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 |
|---|---|---|---|---|---|---|---|---|---|
2. Let the 20 vertices of a regular 20-sided polygon inscribed in the unit circle in the complex plane correspond to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$. Then the number of distinct points corresponding to $z_{1}^{2015}, z_{2}^{2015}, \cdots, z_{20}^{2015}$ is $\qquad$ | 2. 4 .
Assume the vertices corresponding to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$ are arranged in a counterclockwise direction.
Let $z_{1}=\mathrm{e}^{\mathrm{i} \theta}$. Then
$$
\begin{array}{l}
z_{k+1}=\mathrm{e}^{\mathrm{i}\left(\theta+\frac{k}{10}\right)}(k=0,1, \cdots, 19), \\
z_{k+1}^{2015}=\mathrm... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,043 |
4. Let $|a|=1,|b|=2$. If vector $c$ satisfies
$$
|c-(a+b)|=|a-b|,
$$
then the maximum value of $|\boldsymbol{c}|$ is $\qquad$ | 4. $2 \sqrt{5}$.
$$
\begin{array}{l}
\text { Given }|\boldsymbol{c}-(\boldsymbol{a}+\boldsymbol{b})|^{2}=|a-b|^{2} \\
\Rightarrow(\boldsymbol{c}-2 a)(c-2 b)=0 .
\end{array}
$$
Let the starting point of the vectors be the origin $O, \overrightarrow{O A}=2 a$, $\overrightarrow{O B}=2 \boldsymbol{b}, \overrightarrow{O C}... | 2 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,044 |
5. The function $f(x)$ defined on $\mathbf{R}$, for any real number $x$, satisfies
$$
\begin{array}{l}
f(x+3) \leqslant f(x)+3, \\
f(x+2) \geqslant f(x)+2,
\end{array}
$$
and $f(1)=2$. Let $a_{n}=f(n)\left(n \in \mathbf{Z}_{+}\right)$, then
$$
f(2015)=
$$
$\qquad$ | 5.2016.
Notice,
$$
\begin{array}{l}
f(x)+3 \geqslant f(x+3) \\
=f(x+1+2) \geqslant f(x+1)+2 \\
\Rightarrow f(x)+1 \geqslant f(x+1) . \\
\text { Also } f(x)+4 \leqslant f(x+2)+2 \leqslant f(x+4) \\
=f(x+1+3) \leqslant f(x+1)+3 \\
\Rightarrow f(x+1) \geqslant f(x)+1 .
\end{array}
$$
Therefore, $f(x+1)=f(x)+1$.
Thus, $f... | 2016 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,045 |
6. In the quadrilateral pyramid $P-ABCD$, it is known that $AB // CD$, $AB \perp AD$, $AB=4$, $AD=2\sqrt{2}$, $CD=2$, $PA \perp$ plane $ABCD$, $PA=4$. Let $Q$ be a point on the line segment $PB$, and the sine of the angle formed by line $QC$ and plane $PAC$ is $\frac{\sqrt{3}}{3}$. Then
$\frac{PQ}{PB}$ is $\qquad$ | 6. $\frac{7}{12}$.
From $\frac{C D}{A D}=\frac{A D}{A B} \Rightarrow \triangle A C D \backsim \triangle D B A$
$\Rightarrow \angle D A C+\angle A D B=90^{\circ} \Rightarrow A C \perp B D$.
Since $P A \perp$ plane $A B C D \Rightarrow P A \perp B D$.
Therefore, $B D \perp$ plane $P A C$.
Let $B D$ intersect $A C$ at po... | \frac{7}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,046 |
7. Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=n^{4}+6 n^{3}+11 n^{2}+6 n \text {. }
$$
Then the sum of the first 12 terms $S_{12}=$ $\qquad$ | 7. 104832 .
Notice that,
$$
\begin{array}{l}
a_{n}=n^{4}+6 n^{3}+11 n^{2}+6 n \\
=n(n+1)(n+2)(n+3) . \\
\text { Let } f(n)=\frac{1}{5} n(n+1)(n+2)(n+3)(n+4) .
\end{array}
$$
Then $a_{n}=f(n)-f(n-1), S_{n}=f(n)$.
Therefore, $S_{12}=f(12)=104832$. | 104832 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,047 |
Example 7 Proof:
$$
\sum_{k=1}^{m} \frac{m(m-1) \cdots(m-k+1) k}{m^{k+1}}=1 \text {. }
$$
(2012-2013 Hungarian Mathematical Olympiad (Mathematics Special Class Final)) | Proof Consider a number table of length $m+1$, filling each position with any number from $1 \sim m$, the total number of filling methods is $m^{m+1}$.
For a filling method $s$, let $d(s)$ be the largest $k$ such that the first $k$ numbers in $s$ are distinct.
Since $m+1>m$, hence $k \leqslant m$.
Because $k$ is the l... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,048 |
II. (40 points) As shown in Figure 2, in $\triangle ABC$, it is known that $CD$ is the angle bisector of $\angle C$. Take a point $O$ on the line segment $CD$ such that $\odot O$ passes through points $A$ and $B$, and $\odot O$ intersects $AC$ and $BC$ at points $E$ and $F$, respectively. The extensions of $BE$ and $AF... | As shown in Figure 4, draw $B K / / P Q$, intersecting line $C D$ at point $J$ and circle $\odot O$ at point $K$. Connect $C K$, intersecting circle $\odot O$ at point $L$.
Since $C D$ and $P Q$ are the internal and external angle bisectors of $\angle C$, respectively, and $C D \perp P Q, O J \perp B K$, it follows th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,049 |
Example 8 Let $\left\{a_{n}\right\}(n \geqslant 0)$ be a sequence of positive real numbers, and satisfy
$$
\sum_{k=0}^{n} \mathrm{C}_{n}^{k} a_{k} a_{n-k}=a_{n}^{2} .
$$
Prove: $\left\{a_{n}\right\}$ is a geometric sequence.
[4] | Prove that when $n=1$,
$$
a_{0} a_{1}+a_{0} a_{1}=a_{1}^{2} \Rightarrow a_{1}=2 a_{0} \text {. }
$$
When $n=2$,
$$
\begin{array}{l}
a_{2}^{2}=2 a_{0} a_{2}+2 a_{1}^{2}=2 a_{0} a_{2}+8 a_{0}^{2} \\
\Rightarrow\left(a_{2}-4 a_{0}\right)\left(a_{2}+2 a_{0}\right)=0 \\
\Rightarrow a_{2}=4 a_{0} .
