problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Given lines $l_{1}$ and $l_{2}$ and a fixed point $A$ outside the lines, find points $B$ and $C$ on $l_{1}$ and $l_{2}$ respectively, such that $\triangle ABC$ is an equilateral triangle. | As shown in Figure 3, if $l_{1}$ intersects $l_{2}$ at point $E$, draw $AD \parallel l_{1}$ through point $A$, construct a regular $\triangle DEF$, connect $AF$, which intersects line $l_{1}$ at point $B$, and draw $\angle BAC = 60^{\circ}$, where point $C$ is on line $l_{2}$. Then $\triangle ABC$ is a regular triangle... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,682 |
1. Given $\sqrt{x}+\frac{1}{\sqrt{x}}=3$. Then the value of $\frac{x}{x^{2}+2018 x+1}$ is ( ).
(A)2020
(B) $\frac{1}{2020}$
(C) 2025
(D) $\frac{1}{2025}$ | - 1. D.
Squaring both sides of the given equation and rearranging yields $x+\frac{1}{x}=7$.
Therefore, $\frac{x}{x^{2}+2018 x+1}=\frac{1}{7+2018}=\frac{1}{2025}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,683 |
2. In a non-isosceles triangle, one interior angle is equal to the difference of the other two interior angles, and one interior angle is twice another interior angle. If the shortest side of the triangle is 1, then the area of the triangle is ( ).
(A) 1
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{\sqrt{5}}{2}$
(D) 2 | 2. B.
Let the interior angles of the triangle be $\angle A, \angle B, \angle C$, and $\angle B=\angle C-\angle A$. Then
$$
\begin{array}{l}
\angle A+(\angle C-\angle A)+\angle C=180^{\circ} \\
\Rightarrow \angle C=90^{\circ} .
\end{array}
$$
If the largest angle $\angle C$ is twice another interior angle, it is easy ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,684 |
3. As shown in Figure 1, let $P\left(x_{p}, y_{p}\right)$ be a point on the graph of the inverse proportion function $y=\frac{2}{x}$ in the first quadrant of the Cartesian coordinate system $x O y$. Draw lines parallel to the $x$-axis and $y$-axis through point $P$, intersecting the graph of $y=\frac{10}{x}$ in the fir... | 3. B.
Connect $O P$.
Then $S_{\triangle A O B}=S_{\triangle A O P}+S_{\triangle P O B}+S_{\triangle A P B}$
$$
\begin{aligned}
= & \frac{1}{2}\left(\frac{10}{y_{p}}-x_{p}\right) y_{p}+\frac{1}{2}\left(\frac{10}{x_{p}}-y_{p}\right) x_{p}+ \\
& \frac{1}{2}\left(\frac{10}{y_{p}}-x_{p}\right)\left(\frac{10}{x_{p}}-y_{p}\r... | 24 | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,685 |
5. Arrange natural numbers whose digits sum to 11 in ascending order to form a sequence. The $m$-th number is 2018. Then $m$ is ( ).
(A) 134
(B) 143
(C) 341
(D) 413 | 5. A.
Among single-digit numbers, there are no numbers whose digit sum is 11.
Among two-digit numbers, there are 8 numbers: $29, 38, 47, 56, 65, 74, 83, 92$.
For three-digit numbers $\overline{x y z}$, when $x=1$, $y$ can take 9 numbers: $1, 2, \cdots, 9$, and the corresponding $z$ takes $9, 8, \cdots, 1$, a total of... | 134 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,686 |
Example 3 As shown in Figure 6, in $\square A B C D$, $O_{1} 、 O_{2} 、 O_{3}$ are the circumcenters of $\triangle A B E 、 \triangle A D E 、 \triangle C D E$, and $H$ is the orthocenter of $\triangle O_{1} O_{2} O_{3}$. Prove: Point $H$ lies on $A D$. | Proof from the circumcenter we know
$$
\begin{array}{l}
\angle O_{2} O_{1} E=\angle A B E=180^{\circ}-\angle E C D \\
=180^{\circ}-\angle O_{2} O_{3} E
\end{array}
$$
$\Rightarrow O_{1} 、 O_{2} 、 O_{3} 、 E$ are concyclic.
Obviously, $A 、 E$ are symmetric with respect to $O_{2} O_{1}$, and $D 、 E$ are symmetric with res... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,687 |
3. Given that $D$ is a point inside the acute-angled $\triangle A B C$, satisfying $A D=D C, \angle A D C$ $=2 \angle D B C, A B=$ $12, B C=10$. As shown in Figure 2. Then the area of $\triangle B D C$ is $\qquad$ | 3. $5 \sqrt{11}$.
Let $\angle D B C=\theta$. Then $\angle A D C=2 \theta$.
With $D$ as the center of rotation, rotate $\triangle B D C$ to the position of $\triangle A D P$, as shown in Figure 4.
Then $A D=D C, D P=D B, A P=C B=10$, $\angle D P A=\angle D B C=\theta, \angle P D A=\angle B D C$, i.e., $\angle P D C+\an... | 5 \sqrt{11} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,688 |
4. Given that $x_{1}, x_{2}, \cdots, x_{n}$ where $x_{i}(i=1,2, \cdots, n)$ can only take one of the values $-2, 0, 1$, and satisfy
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=-17, \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=37 .
\end{array}
$$
Then $\left(x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}\right)^{2}$ is $\qquad$ | 4. 5041.
Let $x_{1}, x_{2}, \cdots, x_{n}$ have $p$ values of $x_{i}$ equal to 1, $q$ values of $x_{i}$ equal to -2, and the rest of the $x_{i}$ equal to 0. We can obtain
$$
\left\{\begin{array} { l }
{ p - 2 q = - 1 7 } \\
{ p + 4 q = 3 7 }
\end{array} \Rightarrow \left\{\begin{array}{l}
p=1, \\
q=9 .
\end{array}\ri... | 5041 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,689 |
5. Among the $n$ positive integers from 1 to $n$, those with the most positive divisors are called the "prosperous numbers" among these $n$ positive integers. For example, among the positive integers from 1 to 20, the numbers with the most positive divisors are $12, 18, 20$, so $12, 18, 20$ are all prosperous numbers a... | 5. 10080.
First, in the prime factorization of the first 100 positive integers, the maximum number of different prime factors is three. This is because the product of the smallest four primes is $2 \times 3 \times 5 \times 7=210$, which exceeds 100.
Second, to maximize the number of divisors, the prime factors should... | 10080 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,690 |
Three, (10 points) Let positive integers $a, b, c, d$ satisfy $a^{2}-a b+b^{2}=c^{2}-c d+d^{2}$. Prove: $a+b+c+d$ is a composite number.
保留源文本的换行和格式，翻译结果如下：
Three, (10 points) Let positive integers $a, b, c, d$ satisfy $a^{2}-a b+b^{2}=c^{2}-c d+d^{2}$. Prove: $a+b+c+d$ is a composite number. | Let $s=a+b+c+d$.
From the given information, we have
$$
\begin{array}{l}
(a+b)^{2}-(c+d)^{2}=3(a b-c d) \\
\Rightarrow(a+b-c-d) s=3(a b-c d) \\
\quad=3(a b-c(s-a-b-c)) \\
\quad=3(a+c)(b+c)-3 c s \\
\Rightarrow s \mid 3(a+c)(b+c) .
\end{array}
$$
Then each prime factor $p$ of $s$ is a factor of $3(a+c)(b+c)$.
Therefore... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,691 |
Four. (15 points) As shown in Figure 3, there are three isosceles right triangles $\triangle A D C$, $\triangle D P E$, and $\triangle B E C$ with unequal hypotenuses, where,
$$
\begin{array}{c}
A D=C D, \\
D P=E P, \\
B E=C E, \\
\angle A D C \\
=\angle D P E \\
=\angle B E C=90^{\circ} .
\end{array}
$$
Prove: $P$ is... | (1) As shown in Figure 5, extend $DP$ to point $F$ such that $PF = PD$, and connect $FE$, $FB$.
It is easy to see that $DE = EF$, $\angle DEF = 90^\circ$.
Thus, $\angle CED = 90^\circ - \angle CEF = \angle BEF$.
Also, $DE = EF$, $CE = BE$, hence,
$\triangle CED \cong \triangle BEF$.
Therefore, $CD = BF$, $\angle CDE = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,692 |
2. A, B, and C are the recently elected candidates for the student council president. Figure 1 shows the voting results for the three candidates. If B received 36 votes, then a total of $(\quad)$ votes were cast.
(A) 70
(B) 84
(C) 100
(D) 106
(E) 120 | 2. E.
$$
\frac{36}{30 \%}=120 \text {. }
$$ | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,693 |
3. The value of the radical $\sqrt{16 \sqrt{8 \sqrt{4}}}$ is $(\quad)$.
(A) 4
(B) $4 \sqrt{2}$
(C) 8
(D) $8 \sqrt{2}$
(E) 16 | 3. C.
$$
\sqrt{16 \sqrt{8 \sqrt{4}}}=\sqrt{16 \sqrt{8 \times 2}}=\sqrt{16 \times 4}=8
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,694 |
4.0.000315 $\times 7928564$ The approximate value is ( ).
