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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10. (40 points) Write "KOREAIMC" in eight lines as shown in Figure 5, where the first line has one K, the second line has two O's, and so on, with the last line having eight C's.
$$
\begin{array}{llllllll}
\mathbf{K} & & & & & & & \\
\mathbf{O} & \mathbf{O} & & & & & & \\
\mathbf{R} & \mathbf{R} & \mathbf{R} & & & & & ... | 10. First list the number table as shown in Figure 17, where each cell corresponds to a letter, and the number in the cell represents the number of different paths from the letter "K" to that letter. The cell at the very top is filled with 1. Starting from the second row, the number in each cell is the sum of the numbe... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,277 |
1. Let $m, k$ be given non-negative integers, and $p=2^{2^{m}}+1$ be a prime number. Prove:
(1) $2^{2 n+1 p^{k}} \equiv 1\left(\bmod p^{k+1}\right)$;
(2) The smallest positive integer $n$ that satisfies the congruence equation $2^{n} \equiv 1\left(\bmod p^{k+1}\right)$ is $2^{m+1} p^{k}$. (Geng Ping provided the proble... | 1. (1) Prove by mathematical induction: For any non-negative integer $k$, we have
$$
2^{2 m+1 p^{k}}=p^{k+1} t_{k}+1\left(p \nmid t_{k}\right) \text {. }
$$
When $k=0$, from $2^{2 m}=p-1$, we get
$$
2^{2 m+1}=(p-1)^{2}=p(p-2)+1 \text {, }
$$
taking $t_{0}=p-2$ is sufficient.
Assume we already have $2^{2 m+1 p^{k}}=p^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,278 |
2. As shown in Figure 1, given that $A B$ is the diameter of $\odot O$, $C$ and $D$ are two points on the circumference different from points $A$ and $B$ and on the same side of $A B$. Tangents to the circle are drawn through points $C$ and $D$, intersecting at point $E$. The intersection of line segments $A D$ and $B ... | 2. Connect $E O$, $C O$, $D O$, $C A$.
From $\angle C O E=\angle C A F$, we know
Rt $\triangle C O E \backsim \mathrm{Rt} \triangle C A F$.
Therefore, $\frac{C E}{C F}=\frac{C O}{C A}$.
Also, $\angle E C F=90^{\circ}-\angle B C O=\angle O C A$, so $\triangle E C F \backsim \triangle O C A$.
Thus, $\angle C A O=\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,279 |
3. Find all positive integers $n$ such that the set $\{1,2, \cdots, n\}$ has $n$ pairwise distinct three-element subsets $A_{1}, A_{2}, \cdots, A_{n}$, satisfying that for any $k(1 \leqslant i<j \leqslant n)$, we have $\left|A_{i} \cap A_{j}\right| \neq 1$.
(Feng Zhigang) | 3. (1) When $n=4 k\left(k \in \mathbf{N}_{+}\right)$, construct $A_{1}$, $A_{2}, \cdots, A_{4 k}$ as follows:
For any $1 \leqslant i \leqslant k, 0 \leqslant j \leqslant 3$,
$A_{4 i-j}=\{4 i-3,4 i-2,4 i-1,4 i\} \backslash\{4 i-j\}$.
(2) When $n \neq 4 k\left(k \in \mathbf{N}_{+}\right)$, assume the set $\{1,2, \cdots, ... | n = 4k \text{ where } k \in \mathbf{N}_{+} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,280 |
4. Let non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ satisfy the following conditions:
(1) $\sum_{i=1}^{n}\left(a_{i}+b_{i}\right)=1$;
(2) $\sum_{i=1}^{n} i\left(a_{i}-b_{i}\right)=0$;
(3) $\sum_{i=1}^{n} i^{2}\left(a_{i}+b_{i}\right)=10$.
Prove that for any $k(1 \leqslant k ... | For any $k(1 \leqslant k \leqslant n)$, we have
$$
\begin{array}{l}
\left(k a_{k}\right)^{2} \leqslant\left(\sum_{i=1}^{n} i a_{i}\right)^{2}=\left(\sum_{i=1}^{n} i b_{i}\right)^{2} \\
\leqslant\left(\sum_{i=1}^{n} i^{2} b_{i}\right)\left(\sum_{i=1}^{n} b_{i}\right) \leqslant\left(10-k^{2} a_{k}\right)\left(1-a_{k}\rig... | \max \left\{a_{k}, b_{k}\right\} \leqslant \frac{10}{10+k^{2}} | Algebra | proof | Yes | Yes | cn_contest | false | 723,281 |
5. Let $k$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=0, a_{1}=1, \\
a_{n+1}=k a_{n}+a_{n-1}(n=1,2, \cdots) .
\end{array}
$$
Find all $k$ that satisfy the following condition: there exist non-negative integers $l, m (l \neq m)$, and positive i... | When $k=2$, $a_{0}=0, a_{1}=1, a_{2}=2$, then from $a_{0}+2 a_{2}=a_{2}+2 a_{1}=4$, we know that taking $l=0, m=2, p=$ $2, q=1$ is sufficient.
For $k \geqslant 3$, by the recurrence relation, $\left\{a_{n}\right\}$ is a strictly increasing sequence of natural numbers and $k \mid\left(a_{n+1}-a_{n-1}\right)$.
$$
\begin... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,282 |
Example 1 If the greatest common divisor of positive integers $a, b, c$ is 1, and $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$, prove: $a+b$ is a perfect square. | Proof 1 Let $x=(a, b), y=(b, c), z=(c, a)$. Then, by $(a, b, c)=1$, we get that $x, y, z$ are pairwise coprime.
By $x|a, z|a, (x, z)=1$, we can deduce that $xz \mid a$, so we can set $a=Axz$.
Similarly, let $b=Bxy, c=Cyz$, where $A, B, C$ are pairwise coprime. Thus, the original equation can be transformed into
$$
\f... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,283 |
6. As shown in Figure 2, in $\triangle A B C$, $\angle A C B=90^{\circ}$, a circle is drawn with $B$ as the center and $B C$ as the radius. Point $D$ is on side $A C$, and line $D E$ is tangent to $\odot B$ at point $E$. A line perpendicular to $A B$ through point $C$ intersects line $B E$ at point $F$, and $A F$ inter... | 6. Proof 1 As shown in Figure 3, let $A B$ intersect $D E$ and $C F$ at points $K$ and $M$ respectively. Connect $F K$, $A E$, and $M E$.
In $\triangle A B C$,
by the projection theorem, we have
$$
\begin{array}{l}
B M \cdot B A \\
=B C^{2}=B E^{2} .
\end{array}
$$
Therefore, $\triangle B E M \backsim \triangle B A E$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,284 |
7. There are $n(n \geqslant 3)$ players participating in a table tennis tournament, where each pair of players competes in exactly one match with no ties. If player $A$'s defeated opponents are not all defeated by player $B$, then we say $A$ is not inferior to $B$. Determine all possible values of $n$ such that there e... | 7. (1) When $n=3$, a tournament result can be constructed as shown in Figure 4, which clearly meets the requirements.
(2) When $n=4$, assume there exists a tournament result such that each player is not inferior to any other player. It is evident that no player can defeat all others (otherwise, the others cannot be not... | \{n \in \mathbf{N} \mid n \geqslant 3, n \neq 4\} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,285 |
8. Find all integers $k$ such that there exist positive integers $a$ and $b$ satisfying
$$
\frac{b+1}{a}+\frac{a+1}{b}=k .
$$
(Proposed by Yonggao Chen) | 8. For a fixed $k$, among the pairs $(a, b)$ that satisfy the given equation, choose a pair $(a, b)$ such that $b$ is the smallest, then
$$
x^{2}+(1-k b) x+b^{2}+b=0
$$
has one root $x=a$. Let the other root be $x=a^{\prime}$. Then from $a+a^{\prime}=k b-1$, we know $a^{\prime} \in \mathbf{Z}$, and $a a^{\prime}=b(b+1... | k=3,4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,286 |
1. The graph of the function $y=2 \sin \left(3 x-\frac{\pi}{2}\right)$ is shifted to the left by $\varphi(\varphi>0)$ units, and the resulting graph corresponds to an odd function. Then the minimum value of $\varphi$ is ( ).
(A) $\frac{\pi}{3}$
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{6}$
(D) $\frac{\pi}{8}$ | - 1. C.
From the problem, we know that the smallest positive root of the given function is $\frac{\pi}{6}$, and the necessary condition for an odd function is that its value at the origin is 0. Therefore, the minimum value of $\varphi$ is $\frac{\pi}{6}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,287 |
3. Given that $P$ is a point on the plane of $\triangle A B C$, satisfying $\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}=2 \overrightarrow{A B}$. If $S_{\triangle A B C}=6$, then $S_{\triangle P A B}$ $=(\quad)$.
(A) 4
(B) 3
(C) 2
(D) 1 | 3. C.
From $\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}=2 \overrightarrow{A B}=2(\overrightarrow{P B}-\overrightarrow{P A})$, we get $3 \overrightarrow{P A}=\overrightarrow{P B}-\overrightarrow{P C}=\overrightarrow{C B}$.
Therefore, $S_{\triangle P A B}=\frac{1}{3} S_{\triangle A B C}=2$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,289 |
4. If the quadratic function $f(x)=a x^{2}+b x+c$ has a line of symmetry at $x=1$, and its graph passes through the point $(2,0)$, then the value of $\frac{f(-1)}{f(1)}$ is ( ).
