problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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6. Let $f(k)$ be the number of integers $n$ satisfying the following conditions.
(1) $0 \leqslant n < 10^{k}$, $n$ has exactly $k$ digits in decimal notation, and the leading digit can be 0;
(2) The digits of $n$ can be rearranged in some way to form a number that is divisible by 11.
Prove: For each positive integer $m... | Prove: For a fixed positive integer $m$, define sets $A_{0}$ and $B_{0}$ as follows.
$A_{0}$ is the set of all integers $n$ with the following properties:
(i) $0 \leqslant n < 10^{2} m$, i.e., $n$ has $2 m$ digits;
(ii) The $2 m$-digit integer obtained by reordering the leftmost $2 m-1$ digits of $n$ is divisible by 11... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,798 |
1. Let $\angle X O Y=90^{\circ}, P$ be a point inside $\angle X O Y$, and $O P=1, \angle X O P=30^{\circ}$. Draw any line through point $P$ that intersects rays $O X$ and $O Y$ at points $M$ and $N$ respectively. Find the maximum value of $O M+O N-M N$. | 1. First, construct a circle $\odot O_{1}$ passing through point $P$ and tangent to rays $O X$ and $O Y$ (tangent points are $A$ and $B$), with point $P$ on the major arc $\widehat{A B}$.
Establish a Cartesian coordinate system with rays $O X$ and $O Y$ as the $x$-axis and $y$-axis, respectively, as shown in Figure 1.... | \sqrt{3}+1-\sqrt[4]{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,799 |
2. Let $u$ be any given positive integer. Prove: the equation $n!=$ $u^{a}-u^{b}$ has at most finitely many positive integer solutions $(n, a, b)$. | 2. First, prove a lemma.
Lemma Let $p$ be a given odd prime, $p \nmid u$, and let $d$ be the order of $u$ modulo $p$. Suppose $u^{d}-1=p^{v} k$, where $v \geqslant 1$ and $p \nmid k$. Let $m$ be a positive integer, $p \mid m$. Then for any integer $t (t \geqslant 0)$, we have $u^{d m m^{t}}=1+p^{t+v} k_{t}$, where $p ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,800 |
3. Let $n_{1}, n_{2}, \cdots, n_{k}$ be $k(k \geqslant 2)$ positive integers, and $1<n_{1}<n_{2}<\cdots<n_{k}$, positive integers $a$ and $b$ satisfy
$$
\begin{array}{l}
\left(1-\frac{1}{n_{1}}\right)\left(1-\frac{1}{n_{2}}\right) \cdots\left(1-\frac{1}{n_{k}}\right) \\
\leqslant \frac{a}{b}<\left(1-\frac{1}{n_{1}}\rig... | 3. First, prove a lemma.
Lemma If positive integers $n_{1}, n_{2}, \cdots, n_{k}$ and $a, b$ satisfy the inequality in the problem, then there must be an $r(1 \leqslant r \leqslant k)$, such that
$$
n_{1} n_{2} \cdots n_{r} \leqslant\left(2^{r+1} a\right)^{r} \text {. }
$$
Proof of the lemma: We first prove that ther... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,801 |
4. Points $D$, $E$, $F$ are on the sides $BC$, $CA$, $AB$ of acute $\triangle ABC$ (all different from the endpoints), satisfying $EF \parallel BC$. $D_{1}$ is a point on side $BC$ (different from $B$, $D$, $C$). Through $D_{1}$, draw $D_{1} E_{1} \parallel DE$ and $D_{1} F_{1} \parallel DF$, intersecting sides $AC$ an... | 4. As shown in Figure 3, let $P D_{1}$, $D_{1} E_{1}$, and $D_{1} F_{1}$ intersect $E F$ at $D_{2}$, $E_{2}$, and $F_{2}$, respectively. We only need to prove that $E_{1}$, $D_{2}$, and $F_{1}$ are collinear. Since $\triangle E_{1} D_{1} C \sim \triangle E_{1} E_{2} E$, we have
$$
\frac{D_{1} E_{1}}{E_{1} E_{2}}=\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,802 |
1. For the quadratic equation in $x$, $2 a x^{2}-2 x-3 a-2=$ 0, one root is greater than 1, and the other root is less than 1. Then the value of $a$ is ( ).
(A) $a>0$ or $a0$
(D) $-4<a<0$ | (Hint: Utilize the characteristic of the parabola $a f(1)<0$, solving this inequality will yield the result. Or let $y=x-1$, then find $a$ by the roots of the equation in $y$ having opposite signs. Answer: $(\mathrm{A})$ ) | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,803 |
5. Given $p_{1}, p_{2}, \cdots, p_{25}$ as 25 distinct prime numbers not exceeding 2004, find the largest positive integer $T$, such that any positive integer not greater than $T$ can always be expressed as the sum of distinct positive divisors of $\left(p_{1} p_{2} \cdots p_{25}\right)^{2004}$ (for example, $1, p_{1},... | 5. When $p_{1}>2$, 2 cannot be expressed as the sum of distinct positive divisors of $\left(p_{1} p_{2} \cdots p_{25}\right)^{2000}$, at this time $T=1$.
Let $p_{1}=2$, we prove the following more general conclusion:
If $p_{1}, p_{2}, \cdots, p_{k}$ are $k$ distinct primes, $p_{i}2$ when, $T=1$; when $p_{1}=2$, $T=$ $\... | \frac{p_{1}^{2005}-1}{p_{1}-1} \cdot \frac{p_{2}^{2005}-1}{p_{2}-1} \cdots \cdots \frac{p_{25}^{2005}-1}{p_{25}-1} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,804 |
6. Let $a$, $b$, $c$ be the lengths of the three sides of a triangle with a perimeter not exceeding $2 \pi$. Prove: $\sin a$, $\sin b$, $\sin c$ can form the lengths of the three sides of a triangle. | 6. From the given, $00, \sin b>0, \sin c>0$, $|\cos a||\sin \theta|+|\sin (c-\theta)| \\
\geqslant|\sin (\theta+c-\theta)|=\sin c .
\end{array}
$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,805 |
1. If there exists a permutation $a_{1}, a_{2}, \cdots, a_{n}$ of $1,2, \cdots, n$ such that $k + a_{k} \ (k=1,2, \cdots, n)$ are all perfect squares, then $n$ is called a "good number". Which of the numbers in the set $\{11, 13, 15, 17, 19 \}$ are "good numbers" and which are not? Explain your reasoning.
(Su Chun, pro... | 1. All numbers except 11 are "good numbers".
(1) It is easy to see that 11 can only be added to 5 to get $4^{2}$, and 4 can only be added to 5 to get $3^{2}$, therefore, there is no arrangement that satisfies the condition, so 11 is not a "good number".
(2) 13 is a "good number", because in the following arrangement, $... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,806 |
2. Let $a$, $b$, $c$ be positive real numbers. Find the minimum value of
$$
\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c}
$$
(Li Shenghong) | 2. $-17+12 \sqrt{2}$.
Let $x=a+2 b+c, y=a+b+2 c, z=a+b+$
$3 c$, then we have $x-y=b-c, z-y=c$. Therefore, we get
$$
\begin{array}{l}
a+3 c=2 y-x, b=z+x-2 y, c=z-y . \\
\text { Hence } \frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c} \\
=\frac{2 y-x}{x}+\frac{4(z+x-2 y)}{y}-\frac{8(z-y)}{z} \\
=-17+2 \frac... | -17+12 \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,807 |
3. Given an obtuse triangle $\triangle A B C$ with the circumradius of 1. Prove: there exists an isosceles right triangle with hypotenuse length $\sqrt{2}+1$ that can cover $\triangle A B C$.
(Leng Gangsong, problem contributor) | 3. Suppose $\angle C>90^{\circ}$, then $\min \{\angle A, \angle B\}<45^{\circ}$. Suppose $\angle A<45^{\circ}$.
As shown in Figure 1, with $A B$ as the diameter, construct a semicircle $\odot O$ on the same side as vertex $C$, then $C$ is located inside the semicircle $\odot O$. Draw ray $A T$ such that $\angle B A T ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,808 |
4. A deck of tri-color cards, consisting of 32 cards in total, has 10 cards of each color: red, yellow, and blue, numbered $1, 2, \cdots, 10$; there are also a big joker and a small joker, each numbered 0. From this deck, several cards are drawn, and the score is calculated as follows: each card with a number $k$ is wo... | $4.1003^{2}$.
Solution 1: Call the original problem the "Two Kings Problem". If an additional card (called the "Middle King") is added, with the number also being 0, and the same problem is considered, it is called the "Three Kings Problem".