\end{array}
$$
Assume th... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,050 |
Example 1 Let the two foci of the ellipse $\Gamma$ be $F_{1}$ and $F_{2}$, and a line passing through point $F_{1}$ intersects the ellipse $\Gamma$ at points $P$ and $Q$. If $\left|P F_{2}\right|=\left|F_{1} F_{2}\right|$, and $3\left|P F_{1}\right|=4\left|Q F_{2}\right|$, then the ratio of the minor axis to the major ... | According to the first definition of an ellipse, we have
$$
\left|P F_{1}\right|+\left|P F_{2}\right|=2 a \text {. }
$$
Given that $\left|P F_{2}\right|=\left|F_{1} F_{2}\right|=2 c$, we have
$$
\begin{array}{l}
\left|P F_{1}\right|=2 a-2 c . \\
\text { Let } \angle P F_{1} F_{2}=\theta .
\end{array}
$$
Thus, the pol... | \frac{2 \sqrt{6}}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,051 |
Example 1 As shown in Figure $1, \triangle A B C$ has an incircle $\odot I$ that touches sides $C A, A B$ at points $E, F$ respectively, and $B E, C F$ intersect $\odot I$ at points $M, N$. Prove: $M N \cdot E F=3 M F \cdot N E$. | Proof: Let $\odot I$ be tangent to $BC$ at point $D$. Then quadrilaterals $N D F E$ and $M D E F$ are both harmonic quadrilaterals.
By the definition of a harmonic quadrilateral, we have
$$
\begin{array}{l}
N E \cdot D F=D N \cdot E F, \\
M F \cdot D E=M D \cdot E F .
\end{array}
$$
For quadrilateral $N D F E$, by Pto... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,052 |
Example 2 As shown in Figure 2, quadrilateral $ABCD$ is inscribed in $\odot O$, and $M, N$ are the midpoints of $AC, BD$ respectively. If $\angle BMC = \angle DMC$, prove: $\angle AND = \angle CND .{ }^{[1]}$ | Prove that, as shown in Figure 2, draw a line through point $B$ parallel to $AC$, intersecting $\odot O$ at point $B'$. Then quadrilateral $A B' B C$ is an isosceles trapezoid.
Since $M$ is the midpoint of $AC$, thus, $OM \perp AC$.
Therefore, line $OM$ is the axis of symmetry of the isosceles trapezoid $A B' B C$.
Fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,053 |
Example 4 Proof: $\operatorname{gcd}(a, b)=\frac{1}{a} \sum_{m=0}^{a-1} \sum_{n=0}^{a-1} \mathrm{e}^{2 \pi \frac{i m b}{a}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Example 4 Proof: $\operatorname{gcd}(a, b)=\frac{1... | Proof: Let
$$
\operatorname{gcd}(a, b)=d, a=d x, b=d y, \omega=\mathrm{e}^{2 \pi i \frac{y}{x}},
$$
where $a, b, d, x, y \in \mathbf{Z}_{+}$.
Then $\omega^{x}=1$,
$$
\frac{1}{a} \sum_{m=0}^{a-1} \sum_{n=0}^{a-1} \mathrm{e}^{2 \pi i \frac{m b}{a}}=\frac{1}{a} \sum_{m=0}^{a-1} \sum_{n=0}^{a-1} \omega^{m n}.
$$
Notice t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,054 |
Question 2 Proof: For every integer $S$ greater than or equal to 100, there exists an integer $P$ such that the following story must be true:
A mathematician asked a store owner: “How much do this table, this cabinet, and this bookshelf cost in total?”
The owner replied: “The prices of these items are all positive in... | Prove that an integer $S$ satisfying the conditions is called a "good number".
The simplest way to construct equal products is $\{a, 2 b, 3 c\}$ and $\{b, 2 c, 3 a\}$, and then let their sums be equal, yielding
$$
b+c=2 a \text {. }
$$
Therefore, set $b=a-d, c=a+d$.
Then the two arrays are
$$
\begin{array}{l}
\{a, 2(a... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 729,055 |
Question 4 As shown in Figure 2, two circles $\Gamma_{1}$ and $\Gamma_{2}$ are externally separated, and one of their external common tangents touches circles $\Gamma_{1}$ and $\Gamma_{2}$ at points $A$ and $B$, respectively. An internal common tangent touches circles $\Gamma_{1}$ and $\Gamma_{2}$ at points $C$ and $D$... | Proof of a lemma first.
Lemma As shown in Figure 3, let the centers of two externally separated circles be $O_{1}$ and $O_{2}$, $AB$ be the external common tangent of the two circles, and $CD$ be the internal common tangent. Then $O_{1} O_{2}$, $AC$, and $BD$ are concurrent.
Proof As shown in Figure 3, extend $CD$ to ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,056 |
Question 5 As shown in Figure 5, in the acute triangle $\triangle ABC$, $AB > AC$, and $D$, $E$ are the midpoints of sides $AB$, $AC$ respectively. The circumcircle of $\triangle ADE$ intersects the circumcircle of $\triangle BCE$ at point $P$ (other than point $E$), and the circumcircle of $\triangle ADE$ intersects t... | Proof auxiliary line as shown in Figure 5.
It is easy to know that the three radical axes $PE$, $DQ$, and $BC$ of the circumcircles of $\triangle ADE$, $\triangle BCD$, and $\triangle BCE$ intersect at the same point $F$ (the radical center).
By $DE \parallel BC \Rightarrow \angle DFB = \angle EDQ$
$\Rightarrow \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,057 |
Example 3 As shown in Figure $3, O$ is the circumcenter of acute $\triangle A B C$, with $A B<A C, Q$ being the intersection of the external angle bisector of $\angle B A C$ with $B C$. Point $P$ is inside $\triangle A B C$, and $\triangle B P A \backsim$ $\triangle A P C$. Prove: $\angle Q P A+\angle O Q B=90^{\circ}$... | Proof: Let $A P$ intersect $\odot O$ at point $S$.
From $\triangle B P A \backsim \triangle A P C$
$\Rightarrow \angle B C S=\angle B A S=\angle B A P=\angle A C P$,
$\angle C B S=\angle C A S=\angle C A P=\angle A B P$
$\Rightarrow \triangle B P A \backsim \triangle A P C \backsim \triangle B S C$
$\Rightarrow \frac{S... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,058 |
In a cyclic quadrilateral $A B C D$, it is known that the midpoints of $A B, B C, C D, D A$ are $E, F, G, H$ respectively, and the orthocenters of $\triangle A H E, \triangle B E F, \triangle C F G, \triangle D G H$ are $W, X, Y, Z$ respectively. Prove: The area of quadrilateral $A B C D$ is equal to the area of quadri... | Proof: Let the circumradii of $\triangle A H E$, $\triangle C F G$, and quadrilateral $A B C D$ be $R_{1}$, $R_{2}$, and $R$ respectively.
