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
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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5. Given real numbers $a, b, c$ satisfy
$$
a+b+c=2, ab+bc+ca=0 \text{, and } abc=-1 \text{.}
$$
Then the set of real numbers $a, b, c$ is
$\qquad$ | 5. $\left\{\frac{1-\sqrt{5}}{2}, 1, \frac{1+\sqrt{5}}{2}\right\}$.
From the problem, we know
$$
a+b=2-c, a b+c(a+b)=0, a b=-\frac{1}{c},
$$
we get
$$
\begin{array}{l}
-\frac{1}{c}+c(2-c)=0 \\
\Rightarrow c^{3}-2 c^{2}+1=0 \\
\Rightarrow(c-1)\left(c^{2}-c-1\right)=0 \\
\Rightarrow c=\frac{1 \pm \sqrt{5}}{2} \text { or... | \left\{\frac{1-\sqrt{5}}{2}, 1, \frac{1+\sqrt{5}}{2}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,196 |
Example 4 Let the circumcircle of acute triangle $\triangle ABC$ be circle $\Gamma$, and let the tangents to circle $\Gamma$ at points $B$ and $C$ intersect at point $P$. Connect $AP$ to intersect $BC$ at point $D$. Points $E$ and $F$ are on sides $AC$ and $AB$ respectively, such that $DE \parallel BA$ and $DF \paralle... | Proof (1) As shown in Figure 3, to prove that points $F, B, C, E$ are concyclic, it suffices to prove that $A F \cdot A B = A E \cdot A C$.
Given $D E \parallel B A, D F \parallel C A$,
$\Rightarrow A F = D E = A B \cdot \frac{C D}{B C}, A E = F D = A C \cdot \frac{B D}{B C}$.
Thus, it suffices to prove $\frac{B D}{C D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,197 |
II. (15 points) Given that the roots of the quadratic equation $a x^{2}+b x+c=0$ are 2013 times the roots of the quadratic equation $c x^{2}+d x+a=0$. Prove: $b^{2}=d^{2}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given that $a \neq 0, c \neq 0$.
Let $x_{1}, x_{2}$ be the two roots of $c x^{2}+d x+a=0$. Then $2013 x_{1}, 2013 x_{2}$ are both roots of the quadratic equation $a x^{2}+b x+c=0$.
By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=-\frac{d}{c}, x_{1} x_{2}=\frac{a}{c} ; \\
2013 x_{1}+2013 x_{2}=-\frac{b}{a},... | b^{2}=d^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,198 |
(15 points) (1) A positive integer is called a "good integer" if it can be expressed as the arithmetic mean of the squares of several positive integers. For example,
$$
\begin{array}{l}
4=\frac{2^{2}+2^{2}}{2}, 2007=\frac{2^{2}+12^{2}+22^{2}+86^{2}}{4}, \\
2008=\frac{32^{2}+50^{2}+50^{2}}{3},
\end{array}
$$
then $4, 2... | (1) Since each good integer is a positive integer, we know that $M \subseteq \mathbf{Z}_{+}$. For each $n \in \mathbf{Z}_{+}$, we have
$$
n=-\frac{n^{2}+\overbrace{1^{2}+1^{2}+\cdots+1^{2}}^{n}}{n+1} \text {. }
$$
Thus, $n$ is a good integer, i.e., $n \in M$.
Therefore, $\mathbf{Z}_{+} \subseteq M$. Hence, $M=\mathbf{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,199 |
Four. (15 points) In $\triangle A B C$, it is known that $\angle B A C=40^{\circ}$, $\angle A B C=60^{\circ}$. If $D$ and $E$ are points on sides $A C$ and $A B$ respectively, such that $\angle C B D=40^{\circ}$, $\angle B C E=70^{\circ}$, and $F$ is the intersection of $B D$ and $C E$, connect $A F$. Prove: $A F \perp... | As shown in Figure 4, construct the angle bisector of $\angle A B D$ intersecting $E F$ and $A D$ at points $M$ and $N$ respectively. Connect $M D$, $E N$, and $N F$.
Since $\angle A B C=60^{\circ}$, $\angle C B D=40^{\circ}$, therefore,
$$
\begin{array}{l}
\angle A B D=20^{\circ} . \\
\text { Hence } \angle A B M=\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,200 |
Five. (15 points) A mathematics competition for high school freshmen is set up with 35 examination rooms. The team leaders of schools A, B, and C each group the examination rooms with the same number of students from their school. After statistics, school A has $i$ groups, with the number of examination rooms in each g... | Let $n=i+j+k$. Since
$$
a_{1}, a_{2}, \cdots, a_{i}, b_{1}, b_{2}, \cdots, b_{j}, c_{1}, c_{2}, \cdots, c_{k}
$$
includes the integers from $1$ to $14$, we have $n \geqslant 14$.
And $3 \times 35$
$$
\begin{array}{l}
\geqslant a_{1}+a_{2}+\cdots+a_{i}+b_{1}+b_{2}+\cdots+b_{j}+c_{1}+c_{2}+\cdots+c_{k} \\
\geqslant 1+2+... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,201 |
2. Given the function
$$
f(x)=2 \sin \left(\omega x+\frac{\pi}{3}\right)+\cos \left(\omega x-\frac{\pi}{6}\right)(\omega>0)
$$
has the smallest positive period of $\pi$. Then $\omega=(\quad$.
(A) 4
(B) 2
(C) $\frac{1}{2}$
(D) $\frac{1}{4}$ | 2. B.
Notice,
$$
\begin{array}{l}
f(x)=2 \sin \left(\omega x+\frac{\pi}{3}\right)+\cos \left(\omega x+\frac{\pi}{3}-\frac{\pi}{2}\right) \\
=2 \sin \left(\omega x+\frac{\pi}{3}\right)+\sin \left(\omega x+\frac{\pi}{3}\right) \\
=3 \sin \left(\omega x+\frac{\pi}{3}\right) . \\
\text { By } T=\frac{2 \pi}{\omega} \text ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,203 |
5. On a 2x4 grid chessboard, flip the die along one of its edges (the opposite faces are marked with 1 and 6, 2 and 5, 3 and 4). Initially, the die is placed as shown in Figure 2, with the number 2 facing up, and it is finally flipped to position $A$. If the minimum number of flips is required, then the probability tha... | 5. C.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 727,206 |
6. Given the function
$$
f(x)=A \cos \left(\omega x+\frac{\pi}{4} \omega\right)(A>0)
$$
is decreasing on $\left(0, \frac{\pi}{8}\right)$. Then the maximum value of $\omega$ is | Ni,6.8.
Assume $\omega>0$.
To make $f(x)$ a decreasing function in $\left(0, \frac{\pi}{8}\right)$, combining the image of the cosine-type function, we must have
$$
\begin{array}{l}
\frac{T}{2} \geqslant \frac{\pi}{8} \Rightarrow \frac{\pi}{\omega} \geqslant \frac{\pi}{8} \Rightarrow \omega \leqslant 8 . \\
\text { Whe... | 8 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 727,207 |
Example 5 If the product of the opposite sides of a cyclic convex quadrilateral is equal (i.e., it is a harmonic quadrilateral), then the eight points where lines through the intersection of the diagonals, parallel to each side of the quadrilateral, intersect the two adjacent sides are concyclic.
保留源文本的换行和格式如下：
Examp... | Proof As shown in Figure 5, let the harmonic quadrilateral be $ABCD$, with $AC$ and $BD$ intersecting at point $P$. Draw lines through point $P$ parallel to sides $AB$, $BC$, $CD$, and $DA$, intersecting the sides sequentially at points $X_1, X_2, \cdots, X_8$.
Let $BC=a$, $CD=b$, $DA=c$, $AB=d$, $AC=e$, and $BD=f$.
By... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,208 |
7. In the tetrahedron $S-ABC$, it is known that
$$
\begin{array}{l}
\angle SAB = \angle SAC = \angle ACB = 90^{\circ}, \\
AC = 2, BC = \sqrt{13}, SB = \sqrt{29}.
\end{array}
$$
Then the cosine of the angle formed by the lines $SC$ and $AB$ is $\qquad$ | 7. $\frac{\sqrt{17}}{17}$.
As shown in Figure 4, take $A$ as the origin, and the lines $A B$ and $A S$ as the $y$-axis and $z$-axis, respectively, to establish a spatial rectangular coordinate system.
Then the points are $B(0, \sqrt{17}, 0), S(0,0,2 \sqrt{3})$,
$C\left(2 \sqrt{\frac{13}{17}}, \frac{4}{\sqrt{17}}, 0\ri... | \frac{\sqrt{17}}{17} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,209 |
8. Given the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
with four vertices $A, B, C, D$. If the radius of the inscribed circle of the rhombus $A B C D$ is equal to $\frac{\sqrt{6}}{6}$ of the focal distance of the ellipse, then its eccentricity is $\qquad$. | 8. $\frac{\sqrt{2}}{2}$.
From the given, we have
$$
\begin{array}{l}
a b=\frac{\sqrt{6}}{6} \times 2 c \sqrt{a^{2}+b^{2}} \\
\Rightarrow e=\frac{c}{a}=\frac{\sqrt{2}}{2} .
\end{array}
$$ | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,210 |
10. For the sequence $\left\{a_{n}\right\}$, if there exists a sequence $\left\{b_{n}\right\}$, such that for any $n \in \mathbf{Z}_{+}$, we have $a_{n} \geqslant b_{n}$, then $\left\{b_{n}\right\}$ is called a "weak sequence" of $\left\{a_{n}\right\}$. Given
$$
\begin{array}{l}
a_{n}=n^{3}-n^{2}-2 t n+t^{2}\left(n \in... | 10. $\left(-\infty, \frac{1}{2}\right] \cup\left[\frac{3}{2},+\infty\right)$.
From the problem, we know that for any positive integer $n$,
$$
f(n)=a_{n}-b_{n}=n^{2}-(2 t-1) n+t^{2}-\frac{5}{4} \geqslant 0.
$$
Then $\Delta=(2 t-1)^{2}-4\left(t^{2}-\frac{5}{4}\right)=-4 t+6$.