First, consider the "Three Kings Problem". The large, medium, and small three... | 1003^2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,809 |
5. Let $u$, $v$, $w$ be positive real numbers, satisfying the condition $u \sqrt{v w} + v \sqrt{w u} + w \sqrt{u v} \geqslant 1$. Find the minimum value of $u + v + w$.
(Chen Yonggao) | 5. $\sqrt{3}$,
By the AM-GM inequality and the conditions given in the problem, we know
$$
\begin{array}{l}
u \frac{v+w}{2}+v \frac{w+u}{2}+w \frac{u+v}{2} \\
\geqslant u \sqrt{v w}+v \sqrt{u u}+w \sqrt{u v} \geqslant 1,
\end{array}
$$
Thus, $\square$
$$
\begin{array}{l}
u v+v w+w u \geqslant 1 . \\
\text { Hence }(u... | \sqrt{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,810 |
6. Given an acute triangle $\triangle ABC$, point $O$ is its circumcenter, and line $AO$ intersects side $BC$ at point $D$. Moving points $E$ and $F$ are located on sides $AB$ and $AC$ respectively, such that points $A$, $E$, $D$, and $F$ are concyclic. Prove that the length of the projection of segment $EF$ onto side ... | 6. As shown in Figure 2, let the projection of $E F$ on side $B C$ be $E_{0} F_{0}$. Draw $D M \perp A B$ at $M$ and $D N \perp A C$ at $N$. Draw $M M_{0} \perp B C$ at $M_{0}$ and $N N_{0} \perp B C$ at $N_{0}$.
Since $\angle A M D = \angle A N D = 90^{\circ}$, we have
$\angle M D N = 180^{\circ} - \angle B A C$.
Beca... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,811 |
7. Given that $p$ and $q$ are coprime positive integers, and $n$ is a non-negative integer. How many different integers can be expressed in the form $i p + j q$, where $i$ and $j$ are non-negative integers, and $i + j \leqslant n$.
（Li Weiguo, Contributed) | 7. If we denote $A(p, q, n)=|\dot{p}+j q| i, j \geqslant 0, i+j \leqslant n$, then the number of elements in $A(p, q, n)$ is
$$
|A(p, q, n)|=\left\{\begin{array}{ll}
\frac{(n+1)(n+2)}{2}, & n < p; \\
\frac{p(2 n-p+3)}{2}, & n \geqslant p.
\end{array}\right.
$$
According to the definition, we have
$$
\begin{array}{l}
A... | \frac{(n+1)(n+2)}{2}, \text{ for } n < p; \frac{p(2 n-p+3)}{2}, \text{ for } n \geqslant p | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,812 |
8. A square of size $3 \times 3$ with a unit square removed from each corner is called a "cross shape". On a $10 \times 11$ chessboard, what is the maximum number of non-overlapping "cross shapes" (each "cross shape" exactly covers 5 small squares on the chessboard)?
(Supplied by Feng Zuming) | 8. 15.
First, prove that the maximum number of "crosses" that can be placed is 15.
Proof by contradiction. Assume that 16 "crosses" can be placed.
For each "cross", we call the central square the "center" (denoted as *). As shown in Figure 3, remove the outermost layer of squares from the $10 \times 11$ chessboard, re... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,813 |
1. Given that $x, y, z, a, b$ are all non-zero real numbers, and satisfy $\frac{x y}{x+y}=\frac{1}{a^{3}-b^{3}}, \frac{y z}{y+z}=\frac{1}{a^{3}}, \frac{z x}{z+x}=\frac{1}{a^{3}+b^{3}}$, $\frac{x y z}{x y+y z+z x}=\frac{1}{12}$. Then the value of $a$ is ( ).
(A) 1
(B) 2
(C) -1
(D) -2 | $$
-1 . B \text {. }
$$
Taking the reciprocal of each known equation, we get
$$
\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=a^{3}-b^{3}, \text { (1) } \frac{1}{y}+\frac{1}{z}=a^{3}, \\
\frac{1}{z}+\frac{1}{x}=a^{3}+b^{3}, \text { (3) } \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=12 . \\
\frac{(1)+(2)+(3)}{2}, \text { we get } \f... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,815 |
2. As shown in Figure 1, in the isosceles right $\triangle ABC$, $AC = BC$. An equilateral $\triangle ABD$ is constructed with the hypotenuse $AB$ as one side, such that points $C$ and $D$ are on the same side of $AB$; then an equilateral $\triangle CDE$ is constructed with $CD$ as one side, such that points $C$ and $E... | 2.D.
Since $\triangle A B C$ is an isosceles right triangle and $\triangle A B D$ is an equilateral triangle, we have $C D \perp A B$, and
$$
C D=\frac{\sqrt{3}}{2} A B-\frac{1}{2} A B=\frac{\sqrt{3}-1}{2} A B \text {. }
$$
It is easy to see that $\angle A D C=\angle A D E=30^{\circ}$, and $D E=D C$. Therefore, point... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,816 |
3. For any given positive integer $n, n^{6}+3 m$ is the cube of some positive integer, where $m$ is a positive integer. Then, such $m$( ).
(A) is unique
(B) there are only two
(C) there are infinitely many
(D) does not exist
The text above is translated into English, preserving the original text's line breaks and fo... | 3.C.
$$
\begin{array}{l}
\text { Let } n^{6}+3 m=\left(n^{2}+3 p\right)^{3} \\
=n^{6}+3\left(3 n^{4} p+9 n^{2} p^{2}+9 p^{3}\right) .
\end{array}
$$
Take \( m=3 n^{4} p+9 n^{2} p^{2}+9 p^{3} \). Since \( p \) is any positive integer, there are infinitely many such positive integers \( m \). | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,817 |
4. As shown in Figure 2, in the right triangle $\triangle ABC$, $\angle ABC = 90^{\circ}$, $AB = \frac{1}{2} AC$. Take a point $P$ on the line $AB$ or $BC$ such that $\triangle PAC$ is an isosceles triangle. Then, the number of points $P$ that satisfy the condition is $(\quad)$.
(A) 4
(B) 6
(C) 7
(D) 8 | 4.B.
(1) If $AC$ is the leg of isosceles $\triangle PAC$, then draw circles with $A$ and $C$ as centers and the length of segment $AC$ as the radius. They intersect the lines $AB$ and $BC$ at 5 different points
$$
P_{1}, P_{2}, P_{3}, P_{4}, P_{5} \text{ (Note: }
$$
Since $AB = \frac{1}{2} AC$,
the two circles interse... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,818 |
5. Given an acute triangle $\triangle A B C$ with unequal side lengths, $D$ is a point on side $B C$, and $\angle B A D+\angle C=90^{\circ}$. Then, $A D$ must pass through the ( ) of $\triangle A B C$.
(A) Circumcenter
(B) Incenter
(C) Centroid
(D) Orthocenter | 5.A.
As shown in Figure 7, extend $AD$ to intersect the circumcircle $\odot O$ of $\triangle ABC$ at $E$, and connect $BE$. Then $\angle E = \angle C$.
Since $\angle BAD + \angle C = 90^{\circ}$, we have
$$
\begin{array}{l}
\angle BAD + \angle E = 90^{\circ}. \\
\text{Therefore, } \angle ABE = 90^{\circ}.
\end{array}... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,819 |
6. If the positive integer $m$ makes the maximum value of the function $y=x+\sqrt{13-2 m x}(x \geqslant 0)$ also a positive integer, then this maximum value equals ( ).
(A)3
(B) 4
(C) 7
(D) 8 | 6.C.
Let $t=\sqrt{13-2 m x}(t \geqslant 0)$, then $x=\frac{13-t^{2}}{2 m}$. Therefore,
$$
y=\frac{13-t^{2}}{2 m}+t=-\frac{1}{2 m}(t-m)^{2}+\frac{m^{2}+13}{2 m} \text {. }
$$
Since $t \geqslant 0$ and $m$ is a positive integer, when $t=m$, $y$ reaches its maximum value, and the maximum value is $\frac{m^{2}+13}{2 m}$.... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,820 |
1. Given that $a$ and $b$ are positive integers, and satisfy $\frac{a+b}{a^{2}+a b+b^{2}}=\frac{4}{49}$. Then the value of $a+b$ is $\qquad$ ـ. | II. 1. 16.
Let $a+b=4k$ ($k$ is a positive integer), then
$$
a^{2}+a b+b^{2}=49 k,
$$
which means $(a+b)^{2}-a b=49 k$.
Therefore, $a b=16 k^{2}-49 k$.