(A) 210
(B) 240
(C) 2100
(D) 2400
(E) 24000 | 4. D.
$$
\begin{array}{l}
0.000315 \times 7928564 \\
\approx 0.0003 \times 8000000=2400
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,695 |
Example 4 Let quadrilateral $ABCD$ have an inscribed circle, and a line $l$ through point $A$ intersects segments $BC, DC$ at points $M, N$ respectively. $I, J, K$ are the incenters of $\triangle ABM, \triangle MNC, \triangle NDA$ respectively. Prove: The orthocenter of $\triangle IJK$ lies on line $l$.
untranslated ... | Prove as shown in Figure 7, connect auxiliary lines. Let $CT$ be tangent to $\odot I$ at point $T$, and intersect $AM$ at point $G$.
By the problem statement,
$$
\begin{array}{l}
DA - DC = AB - BC = AX - CY \\
= AZ - CT = AG - CG
\end{array}
$$
$\Rightarrow CT$ is tangent to $\odot K$
$$
\begin{aligned}
\Rightarrow & \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,696 |
5. $\frac{1 \times 2 \times \cdots \times 8}{1+2+\cdots+8}$ 的值为 $(\quad)$.
(A) 1020
(B) 1120
( C) 1220
(D) 2400
(E) 3360
5. The value of $\frac{1 \times 2 \times \cdots \times 8}{1+2+\cdots+8}$ is $(\quad)$.
(A) 1020
(B) 1120
( C) 1220
(D) 2400
(E) 3360 | 5. B. Original expression $=\frac{40320}{36}=1120$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,697 |
10. There are five cards in a box, marked with $1$, $2$, $3$, $4$, $5$ respectively. Three cards are drawn without replacement, and the probability that the largest one is 4 is ( ).
(A) $\frac{1}{10}$
(B) $\frac{1}{5}$
(C) $\frac{3}{10}$
(D) $\frac{2}{5}$
(E) $\frac{1}{2}$ | 10. C.
$$
P(A)=\frac{\mathrm{C}_{3}^{2}}{\mathrm{C}_{5}^{3}}=\frac{3}{10} .
$$ | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 730,698 |
14. One evening, students Chloe and Zoe first independently solved half of the problems in their homework, and then worked together to solve the other half of the problems. Chloe's correct rate for solving problems independently is $80 \%$, and her overall correct rate is $88 \%$. It is known that Zoe's correct rate fo... | 14. C.
Let there be $x$ questions in total, and the correct rate of the two people working together is $y$. According to the problem, we have
$$
\begin{array}{l}
\frac{\frac{1}{2} x \times 80 \%+\frac{1}{2} x y}{x} \times 100 \%=88 \% \\
\Rightarrow y=96 \% .
\end{array}
$$
Therefore, Zoe's total correct rate for sol... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,699 |
Example 5 Given that line $l$ intersects the sides $BC$, $CA$, $AB$ of $\triangle ABC$ at points $D$, $E$, $F$, and $G$, $H$, $I$ are the circumcenters of $\triangle AEF$, $\triangle BDF$, $\triangle CDE$ respectively. Prove: The circumcenter of $\triangle GHI$ lies on line $l$.
untranslated text remains unchanged. | Proof As shown in Figure 8, let $\odot H$ and $\odot I$ intersect at point $M$.
Then $\angle C E M=\angle C D M$
$\Rightarrow C 、 E 、 D 、 M$ are concyclic
$$
\begin{aligned}
\Rightarrow & \angle G I M=\angle E C M=\angle F D M \\
& =180^{\circ}-\angle G H M
\end{aligned}
$$
$\Rightarrow G 、 H 、 M 、 I$ are concyclic.
Th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,700 |
15. In the figure 2 composed of numbers and letters, there are ( ) different paths to spell out AMC8. Requirement: Each path starts from the letter A in the middle of the figure, and allows passing through adjacent letters or numbers once (can move up, down, left, or right, but not diagonally). For example, figure 2 ma... | 15. D.
By the multiplication principle, we know that the desired result is
$$
1 \times 4 \times 3 \times 2=24 \text {. }
$$ | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 730,701 |
17. Initially, there were some gold coins and empty boxes to hold the gold coins. If each box is filled with 9 gold coins, there will be two empty boxes left; if each box is filled with 6 gold coins, there will be 3 gold coins left. Therefore, there are ( ) gold coins.
(A) 9
(B) 27
(C) 45
(D) 63
(E) 81 | 17. C.
Let there be $x$ boxes in total. According to the problem, $9 x-18=6 x+3 \Rightarrow x=7$. Therefore, the number of gold coins is $9 \times 7-18=45$. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,702 |
19. For any positive integer $M$, let
$$
M!=M \cdot(M-1) \cdots \cdot 2 \cdot 1 \text {, }
$$
$x^{n} \| y$ denotes $x^{n} \mid y$, and $x^{n+1} \nmid y\left(x, y \in \mathbf{Z}_{+}\right)$.
Then the largest integer $n$ such that $5^{n} \|(98!+99!+100!)$ is ( ).
(A) 23
(B) 24
(C) 25
(D) 26
(E) 27 | 19. D.
Notice,
$$
\begin{array}{l}
98!+99!+100!=98!(1+99+100 \times 99) \\
=98! \times 2^{4} \times 5^{4}.
\end{array}
$$
Let $[x]$ denote the greatest integer not exceeding the real number $x$.
Also, $\left[\frac{98}{5}\right]+\left[\frac{98}{5^{2}}\right]=19+3=22$, so the largest integer $n=22+4=26$. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,703 |
20. From 1000 to 9999, the probability of randomly selecting a four-digit number where all digits are different and the unit digit is odd is $(\quad)$.
(A) $\frac{14}{75}$
(B) $\frac{56}{225}$
(C) $\frac{107}{400}$
(D) $\frac{7}{25}$
(E) $\frac{9}{25}$ | 20. B.
From 1000 to 9999, there are 9000 ways to choose a four-digit number. Since the four-digit number to be chosen is odd, the unit digit must be chosen from 1, 3, 5, 7, 9, which gives 5 ways; the first digit must be chosen from $1,2, \cdots, 9$ excluding the unit digit, leaving 8 choices; the hundredth and tenth d... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 730,704 |
23. In Linda's four-day trip, she has to walk the same distance every day. On the first day, she spent one hour walking. It is known that she spends an integer number of minutes per mile, and after that, due to fatigue, her speed decreases each day, spending 5 more minutes per mile than the previous day, and the total ... | 23. C.
Let Linda use $x\left(x \in \mathbf{Z}_{+}\right)$ minutes per mile on the first day. Then the distance walked on the first day is $\frac{60}{x}$, the distance walked on the second day is $\frac{60}{x+5}$, the distance walked on the third day is $\frac{60}{x+10}$, and the distance walked on the fourth day is $\... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,705 |
2. Given the tetrahedron $P-ABC$ with the base $\triangle ABC$ being an equilateral triangle with side length 3, $PA=3, PB=4, PC=5$. Then the volume of the tetrahedron $P-ABC$ is ( ).
(A) 3
(B) $\sqrt{10}$
(C) $\sqrt{11}$
(D) $2 \sqrt{3}$ | 2. C.
$$
\begin{array}{l}
V_{\text {tetrahedron } P-A B C}=V_{\text {tetrahedron } A-P B C} \\
=\frac{1}{3}\left(\frac{1}{2} \times 3 \times 4\right) \times \sqrt{3^{2}-\left(\frac{5}{2}\right)^{2}}=\sqrt{11} .
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,706 |
3. Given the function $f(x)$ satisfies:
$$
\begin{array}{l}
f(1)=\frac{1}{4}, \\
4 f(x) f(y)=f(x+y)+f(x-y)(x, y \in \mathbf{R}) .
\end{array}
$$
Then $f(2019)=$ ( ).
(A) $\frac{1}{2}$
(B) $-\frac{1}{2}$
(C) $\frac{1}{4}$
(D) $-\frac{1}{4}$ | 3. B.
Let $x=1, y=0$, we get $f(0)=\frac{1}{2}$.
Let $x=1, y=1$, we get $4 f^{2}(1)=f(2)+f(0)$.
Thus, $f(2)=-\frac{1}{4}$.
Let $x=2, y=1$, we get $4 f(1) f(2)=f(3)+f(1)$.
Thus, $f(3)=-\frac{1}{2}$.
Let $x=n, y=1$, we get $f(n)=f(n+1)+f(n-1)$.
Similarly, $f(n+1)=f(n+2)+f(n)$.
Combining these, we get $f(n+2)=-f(n-1)$.
T... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,707 |
6. Let $x, y, z > 0$, satisfying $x + y = xy$ and $x + y + z = xyz$. Then the range of $z$ is ( ).
(A) $(0, \sqrt{3}]$
(B) $(1, \sqrt{3}]$
(C) $\left(0, \frac{4}{3}\right]$
(D) $\left(1, \frac{4}{3}\right]$ | 6. D.
Given $y=\frac{x}{x-1}$, we have
$$
\begin{aligned}
z & =\frac{x+y}{x y-1}=\frac{x y}{x y-1}=\left(1-\frac{1}{x y}\right)^{-1} \\
& =\left(1-\frac{1}{x}+\frac{1}{x^{2}}\right)^{-1} .