(1) When $\Delta \leqslant 0$, i.e., $t \ge... | \left(-\infty, \frac{1}{2}\right] \cup\left[\frac{3}{2},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,212 |
11. Given that the radius of $\odot O$ is $1$, and $P$ is a point on the circumference. As shown in Figure 3, a square with side length $1$ (indicated by the solid line, with vertex $A$ coinciding with $P$) rolls along the circumference in a clockwise direction. After several rolls, point $A$ returns to point $P$ for t... | 11. $\frac{2+\sqrt{2}}{2} \pi$.
Point $A$ travels a path consisting of nine minor arcs, each with a central angle of $\frac{\pi}{6}$. Six of these arcs are from circles with a radius of 1, and three arcs are from circles with a radius of $\sqrt{2}$. Therefore, the length of the path traveled by point $A$ is
$$
\frac{\... | \frac{2+\sqrt{2}}{2} \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,213 |
12. (15 points) Given a sequence of positive numbers $\left\{a_{n}\right\}$, the sum of the first $n$ terms is $S_{n}$, and it satisfies $a_{n}^{2}+a_{n}-2 S_{n}=0$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) If $b_{1}=1,2 b_{n}-b_{n-1}=0\left(n \geqslant 2, n \in \mathbf{Z}_{+}\righ... | Three, 12. (1) From the given,
$$
\begin{array}{l}
a_{n-1}^{2}+a_{n-1}-2 S_{n-1}=0 \\
\Rightarrow\left(a_{n}-a_{n-1}\right)\left(a_{n}+a_{n-1}\right)+a_{n}-a_{n-1}-2 a_{n}=0 \\
\Rightarrow\left(a_{n}+a_{n-1}\right)\left(a_{n}-a_{n-1}-1\right)=0 \\
\Rightarrow a_{n}-a_{n-1}=1 .
\end{array}
$$
Let $n=1$.
Then $a_{1}^{2}... | 4-(2 n+4)\left(\frac{1}{2}\right)^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,214 |
13. (25 points) Given that $a$, $b$, and $c$ are the sides opposite to the internal angles $\angle A$, $\angle B$, and $\angle C$ of $\triangle ABC$, respectively, and
\[ b \cos C + \sqrt{3} b \sin C - a - c = 0 \].
(1) Prove that $\angle A$, $\angle B$, and $\angle C$ form an arithmetic sequence;
(2) If $b = \sqrt{3}$... | 13. (1) From the given condition, we have
$$
\sin B \cdot \cos C+\sqrt{3} \sin B \cdot \sin C-\sin A-\sin C=0 \text {. (1) }
$$
Since $\angle A+\angle B+\angle C=\pi$, then $\sin A=\sin (B+C)$.
Substituting the above into equation (1) yields
$$
\begin{array}{l}
\sin B \cdot \cos C+\sqrt{3} \sin B \cdot \sin C- \\
(\si... | 2 \sqrt{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,215 |
14. (25 points) (1) If $0 \leqslant x \leqslant 1$, prove:
$$
x-\frac{3}{8} \leqslant 2 x^{4} \leqslant 2 x \text {; }
$$
(2) If $x, y, z \geqslant 0$, and $x+y+z=1$, find
$$
f(x, y, z)=2 x^{4}+y^{2}+z
$$
the maximum and minimum values. | 14. (1) Since $0 \leqslant x \leqslant 1$, we have
$$
\begin{array}{l}
2 x^{4}-2 x=2 x(x-1)\left(x^{2}+x+1\right) \leqslant 0, \\
2 x^{4}-\left(x-\frac{3}{8}\right)=\frac{1}{8}(2 x-1)^{2}\left(4 x^{2}+4 x+3\right) \geqslant 0 .
\end{array}
$$
Therefore, $x-\frac{3}{8} \leqslant 2 x^{4} \leqslant 2 x$.
[Note] The resul... | \frac{3}{8} \text{ and } 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,216 |
15. (25 points) Find all non-negative integer solutions to the equation
$$
x^{3}+y^{3}-x^{2} y^{2}-(x+y)^{2} z=0
$$ | 15. (1) When $x=0$, the equation becomes
$$
y^{3}-y^{2} z=0 \Rightarrow y^{2}(y-z)=0
$$
$\Rightarrow y=0$ or $y=z$.
Thus, $(0,0, m),(0, m, m)\left(m \in \mathbf{Z}_{+}\right)$ are solutions that satisfy the problem.
(2) When $y=0$, similarly,
$$
(0,0, m),(m, 0, m)\left(m \in \mathbf{Z}_{+}\right)
$$
are solutions that... | (2,2,0),(0,0, m),(0, m, m),(m, 0, m) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,217 |
1. For the real number $x$, the functions are
$$
f(x)=\sqrt{3 x^{2}+7}, g(x)=x^{2}+\frac{16}{x^{2}+1}-1,
$$
then the minimum value of the function $g(f(x))$ is . $\qquad$ | $-, 1.8$.
From the problem, we have
$$
g(f(x))=3 x^{2}+7+\frac{16}{3 x^{2}+8}-1.
$$
Let $t=3 x^{2}+8(t \geqslant 8)$. Then
$$
h(t)=g(f(x))=t+\frac{16}{t}-2 \text{. }
$$
It is easy to see that $h(t)$ is a monotonically increasing function on the interval $[8,+\infty)$. Therefore, $h(t) \geqslant h(8)=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,218 |
Example 1 For any $n(n \geqslant 2)$ points given in space. If the midpoint of each pair of points connected by a line segment is painted red, find the minimum number of red points. | Let the minimum number of distinct red points (midpoints) in the case of $n$ points be denoted as $f(n)$.
Obviously, $f(2)=1, f(3)=3, f(4)=5$.
Based on this, we conjecture that generally $f(n)=2 n-3$.
(1) Consider the case where $n$ points are collinear.
Take the line they share as the number line, from left to right, ... | 2n-3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,219 |
2. In the region $\left\{\begin{array}{l}0 \leqslant x \leqslant 2 \pi, \\ 0 \leqslant y \leqslant 3\end{array}\right.$, a point $P(a, b)$ is randomly taken, the probability that $b>\left(\sin \frac{a}{2}+\cos \frac{a}{2}\right)^{2}$ is $ـ$. $\qquad$ | 2. $\frac{2}{3}$.
Consider the function
$$
y=\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}=1+\sin x,
$$
and the region $\left\{\begin{array}{l}0 \leqslant x \leqslant 2 \pi, \\ 0 \leqslant y \leqslant 3\end{array}\right.$ with area $6 \pi$.
By symmetry and cutting and pasting, we know that the area of points $P(... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,220 |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If $\left[x-\frac{1}{2}\right]\left[x+\frac{1}{2}\right]$ is a prime number, then the range of values for the real number $x$ is | 3. $-\frac{3}{2} \leqslant x<-\frac{1}{2}$ or $\frac{3}{2} \leqslant x<\frac{5}{2}$.
Since $\left[x-\frac{1}{2}\right]$ and $\left[x+\frac{1}{2}\right]$ are both integers, to make $\left[x-\frac{1}{2}\right]\left[x+\frac{1}{2}\right]$ a prime number, one of $\left[x-\frac{1}{2}\right]$ or $\left[x+\frac{1}{2}\right]$ ... | -\frac{3}{2} \leqslant x<-\frac{1}{2} \text{ or } \frac{3}{2} \leqslant x<\frac{5}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,221 |
4. If $x, y$ are two different real numbers, and
$$
x^{2}=2 x+1, y^{2}=2 y+1 \text {, }
$$
then $x^{6}+y^{6}=$ $\qquad$ . | 4. 198 .
$$
\text { Let } S_{n}=x^{n}+y^{n} \text {. }
$$
From $x^{2}=2 x+1, y^{2}=2 y+1$, we get
$$
x^{n+2}=2 x^{n+1}+x^{n}, y^{n+2}=2 y^{n+1}+y^{n} \text {. }
$$
Thus, $S_{n+2}=2 S_{n+1}+S_{n}$.
Also, $S_{1}=2, S_{2}=6$, so
$$
S_{3}=14, S_{4}=34, S_{5}=82, S_{6}=198
$$ | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,222 |
5. Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C: \frac{x^{2}}{19}+\frac{y^{2}}{3}=1$, and point $P$ is on the ellipse $C$. If $S_{\triangle P F_{1} F_{2}}=\sqrt{3}$, then $\angle F_{1} P F_{2}=$ $\qquad$ | 5. $60^{\circ}$.
Let $\angle F_{1} P F_{2}=\theta$. Then
$$
\left\{\begin{array}{l}
P F_{1}+P F_{2}=2 \sqrt{19}, \\
P F_{1}^{2}+P F_{2}^{2}-2 P F_{1} \cdot P F_{2} \cos \theta=64 .
\end{array}\right.
$$
Thus, $P F_{1} \cdot P F_{2}=\frac{6}{1+\cos \theta}$.
And $S_{\triangle P F_{1} F_{2}}=\frac{1}{2} P F_{1} \cdot P... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,223 |
6. Given four points $A, B, C, D$ on a sphere with radius 3. If $AB=3, CD=4$, then the maximum volume of tetrahedron $ABCD$ is $\qquad$ . | 6. $2 \sqrt{5}+3 \sqrt{3}$.
Take the common perpendicular segment $M N$ of the skew lines $A B$ and $C D$, and let the angle between the skew lines $A B$ and $C D$ be $\theta \in\left(0, \frac{\pi}{2}\right]$.
Then $V_{\text {tetrahedron } A B C D}=\frac{1}{6} A B \cdot C D \cdot M N \sin \theta \leqslant 2 M N$.