It is easy to see that $a, b$ are the two positive integer roots of the quadratic equation in $x$
$$
x^{2}-4 k x+\left(16 k^{2}-49 k\right)=0
$$
By $\Delta=16 k^{2}-4... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,821 |
3. Given that $a, b$ are the real roots of the equation $x^{2}+2 p x+3=0$, $c, d$ are the real roots of the equation $x^{2}+2 q x+3=0$, and $e, v$ are the real roots of the equation $x^{2}+2\left(p^{2}-q^{2}\right) x+3=0$, where $p, q$ are real numbers. Then, the value of $\frac{(a-c)(b-c)(a+d)(b+d)}{e+v}$ is $\qquad$ ... | 3.6 .
From the relationship between roots and coefficients, we get
$$
\begin{array}{l}
a+b=-2 p, a b=3 ; c+d=-2 q, c d=3 ; \\
e+f=-2\left(p^{2}-q^{2}\right), e f=3 . \\
\text { Therefore } \frac{(a-c)(b-c)(a+d)(b+d)}{e+f} \\
=\frac{\left[a b-(a+b) c+c^{2}\right]\left[a b+(a+b) d+d^{2}\right]}{e+f} \\
=\frac{\left(c d+... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,823 |
4. As shown in Figure 4, the line $y = -\frac{\sqrt{3}}{3} x + 1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. An isosceles right triangle $\triangle ABC$ with $\angle BAC = 90^{\circ}$ is constructed in the first quadrant with segment $AB$ as one of its legs. If there exists a point $P\lef... | 4. $\frac{\sqrt{3}-8}{2}$.
It is easy to know that the intersection points of the line $y=-\frac{\sqrt{3}}{3} x+1$ with the $x$-axis and $y$-axis are $A(\sqrt{3}, 0)$ and $B(0,1)$, respectively, i.e., $O A=\sqrt{3}, O B=1$, then $A B=2$. Therefore, $S_{\triangle A B C}=2$. Thus, $S_{\triangle A B P}=2$.
As shown in F... | \frac{\sqrt{3}-8}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,824 |
3. For what value of $k$ does the equation $x^{2}-k x+k+3=0$ have two real roots, both of which are in the interval $(1,4)$?
When $k$ is what value, the equation $x^{2}-k x+k+3=0$ has two real roots, and both roots are within the interval $(1,4)$? | (Prompt: Example 4. Answer: $6 \leqslant k<\frac{19}{3}$ ) | 6 \leqslant k<\frac{19}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,825 |
One, (20 points) In the quadratic function $f(x)=a x^{2}+b x$ $+c$, $a$ is a positive integer, $a+b+c \geqslant 1, c \geqslant 1$, and the equation $a x^{2}+b x+c=0$ has two distinct positive roots less than 1. Find the minimum value of $a$.
保留源文本的换行和格式，直接输出翻译结果如下：
```
One, (20 points) In the quadratic function $f(x)... | Given the equation $a x^{2}+b x+c=0$ has two roots $x_{1} 、 x_{2}(0<x_{1}<x_{2})$, and $x_{1} x_{2}=\frac{c}{a}=\frac{3}{5}$, we have $a x_{1} x_{2}=3$.
Since $x_{1}+x_{2}=\frac{11}{10}$, we have $a(x_{1}+x_{2})=\frac{11}{10}a>4.4$.
Given $a>4$, and $a$ is a positive integer, so, $a \geqslant 5$.
Taking $f(x)=5\left... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,826 |
1. Given $a, b \in \mathbf{N}, a^{100}$ is a 120-digit number, and $a^{b}$ is a 10-digit number. Then the value of $b$ is ( ).
(A) 7
(B) 8
(C) 9
(D) 10 | $-、 1 . B$.
From the given, we have $10^{119} \leqslant a^{100}<10^{120}, 10^{9} \leqslant a^{6}<10^{10}$. Taking the logarithm, we get $\lg 10^{119} \leqslant \lg a^{100}<\lg 10^{120}$, simplifying to $1.19 \leqslant \lg a<1.2$.
Therefore, $1.19 b<b \lg a<1.2 b$.
Also, $9 \leqslant b \lg a<10$, so $b=8$. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,828 |
2. Given that for every pair of real numbers $x, y$, the function $f$ satisfies $f(x)+f(y)=f(x+y)-x y-1$. If $f(1)=1$, then the number of integers $n$ that satisfy $f(n)=n$ is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) infinitely many | 2.B.
Let $y=1$ we get $f(x)+f(1)=f(x+1)-x-1$, i.e., $f(x+1)=f(x)+x+2$.
Let $x=0$ we get $f(1)=f(0)+2$.
From $f(1)=1$ we know $f(0)=-1$.
When $n \in \mathbf{N}_{+}$,
$$
\begin{array}{l}
f(n)=\sum_{k=1}^{n}[f(k)-f(k-1)]+f(0) \\
=\sum_{k=1}^{n}(k+1)+f(0)=\frac{n(n+3)}{2}-1 .
\end{array}
$$
Similarly, $f(-n)=-\frac{n(-n+... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,829 |
3. When drawing the graph of $f(x)=a x^{2}+b x+c(a \neq 0)$, a table is first created. When the values of $x$ in the table increase at equal intervals, the corresponding values of the function $f(x)$ are: $3844, 3969, 4096, 4227, 4356, 4489$, 4624, 4761. One of these values is incorrect, and this value is ( ).
(A) 4096... | 3.C.
Since the interval between the values of $x$ is equal, it is known that the values of $x$ form an arithmetic sequence $\left\{a_{n}\right\}$, with a common difference of $d$. After calculation, it is easy to obtain
$$
\begin{array}{l}
{\left[f\left(a_{n}\right)-f\left(a_{n-1}\right)\right]-\left[f\left(a_{n-1}\ri... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,830 |
4. In $\triangle A B C$, $M=\sin A+\sin B+$ $\sin C, N=\cos \frac{A}{2}+\cos \frac{B}{2}+\cos \frac{C}{2}$. The size relationship between $M$ and $N$ is ( ).
(A) $M=N$
(B) $M \leqslant N$
(C) $M \geqslant N$
(D) Cannot be determined | 4.B.
In $\triangle A B C$,
$$
\sin A+\sin B=2 \cos \frac{C}{2} \cdot \cos \frac{A-B}{2} \leqslant 2 \cos \frac{C}{2} \text{. }
$$
Similarly, $\sin B+\sin C \leqslant 2 \cos \frac{A}{2}$,
$$
\sin C+\sin A \leqslant 2 \cos \frac{B}{2} \text{. }
$$
Adding the three inequalities, we get
$$
\sin A+\sin B+\sin C \leqslant... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 714,831 |
5. In a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, among the six diagonal lines on the lateral faces $A B_{1} 、 A_{1} B 、 B C_{1} 、 B_{1} C 、 C_{1} A 、 C A_{1}$, if one pair $A B_{1} \perp B C_{1}$, how many other pairs, denoted as $k$, are also perpendicular to each other? Then $k$ equals ( ).
(A)2
(B) 3
(C) ... | 5.D.
As shown in Figure 1, take the midpoint $M$ of $B C$, and connect $B_{1} M$. Then the projection of $A B_{1}$ on the plane $B B_{1} C_{1} C$ is $B_{1} M$. Given that $A B_{1} \perp B C_{1}$, we know that $B_{1} M \perp B C_{1}$.
Next, take the midpoint $N$ of $B_{1} C_{1}$, we have $C N \perp B C_{1}$, thus $A_{... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,832 |
6. A line $l$ is drawn through point $P(2,1)$, intersecting the two coordinate axes at points $A$ and $B$. When the area $S$ of $\triangle A O B$ takes values in $(0,+\infty)$, the number of lines $l$ that can be drawn is ( ) lines.
(A) 2
(B) 3
(C) 2 or 3
(D) 2 or 3 or 4 | 6.D.
The line $l$ passes through $P(2,1)$ and forms a triangle with the two coordinate axes in the first quadrant, the minimum area of which is 4. In fact, let the equation of $l$ be
$$
\frac{x}{a}+\frac{y}{b}=1(a>0, b>0),
$$
then $\frac{2}{a}+\frac{1}{b}=1 \geqslant 2 \sqrt{\frac{2}{a} \cdot \frac{1}{b}}, a b \geqsl... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,833 |
1. The set $A=\left\{x \mid x \in \mathbf{Z}\right.$ and $\left.\frac{600}{5-x} \in \mathbf{Z}\right\}$ has $\qquad$ elements, and the sum of all elements is $\qquad$ . | 二、1. 48,240.
Since $600=2^{3} \times 3 \times 5^{2}$, the number of positive divisors of 600 is $(3+1)(1+1)(2+1)=24$.
Therefore, the set $A$ contains $2 \times 24=48$ elements.
Considering that the sum of the elements in $A$ corresponding to two opposite divisors of 600 is 10, the sum of the elements in $A$ is $24 \tim... | 240 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,834 |
2. Let the set $A=\{1,2,3,4,5,6\}$, and the mapping $f: A$
$\rightarrow A$ satisfies $f(f(x))=x$. Then the number of mappings $f$ is
$\qquad$ ـ. | 2.76.
