610 people go to the bookstore to buy books, it is known that
(1) each person bought three books;
(2) any two people have at least one book in common.
How many people at most could have bought the book that was purchased by the fewest people?
(1993, China Mathematical Olympiad) | Solution: Let's assume that person A bought three books, and since A has at least one book in common with each of the other 9 people, among A's three books, the book purchased by the most people is bought by no less than 4 people.
If the book purchased by the most people is bought by 4 people, then all three books bou... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,537 |
Example 7 Let $S=\{1,2,3,4\}$. An $n$-term sequence: $q_{1}, q_{2}, \cdots, q_{n}$ has the following property: For any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted by $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
(1997, Shan... | Solution: First, each number in $S$ appears at least twice in the sequence $q_{1}, q_{2}$, $\cdots, q_{n}$, otherwise, since there are 3 binary subsets containing a certain number, but in the sequence, the number of adjacent pairs containing this number is at most 2, therefore, $n \geqslant 8$.
Moreover, the 8-term se... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,538 |
Example 8 In a $100 \times 25$ rectangular table, each cell is filled with a non-negative real number, the number in the $i$-th row and $j$-th column is denoted as $x_{ij}$, as shown in Table 1. Then, the numbers in each column of Table 1 are rearranged in non-increasing order as $x_{1j}^{\prime} \geqslant x_{2j}^{\pri... | Solution: First, consider a row $x_{r 1}, x_{r 2}$, $\cdots, x_{r 25}$ in Table 1, which must appear in the first few items of Table 2, because $100 \times 24 = 2400$, while $97 \times 25 = 2425$. Therefore, a row in Table 1 must appear in the first 97 items of Table 2.
Thus, when $i \geqslant 97$, $x_{i j}^{\prime} \... | 97 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,539 |
Example 9 Let $S=\{1,2,3, \cdots, 280\}$. Find the smallest natural number $n$ such that every subset of $S$ with $n$ elements contains 5 pairwise coprime numbers.
(32nd IMO) | Solution: Let $A_{1}=\{S$ where the numbers are divisible by 2 $\}, A_{2}=$ $\{S$ where the numbers are divisible by 3 $\}, A_{3}=\{S$ where the numbers are divisible by 5 $\}, A_{4}=\{S$ where the numbers are divisible by 7 $\}$, and denote $A=$ $A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$. It is easy to see that $|A|=216... | 217 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,540 |
Example 1 Proof: For any integer $n$, $19 \times 8^{n}+17$ is a composite number. | To prove that $A(n)=19 \times 8^{n}+17$ is a composite number, we need to consider which prime number can divide $A(n)$. For different $n$, we need to choose an appropriate prime modulus for discussion. This requires starting with some simple numbers $n$ for experimentation.
For $n=0$, $A(0)=36$, clearly $2 \mid A(0)$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,541 |
Example 3 Given: As shown in Figure 3, $AB = BC = CA = AD$, $AH \perp CD$ at $H$, $CP \perp BC$, and $CP$ intersects $AH$ at $P$. Prove: The area $S$ of $\triangle ABC$ is $S = \frac{\sqrt{3}}{4} AP \cdot BD$.
$(1984$, National Junior High School Mathematics League $)$ | Analysis: Since $S_{\triangle A B C}=\frac{\sqrt{3}}{4} B C^{2}=\frac{\sqrt{3}}{4} A C \cdot B C$, it is sufficient to prove that $A C \cdot B C=A P \cdot B I$, which can be transformed into proving $\triangle A P C \sim \triangle B C D$. This implies that $A, B, C, Q$ are concyclic (where $Q$ is the intersection of $B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,544 |
Example 1 Given $a_{k} \geqslant 0, k=1,2, \cdots, n$, and it is defined that $a_{n \cdot 1}=n_{1}$.