Let ... | 2 \sqrt{5}+3 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,224 |
7. Given $a_{1}, a_{2}, \cdots, a_{10}$ and $b_{1}, b_{2}, \cdots, b_{10}$ are 20 distinct real numbers. If the equation
$$
\begin{array}{l}
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{10}\right| \\
=\left|x-b_{1}\right|+\left|x-b_{2}\right|+\cdots+\left|x-b_{10}\right|
\end{array}
$$
has a finite numb... | 7.9.
$$
\text { Let } \begin{aligned}
f(x)= & \left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{10}\right|- \\
& \left|x-b_{1}\right|-\left|x-b_{2}\right|-\cdots-\left|x-b_{10}\right| .
\end{aligned}
$$
Thus, by the problem statement, $f(x)=0$.
Let $c_{1}<c_{2}<\cdots<c_{20}$ be the elements of the set
$$
\l... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,225 |
8. If the remainder of $\underbrace{11 \cdots 1}_{n+1 \uparrow} 1$ divided by 3102 is 1, then the smallest positive integer $n$ is $\qquad$ . | 8. 138 .
Notice that, $3102=2 \times 3 \times 11 \times 47$.
From $\underbrace{11 \cdots 1}_{n+1 \uparrow}=3102 k+1(k \in \mathbf{Z})$, we know $\underbrace{11 \cdots 10}_{n \uparrow}=3102 k$.
Thus, $\underbrace{11 \cdots 10}_{n \uparrow}$ is divisible by $2, 3, 11, 47$.
(1) For any positive integer $n$, obviously, $\... | 138 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,226 |
II. (16 points) Given the sequence $\left\{F_{n}\right\}$ satisfies
$$
\begin{array}{l}
F_{1}=F_{2}=1, \\
F_{n+2}=F_{n+1}+F_{n}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$
If $F_{a} 、 F_{b} 、 F_{c} 、 F_{d}(a<b<c<d)$ are the side lengths of a convex quadrilateral, find the value of $d-b$. | From the given, we know that $F_{a}+F_{b}+F_{c}>F_{d}$.
If $c \leqslant d-2$, then
$$
F_{a}+\left(F_{b}+F_{c}\right) \leqslant F_{a}+F_{d-1} \leqslant F_{d},
$$
which is a contradiction.
Therefore, $c=d-1$.
Thus, the side lengths of the quadrilateral are $F_{a} 、 F_{b} 、 F_{d-1} 、 F_{d}$.
If $b \leqslant d-3$, then
$$... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,227 |
Three. (20 points) Let the moving point $P$ be on the line $l_{1}: y=x-4$. Draw two tangents $PA$ and $PB$ from $P$ to the circle $\odot C: x^{2}+y^{2}=1$, where $A$ and $B$ are the points of tangency. Find the equation of the trajectory of the midpoint $M$ of segment $AB$. | Three, let point $P\left(x_{0}, y_{0}\right)$, tangent points $A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right)$. Then the equations of the tangent lines $P A$ and $P B$ are
$$
l_{P A}: x_{A} x+y_{A} y=1, l_{B P}: x_{B} x+y_{B} y=1 \text {. }
$$
Since $P$ is the intersection point of the two tangent lines, we ha... | x^{2}+y^{2}-\frac{x}{4}+\frac{y}{4}=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,228 |
Four, (20 points) If real numbers $a, b, c$ satisfy $a^{2}+b^{2} \leqslant c \leqslant 1$, find the maximum and minimum values of $a+b+c$.
| Let $a=r \cos \theta, b=r \sin \theta(\theta \in \mathbf{R})$. Then $r=\sqrt{a^{2}+b^{2}}(0 \leqslant r \leqslant \sqrt{c} \leqslant 1)$.
Thus, $a+b+c=r(\cos \theta+\sin \theta)+c$
$$
=\sqrt{2} r \sin \left(\theta+\frac{\pi}{4}\right)+c \text {. }
$$
By $-1 \leqslant \sin \left(\theta+\frac{\pi}{4}\right) \leqslant 1$... | -\frac{1}{2} \leqslant a+b+c \leqslant 1+\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,229 |
Example 2 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{2}=7, a_{n}=3 a_{n-1}+2 a_{n-2}(n \geqslant 3) \text {. }
$$
Prove: For any $n \in \mathbf{Z}_{+}, a_{2 n-1}$ can be expressed as the sum of squares of two positive integers. | Proof from the given conditions:
$$
\begin{array}{l}
a_{3}=25, a_{4}=89, a_{5}=317, \\
a_{6}=1129, a_{7}=4021, \cdots \cdots
\end{array}
$$
Now consider the intrinsic relationship between these information and the inherent terms of the sequence.
Notice that, $a_{1}=2=1^{2}+1^{2}, a_{3}=25=3^{2}+4^{2}$, and
$$
\begin{a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,230 |
一、(40 points) As shown in Figure 1, $PA$ and $PB$ are tangent to $\odot O$ at points $A$ and $B$, respectively. A secant line through point $P$ intersects $\odot O$ at points $C$ and $D$. $M$ is the midpoint of $PA$, and $CM$ intersects $AB$ at point $E$. Prove: $DE \parallel PA$. | I. Method of the Same.
As shown in Figure 2, draw $D E^{\prime} / / P A$ intersecting $A B$ at point $E^{\prime}$, and connect $C E^{\prime}$ and extend it to intersect $P A$ at point $M^{\prime}$.
It is sufficient to prove that $P M^{\prime}=M^{\prime} A$, which implies that point $M$ coincides with $M^{\prime}$.
Conn... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,231 |
II. (40 points) Let positive real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a+b=\sqrt{a b+9}, \\
b+c=\sqrt{b c+16}, \\
c+a=\sqrt{c a+25} .
\end{array}
$$
Find $a+b+c$. | $$
\begin{array}{l}
a^{2}+b^{2}-2 a b \cos 120^{\circ}=9, \\
b^{2}+c^{2}-2 b c \cos 120^{\circ}=16, \\
c^{2}+a^{2}-2 c a \cos 120^{\circ}=25 .
\end{array}
$$
From the cosine rule, we can construct the following geometric model.
In the plane, line segments $PA$, $PB$, and $PC$ share a common endpoint $P$ and the angles... | \sqrt{25+12 \sqrt{3}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,232 |
Four. (50 points) Let $p$ be an odd prime, and integers $a_{1}, a_{2}, \cdots, a_{p-1}$ all be coprime with $p$. If for $k=1,2, \cdots, p-2$ we have $a_{1}^{k}+a_{2}^{k}+\cdots+a_{p-1}^{k} \equiv 0(\bmod p)$,
prove: The remainders of $a_{1}, a_{2}, \cdots, a_{p-1}$ when divided by $p$ are distinct. | Let $a_{i}$ modulo $p$ be $r_{i}$, where $1 \leqslant i \leqslant p-1\left(i \in \mathbf{Z}_{+}\right)$. Then $1 \leqslant r_{i} \leqslant p-1$.
Therefore, for $k=1,2, \cdots, p-2$, we have
$$
r_{1}^{k}+r_{2}^{k}+\cdots+r_{p-1}^{k} \equiv 0(\bmod p).
$$
To prove that $r_{1}, r_{2}, \cdots, r_{p-1}$ are distinct, it su... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,234 |
1. Given the sets
$$
\begin{array}{l}
A=\left\{x \mid x^{2}-3 x-10 \leqslant 0\right\}, \\
B=\{x \mid m+1 \leqslant x \leqslant 2 m-1\} .
\end{array}
$$
If $A \cap B=\varnothing$, then the range of real number $m$ is ( ).
(A) $24$
(C) $-\frac{1}{2}4$
Translate the above text into English, please retain the original t... | -1 . B.
From the problem, we know $A=\{x \mid-2 \leqslant x \leqslant 5\}$.
Also, $A \cap B=\varnothing$, then
(1) When $B=\varnothing$, we have
$$
2 m-15 \Rightarrow m>4 .
$$ | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,235 |
2. The line $l$ passing through the origin intersects the hyperbola $x y=-2 \sqrt{2}$ at points $P$ and $Q$, with point $P$ in the second quadrant. If the lower half-plane is folded along the $x$-axis to form a right dihedral angle with the upper half-plane, then the shortest length of segment $P Q$ is ( ).
(A) $2 \sqr... | 2. D.
Let point $P(x, y)$. By symmetry, point $Q(-x, -y)$. Draw perpendiculars from points $P$ and $Q$ to the $x$-axis, with feet at $P_{1}(x, 0)$ and $Q_{1}(-x, 0)$, respectively.
After folding into a right dihedral angle,
$$
\begin{array}{l}
|P Q|^{2}=\left|P P_{1}\right|^{2}+\left|P_{1} Q_{1}\right|^{2}+\left|Q_{1}... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,236 |
3. Let $a$, $b$, and $c$ be non-zero complex numbers, and let $\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$. If $\frac{a}{b} = \frac{b}{c} = \frac{c}{a}$, then $\frac{a+b-c}{a-b+c} = (\quad)$.
(A) 1
(B) $\pm \omega$
(C) $1, \omega, \omega^{2}$
(D) $1, -\omega, \omega^{2}$ | 3. C.
Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k$. Then
$$
\begin{array}{l}
k^{3}=1 \Rightarrow k=1, \omega, \omega^{2} . \\
\text { Hence } \frac{a+b-c}{a-b+c}=\frac{b k+b-b k^{2}}{b k-b+b k^{2}}=\frac{k+1-k^{2}}{k\left(1-k^{2}+k\right)} \\
=\frac{1}{k}=\left\{\begin{array}{ll}
1, & k=1 ; \\
\omega^{2}, & k=\omega ; ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,237 |
4. Given that $f(x)$ is a monotonic function on $(0,+\infty)$, and for any $x \in(0,+\infty)$, we have $f\left(f(x)-\log _{2} x\right)=6$. If $x_{0}$ is a solution to the equation $f(x)-f^{\prime}(x)=4$, and $x_{0} \in (a-1, a)\left(a \in \mathbf{Z}_{+}\right)$, then $a=(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4 | 4. B.