Auxiliary lines as shown in Figure 1.
From $E$, $H$ being the midpoints of sides $A B$ and $D A$ respectively, we get
$$
\begin{array}{l}
E H / / B D \text {, and } E H=\frac{1}{2} ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,059 |
1. Let $A, B, C$ be three sets. Then “$B, C$ are both subsets of $A$” is a ( ) condition for “$(A \cap B) \cup (A \cap C) = B \cup C$” to hold.
(A) Sufficient but not necessary
(B) Necessary but not sufficient
(C) Sufficient and necessary
(D) Neither sufficient nor necessary | $-1 . C$.
If $B$ and $C$ are both subsets of $A$, then
$$
(A \cap B) \cup (A \cap C) = B \cup C \text{. }
$$
Conversely, if equation (1) holds, then since $A \cap B$ and $A \cap C$ are both subsets of $A$, it follows that $B \cup C$ is a subset of $A$, i.e., $B$ and $C$ are both subsets of $A$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,060 |
2. The curve represented by the equation $|y|=1+\sqrt{2 x-x^{2}}$ is ( ).
(A) a circle
(B) two semicircles
(C) an ellipse
(D) none of the above conclusions is correct | 2. B.
From the given equations, we have
$$
y=1+\sqrt{2 x-x^{2}} \text { and } y=-1-\sqrt{2 x-x^{2}},
$$
which represent the upper semicircle with center at $(1,1)$ and radius 1, lying between the points $(0,1)$ and $(2,1)$, and the lower semicircle with center at $(1,-1)$ and radius 1, lying between the points $(0,-1... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,061 |
2. Let $A_{n}$ and $B_{n}$ be the sums of the first $n$ terms of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively. If $\frac{A_{n}}{B_{n}}=\frac{5 n-3}{n+9}$, then $\frac{a_{8}}{b_{8}}=$ $\qquad$ | 2.3.
Let $\left\{a_{n}\right\}, \left\{b_{n}\right\}$ have common differences $d_{1}, d_{2}$, respectively. Then $\frac{A_{n}}{B_{n}}=\frac{a_{1}+\frac{1}{2}(n-1) d_{1}}{b_{1}+\frac{1}{2}(n-1) d_{2}}=\frac{5 n-3}{n+9}$.
Let $d_{2}=d$. Then $d_{1}=5 d$.
Thus, $a_{1}=d, b_{1}=5 d$.
Therefore, $\frac{a_{8}}{b_{8}}=\frac{... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,062 |
3. Let $O$ be the origin, $A$ be a moving point on the parabola $x=\frac{1}{4} y^{2}+1$, and $B$ be a moving point on the parabola $y=x^{2}+4$. Then the minimum value of the area of $\triangle O A B$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output... | 3.2.
Let $A\left(s^{2}+1,2 s\right), B\left(t, t^{2}+4\right)$. Then $l_{O B}:\left(t^{2}+4\right) x-t y=0$.
Let the distance from point $A$ to line $O B$ be $h$, we have
$$
h=\frac{\left|\left(t^{2}+4\right)\left(s^{2}+1\right)-t \cdot 2 s\right|}{\sqrt{\left(t^{2}+4\right)^{2}+t^{2}}} \text {. }
$$
Therefore, $S_{\... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,063 |
Example 4 As shown in Figure $4, P B, P D$ are tangents to $\odot O$, $P C A$ is a secant of $\odot O$, the tangent to $\odot O$ at $C$ intersects $P D, A D$ at points $Q, R$ respectively, and $A Q$ intersects $\odot O$ again at $E$. Prove that $B, E, R$ are collinear. ${ }^{[2]}$ | Prove as shown in Figure 4, let $K$ be a point on the extension of $A D$ such that $A D = D K$.
Since $Q D, Q C$ are tangents to $\odot O$, quadrilateral $E D A C$ is a harmonic quadrilateral.
Thus, $\frac{E C}{C A} = \frac{E D}{D A} = \frac{E D}{D K}$.
Also, $\angle E C A = \angle E D K$, so
$\triangle E C A \backsim... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,064 |
5. Let the complex number $z=\cos \frac{4 \pi}{7}+\mathrm{i} \sin \frac{4 \pi}{7}$. Then
$$
\left|\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}}\right|
$$
is equal to $\qquad$ (answer with a number). | 5. 2 .
Given that $z$ satisfies the equation $z^{7}-1=0$.
From $z^{7}-1=(z-1) \sum_{i=0}^{6} z^{i}$, and $z \neq 1$, we get $z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1=0$. After combining the fractions of $\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}}$, the denominator is
$$
\begin{array}{l}
\left(1+z^{2}\right... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,065 |
6. Let $a, b, c, d$ be real numbers, satisfying
$$
a+2 b+3 c+4 d=\sqrt{10} \text {. }
$$
Then the minimum value of $a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}$ is $\qquad$ | 6. 1 .
From the given equation, we have
$$
\begin{array}{l}
(1-t) a+(2-t) b+(3-t) c+(4-t) d+ \\
t(a+b+c+d)=\sqrt{10} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
{\left[(1-t)^{2}+(2-t)^{2}+(3-t)^{2}+(4-t)^{2}+t^{2}\right] .} \\
{\left[a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}\right] \geqs... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,066 |
For any real number $x$, we have
$$
|x+a|-|x+1| \leqslant 2 a \text{. }
$$
Then the minimum value of the real number $a$ is $\qquad$ | 1. $\frac{1}{3}$.
From $|x+a|-|x+1|$
$$
\leqslant|(x+a)-(x-1)|=|a-1|,
$$
we know that the maximum value of $f(x)=|x+a|-|x-1|$ is $|a-1|$. According to the problem, $|a-1| \leqslant 2 a \Rightarrow a \geqslant \frac{1}{3}$.
Therefore, the minimum value of the real number $a$ is $\frac{1}{3}$. | \frac{1}{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,067 |
2. Assign five college students to three villages in a certain town. If each village must have at least one student, then the number of different assignment schemes is $\qquad$ . | 2. 150 .