Prove: $\sum_{k=1}^{n} \sqrt{a_{k}^{2}-a_{k} a_{k+1}+a_{k+1}^{2}} \geqslant \sum_{k=1}^{n} a_{k}$. | Prove: Given
$$
\sqrt{a_{k}^{2}-a_{k} a_{k+1}+a_{k+1}^{2}}=\sqrt{\left(a_{k}-\frac{1}{2} a_{k+1}\right)^{2}+\left(\frac{\sqrt{3}}{2} a_{k+1}\right)} i \text {, }
$$
construct the complex number
$$
\begin{array}{l}
z_{k}=\left(a_{k}-\frac{1}{2} a_{k+1}\right)+\left(\frac{\sqrt{3}}{2} a_{k+1}\right) i, \\
k=1,2, \cdots,... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,545 |
For $n \in \mathbf{N}$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=17$. If there exists a unique $n$ such that $S_{n}$ is also an integer, find the value of $n$. | Solution: Based on the structure of $\sqrt{(2 k-1)^{2}+a_{k}^{2}}$, we construct the complex number $z_{k}=(2 k-1)+\sigma_{k} i, k=1,2, \cdots, n$.
Then $\sqrt{(2 k-1)^{2}+a_{k}^{2}}=\left|(2 k-i)+a_{k} i\right|$.
Thus $\sum_{k=1}^{n} \vee \sqrt{(2 k-1)^{2}+a_{k}^{2}}$
$=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} i\right|$
$=\... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,546 |
Example 3 Let the complex numbers $z_{1}$ and $z_{2}$ satisfy
$$
\left|z_{1}\right|=\left|z_{1}+z_{2}\right|=3,\left|z_{1}-z_{2}\right|=3 \sqrt{3} \text {. }
$$
Find the value of $\log _{3}\left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|$.
(1991, National High School Mathematics... | Solution: From the given, we have
$$
\begin{aligned}
9 & =\left|z_{1}+z_{2}\right|^{2}=\left|z_{1}\right|^{2} \\
& =\left(z_{1}+z_{2}\right)\left(\overline{z_{1}+z_{2}}\right) \\
& =\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+\left(z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}\right), \\
27 & =\left|z_{1}-z_{2}\right|^{2}=\lef... | 4000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,547 |
Example 4 Let Ming: For any point $P$ on the circumcircle of a regular polygon $A_{1} A_{2} \cdots A_{n}$, the sum of the squares of the distances from $P$ to each vertex is a constant. | Prove: By establishing an appropriate complex coordinate system, the $n$ vertices of a regular polygon can be represented by the $n$ roots of $z^{n}-1=0$. Let the complex number representing point $P$ be $z$, then $|z|=1$.
Thus, $\sum_{k=1}^{n}\left|P A_{k}\right|^{2}=\sum_{k=1}^{n}\left|z-\varepsilon^{k}\right|^{2}$
$... | 2n | Geometry | proof | Yes | Yes | cn_contest | false | 710,548 |
Example 5 Given that the complex number $z$ satisfies $10 z^{10}+10 i z^{9}+$ $10 i z-11=0$. Prove: $|z|=1$.
untranslated text retained the original line breaks and formatting. | Proof: From the given, we have $z^{9}=\frac{11-10 i z}{11 z+10 i}$. Let $z = a + bi$, then taking the modulus on both sides, we get
$$
\begin{array}{l}
|z|^{9}=\left|\frac{11-10 i z}{11 z+10 i}\right| \\
=\sqrt{\frac{11^{2}+220 b+10^{2}\left(a^{2}+b^{2}\right)}{11^{2}\left(a^{2}+b^{2}\right)+220 b+10^{2}}} \text {. } \... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,549 |
Example 6 Let the complex numbers $a_{1}, a_{2}, \cdots, a_{n}$ all lie on a circle centered at the origin with radius $r$, and let $n T_{s}$ denote the sum of the products of every $s$ of these $n$ numbers, where $s<n$. Prove: $\left|\frac{n T_{s}}{n T_{n-s}}\right|=r^{2 s-n}$.