(A) -3
(B) -2
(C) 2
(D) 3 | 4. A.
From the problem, we know the curve passes through the point $(0,0)$, so $c=f(0)=0$.
Thus, $f(x)=a x^{2}+b x$.
Also, the axis of symmetry of the graph is $x=\frac{-b}{2 a}=1$, i.e., $b=$ $-2 a$, then $\frac{f(-1)}{f(1)}=\frac{a-b}{a+b}=-3$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,290 |
5. In $\triangle A B C$, the median $A D$ is perpendicular to the median $B E$ at point $G$. Then the maximum value of $\sin C$ is ( ).
(A) $\frac{2}{5}$
(B) $\frac{3}{5}$
(C) $\frac{2}{3}$
(D) $\frac{4}{5}$ | 5. B.
In $\triangle A B C$, let $B C = a, C A = b, A B = c, G D = x, G E = y$.
Then
$$
\begin{array}{l}
G A = 2 x, \\
G B = 2 y .
\end{array}
$$
In the right triangles $\triangle A G B$, $\triangle A G E$, and $\triangle B G D$, applying the Pythagorean theorem respectively, we get
$$
4 x^{2} + 4 y^{2} = c^{2}, \quad... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,291 |
6. Given that the domain of the function $f(x)$ is $(-\infty, 0)$ $\cup(0,+\infty)$, and it satisfies $f\left(\frac{1}{x}\right)+\frac{1}{x} f(-x)=4 x$. Then the value of $f(2) f\left(-\frac{1}{2}\right)$ is ( ).
(A) 31.5
(B) 30.5
(C) -30.5
(D) -31.5 | 6. D.
Substitute $x=-2, x=\frac{1}{2}$ into the given equation, we get $f(2) f\left(-\frac{1}{2}\right)=-31.5$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,292 |
1. Given the function $f(x)=x^{2}-1$ with domain $D$, and the range is $\{-1,0,1,3\}$. Determine the maximum number of such sets $D$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 2, 1.27.
Since $f(0)=-1, f( \pm 1)=0$,
$$
f( \pm \sqrt{2})=1, f( \pm 2)=3 \text {, }
$$
Therefore, $0 \in D$; at least one from each of the sets $\{-1,1\}$, $\{-\sqrt{2}, \sqrt{2}\}$, and $\{-2,2\}$ belongs to $D$.
Thus, there are $3 \times 3 \times 3=27$ such $D$'s. | 27 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,293 |
Example 2 Let $A$ and $B$ be two unequal $2n (n \in \mathbf{N}_{+})$-digit numbers, the first $n$ digits of $A$ equal the last $n$ digits of $B$, and the last $n$ digits of $A$ equal the first $n$ digits of $B$, and $A \mid B$. Find the value of $A+B$. | Let $x$ be the positive integer corresponding to the first $n$ digits of $A$, and $y$ be the positive integer corresponding to the last $n$ digits of $A$. Then
$$
A=10^{n} x+y, B=10^{n} y+x \text {. }
$$
Let $B=k A(2 \leqslant k \leqslant 9)$, i.e.,
$$
10^{n} y+x=k\left(10^{n} x+y\right) \text {. }
$$
Thus, $A+B=(x+y... | 10^{2 n}-1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,294 |
2. Find the value of $\frac{\log _{2} \frac{2}{3}+\log _{2} \frac{3}{4}+\log _{2} \frac{4}{5}+\log _{2} \frac{5}{6}+\log _{2} \frac{6}{7}+\log _{2} \frac{7}{8}}{\log _{3} 3 \cdot \log _{4} 3 \cdot \log _{5} 4 \cdot \log _{8} 5 \cdot \log _{7} 6 \cdot \log _{8} 7}$. | 2. -6 .
$$
\begin{array}{l}
\text { Original expression }=\frac{\log _{2}\left(\frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7} \cdot \frac{7}{8}\right)}{\frac{\lg 2}{\lg 3} \cdot \frac{\lg 3}{\lg 4} \cdot \frac{\lg 4}{\lg 5} \cdot \frac{\lg 5}{\lg 6} \cdot \frac{\lg 6}{\lg 7} \cdot \... | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,295 |
3. As shown in Figure 2, in rectangle $A B C D$, $E$ is a point on side $A B$, $A B=14, C E=13, D E$ $=15, C F \perp D E$ at point $F$, connect $A F$ and $B F$. Find the area of $\triangle A B F$. | 3. 36.96 .
Let $B E=x$. Then $A E=14-x$.
In Rt $\triangle A D E$ and Rt $\triangle B C E$, applying the Pythagorean theorem respectively, we get
$$
D E^{2}-A E^{2}=A D^{2}=B C^{2}=C E^{2}-B E^{2} \text {, }
$$
which is $15^{2}-(14-x)^{2}=13^{2}-x^{2}$.
Solving for $x$ gives $x=5$.
Then $A D=B C=13^{2}-x^{2}=12$.
Ther... | 36.96 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,296 |
4. In the same Cartesian coordinate system, it is known that the line $y=k x$ intersects the graph of the function
$$
y=\left\{\begin{array}{ll}
2 x+4, & x<3 \\
x+7, & x \geq 3
\end{array}\right.
$$
at exactly three different points. Determine the range of values for $k$. | 4. $\left(\frac{2}{3}, 2\right)$.
Draw a rough sketch of the function in the coordinate system, i.e., the thick black broken line in Fig. 4.
The line $l_{1}$: $y=2 x$ intersects the broken line at only one point;
The line $l_{2}$: $y=\frac{2}{3} x$ intersects the broken line at only two points. | \left(\frac{2}{3}, 2\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,297 |
5. Given that $P$ is a point inside $\triangle A B C$ and satisfies
$$
2 \overrightarrow{P A}+3 \overrightarrow{P B}+6 \overrightarrow{P C}=0 \text {. }
$$
Determine the ratio of the areas of $\triangle P A B$, $\triangle P B C$, and $\triangle P C A$. | 5. 6: 2:3.
As shown in Figure 5, take points $A_{1}, B_{1}, C_{1}$ such that $\overrightarrow{P A_{1}}=2 \overrightarrow{P A}, \overrightarrow{P B_{1}}=3 \overrightarrow{P B}, \overrightarrow{P C_{1}}=6 \overrightarrow{P C}$. Then $\overrightarrow{P A_{1}}+\overrightarrow{P B_{1}}+\overrightarrow{P C_{1}}=\overrightar... | 6:2:3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,298 |
6. In a convex quadrilateral $ABCD$, $\angle BAD + \angle ADC = 240^{\circ}$, $E$ and $F$ are the midpoints of sides $AD$ and $BC$, respectively, and $EF = \sqrt{7}$. If two squares $A_1$ and $A_2$ are drawn with sides $AB$ and $CD$, respectively, and a rectangle $A_3$ is drawn with length $AB$ and width $CD$, find the... | 6.28 .
As shown in Figure 6, extend $BA$ and $CD$ to intersect at point $P$.
Given $\angle BAD + \angle ADC = 240^{\circ}$, we have
$$
\angle BPC = 60^{\circ}.
$$
Connect $BD$, and take the midpoint $G$ of $BD$, then connect $EG$ and $FG$.
By the Midline Theorem of a triangle, we know
$$
EG \perp \frac{1}{2} AB, FG \... | 28 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,299 |
7. Find the value of $\sin ^{4} \frac{\pi}{8}+\cos ^{4} \frac{3 \pi}{8}+\sin ^{4} \frac{5 \pi}{8}+\cos ^{4} \frac{7 \pi}{8}$. | $\begin{array}{l}\text { 7. } \frac{3}{2} \\ \sin ^{4} \frac{\pi}{8}+\cos ^{4} \frac{3 \pi}{8}+\sin ^{4} \frac{5 \pi}{8}+\cos ^{4} \frac{7 \pi}{8} \\ =\sin ^{4} \frac{\pi}{8}+\sin ^{4} \frac{\pi}{8}+\cos ^{4} \frac{\pi}{8}+\cos ^{4} \frac{\pi}{8} \\ =\frac{1}{2}\left[\left(1-\cos \frac{\pi}{4}\right)^{2}+\left(1+\cos \... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,300 |
8. In $\triangle A B C$, $\angle A B C=100^{\circ}, \angle A C B$ $=65^{\circ}$, points $M, N$ are taken on sides $A B, A C$ respectively, such that $\angle M C B=55^{\circ}, \angle N B C=80^{\circ}$. Determine the measure of $\angle N M C$. | 8. $25^{\circ}$.
It is easy to know that $\angle B A C=15^{\circ}$.
As shown in Figure 7, construct the circumcircle of $\triangle M C B$ intersecting the extension of $B N$ at point $M_{1}$. In this circle, the arcs $\overparen{C M_{1}}$ and $\overparen{C M}$ subtend complementary angles. Therefore, $C M_{1}=C M$.
$$... | 25^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,301 |
1. Let real numbers $a$, $b$, $c$ satisfy
$$
\begin{array}{l}
a^{2}-b c-2 a+10=0, \\
b^{2}+b c+c^{2}-12 a-15=0 .
\end{array}
$$
Then the range of values for $a$ is | 1. $1 \leqslant a \leqslant 5$.
From $b c=a^{2}-2 a+10$,
$$
(b+c)^{2}=b c+12 a+15 \text {, }
$$
we get $(b+c)^{2}=a^{2}+10 a+25=(a+5)^{2}$.