Let $f(x)=a$, then $f(a)=x$, i.e., $x \rightarrow a, a \rightarrow x$.
If $a=x$, i.e., $x \rightarrow x$ is called self-correspondence;
If $a \neq x$, i.e., $x \rightarrow a, a \rightarrow x$ is called a pair of cyclic correspondence.
It is also known that the number of self-correspondences should be even, so th... | 76 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,835 |
Example 1 In quadrilateral $A B C D$, $A B=30, A D$ $=48, B C=14, C D=40, \angle A B D+\angle B D C=$ $90^{\circ}$. Find the area of quadrilateral $A B C D$. | Solution: As shown in Figure 2, we have
$$
\begin{array}{l}
S_{\triangle A B D}=S_{\triangle A_{1} D B}, \\
A_{1} D=A B=30, \\
A_{1} B=A D=48, \\
\angle A_{1} D B=\angle A B D .
\end{array}
$$
Thus, we have
$$
\begin{array}{l}
\angle A_{1} D C=\angle A_{1} D B+\angle B D C \\
=\angle A B D+\angle B D C=90^{\circ} .
\e... | 936 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,836 |
Example 2 In $\triangle A B C$, $A B=A C, \angle B A C=$ $80^{\circ}, O$ is a point inside the triangle, $\angle O B C=10^{\circ}, \angle O C B=$ $30^{\circ}$. Find the degree measure of $\angle B A O$. | Solution: As shown in Figure 3, draw $A H \perp B C$ at $H$. Also, $A H$ is the angle bisector of $\angle B A C$, so $\angle B A H=\angle C A H=40^{\circ}$.
Extend $\mathrm{CO}$ to intersect $A H$ at $P$, then $\angle B O P=40^{\circ}=\angle B A P$. Connect $B P$, by symmetry we have $\angle P B C=\angle P C B=30^{\ci... | 70^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,837 |
Example 6 If numbers $a_{1}, a_{2}, a_{3}$ are taken in ascending order from the set $1,2, \cdots, 14$, such that both $a_{2}-a_{1} \geqslant 3$ and $a_{3}-a_{2} \geqslant 3$ are satisfied. Then, the number of all different ways to select the numbers is $\qquad$ kinds. | Solution: Given $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$, $14-a_{3} \geqslant 0$, let $a_{1}=x_{1}, a_{2}-a_{1}=x_{2}, a_{3}-a_{2}=x_{3}, 14-a_{3}=x_{4}$, then
$$
x_{1}+x_{2}+x_{3}+x_{4}=14 .
$$
Thus, the problem is transformed into finding the number of integer solutions to the equation u... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,838 |
One. (20 points) A factory needs to produce two types of products, A and B. According to the process specifications, each unit of product A requires 2 hours, 3 hours, and 4 hours of processing on three different machines A, B, and C, respectively. Each unit of product B requires 4 hours, 4 hours, and 3 hours of process... | Let the daily production of product A be $x$ units, and the production of product B be $y$ units. Then we have:
$$
\left\{\begin{array}{l}
2 x \leqslant 12, \\
3 x+4 y \leqslant 15, \\
4 x+4 y \leqslant 16, \\
3 y \leqslant 24 .
\end{array}\right.
$$
The profit obtained is
$$
W=700 x+800 y=100(7 x+8 y) \text {. }
$$
... | 1 \text{ unit of product A and 3 units of product B} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,839 |
II. (25 points) As shown in Figure 2, circles $\odot O_{1}$ and $\odot O_{2}$ intersect at points $M$ and $N$. $A$ is a point on $\odot O_{1}$, $AM$ intersects $\odot O_{2}$ at point $B$, and $AN$ intersects $\odot O_{2}$ at point $C$. $O$ is the circumcenter of $\triangle ABC$. Prove:
$$
AO = O_{1}O_{2}.
$$ | As shown in Figure 6, connect $M N$, and draw the tangent $A T$ of $\odot O$ through $A$, then
$$
\begin{array}{l}
\angle T A C=\angle A B C \\
=\angle A N M .
\end{array}
$$
Therefore, $A T \parallel M N$.
Since $O A \perp A T$, then $O A \perp M N$.
Also, since $O_{1} O_{2} \perp M N$, it follows that $O_{1} O_{2} \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,840 |
Three, (25 points) Try to determine, for any $n$ positive integers, the smallest positive integer $n$ such that at least 2 of these numbers have a sum or difference that is divisible by 21.
| Three, let $A_{i}$ represent all numbers that leave a remainder of $i$ when divided by 21; $i=0,1$, $\cdots, 20$; let $B_{j}=A_{j} \cup A_{21-j}, j=1,2, \cdots, 10, B_{11}=A_{0}$.
By taking 12 numbers, there will be at least 2 numbers belonging to the same category among the 11 categories $B_{1}, B_{2}$, $\cdots, B_{1... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,841 |
1. Given the sets $M=\left\{x \mid x^{2}+9 y^{2}=9, x, y \in\right.$ $\mathbf{R}\}, N=\left\{x\left|x^{2}+y^{2}-2 a x=0, x, y \in \mathbf{R},\right| a \mid\right.$ $\leqslant 1$, and $a$ is a constant $\}$. Then $M \cap N=(\quad)$.
(A) $\{x|| x \mid \leqslant 1\}$
(B) $\{x|| x|\leqslant| a \mid\}$
(C) $\{x|a-| a|\leqsl... | -、1.C.
It is easy to get $M=\{x|-3 \leqslant x \leqslant 3\}, N=\{x|a-|a| \leqslant x \leqslant a+|a|\}$. Also, $|a| \leqslant 1$, so $N \subset M$. Therefore, $M \cap N=N$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,842 |
2. The equation $(a-1)(\sin 2 x+\cos x)+(a+1) \cdot$ $(\sin x-\cos 2 x)=0$ (parameter $a<0$) has $(\quad)$ solutions in the interval $(-\pi$, $\pi)$.
(A) 2
(B) 3
(C) 4
(D) 5 or more than 5 | 2.C.
The original equation can be transformed into
$$
\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)\left[\cos \left(\frac{3 x}{2}-\frac{\pi}{4}\right)+\frac{a+1}{a-1} \sin \left(\frac{3 x}{2}-\frac{\pi}{4}\right)\right]=0 \text {. }
$$
(1) The equation $\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)=0$ has a solution $x=-\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,843 |
3. A student goes upstairs from the first floor to the third floor for class, with a total of 30 steps. However, he moves forward and backward randomly on the stairs. When he stops after moving from the 1st step to the 9th step, he thinks: If all the backward steps were turned into forward steps, he would be
(1) possib... | 3. B.
Let the total grade improvement of the student be $x$, and the total grade decline be $y$, then $x-y=9$. Therefore, $x+y=9+2y$.
Since $9+2y$ is an odd number, $x+y$ is also an odd number.
Thus, we know that the sum of the student's improvement and decline is an odd number, so we can eliminate (A), (C), (D). | B | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 714,844 |
4. The range of the function $y=\sqrt{2004-x}+\sqrt{x-2003}$ is ( ).
(A) $[1,1.5]$
(B) $[1, \sqrt{2})$
(C) $[1, \sqrt{2}]$
(D) $[1,2)$ | 4.C.
It is known that the domain of the function is $[2003,2004], y>0$.
Also, $y^{2}=1+2 \sqrt{(2004-x)(x-2003)}$, and
$$
\begin{array}{l}
0 \leqslant(2004-x)(x-2003) \\
\leqslant\left(\frac{2004-x+x-2003}{2}\right)^{2}=\frac{1}{4},
\end{array}
$$
thus $1 \leqslant y^{2} \leqslant 2 \Rightarrow 1 \leqslant y \leqslan... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,845 |
5. Given $z \in \mathbf{C}, a>0$. Then $1-\frac{2 a}{z+a}$ is a pure imaginary number if and only if ( ).
(A) $|z|=a$
(B) $|z|=a$ and $z \neq-a$
(C) $|z|=a$ or $z \neq \pm a$
(D) $|z|=a$ and $z \neq \pm a$ | 5.D.
$$
1-\frac{2 a}{z+a}=\frac{z-a}{z+a} \text {. }
$$
As shown in Figure 2, let $A(-a, 0)$, $B(a, 0)$, and the point corresponding to the complex number $z$ be $P$, then $AP = z + a$,
$$
\begin{array}{l}
BP = z - a \text {. } \\
\frac{z-a}{z+a} \text { is a pure imaginary number } \\
\Leftrightarrow \arg \frac{z-a}{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,846 |
6. The integer $M=2003^{2004}{ }^{2005} \underbrace{2008}$ when divided by 80 leaves a remainder of ( ).
(A) 1
(B) 11
(C) 31
(D) 79 | 6.A.