\end{aligned}
$$
Since $x>1$, i.e., $0<\frac{1}{x}<1$, we have
$$
\frac{3}{4} \leqslant 1-\frac{1}{x}+\frac{1}{x^{2}}<1 \text {. ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,708 |
8. Given the equation of circle $\odot C$ is $x^{2}+y^{2}-8 x+15=0$. If there exists at least one point on the line $y=k x-2(k \in \mathbf{R})$, such that a circle with this point as its center and 1 as its radius has a common point with $\odot C$, then the maximum value of $k$ is . $\qquad$ | 8. $\frac{4}{3}$.
From the problem, we know the center of the circle $C(4,0)$, with a radius of 1.
If the line $y=k x-2$ has at least one point $A\left(x_{0}, k x_{0}-2\right)$, such that the circle $\odot A$ with radius 1 intersects with $\odot C$, i.e., $|A C| \leqslant 1+1=2$, then
$$
|A C|_{\min } \leqslant 2 \tex... | \frac{4}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,709 |
10. Given point $P$ is on the line $x+2 y-1=0$, point $Q$ is on the line $x+2 y+3=0$, the midpoint of $P Q$ is $M\left(x_{0}, y_{0}\right)$, and $y_{0}>x_{0}+2$. Then the range of $\frac{y_{0}}{x_{0}}$ is | 10. $\left(-\frac{1}{2},-\frac{1}{5}\right)$.
Notice that, the two lines are parallel.
Therefore, the locus of point $M$ is a line parallel to the two lines and equidistant from them, with the equation $x+2 y+1=0$, i.e., $M\left(x_{0}, y_{0}\right)$ satisfies $x_{0}+2 y_{0}+1=0$.
The points satisfying the inequality ... | \left(-\frac{1}{2},-\frac{1}{5}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,710 |
11. Given real numbers $a, b$ satisfy
$$
\left\{\begin{array}{l}
a+b-2 \geqslant 0, \\
b-a-1 \leqslant 0, \\
a \leqslant 1 .
\end{array}\right.
$$
Then the maximum value of $\frac{a+2 b}{2 a+b}$ is $\qquad$ | 11. $\frac{7}{5}$.
Draw the feasible region represented by the constraint conditions, as shown in Figure 2, which is the interior and boundary of $\triangle ABC$, where $A\left(\frac{1}{2}, \frac{3}{2}\right)$, $B(1,1)$, and $C(1,2)$.
Let $P(a, b)$ be any point in the feasible region.
Then $k_{O P}=\frac{b}{a}$ has a ... | \frac{7}{5} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,711 |
12. In the sequence $\left\{a_{n}\right\}$, if $a_{n}^{2}-a_{n-1}^{2}=p(n \geqslant 2, n \in \mathbf{N}, p$ is a constant $)$, then $\left\{a_{n}\right\}$ is called a "sequence of equal variances". The following are judgments on sequences of equal variances:
(1) The sequence $\left\{(-1)^{n}\right\}$ is a sequence of e... | 12. (1)(2)(3)(4).
(1) Since $\left((-1)^{n}\right)^{2}-\left((-1)^{n-1}\right)^{2}=0$, therefore, $\left\{(-1)^{n}\right\}$ satisfies the definition.
(2) According to the definition, it is obvious that $\left\{a_{n}^{2}\right\}$ is an arithmetic sequence.
$$
\text { (3) } a_{k n}^{2}-a_{k(n-1)}^{2}=\sum_{i=1}^{k}\left(... | (1)(2)(3)(4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,712 |
15. (25 points) As shown in Figure 1, the parabola $y=a x^{2}$ passes through the point $P(-1,1)$. A line $l$ with a positive slope is drawn through the point $Q\left(-\frac{1}{2}, 0\right)$, intersecting the parabola at points $M$ and $N$ (point $M$ is between $Q$ and $N$). A horizontal line is drawn through $M$, inte... | 15. Given that the parabola $y=a x^{2}$ passes through the point $P(-1,1)$, we get $a=1$.
Thus, the equation of the parabola is $y=x^{2}$.
Let the line $l: y=k\left(x+\frac{1}{2}\right)(k>0)$.
By combining the equation of line $l$ with the parabola, we get
$$
2 x^{2}-2 k x-k=0 \text {. }
$$
Let $M\left(x_{1}, y_{1}\ri... | S_{1}>3 S_{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,713 |
1. Divide the set of positive even numbers $\{2,4, \cdots\}$ into groups in ascending order, with the $n$-th group containing $3 n-2$ numbers:
$$
\{2\},\{4,6,8,10\},\{12,14, \cdots, 24\}, \cdots \text {. }
$$
Then 2018 is in the group. | - 1. 27 .
Let 2018 be in the $n$-th group. Since 2018 is the 1009th positive even number and according to the problem, we have
$$
\begin{array}{l}
\sum_{i=1}^{n-1}(3 i-2)<1009 \leqslant \sum_{i=1}^{n}(3 i-2) \\
\Rightarrow \frac{3(n-1)^{2}-(n-1)}{2}<1009 \leqslant \frac{3 n^{2}-n}{2} \\
\Rightarrow n=27 .
\end{array}
... | 27 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,714 |
2. In $\triangle A B C$, the internal angles $\angle A, \angle B, \angle C$ are opposite to the sides $a, b, c$ respectively. If $a=2, b=3, \angle C=$ $2 \angle A$, then $\cos C=$ $\qquad$ | 2. $\frac{1}{4}$.
From the problem, we know
$$
\begin{array}{l}
\sin C=\sin 2 A=2 \sin A \cdot \cos A \\
\Rightarrow c=2 a \cos A=2 a \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c} \\
\quad=4 \times \frac{9+c^{2}-4}{6 c} \\
\Rightarrow c^{2}=10 \\
\Rightarrow \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{1}{4} .
\end{array}
$$ | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,715 |
8. Recently, an article on the internet has gone viral. It originates from a common question, as shown in Figure 1. This seemingly easy-to-solve problem actually contains profound wisdom.
Let $a, b \in\{2,3, \cdots, 8\}$. Then the maximum value of $\frac{a}{10 b+a}+\frac{b}{10 a+b}$ is $\qquad$ | 8. $\frac{89}{287}$.
Assume $a \geqslant b, x=\frac{a}{b}$. Then $1 \leqslant x \leqslant 4$,
and $\square$
$$
\begin{array}{l}
\frac{a}{10 b+a}+\frac{b}{10 a+b}=\frac{x}{10+x}+\frac{1}{10 x+1} \\
=\frac{10 x^{2}+2 x+10}{10 x^{2}+101 x+10} \\
=1-\frac{99 x}{10 x^{2}+101 x+10} \\
=1-\frac{99}{10\left(x+\frac{1}{x}\rig... | \frac{89}{287} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,716 |
10. Given the function $f(x)$ satisfies for any real numbers $x, y$,
$$
f(x+y)=f(x)+f(y)+6xy,
$$
and $f(-1) f(1) \geqslant 9$.
Then $f\left(\frac{2}{3}\right)=$ . $\qquad$ | 10. $\frac{4}{3}$.
In the given expression, let $x=y=0$, we get $f(0)=0$. Let $x=-1, y=1$, we get $f(1)+f(-1)=6$. Also, $f(-1) f(1) \geqslant 9=\left(\frac{f(1)+f(-1)}{2}\right)^{2}$, then
$$
\begin{array}{l}
f(1)=f(-1)=3 . \\
\text { Also } f\left(\frac{2}{3}\right)=f\left(\frac{1}{3}+\frac{1}{3}\right)=2 f\left(\fra... | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,717 |
11. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms $S_{n}$ satisfies $2 S_{n}-n a_{n}=n\left(n \in \mathbf{Z}_{+}\right)$, and $a_{2}=3$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{1}{a_{n} \sqrt{a_{n+1}}+a_{n+1} \sqrt{a_{n}}}$, and $T_{n}$ ... | (1) From $2 S_{n}-n a_{n}=n$, we get
$$
2 S_{n+1}-(n+1) a_{n+1}=n+1 \text {. }
$$
Subtracting the above two equations yields
$$
2 a_{n+1}-(n+1) a_{n+1}+n a_{n}=1 \text {. }
$$
Thus, $n a_{n}-(n-1) a_{n+1}=1$,
$$
(n+1) a_{n+1}-n a_{n+2}=1 \text {. }
$$
Subtracting (2) from (1) and rearranging gives
$$
a_{n}+a_{n+2}=2 ... | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,718 |
15. Let $M$ be a set composed of a finite number of positive integers
$$
\begin{array}{l}
\text { such that, } M=\bigcup_{i=1}^{20} A_{i}=\bigcup_{i=1}^{20} B_{i}, \\
A_{i} \neq \varnothing, B_{i} \neq \varnothing(i=1,2, \cdots, 20),
\end{array}
$$
and satisfies:
(1) For any $1 \leqslant i<j \leqslant 20$,
$$
A_{i} \... | 15. Let $\min _{1 \leqslant i \leqslant 20}\left\{\left|A_{i}\right|,\left|B_{i}\right|\right\}=t$.
Assume $\left|A_{1}\right|=t$,
$$
\begin{array}{l}
A_{1} \cap B_{i} \neq \varnothing(i=1,2, \cdots, k) ; \\
A_{1} \cap B_{j}=\varnothing(j=k+1, k+2, \cdots, 20) .