From the monotonicity of $f(x)$ and $f\left(f(x)-\log _{2} x\right)=6$, we know that $f(x)-\log _{2} x=c$ (constant) $\Rightarrow f(c)=6$.
$$
\begin{array}{l}
\text { Hence } f(c)-\log _{2} c=c \Rightarrow \log _{2} c=6-c \\
\Rightarrow c=4 \Rightarrow f(x)=\log _{2} x+4 . \\
\text { From } f(x)-f^{\prime}(x)=\l... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,238 |
5. The maximum number of small balls with a diameter of 2 that can be placed in a cylindrical container with an inner diameter of $\frac{4 \sqrt{3}}{3}+2$ and a height of 20 is ( ).
(A) 30
(B) 33
(C) 36
(D) 39 | 5. C.
The bottom of the cylinder can accommodate three small balls that are pairwise tangent. The next layer also holds three small balls that are pairwise tangent, with each ball touching two balls from the layer below. This method is repeated layer by layer, with a total of \( k \) layers. The total height is given ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,239 |
6. Let real numbers $x, y$ satisfy
$$
17\left(x^{2}+y^{2}\right)-30 x y-16=0 \text {. }
$$
Then the maximum value of $\sqrt{16 x^{2}+4 y^{2}-16 x y-12 x+6 y+9}$ is ( ).
(A) 7
(B) $\sqrt{29}$
(C) $\sqrt{19}$
(D) 3 | 6. A.
$$
\begin{array}{l}
\text { Given } 17\left(x^{2}+y^{2}\right)-30 x y-16=0 \\
\Rightarrow(x+y)^{2}+16(x-y)^{2}=16 \\
\Rightarrow\left(\frac{x+y}{4}\right)^{2}+(x-y)^{2}=1 \text {. } \\
\text { Let }\left\{\begin{array}{l}
x+y=4 \cos \theta, \\
x-y=\sin \theta
\end{array}(\theta \in \mathbf{R})\right. \text {. The... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,240 |
Example 3 A stack of cards consists of $n$ cards, numbered from top to bottom as $1,2, \cdots, n$. Now perform the following operation: First, discard the top card, then place the next card at the bottom, then discard the top card of the current hand, and place the next card at the bottom, and so on, until only one car... | (1) First, consider a simple case.
When $n$ is even and each card is moved once (either discarded or placed at the bottom) in sequence, obviously, after such a round of operations, $\frac{n}{2}$ cards are discarded, and the bottom card among the remaining $\frac{n}{2}$ cards is the last one placed at the bottom, which ... | a_{n}=2 n-2^{-\left[-\log _{2} n\right]} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,241 |
7. If $2^{a}+2^{b}=2^{a+b}, 2^{a}+2^{b}+2^{c}=2^{a+b+c}$, then the maximum value of $2^{c}$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 7. $\frac{4}{3}$.
From $2^{a+b}=2^{a}+2^{b} \geqslant 2 \times 2^{\frac{a+b}{2}}=2^{1+\frac{a+b}{2}}$ $\Rightarrow a+b \geqslant 1+\frac{a+b}{2} \Rightarrow a+b \geqslant 2$.
And $2^{a+b}+2^{c}=2^{a+b+c}$, so $2^{c}=\frac{2^{a+b}}{2^{a+b}-1}=1+\frac{1}{2^{a+b}-1} \leqslant 1+\frac{1}{2^{2}-1}=\frac{4}{3}$. The equality... | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,242 |
8. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that $A B=$ $A A_{1}=4, A D=3$. Then the distance between the skew lines $A_{1} D$ and $B_{1} D_{1}$ is $\qquad$ | 8. $\frac{6 \sqrt{34}}{17}$.
From $B_{1} D_{1} / / B D$, we know $B_{1} D_{1} / /$ plane $A_{1} B D$. Therefore, the distance from line $B_{1} D_{1}$ to $A_{1} D$ is equal to the distance from $B_{1} D_{1}$ to plane $A_{1} B D$, denoted as $h$.
Given that $A_{1} D=B D=5, A_{1} B=4 \sqrt{2}$. Then
Thus, $h=\frac{6 \sqr... | \frac{6 \sqrt{34}}{17} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,243 |
9. Given the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
with an eccentricity of $\frac{\sqrt{3}}{2}$, a line with a slope of 1 passing through the point $M(b, 0)$ intersects the ellipse at points $A$ and $B$. Let $O$ be the origin. If
$$
\overrightarrow{O A} \cdot \overrightarrow{O B}=\frac{32}{5} ... | 9. $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$.
From $e=\frac{\sqrt{3}}{2}=\frac{c}{a}$, we know $a=2 b, c=\sqrt{3} b$.
From $\left\{\begin{array}{l}y=x-b, \\ x^{2}+4 y^{2}=4 b^{2},\end{array}\right.$ we get $B(0,-b), A\left(\frac{8 b}{5}, \frac{3 b}{5}\right)$.
Also, $\angle A O B=\frac{\pi}{2}+\angle A O X$, so $\cot \angl... | \frac{x^{2}}{16}+\frac{y^{2}}{4}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,244 |
10. Place 11 identical balls into six distinct boxes so that at most three boxes are empty. The number of ways to do this is $\qquad$. | 10. 4212 .
If there are no empty boxes, there are $\mathrm{C}_{10}^{5}$ ways to place them; if there is one empty box, there are $\mathrm{C}_{6}^{1} \mathrm{C}_{10}^{4}$ ways to place them; if there are two empty boxes, there are $\mathrm{C}_{6}^{2} \mathrm{C}_{10}^{3}$ ways to place them; if there are three empty box... | 4212 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,245 |
11. Given the function
$$
f(x)=\left\{\begin{array}{ll}
2^{x}-1, & x \leqslant 0 ; \\
f(x-1)+1, & x>0 .
\end{array}\right.
$$
Let the sum of all real roots of the equation $f(x)=x$ in the interval $(0, n]$ be $S_{n}$. Then the sum of the first $n$ terms of the sequence $\left\{\frac{1}{S_{n}}\right\}$ is $T_{n}=$ $\qq... | 11. $\frac{2 n}{n+1}$.
When $x \in(-1,0]$, we have
$$
(f(x)-x)^{\prime}=2^{x} \ln 2-10$$. Noting that $f(x)=f(x-1)+1$, then on $(k-1, k]$ $f(x)=x$ has only one real root $x=k$. Therefore, $f(x)=x$ has all the real roots $1,2, \cdots, n$ in the interval $(0, n]$, and
$$
\begin{array}{l}
S_{n}=\frac{n(n+1)}{2} \Rightarr... | \frac{2 n}{n+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,246 |
12. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=6, a_{n}-2 a_{n-1}=\frac{2 a_{n-1}}{n}+n+1(n \geqslant 2) \text {. }
$$
Then the general term formula of this sequence $a_{n}=$ | 12. $(n+1)\left(2^{n+1}-1\right)$.
From the given condition, we have $\frac{a_{n}}{n+1}=2 \cdot \frac{a_{n-1}}{n}+1(n \geqslant 2)$.
Let $u_{n}=\frac{a_{n}}{n+1}$. Then
$$
u_{n}=2 u_{n-1}+1 \Rightarrow u_{n}+1=2\left(u_{n-1}+1\right)(n \geqslant 2) \text {. }
$$
Thus, $u_{n}+1=2^{n-1}\left(u_{1}+1\right), u_{1}=\frac... | (n+1)\left(2^{n+1}-1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,247 |
13. Let the equation $x^{2}-m x-1=0$ have two real roots $\alpha, \beta (\alpha<\beta)$, and the function $f(x)=\frac{2 x-m}{x^{2}+1}$.
(1) Find the value of $\alpha f(\alpha)+\beta f(\beta)$;
(2) Determine the monotonicity of $f(x)$ in the interval $(\alpha, \beta)$, and provide a proof;
(3) If $\lambda, \mu$ are both... | Three, 13. (1) Given that $\alpha, \beta$ are the roots of the equation $x^{2}-m x-1=0$, we know
$$
\begin{array}{l}
\alpha+\beta=m, \alpha \beta=-1 . \\
\text { Then } f(\alpha)=\frac{2 \alpha-m}{\alpha^{2}+1}=\frac{2 \alpha-(\alpha+\beta)}{\alpha^{2}-\alpha \beta} \\
=\frac{\alpha-\beta}{\alpha(\alpha-\beta)}=\frac{1... | 2 | Algebra | proof | Yes | Yes | cn_contest | false | 727,248 |
14. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=3, a_{n+1}=a_{n}^{2}-n a_{n}+\lambda\left(n \in \mathbf{Z}_{+}, \lambda \in \mathbf{R}\right) \text {. }
$$
(1) If $a_{n} \geqslant 2 n$ always holds, find the range of $\lambda$;
(2) If $\lambda=-2$, prove:
$$
\frac{1}{a_{1}-2}+\frac{1}{a_{2}-2}+\cdots+\... | 14. (1) When $n=2$, we have
$$
a_{2}=6+\lambda \geqslant 2 \times 2 \Rightarrow \lambda \geqslant-2 \text {. }
$$
Next, we prove that when $\lambda \geqslant-2$, we have $a_{n} \geqslant 2 n$.
When $n=2$, it is clearly true.
Assume that when $n=k(k \geqslant 2)$, $a_{k} \geqslant 2 k$ holds.
Then when $n=k+1$,
$$
\beg... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,249 |
15. As shown in Figure 1, in the acute triangle $\triangle ABC$, it is known that $AB < AC$, and points $D$ and $E$ are on side $BC$, satisfying $BD = CE$. If there exists a point $P$ inside $\triangle ABC$ such that $PD \parallel AE$ and $\angle PAB = \angle EAC$, prove: $\angle PBA = \angle PCA$. | 15. As shown in Figure 2, construct $\square A B F C$ and $\square A B G P$.
Then $A C=F B, \angle A C E=\angle F B D$.
Also, $B D=C E$, thus,
$\triangle A E C \cong \triangle F D B \Rightarrow \angle B D F=\angle A E C$.