Let $m=2004^{2000^{2008}}, p=2005^{2000^{2002^{2008}}}$.
Then $m=2004^{p}=4^{p} \times 501^{p}=4 n, n=4^{p-1} \times 501^{p}$.
Thus, by the binomial theorem we have
$$
\begin{array}{l}
M=2003^{m}=(2000+3)^{m} \\
=2000^{m}+\mathrm{C}_{m}^{1} \cdot 2000^{m-1} \cdot 3+\cdots+ \\
\mathrm{C}_{m}^{m-1} \cdot 2000 \cdot... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,847 |
1. Given points $A(1,1)、B(3,2)、C(2,3)$ and line $l: y=k x(k \in \mathbf{R})$. Then the maximum value of the sum of the squares of the distances from points $A、B、C$ to line $l$ is $\qquad$ | Let $A$, $B$, $C$ be three points, and the sum of the squares of their distances to the line $k x - y = 0$ is $d$. Then,
$$
\begin{aligned}
d & =\frac{(k-1)^{2}}{k^{2}+1}+\frac{(3 k-2)^{2}}{k^{2}+1}+\frac{(2 k-3)^{2}}{k^{2}+1} \\
& =\frac{14 k^{2}-26 k+14}{k^{2}+1} .
\end{aligned}
$$
Therefore, $(d-14) k^{2}+26 k+(d-1... | 27 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,848 |
Example 7 There are $n$ squares arranged in a row, each square is to be colored with one of the three colors: red, yellow, or blue, with each square being one color. It is required that any adjacent squares must be of different colors, and the first and last squares must also be of different colors. How many ways are t... | Solution: Let there be $a_{n}$ different ways of coloring. It is easy to get $a_{1}=3$, $a_{2}=6, a_{3}=6$, and when $n \geqslant 4$, after numbering the $n$ squares in sequence, square 1 and square $(n-1)$ are not adjacent.
(1) If square $(n-1)$ and square 1 are colored differently, at this time, square $n$ has only 1... | a_{n}=2^{n}+2(-1)^{n}(n \geqslant 2) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,849 |
2. Consider the binomial coefficients $\mathrm{C}_{n}^{0}, \mathrm{C}_{n}^{1}, \cdots, \mathrm{C}_{n}^{n}$ as a sequence. When $n \leqslant 2004\left(n \in \mathbf{N}_{+}\right)$, the number of sequences in which all terms are odd is $\qquad$ groups. | 2.10.
From $\mathrm{C}_{n}^{r}=\mathrm{C}_{n-1}^{r-1}+C_{n-1}^{r}$, we know that the 2004 numbers $n=1,2, \cdots, 2004$ can be arranged in the shape of Pascal's Triangle, and the even and odd numbers can be represented by 0 and 1, respectively, as shown in Figure 3.
The pattern of the changes in the parity of each ter... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,850 |
3. Place 7 goldfish of different colors into 3 glass fish tanks numbered $1,2,3$. If the number of fish in each tank must be no less than its number, then the number of different ways to place the fish is $\qquad$ kinds. | 3.455.
(1) If the number of goldfish placed in tanks 1, 2, and 3 are $2, 2, 3$ respectively, then there are $C_{5}^{2} \cdot C_{5}^{2}=210$ different ways;
(2) If the number of goldfish placed in tanks 1, 2, and 3 are $1, 3, 3$ respectively, then there are $\mathrm{C}_{6}^{1} \cdot \mathrm{C}_{6}^{3}=140$ different way... | 455 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,851 |
4. Let the $n$ roots of the equation $x^{n}=1$ (where $n$ is an odd number) be $1, x_{1}, x_{2}, \cdots, x_{n-1}$. Then $\sum_{i=1}^{n-1} \frac{1}{1+x_{i}}=$ $\qquad$ . | 4. $\frac{n-1}{2}$.
From $x^{n}=1$ we know $\bar{x}^{n}=1$, meaning the conjugate of the roots of $x^{n}=1$ are also its roots. Therefore, 1, $\bar{x}_{1}, \bar{x}_{2}, \cdots, \bar{x}_{n-1}$ are also the $n$ roots of $x^{n}=1$, and $\left|x_{i}\right|=1$. Thus,
$$
\begin{array}{l}
x_{i} \bar{x}_{i}=1, i=1,2, \cdots, ... | \frac{n-1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,852 |
5. Given a tetrahedron $S-ABC$ with the base being an equilateral triangle, point $A$'s projection $H$ on the face $SBC$ is the orthocenter of $\triangle SBC$, and $SA=a$. Then the maximum volume of this tetrahedron is $\qquad$
Translate the above text into English, please retain the original text's line breaks and fo... | 5. $\frac{1}{6} a^{3}$.
From the given, $A H \perp$ plane $S B C$. Let $B H \perp S C$ at $E$. By the theorem of three perpendiculars, we know $S C \perp A E$ and $S C \perp A B$. Therefore, $S C \perp$ plane $A B E$. Draw $S O \perp$ plane $A B C$ at $O$, then $O C$ is the projection of $S C$ in plane $A B C$, so $C ... | \frac{1}{6} a^{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,853 |
6. Given that $x, y, z$ are positive real numbers, and $x+y+z=1$. If $\frac{a}{x y z}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-2$, then the range of the real number $a$ is . $\qquad$ | 6. $\left(0, \frac{7}{27}\right]$.
It is easy to see that $x, y, z \in (0,1)$, and $1-x, 1-y, 1-z \in (0,1)$.
On one hand, it is easy to get $a = xy + yz + zx - 2xyz > xy > 0$.
When $x, y \rightarrow 0, z \rightarrow 1$, $a \rightarrow 0$. On the other hand, it is easy to get $a = (1-x)x + (1-2x)yz$.
(1) When $0 < x <... | \left(0, \frac{7}{27}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,854 |
Three. (20 points) Let real numbers $a, b$ satisfy $a^{2}+b^{2} \leqslant \frac{4}{5}$. Determine whether the equation $x^{4}+a x^{3}+b x^{2}+a x+1=0$ has real solutions for $x$.
| Three, if the equation has a real root $x_{0}\left(x_{0} \neq 0\right)$, substituting it into the equation, we get
$$
\left(x_{0}+\frac{1}{x_{0}}\right) a+b+\left(x_{0}^{2}+\frac{1}{x_{0}^{2}}\right)=0 .
$$
Considering $a, b$ as variables, the problem transforms into:
whether the line $\left(x_{0}+\frac{1}{x_{0}}\righ... | x = 1, -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,855 |
Four, (20 points) If the inequality $x|x-a|+b<0$ (where $b$ is a constant) holds for any value of $x$ in $[0,1]$. Find the range of real number $a$.
---
The above text has been translated into English, preserving the original text's line breaks and format. | Obviously, $b1+b$.
In summary, when $-1 \leqslant b<2 \sqrt{2}-3$, the range of $a$ is $(1+b, 2 \sqrt{-b})$; when $b<-1$, the range of $a$ is $(1+b, 1-b)$. Clearly, $b$ cannot be greater than or equal to $2 \sqrt{2}-3$. | not found | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,856 |
Five. (20 points) Given the function
$$
y=f(x)=\cos m x \cdot \cos ^{m} x-\sin m x \cdot \sin ^{m} x-\cos ^{m} 2 x
$$
whose value is independent of the real number $x$. Draw tangents $P A$ and $P B$ from any point $P$ on its graph to the circle $x^{2}+(y-2)^{2}=1$, with the points of tangency being $A$ and $B$.
(1) Fi... | (1) Let $x=0$, for all $m$. Then $f(0)=0$. Since the value of $y=f(x)$ is independent of $x$, we have $f\left(\frac{\pi}{m}\right)=f(0)=0$, which gives
$$
-\cos ^{m} \frac{\pi}{m}=\cos ^{m} \frac{2 \pi}{m} \text {. }
$$
From the above equation and $m \in \mathbf{N}_{+}$, it is easy to see that $m$ is a positive odd nu... | \frac{3 \sqrt{3}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,857 |
One. (50 points) As shown in Figure 1, given that trapezoid $ABCD$ is inscribed in $\odot O$, with $AB \parallel CD$. A tangent line is drawn through point $D$ intersecting the extension of $CA$ at point $F$, and $DF \parallel BC$. If $CA = 5, BC = 4$, find the length of segment $AF$. | Extend $BA$ to intersect $DF$ at $E$, and connect $BD$, then
$$
\angle DAC = \angle DBC.
$$
Since $DF$ is tangent to $\odot O$ at point $D$, we have
$$
\angle ADF = \angle ACD.
$$
From $AB \parallel CD$, we know
$$
\overparen{AD} = \overparen{BC} \Rightarrow \angle BDC = \angle ACD.
$$
Thus, $\angle ADF = \angle BD... | \frac{80}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,858 |
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=a_{n}^{2}+2 a_{n}+3, n \in \mathbf{N}_{+}$, try to find
$$
M=a_{m}^{2}+b_{n}^{2}+m^{2}+n^{2}-2\left(a_{m} b_{n}+m n\right)
$$
$\left(m 、 n \in \mathbf{N}_{+}\right)$ the minimum value. | (1) It is easy to know that when $n \geqslant 2, n \in \mathbf{N}_{+}$, we have
$4\left(S_{n}-S_{n-1}\right)=\left(a_{n}+1\right)^{2}-\left(a_{n-1}+1\right)^{2}$.