Then $(b-c)^{2}=(a+5)^{2}-4\left(a^{2}-2 a+10\right) \geqslant 0$,
which means $a^{2}-6 a+5 \leqslant 0$.
Solving this, we get $1 \leqslant a \leqslant 5$. | 1 \leqslant a \leqslant 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,302 |
2. The set of positive integer solutions $(a, b)$ that satisfy $a^{2}+a b+b^{2}=2010$ is $\qquad$ | 2. $\varnothing$.
If $a$ and $b$ are both odd or one is odd and the other is even, then $a^{2}+a b+b^{2}$ is odd, which contradicts the fact that 2010 is even;
If $a$ and $b$ are both even, then $a^{2}+a b+b^{2}$ is a multiple of 4, which contradicts the fact that 2010 is congruent to 2 modulo 4.
Therefore, the origi... | \varnothing | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,303 |
3. Given that $x_{1}, x_{2}, \cdots, x_{2010}$ are all positive real numbers. Then
$$
x_{1}+\frac{x_{2}}{x_{1}}+\frac{x_{3}}{x_{1} x_{2}}+\cdots+\frac{x_{2010}}{x_{1} x_{2} \cdots x_{200}}+\frac{4}{x_{1} x_{2} \cdots x_{2010}}
$$
the minimum value is $\qquad$ | 3.4.
Starting from the last two terms, repeatedly applying the AM-GM inequality, we get
$$
\begin{array}{l}
\text { Original expression }=\sum_{i=1}^{2010} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\frac{4}{\prod_{j=1}^{2010} x_{j}} \\
=\sum_{i=1}^{2009} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\left(\frac{x_{2010}}{\prod_{j... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,304 |
Example 3 Find positive integers $x, y$, such that $x+y^{2}+(x, y)^{3}=x y(x, y)$. | 【Analysis】Let $z=(x, y), x=a z, y=b z$. Then $(a, b)=1$.
The equation transforms into $a+b^{2} z+z^{2}=a b z^{2}$. Hence $z \mid a$. Let $a=c z$, i.e., $x=c z^{2}$. Then the equation becomes $c+b^{2}+z=b c z^{2} \Rightarrow c=\frac{b^{2}+z}{b z^{2}-1}$.
We now discuss the variable $z$.
(1) If $z=1$, then
$$
c=\frac{b^{... | (x, y)=(5,2),(5,3),(4,2),(4,6) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,305 |
4. Given that the circumcenter, incenter, and orthocenter of a non-isosceles acute $\triangle ABC$ are $O, I, H$ respectively, and $\angle A=60^{\circ}$. If the altitudes of $\triangle ABC$ are $AD, BE, CF$, then the ratio of the circumradius of $\triangle OIH$ to the circumradius of $\triangle DEF$ is $\qquad$ . | 4. 2 .
From $\angle B O C=\angle B I C=\angle B H C=120^{\circ}$, we know that $O, I, H, B, C$ are concyclic.
Let the circumradii of $\triangle A B C$ and $\triangle O I H$ be $R$ and $r$, respectively.
By the Law of Sines, we have
$2 R \sin A=B C=2 r \sin \angle B O C$.
Thus, $r=R$.
Let the circumradius of $\triangl... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,306 |
5. In a room, the floor is a square with a side length of $6 \mathrm{~m}$, and its center is set as $O$. A lamp $V$ is to be installed on the ceiling directly above $O$. It is known that the angle of illumination of the lamp is $90^{\circ}$ (the maximum angle between the boundaries of the light rays emitted by $V$, i.e... | 5. $3 \sqrt{2}$.
Let the square $A B C D$ have a side length of $6 \mathrm{~m}$.
When $\triangle V A C$ is an isosceles right triangle, the altitude from $V$ to the hypotenuse $A C$, which is $3 \sqrt{2}$, is the minimum value of $V O$. | 3 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,307 |
6. If the function $f(x)=|x-[x+a]|$ has a maximum value $M(a)$, then the range of positive real number $a$ is $\qquad$ ( $[y]$ denotes the greatest integer not exceeding the real number $y$).
保留了原文的换行和格式。 | 6. $\left[\frac{1}{2},+\infty\right)$.
Let $x+a=N+\alpha$ (where $N$ is an integer, $0 \leqslant \alpha|-a|$, so, $f(x)$ has no maximum value;
When $a \geqslant \frac{1}{2}$, $-a \leqslant \alpha-a<1-a$, since $|-a| \geqslant|1-a|$, so, $f(x)$ has a maximum value, and when $\alpha=0$, i.e., $x+a$ is an integer, the m... | \left[\frac{1}{2},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,308 |
7. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, the right focus is $F$. The line $l: y=k x+d$ does not pass through point $F$, and intersects the right branch of the hyperbola at points $P$ and $Q$. If the external angle bisector of $\angle P F Q$ intersects $l$ at point $A$, then the x-coordinate of ... | 7. $\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}$.
Draw a perpendicular line $h$ from point $A$ to the $x$-axis.
Let the projections of $P$ and $Q$ on line $h$ be $P^{\prime}$ and $Q^{\prime}$, respectively. Then $\frac{P P^{\prime}}{Q Q^{\prime}}=\frac{A P}{A Q}=\frac{F P}{F Q}$, which means $\frac{P F}{P P^{\prime}}=\frac{Q F}{... | \frac{a^{2}}{\sqrt{a^{2}+b^{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,309 |
8. Given that $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ is a permutation of $1,2,3,4,5$, and satisfies $\left|a_{i}-a_{i+1}\right| \neq 1(i=1,2,3,4)$. Then the number of permutations $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ that meet the condition is $\qquad$. | 8. 14 .
The permutations that satisfy the conditions are:
$$
\begin{array}{l}
1,3,5,2,4 ; 1,4,2,5,3 ; 2,4,1,3,5 ; \\
2,4,1,5,3 ; 2,5,3,1,4 ; 3,1,4,2,5 ; \\
3,1,5,2,4 ; 3,5,1,4,2 ; 3,5,2,4,1 ; \\
4,1,3,5,2 ; 4,2,5,1,3 ; 4,2,5,3,1 ; \\
5,2,4,1,3 ; 5,3,1,4,2 .
\end{array}
$$ | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,310 |
9. (14 points) As shown in Figure 1, let the circumcenter and incenter of $\triangle ABC$ be $O$ and $I$, respectively, and the excenters opposite to $\angle A, \angle B, \angle C$ be $I_{1}, I_{2}, I_{3}$. Prove:
(1) $\triangle I_{1} I_{2} I_{3}$
is an acute triangle;
(2) If the circumcenter of $\triangle I_{1} I_{2}... | (1) From $\angle I_{3} I_{1} I_{2}=90^{\circ}-\frac{\angle A}{2}<90^{\circ}$, we get $\angle I_{3} I_{1} I_{2}$ is an acute angle.
Similarly, $\angle I_{1} I_{2} I_{3}$ and $\angle I_{2} I_{3} I_{1}$ are also acute angles.
Therefore, $\triangle I_{1} I_{2} I_{2}$ is an acute triangle.
(2) As shown in Figure 3, let $A I... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,311 |
10. (15 points) Given a sequence of rational numbers $\left\{a_{n}\right\}(n=0$, $1, \cdots)$ satisfying
$$
a_{n}=\alpha x_{1}^{n}+\beta x_{2}^{n} \neq 0(n=0,1, \cdots),
$$
where $\alpha, \beta \in \mathbf{R}, x_{1}, x_{2} \in \mathbf{C}$, and $x_{1} x_{2}=1$. Prove:
(1) $x_{1}+x_{2}$ is a rational number;
(2) If $x_{... | 10. (1) Since $a_{0}=\alpha+\beta, a_{1}=\alpha x_{1}+\beta x_{2}$, $a_{2}=\alpha x_{1}^{2}+\beta x_{2}^{2}$,
therefore, $a_{0} x_{1}-a_{1}=\beta\left(x_{1}-x_{2}\right)$, $a_{1} x_{1}-a_{2}=\beta x_{2}\left(x_{1}-x_{2}\right)$.
Thus, $a_{1} x_{1}-a_{2}=x_{2}\left(a_{0} x_{1}-a_{1}\right)$, i.e., $a_{2}-\left(x_{1}+x_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,312 |
11. (15 points) In a regular pentagon $A B C D E$, the diagonal $B E$ intersects diagonals $A D$ and $A C$ at points $F$ and $G$, respectively. The diagonal $B D$ intersects diagonals $C A$ and $C E$ at points $H$ and $I$, respectively. The diagonal $C E$ intersects diagonal $A D$ at point $J$. Let the set of isosceles... | 11. (1) Since all triangles formed by 10 points $A, B, C, D, E, F, G, H, I, J$ and line segments in Figure 2 are isosceles triangles,
$$
|M|=\mathrm{C}_{5}^{3}+4 \mathrm{C}_{5}^{4}+5 \mathrm{C}_{5}^{5}=35 \text{. }
$$
(2) By the pigeonhole principle, we know that among $A, B, C, D, E$, there must be three points of the... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,313 |
1. Xiao Pei and Xiao Fo each have an integer amount of RMB. Xiao Pei says to Xiao Fo: "If you give me 2 yuan, my money will be $n$ times yours." Xiao Fo says to Xiao Pei: "If you give me $n$ yuan, my money will be 2 times yours." Here, $n$ is a positive integer. Then the number of possible values for $n$ is ( ).