According to the number of college students allocated to each village, the only two types that meet the requirements are $1,1,3$ and $1, 2, 2$. Therefore, the number of different allocation schemes is
$$
C_{3}^{1} C_{5}^{3} C_{2}^{1}+C_{3}^{1} C_{5}^{2} C_{3}^{2}=60+90=150
$$ | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,068 |
3. If $\left(x^{2}-x-2\right)^{3}=a_{0}+a_{1} x+\cdots+a_{6} x^{6}$, then $a_{1}+a_{3}+a_{5}=$ | 3. -4 .
Let $x=0, x=1$, we get
$$
\begin{array}{l}
a_{0}=-8, \\
a_{0}+a_{1}+\cdots+a_{6}=\left(1^{2}-1-2\right)^{3}=-8 .
\end{array}
$$
Thus, $a_{1}+a_{2}+\cdots+a_{6}=0$.
Let $x=-1$, we get
$$
\begin{array}{l}
a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+a_{6} \\
=\left[(-1)^{2}-(-1)-2\right]^{3}=0 .
\end{array}
$$
Thus, $-... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,069 |
4. Given an isosceles triangle with a vertex angle of $20^{\circ}$ and a base length of $a$, the length of the legs is $b$. Then the value of $\frac{a^{3}+b^{3}}{a b^{2}}$ is $\qquad$ | 4.3.
Given $a=2 b \sin 10^{\circ}$.
Thus $a^{3}+b^{3}=8 b^{3} \sin ^{3} 10^{\circ}+b^{3}$
$$
\begin{array}{l}
=8 b^{3} \cdot \frac{1}{4}\left(3 \sin 10^{\circ}-\sin 30^{\circ}\right)+b^{3}=6 b^{3} \sin 10^{\circ} \\
\Rightarrow \frac{a^{3}+b^{3}}{a b^{2}}=\frac{6 b^{3} \sin 10^{\circ}}{2 b \sin 10^{\circ} \cdot b^{2}}... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,070 |
5. Let $a_{n}=2^{n}, b_{n}=5 n-1\left(n \in \mathbf{Z}_{+}\right)$,
$$
S=\left\{a_{1}, a_{2}, \cdots, a_{2015}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{a_{2015}}\right\} \text {. }
$$
Then the number of elements in the set $S$ is | 5.504.
Since the set $\left\{b_{1}, b_{2}, \cdots, b_{2015}\right\}$ contains $2^{2015}$ elements, forming an arithmetic sequence with a common difference of 5, and each term has a remainder of 4 when divided by 5, we only need to consider the number of terms in the set $\left\{a_{1}, a_{2}, \cdots, a_{2015}\right\}$ ... | 504 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,071 |
6. Given point $P$ is in the plane of Rt $\triangle A B C$, $\angle B A C=90^{\circ}, \angle C A P$ is an acute angle, and
$$
|\overrightarrow{A P}|=2, \overrightarrow{A P} \cdot \overrightarrow{A C}=2, \overrightarrow{A P} \cdot \overrightarrow{A B}=1 \text {. }
$$
When $| \overrightarrow{A B}+\overrightarrow{A C}+\o... | 6. $\frac{\sqrt{2}}{2}$.
Let $\angle C A P=\alpha$.
By the problem, $\angle B A P=\frac{\pi}{2}-\alpha$.
$$
\begin{array}{l}
\text { Given }|\overrightarrow{A P}|=2, \overrightarrow{A P} \cdot \overrightarrow{A C}=2, \overrightarrow{A P} \cdot \overrightarrow{A B}=1 \\
\Rightarrow|\overrightarrow{A C}|=\frac{1}{\cos ... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,072 |
7. Given a regular tetrahedron $P-ABC$ with the side length of the base being 6 and the side length of the lateral edges being $\sqrt{21}$. Then the radius of the inscribed sphere of the tetrahedron is $\qquad$ | 7.1 .
Let $P O \perp$ plane $A B C$ at point $O$. Then $O$ is the center of the equilateral $\triangle A B C$. Connect $A O$ and extend it to intersect $B C$ at point $D$, and connect $P D$. Thus, $D$ is the midpoint of $B C$.
It is easy to find that $P D=2 \sqrt{3}, O D=\sqrt{3}, P O=3$.
Let the radius of the inscrib... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,073 |
9. Given $F_{1} 、 F_{2}$ are the two foci of the ellipse $\frac{x^{2}}{4}+y^{2}=1$, $A 、 B$ are the left vertex and the upper vertex of the ellipse, respectively, and point $P$ lies on the line segment $A B$. Then the minimum value of $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ is | 9. $-\frac{11}{5}$.
Let $P(x, y), F_{1}(-c, 0), F_{2}(c, 0)$. Then
$$
\begin{array}{l}
\overrightarrow{P F_{1}}=(-c-x,-y), \overrightarrow{P F_{2}}=(c-x,-y) \\
\Rightarrow \overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=x^{2}+y^{2}-c^{2} .
\end{array}
$$
When $\overrightarrow{O P} \perp A D$, $x^{2}+y^{2}$ re... | -\frac{11}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,074 |
10. The largest prime $p$ such that $\frac{p+1}{2}$ and $\frac{p^{2}+1}{2}$ are both perfect squares is $\qquad$. | 10.7.
Let $\frac{p+1}{2}=x^{2} ， \frac{p^{2}+1}{2}=y^{2}\left(x, y \in \mathbf{Z}_{+}\right)$.
Obviously, $p>y>x, p>2$.
From $p+1=2 x^{2}, p^{2}+1=2 y^{2}$, subtracting the two equations gives $p(p-1)=2(y-x)(y+x)$.
Since $p(p>2)$ is a prime number and $p>y-x$, then $p \mid(y+x)$.
Because $2 p>y+x$, so $p=y+x$.
Thus, $... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,075 |
12. Let $T_{n}$ be the product of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, satisfying
$$
T_{n}=1-a_{n}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Let $S_{n}=T_{1}^{2}+T_{2}^{2}+\cdots+T_{n}^{2}$, prove:
$$
a_{n+1}-\fra... | 12. (1) It is easy to know, $T_{1}=a_{1}=\frac{1}{2}, T_{n} \neq 0, a_{n} \neq 1$.
From $T_{n+1}=1-a_{n+1}, T_{n}=1-a_{n}$
$$
\begin{array}{l}
\Rightarrow a_{n+1}=\frac{T_{n+1}}{T_{n}}=\frac{1-a_{n+1}}{1-a_{n}} \\
\Rightarrow \frac{1}{1-a_{n+1}}-\frac{1}{1-a_{n}}=1 \\
\Rightarrow \frac{1}{1-a_{n}}=\frac{1}{1-a_{1}}+n-1... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,076 |
1. The function
$$
f(x)=|x+1|+|x+3|+\mathrm{e}^{-x}(x \in \mathbf{R})
$$
has a minimum value of $\qquad$ | $-1.6-2 \ln 2$.