Thus, $\left(a_{n}+a_{n-1}\right)\left(a_{n}-a_{n-1}-2\right)=0$.
Since $a_{n}+a_{n-1}>0$, it follows that,
$$
a_{n}-a_{n-1}=2 \text {. }
$$
It is easy to s... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,859 |
1. If: (1) $a, b, c, d$ all belong to $\{1,2,3,4\}$;
(2) $a \neq b, b \neq c, c \neq d, d \neq a$;
(3) $a$ is the smallest number among $a, b, c, d$.
Then, the number of different four-digit numbers $\overline{a b c d}$ that can be formed is $\qquad$. | (提示: $\overline{a b c d}$ consists of 2 different digits, then it must be that $a=$ $c, b=d$. Therefore, there are $\mathrm{C}_{4}^{2}=6$ such numbers. $\overline{a b c d}$ consists of 3 different digits, then $a$ is uniquely determined, when $c=a$, $b, d$ have 2 ways to be chosen; when $c \neq$ $a$, $c$ has 2 ways to ... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,860 |
Three, (50 points) On a plane, there are 12 points, and no three points are collinear. Using any one of these points as the starting point and another as the endpoint, draw all possible vectors. A triangle whose three side vectors sum to the zero vector is called a "zero triangle." Find the maximum number of "zero tria... | Three, let these 12 points be $P_{1}, P_{2}, \cdots, P_{12}$. The number of triangles determined by these 12 points is $\mathrm{C}_{12}^{3}$. Let the number of vectors starting from $P_{1}(i=1,2, \cdots, 12)$ be $x_{i}, 0 \leqslant x_{i} \leqslant 11$. If a triangle with 3 points as vertices is a "non-zero triangle", t... | 70 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,861 |
In the convex quadrilateral $ABCD$, all four sides are tangent to $\odot O$, and the lengths of the sides are $a, b, c, d$ respectively. Prove:
$$
O A \cdot O C + O B \cdot O D = \sqrt{a b c d} \text{. }
$$ | Proof: As shown in Figure 1, mark the equal angles as $\angle 1, \angle 2, \angle 3, \angle 4$. Construct $\triangle M A B \backsim \triangle O D C, \triangle N D C \backsim \triangle O A B$.
It is easy to see that $\angle A O B$ and $\angle C O D$ are supplementary
$$
\Rightarrow \angle A O B + \angle A M B = \angle C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,862 |
Given $ABCD$ is a square, $P$ is a point on side $BC$, and line $DP$ intersects the extension of $AB$ at point $Q$. If $DP^2 - BP^2 = BP \cdot BQ$, find the degree measure of $\angle CDP$ (do not use inverse trigonometric functions in your answer).
---
Translate the above text into English, preserving the original te... | Solution: As shown in Figure 2, let $\angle C D P=\angle P Q B=\theta$. Take a point $T$ on side $A B$ such that $B T=B P$, and connect $D T$ and $D B$. By symmetry, it is easy to see that
$$
D T=D P, \angle B D T=\angle B D P, \angle A D T=\angle C D P=\theta.
$$
From the given conditions, we have
$$
\begin{array}{l}... | 22.5^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,863 |
143 Let $H$ be the orthocenter of $\triangle ABC$ located inside the triangle, and $O$ be the circumcenter of $\triangle ABC$. If $O H \perp O A$, then $\frac{\pi}{4}<\angle A<\frac{\pi}{3}$. | Proof: As shown in Figure 3, let $H$ be the intersection of the altitudes $AD$ and $BE$, and connect $AO$ and extend it to intersect side $BC$ at $M$.
(1) First prove $\angle AOA \\
\Rightarrow & 2R \cos A > R \Rightarrow \cos A > \frac{1}{2} \Rightarrow \angle A < \frac{\pi}{3}$.
By the fact that $O, H, D, M$ are con... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,864 |
144 Given that $a$, $b$, and $c$ are positive real numbers. Prove:
$$
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}}<2 .
$$ | Prove: Let $x=\frac{b c}{a^{2}}, y=\frac{c a}{b^{2}}, z=\frac{a b}{c^{2}}$.
Then the original inequality is equivalent to
$$
\frac{1}{\sqrt{1+8 x}}+\frac{1}{\sqrt{1+8 y}}+\frac{1}{\sqrt{1+8 z}} \leq 2,
$$
if $x \geq 64$, then
$$
\begin{array}{l}
\frac{1}{\sqrt{1+8 y}}+\frac{1}{\sqrt{1+8 z}}-2 \\
\leqslant \sqrt{2+\frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,865 |
2. Among the 27 points consisting of the 8 vertices, the midpoints of the 12 edges, the centers of the 6 faces, and the center of the cube, the number of groups of three collinear points is ( ).
(A) 57
(B) 49
(C) 43
(D) 37 | (提示: The number of collinear triplets with both endpoints being vertices is $\frac{8 \times 7}{2}$ $=28$; The number of collinear triplets with both endpoints being the centers of faces is $\frac{6 \times 1}{2}=3$; The number of collinear triplets with both endpoints being the midpoints of edges is $\frac{12 \times 3}{... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,866 |
3. Color each vertex of a square pyramid with one color, and make the endpoints of the same edge have different colors. If only 5 colors are available, the number of different coloring methods is $\qquad$ . | (Given that there are 5 colors, 1, 2, 3, 4, 5, to color the pyramid $S-ABCD$. Since $S$, $A$, and $B$ are colored differently, there are $5 \times 4 \times 3=60$ coloring methods. Assuming $S$, $A$, and $B$ are already colored, we then categorize and discuss the coloring of $C$ and $D$, which have 7 coloring methods. T... | 420 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,867 |
4. Before the World Cup, the coach of country $F$ plans to evaluate 7 players, $A_{1}$, $A_{2}, \cdots, A_{7}$, by letting them play in 3 training matches (each 90 minutes). Assuming that at any moment during the matches, exactly one of these players is on the field, and the total playing time (in minutes) of $A_{1}, A... | (Tip: Let the playing time of the $i$-th player be $x_{i}$ minutes $(i=1,2, \cdots, 7)$. The problem is to find the number of positive integer solutions to the indeterminate equation $x_{1}+x_{2}+\cdots+x_{7}=$ 270, under the conditions $7 \mid x_{i}(1 \leqslant i \leqslant 4)$, and $13 \mid x_{j}(5 \leqslant j \leqsla... | 42244 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,868 |
In the Cartesian coordinate system $x O y$, the sequence of points $\left\{A_{n}\right\}$ on the positive $y$-axis and the sequence of points $\left\{B_{n}\right\}$ on the curve $y=\sqrt{2 x}$ $(x \geqslant 0)$ satisfy $\left|O A_{n}\right|=\left|O B_{n}\right|$ $=\frac{1}{n}$, the intercept of the line $A_{n} B_{n}$ o... | Proof 1: (1) Since $\left|O B_{n}\right|=\frac{1}{n}$, we have $b_{n}^{2}+2 b_{n}=\frac{1}{n^{2}}$. Solving this, we get $b_{n}=\sqrt{1+\frac{1}{n^{2}}}-1$. Therefore, for $n \in \mathbf{N}_{+}$, we have $b_{n}>b_{n+1}>0$.