Therefore, $F D \parallel A E$.
Since $P D \parallel A E$, points $P, D, F$ are collinear.
Hence ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,250 |
16. Let $P$ be a moving point on the circle $C_{1}: x^{2}+y^{2}=2$, and let $Q$ be the foot of the perpendicular from $P$ to the $x$-axis. Point $M$ satisfies $\sqrt{2} \overrightarrow{M Q}=\overrightarrow{P Q}$.
(1) Find the equation of the trajectory $C_{2}$ of point $M$;
(2) Draw two tangents from a point $T$ on the... | 16. (1) Let point $M(x, y)$.
From $\sqrt{2} \overrightarrow{M Q}=\overrightarrow{P Q}$, we know point $P(x, \sqrt{2} y)$.
Since point $P$ is on the circle $C_{1}: x^{2}+y^{2}=2$, thus, $x^{2}+2 y^{2}=2$,
which means the trajectory equation of point $M$ is $\frac{x^{2}}{2}+y^{2}=1$.
(2) Let point $T(2, t)$. Then the eq... | \left[\frac{\sqrt{2}}{2}, 1\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,251 |
Example 4: There are 2013 balls placed around a circle, numbered $1, 2, \cdots, 2013$ in a clockwise direction. Starting from a certain ball $a$, and then taking every other ball in a clockwise direction, this process continues until only one ball remains on the circle. For what value of the number of ball $a$ will the... | First, consider a special case.
When the ball-picking procedure starts with the ball numbered 1, then when the person picking the balls reaches the next ball to be taken (numbered 3), it is essentially as if the ball numbered 2 has been placed at the farthest point in the direction of his movement. And when he takes th... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,252 |
1. The number of positive integers less than $\sqrt[3]{7+5 \sqrt{2}}-\sqrt[3]{7-5 \sqrt{2}}$ is ( ) .
(A) 1
(B) 2
(C) 3
(D) 4 | - 1. B.
Let $x=\sqrt[3]{7+5 \sqrt{2}}-\sqrt[3]{7-5 \sqrt{2}}$. Then
$$
\begin{array}{l}
x^{3}=10 \sqrt{2}-3 x \sqrt[3]{7^{2}-(5 \sqrt{2})^{2}} \\
\Rightarrow x^{3}-3 x-10 \sqrt{2}=0 .
\end{array}
$$
Let $x=\sqrt{2} y$. Then
$$
\begin{array}{l}
2 y^{3}-3 y-10=0 \\
\Rightarrow(y-2)\left(2 y^{2}+4 y+5\right)=0 \\
\Right... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,253 |
2. For a positive integer $n$, let $1 \times 2 \times \cdots \times n=n$!. If $M=1!\times 2!\times \cdots \times 10!$, then the number of perfect cubes among the positive divisors of $M$ is ( ).
(A) 468
(B) 684
(C) 846
(D) 648 | 2. A.
Notice that,
$$
\begin{array}{l}
M=1! \times 2! \times \cdots \times 10! \\
=2^{9} \times 3^{8} \times 4^{7} \times 5^{6} \times 6^{5} \times 7^{4} \times 8^{3} \times 9^{2} \times 10 \\
=2^{9+2 \times 7+5+3 \times 3+1} \times 3^{8+5+2 \times 2} \times 5^{6+1} \times 7^{4} \\
=2^{38} \times 3^{17} \times 5^{7} \... | 468 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,254 |
3. Calculate:
$$
\begin{aligned}
A= & \frac{1-2}{1^{2}-2^{2}}+\frac{1-2+3}{1^{2}-2^{2}+3^{2}}+\frac{1-2+3-4}{1^{2}-2^{2}+3^{2}-4^{2}}+ \\
& \cdots+\frac{1-2+3-4+\cdots+9}{1^{2}-2^{2}+3^{2}-4^{2}+\cdots+9^{2}}=(\quad) .
\end{aligned}
$$
(A) $\frac{1301}{1260}$
(B) $\frac{3349}{1260}$
(C) $\frac{3347}{1260}$
(D) $\frac{1... | 3. B.
Notice,
$$
\begin{array}{l}
\frac{1-2+3-\cdots-2 k}{1-2^{2}+3^{2}-\cdots-(2 k)^{2}} \\
=\frac{-k}{-(1+2+\cdots+2 k)}=\frac{1}{2 k+1}, \\
\frac{1-2+3-\cdots-2 k+(2 k+1)}{1-2^{2}+3^{2}-\cdots-(2 k)^{2}+(2 k+1)^{2}} \\
=\frac{1+k}{1+[2+3+\cdots+(2 k+1)]}=\frac{1}{2 k+1} . \\
\text { Therefore, } A=\frac{1}{3}+\frac... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,255 |
4. Equation
$$
\left(x^{3}-3 x^{2}+4 x-3\right)\left(x^{3}+2 x^{2}-x-1\right)+5 x^{2}-5 x+1=0
$$
The distinct real roots of the equation are ( ).
(A) $1,-1$
(B) $1,-1,-2$
(C) $1,-1,-2,2$
(D) None of the above | 4. B.
Let $x^{3}-3 x^{2}+4 x-3=A, x^{3}+2 x^{2}-x-1=B$.
It is easy to see that $5 x^{2}-5 x+1=(B-A)-1$.
Therefore, the original equation can be transformed into
$$
A B+(B-A)-1=0 \Rightarrow(A+1)(B-1)=0 \text {. }
$$
If $A+1=0$, then
$$
\begin{array}{l}
x^{3}-3 x^{2}+4 x-2=0 \\
\Rightarrow(x-1)\left(x^{2}-2 x+2\right)... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,256 |
5. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
\frac{2 x+5}{3}-t>5 \\
\frac{x+3}{2}-t>x
\end{array}\right.
$$
has exactly three integer solutions. Then the range of $t$ is ( ).
(A) $-\frac{12}{7} \leqslant t<-\frac{8}{7}$
(B) $-\frac{12}{7} \leqslant t<-\frac{3}{2}$
(C) $-\frac{3}{2} \leqslan... | 5. C.
Solving the system of inequalities yields $\frac{3}{2} t+5<x<3-2 t$.
Since the system of inequalities has exactly three integer solutions, we have
$$
\begin{array}{l}
2<(3-2 t)-\left(\frac{3}{2} t+5\right) \leqslant 4 \\
\Rightarrow-\frac{12}{7} \leqslant t<-\frac{8}{7} .
\end{array}
$$
When $-\frac{12}{7} \leq... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 727,257 |
6. As shown in Figure 1, in rhombus $A B C D$, let $A E \perp B C$ at point $E, \cos B=\frac{4}{5}, E C=2$, and $P$ is a moving point on side $A B$. Then the minimum value of $P E+P C$ is ( ).
(A) $\frac{2 \sqrt{741}}{5}$
(B) $\frac{2 \sqrt{743}}{5}$
(C) $\frac{2 \sqrt{745}}{5}$
(D) $\frac{2 \sqrt{747}}{5}$ | 6. C.
Let $AB = x$. Then $BE = x - 2$.
In the right triangle $\triangle ABE$,
$$
\cos B = \frac{BE}{AB} = \frac{x-2}{x} = \frac{4}{5} \Rightarrow x = 10.
$$
By the Pythagorean theorem, it is easy to know that $AE = 6$.
As shown in Figure 5, construct the symmetric point $C'$ of point $C$ with respect to $AB$, and con... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,258 |
1. Given non-negative real numbers $x, y, z$ satisfy
$$
x+2 y+3 z=3, \quad 3 x+2 y+z=4 \text {. }
$$
Then the maximum value of $M=3 x^{2}-4 y^{2}+5 z^{2}$ is $\qquad$, and the minimum value is $\qquad$. | 2. 1. $\frac{23}{4},-\frac{11}{2}$.
Then $M=3 x^{2}-4 y^{2}+5 z^{2}$
$=3\left(\frac{1}{2}+z\right)^{2}-4\left(\frac{5}{4}-2 z\right)^{2}+5 z^{2}$
$=-8 z^{2}+23 z-\frac{11}{2}$
$=-8\left(z-\frac{23}{16}\right)^{2}+\frac{353}{32}$.
Since $x$, $y$, and $z$ are all non-negative real numbers, we have
$$
\left\{\begin{array... | \frac{23}{4},-\frac{11}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,259 |
2. The number of all positive integer solutions $(x, y, z)$ for the equation $x y+1000 z=2012$ is. $\qquad$ | 2. 18 .
First consider the positive integer solutions for $x \leqslant y$.
When $z=1$, $x y=1012=4 \times 253=2^{2} \times 11 \times 23$. Hence, $(x, y)=(1,1012),(2,506),(4,253)$,
$$
(11,92),(22,46),(23,44) \text {. }
$$
When $z=2$, $x y=12=2^{2} \times 3$. Hence, $(x, y)=(1,12),(2,6),(3,4)$.
Next, consider the posit... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,260 |
3. Xiao Li and Xiao Zhang are running at a constant speed on a circular track. They start from the same place at the same time. Xiao Li runs clockwise and completes a lap every 72 seconds, while Xiao Zhang runs counterclockwise and completes a lap every 80 seconds. At the start, Xiao Li has a relay baton, and each time... | 3. 720 .
The time taken from the start to the first meeting is
$$
\frac{1}{\frac{1}{72}+\frac{1}{80}}=\frac{720}{19}
$$
seconds, during which Xiao Li runs $\frac{10}{19}$ laps, and Xiao Zhang runs $\frac{9}{19}$ laps.
Divide the circular track into 19 equal parts, and number the points clockwise from the starting poin... | 720 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 727,261 |
Example 1: At each vertex of a regular 2009-gon, a non-negative integer not exceeding 100 is placed. Adding 1 to the numbers at two adjacent vertices is called an operation on these two adjacent vertices. For any given two adjacent vertices, the operation can be performed at most $k$ times. Find the minimum value of $k... | Solve $k_{\min }=100400$.