\end{array}
$$
Let $a_{i} \in A_{1} \cap B_{i}(i=1,2, \... | 180 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,719 |
1. In a certain sports association, each team can choose at most $t$ distinctive colors for their team uniforms. For a set $S$ of teams, if each team in $S$ can be assigned one of their distinctive colors such that no team in $S$ is assigned any distinctive color of another team in $S$, then the set $S$ is called "colo... | 1. The maximum value of $g(n, t)$ is $\left\lceil\frac{n}{t}\right\rceil$, where $\lceil x\rceil$ denotes the smallest integer not less than the real number $x$.
Assume the set $S$ is recognizable by colors, and $|S|$ is minimized. If each color is a distinctive color of a team, and each team in $S$ has $t$ distinctiv... | \left\lceil\frac{n}{t}\right\rceil | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,720 |
2. Given a scalene acute triangle $\triangle ABC$ with circumcircle $\odot O$, $T$ is a point on line $BC$ such that $\angle TAO = 90^{\circ}$. The circle with diameter $AT$ intersects the circumcircle of $\triangle BOC$ at points $A_1$ and $A_2$, where $OA_1 < OA_2$. Similarly, define points $B_1, B_2, C_1, C_2$. Prov... | 2. Draw the tangents to $\odot O$ through points $A, B, C$, intersecting to form $\triangle X Y Z$, as shown in Figure 1.
Then $\odot O$ is the incircle of $\triangle X Y Z$.
Let the circumcircle of $\triangle X Y Z$ be $\Gamma$, and the symmedian of $\triangle A B C$ through $A$ intersect $\odot O$ again at $D$, with ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,721 |
3. Given non-constant real-coefficient polynomials $P(x)$ and $Q(x)$ that are coprime. Prove: There exist at most three real numbers $\lambda$ such that $P+\lambda Q$ is a square of a polynomial. | 3. This is an algebraic geometry problem. The conclusion also holds if $\mathbf{R}$ is replaced by $\mathbf{C}$.
Now, let's generalize the problem.
Given non-constant complex-coefficient polynomials $P(x)$ and $Q(x)$ that are coprime. If there exist four distinct ratios $\alpha: \beta$, such that $\alpha P + \beta Q$ i... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,722 |
4. Participant A cheats in a Q&A competition, where for each question, A can peek at the guesses of $n(n>1)$ other participants before writing down their own answer. For each question, after all guesses are submitted, the host announces the correct answer. A correct guess scores 0 points; an incorrect guess scores -2 p... | 4. For simplicity, use $+1, -1, 0$ to replace $0, -2, -1$ respectively, which clearly does not affect the conclusion.
Without loss of generality, assume that at the beginning, all participants except for A have 0 points, and A's score exceeds $2^{n-2}$. We do not consider rounds where all answers are the same, nor do ... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 730,723 |
5. Given that $A E$ is the altitude of $\triangle A B C$, the excircle $\odot J$ of $\angle A$ touches $B C$ at point $D$, and intersects the circumcircle $\odot O$ of $\triangle A B C$ at points $F$ and $G$. Prove: There exist points $V$ and $N$ on lines $D G$ and $D F$ respectively, such that quadrilateral $E V A N$ ... | 5. As shown in Figure 2, let the incenter of $\triangle ABC$ be $I$, and the midpoint of $AE$ be $L$. Draw a tangent line to $\odot J$ through point $I$, with the points of tangency being $Y$ and $Z$.
Then $B, I, C, Y, J, Z$ are concyclic, denoted as $\odot O_{1}$.
For $\odot O_{1}, \odot O, \odot J$, by Monge's theore... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,724 |
6. Prove: There exist infinitely many integer triples $(a, b, p)$ such that $p$ is a prime, $0<a \leqslant b<p$, and
$$
p^{5} \text { divides }\left((a+b)^{p}-a^{p}-b^{p}\right) .
$$ | 6. Just prove: If $p \equiv 1(\bmod 3)$, then
$$
p\left(x^{2}+x y+y^{2}\right)^{2} \mid \left((x+y)^{p}-x^{p}-y^{p}\right) \text {, }
$$
where $p\left(x^{2}+x y+y^{2}\right)$ and $(x+y)^{p}-x^{p}-y^{p}$ are quadratic polynomials in $x, y$.
By a well-known conclusion, there exist positive integers $a, b$, such that
$$
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,725 |
1. In $\triangle A B C$, $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ respectively, satisfying $a^{2}+b^{2}=4-\cos ^{2} C, a b=2$. Then $S_{\triangle A B C}=$ $\qquad$ | $-1.1$
From the problem, we have $(a-b)^{2}+\cos ^{2} C=0$.
Solving, we get $a=b=\sqrt{2}, \cos C=0$.
Therefore, $S_{\triangle A B C}=1$. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,726 |
2. A crystal sphere is cut into a decahedron as shown in Figure 1, with all edges being 2 units long, and the top face $ABCD$ and the bottom face $EFGH$ being parallel squares. Then the minimum surface area of the crystal sphere is $\qquad$ .
Translate the above text into English, please retain the original text's lin... | 2. $(2 \sqrt{2}+8) \pi$.
To minimize the surface area of the crystal ball, the ball should be the circumsphere of the decahedron. According to the conditions and symmetry, the projection of point $A$ on the bottom face $E F G H$ lies on the small circle of the sphere cut by the plane $E F G H$, and it is the midpoint ... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 730,727 |
3. Given $x, y>0$. If
$$
f(x, y)=\left(x^{2}+y^{2}+2\right)\left(\frac{1}{x+y}+\frac{1}{x y+1}\right) \text {, }
$$
then the minimum value of $f(x, y)$ is | 3.4.
By completing the square, we get
$$
x^{2}+y^{2}+2 \geqslant(x+y)+(x y+1) \text {. }
$$
Then $f(x, y)$
$$
\begin{array}{l}
\geqslant((x+y)+(x y+1))\left(\frac{1}{x+y}+\frac{1}{x y+1}\right) \\
\geqslant(1+1)^{2}=4 .
\end{array}
$$
Equality holds if and only if $x=y=1$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,728 |
5. Given that $A$ and $B$ are the left and right vertices of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, $F_{1}$ and $F_{2}$ are the left and right foci respectively, and $C$ is a point on the hyperbola different from $A$ and $B$. If the lines $A C$ and $B C$ intersect the right directrix of the hyperbol... | 5. $-2 b^{2}$.
By symmetry, without loss of generality, assume point $C\left(x_{C}, y_{C}\right)$ is in the first quadrant.
Thus, $M\left(\frac{a^{2}}{c}, y_{M}\right), N\left(\frac{a^{2}}{c}, y_{N}\right)$.
From the collinearity of three points, we get
$$
\begin{array}{l}
y_{M}=\frac{\frac{a^{2}}{c}+a}{x_{C}+a} y_{C}... | -2 b^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,729 |
6. In the complex plane, the complex numbers $z_{1}, z_{2}, z_{3}$ satisfy:
$$
\begin{array}{l}
z_{1}=\lambda z_{2}(\lambda<0),\left|z_{1}-z_{2}\right|=13, \\
\left|z_{1}\right|^{2}+\left|z_{3}\right|^{2}+\left|z_{1} z_{3}\right|=144, \\
\left|z_{2}\right|^{2}+\left|z_{3}\right|^{2}-\left|z_{2} z_{3}\right|=25 .
\end{a... | 6. $\frac{40 \sqrt{3}}{13}$.
From the problem, we can construct a right triangle $\triangle ABC$ such that
$$
|OA| = |z_1|, |OB| = |z_2|, |OC| = |z_3|.
$$
Then, $\angle ACB = 90^\circ$, $\angle AOC = 120^\circ$,
$$
\angle BOC = 60^\circ, |BC| = 5, |AC| = 12.
$$
Thus, $|OA| + |OB| = 13$.
By $S_{\triangle ABC} = S_{\t... | \frac{40 \sqrt{3}}{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,730 |
7. For any $n$-element set $S_{n}$, if its subsets $A_{1}$, $A_{2}, \cdots, A_{k}$ satisfy $\bigcup_{i=1}^{k} A_{i}=S_{n}$, then the unordered set group $\left(A_{1}, A_{2}, \cdots, A_{k}\right)$ is called a “$k$-stage partition” of set $S_{n}$. Therefore, the number of 2-stage partitions of $S_{n}$ is $\qquad$. | 7. $\frac{1}{2}\left(3^{n}+1\right)$.
Let $F_{n}$ be the number of second-order partitions of an $n$-element set $S_{n}=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$.
Obviously, $F_{0}=1, F_{1}=2$.
Suppose $\left(A_{i}, B_{i}\right)\left(i=1,2, \cdots, F_{n}\right)$ is any partition of $S_{n}$. To obtain the partition o... | \frac{1}{2}\left(3^{n}+1\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,731 |
11. (20 points) Given a positive sequence $\left\{a_{n}\right\}$ and $k$ is a positive integer, satisfying $a_{n+1} \leqslant a_{n}^{k}\left(1-a_{n}\right)$. Prove: For all natural numbers $n$ not less than 2, we have
$$
\frac{1}{a_{n}} \geqslant \frac{(k+1)^{k+1}}{k^{k}}+(n-2) .
$$ | 11. From the problem, we know $0 < a_{n} < 1$, and
$$
\begin{array}{l}
a_{2} \leqslant \frac{1}{k} a_{1}^{k}\left(k - k a_{1}\right) \\
\leqslant \frac{1}{k}\left(\frac{k a_{1} + \left(k - k a_{1}\right)}{k+1}\right)^{k+1} \\
= \frac{k^{k}}{(k+1)^{k+1}}, \\
\end{array}
$$
which implies
$$
\frac{1}{a_{2}} \geqslant \fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,732 |
一、(40 points) Given $H$ and $O$ are the orthocenter and circumcenter of acute $\triangle ABC$ respectively, $D$ is the midpoint of the minor arc $\overparen{BC}$, connect $AD$, and draw $OE \parallel AD$ through point $O$, intersecting $BC$ at point $E$. Prove:
$$
\angle ADH = \angle ODE.
$$ | As shown in Figure 3, connect $A O$ and extend it to intersect $\odot O$ at point $S$, connect $H S$, and let it intersect $B C$ at point $M$.