(A) 1
... | - 1. D.
Let $x$ yuan and $y$ yuan represent the amounts of money Xiao Pei and Xiao Fo have, respectively. Thus,
$$
\left\{\begin{array}{l}
x+2=n(y-2), \\
y+n=2(x-n) .
\end{array}\right.
$$
From equation (1), we get $x=n y-2 n-2$.
Substituting the above equation into equation (2) yields
$$
(2 y-7) n=y+4 \text {, }
$$
... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,314 |
2. Given weights of $7 \mathrm{~g}$ and $m \mathrm{~g}(m$ being a positive integer) are available, and to weigh an object of $1 \mathrm{~g}$ on a balance, 7 of these weights are needed. Then the number of possible values for $m$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 2. C.
Let the number of $7 \mathrm{~g}$ weights be $n$. Then the number of $m \mathrm{~g}$ weights is $7-n$. According to the problem,
$(7-n) m-7 n=1$ or $7 n-(7-n) m=1$.
If $(7-n) m-7 n=1$, then
$$
m=\frac{7 n+1}{7-n}=\frac{7 n-49+50}{7-n}=\frac{50}{7-n}-7 .
$$
Since $0<7-n<7$, we have $7-n=1,2,5$.
Accordingly, $m=4... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,315 |
Example 4 Let $n \in \mathbf{N}_{+}, f(n)$ be the product of the digits of $n$ in decimal notation. Solve the equation $f(n)=\frac{2}{3} n+8$.
| Let $n$ be a $k+1$-digit number, i.e., $n=\overline{x_{k} x_{k-1} \cdots x_{1} x_{0}}$. By estimating through inequalities, study the range of $k$.
$$
\begin{array}{l}
n=\sum_{i=0}^{k} 10^{i} x_{i} \geqslant 10^{k} x_{k}=9^{k}\left(\frac{10}{9}\right)^{k} x_{k} \\
\geqslant\left(\frac{10}{9}\right)^{k} \prod_{i=0}^{k} ... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,316 |
3. As shown in Figure 1, given that $E$ is the midpoint of side $AD$ of rectangle $ABCD$, $BE \perp AC$ at point $F$, and $AF=2$. Then $DF=$ ( ).
(A) $2 \sqrt{3}$
(B) 3
(C) $3 \sqrt{3}$
(D) $2 \sqrt{2}$ | 3. A.
Since $\triangle A E F \sim \triangle C B F$, we have $C F=2 A F=4$.
In Rt $\triangle A B C$ and Rt $\triangle A B E$, by the projection theorem, we get
$$
\begin{array}{l}
B F^{2}=A F \cdot C F=8, B F=2 \sqrt{2}, \\
E F=\frac{1}{2} B F=\sqrt{2} ; \\
A E^{2}=E F \cdot E B, A E=\sqrt{6} . \\
\text { Therefore, } ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,317 |
4. Let the three sides of a triangle be $a, b, c$, and the radius of its circumcircle be $R, a \sqrt{bc} \geqslant (b+c) R$. Then the degree measure of the largest interior angle of the triangle is ( ).
(A) $150^{\circ}$
(B) $120^{\circ}$
(C) $90^{\circ}$
(D) $135^{\circ}$ | 4. C.
From $b+c \geqslant 2 \sqrt{b c}$
$$
\Rightarrow a \sqrt{b c} \geqslant(b+c) R \geqslant 2 \sqrt{b c} R \Rightarrow a \geqslant 2 R \text {. }
$$
Since the side length of a triangle is no greater than the diameter of its circumcircle, then $a \leqslant 2 R$.
Therefore, $a=2 R$.
Thus, the triangle is a right tri... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,318 |
5. Given that $\alpha$ is an acute angle. Then the situation of the roots of the equation about $x$
$$
x^{3}-x^{2}+(\sin \alpha-3) x+1=0
$$
is ( ).
(A) Only one positive root
(B) Three positive roots
(C) One positive root, two negative roots
(D) Two positive roots, one negative root | 5. D.
Obviously, $x \neq 0$. Thus, the original equation can be rewritten as $x^{2}-x+\sin \alpha-3=-\frac{1}{x}$. Let $y_{1}=x^{2}-x+\sin \alpha-3, y_{2}=-\frac{1}{x}$.
The graphs are shown in Figure 4.
The vertex of the parabola $y_{1}$ is
$$
A\left(\frac{1}{2}, \sin \alpha-\frac{13}{4}\right) .
$$
When $x=\frac{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,319 |
6. As shown in Figure 2, given that $C$ is a point on the semicircle $\odot O$ with diameter $A B$, $I$ is the incenter of $\triangle A B C$, the extensions of $A I$ and $B I$ intersect the semicircle $\odot O$ at points $D$ and $E$ respectively, and $A B=6$. Then the length of $D E$ is ( ).
(A)3
(B) $3 \sqrt{2}$
(C) $... | 6. B.
It is known that $\angle C=\angle A E B=90^{\circ}$,
$$
\begin{array}{l}
\angle A I E=\angle I A B+\angle A B I \\
=\frac{1}{2}(\angle C A B+\angle A B C)=45^{\circ} .
\end{array}
$$
Therefore, $\angle E A D=45^{\circ}$.
In $\triangle A D E$, by the Law of Sines we get
$$
D E=A B \sin \angle E A D=6 \sin 45^{\c... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,320 |
1. Given that $P$ is a point inside the equilateral $\triangle A B C$, $P A=1$, $P B=2$, $\angle A P B=150^{\circ}$. Then $P C=$ $\qquad$ | $1 . \sqrt{5}$.
As shown in Figure 5, $\triangle P A B$ is rotated $60^{\circ}$ clockwise around point $B$ to $\triangle Q C B$, and $P Q$ is connected. Then $\triangle P Q B$ is an equilateral triangle,
$$
\begin{array}{c}
P Q=P B=2, \\
C Q=P A=1, \\
\angle P Q C=\angle B Q C-\angle P Q B \\
=\angle A P B-\angle P Q B... | \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,321 |
2. Let $m$ be a real number, and consider the quadratic equation in $x$
$$
(m-2) x^{2}+(2 m-1) x+(m+2)=0
$$
The maximum value of the sum of the reciprocals of the two real roots is 8. Then the range of values for $m$ is $\qquad$ | 2. $m<-2$ or $-\frac{3}{2} \leqslant m \leqslant \frac{17}{4}$ and $m \neq 2$.
Let the two real roots of the equation be $x_{1}$ and $x_{2}$. By the problem, we have
$$
\left\{\begin{array}{l}
m-2 \neq 0, \\
\Delta=(2 m-1)^{2}-4(m-2)(m+2) \geqslant 0 .
\end{array}\right.
$$
Solving this, we get $m \leqslant \frac{17}... | m<-2 \text{ or } -\frac{3}{2} \leqslant m \leqslant \frac{17}{4} \text{ and } m \neq 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,322 |
3. Given real numbers $a, b, c$ satisfy
$$
a^{2}+b^{2}=4, b^{2}+c^{2}=8 \text {. }
$$
Then the minimum value of $a b+b c+\frac{\sqrt{2}}{2} a c$ is $\qquad$ | 3. $-3 \sqrt{2}$.
Notice that
$$
\begin{aligned}
(a+ & \sqrt{2} b+c)^{2} \\
& =a^{2}+2 b^{2}+c^{2}+2(\sqrt{2} a b+\sqrt{2} b c+a c) \\
& =\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)+2 \sqrt{2}\left(a b+b c+\frac{\sqrt{2}}{2} a c\right) .
\end{aligned}
$$
$$
\begin{array}{l}
\text { Hence } a b+b c+\frac{\sqrt{2}... | -3 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,323 |
4. Divide a wire of length $11 \mathrm{~cm}$ into several segments of integer centimeters, such that any three segments can form the sides of a triangle. Then the number of different ways to divide the wire is $\qquad$ (ways with the same number of segments and corresponding equal lengths are considered the same way). | 4.8.
Let the shortest two segments be $a \mathrm{~cm}, b \mathrm{~cm}$, and take any other segment $c \mathrm{~cm} (a \leqslant b \leqslant c)$. Then $b \leqslant c < a + b$.
(1) If $a = 1$, then $1 \leqslant b \leqslant c < b + 1$. Thus, $b = c$. Therefore, $b \in (11 - 1), 2b < 11 - 1$. Hence, $b$ can be $1, 2, 5$, ... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,324 |
One, (20 points) A competitive event involves several teams from two cities, A and B. It is known that city B has 8 more teams than city A, and any two teams play exactly one match. The event rules state: the winner gets 1 point, the loser gets 0 points, and there are no ties. In the end, the total score of all teams f... | Let city A have $x$ teams participating in the competition. Then city B has $x+8$ teams. When the teams from cities A and B compete against each other, city A scores a total of $y$ points, and city B scores a total of $x(x+8)-y$ points.
From the problem, we have
$$
\begin{array}{l}
\frac{1}{2} x(x-1)+y+4 \\
=\frac{1}{2... | 19 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,325 |
II. (25 points) As shown in Figure 3, given an acute triangle $\triangle ABC$ with the incircle $\odot I$ touching sides $AB$ and $AC$ at points $D$ and $E$ respectively, the extensions of $BI$ and $CI$ intersect line $DE$ at points $F$ and $G$ respectively, with $AB < AC$ and $BC = 2FG$. Prove that $\triangle ADE$ is ... | II. As shown in Figure 6, connect $A I$, $D I$, $E I$, $B G$, and $C F$.