Let the numbers at vertices $A_{1}, A_{2}, \cdots, A_{2009}$ be $a_{1}, a_{2}, \cdots, a_{2009}$ respectively.
First, take $a_{2}=a_{4}=a_{2008}=100, a_{1}=a_{3}=a_{2009}=0$.
For each operation, assign
$$
S=\left(a_{2}-a_{3}\right)+\left(a_{4}-a_{5}\right)+\cdots+\left(a_{2008}-a_{2009}\right)... | 100400 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,263 |
One. (20 points) As shown in Figure 3, given an acute triangle $\triangle ABC$, the perpendicular from vertex $C$ to side $AB$ is $CD$, with foot $D$. Point $E$ is on segment $CD$. Draw perpendiculars from point $D$ to lines $AC$, $AE$, $BE$, and $BC$, with feet $P$, $Q$, $R$, and $S$ respectively.
Prove: (1) $C, P, D... | (1) From $D P \perp A C, D S \perp B C$, we know
$\angle C P D=\angle C S D=90^{\circ}$.
Therefore, points $C, P, D, S$ are concyclic.
Similarly, points $E, Q, D, R$ are concyclic.
(2) Similarly to (1), points $P, A, D, Q$ are concyclic. Thus, $\angle C P Q=\angle A D Q$.
Also, $\angle A Q D=\angle A D E=90^{\circ}$, s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,264 |
II. (25 points) As shown in Figure 4, in the Cartesian coordinate system $x O y$, the vertices of the polygon $O A B C D E$ are
$$
\begin{array}{l}
O(0,0), A(0,6), B(4,6), \\
C(4,4), D(6,4), E(6,0) .
\end{array}
$$
It is known that the line $l$ passes through point $M$, intersects sides $O A$ and $D E$, and divides th... | II. As shown in Figure 7, extend $BC$ to intersect the $x$-axis at point $F$, connect $OB$ and $AF$ to intersect at point $P$, and connect $CE$ and $DF$ to intersect at point $Q$.
It is easy to see that $P(2,3)$ and $Q(5,2)$ are the centers of rectangles $OFBA$ and $FEDC$, respectively.
Therefore, a line through poin... | y=-\frac{1}{3} x+\frac{11}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,265 |
Three. (25 points) Given that $n$ is a positive integer. If in any permutation of $1,2, \cdots, n$, there always exists a sum of six consecutive numbers greater than 2013, then $n$ is called a "sufficient number". Find the smallest sufficient number.
---
Please note that the term "温饱数" is translated as "sufficient nu... | Three, the smallest well-fed number is 671.
First, prove: 671 is a well-fed number.
For any permutation $b_{1}, b_{2}, \cdots, b_{671}$ of $1,2, \cdots, 671$, divide it into 112 groups, where each group contains six numbers, and the last two groups share one number, i.e.,
$$
\begin{array}{l}
A_{1}=\left(b_{1}, b_{2}, \... | 671 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,266 |
2. From $1,2, \cdots, 100$ choose three different numbers such that they cannot form the three sides of a triangle. The number of different ways to do this is. | 2. 82075.
Let these three numbers be $i, j, k(1 \leqslant i<j<k \leqslant 100)$. Then $k \geqslant i+j$.
Therefore, the number of different ways to choose these three numbers is
$$
\begin{array}{l}
\sum_{i=1}^{100}\left[\sum_{j=i+1}^{100}\left(\sum_{k=i+j}^{100} 1\right)\right]=\sum_{i=1}^{100}\left[\sum_{j=i+1}^{100-... | 82075 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,268 |
3. Given the set
$$
A=\left\{x \mid x=a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}\right\} \text {, }
$$
where, $a_{i} \in\{1,2, \cdots, 6\}(i=0,1,2,3)$. If positive integers $m, n \in A$, and $m+n=2014(m>n)$, then the number of pairs of positive integers $(m, n)$ that satisfy the condition is $\qquad$.
| 3. 551.
Notice that, $2014=5 \times 7^{3}+6 \times 7^{2}+5$.
Given $m, n \in A$, let
$$
\begin{array}{l}
m=a \times 7^{3}+b \times 7^{2}+c \times 7+d, \\
n=a^{\prime} \times 7^{3}+b^{\prime} \times 7^{2}+d^{\prime},
\end{array}
$$
where $a, b, c, d, a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime} \in\{1,2, \cdots, 6\}... | 551 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,269 |
4. As shown in Figure 2, let $P$ and $Q$ be moving points on two concentric circles (with radii 6 and 4, respectively). When $P$ and $Q$ move along the circles, the area of the region formed by the midpoint $M$ of line segment $PQ$ is $\qquad$ | 4. $24 \pi$.
Draw $O H \perp P Q$ through the center $O$, with the foot of the perpendicular at $H$. Let $O P=R$, $O Q=r, O M=m, O H=h$. Then
$$
\begin{array}{l}
P M=\sqrt{R^{2}-h^{2}}-\sqrt{m^{2}-h^{2}}, \\
Q M=\sqrt{r^{2}-h^{2}}+\sqrt{m^{2}-h^{2}} .
\end{array}
$$
Since $M$ is the midpoint of $P Q$, we have
$$
\beg... | 24 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,270 |
5. Function
$$
\begin{aligned}
f(x, y)= & x^{4}+y^{4}+2 x^{2} y-2 x y^{2}+3 x^{2}+ \\
& 3 y^{2}-2 x+2 y+2
\end{aligned}
$$
The minimum value of the function is | 5. $\frac{9}{8}$.
Notice,
$$
\begin{array}{l}
f(x, y) \\
= x^{4}+y^{4}+2 x^{2} y-2 x y^{2}+3 x^{2}+3 y^{2}-2 x+2 y+2 \\
= x^{4}+(2 y+3) x^{2}-2\left(y^{2}+1\right) x+y^{4}+3 y^{2}+2 y+2 \\
= x^{4}+2(y+1) x^{2}+(y+1)^{2}+y^{4}- \\
2(x-1) y^{2}+(x-1)^{2} \\
=\left(x^{2}+y+1\right)^{2}+\left(y^{2}+1-x\right)^{2} .
\end{... | \frac{9}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,271 |
6. Calculate:
$$
\sum_{x=1}^{9} C_{9}^{x}\left[\sum_{y=0}^{9-x} C_{9-x}^{y}\left(\sum_{z=0}^{9-x-y} C_{9-x-y}^{z}\right)\right]=
$$
$\qquad$ | 6. $4^{9}-3^{9}$.
Consider the polynomial
$$
\begin{array}{l}
(a+b+c+d)^{9} \\
=\sum_{x=0}^{9}\left[\sum_{y=0}^{9-x}\left(\sum_{z=0}^{9-x-y} \mathrm{C}_{9}^{x} \mathrm{C}_{9-x}^{y} \mathrm{C}_{9-x-y}^{z} a^{x} b^{y} c^{z} d^{9-x-y-z}\right)\right] .
\end{array}
$$
Let $a=b=c=d=1$. Then
$$
\sum_{x=0}^{9} \mathrm{C}_{9... | 4^{9}-3^{9} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,272 |
7. In the tetrahedron $O-ABC$, it is known that
$$
OA=OB=OC=\sqrt{3}, AC=2\sqrt{2}, AB=2,
$$
and $OB \perp AC$. A sphere with radius 1 is constructed with $O$ as the center. Then the volume of the part of the tetrahedron $O-ABC$ that is not inside the sphere is $\qquad$ | 7. $\frac{2}{3}-\frac{\pi}{9}$.
Consider a cube with edge length 2. Place point $O$ at the center of the cube, and points $A$, $B$, and $C$ at three vertices of the cube, all on the same face.
Therefore, the part where the sphere intersects with the tetrahedron is $\frac{1}{12}$ of the sphere.
Thus, the required volum... | \frac{2}{3}-\frac{\pi}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,273 |
Example 2 Color all the small squares of an $m \times n$ chessboard (consisting of $m$ rows and $n$ columns of squares, $m \geqslant 3, n \geqslant 3$) with one of two colors, red or blue. If two adjacent (sharing a common edge) small squares are of different colors, then these two small squares are called a “standard ... | Divide all the squares into three categories: the first category of squares are located at the four corners of the chessboard, the second category of squares are located on the boundary of the chessboard (excluding the four corners), and the remaining squares are the third category.
Fill all the red squares with the n... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,274 |
8. Roll a die three times, the obtained points are $m$, $n$, $p$. Then the function $y=\frac{2}{3} m x^{3}-\frac{n}{2} x^{2}-p x+1$ is an increasing function on $[1,+\infty)$ with a probability of $\qquad$ | 8. $\frac{11}{24}$.
Notice,
$$
f(x)=\frac{2}{3} m x^{3}-\frac{n}{2} x^{2}-p x+1
$$
being an increasing function on $[1,+\infty)$ is equivalent to
$$
f^{\prime}(x)=2 m x^{2}-n x-p>0
$$
holding for all $x$ in $[1,+\infty)$, which is equivalent to
$f^{\prime}(1)>0$, i.e., $2 m>n+p$.
When $m=2$, $n+p \leqslant 3$, there... | \frac{11}{24} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,275 |
9. (16 points) Given the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$. Find the real number pair $(a, b)$, such that for any line $l_{a}$ with slope $a$ intersecting the ellipse at points $A$ and $B$, and the line $x=b$ intersecting the upper half of the ellipse at point $P$, the triangle $\triangle P A B$ always has i... | $$
\begin{array}{l}
\text { Let points } A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right), P\left(x_{P}, y_{P}\right), \text { and } l_{a}: y=a x+m.