Obviously, $M$ is the midpoint of side $B C$. Extend $D H$ to intersect $\odot O$ at point $F$, connect $F O$ and $F S$, take the midpoint $N$ of $F S$, and connect $O N$.
It is... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,733 |
Example 1 In an $8 \times 8$ chessboard, how many ways are there to select 56 squares such that: all the black squares are selected, and each row and each column has exactly seven squares selected? ? ${ }^{[1]}$
(2014, Irish Mathematical Olympiad) | The problem is equivalent to selecting eight white squares on the chessboard, with exactly one square selected from each row and each column.
The white squares on the chessboard are formed by the intersections of rows $1, 3, 5, 7$ and columns $1, 3, 5, 7$, resulting in a $4 \times 4$ submatrix, as well as the intersec... | 576 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,734 |
Example 2 Consider a $19 \times 2015$ grid. A "block" is formed by a $10 \times 10$ grid and a cell to the right of its top-right cell. It is known that the cells of different blocks do not overlap, and the cells of the blocks coincide with the cells of the grid. If blocks cannot be rotated, find the number of ways to ... | Let the number of blocks that can be placed be $n$. Color the top-right corner of each block black. Since each block occupies 10 rows, and the grid has only 19 rows, the black cells must be in different columns. Therefore, we can sort the black cells from left to right. For $i=1,2, \cdots, n$, let the $i$-th black cell... | \mathrm{C}_{250}^{199} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,735 |
4. How many positive integers satisfy that in their decimal representation they have exactly 150 digits, and any two adjacent digits differ by $\pm 2$? (2013-2014 Hungarian Mathematical Olympiad) | Let $a_{i, j}$ denote the number of $i$-digit numbers whose first digit is $j$ (which can be 0) and where the difference between any two adjacent digits is $\pm 2$.
Use mathematical induction to prove: For any positive integer $k$, we have
$$
\begin{array}{l}
a_{2 k, 0}=a_{2 k, 1}=a_{2 k, 8}=a_{2 k, 9}=3^{k-1}, \\
a_{... | 5 \times 3^{75} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,736 |
Question 1 Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1} \in(0,1), a_{n+1}=a_{n}+\frac{a_{n}^{2}}{n^{2}} .
$$ | Prove: There exists a constant $M$, such that for any positive integer $n$, we have $a_{n} \leqslant M$.
(2014, Nankai University Independent Enrollment Mathematics Exam) This article generalizes it to a more general proposition. | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,737 |
Example 1 Let $n$ be a positive integer, $X$ a finite set, and the mapping $f: X \rightarrow X$ satisfies that for any $x \in X$,
$$
\begin{array}{l}
f^{(n)}(x)=x, \text { where } f^{(1)}(x)=f(x), \\
f^{(i)}(x)=f\left(f^{(i-1)}(x)\right)(i \geqslant 2) .
\end{array}
$$
Let $m_{j}$ be the number of elements in the set ... | 【Analysis】This problem appears to be about set mapping, but the conclusion involves trigonometric functions, seemingly unrelated to graph theory. In fact, the relationship between the preimage and image of the mapping $f$ can be imagined as edges in graph theory.
Proof: Construct a graph theory model: Use the elements... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,738 |
For the four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b 、 c$ 、 $d \leqslant 9)$ : if $a>b, bd$, then $\overline{a b c d}$ is called a $P$ class number; if $ac, c<d$, then $\overline{a b c d}$ is called a $Q$ class number. Let $N(P)$ and $N(Q)$ represent the number of $P$ class numbers and ... | Let the set of all numbers of type $P$ and type $Q$ be denoted as $A$ and $B$, respectively. Further, let the set of all numbers of type $P$ that end in zero be denoted as $A_{0}$, and the set of all numbers of type $P$ that do not end in zero be denoted as $A_{1}$.
For any four-digit number $\overline{a b c d} \in A_... | 285 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,739 |
Example 3 Given $n$ cards, on each side of every card, a natural number is written, chosen from $1,2, \cdots, n$, and each number appears exactly twice on these $n$ cards. Prove: It is possible to lay these $n$ cards on the table so that the numbers $1,2, \cdots, n$ all appear on the top side. | Prove the construction of a graph model: take the numbers $1,2, \cdots, n$ as vertices, and connect an edge for each card. If both sides of a card are the same number, then the point corresponding to that number is connected to itself (i.e., a loop), making that point an isolated point; if the numbers on both sides of ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,740 |
Let $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=1$. Prove:
$$
\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+z x}}+\frac{z x}{\sqrt{z x+x y}} \leqslant \frac{\sqrt{2}}{2} .
$$
(2006, China National Training Team Test) | Text [1] has sufficiently discussed the proof and strengthening of the upper bound of this problem. This paper provides a stronger estimate in the opposite direction:
$$
\begin{array}{l}
\sqrt{\frac{x y}{x+y}+\frac{y z}{y+z}+\frac{z x}{z+x}} \\
\leqslant \frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+z x}}+\frac{z x}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,741 |
For the set $N=\{1,2, \cdots, n\}$, select $m$ numbers such that none of them is $p$ times another. Find the maximum value of $m$.
| For this problem, we can obtain a formula:
$$
m_{\max }=\sum_{i=0}^{\infty}(-1)^{i}\left[\frac{n}{p^{i}}\right] \text {, }
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Proof: For convenience, we introduce the notations “+” and “-”.
For sets $A$ and $B$, we denote
$A+B=A \cup B$,
$A+(-... | m_{\max }=\sum_{i=0}^{\infty}(-1)^{i}\left[\frac{n}{p^{i}}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,742 |
2. Find all integers $n(n \geqslant 3)$, such that there exist real numbers $a_{1}, a_{2}, \cdots, a_{n+2}$, satisfying $a_{n+1}=a_{1}, a_{n+2}=a_{2}$, and for $i=1,2, \cdots, n$, we have $a_{i} a_{i+1}+1=a_{i+2}$. | 2. The required $n$ is all multiples of 3.
On one hand, if $n$ is a multiple of 3, let $n=3k$, and take
$$
\begin{array}{l}
a_{3i-2}=a_{3i-1}=-1, a_{3i}=2(i=1,2, \cdots, k), \\
a_{n+1}=a_{n+2}=-1 .
\end{array}
$$
It is easy to verify that for $i=1,2, \cdots, n$, we have
$$
a_{i} a_{i+1}+1=a_{i+2}
$$
satisfying the r... | 3 \mid n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,743 |
3. The "anti-Pascal triangle" is an array of numbers arranged in an equilateral triangular shape, where each number not in the last row is exactly the absolute difference of the two numbers directly below it. For example, Figure 1 is an anti-Pascal triangle with four rows, and it contains each integer from 1 to 10 exac... | 3. Does not exist.
Proof by contradiction. Suppose there exists an anti-Pascal triangle that satisfies the problem's conditions. Let \( N = 1 + 2 + \cdots + 2018 \), and denote the \( j \)-th number from the left in the \( i \)-th row as \( a_{i, j} \). Let \( a_1 = a_{1,1} \). Denote the larger of the two numbers be... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,744 |
1. Let positive integers $a_{1}, a_{2}, \cdots, a_{n}, k$ and $M$ satisfy $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=k, a_{1} a_{2} \cdots a_{n}=M$. If $M>1$, prove: the polynomial $P(x)=M(x+1)^{k}-\prod_{i=1}^{n}\left(x+a_{i}\right)$
has no positive roots. | 1. First, prove: For any $x>0$, we have
$$
a_{i}(x+1)^{\frac{1}{a_{i}}} \leqslant x+a_{i} \text {, }
$$
with equality holding if and only if $a_{i}=1$.
When $a_{i}=1$, the equality in (1) clearly holds.
If $a_{i}>1$, by the AM-GM inequality, we get
$$
\begin{array}{l}
\frac{(x+1)+\overbrace{1+1+\cdots+1}^{a_{i}-1 \upa... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,746 |
Example 4 Let $f(x)=\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{2013!}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$). For an integer $n$, if the equation $f(x)=n$ has a real solution, then $n$ is called a "good number". Find the number of good numbers... | First, point out two obvious conclusions:
(1) If $m \in \mathbf{Z}_{+}, x \in \mathbf{R}$, then $\left[\frac{x}{m}\right]=\left[\frac{[x]}{m}\right]$;
(2) For any integer $l$ and positive even number $m$, we have
$$
\left[\frac{2 l+1}{m}\right]=\left[\frac{2 l}{m}\right] \text {. }
$$
In (1), let $m=k!(k=1,2, \cdots, ... | 587 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,747 |
2. Given that $q$ is a real number, person A has a napkin with 10 different real numbers written on it. He writes three rows of real numbers on the blackboard in the following way:
The first row, A writes each number of the form $a-b$, where $a$ and $b$ are numbers from the napkin (they can be the same); The second ro... | 2. $q=-2,0,2$.