Let $\angle B A C = 2 \alpha$.
Obviously, $A$, $D$, $I$, and $E$ are concyclic.
Thus, $\angle I D E = \frac{1}{2} \angle B A C = \alpha$.
Therefore, $\angle B D F = \angle B D I + \angle I D E = 90^{\circ} + \alpha$.
Since $\angle B I C = 180^{\ci... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,326 |
Example 1 Given that the three sides of $\triangle ABC$ are $a, b, c$ which are positive integers (the greatest common divisor of $a, b, c$ is 1), and
$$
\angle A: \angle B: \angle C=4: 2: 1 \text {. }
$$
Prove: $a+b, a-c, b-c$ are all perfect squares. | Prove as shown in Figure 1, draw $BD$ to bisect $\angle ABC$. Then
$$
\begin{array}{l}
\angle DBC = \angle C, \\
\angle BDA = 2 \angle C. \\
\text{Hence } \triangle ABC \backsim \triangle ADB \\
\Rightarrow \frac{b}{c} = \frac{c}{AD}.
\end{array}
$$
By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
\frac{a}{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,327 |
Example 2: With $2009^{12}$ as one of the legs, and all three sides being integers, the number of different right-angled triangles (congruent triangles are considered the same) is $\qquad$.
(2009, International Mathematics Tournament of the Cities for Young Mathematicians) | Let $a$, $2009^{12}$, and $c$ be the sides of a right triangle, with $c$ being the hypotenuse. Then,
$$
(c+a)(c-a)=2009^{24}=41^{24} \times 7^{48}.
$$
Since $c+a > c-a$ and they have the same parity, both must be odd.
Also, $2009^{24}=41^{24} \times 7^{48}$ has $25 \times 49$ different factors, which can be paired in... | 612 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,328 |
Example 3 There is a railway network between six cities, such that there is a direct railway between any two cities. On Sundays, some railways will be closed for maintenance. The railway department stipulates: after closing several sections of the railway, it must still be possible to travel by rail between any two cit... | 【Analysis】The problem is equivalent to finding the number of all possible ways to connect a graph composed of six points and several edges, such that any two points are connected.
Let $f(n)$ denote the number of all possible ways to connect $n$ points such that any two points are connected. Then
$$
f(1)=1, f(2)=1 \tex... | 26703 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,329 |
8. In the geometric sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{2010}=4$, the function
$$
f(x)=x\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{2010}\right) .
$$
Then the equation of the tangent line to the function $y=f(x)$ at the point $(0,0)$ is _. $\qquad$ | 8. $y=2^{2010} x$.
$$
\text { Let } g(x)=(x-a_{1})(x-a_{2}) \cdots(x-a_{2010}) \text {. }
$$
Then $f(x)=x g(x)$.
Since $f^{\prime}(x)=g(x)+x g^{\prime}(x)$, we have,
$$
\begin{array}{l}
f^{\prime}(0)=g(0)=a_{1} a_{2} \cdots a_{2010} \\
=\left(a_{1} a_{2010}\right)^{\frac{2010}{2}}=2^{2010} .
\end{array}
$$
Therefore,... | y=2^{2010} x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,330 |
10. Given that $y=f(x)$ is a continuous function on the interval $[0,1]$, and always $0 \leqslant f(x) \leqslant 1$, the integral $\int_{0}^{1} f(x) \mathrm{d} x$ can be approximately calculated using a random simulation method. First, generate two sets (each set containing $N$ numbers) of uniformly distributed random ... | 10. $\frac{N_{1}}{N}$.
Let $\int_{0}^{1} f(x) \mathrm{d} x=S$.
If $N$ points are taken uniformly at random in the region $x \in [0,1]$, $y \in [0,1]$ with an area of 1, and the number of points within the integration region is $N_{1}$, then $\frac{S}{1}=\frac{N_{1}}{N}$.
Therefore, the approximate value of $S$ is $\fr... | \frac{N_{1}}{N} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 723,332 |
11. Let $a_{n}$ be the coefficient of $x$ in the binomial expansion of $(3-\sqrt{x})^{n}(n=2,3, \cdots)$. Then
$$
\sum_{n=2}^{18} \frac{3^{n}}{a_{n}}=
$$
$\qquad$ | 11. 17.
Since $a_{n}=3^{n-2} C_{n}^{2}$, therefore,
$$
\frac{3^{n}}{a_{n}}=3^{2} \times \frac{2}{n(n-1)}=\frac{18}{n(n-1)} \text {. }
$$
Thus, $\sum_{n=2}^{18} \frac{3^{n}}{a_{n}}=18 \sum_{n=2}^{18} \frac{1}{n(n-1)}=17$. | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,333 |
12. If three non-zero real numbers
$$
x(y-z) 、 y(z-x) 、 z(y-x)
$$
form a geometric sequence, then the common ratio $q=$ $\qquad$ | 12. $\frac{1 \pm \sqrt{5}}{2}$.
Notice that
$$
x(y-z)+y(z-x)=z(y-x) \text {. }
$$
Therefore, $1+q=q^{2}$.
Solving for $q$ gives $q=\frac{1 \pm \sqrt{5}}{2}$. | \frac{1 \pm \sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,334 |
13. Let the function be
$$
f(x)=4 \sin x \cdot \sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)+\cos 2 x \text {. }
$$
If $|f(x)-m|<2$ holds for $\frac{\pi}{6} \leqslant x \leqslant \frac{2 \pi}{3}$, then the range of the real number $m$ is | 13. $(1,4)$.
Notice
$$
\begin{array}{l}
f(x)=2 \sin x\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}+\cos 2 x \\
=2 \sin x(1+\sin x)+1-2 \sin ^{2} x \\
=1+2 \sin x . \\
\text { When } \frac{\pi}{6} \leqslant x \leqslant \frac{2 \pi}{3} \text {, }|f(x)-m|<2 \text { always holds }
\end{array}
$$
When $\frac{\pi}{6} ... | (1,4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,335 |
14. In space, there are five points, no four of which are coplanar. If several line segments are drawn such that no tetrahedron exists in the graph, then the maximum number of triangles in the graph is $\qquad$. | 14.4.
First, construct graph 6. It is easy to see that it meets the conditions and has exactly four triangles.
Now assume there exists some configuration where the number of triangles is no less than five.
If only two line segments are not connected, then these two line segments must have no common endpoints (as sho... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,336 |
15. (15 points) Given that when $x \in[1, \mathrm{e}]$, the inequality
$$
a \ln x \leqslant-\frac{1}{2} x^{2}+(a+1) x
$$
always holds. Try to find the range of the real number $a$. | Three, 15. The inequality can be transformed into
$$
a(x-\ln x) \geqslant \frac{x^{2}}{2}-x \text {. }
$$
Since $x \in[1, \mathrm{e}]$, we have $x-\ln x>0$.
Thus, the inequality becomes $a \geqslant \frac{\frac{x^{2}}{2}-x}{x-\ln x}$.
Let $g(x)=\frac{\frac{x^{2}}{2}-x}{x-\ln x}(x \in[1, \mathrm{e}])$. Note that
$$
g^{... | \frac{e^2 - 2e}{2(e - 1)} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,337 |
16. (15 points) As shown in Figure 4, $\odot O_{1}$ and $\odot O_{2}$ roll inside $\odot O$ and always remain internally tangent to $\odot O$, with points of tangency at $P$ and $Q$, respectively. $MN$ is the external common tangent of $\odot O_{1}$ and $\odot O_{2}$. Given that the radii of $\odot O_{1}$, $\odot O_{2}... | 16. It is known that $O_{1}$ and $O_{2}$ lie on segments $O P$ and $O Q$, respectively, and $O_{1} M \perp M N, O_{2} N \perp M N$. Then
$$
M N^{2}=O_{1} O_{2}^{2}-\left(r_{1}-r_{2}\right)^{2} \text {. }
$$
In $\triangle \mathrm{O}_{1} \mathrm{OO}_{2}$, by the cosine rule we have
$$
\begin{array}{l}
O_{1} O_{2}^{2}=\l... | \frac{\left(R-r_{1}\right)\left(R-r_{2}\right)}{R^{2}} | Geometry | proof | Yes | Yes | cn_contest | false | 723,338 |
17. (17 points) Let the ellipses be
$$
C_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \quad C_{2}: \frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1,
$$
Draw rays from the origin $O$ intersecting the ellipses $C_{1}$ and $C_{2}$ at points $A$ and $B$, respectively, and let $P$ be a point on the line segment $AB$.
(1) Prove... | 17. (1) Let the parametric equation of ray $O A$ be
$$
\left\{\begin{array}{l}
x=t \cos \theta, \\
y=t \sin \theta
\end{array}(0 \leqslant \theta \leqslant 2 \pi, t>0) .\right.
$$
Let $A\left(t_{1} \cos \theta, t_{1} \sin \theta\right), B\left(t_{2} \cos \theta, t_{2} \sin \theta\right)$,
$$
P\left(t_{3} \cos \theta, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,339 |
Example 4 A country has 1001 cities, and there is a one-way road between every two cities. Each city has exactly 500 outgoing roads and exactly 500 incoming roads. A new independent country splits off, consisting of 668 cities. Prove that from each city in the new country, one can reach any other city in the new countr... | 【Analysis】The background of this problem mainly examines the properties of tournament graphs.