---
The translation preserves the orig... | Analysis: For $n=3, s=2, a_{1}, a_{2}, a_{3}$, prove the conclusion in this case, and find a method suitable for the general case in the problem.
$$
\begin{array}{l}
\text { Let } a_{j}=r e^{i \theta_{j}}, j=1,2,3 \text {. Then } \\
\left|\frac{a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}}{a_{1}+a_{2}+a_{3}}\right| \\
=r\left|\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,550 |
Example 7 Let $z_{1}, z_{2}, \cdots, z_{n}$ be complex numbers, and satisfy $\sum_{k=1}^{n}\left|z_{k}\right|=1$. Prove: Among the above $n$ complex numbers, there must exist some complex numbers, the modulus of whose sum is not less than $\frac{1}{6}$. | Analysis: Let $z=a+bi$, then we must have
$$
\begin{array}{l}
\max \{\operatorname{Re}(z), \operatorname{Im}(z)\} \leqslant|z| \\
\leqslant|a|+|b| .
\end{array}
$$
Therefore, we can find several complex numbers based on the condition, such that the real part or the imaginary part of their sum is not less than $\frac{1... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,551 |
Example 8 Let $a_{0}>a_{1}>a_{2}>\cdots>a_{n}>0$. Prove: $f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}$ $=0$ all roots lie inside the unit circle (i.e., the modulus of all its complex roots is less than 1). | Analysis: The key to this problem is clearly to prove:
When $|c|>1$, $|f(c)|>0$, and when $|c|=1$, $|f(c)|$ is also greater than zero.
However, proving these directly is not easy, so consider multiplying $f(x)$ by a polynomial $g(x)$. We can let $g(x) = 1 + x$ or $1 - x$, to establish a connection with "1".
Proof:
$$
... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,552 |
Theorem: In $\triangle A B C$, three points $D, E, F$ satisfy $\angle B A E=\angle C A F, \angle A B D=\angle C B F$, and $A D$, $B E$, $C F$ are concurrent at point $P$, then $\angle A C D=\angle B C E$. Conversely, if $\angle A C D=\angle B C E$, then $A D$, $B E$, $C F$ are concurrent. | $$
\begin{array}{r}
\text { Proof: Let } \\
\angle B A E=\angle C A F \\
=\alpha, \angle A B D= \\
\angle C B F=\beta, \\
\angle A C D=x, \\
\angle B C E=y \text {. Let } M \text {, }
\end{array}
$$
Proof: Let
$N, K$ be the intersections of $A D, B E, C F$ with the sides of $\triangle A B C$.
Since $A D, B E, C F$ ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,554 |
Example 4 As shown in Figure 4, in quadrilateral $ABCD$, $AB // CD, AD=$ $DC=DB=p, BC=q$. Find the length of diagonal $AC$.
(Mathematical Olympiad Junior Training Questions, Issue 6, 1996) | Analysis: From “$A D=D C=D B=p$”, we know that $A$, $B$, and $C$ are on the circle $\odot D$ with radius $p$. Using the properties of the circle, we can find the relationship between $A C$ and $p$, $q$.
Solution: Extend $C D$ to intersect the circle $\odot D$ with radius $p$ at point $E$, and connect $A E$. Clearly, $... | \sqrt{4 p^{2}-q^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,555 |
Given a real number $a$, let the sequence of real polynomials $\left\{f_{n}(x)\right\}$ satisfy
$$
\left\{\begin{array}{l}
f_{0}(x)=1, \\
f_{n+1}(x)=x f_{n}(x)+f_{n}(a x), n=0,1,2, \cdots
\end{array}\right.
$$
(1) Prove that
$$
f_{n}(x)=x^{n} \int_{n}\left(\frac{1}{x}\right), n=0,1,2, \cdots \text {. }
$$
(2) Find the ... | Proof: (i) By convention, we denote
$$
\begin{aligned}
F_{k}(x)= & (x-1) f_{k}(x)+f_{k}(a x) \\
& -a^{k} x f_{k}\left(\frac{x}{a}\right) .