\text { Substituting the equation of line } l_{a} \text { into the ellipse equation, we get } \\
x^{2}+9(a x+m)^{2}-36=0 \\
\Rightarrow\left(9 a^{2}+1\right) x^{2}+1... | (a, b)=\left(a, \frac{18 a}{\sqrt{9 a^{2}+1}}\right)(a \in \mathbf{R}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,276 |
10. (20 points) The numbers $1, 2, \cdots, 2014$ are written on a blackboard. The following operation is performed: in the first step, the first two numbers $1$ and $2$ are erased, and their sum $3$ is written at the end of the sequence; in the second step, the first three numbers $3$, $4$, and $5$ are erased, and thei... | 10. Since after the $k$-th step, the number on the blackboard decreases by $k$, hence
$$
\begin{array}{l}
1+2+\cdots+(t-1)2014 .
\end{array}
$$
Therefore, the numbers written in the first 14 steps are all less than 2014, and the numbers written afterward are all greater than 2014.
The numbers erased in the 62nd step ... | 2062 \text{ (numbers)}, 4051073 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,277 |
11. (20 points) In order to prevent the Decepticons from getting the Allspark, Optimus Prime plans to destroy it with a laser blade. Optimus Prime's method is: each cut can divide the Allspark into two cuboids with a volume ratio of $2: 7$, each cuboid exactly containing one face of the Allspark, and any two cuts resul... | 11. Clearly, each section obtained by cutting is a parallelogram, symmetric about its center $I$.
According to the operation rules, point $I$ lies on the line segment connecting the centers of a pair of opposite faces of the cube, and divides this line segment into a ratio of $2:7$.
The above $I$ is a fixed point, an... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 727,278 |
One, (40 points) As shown in Figure $3, AB$ is a tangent of $\odot O$, satisfying $BD \perp AO, AB$ intersects the radius $OC$ of $\odot O$ at point $E, K$ is a point on line segment $AE$, and $AL \parallel OK$ intersects $CK$ at point $L$. Prove: $CK=KL$ if and only if $CK$ is tangent to $\odot O$. | Given $K L$ intersects $O A$ at point $M$ as shown in Figure 4.
Since $A L / / O K$, we have
$$
\frac{K M}{M L}=\frac{O M}{M A} \Rightarrow \frac{K M}{K L}=\frac{O M}{O A} \text {. }
$$
For $\triangle O C M$ and the secant $A K E$, by Menelaus' theorem, we know
$$
\frac{C K}{K M} \cdot \frac{M A}{A O} \cdot \frac{O E}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,279 |
$$
\begin{array}{l}
\text { II. (40 points) Given } x_{i} \in\{\sqrt{2}-1, \sqrt{2}+1\} \\
(i=1,2, \cdots, 2013) \text {, let } \\
S=x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{2012} x_{2013} .
\end{array}
$$
Find the number of different integer values that $S$ can take. | Because $(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$,
$$
(\sqrt{2}+1)^{2}=3+2 \sqrt{2}, (\sqrt{2}-1)(\sqrt{2}+1)=1,
$$
Therefore, $x_{i} x_{i+1} \in\{3-2 \sqrt{2}, 3+2 \sqrt{2}, 1\}$.
Let the sum $S$ contain $a$ instances of $3+2 \sqrt{2}$, $b$ instances of $3-2 \sqrt{2}$, and $c$ instances of 1. Then $a, b, c \in \mathbf{N}$, and... | 1005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,280 |
Three, (50 points) Given a positive integer $n$ satisfying
$$
\begin{array}{l}
n>2014, (n, 2014)=1. \\
\text { Let } A_{n}=\{k \in \mathbf{N} \mid 1 \leqslant k \leqslant n, (n, k)=1\}, \\
B_{n}=\left\{k \in A_{n} \mid k+1 \notin A_{n}\right\}, \\
C_{n}=\left\{k \in A_{n} \mid k-1 \notin A_{n}\right\}.
\end{array}
$$
... | Obviously, $n$ is odd, and $|A_{n}|=\varphi(n)$ is even, where $\varphi(n)$ is the Euler's totient function.
By Euler's theorem, for any $k \in A_{n}$, we have
$$
k^{\varphi(n)} \equiv 1(\bmod n) \Rightarrow S_{k}=\left[\frac{k^{\varphi(n)}}{n}\right]=\frac{k^{\varphi(n)}-1}{n} \text {. }
$$
Thus, for any subset $A$ o... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,281 |
Four. (50 points) A country has built a time machine, which resembles a circular subway track, with 2014 stations (numbered $1, 2, \cdots, 2014$) evenly distributed along it, each corresponding to a year. The starting and ending stations are both the 1st station (corresponding to the year 2014). To save costs, the mach... | Let $2 n=2014$. Suppose there are $f(n)$ different ways of stopping.
First, for each way of stopping, define the stations where the machine stops as class $a$ stations, and the stations where it does not stop as class $b$ stations. Then, class $a$ stations and class $b$ stations are paired one-to-one, forming diametri... | \frac{1}{2}\left[1+\left(\frac{1+\sqrt{5}}{2}\right)^{1007}+\left(\frac{1-\sqrt{5}}{2}\right)^{1007}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,282 |
Given $x 、 y 、 z \in[0,1]$. Find
$$
A=\sqrt{3|x-y|}-\sqrt{|y-z|}-\sqrt{3|z-x|}
$$
the maximum and minimum values. | Prove (1) If $0 \leqslant x \leqslant y \leqslant z \leqslant 1$, then
$$
A=\sqrt{3(y-x)}-\sqrt{z-y}-\sqrt{3(z-x)} .
$$
Notice that, $(y-x)+(z-y)=z-x$.
Let $a^{2}=y-x, b^{2}=z-y, c^{2}=z-x$. Then $a^{2}+b^{2}=c^{2}(a, b, c \in[0,1])$.
Let $a=c \cos \theta, b=c \sin \theta\left(\theta \in\left[0, \frac{\pi}{2}\right]\r... | A_{\text {max }}=\sqrt{3}-1, A_{\text {min }}=-\sqrt{3}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,283 |
Given $A B 、 C D$ are two perpendicular chords of $\odot O$, $P$ is any point on the minor arc $\overparen{B D}$, $P H \perp A D$ at point $H$, $P H$ intersects $C D$ and $P B$ intersects $D C$ at points $F 、 E$ respectively. Prove:
$$
P H^{2}-F H^{2}=C F \cdot F E \text {. }
$$ | Prove as shown in Figure 2, let $P^{\prime}$ be the reflection of $P$ over $AD$, $K$ be the reflection of $P$ over $AD$, $G$ be the projection of $P$ onto $CD$, and $H_{ACD}$ be the orthocenter of $\triangle ACD$. Connect $P^{\prime}H_{ACD}$, $KE$, $KD$, $GP$, and $GH$.
Then $P^{\prime}H_{ACD}$ is the Steiner line of ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,284 |
Example 1 As shown in Figure 5, let $D, E, F$ be the midpoints of the three sides of $\triangle ABC$, $AD', BE', CF'$ be the three altitudes, and the corresponding sides of $\triangle DEF$ and $\triangle D'E'F'$ intersect at points $A', B', C'$. Prove: $AA' \parallel BB' \parallel CC'$, and the direction of the paralle... | Note that, the circumcenter $O$ and the orthocenter $H$ are a pair of isogonal conjugates in a triangle.
This problem can be simplified to proving the following equivalent proposition in each angle:
As shown in Figure 6, let $P$ and $Q$ be the isogonal points of $\angle AOB$. Draw perpendiculars from points $P$ and $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,285 |
Example 2 Let $P$ be any point inside $\triangle ABC$, and $O, O_A, O_B, O_C$ be the circumcenters of $\triangle ABC, \triangle PBC, \triangle PCA, \triangle PAB$ respectively. Prove that $O, P$ are isogonal conjugates with respect to $\triangle O_A O_B O_C$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 2 Let $P$ be any point ins... | Prove as shown in Figure 7, connect $O_{A} B, O_{A} P, O_{A} C$.
Then $O_{A} B=O_{A} P=O_{A} C$.
Since points $O_{A}, O$ are both on the perpendicular bisector of $B C$, therefore, $O_{A} O$ bisects $\angle B O_{A} C$.
Let $\angle P O_{A} O_{C}=\alpha, \angle P O_{A} O_{B}=\beta, \angle O O_{A} O_{B}=\alpha^{\prime}$.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,286 |
Example 3 As shown in Figure 3, given that the incircle of $\triangle ABC$ touches $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively, and line segments $BE$, $CF$ intersect the incircle again at points $P$, $Q$. If line $FE$ intersects $BC$ at a point $R$ outside the circle, prove that points $P$, $Q$, $R$ are coll... | By the tangent-secant theorem, we know that $A E = A F$.
For $\triangle A B C$ and the transversal $E F R$, applying Menelaus' theorem, we get
$$
\text {, } \frac{A F}{F B} \cdot \frac{B R}{R C} \cdot \frac{C E}{E A}=1 .
$$
Therefore, $\frac{B R}{R C}=\frac{E A}{C E} \cdot \frac{F B}{A F}=\frac{F B}{C E}$.
As shown in... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,287 |
In a 376-dimensional space, for a finite set of points $M$ not all lying on the same plane, the following property holds: for any $A, B \in M$, there exist $C, D \in M$ such that the lines $AB$ and $CD$ are two distinct parallel lines. Prove: for any positive integer $n \geqslant 10$, there always exists a set $M$ cont... | First, provide the construction when $|M|=10$.
Take the 8 vertices of a cube, then take the points symmetric to the center of the cube about the top and bottom faces, for a total of $8+2=10$ points. It is easy to prove that any two points chosen from these 10 points can find another two points that satisfy the conditio... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,288 |
Example 4 As shown in Figure 4, let $D$, $E$, and $F$ be points on the sides $AB$, $BC$, and $CA$ of $\triangle ABC$, respectively, such that quadrilateral $CFDE$ is a parallelogram. $AE$ intersects $DF$ and $BF$ intersects $DE$ at points $P$ and $Q$, respectively, and $AE$ intersects $BF$ at point $R$. Let $S$ be a mo... | Proof: Let $BC=a, AC=b, AB=c, CE=d$.