Since there is a one-way road between any two of the 1001 cities, these cities can be regarded as points, and the one-way roads connecting them as directed edges. For any point $v$, we have
$$
d^{+}(v)=d^{-}(v)=500 .
$$
Thus... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,340 |
18. (17 points) Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of the integers 1, 2, $\cdots, n$, and satisfy the following conditions:
(1) $a_{1}=1$;
(2) $\left|a_{i}-a_{i+1}\right| \leqslant 2(i=1,2, \cdots, n-1)$.
Let the number of such permutations be $f(n)$. Find the remainder when $f(2010)$ is divided by 3. | 18. Verifiable
$$
f(1)=1, f(2)=1, f(3)=2 \text {. }
$$
Let $n \geqslant 4$. Then $a_{2}=2$ or 3.
For $a_{2}=2$, the number of permutations is $f(n-1)$. This is because by removing the first term and reducing all subsequent terms by 1, a one-to-one correspondence can be established.
For $a_{2}=3$, if $a_{3}=2$, then $... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,341 |
1. If $y=(x+1)(x+2)(x+3)(x+4)+5$, where $-3 \leqslant x \leqslant 3$, then the minimum value of $y$ is ( ).
(A) 3
(B) 4
(C) 29
(D) 5.5625 | - 1. B.
Notice
$$
\begin{aligned}
y & =(x+1)(x+2)(x+3)(x+4)+5 \\
& =\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)+5 \\
& =\left(x^{2}+5 x+4\right)^{2}+2\left(x^{2}+5 x+4\right)+5 .
\end{aligned}
$$
Let $z=x^{2}+5 x+4$.
From $-3 \leqslant x \leqslant 3$, we get
$$
-2.25 \leqslant z \leqslant 28, y=(z+1)^{2}+4 \text... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,342 |
2. From the numbers 2, 3, 4, 5, take any two numbers $p, q (p \neq q)$ to form the functions $y = px - 2$ and $y = x + q$, and ensure that the intersection point of these two function graphs is to the right of the line $x = 2$. Then the number of such ordered pairs $(p, q)$ is ( ).
(A) 12
(B) 6
(C) 5
(D) 3 | 2. B.
From the problem, we have
$$
p x-2=x+q \Rightarrow x=\frac{q+2}{p-1} \text {. }
$$
Since the intersection point is to the right of the line $x=2$, we have
$$
\frac{q+2}{p-1}>2 \Rightarrow q>2(p-2) \text {. }
$$
Substituting $p, q=2,3,4,5$ into the above inequality, only $(2,3),(2,4),(2,5),(3,4),(3,5)$, and $(4... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,343 |
4. As shown in Figure 2, in the right trapezoid $A B C D$, $A D / /$ $B C, \angle A B C=90^{\circ}, A B=B C, E$ is a point on side $A B$, $\angle B C E=15^{\circ}$, and $A E=A D$. Connect $D E$ and diagonal $A C$ intersect at point $H$, and connect $B H$. Given the following conclusions:
(1) $\triangle A C D \cong \tri... | 4. B.
It is known that $\triangle A B C$ and $\triangle A D E$ are both isosceles right triangles.
Therefore, $\angle B A C=45^{\circ}$, which means $A H$ bisects $\angle E A D$.
Thus, $\triangle A C D \cong \triangle A C E$, so conclusion (1) is correct.
Since $\angle A C B=45^{\circ}$ and $\angle B C E=15^{\circ}$, ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,345 |
5. Let $a, b$ be integers greater than 1,
$$
m=\frac{a^{2}-1}{b+1}, n=\frac{b^{2}-1}{a+1} \text {. }
$$
If $m+n$ is an integer, then the integer(s) among $m, n$ is(are) ).
(A) $m, n$
(B) $m$
(C) $n$
(D) none | 5. A.
Since $q=m n=(a-1)(b-1)$ is an integer, and $p=m+n$ is an integer, therefore, $m$ and $n$ are the two rational roots of the equation $x^{2}-p x+q=0$.
Thus, $\Delta=p^{2}-4 q$ is a perfect square.
Solving the equation gives $m, n=\frac{p \pm \sqrt{p^{2}-4 q}}{2}$.
Since $p^{2}-4 q$ has the same parity as $p$, the... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,346 |
6. Let $n$ be a positive integer, and $3n+1$, $5n-1$ are both perfect squares. Consider the following two propositions:
(1) $7n+13$ must be a composite number;
(2) $8(17n^2+3n)$ must be the sum of two perfect squares.
Which of the following is correct?
(A)(1)
(B)(2)
(C)(1)(2)
(D) None | 6. C.
Let $3 n+1=a^{2}, 5 n-1=b^{2} (a, b$ are positive integers). Then
$$
\begin{array}{l}
7 n+13=9(3 n+1)-4(5 n-1) \\
=(3 a)^{2}-(2 b)^{2} \\
=(3 a-2 b)(3 a+2 b) .
\end{array}
$$
Therefore, $3 a-2 b$ is a positive integer.
If $3 a-2 b=1$, then
$$
27 n+9=(3 a)^{2}=(2 b+1)^{2},
$$
which means $27 n=4\left(b^{2}+b-2\... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,347 |
1. Let $a_{n}=\frac{2^{n}}{2^{2 n+1}-2^{n+1}-2^{n}+1}(n$ be a positive integer). Then the value of $a_{1}+a_{2}+\cdots+a_{2011}$ is $\qquad$ 1 (fill in “greater than”, “equal to”, or “less than”). | $$
\begin{array}{l}
\text { Given } a_{n}=\frac{2^{n}}{\left(2^{n}-1\right)\left(2^{n+1}-1\right)} \\
=\frac{1}{2^{n}-1}-\frac{1}{2^{n+1}-1},
\end{array}
$$
$$
\begin{array}{l}
\text { we get } a_{1}+a_{2}+\cdots+a_{2011} \\
=\left(1-\frac{1}{2^{2}-1}\right)+\left(\frac{1}{2^{2}-1}-\frac{1}{2^{3}-1}\right)+\cdots+ \\
\... | 1-\frac{1}{2^{2012}-1}<1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,348 |
2. Given a square $ABCD$ with side length $5$, $P$ is a point inside the square, and $PA=\sqrt{5}, PC=5$. Then $PB=$ $\qquad$ | 2. $\sqrt{10}$ or $2 \sqrt{5}$.
As shown in Figure 4, draw $P E \perp$ $A B$ at point $E$, and $P F \perp B C$ at point $F$.
Let $P E=m, P F=n$.
In the right triangles $\triangle P A E$ and $\triangle P C F$, by the Pythagorean theorem, we get
$$
\left\{\begin{array}{l}
m^{2}+(5-n)^{2}=5, \\
(5-m)^{2}+n^{2}=25 .
\end{... | \sqrt{10} \text{ or } 2 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,349 |
3. Given real numbers $a$ and $b$ satisfy $a^{3}-b^{3}-3 a b=1$. Then $a-b=$ $\qquad$ . | 3.1 or -2.
Let $a-b=k$. Then from the condition we know
$$
\begin{array}{l}
k\left(k^{2}+3 a b\right)-3 a b-1=0 \\
\Rightarrow k^{3}+3 k a b-3 a b-1=0 \\
\Rightarrow(k-1)\left(k^{2}+k+1\right)+3 a b(k-1)=0 \\
\Rightarrow(k-1)\left(k^{2}+k+1+3 a b\right)=0 .
\end{array}
$$
Thus, $k-1=0$ or $k^{2}+k+1+3 a b=0$.
When $k... | 1 \text{ or } -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,350 |
Example 5 In a sports competition with $n(n \geqslant 3)$ participants, every two participants have played against each other once, and there were no ties. Furthermore, no participant has defeated all the others. Prove: there must exist three participants $A$, $B$, and $C$, such that the results between them are $A$ be... | Prove that if player $A$ beats player $B$, denoted as $A \rightarrow B$, this results in a tournament graph.
Since no one has achieved a perfect record, assume without loss of generality that $A$ is the player with the most wins. Therefore, there exists $A \rightarrow B$, and among the players $A$ has beaten, there mu... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,351 |
4. In Rt $\triangle A B C$, $C D$ is the altitude on the hypotenuse $A B$, the three sides $a, b, c$ of $\triangle A B C$ are all positive integers, $B D$ $=27$. Then $\cos B=$ $\qquad$ | 4. $\frac{3}{5}$.
Let $(a, c)=d, a=a_{0} d, c=c_{0} d,\left(a_{0}, c_{0}\right)=1$.
By the projection theorem, we know $B C^{2}=B D \cdot B A$
$$
\Rightarrow a^{2}=27 c \Rightarrow d a_{0}=27 c_{0}
$$
$\Rightarrow a_{0}^{2} \mid 27 \Rightarrow a_{0}=1$ or 3:
If $a_{0}=1$, then
$$
\begin{array}{l}
b=\sqrt{c^{2}-a^{2}}=... | \frac{3}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,352 |
For each positive integer $n$, let $f(n)$ denote the last digit of $1+2+\cdots+n$ (for example, $f(1)=1$, $f(2)=3$, $f(3)=6$). Calculate the value of $f(1)+f(2)+\cdots+f(2011)$. | One, because the last digit of the sum of any 20 consecutive positive integers is 0, so,
$$
f(n+20)=f(n) .
$$
Thus, it is only necessary to calculate the $f(n)$ corresponding to 1 to 20, as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$f... | 7046 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,353 |
II. (25 points) As shown in Figure 3, from a point $P$ outside the circle, draw two tangents $PA$ and $PB$, with $A$ and $B$ being the points of tangency. Draw a secant line through $P$ that intersects the circle at points $C$ and $D$. Through any point $Q$ on $AB$, draw a line parallel to $PA$ that intersects lines $A... | II. As shown in Figure 5, connect $BD$, $BC$.
$$
\begin{array}{l}
\text { Then } \angle ABC \\
=\angle PAC \\
=\angle E.