\end{aligned}
$$
First, we point out that
$$
F_{k+1}(x)=x F_{k}(x)+F_{k}(a x) \text {. }
$$
In fact,
$$
\begin{aligned}
& F_{k+1}(x)-x F_{k}(x) \\
= & (x-1) f_{k+1}(x)+f_{k+1}(a x... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,557 |
Let $m$ be a given integer. Prove that there exist integers $a, b$ and $k$, where $a, b$ are both not divisible by 2, and $k \geqslant 0$, such that
the above text is translated into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
2 m=a^{19}+b^{99}+k \cdot 2^{1999} \text {. }
$$
Prove: (i) Let $r$ and $s$ be positive integers, where $r$ is odd. If $x$ and $y$ are odd numbers that are not congruent modulo $2^{r}$, then $x^{s}$ and $y^{s}$ are also not congruent modulo $2^{r}$. This is because
$$
x^{s}-y^{s}=(x-y)\left(x^{s-1}+x^{s-2} y+\cdots... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,559 |
Five, find the largest real number $\lambda$ such that for the real-coefficient polynomial $f(x)=x^{3}+a x^{2}+b x+c$ with all roots being non-negative real numbers, if $x \geqslant 0$, then
$$
f(x) \geqslant \lambda(x-a)^{3} .
$$
And ask when the equality holds in the above formula. | Solution: Let the three roots of $f(x)$ be $\alpha, \beta, \gamma$, and assume $0 \leqslant \alpha \leqslant \beta \leqslant \gamma$, then we have
$$
\begin{array}{l}
x-a=x+\alpha+\beta+\gamma, \\
f(x)=(x-\alpha)(x-\beta)(x-\gamma).
\end{array}
$$
(i) When $0 \leqslant x \leqslant \alpha$, since
$$
-f(x)=(\alpha-x)(\be... | -\frac{1}{27} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,560 |
Six, let a large cube of $4 \times 4 \times 4$ be composed of 64 unit cubes. Select 16 of these unit cubes to be painted red, such that in the large cube, each $1 \times 1 \times 4$ small rectangular prism composed of 4 unit cubes contains exactly 1 red cube. How many different ways are there to select the 16 red cubes... | Solution: Choose a $4 \times 4$ side of the $4 \times 4 \times 4$ large cube as the base plane. Divide the large cube into four layers parallel to the base plane (each layer being $4 \times 4 \times 1$ in size), and number these layers as $1, 2, 3, 4$. Project each red unit cube onto the base plane and write the layer ... | 576 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,561 |
Given $\lambda>1$, let point $P$ be a moving point on the arc $B A C$ of the circumcircle of $\triangle A B C$. On the lines $B P$ and $C P$, take points $U$ and $V$ respectively, such that $B U=\lambda B A, C V=$ $\lambda U V$. Find the locus of point $\Omega$. | Lead: Lian Wei
$A U, A V, A Q$.
On the extension of $B C$,
take point $D$, such that $B D$
$=\therefore C$. Connect
$A D$. (D)
$$
\begin{array}{l}
\because C V= \\
\lambda C A, \quad B U= \\
\lambda B A,
\end{array}
$$
$\lambda B A$,
$$
\begin{array}{l}
\angle A C V=\angle A B, \\
\therefore \triangle A V C \backsim ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,562 |
Play a match. The winning team gets 3 points, the losing team gets 0 points, and in case of a draw, each team gets 1 point. How many points must a team score at least to ensure that no more than $k-1$ teams score no less than this team, where $2 \leqslant$ $k \leqslant n-1$? | Solution: Clearly, the worst-case scenario is that $k+1$ teams have the same score and all have the highest score.