When $x \leqslant-3$, $f(x)=-2 x-4+\mathrm{e}^{-x}$, $f(x)$ is monotonically decreasing.
When $-3 \leqslant x \leqslant-1$, $f(x)=2+\mathrm{e}^{-x}$, at this time, $f(x)$ is also monotonically decreasing.
When $x \geqslant-1$,
$$
\begin{array}{l}
f(x)=2 x+4+\mathrm{e}^{-x}, f^{\prime}(x)=2-\mathrm{e}^{... | -1.6-2 \ln 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,077 |
3. Let planar vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ satisfy
$$
1 \leqslant|\boldsymbol{\alpha}|, |\boldsymbol{\beta}|, |\boldsymbol{\alpha}+\boldsymbol{\beta}| \leqslant 3 \text {. }
$$
Then the range of $\boldsymbol{\alpha} \boldsymbol{\beta}$ is $\qquad$ | 3. $\left[-\frac{17}{2}, \frac{9}{4}\right]$.
Notice,
$$
\begin{array}{l}
\boldsymbol{\alpha} \cdot \boldsymbol{\beta}=\frac{1}{2}\left(|\boldsymbol{\alpha}+\boldsymbol{\beta}|^{2}-|\boldsymbol{\alpha}|^{2}-|\boldsymbol{\beta}|^{2}\right) \\
\geqslant \frac{1-9-9}{2}=-\frac{17}{2}, \\
\boldsymbol{\alpha} \cdot \boldsy... | \left[-\frac{17}{2}, \frac{9}{4}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,078 |
5. Let $a$ be a real number, and the equation with respect to $x$
$$
(a+\cos x)(a-\sin x)=1
$$
has real roots. Then the range of values for $a$ is | 5. $\left[-1-\frac{\sqrt{2}}{2},-1+\frac{\sqrt{2}}{2}\right] \cup\left[1-\frac{\sqrt{2}}{2}, 1+\frac{\sqrt{2}}{2}\right]$. Let $u=a+\cos x, v=a-\sin x$.
The equation has real roots
$\Leftrightarrow$ the hyperbola $w v=1$ intersects the circle $(u-a)^{2}+(v-a)^{2}=1$.
Notice that, the center of the circle lies on the l... | \left[-1-\frac{\sqrt{2}}{2},-1+\frac{\sqrt{2}}{2}\right] \cup\left[1-\frac{\sqrt{2}}{2}, 1+\frac{\sqrt{2}}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,079 |
7. Let $z=x+y \mathrm{i} (x, y \in \mathbf{R}, \mathrm{i}$ be the imaginary unit), the imaginary part of $z$ and the real part of $\frac{z-\mathrm{i}}{1-z}$ are both non-negative. Then the area of the region formed by the points $(x, y)$ on the complex plane that satisfy the conditions is $\qquad$ . | 7. $\frac{3 \pi+2}{8}$.
Notice that,
$$
\begin{array}{l}
\operatorname{Re} \frac{z-\mathrm{i}}{1-z}=\operatorname{Re} \frac{x+(y-1) \mathrm{i}}{1-x-y \mathrm{i}} \\
=\frac{x(1-x)-(y-1) y}{(1-x)^{2}+y^{2}} \geqslant 0 \\
\Leftrightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2} ... | \frac{3 \pi+2}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,080 |
10. (22 points) In the polyhedron $A B C D E F$ shown in Figure 1, it is known that $A D, B E, C F$ are all perpendicular to the plane $A B C$.
Let $A D=a, B E=b, C F$ $=c, A B=A C=B C=$
B
1. Find the volume of the common part of the tetrahedron $A B C E$ and
the tetrahedron $B D E F$ (expressed in terms of $a, b, c$) | 10. As shown in Figure 2, let $A E$ intersect $B D$ and $B F$ intersect $C E$ at points $G$ and $H$, respectively. Then the tetrahedron $B E G H$ is the desired common part.
The calculations yield:
The distance from point $G$ to line $A B$ is $d_{1}=\frac{a b}{a+b}$, the distance from point $G$ to plane $B C F E$ is $d... | \frac{\sqrt{3} b^{3}}{12(a+b)(b+c)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,081 |
1. Find all positive integers $n$ such that for any two $n$-degree polynomials $P(x), Q(x)$, there exist monomials $a x^{k}, b x^{l}$ (where $a, b$ are real numbers, and integers $k, l$ satisfy $0 \leqslant k, l \leqslant n$), such that the graphs of $P(x)+a x^{k}$ and $Q(x)+b x^{l}$ have no intersection points. | 1. The original proposition can be rewritten as: For any polynomial $R(x)$ of degree not exceeding $n$, there exist monomials $a x^{k}$ and $b x^{l}$ (where $a, b$ are real numbers, and integers $k, l$ satisfy $0 \leqslant k < l \leqslant n$), such that $R(x) + a x^{k} + b x^{l}$ always has no roots.
If $n \geqslant 3... | n = 1 \text{ or } n \text{ is even} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,082 |
2. On a plane, there is an $8 \times 8$ grid colored in a black and white checkerboard pattern. Basil arbitrarily selects one of the cells. Each turn, Peter draws a polygon (which can be concave but not self-intersecting) on the grid, and Basil will honestly inform Peter whether the selected cell is inside or outside t... | 2. If the polygon drawn by Peter includes only all the cells of a certain color, then this polygon must intersect itself. Therefore, Peter cannot determine the color of the cell chosen by Basil in just one round.
Next, two strategies are given that can determine the color of the selected cell in two rounds.
【Strategy... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 729,083 |
3. Given that the center of equilateral $\triangle A B C$ is point $O$, a line passing through point $C$ intersects the circumcircle of $\triangle A O B$ at points $D$ and $E$. Prove: Points $A, O$ and the midpoints of segments $B D$ and $B E$ are concyclic. | 3. Let $C^{\prime}$ be the midpoint of $B C$, point $E$ lies between $C$ and $D$, $E^{\prime}$ be the midpoint of $B E$, and $D^{\prime}$ be the midpoint of $B D$, as shown in Figure 4.
In $\triangle B D C$, by the midline theorem, we know that point $E^{\prime}$ lies on the midline $C^{\prime} D^{\prime}$.