If each number in the second row unconditionally belongs to the third row, then $q$ is called "good".
First, prove: $0, \pm 2$ are good.
Since the third row must include 0, 0 is good.
For any two numbers $a, b$ in the first row, let
$$
a=x-y, b=u-v,
$$
where $x, y, u, v$ are numbers written on the napk... | q=-2,0,2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,748 |
5. Given a positive integer $n \geqslant 3$, for an $n$-tuple of real numbers $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$, if every permutation $y_{1}, y_{2}, \cdots$, $y_{n}$, satisfies $\sum_{i=1}^{n-1} y_{i} y_{i+1} \geqslant-1$, then the real number tuple $\left(x_{1}, x_{2}\right.$,
$\left.\cdots, x_{n}\right)$ is ... | 5. $C=-\frac{n-1}{2}$.
Let $t$ be a positive real number, and take
$$
x_{1}=-\frac{1}{2 t}, x_{2}=x_{3}=\cdots=x_{n}=t \text {. }
$$
Then $x_{i} x_{j}=t^{2}$ or $-\frac{1}{2}(i \neq j)$.
Thus, for each permutation $y_{1}, y_{2}, \cdots, y_{n}$, we have
$$
\sum_{i=1}^{n-1} y_{i} y_{i+1} \geqslant(n-3) t^{2}-1 \geqslan... | -\frac{n-1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,750 |
7. Given $a_{0}, a_{1}, \cdots$ as an integer sequence, and $b_{0}, b_{1}, \cdots$ as a sequence of positive integers, satisfying $a_{0}=0, a_{1}=1$, and for any positive integer $n$,
$$
a_{n+1}=\left\{\begin{array}{ll}
a_{n} b_{n}+a_{n-1}, & b_{n-1}=1 \\
a_{n} b_{n}-a_{n-1}, & b_{n-1}>1
\end{array} .\right.
$$
Prove:... | 7. Since $a_{0}=0$, $b_{0}$ can be any positive integer, let's assume $b_{0}=1$.
First, we prove a lemma.
Lemma For all positive integers $n$, we have $a_{n} \geqslant 1$.
Proof By contradiction.
Assume there exists a positive integer $n$ such that $a_{n} \leqslant 0$, and $n$ is the smallest. Then $n \geqslant 2$.
Sin... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,751 |
1. The range of the function $f(x)=\sqrt{x-2}+\sqrt{3-x}$ is
$\qquad$ | $-1 .[1, \sqrt{2}]$.
It is easy to know that the domain of $f(x)$ is $[2,3]$.
Let $x=2+\sin ^{2} \alpha\left(0 \leqslant \alpha \leqslant \frac{\pi}{2}\right)$.
Then $f(x)=\sin \alpha+\cos \alpha=\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right)$.
Therefore, the range of $f(x)$ is $[1, \sqrt{2}]$. | [1, \sqrt{2}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,753 |
2. Equation
$$
x^{2}-31 x+220=2^{x}\left(31-2 x-2^{x}\right)
$$
The sum of the squares of all real roots is $\qquad$ . | 2. 25 .
Let $y=x+2^{x}$. Then the original equation is equivalent to
$$
\begin{array}{l}
y^{2}-31 y+220=0 \\
\Rightarrow y_{1}=11, y_{2}=20 \\
\Rightarrow x_{1}+2^{x_{1}}=11 \text { and } x_{2}+2^{x_{2}}=20 .
\end{array}
$$
Since $f(x)=x+2^{x}$ is a monotonically increasing function, each equation has at most one rea... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,754 |
3. As shown in Figure 1, in a regular hexagon $A B C D E F$ with side length 10, $H$ is the midpoint of side $D E$, and $G$ is a point on side $B C$ such that $\angle A G B = \angle C G H$. Then the area of pentagon $A F E H G$ is $\qquad$ | 3. $\frac{205 \sqrt{3}}{2}$.
As shown in Figure 3, draw a line through point $H$ parallel to $B E$, intersecting $G C$ at point $K$.
It is easy to see that $K H=15$,
$B K=K C=5$.
Let $B G=x$. Then
$$
G K=5-x \text {. }
$$
Since $\triangle K G H \backsim \triangle B G A$
$$
\begin{array}{l}
\Rightarrow \frac{K H}{A B}... | \frac{205 \sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,755 |
Example 5 Given an integer $n \geqslant 2$. For the set of integer points in the Cartesian coordinate system
$$
A=\{(a, b) \mid a, b \in\{1,2, \cdots, n\}\}
$$
each point is colored one of three colors: red, yellow, or blue, and it satisfies: for any $a, b \in\{1,2, \cdots, n-1\}$, if $(a, b)$ and $(a+1, b)$ are the s... | For $k=1,2, \cdots, n$, define the set of integer points
$A_{k}=\{(k, b) \mid b \in\{1,2, \cdots, n\}\}$.
Color $A_{1}, A_{2}, \cdots, A_{n}$ in sequence (to complete the coloring of $A$). Let the number of ways to color $A_{k}$ during these $n$ steps be denoted as $N_{k}(k=1,2, \cdots, n)$.
First, color $A_{1}$: Sinc... | 3^{n}\left(2^{n+1}-1\right)^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,756 |
5. In the plane, there are 200 points, no three of which are collinear, and each point is labeled with one of the numbers $1, 2, 3$. All pairs of points labeled with different numbers are connected by line segments, and each line segment is labeled with a number 1, 2, or 3, which is different from the numbers at its en... | 5.199.
Let the points labeled with $1, 2, 3$ be $a, b, c$ respectively.
Thus, $a+b+c=200$, and the number of line segments labeled with $1, 2, 3$ are $bc, ca, ab$ respectively.
Then $n=a+bc=b+ca=c+ab$.
Therefore, $(a+bc)-(b+ca)=(a-b)(1-c)=0$.
Similarly, $(b-c)(1-a)=(c-a)(1-b)=0$.
If at least two of $a, b, c$ are not 1... | 199 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,757 |
Three. (15 points) As shown in Figure 2, $D$ is a point inside the acute $\triangle ABC$ such that $\angle ADB = \angle ACB + 90^\circ$, and $AC \cdot BD = AD \cdot BC$. Extend $AD$, $BD$, and $CD$ to intersect the circumcircle $\Gamma$ of $\triangle ABC$ at points $G$, $E$, and $F$ respectively.
Prove: (1) $EF = FG$;
... | (1) Notice that,
$\triangle E F D \backsim \triangle C B D \Rightarrow \frac{E F}{F D}=\frac{B C}{B D}$,
$\triangle F D G \backsim \triangle A D C \Rightarrow \frac{F G}{F D}=\frac{A C}{A D}$,
$A C \cdot B D=A D \cdot B C \Rightarrow \frac{B C}{B D}=\frac{A C}{A D}$.
Therefore, $\frac{E F}{F D}=\frac{F G}{F D} \Rightar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,758 |
(15 points) (1) Prove: 2018 can be expressed as the sum of the squares of two positive integers;
(2) Prove: there exists a triangle that can be divided into 2018 congruent triangles. | (1) If there exist positive integers $x, y$, such that
$$
2018=x^{2}+y^{2} \text {, }
$$
Since 2018 leaves a remainder of 2 when divided by 4, then $x, y$ must both be odd.
Furthermore, the last digit of the square of an odd number must be one of $1, 5, 9$. Therefore, the last digits of $x^{2}$ and $y^{2}$ must both b... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,759 |
2. Let $a, b, c, d > 0$. Prove:
$$
a^{\frac{1}{3}} b^{\frac{1}{3}} + c^{\frac{1}{3}} d^{\frac{1}{3}} \leqslant (a+b+c)^{\frac{1}{3}}(a+c+d)^{\frac{1}{3}},
$$
and find the necessary and sufficient conditions for equality. | 2. By the AM-GM inequality, we have
$$
\begin{array}{l}
3\left(\frac{a b}{(a+b+c)(a+c+d)}\right)^{\frac{1}{3}} \\
=3\left(\frac{a}{a+c} \cdot \frac{b}{a+b+c} \cdot \frac{a+c}{a+c+d}\right)^{\frac{1}{3}} \\
\leqslant \frac{a}{a+c}+\frac{b}{a+b+c}+\frac{a+c}{a+c+d}, \\
3\left(\frac{c d}{(a+b+c)(a+c+d)}\right)^{\frac{1}{3... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,760 |
Example 6 Let $n$ be a given positive integer. Consider the set of points in the Cartesian coordinate system $x O y$
$$
T=\{(x, y)|| x|=| y \mid \leqslant n, x, y \in \mathbf{Z}\} .
$$
For two points $P, Q$ in $T$, $P, Q$ are called "adjacent" if and only if $|P Q|=\sqrt{2}$ or $P Q$ is parallel to one of the coordina... | $$
\begin{array}{l}
\text { For } k=1,2, \cdots, n, \text { let } \\
A_{k}(-k, k), B_{k}(k, k), C_{k}(k,-k), D_{k}(-k,-k), \\
\text { and denote } S_{k}=\left\{A_{k}, B_{k}, C_{k}, D_{k}\right\}, \\
T_{k}=\{O\} \cup S_{1} \cup S_{2} \cup \cdots \cup S_{k}, \\
\text { in particular, } T_{n}=T .