(1) When $k$ is even, let $k=2 m$. Represent the $2 m+1$ teams with $2 m+1$ equally spaced points on a circle. Each team defeats the $m$ teams that follow it in a clockwise direction and loses to the other... | 3 n-\left[\frac{3 k+1}{2}\right]-2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,563 |
Three, prove that there exists a natural number $m$, such that there is an integer sequence $\left\{a_{n}\right\}$, satisfying:
(1) $a_{0}=1, a_{1}=337$;
(2) $\left(a_{n+1} a_{n-1}-a_{n}^{2}\right)+\frac{3}{4}\left(a_{n+1}+a_{n-1}-2 a_{n}\right)=m, \forall n \geqslant 1$;
(3) $\frac{1}{6}\left(a_{n}+1\right)\left(2 a_{... | (2) and (3). Let $b_{n}=a_{n}+\frac{3}{4}, n=0,1,2, \cdots$, then
$$
b_{0}=1+\frac{3}{4}, b_{1}=337+\frac{3}{4}
$$
and $\quad(2) \Leftrightarrow b_{n+1} b_{n-1}-b_{n}^{2}=m, n=1,2, \cdots$
By induction, it is easy to know that the sequence $\left\{b_{n}\right\}$ is a strictly increasing positive sequence.
So, $b_{n+1}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,564 |
Find all real-coefficient polynomials $f(x)$ that satisfy the following conditions:
(1) $f(x)=a_{0} \cdot x^{2 n}+a_{2} x^{2 n-2}+\cdots+$ $a_{2 n-2} x+a_{2 n}, a_{0}>0$;
(2) $\sum_{j=0}^{n} a_{2,} a_{2 n-2 j} \leqslant \mathrm{C}_{2 n}^{n} a_{0} a_{2 n}$;
(3) All $2 n$ roots of $f(x)$ are purely imaginary.
(Zhang Zhus... | Solution: First, let $g(i)=u_{0} t^{n}-a_{2} t^{n-1}+\cdots+(-1)^{\prime} a_{2}, t^{\prime} i+\cdots+(-1)^{n} \rho_{2 \pi}$.
We see $f(x)=(-1)^{n} g\left(-x^{2}\right)$.
Suppose $\pm i \beta_{1}, \cdots, \pm i \beta_{n}$ are the $2n$ roots of the polynomial $f(x)$ (assuming $\left.\beta_{j}>0, j=1, \cdots, n\right)$.
... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,565 |
Example 5 Given that the parabola $y=-x^{2}+2 x+8$ intersects the $x$-axis at points $B$ and $C$, and point $D$ bisects $BC$. If point $A$ is a moving point on the parabola above the $x$-axis, and $\angle BAC$ is acute, then the range of $A i$ is $\qquad$
(1986, National Junior High School Mathematics League) | Analysis: From “$\angle B A C$ is an acute angle”, we know that point $A$ is outside the circle with the fixed line segment $B C$ as its diameter. Since point $A$ is above the $x$-axis, we can determine the range of the moving point $A$, and thus determine the range of $A D$.
Solution: As shown in Figure 5, the vertex... | 3<A D \leqslant 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,566 |
2. During the following time periods, the hour and minute hands of a clock will overlap at ( ).
(A) $5: 25 \sim 5: 26$
(B) $5: 26 \sim 5: 27$
(C) $5: 27 \sim 5: 28$
(D) $5: 28 \sim 5: 29$ | 2. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 710,570 |
3. Given $A=(2-x)-(\sqrt{2} x-5) \sqrt{2}$, the number of natural numbers $x$ that make $A$ positive is ( ).
(A) 1
(B) 2
(C) more than 2 but finite
(D) infinitely many
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 710,571 |
5. In $\triangle A B C$, $\angle C=90^{\circ}, \angle A=15^{\circ}$, $A B=12$. Then the area of $\triangle A B C$ is $(\quad)$.