In the equ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,084 |
Example 6 Let $n$ be a positive integer, and the sequence $A: a_{1}, a_{2}, \cdots$, $a_{n}$ be a sequence composed of 0 or 1, that is,
$a_{k}=0$ or $1(1 \leqslant k \leqslant n)$.
Find the number of sequences such that $\sum_{i=1}^{n} a_{i}$ leaves a remainder of 3 when divided by 4. | Let $f(x)=(1+x)^{n}=\sum_{i=0}^{n} b_{i} x^{i}$.
It is easy to see that the number of $A$ is $S=b_{3}+b_{7}+b_{11}+\cdots$. Let $\omega$ be a fourth root of unity, and without loss of generality, let $\omega=\mathrm{i}, 1+\omega+\omega^{2}+\omega^{3}=0$.
Then $x f(x)=\sum_{i=0}^{n} b_{i} x^{i+1}$.
Substituting $1, \ome... | S=\begin{cases}
\frac{2^{4k}}{4}, & \text{if } n=4k \\
\frac{2^{4k}-(-4)^{k}}{2}, & \text{if } n=4k+1 \\
2^{4k}-(-4)^{k}, & \text{if } n=4k+2 \\
2^{4k+1}-(-4)^{k}, & | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,086 |
2. Let the complex numbers be
$$
\begin{array}{l}
z_{1}=(6-a)+(4-b) \mathrm{i}, \\
z_{2}=(3+2 a)+(2+3 b) \mathrm{i}, \\
z_{3}=(3-a)+(3-2 b) \mathrm{i},
\end{array}
$$
where, $a, b \in \mathbf{R}$.
When $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ reaches its minimum value, $3 a+4 b$ $=$ | 2. 12 .
Notice that,
$$
\begin{array}{l}
\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right| \\
=|12+9 \mathrm{i}|=15 .
\end{array}
$$
Equality holds if and only if
$$
\frac{6-a}{4-b}=\frac{3+2 a}{2+3 b}=\frac{3-a}{3-2 b}=\frac{12}{9} \text {, }
$$
i.e., when $a=\frac{7}{... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,087 |
3. Let the line $l$ passing through the origin intersect the graph of the function $y=|\sin x|$ $(x \geqslant 0)$ at exactly three points, with $\alpha$ being the largest of the x-coordinates of these intersection points. Then
$$
\frac{\left(1+\alpha^{2}\right) \sin 2 \alpha}{2 \alpha}=
$$
$\qquad$ . | 3. 1 .
As shown in Figure 2, let the line $l$ be tangent to the function $y=|\sin x|(x \geqslant 0)$
at point $P(\alpha,-\sin \alpha)$, and $k_{l}=-\cos \alpha$.
Then the line $l: y+\sin \alpha=-(x-\alpha) \cos \alpha$.
Substituting $(0,0)$, we get $\alpha=\tan \alpha$.
Therefore, $\frac{\left(1+\alpha^{2}\right) \sin... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 729,088 |
4. Given $P(1,4,5)$ is a fixed point in the rectangular coordinate system $O-x y z$, a plane is drawn through $P$ intersecting the positive half-axes of the three coordinate axes at points $A$, $B$, and $C$ respectively. Then the minimum value of the volume $V$ of all such tetrahedrons $O-A B C$ is $\qquad$ | 4. 90 .
Let the plane equation be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where the positive numbers $a$, $b$, and $c$ are the intercepts of the plane on the $x$-axis, $y$-axis, and $z$-axis, respectively.
Given that point $P$ lies within plane $ABC$, we have $\frac{1}{a}+\frac{4}{b}+\frac{5}{c}=1$.
From $1=\frac{1}{... | 90 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,089 |
5. Let $\Gamma$ be the region formed by the points $(x, y)$ satisfying
$$
\left\{\begin{array}{l}
x \geqslant 0, \\
y \geqslant 0, \\
x+y+[x]+[y] \leqslant 5
\end{array}\right.
$$
The area of the region $\Gamma$ is
$\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 5. $\frac{9}{2}$.
When $x+y>3$,
$$
[x]+[y]+\{x\}+\{y\}>3 \text {. }
$$
and $\{x\}+\{y\}\leq1 \Rightarrow[x]+[y] \geqslant 2} \\
\Rightarrow x+y+[x]+[y]>5,
\end{array}
$$
which contradicts the condition.
Therefore, the area of region $\Gamma$ is $\frac{9}{2}$. | \frac{9}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,090 |
6. A function of the form $y=\frac{b}{|x|-a}(a, b>0)$ has a graph resembling the Chinese character “圆”, and is called a “Pei function”. The point symmetric to the intersection of the function with the $y$-axis about the origin is called the “fen point”. A circle centered at the fen point that has at least one common po... | $6.3 \pi$.
As shown in Figure 3, when $a=b=1$, the analytical expression of the function is $y=\frac{1}{|x|-1}$, and its intersection with the $y$-axis is $(0,-1)$. Thus, the point $C(0,1)$, and the distance between them is 2.
Taking any point $(x, y)$ on the graph of the function in the first quadrant, the distance t... | 3 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,091 |
Given a positive number $k(k>2)$ and a positive integer $n(n \geqslant 3)$. Find the maximum positive number $\lambda$, such that if positive numbers $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)<\lambda,
$$
then it m... | $$
\begin{array}{l}
\text { II. } \lambda_{\max }=\left(\sqrt{k+\frac{4}{k}+5}+n-3\right)^{2} . \\
\text { If } \lambda>\left(\sqrt{k+\frac{4}{k}+5}+n-3\right)^{2}, \text { let } \\
a_{1}=a_{2}=k, a_{3}=2, \\
a_{4}=a_{5}=\cdots=a_{n}=\sqrt{\frac{4 k(k+1)}{k+4}} . \\
\text { Then }\left(\sum_{i=1}^{n} a_{i}\right)\left(... | \left(\sqrt{k+\frac{4}{k}+5}+n-3\right)^{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,092 |
Given $x_{i}>0(i=1,2, \cdots, n, n \geqslant 2)$ and integer $k \geqslant 2$, and $\sum_{i=1}^{n} \frac{1}{x_{i}^{k}+1}=\frac{n}{2}$. Prove:
$$
\sum_{i=1}^{n} \frac{x_{i}}{x_{i}^{k}+1} \leqslant \frac{n}{2} .