\end{array}
$$
When colo... | t_{n}=\frac{2 \sqrt{33}}{11}\left(\left(\frac{7+\sqrt{33}}{2}\right)^{n}-\left(\frac{7-\sqrt{33}}{2}\right)^{n}\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,761 |
6. Given the function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, which satisfies for any $x \in \mathbf{Z}_{+}$,
$$
f(x)+f(x+2) \leqslant 2 f(x+1) .
$$
Prove: there exists an infinite set of positive integers $M$, such that for any $i, j, k \in M$, we have
$$
(i-j) f(k)+(j-k) f(i)+(k-i) f(j)=0 \text {. }
$$ | 6. Consider the sequence of points $A_{n}=(n, f(n))\left(n \in \mathbf{Z}_{+}\right)$. Note that the condition in the problem is equivalent to
$$
f(x+1)-f(x) \geqslant f(x+2)-f(x+1) \text {. }
$$
Then $k_{A_{n} A_{n+1}}=f(n+1)-f(n)$
$$
\geqslant f(n+2)-f(n+1)=k_{A_{n+1} A_{n+2}} \text {. }
$$
Thus, the slopes $\left\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,762 |
7. If $n$ is a perfect square or the distance from $n$ to its nearest perfect square is a perfect cube, then $n$ is called a "good number". For example, $145^{2}-2017 \mid=2^{3}$, so 2017 is a good number. For a positive integer $N$, let $S(N)$ be the number of good numbers in the set $\{1$, $2, \cdots, N\}$.
(1) Find ... | 7. (1) For positive integer $n, n^{2}$, the perfect square to the left of $n^{2}$ is $n^{2}-2 n+1$, and the perfect square to the right of $n^{2}$ is $n^{2}+2 n+1$.
Thus, the nearest perfect square to the integers in the interval $\left[n^{2}-n+1, n^{2}+n\right]$ is $n^{2}$.
Let $[x]$ denote the greatest integer not ex... | \alpha=\frac{2}{3}, \beta=\frac{3}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,763 |
2. Consider the set $A=\left\{\left.1+\frac{1}{k} \right\rvert\, k=1,2, \cdots\right\}$.
(1) Prove: Every integer $x \geqslant 2$ can be expressed as the product of at least one element in $A$ (elements can be equal).
(2) For all integers $x \geqslant 2$, let $f(x)$ be the smallest positive integer such that $x$ can be... | 2. (1) Notice that, $x=\prod_{k=1}^{x-1} \frac{k+1}{k}=\prod_{k=1}^{x-1}\left(1+\frac{1}{k}\right)$. The proposition is obviously true.
(2) First, prove that $(x, y)=(7,7)$ satisfies equation (1).
In fact, by $2^{3}>7$ and $2^{2} \times \frac{3}{2}M$ or $y>M$, then
$$
f(x y)=f(x)+f(y) \text {. }
$$
Take $n=M+1>M, x=y... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,764 |
3. In a certain competition, there are $n$ contestants $C_{1}, C_{2}, \cdots, C_{n}$. After the competition, all contestants queue up at the restaurant entrance to wait for their meal according to the following rules:
(i) The organizing committee arranges the initial positions of each contestant in the queue;
(ii) Ever... | 3. (1) The queue order of the contestants can be regarded as a permutation $\sigma$ of the set $\{1,2, \cdots, n\}$. In other words, contestant $C_{\sigma(i)}$ stands in the $i$-th position in the queue.
$$
\begin{array}{l}
\text { Let } R(\sigma)=\{(i, j) \mid 1 \leqslant i\sigma(j)\}, \\
W(\sigma)=\sum_{(i, j) \in R(... | 2^{n}-n-1 | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,765 |
1. The solution set of the inequality $\frac{4 x^{2}}{(1-\sqrt{1+2 x})^{2}}<2 x+9$ is | $$
\text { - 1. }\left[-\frac{1}{2}, 0\right) \cup\left(0, \frac{45}{8}\right) \text {. }
$$
From $1-\sqrt{1+2 x} \neq 0$
$$
\Rightarrow x \geqslant-\frac{1}{2}, x \neq 0 \text {. }
$$
The original inequality becomes
$$
\begin{array}{l}
(1+\sqrt{1+2 x})^{2}<2 x+9 \\
\Rightarrow x<\frac{45}{8} .
\end{array}
$$
Theref... | \left[-\frac{1}{2}, 0\right) \cup\left(0, \frac{45}{8}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,766 |
2. If the three lateral edges of the tetrahedron $S-ABC$ are pairwise perpendicular, and $O$ is a point within the base $\triangle ABC$, then
$$
\mu=\tan \angle O S A \cdot \tan \angle O S B \cdot \tan \angle O S C
$$
the range of values for $\mu$ is $\qquad$ | 2. $\mu \geqslant 2 \sqrt{2}$.
From $\cos ^{2} \angle O S A+\cos ^{2} \angle O S B+\cos ^{2} \angle O S C=1$
$$
\begin{array}{l}
\Rightarrow \sin ^{2} \angle O S C=\cos ^{2} \angle O S A+\cos ^{2} \angle O S B \\
\quad \geqslant 2 \cos \angle O S A \cdot \cos \angle O S B .
\end{array}
$$
Similarly, we can obtain the... | \mu \geqslant 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,767 |
4. Given that the complex number $z$ satisfies $z^{3}+z=2|z|^{2}$. Then all possible values of $z$ are $\qquad$ | 4. $0, 1, -1 \pm 2$ i.
$$
\begin{array}{l}
\text { Given } z^{3}+z=2|z|^{2}=2 z \bar{z} \\
\Rightarrow z\left(z^{2}+1-2 \bar{z}\right)=0 .
\end{array}
$$
When $z=0$, the condition is satisfied.
When $z \neq 0$, $z^{2}+1-2 \bar{z}=0$.
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R})$. Then
$$
\begin{array}{l}
\left\{\begin{a... | 0, 1, -1 \pm 2i | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,768 |
9. (16 points) For positive integers $n(n \geqslant 2)$, let
$$
a_{n}=\sum_{k=1}^{n-1} \frac{n}{(n-k) 2^{k-1}} \text {. }
$$
Find the maximum value in the sequence $\left\{a_{n}\right\}$. | Sure, here is the translated text:
```
9. Calculate
$$
a_{2}=2, a_{3}=3, a_{4}=a_{5}=\frac{10}{3} \text {. }
$$
Below, we use mathematical induction to prove:
When $n \geqslant 5$, $a_{n} \leqslant \frac{10}{3}$.
Assume $a_{n} \leqslant \frac{10}{3}(n \geqslant 5)$.
Then $a_{n+1}=\sum_{k=1}^{n} \frac{n+1}{(n+1-k) 2^{... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,769 |
11. (20 points) Given the function
$$
f(x)=a(|\sin x|+|\cos x|)-3 \sin 2 x-7,
$$
where $a$ is a real parameter. Find all pairs $(a, n)(n \in \mathbf{Z}_{+})$ such that the function $y=f(x)$ has exactly 2019 zeros in the interval $(0, n \pi)$. | 11. First, the function $f(x)$ has a period of $\pi$ and is symmetric about $x=\frac{k \pi}{2}+\frac{\pi}{4}(k \in \mathbf{Z})$.
Thus, the number of zeros of $f(x)$ in the intervals $\left(\frac{k \pi}{2}, \frac{k \pi}{2}+\frac{\pi}{4}\right)$ and $\left(\frac{k \pi}{2}+\frac{\pi}{4}, \frac{k \pi}{2}+\frac{\pi}{2}\rig... | (7,505),(5 \sqrt{2}, 2019),(2 \sqrt{2}, 2019) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,770 |
Example 8 Let the set of positive integers be
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{1000}\right\},
$$
where, $a_{1}<a_{2}<\cdots<a_{1000} \leqslant 2014$, and
$$
S_{A}=\left\{a_{i}+a_{j} \mid 1 \leqslant i, j \leqslant 1000, i+j \in A\right\}
$$
is a subset of $A$. Find the number of sets $A$ that satisfy the conditio... | Define a set $A$ that satisfies the conditions as a "good set".
First, prove a lemma.
Lemma: $A$ is a good set if and only if $A = B \cup C$, where,
$$
\begin{array}{l}
C \subseteq \{2001, 2002, \cdots, 2014\}, \\
B = \{1, 2, \cdots, 1000 - |C|\}.
\end{array}
$$
Proof: On the one hand, suppose $A = B \cup C$, where $B... | 2^{14} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,771 |
Three, (50 points) Let $n, k \in \mathbf{Z}_{+}, 3 \nmid n, k \geqslant n$. Prove: There exists a positive integer $m$, the sum of whose digits is $k$, and $n \mid m$.
---
The translation maintains the original text's formatting and line breaks. | Let $n=2^{a} 5^{b} t\left(a, b \in \mathbf{N}, t \in \mathbf{Z}_{+}\right.$, and $(t, 10)=1)$.
It is sufficient to prove: there exists a positive integer $M$, such that $t \mid M$, and the sum of the digits of $M$ is $k$.
In fact, it is sufficient to take $m=10^{c} M(c=\max \{a, b\})$.