(A) 16
(B) 18
(C) $12 \sqrt{2}$
(D) $12 \sqrt{3}$ | 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 | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,573 |
For example, $6 A D$ is the altitude on the hypotenuse $B C$ of Rt $\triangle A B C$, the angle bisector of $\angle B$ intersects $A D$ at $M$ and $A C$ at $N$. Prove: $A B^{2}-A N^{2}=B M \cdot B N$. | Analysis: Since $A B^{2}-A N^{2}=(A B+A N)(A B-A N)=B M \cdot B N$, and from the given conditions, it is easy to know that $A M=A N$. By associating with the secant theorem, constructing an auxiliary circle can lead to the conclusion.
Proof: As shown in Figure 6,
$$
\begin{array}{c}
\because \angle 2+\angle 3=\angle 4+... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,577 |
7. The two diagonals of a trapezoid are perpendicular to each other, one of the diagonals is 5 cm long, and the height of the trapezoid is 4 cm. Then, the area of the trapezoid is | 7.
$$
\frac{50}{3}
$$ | \frac{50}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,582 |
2. As shown in Figure 1, in the isosceles right triangle $ABC$, $D$ is the midpoint of the hypotenuse $AB$, and $E$ and $F$ are points on the legs $AC$ and $BC$ respectively (excluding the endpoints). $DE \perp IF$, $AB=10$. Let $x=|DE|+|IF|$, then the range of values for $x$ is | 2. $5 \sqrt{2} \leqslant x<10$ | 5 \sqrt{2} \leqslant x<10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,585 |
3. Real numbers $a, b$ satisfy $\sqrt{a^{2}}-2 a+1+$ $\sqrt{36-12 a+a^{2}}=10-|b+3|-|b-2|$. Then the maximum value of $a^{2}+b^{2}$ is $\qquad$ | 3. 45 | 45 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,586 |
Example 7 As shown in Figure 7, ABCD is a cyclic quadrilateral inscribed in $\odot O$, extending $A B$ and $D C$ to intersect at $E$, extending $A D$ and $B C$ to intersect at $F, E P$ and $F Q$ are tangents to $\odot O$ at $P$ and $Q$ respectively. Prove: $E P^{2}+F Q^{2}=E F^{2}$. | Analysis: Since $E P$ and $F Q$ are tangents to $\odot O$, the conclusion leads us to think of the secant-tangent theorem, constructing an auxiliary circle to transform $E P$ and $F Q$ into $E F$.
Proof: As shown in Figure 7, construct the circumcircle of $\triangle B C E$ intersecting $E F$ at $G$, and connect $C G$.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,588 |
II. (Full marks 15 points) Now we have three types of salt water, A, B, and C, with concentrations of $5\%, 8\%$, and $9\%$ respectively. They are 60 grams, 60 grams, and 47 grams each. We need to prepare 100 grams of salt water with a concentration of $7\%$. How many grams of type A salt water can be used at most? How... | Let's take $x$ grams, $y$ grams, and $z$ grams of salt water from solutions A, B, and C respectively, to mix a 100-gram salt water solution with a concentration of $7\%$. According to the problem, we have
$$
\left\{\begin{array}{l}
x+y+z=100 \\
5 x+8 y+9 z=700 .
\end{array}\right.
$$
where, $0 \leqslant x \leqslant 60... | 49 \text{ grams (maximum)}, 35 \text{ grams (minimum)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,590 |
Three, (Full marks 15 points) For the rectangle $\triangle B C(1)$, $A B=20$ cm, $B C=10$ cm. If points $M, N$ are taken on $A C$ and $A B$ (as shown in Figure 2), to make the value of $B M+M N$ the smallest, find this minimum value. | Three, as shown in Figure 1, construct the symmetric point $B'$ of $B$ with respect to $AC$, and connect $AB'$. Then the symmetric point of $N$ with respect to $AC$ is the point $N'$ on $AB$. At this time, the minimum value of $B$ and $M$ to $V$ is equal to the minimum value of $B \rightarrow M \rightarrow N'$, which i... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,591 |
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