$$ | Prove that
$$
\sum_{i=1}^{n} \frac{1}{x_{i}^{k}+1}=\frac{n}{2} \Leftrightarrow \sum_{i=1}^{n} \frac{x_{i}^{k}}{x_{i}^{k}+1}=\frac{n}{2} \text {. }
$$
By the $k$-th mean inequality, we have
$$
\begin{array}{l}
\frac{k n}{2}=\sum_{i=1}^{n} \frac{x_{i}^{k}}{x_{i}^{k}+1}+(k-1) \sum_{i=1}^{n} \frac{1}{x_{i}^{k}+1} \\
=\sum... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,093 |
Given a positive integer $n(n \geqslant 3)$. Try to find the largest constant $\lambda(n)$, such that for any $a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}_{+}$, we have
$$
\prod_{i=1}^{n}\left(a_{i}^{2}+n-1\right) \geqslant \lambda(n)\left(\sum_{i=1}^{n} a_{i}\right)^{2} .
$$ | Prove that if $a_{1}=a_{2}=\cdots=a_{n}=1$, then $\lambda(n) \leqslant n^{n-2}$.
It is easy to prove the following lemma by mathematical induction.
Lemma Let $n \in \mathbf{Z}_{+}, b_{1}, b_{2}, \cdots, b_{n} \in(-1$, $+\infty)$, and these $n$ numbers have the same sign. Then
$$
\prod_{i=1}^{n}\left(1+b_{i}\right) \geq... | n^{n-2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,094 |
Given $n_{1}>n_{2}>m \in \mathbf{Z}_{+}$. Prove:
$$
\begin{array}{l}
\frac{n_{1}^{m}}{n_{1}!}+\sum_{i=1}^{n_{1}} \sum_{j=0}^{i}(-1)^{j}\left(n_{1}-i\right)^{m} \frac{1}{(j!)\left[\left(n_{1}-i\right)!\right]} \\
=\frac{n_{2}^{m}}{n_{2}!}+\sum_{i=1}^{n_{2}} \sum_{j=0}^{i}(-1)^{j}\left(n_{2}-i\right)^{m} \frac{1}{(j!)\le... | Proof Note
$$
f(n)=\frac{n^{m}}{n!}+\sum_{i=1}^{n} \sum_{j=0}^{i}(-1)^{j}(n-i)^{m} \frac{1}{(j!)[(n-i)!]} .
$$
We will use induction on $n_{1}-n_{2}$.
When $n_{1}-n_{2}=1$,
in equation (1),
left side - right side
$$
=\frac{\left(n_{2}+1\right)^{m}}{\left(n_{2}+1\right)!}-\frac{n_{2}^{m}}{n_{2}!}+
$$
$$
\begin{array}{l... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,095 |
Given $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$
\frac{1}{a^{2}+b+1}+\frac{1}{b^{2}+c+1}+\frac{1}{c^{2}+a+1} \leqslant 1 .
$$ | Prove that from a well-known conclusion,
$$
\sum \frac{1}{a b+a+1}=1,
$$
where, “ $\sum$ ” denotes the cyclic symmetric sum.
Thus, it suffices to prove
$$
\begin{array}{l}
\sum \frac{1}{a^{2}+b+1} \leqslant \sum \frac{1}{a b+a+1} \\
\Leftrightarrow \sum \frac{1}{a b+a+1}-\sum \frac{1}{a^{2}+b+1} \geqslant 0 \\
\Leftri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,096 |
Example 1 Given that $\angle A P B$ has an incircle tangent to the sides at points $A, B$, and $P C D$ is any secant line, intersecting the circle at points $C, D$. Point $Q$ is on $C D$, and $\angle Q A D=\angle P B C$. Prove:
$$
\angle P A C=\angle Q B D \text {. }
$$
(2003, National High School Mathematics League Co... | Prove: As shown in Figure 9, connect $A B$.
By Property 3, we know that $A B$ is the inner conjugate median of $\triangle A C D$.
Given $\angle Q A D=\angle P B C=\angle C A B$, we know that $Q$ is the midpoint of $C D$. Therefore, for $\triangle B C D$, we have
$$
\angle Q B D=\angle C B A=\angle P A C,
$$
Note that ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,097 |
Let $m, n \in \mathbf{Z}_{+}, 1 \leqslant i \leqslant n, 1 \leqslant j \leqslant m$, and $i, j \in \mathbf{Z}_{+}, x_{i}, \alpha_{i}, y_{j}, \beta_{j} \in \mathbf{R}$, and $0 \leqslant x_{i}<\alpha_{i}<\frac{1}{2}, 0 \leqslant y_{j}<\beta_{j}<\frac{1}{2}$. If
$$
\begin{array}{l}
{\left[\prod_{i=1}^{n}\left(1+x_{i}\righ... | Prove: $\left[\prod_{i=1}^{n}\left(1+2 \alpha_{i}\right)\right]\left[\prod_{j=1}^{m}\left(1-2 \beta_{j}\right)\right]$
$$
<\left[\prod_{i=1}^{n}\left(1+2 x_{i}\right)\right]\left[\prod_{j=1}^{m}\left(1-2 y_{j}\right)\right] \text {. }
$$
Proof First, prove two local inequalities:
$\frac{1+2 \alpha_{i}}{1+2 x_{i}} \leq... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,098 |
Given 463 Let $x_{i}, y_{i} \in [a, b] (0 < a < b, i = 1,2, \cdots, n), \sum_{i=1}^{n} x_{i}^{2} = \sum_{i=1}^{n} y_{i}^{2}$. Prove:
$$
\sum_{i=1}^{n} \frac{x_{i}^{3}}{y_{i}} + \sum_{i=1}^{n} \frac{y_{i}^{3}}{x_{i}} + 2 \sum_{i=1}^{n} x_{i} y_{i} < 4 n \cdot \frac{a^{4} b + b^{5}}{a^{3} + a b^{2}} \text{. }
$$ | $$
\begin{array}{l}
\frac{a}{b} \leqslant \frac{x_{i}}{y_{i}} \leqslant \frac{b}{a} \\
\Rightarrow\left(x_{i}-\frac{a}{b} y_{i}\right)\left(x_{i}-\frac{b}{a} y_{i}\right) \leqslant 0 \\
\Rightarrow x_{i}^{2}-\frac{a^{2}+b^{2}}{a b} x_{i} y_{i}+y_{i}^{2} \leqslant 0 \\
\Rightarrow\left(\frac{x_{i}}{y_{i}}+\frac{a b}{a^{... | 4 n \cdot \frac{a^{4} b+b^{5}}{a^{3}+a b^{2}} | Inequalities | proof | Yes | Yes | cn_contest | false | 729,099 |
464 Let non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $\sum_{i=1}^{n} x_{i}=n$. Find the maximum value of $\sum_{k=1}^{n-1} x_{k} x_{k+1}+x_{1}+x_{n}$. | (1) When $n$ is odd,
$$
\begin{array}{l}
\sum_{k=1}^{n-1} x_{k} x_{k+1}+x_{1}+x_{n} \\
\leqslant\left(x_{1}+x_{3}+\cdots+x_{n}\right)\left(x_{2}+x_{4}+\cdots+x_{n-1}\right)+\left(x_{1}\right. \\
\leqslant\left(x_{1}+x_{3}+\cdots+x_{n}\right)\left(x_{2}+x_{4}+\cdots+x_{n-1}\right. \\
\leqslant\left(\frac{n+1}{2}\right)^... | \left(\frac{n+1}{2}\right)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,100 |
Example 6 Find the number of solutions to the equation
$$
|| \cdots|||x|-1|-2| \cdots|-2011|=2011
$$
The number of solutions. ${ }^{[4]}$ | 【Analysis】Remove the absolute value symbols from outside to inside, step by step.