From the parallelogram $CFDE$, we know
$$
\frac{DE}{AC}=\frac{BE}{BC}=\frac{a-d}{a} \text{. }
$$
Therefore, $CF=DE=\frac{b(a-d)}{a}$.
Applying Menelaus' theorem to $\triangle BDF$ and the transversal $APR$, we have
$$
\frac{FR}{RB} \cdot \frac{BA}{AD} \cdot \frac{DP}{PF}=1 \text{. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,289 |
Example 5 As shown in Figure 5, in the right triangle $\triangle ABC$, it is known that $\angle A = 90^{\circ}, \angle B > \angle C, O$ is the center of the circumcircle of $\triangle ABC$, the lines $l_{A}$ and $l_{B}$ are tangent to $\odot O$ at points $A$ and $B$ respectively, $BC$ intersects with lines $l_{A}$ and ... | Prove as shown in Figure 5, let the extension of $BA$ intersect the tangent of $\odot O$ (passing through point $C$) at point $E'$.
By Pascal's theorem, points $S$, $D$, and $E'$ are collinear.
Thus, point $E'$ coincides with $E$.
By the power of a point theorem, we have
$$
T A^{2}=T R \cdot T Q, S A^{2}=S B \cdot S C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,290 |
Example 6 As shown in Figure 6, in trapezoid $ABCD$, it is known that $BC$ and $AD$ are the upper and lower bases respectively, $F$ is a point on the leg $CD$, $AF$ intersects $BD$ at point $E$, $G$ is a point on side $AB$ such that $EG$ // $AD$, $CG$ intersects $BD$ at point $H$, and $FH$ intersects $AB$ at point $I$.... | Proof As shown in Figure 6, let $A B$ and $D C$, $A F$ and $D G$ intersect at points $S$, $T$ respectively.
First, we prove that $S$, $H$, $T$ are collinear.
Since $E G / / A D, A D / / B C$, we have
$\triangle A T D \backsim \triangle E T G, \triangle G H E \backsim \triangle C H B$,
$\triangle A S D \backsim \triangl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,291 |
A password lock is set by assigning one of the two numbers 0 and 1 to each vertex of a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, and coloring each vertex in one of two colors, red or blue, such that for any two adjacent vertices, at least one of the number or color is the same. How many different password s... | Using 2 to represent red and 3 to represent blue, the problem is equivalent to:
In an $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, label the vertices $A_{i}(i=1,2, \cdots, n)$ with ordered pairs $\left(x_{i}, y_{i}\right)\left(x_{i}=0\right.$ or $1, y_{i}=2$ or 3$)$, such that for any $i(1 \leqslant i \leqslant n)$, $... | 3^{n}+2+(-1)^{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,292 |
For integers $n$ greater than 1, define the set
$$
D(n)=\left\{a-b \mid n=a b, a 、 b \in \mathbf{Z}_{+}, a>b\right\} \text {. }
$$
Prove: For any integer $k$ greater than 1, there always exist $k$ distinct integers $n_{1}, n_{2}, \cdots, n_{k}$, all greater than 1, such that
$$
D\left(n_{1}\right) \cap D\left(n_{2}\ri... | 【Analysis】Let $\{p_{n}\}$ be a strictly increasing sequence of odd prime numbers, where $p_{1}=3, p_{2}=5, p_{3}=7, \cdots$.
For any integer $k$ greater than 1, define
$$
\begin{array}{l}
A_{k}=2^{k} p_{1} p_{2} \cdots p_{k}+1, \\
x_{j}=\frac{A_{k}^{2}-1}{4 p_{j}}-p_{j}-1,
\end{array}
$$
and $\square$
$$
n_{j}=x_{j}\l... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,293 |
Question As shown in Figure 1, two rectangles $R$ and $S$ are placed inside an acute triangle $T$. Let $A(X)$ denote the area of polygon $X$. Find the maximum value of $\frac{A(R)+A(S)}{A(T)}$ or prove that the maximum value does not exist. | As shown in Figure 1, let the heights of rectangles $R$ and $S$ be $a$ and $b$, respectively, the height of the triangle be $h$, and the distance from the vertex opposite the base to rectangle $S$ be $c$. Then $a+b+c=h$.
In reference [1], the ratio of areas is expressed using similar triangles as
$$
\frac{A(R)+A(S)}{A(... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,294 |
Proposition 1 As shown in Figure 2, place three rectangles $X_{1}, X_{2}, X_{3}$ in an acute triangle $T$. Let $A(X)$ denote the area of rectangle $X$. Question: Does the maximum value of $\frac{A\left(X_{1}\right)+A\left(X_{2}\right)+A\left(X_{3}\right)}{A(T)}$ exist? If it exists, find the maximum value. | Solve as shown in Figure 2, let the side of rectangle $X_{1}$ that lies along the base of the triangle be denoted as the base of the triangle, with the base of the triangle being $a_{1}$, and the height of the triangle being $h$. The lengths of the sides of rectangles $X_{1}$, $X_{2}$, and $X_{3}$ parallel to the base ... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,295 |
Example 1 Given an acute triangle $\triangle ABC$, $CD$ is the altitude from point $C$, $M$ is the midpoint of side $AB$, a line through $M$ intersects $CA$ and $CB$ at points $K$ and $L$ respectively, and $CK=CL$. If the circumcenter of $\triangle CKL$ is $S$, prove: $SD=SM$. ${ }^{[1]}$
(54th Polish Mathematical Olym... | 【Analysis】This problem has many proofs, but using the Simson line theorem is one of the more concise methods.
Proof As shown in Figure 1, construct the circumcircle of $\triangle A B C$, extend $C S$ to intersect the circumcircle at point $T$, connect $T M$, and draw $T K^{\prime} \perp A C$ at point $K^{\prime}$, $T ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,297 |
Example 3 Given that point $P$ has projections $S_{1}, S_{2}, S_{3}$ on the sides $BC, CA, AB$ of $\triangle ABC$, and projections $T_{1}, T_{2}, T_{3}$ on the altitudes $AD, BE, CF$ respectively. Prove: $S_{1} T_{1}, S_{2} T_{2}, S_{3} T_{3}$ are concurrent. | Simplify the original problem to the following equivalent form:
Point $P$ has projections $S_{1}, S_{2}, S_{3}$ on the external angle bisectors of $\triangle D E F$, and projections $T_{1}, T_{2}, T_{3}$ on the internal angle bisectors. Prove that $S_{1} T_{1}, S_{2} T_{2}, S_{3} T_{3}$ are concurrent.
Proof As shown i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,298 |
Example 2 Prove the Simson line property.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Proof As shown in Figure 2, let $AD$ be the altitude on side $BC$, extend $AD$ to intersect the circumcircle of $\triangle ABC$ at point $F$, connect $HL, PH$, and let $PH, PF$ intersect the Simson line at points $S, Q$ respectively, $PF$ intersects $BC$ at point $G$, and connect $HG$.
Since $P, C, L, M, A, F, C, P$ ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,299 |
Example 3 As shown in Figure 3, from vertex $A$ of $\triangle ABC$, draw perpendiculars to the internal and external angle bisectors of vertices $B$ and $C$, with the feet of the perpendiculars being $F$, $G$, $E$, and $D$. Prove: $F$, $G$, $E$, and $D$ are collinear, and this line coincides with the midline of $\trian... | Proof As shown in Figure 4, it is easy to know that $I$ is the incenter of $\triangle ABC$. Connect $AI$, extend $BF$ and $CG$ to intersect at point $L$, and $BE$ and $CD$ to intersect at point $K$.
Since $BF, BE, CK, CG$ are the external and internal angle bisectors of $\angle ABC$ and $\angle ACB$, respectively, we ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,300 |
Example 4 Draw three parallel lines through the three vertices of $\triangle ABC$, and let their intersections with the circumcircle of $\triangle ABC$ be $A'$, $B'$, and $C'$. Take any point $P$ on the circumcircle of $\triangle ABC$, and let $PA'$, $PB'$, and $PC'$ intersect $BC$, $CA$, and $AB$ or their extensions a... | 【Analysis】When encountering problems of proving points are collinear or intersection points lie on a circle, one should first think of the Simson line theorem or Carnot's theorem.
Proof As shown in Figure 6, let $P B^{\prime}$ intersect $A B$ at point $X$.
Since $B B^{\prime} / / C C^{\prime}$, we have $B^{\prime} C^{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,302 |
2. For integers $n$ greater than 1, define the set
$$
D(n)=\left\{a-b \mid n=a b, a 、 b \in \mathbf{Z}_{+}, a>b\right\} \text {. }
$$
Prove: For any integer $k$ greater than 1, there always exist $k$ distinct integers $n_{1}, n_{2}, \cdots, n_{k}$, all greater than 1, such that
$$
D\left(n_{1}\right) \cap D\left(n_{2}... | 2. Let $a_{1}, a_{2}, \cdots, a_{k+1}$ be $k+1$ distinct positive odd numbers, and any one of these numbers is less than the product of the other $k$ numbers (for example, $a_{1}=3, a_{2}=5, \cdots, a_{k+1}=2 k+3$).
Let $N=a_{1} a_{2} \cdots a_{k+1}$.
For $i=1,2, \cdots, k+1$, take
$$
x_{i}=\frac{1}{2}\left(\frac{N}{a_... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,304 |
3. Prove: There exists a unique function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$ satisfying
$$
\begin{array}{l}
f(1)=f(2)=1, \\
f(n)=f(f(n-1))+f(n-f(n-1))(n \geqslant 3),(1)
\end{array}
$$
and for each integer $m \geqslant 2$, find the value of $f\left(2^{m}\right)$. | 3. From $f(1)=1$, we know $\frac{1}{2} \leqslant f(1) \leqslant 1$.
First, we prove by mathematical induction: For any integer $n>1$, $f(n)$ can be uniquely determined by the values of $f(1), f(2), \cdots, f(n-1)$, and
$$
\frac{n}{2} \leqslant f(n) \leqslant n \text {. }
$$
When $n=2$, $f(2)=1$, the conclusion holds.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,305 |
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