The equation of the line $A_{n} B_{n}$ is
$$
\left(\sqrt{2 b_{n}}-\frac{1}{n}\right) x=b_{n}\left... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,872 |
Example 3 As shown in Figure 4, in $\triangle A B C$, $A C=B C$, $\angle A C B=90^{\circ}, P, Q$ are two points on side $A B$, and $\angle P C Q=45^{\circ}$. Prove: $A P^{2}+B Q^{2}=P Q^{2}$ | Proof: As shown in Figure 5, construct $\angle M C Q = \angle B C Q$, and cut $C M = C B$, then connect $M Q$. It is easy to see that $\triangle M C Q \cong \triangle B C Q$. Therefore,
$$
\begin{array}{l}
M Q = Q B, \\
\angle C M Q = \angle B = 45^{\circ}. \\
\text{Since } \angle A C P + \angle B C Q = 90^{\circ} - 45... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,873 |
For the integer $n(n \geqslant 4)$, find the smallest integer $f(n)$, such that for any positive integer $m$, in any $f(n)$-element subset of the set $\{m, m+1, \cdots, m+n-1\}$, there are at least 3 pairwise coprime elements. | 1 \1)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,874 |
Example 4 Given $a>0$, and
$$
\sqrt{b^{2}-4 a c}=b-2 a c \text{. }
$$
Find the minimum value of $b^{2}-4 a c$. | Solution 1: The given condition is
$$
\frac{-b+\sqrt{b^{2}-4 a c}}{2 a c}=-1(a c \neq 0),
$$
This indicates that the quadratic equation
$$
a c x^{2}+b x+1=0
$$
has a real root $x=-1$. Substituting into equation (1) yields
$$
a c=b-1 \text {. }
$$
Transform equation (3) into the overall structure of the discriminant
... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,875 |
Example 5 Given that $b^{2}-4 a c$ is a real root of the quadratic equation $a x^{2}+b x+c=0(a \neq 0)$. Then the range of values for $a b$ is ( ).
(A) $a b \geqslant \frac{1}{8}$
(B) $a b \leqslant \frac{1}{8}$
(C) $a b \geqslant \frac{1}{4}$
(D) $a b \leqslant \frac{1}{4}$ | Given $x_{0}=b^{2}-4 a c \geqslant 0$ is a real root of the equation, we have
$$
a x_{0}^{2}+b x_{0}+c=0 \text {. }
$$
Rearranging into the discriminant form (i.e., equation (2)) and substituting $x_{0}$, we get
$$
\left(2 a x_{0}+b\right)^{2}-x_{0}=0,
$$
which can be factored as $\left[\left(2 a x_{0}+b\right)-\sqrt... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,876 |
Example 6 Given that the roots of the equation $x^{2}-6 x-4 n^{2}-32 n=0$ are integers. Find the integer value of $n$.
---
The translation maintains the original text's line breaks and format as requested. | Solution: Completing the square for $x$, we get
$$
(x-3)^{2}-\left(4 n^{2}+32 n+9\right)=0 \text {. }
$$
Then completing the square for $n$ (both steps of completing the square can be done at once), we get
$$
(x-3)^{2}-(2 n+8)^{2}+55=0 \text {. }
$$
Rearranging and factoring, we get
$$
(2 n+8+|x-3|)(2 n+8-|x-3|)=55 \... | -18, -8, 0, 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,877 |
1. Find all positive integers $n$ that make $\frac{n^{2}}{200 n-999}$ a positive integer. | 1. Let $\frac{n^{2}}{200 n-999}=k, k$ be a positive integer. Then
$$
n^{2}-200 k n+999 k=0 \text {. }
$$
Assume the equation (1) has a positive integer root $n_{1}$, and the other root is $n_{2}$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
n_{1}+n_{2}=200 k, \\
n_{1} n_{2}=999 k .
\end{array}\right.
$$
T... | 5, 4995 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,878 |
2. Given that $B E$ and $C F$ are altitudes of acute $\triangle A B C$. Prove that the angle bisector of $\angle A B E$, the angle bisector of $\angle A C F$, and the perpendicular bisector of segment $E F$ intersect at one point. | 2. As shown in Figure 2, let the angle bisector of $\angle A B E$ intersect the angle bisector of $\angle A C F$ at point $N$. Connect $N E$ and $N F$. Since points $B, C, E, F$ are concyclic, then
$$
\begin{array}{l}
\angle A B E=\angle A C F, \\
\angle F B N=\angle F C N .
\end{array}
$$
Therefore, points $B, C, N, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,879 |
3. In the Cartesian coordinate system, find the coordinates of the point that satisfies the following two conditions simultaneously:
(1) The line $y=-2 x+3$ passes through such a point;
(2) Regardless of the value of $m$, the parabola
$$
y=m x^{2}+\left(m-\frac{2}{3}\right) x-\left(2 m-\frac{3}{8}\right)
$$ | 3. From (2) we know $m \neq 0$.
Let the point $\left(x_{0}, y_{0}\right)$ satisfy (1) and (2), then
$$
y_{0}=-2 x_{0}+3 \text {, }
$$
and for any non-zero real number $m$, we have
$$
y_{0} \neq m x_{0}^{2}+\left(m-\frac{2}{3}\right) x_{0}-\left(2 m-\frac{3}{8}\right) .
$$
Substituting equation (1) into equation (2) ... | (1,1),(-2,7),\left(\frac{63}{32},-\frac{15}{16}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,880 |
2. The sum of all values of $a$ that make the equation
$$
\frac{x+1}{x-1}+\frac{x-1}{x+1}+\frac{2 x+a+2}{x^{2}-1}=0
$$
have only one real root is
$\qquad$ | 2. -15.5 | -15.5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,882 |
3. To make the equation
$$
x^{4}+(m-4) x^{2}+2(1-m)=0
$$
have exactly one real root not less than 2, the range of values for $m$ is $\qquad$ | 3. $m \leqslant -1$ | m \leqslant -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,883 |
8. In Rt $\triangle A B C$, $A B=3, B C=4, \angle B$ $=90^{\circ}, A D 、 B E 、 C F$ are the three angle bisectors of $\triangle A B C$. Then, the area of $\triangle D E F$ is $\qquad$ | 8. $\frac{10}{7}$ | \frac{10}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,889 |
1. If $|1-x|=1+|x|$, then $\sqrt{(x-1)^{2}}$ equals ( ).
(A) $x-1$
(B) $1-x$
(C) 1
(D) -1 | -1.B.
If $x>1$, then $|1-x|=x-1,|x|+1=x+1$. Contradiction, hence $x \leqslant 1$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,892 |
2. In $\triangle A B C$, if $\angle A=50^{\circ}, A B>B C$, then the range of $\angle B$ is ( ).
(A) $0^{\circ}<\angle B<80^{\circ}$
(B) $50^{\circ}<\angle B<80^{\circ}$
(C) $50^{\circ}<\angle B<130^{\circ}$
(D) $80^{\circ}<\angle B<130^{\circ}$ | 2.A.
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 | 714,893 |
3. As shown in Figure 1, in $\triangle A B C$, $D$ is the midpoint of $A C$, $E$ and $F$ are the trisection points of $B C$, and $A E$ and $A F$ intersect $B D$ at points $M$ and $N$ respectively. Then $B M: M N$ :
$$
N D=(\quad) \text {. }
$$
(A) $3: 2: 1$
(B) $4: 2: 1$
(C) $5: 2: 1$
(D) $5: 3: 2$ | 3.D.
Connect $D F$, then $B M=M D$. Let $E M=a$, then $D F=2 a$, $A E=4 a$.
Therefore, $A M: D F=3: 2$, which means $M N: N D=3: 2$.
Hence, $B M: M N: N D=5: 3: 2$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,894 |
Example 5 Given that point $M$ is the midpoint of side $BC$ of quadrilateral $ABCD$, and $\angle AMD=120^{\circ}$. Prove:
$$
AB+\frac{1}{2} BC+CD \geqslant AD \text {. }
$$ | Proof: As shown in Figure 8, with $AM$ as the axis of symmetry, construct the symmetric point $B_1$ of point $B$ with respect to $AM$, and connect $AB_1$, $MB_1$: Then
$AB_1 = AB, MB_1 = MB$.
Therefore, $\triangle AB_1M \cong \triangle ABM$.
Thus, $\angle B_1MA = \angle BMA$.
Next, with $DM$ as the axis of symmetry, co... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,895 |
4. Simplify $\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}}$, the result is $(\quad)$.
(A) $1+\frac{1}{n}+\frac{1}{n+1}$
(B) $1-\frac{1}{n}+\frac{1}{n+1}$
(C) $1+\frac{1}{n}-\frac{1}{n+1}$
(D) $1-\frac{1}{n}-\frac{1}{n+1}$ | 4.C.
$$
\begin{array}{l}
\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(1+\frac{1}{n}\right)^{2}-\frac{2}{n}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(1+\frac{1}{n}\right)^{2}-2 \cdot \frac{n+1}{n} \cdot \frac{1}{n+1}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^{2}}=1+\frac{1}{n}-\f... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,896 |
5. As shown in Figure 2, in rectangle $A B C D$, $A B=8, B C=4$. If rectangle $A B C D$ is folded along $A C$, then the area of the overlapping part $\triangle A F C$ is ( ).
(A) 12
(B) 10
(C) 8
(D) 6 | 5.B.