Since $(t, 10)=1$, by Euler's ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,772 |
Let $S$ be a finite set, and let $A_{1}, A_{2}, \cdots, A_{n} \subseteq S$. Denote $|A|$ as the number of elements in the set $A$. Prove: There exists a real-valued mapping $F$ defined on the power set of $X=\{1,2, \cdots, n\}$ (i.e., the set of all subsets of $X$) such that for any $I \subseteq X$, we have
$$
\sum_{J ... | Notice that, $\complement_{s}\left(\bigcap_{j \in J} A_{j}\right)=\bigcup_{j \in J} \complement_{S} A_{j}$. Define $F(I)=\sum_{J \subseteq I}(-1)^{|I|-|J|}\left|\bigcap_{j \in J} \bigcap_{s} A_{j}\right|$. By the principle of inclusion-exclusion, we have
$$
\begin{array}{l}
\sum_{J \leq I} F(J)=\sum_{J \subseteq I} \su... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,773 |
High $\mathbf{5 8 7}$ As shown in Figure 1, line segment $A D$ intersects $\odot O$ at points $B$ and $D$, line segment $A E$ intersects $\odot O$ at points $C$ and $E$ and passes through the center $O$, $B E$ intersects $C D$ at point $H$, take the midpoint $F$ of $C D$, connect $A F$, which intersects $B C$ at point ... | Prove as shown in Figure 1, draw two tangents from point $A$ to $\odot O$, which touch $\odot O$ at points $L$ and $M$.
By the properties of harmonic point series, we know that $L, H, M$ and $L, G, M$ are collinear respectively.
By symmetry, point $J$ also lies on this line segment.
Therefore, $L, G, H, J, M$ are coll... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,774 |
Does there exist an increasing sequence of positive integers $\left\{a_{n}\right\}$ such that for any $i, j \in \mathbf{Z}_{+}, i \neq j$, we have
$$
\left(i a_{j}, j a_{i}\right)=(i, j) .
$$ | Exist.
Let all prime numbers be arranged in ascending order, and let $a_{n}$ be the $n$-th prime number.
The following proves that $\left\{a_{n}\right\}$ satisfies the condition.
For any positive integer $n$, since $a_{n}>n$ and $a_{n}$ is a prime number, we have $\left(a_{n}, n\right)=1$.
For different positive integ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,775 |
1. The sequence $\left\{a_{n}\right\}$ has nine terms, $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in\left\{2,1,-\frac{1}{2}\right\}$. The number of such sequences is $\qquad$
(2013, National High School Mathematics League Competition) | Let $b_{i}=\frac{a_{i+1}}{a_{i}}(1 \leqslant i \leqslant 8)$, mapping each sequence $\left\{a_{n}\right\}$ that meets the conditions to a unique eight-term sequence $\left\{b_{n}\right\}$, where $\prod_{i=1}^{8} b_{i}=\frac{a_{9}}{a_{1}}=1$, and $b_{i} \in\left\{2,1,-\frac{1}{2}\right\}(1 \leqslant i \leqslant 8)$.
Fro... | 491 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,776 |
2. Let $n$ be a three-digit positive integer without the digit 0. If the digits of $n$ in the units, tens, and hundreds places are permuted arbitrarily, the resulting three-digit number is never a multiple of 4. Find the number of such $n$.
(54th Ukrainian Mathematical Olympiad) | Hint: Classify by the number of even digits (i.e., $2,4,6,8$) appearing in the three-digit code of $n$. It is known from the discussion that the number of $n$ satisfying the condition is $125+150+0+8=283$.
| 283 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,777 |
For a given positive integer $k>1, S$ is a set composed of certain positive integers. If there exists a $k$-coloring of all positive integers such that no number in the set $S$ can be expressed as the sum of two different numbers of the same color, then $S$ is called "good". Find the maximum positive integer $t$, such ... | 【Analysis】 $t_{\max }=2 k-2$.
【Analysis】 $t_{\max }=2 k-2$.
Note: The original text appears to be a mathematical expression with a label "Analysis". The translation retains the format and content as requested. However, the repetition in the English version is due to the direct translation of the provided text, which... | 2k-2 | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,778 |
Example 3 Let $H$ be the orthocenter of acute $\triangle A B C$, $O^{\prime}$ be the circumcenter of $\triangle B H C$, $N$ be the midpoint of segment $A O^{\prime}$, and $D$ be the reflection of $N$ over side $B C$. Prove that $A, B, D, C$ are concyclic if and only if $b^{2}+c^{2}-a^{2}=3 R^{2}$, where $a=BC$, $b=CA$,... | 【Analysis】Let the circumradius of $\triangle B H C$ be $R^{\prime}, O$ be the circumcenter of $\triangle A B C$, the extension of $B H$ intersects $A C$ at point $S$, the extension of $C H$ intersects $B A$ at point $T$, and $O O^{\prime}$ intersects $B C$ at point $M$. Then
$\angle B H C=\pi-\angle C A B$.
By the Law ... | b^{2}+c^{2}-a^{2}=3 R^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 730,779 |
Example 4 As shown in Figure 5, the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ intersect at point $P$, the midpoints of sides $AB$, $BC$, $CD$, and $DA$ are $E$, $F$, $G$, and $H$ respectively, and the circumcenters of $\triangle PHE$, $\triangle PEF$, $\triangle PFG$, and $\triangle PGH$ are $O_1$, $O_... | 【Analysis】From the problem, we know
$O_{1} O_{2} \perp P E, O_{2} O_{3} \perp P F$,
$O_{3} O_{4} \perp P G, O_{4} O_{1} \perp P H$.
Therefore, $O_{1}, O_{2}, O_{3}, O_{4}$ are concyclic
$\Leftrightarrow \angle O_{2} O_{1} O_{4}+\angle O_{2} O_{3} O_{4}=\pi$
$\Leftrightarrow \angle H P E+\angle F P G=\pi$
$\Leftrightarr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,780 |
Example 5 As shown in Figure 7, in the acute triangle $\triangle ABC$, $AB < AC$, $AD \perp BC$ at point $D$, the incircle $\odot I$ touches $BC$ at point $E$, point $F$ is on $\odot I$ such that the circumcircle of $\triangle BCF$ is internally tangent to $\odot I$. The line segment $EF$ intersects $AD$ at point $G$, ... | 【Analysis】Let the extension of $F E$ intersect the circumcircle of $\triangle B C F$ at point $L$. Then
$$
\begin{array}{l}
\angle F E B=\frac{1}{2} \widehat{F E}^{\circ}=\frac{1}{2} \overparen{F B L}^{\circ} \\
=\angle F C L=\pi-\angle F B L \\
\Rightarrow \angle F B L=\pi-\angle F E B=\angle B E L \\
\Rightarrow \tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,781 |
Example 2 Given a convex hexagon $A B C D E F$, any two opposite sides have the following property: the distance between their midpoints is equal to $\frac{\sqrt{3}}{2}$ times the sum of their lengths. Prove: all the interior angles of the hexagon are equal.
Note: Convex hexagon $A B C D E F$ has three pairs of opposi... | Let $\overrightarrow{A B}=a, \overrightarrow{B C}=b, \overrightarrow{C D}=c$,
$\overrightarrow{D E}=d, \overrightarrow{E F}=e, \overrightarrow{F A}=f$.
Let $G$ and $H$ be the midpoints of $A B$ and $D E$, respectively. Then
$\overrightarrow{G H}=\frac{1}{2} a+b+c+\frac{1}{2} d$,
$\overrightarrow{G H}=-\frac{1}{2} a-f-e... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,782 |
Let $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} \in \mathbf{Z}_{+}$. Prove: there exist not all zero numbers $\lambda_{1}, \lambda_{2}, \lambda_{3} \in\{0,1,2\}$, such that $\lambda_{1} a_{1}+\lambda_{2} a_{2}+\lambda_{3} a_{3}$ and $\lambda_{1} b_{1}+\lambda_{2} b_{2}+\lambda_{3} b_{3}$ are simultaneously divisible by 3... | For given positive integers $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$, let $l_{1}, l_{2}, l_{3} \in \mathbf{Z}$, construct the binary array
$$
\left(\sum_{i=1}^{3} l_{i} a_{i}, \sum_{i=1}^{3} l_{i} b_{i}\right).
$$
On one hand, since each term $\sum_{i=1}^{3} l_{i} a_{i}$ and $\sum_{i=1}^{3} l_{i} b_{i}$ has 3 cases ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,783 |
Given an infinite sequence $\left\{a_{n}\right\}$ with all terms being integers, satisfying:
$$
\begin{array}{l}
p, q, r \in \mathbf{Z}, a_{1}=p^{2}+q^{2}+r^{2}+1, \\
a_{2}=p^{2}-q^{2}+r^{2}-2, \\
a_{3}=p^{2}-5 q^{2}-6 r^{2}+3, \\
a_{4}=3 p^{2}-11 q^{2}-18 r^{2}+15, \\
a_{n}=3 a_{n-1}-2 a_{n-2}+2 a_{n-3}+a_{n-4}(n>4).
... | Consider the sequence $\left\{a_{n}\right\}$ and its modulo $m$ remainder sequence $\left\{b_{n}\right\}$. Using the terms of the sequence $\left\{b_{n}\right\}$, we form the quadruples:
$$
\left\{b_{i}, b_{i+1}, b_{i+2}, b_{i+3}\right\}\left(i=1,2, \cdots, m^{4}+1\right).
$$
By the multiplication principle, we know t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,784 |
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