\end{array}
$$
Thus, $\triangle ABC$
$$
\sim \triangle AEQ.
$$
Therefore, $\frac{EQ}{BC}=\frac{AQ}{AC}$, which means
$$
EQ=\frac{AQ \cdot BC}{AC}.
$$
Similarly, $QF=\frac{AQ \cdot BD}{AD}$.
On the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,354 |
Three. (25 points) Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there are always three numbers, the sum of any two of which is irrational. | Three, take four irrational numbers $\{\sqrt{2}, \sqrt{3},-\sqrt{2},-\sqrt{3}\}$, obviously they do not satisfy the condition, hence $n \geqslant 5$.
Consider five irrational numbers $a, b, c, d, e$, viewed as five points.
If the sum of two numbers is a rational number, then connect the corresponding two points with a ... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,355 |
1. Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies
$$
f(f(x)+f(y))=f(x)+y(\forall x, y \in \mathbf{R}) \text {. }
$$
then $f(2011)=$ $\qquad$ . | $-1.2011$.
Let $x=0$ in equation (1), we get
$$
f(f(0)+f(y))=f(0)+y,
$$
which shows that $f$ is surjective.
Let $x$ be the zero root $a$ of $f(x)$ in equation (1). Then $f(f(y))=y$.
Replace $x$ with $f(x)$ in equation (1) to get
$$
f(x+f(y))=x+y \text {. }
$$
Let $y=a$ in equation (2) to get $f(x)=x+a$.
Substitute in... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,356 |
3. Let $a_{1}, a_{2}, \cdots, a_{2011}$ be positive real numbers, $S=\sum_{i=1}^{2011} a_{i}$, and
$$
\begin{array}{l}
(S+2011)\left(S+a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2011} a_{1}\right) \\
=4\left(\sqrt{a_{1} a_{2}}+\sqrt{a_{2} a_{3}}+\cdots+\sqrt{a_{2011} a_{1}}\right)^{2} .
\end{array}
$$
Then $S=$ . $\qquad$ | 3. 2011.
By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
& (S+2011)\left(S+a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2011} a_{1}\right) \\
= & \left(a_{1}+a_{2}+\cdots+a_{2011}+\frac{1+\cdots+1}{2011}\right) . \\
& \left(a_{2}+a_{3}+\cdots+a_{1}+a_{1} a_{2}+\cdots+a_{2011} a_{1}\right) \\
\geqslant & {\left[2\lef... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,358 |
4. Inside an irregular convex cube, there is a point $Q$. A line $l$ passing through $Q$ intersects the surface of the cube at points $M$ and $N$. The number of lines $l$ for which $Q M = Q N$ holds is $\qquad$. | 4. Infinitely many.
Let $Q$ be any point, and let $T_{1}$ be any plane passing through $Q$ that intersects the surface of the cube in a closed curve $\Gamma_{1}$, with point $Q$ inside the curve.
Let $N_{0}$ be the point on curve $\Gamma_{1}$ that is closest to $Q$. Extend $N_{0} Q$ to intersect curve $\Gamma_{1}$ at... | Infinitely many | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,359 |
6. From the set $\{1,2, \cdots, 2011\}$, randomly select 1005 different numbers such that their sum is 1021035. Then, there are at least some odd numbers among them. | 6.5.
From 1 to 2011, taking 1005 even numbers, their sum is $\frac{1005(2+2010)}{2}=1011030$.
Replacing $2,4,6,8,10$ with 2003, 2005, $2007,2009,2011$, increases the sum by
$$
2001 \times 5=10005,
$$
making the total sum 1021035.
Thus, there are at least 5 odd numbers. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,361 |
Example 6 In a competition with $2 n+1$ teams, each team played a match against every other team, and each match had a winner. If $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$, then the set of three teams $\{A, B, C\}$ is called "cyclic." Find the maximum and minimum number of cyclic sets. ${ }^{[1]}$
(2007, Italian ... | The maximum value is $\frac{n(n+1)(2 n+1)}{6}$, and the minimum value is 0.
Since any two teams must play a match, the total number of matches is $\mathrm{C}_{2 n+1}^{2}$.
To obtain the minimum number of cycles, let the number of wins for each team be $0,1, \cdots, 2 n$. In this case, there are no cyclic sets, i.e., ... | \frac{n(n+1)(2 n+1)}{6} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,362 |
7. Fill the 16 positive integers from $1 \sim 16$ randomly into the 16 squares of a $4 \times 4$ chessboard (one number per square). The probability that the sum of the numbers in each row and each column is even is $\qquad$ | 7. $\frac{41}{2145}$.
First, color a $4 \times 4$ chessboard with black and white such that there are 8 black and 8 white squares, and each row (or column) contains an even number of squares of the same color.
We consider three cases:
(1) If the first column has two black and two white squares, then there are $\mathrm... | \frac{41}{2145} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,363 |
8. Given that a series of non-overlapping disks with radius $R$ can cover all lattice points (in the Cartesian coordinate system, points with both integer horizontal and vertical coordinates are lattice points) on the plane. Then $R_{\max }$ $\qquad$ 4 (fill in “greater than”, “less than”, or “equal to”).
| 8. Less than.
Place three disks $G\left(O_{i}, R\right)(i=1,2,3)$, each with radius $R$, such that they are pairwise externally tangent within a grid where the vertices are lattice points. Then, there must be a unit grid whose vertices $A, B, C, D$ are each contained in one of the three disks.
[Note] If a unit grid's... | less than | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,364 |
9. (16 points) Given $\odot O: x^{2}+y^{2}=r^{2}$, and ellipse $C$: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, where $0<b<a<r$. From any point on $\odot O$, draw two tangent lines to ellipse $C$, and the line segment connecting the two points of tangency is called the corresponding chord of tangency. Try to find the ... | 9. Let any point on $\odot O$ be $A(r \cos \alpha, r \sin \alpha)(\alpha \in[0,2 \pi))$.
Then the equation of the chord of the tangent points corresponding to $A$ on the ellipse is
$$
\frac{x r \cos \alpha}{a^{2}}+\frac{y r \sin \alpha}{b^{2}}=1 \text {. }
$$
Consider an ellipse $\frac{x^{2}}{c^{2}}+\frac{y^{2}}{d^{2}... | \frac{r^{2} x^{2}}{a^{4}}+\frac{r^{2} y^{2}}{b^{4}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,365 |
10. (20 points) There are 100 people standing in a circle on the playground, labeled as $C_{1}, C_{2}, \cdots, C_{100}$ in a clockwise direction. The host distributes souvenirs numbered $1,2, \cdots, 50$ to the people in the following manner: First, the souvenir numbered 1 is given to $C_{1}$; then, skipping 1 person i... | 10. Let the $s$-th and $t$-th (where $1 \leqslant s < t \leqslant 50$) souvenirs be obtained by the same person. Then
$$
100 \left\lvert\,\left[\frac{t(t+1)}{2}-\frac{s(s+1)}{2}\right]\right.,
$$
which is $200 I (t-s)(t+s+1)$.
Clearly, $t+s+1 \leqslant 100$, and $t-s$ and $t+s+1$ have different parities.
(1) If $81(t+... | (2,37,42) \text{ and } (7,32,47) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,366 |
11. (20 points) Inside a large sphere with a radius of 4, 24 cubes with edge lengths of 1 have been placed arbitrarily. Prove: At least 4 small spheres with a radius of $\frac{1}{2}$ can still be placed inside the large sphere, such that these small spheres and the cubes are all within the large sphere and do not overl... | 11. To place a small ball with a radius of $\frac{1}{2}$ completely inside the large ball $D(O, 4)$, the distance from its center to the surface of the large ball should be no less than $\frac{1}{2}$, meaning the center of the small ball should be within the ball $D\left(0,4-\frac{1}{2}\right)$. The volume of this ball... | 4 | Geometry | proof | Yes | Yes | cn_contest | false | 723,367 |
One, (40 points) Given an acute triangle $\triangle ABC (AB \neq AC)$, the circle $\odot O$ passing through points $B$ and $C$ intersects sides $AC$ and $AB$ at points $D$ and $E$ respectively, and $P$ is a point on the arc $\overparen{BAC}$ of the circumcircle of $\triangle ABC$. Prove that $BD$, $CE$, and $OP$ are co... | I. Necessity.
As shown in Figure 2, let $B D$, $C E$, and $O P$ intersect at point $F$, and let $O P$ intersect $\odot O$ at point $I$, with the radius of $\odot O$ being $r$. On the ray $O F$, take point $P'$ such that
$$
O F \cdot O P' = r^2.
$$
Then $O F \cdot F P'$
$$
\begin{array}{l}
= O F \left( O P' - O F \righ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,368 |
Given positive real numbers $\alpha, \beta, \gamma$ satisfying $\alpha+\beta+\gamma=1$. Prove: For any positive real numbers $x, y, z$, we have
$$
\begin{array}{l}
\alpha(y+z) \sqrt{\frac{x(7 x+4 y+10 z)}{7}+\frac{(2 y-5 z)^{2}}{49}}+ \\
\beta(z+x) \sqrt{4 x(x+y+z)+(y-z)^{2}}+ \\
\gamma(x+y) \sqrt{\frac{5 x(5 x+2 y+8 z... | II. Proof by Construction.