Since $\triangle C D^{\prime} A \cong \triangle A B C$, therefore,
$S_{\triangle A D F}=S_{\triangle C B F}$.
And $\triangle A D^{\prime} F \backsim \triangle C B F$, so,
$\triangle A D^{\prime} F \cong \triangle C B F$.
Let $B F=x$, then $\sqrt{x^{2}+4^{2}}=8-x$.
Solving for $x$ gives $x=3$. Hence $S_{\triangle ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,897 |
6. If $2 x+5 y+4 z=6,3 x+y-7 z=-4$, then the value of $x+y-z$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 4 | 6. B.
$$
\begin{array}{l}
\text { From }\left\{\begin{array}{l}
2(x+y-z)+3(y+2 z)=6, \\
3(x+y-z)-2(y+2 z)=-4
\end{array}\right. \text { we get } \\
x+y-z=0 .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,898 |
7. As shown in Figure 3, in
$\square A B C D$, $B C$ $=2 A B, C E \perp A B$, $E$ is the foot of the perpendicular, $F$ is the midpoint of $A D$. If $\angle A E F=54^{\circ}$, then $\angle B=$ ( ).
$\begin{array}{lll}\text { (A) } 54^{\circ} & \text { (B) } 60^{\circ}\end{array}$
(C) $66^{\circ}$
(D) $72^{\circ}$ | 7.D.
As shown in Figure 6, take the midpoint $G$ of $BC$, and connect $FG$ to intersect $EC$ at $K$. It is easy to see that quadrilateral $FGCD$ is a rhombus, thus
$$
\angle 2=\angle 3 \text {. }
$$
Since $EK=KC$, and $FK \perp EC$, we have $\angle 1=\angle 2$. Therefore, $\angle A=2 \angle AEF$. Hence, $\angle B=72^... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,899 |
8. In a right trapezoid $A B C D$, $A D / / B C, \angle A$ $=90^{\circ}, A B=7, A D=2, B C=3, E$ is on the line segment $A B$, and $\triangle E A D$ is similar to $\triangle E B C$. Then the number of such points $E$ ( ).
(A) is exactly one
(B) is two
(C) is three
(D) is more than three | 8.C.
Let $A E=x$, then $E B=7-x$. From the condition $\frac{x}{2}=\frac{7-x}{3}$, we get $x=\frac{14}{5}$. Also, from the condition $\frac{x}{2}=\frac{3}{7-x}$, we get two solutions for $x$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,900 |
9. If $x-1=\frac{y+1}{2}=\frac{z-2}{3}$, then the minimum value of $x^{2}+y^{2}+z$ is ( ).
(A) 3
(B) $\frac{59}{14}$
(C) $\frac{9}{2}$
(D) 6 | 9. B.
$$
\begin{array}{l}
\text { Let } x-1=\frac{y+1}{2}=\frac{z-2}{3}=k \text {, then } \\
x=k+1, y=2 k-1, z=3 k+2 . \\
x^{2}+y^{2}+z^{2}=14 k^{2}+10 k+6 \\
=14\left(k+\frac{5}{14}\right)^{2}+\frac{59}{14} .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,901 |
10. Let the distances from $P$ to two vertices $A, B$ of the equilateral $\triangle A B C$ be $2, 3$ respectively. Then the maximum value that $P C$ can reach is ( ).
(A) $\sqrt{5}$
(B) $\sqrt{13}$
(C) 5
(D)6 | 10.C.
As shown in Figure 7, rotate $AB$ so that $AB$ coincides with $AC$, and $P$ rotates to $P^{\prime}$. Therefore,
$$
\begin{array}{r}
\angle P A B = \angle P^{\prime} A C . \\
\text{Thus, } \angle P A P^{\prime} = 60^{\circ} . \\
\text{Hence, } P P^{\prime} = P A = P^{\prime} A \\
= 2, P^{\prime} C = P B = 3 \text... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,902 |
11. Given that a regular $n$-sided polygon has $n+3$ diagonals, its perimeter is $x$, and the sum of the lengths of the diagonals is $y$. Then $\frac{y}{x}=$ $\qquad$ | $$
=.11 .1+\sqrt{3} \text {. }
$$
The number of diagonals is $\frac{n(n-3)}{2}=n+3$. Therefore,
$$
n^{2}-5 n-6=0 \text {. }
$$
Solving gives $n=6$ or $n=-1$ (discard).
$$
\text { Hence } \frac{y}{x}=1+\sqrt{3} \text {. }
$$ | 1+\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,903 |
12. If the product of 3 prime numbers is 5 times their sum, then these 3 prime numbers are $\qquad$ . | 12.7,5,2.
Let these 3 prime numbers be $x, y, z$, then
$$
5(x+y+z)=x y z \text{. }
$$
Since $x, y, z$ are prime numbers, we can assume $x=5$, then
$$
5+y+z=y z \text{. }
$$
Therefore, $(y-1)(z-1)=6$.
Assuming $y \geqslant z$, we have
$$
\left\{\begin{array} { l }
{ y - 1 = 3 , } \\
{ z - 1 = 2 }
\end{array} \text {... | 7,5,2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,904 |
Example 1: From the 10 numbers $0,1,2,3,4,5,6,7,8,9$, select 3 numbers such that their sum is an even number not less than 10. The number of different ways to do this is $\qquad$ | Solution: To select 3 numbers from 10 numbers such that their sum is even, the 3 numbers must all be even or 1 even and 2 odd.
When all 3 numbers are even, there are $\mathrm{C}_{5}^{3}$ ways to choose;
When there is 1 even and 2 odd numbers, there are $\mathrm{C}_{5}^{1} \mathrm{C}_{5}^{2}$ ways to choose. Since their... | 51 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,906 |
15. As shown in Figure 5, in $\triangle ABC$, $D$ and $E$ are the midpoints of $BC$ and $AC$ respectively, and $AD$ and $BE$ intersect at $P$. If $\angle BPD = \angle C$, prove that the triangle formed by the three medians of $\triangle ABC$ is similar to $\triangle ABC$. | Three, 15. As shown in Figure 11, extend $PD$ to $F$ such that $PD = DF$, and connect $BF$, $FC$, $PC$, and $DE$. Then quadrilateral $BFCP$ is a parallelogram.
Since $\angle BPD = \angle C$, then
$\triangle BPD \sim \triangle BCE$.
Therefore, $\frac{BP}{BC} = \frac{BD}{BE}$.
Thus, $\triangle BPC \sim \triangle BDE$.
He... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,908 |
16. Rearrange the digits of any three-digit number to form the largest and smallest numbers possible, and their difference will form another three-digit number (allowing the hundreds digit to be $0$). Repeat this process. What is the number obtained after 2003 repetitions? Prove your conclusion. | 16. (1) If all three digits are the same, the resulting number is 0.
(2) If the three digits are not all the same, the resulting number is 495.
(1) This is obviously true.
Below is the proof of (2).
If the three digits are not all the same, let's assume this three-digit number is $\overline{a b c}$, where $a \geqslant ... | 495 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,909 |
3. Given the inequality
$$
m^{2}+\left(\cos ^{2} \theta-5\right) m+4 \sin ^{2} \theta \geqslant 0
$$
always holds. Then the range of real number $m$ is ( ).
(A) $0 \leqslant m \leqslant 4$
(B) $1 \leqslant m \leqslant 4$
(C) $m \geqslant 4$ or $m \leqslant 0$
(D) $m \geqslant 1$ or $m \leqslant 0$ | 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 | Inequalities | MCQ | Yes | Yes | cn_contest | false | 714,912 |
4. In a circular cone with a slant height of $\sqrt{6}$, the radius of the base circle of the cone with the maximum volume is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 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 | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,913 |
5. The dihedral angle between adjacent lateral faces of a regular triangular pyramid is twice the dihedral angle between a lateral face and the base. Then the ratio of the lateral edge to the base edge is ( ).
(A) $\frac{3}{2}$
(B) $\frac{4}{3}$
(C) $\frac{3}{4}$
(D) $\frac{2}{3}$ | 5.D
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 | 714,914 |
Example 2 Let ABCDEF be a regular hexagon. A frog starts at vertex $A$, and each time it can randomly jump to one of the two adjacent vertices. If it reaches point $D$ within 5 jumps, it stops jumping; if it cannot reach point $D$ within 5 jumps, it also stops after 5 jumps. Then, the number of different possible jumpi... | Solution: As shown in Figure 1, the frog cannot reach point $D$ by jumping 1 time, 2 times, or 4 times. Therefore, the frog's jumping methods are only in the following two cases:
(1) The frog jumps 3 times to reach point $D$, with two jumping methods: $A B C D$ and $A F E D$.
(2) The frog stops after a total of 5 jumps... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,917 |
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