As shown in Figure 3, construct $\triangle ABC$ such that
$$
\begin{array}{l}
AB = x + y, \\
BC = y + z, \\
CA = z + x.
\end{array}
$$
On the lines $BC$, $CA$, and $AB$, take points $D$, $E$, and $F$ respectively, satisfying
$$
\frac{\overrightarrow{DC}}{\overrightarrow{DC}} = \frac{5}{2}, ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,369 |
Three, (50 points) Try to find the last two non-zero digits of 2011!.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let the last two non-zero digits of a positive integer $k$ be denoted as $f(k)$. Below, we discuss under modulo 100.
Clearly, $5 \times f(k!)$, and if $5 \times f(k), 5 \times f(l)$, then $f(k l) \equiv f(k) f(l)$. Therefore,
$$
\begin{array}{l}
f(2011!) \equiv f\left(\frac{2011!}{2000!}\right) f(2000!) \\
\equiv 68 f(... | 44 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,370 |
Initially, the level markers for internet users are stars, moons, and suns. The rule is: if the cumulative online time from 0:00 to 24:00 in a day is full 2 hours, it counts as 1 day of internet use. Starting from the first day of internet use, after 5 days, the level marker is 1 star; after another 7 days, the level m... | Solving, given that July 1, 2005, is the 1st day of internet usage, we know that January 1, 2011, is the
$$
\begin{array}{l}
185+365 \times 5+1 \text { (2008 is a leap year, with } 366 \text { days) } \\
=2011 \text { days. }
\end{array}
$$
Let the level marker on January 1, 2011, be $n$ stars. We know that $n$ is the... | 2 \text{ suns, 2 moons, and 2 stars; 10 days} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,372 |
As shown in Figure 2, given that $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$ intersect at points $M$ and $N$, $MA$ is a tangent to $\odot O_{2}$ and intersects $\odot O_{1}$ at point $A$, $MB$ is a tangent to $\odot O_{1}$ and intersects $\odot O_{2}$ at point $B$, extend $MN$ to point $P$ such that $MN = NP$, $P... | Prove as shown in Figure 2, connect $A B, N A, N B, M C$.
Since $M B, M A$ are the tangents of $\odot O_{1}, \odot O_{2}$ respectively, therefore,
$$
\begin{array}{l}
\angle M A N=\angle B M N, \angle A M N=\angle M B N \\
\Rightarrow \triangle M A N \backsim \triangle B M N \\
\Rightarrow \frac{M A}{M B}=\frac{M N}{N ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,374 |
For any positive integer $n(n \geqslant 2)$, try to find:
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right]+ \\
\frac{1}{2} \log _{\frac{3}{2}}\left[1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}\right]
\end{array}
$$
the value. | Notice
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right] \\
=\sum_{k=2}^{n} \log _{\frac{3}{2}} \frac{k^{3}+1}{k^{3}-1}=\log _{\frac{3}{2}} \prod_{k=2}^{n} \frac{k^{3}+1}{k^{3}-1} \\
=\log _{\frac{3}{2}} \prod_{k=2}^{n} \frac{(k+1)\left(k^{2}-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,375 |
Given $x_{1} 、 x_{2}$ are the two roots of the quadratic equation $x^{2}-b x+c=0$ with real coefficients. Find $x_{1}^{n}+x_{2}^{n}\left(n \in \mathbf{N}_{+}\right)$. | By Vieta's formulas, we have
$$
x_{1}+x_{2}=b, x_{1} x_{2}=c \text {. }
$$
Then $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=b^{2}-2 c$. Let $x_{1}^{n}+x_{2}^{n}=S_{n}$. Then $S_{n}=b S_{n-1}-c S_{n-2}$. Therefore,
$$
S_{3}=b S_{2}-c S_{1}=b\left(b^{2}-2 c\right)-c b=b^{3}-3 b c \text {. }
$$
Simil... | x_{1}^{n}+x_{2}^{n}=\sum_{i=0}^{\left[\frac{n}{2}\right]}(-1)^{i}\left(\mathrm{C}_{n+1-i}^{i}-\mathrm{C}_{n-1-i}^{i-2}\right) b^{n-2 i} c^{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,376 |
2. In a graph, it is known that the out-degree of each vertex does not exceed 6. Prove: It is possible to color each vertex with one of 13 colors such that no two vertices of the same color are connected by an edge. ${ }^{[s]}$ | Consider an $n$-order tournament graph.
Since $\sum_{v \in V} d^{+}(v)=\sum_{v \in V} d^{-}(v)$, and all $d^{+}(v) \leqslant 6$, thus, $\sum_{v \in V} d^{-}(v) \leqslant 6 n$.
Therefore, there must be a point $A$ with in-degree $d^{-}(A) \leqslant \frac{6 n}{n}=6$. Remove $A$ and the edges with $A$ as a vertex, and us... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,377 |
3. In a certain sports competition, every two players have a match to determine the winner, and through the matches, an outstanding player $A$ is determined. The selection criteria for the outstanding player $A$ is that for any other player $B$, either $A$ beats $B$, or there exists a player $C$ such that $C$ beats $B$... | On the one hand, to prove that there must exist an excellent player.
Let the player with the most wins be $A$. If $A$ is not an excellent player, then there must exist a player $B$, such that $B$ beats $A$, and player $B$ has defeated all the players that player $A$ has defeated. Thus, player $B$ has more wins than pla... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 723,378 |
4. Among 100 types of insects, for every two types, one can eliminate the other. Prove: It is possible to arrange these 100 types of insects in a row such that each type can eliminate the type immediately following it. ${ }^{[s]}$ | Make a graph $G$, where each vertex $v_{i}$ represents a kind of insect. If insect $v_{i}$ can eliminate insect $v_{j}$, then connect $v_{i}$ and $v_{j}$ with a directed edge, thus forming a tournament graph. Therefore, the problem is reduced to proving that a Hamiltonian path must exist in a tournament graph. The proo... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 723,379 |
Example $2 F$ is a focus of the non-degenerate quadratic curve $G$.
(1) $P$ is a point outside the curve $G$, the polar line $l$ of $P$ intersects the curve $G$ at points $T_{1}$ and $T_{2}$, $T_{1} T_{2}$ intersects the directrix $l_{1}$ of the curve $G$ at point $T$, then $P F$ bisects $\angle T_{1} F T_{2}$, and $T ... | Proof (1) Let the projections of points $T_{1}$ and $T_{2}$ on the directrix $l_{1}$ be $H_{1}$ and $H_{2}$, respectively, and let $P F$ intersect $T_{1} T_{2}$ at point $S$.
(i) If $T_{1} T_{2} \parallel l_{1}$, then $\triangle F T_{1} T_{2}$ is an isosceles triangle, and $F S$ is the altitude on the base $T_{1} T_{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,380 |
Example 3 As shown in Figure 7, let $D$ be a point on side $BC$ of $\triangle ABC$, satisfying $\angle CAD = \angle CBA$. $\odot O$ passes through $B, D$, and intersects line segments $AB, AD$ at points $E, F$ respectively. $BF$ intersects $DE$ at point $G$, and $M$ is the midpoint of $AG$. Prove: $CM \perp AO{ }^{[6]}... | 【Analysis】Connect $E F$ and extend it to intersect $B C$ at point $P$, connect $G P$ with $A D$ intersecting at point $K$, and intersecting the extension of $A C$ at point $L$.
Consider the complete quadrilateral BDPFAE and $\odot O$.
By Proposition 2, it can be proven that line $G P$ is the polar of point $A$, i.e., t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,381 |
Example 3 A right-angled triangle with integer side lengths, if the lengths of its two legs are the roots of the equation
$$
x^{2}-(k+2) x+4 k=0
$$
find the value of $k$ and determine the lengths of the three sides of the right-angled triangle. ${ }^{[2]}$
(2010, National Junior High School Mathematics League, Jiangxi... | 【Analysis】From the known conditions, first convert the relationship between the roots and coefficients of a quadratic equation into the difference of squares formula, then solve by factorization.
Solution Let the lengths of the legs be $a, b(a \leqslant b)$. Then
$$
\begin{array}{l}
a+b=k+2, a b=4 k \\
\Rightarrow a b-... | (a, b)=(5,12) \text{ or } (6,8) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,382 |
Example 4: For a central conic section $G$, the axis of symmetry intersects the curve $G$ at points $B$ and $C$. The sides $AB$ and $AC$ of $\triangle ABC$ intersect the curve $G$ at points $E$ and $F$, respectively. The tangents at points $E$ and $F$ intersect at point $D$. Prove that $AD \perp BC$. | Prove that with the center of curve $G$ as the origin and $BC$ as the $x$-axis, establish a Cartesian coordinate system.
Let $B(-1,0)$ and $C(1,0)$. Then the equation of curve $G$ is
$$
x^{2}+B y^{2}=1(B \neq 0).
$$
Let $l_{A B}: x=t_{1} y-1$,
$$
l_{A C}: x=t_{2} y+1.
$$
(2) $\times t_{2} - (3) \times t_{1}$ gives $x... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,383 |
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