From the original equation, we get
$$
|| \cdots|||x|-1|-2| \cdots|-2010|=0
$$
or ||$\cdots|||x|-1|-2| \cdots|-2010|=4002$.
For equation (1), we have
$$
\begin{array}{l}
|| \cdots|||x|-1|-2| \cdots|-2009|=2010 \\
\Rightarrow|| \cdots|||x|... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,101 |
1. Let $i_{1}, i_{2}, \cdots, i_{10}$ be a permutation of $1,2, \cdots, 10$. Define $S=\left|i_{1}-i_{2}\right|+\left|i_{3}-i_{4}\right|+\cdots+\left|i_{9}-i_{10}\right|$. Find all possible values of $S$.
[2] | Since $S \geqslant 1+1+1+1+1=5$, $S \leqslant 6+7+\cdots+10-(1+2+\cdots+5)=25$,
and $S \equiv \sum_{k=1}^{10} k(\bmod 2) \equiv 1(\bmod 2)$,
therefore, it only needs to be proven that $S$ can take all odd numbers from 5 to 25. | 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,102 |
3. Given that $n$ is a positive integer, and the real number $x$ satisfies
$$
|1-| 2-|3-\cdots|(n-1)-|n-x||\cdots|||=x \text {. }
$$
Find the value of $x$. | Hint: By giving $n$ special values, we can find the pattern: when $n=4 k+1$ or $n=4 k+2(k \in \mathbf{N})$, the solution set of the equation is $\left\{\frac{1}{2}\right\}$; when $n=4 k$ or $n=4 k+3(k \in \mathbf{N})$, the solution set of the equation is $\{x \in \mathbf{R} \mid 0 \leqslant x \leqslant 1\}$. | \left\{\frac{1}{2}\right\} \text{ or } \{x \in \mathbf{R} \mid 0 \leqslant x \leqslant 1\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,104 |
5. Let $n$ be a given positive integer, and the sum
$$
S=\sum_{1 \leqslant i<j \leqslant n}\left|x_{i}-x_{j}\right|\left(0<x_{i}<1, i=1,2, \cdots, n\right) \text {. }
$$
Find the maximum possible value of the sum. | Notice that,
$$
\begin{array}{l}
S=\sum_{k=1}^{n}\left((k-1) x_{k}-(n-k) x_{k}\right) \\
=\sum_{k=1}^{n}\left(x_{k}(2 k-n-1)\right) \\
\leqslant \sum_{k \geq \frac{n+1}{2}}\left(x_{k}(2 k-n-1)\right) .
\end{array}
$$
When $n$ is odd,
$$
\begin{aligned}
S & \leqslant \sum_{k=\frac{n+1}{2}}^{n}\left(x_{k}(2 k-n-1)\right... | \left[\frac{n^{2}}{4}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,105 |
Question 1 Let $n$ be a positive integer, $p$ a prime number, satisfying $p^{m} \mid n$, and $p^{m+1} \nmid n$. Let $\operatorname{ord}_{p} n$ denote the exponent of the prime $p$ in the factorization of $n$. Define $\operatorname{ord}_{p} n=m$, and let $S_{p}(n)$ be the sum of the digits of $n$ in its $p$-ary represen... | (1) Proof: Let in base $p$,
$$
n=\sum_{i=0}^{k} a_{i} p^{i}\left(a_{i} \in\{0,1, \cdots, p-1\}\right) \text {. }
$$
Then the exponent of $p$ in the prime factorization of $n!$ is
$$
\begin{array}{l}
\operatorname{ord}_{p} n!=\sum_{i=1}^{+\infty}\left[\frac{n}{p^{i}}\right] \\
=\sum_{i=1}^{+\infty} \sum_{j=i}^{k} a_{j}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,106 |
Question 3 Let $a, b, n$ and $\frac{n!}{(a!)(b!)}$ all be positive integers. Prove:
$$
a+b \leqslant n+1+2 \log _{2} n
$$ | 【Analysis】Obviously, for positive integers $a, b \leqslant n$, we have
$$
\begin{array}{l}
\operatorname{ord}_{2} n!=n-S_{2}(n), \\
\operatorname{ord}_{2} a!=a-S_{2}(a), \\
\operatorname{ord}_{2} b!=b-S_{2}(b).
\end{array}
$$
Using $\frac{n!}{(a!)(b!)}$ as a positive integer, we get $\operatorname{ord}_{2} n!-\operato... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,108 |
Example 3 As shown in Figure $11, A$ and $B$ are two fixed points on $\odot O$, $C$ is the midpoint of the major arc $\overparen{A B}$, $D$ is any point on the minor arc $\overparen{A B}$, and a tangent line is drawn through $D$ to $\odot O$, intersecting the tangents to $\odot O$ at points $A$ and $B$ at points $E$ an... | Prove as shown in Figure 11, connect $AC$, $AD$, $CD$, and let $CD$ intersect $AB$ at point $K$. Draw the tangent line $TS$ of $\odot O$ through point $C$.
Since $C$ is the midpoint of the major arc $\overparen{AB}$, we know $TS \parallel AB$.
By Property 2, we know that $CE$ and $CT$ are the internal and external sy... | GH = \frac{1}{2} AB | Geometry | proof | Yes | Yes | cn_contest | false | 729,109 |
If $a, b, c > 0$, prove:
$$
\begin{array}{l}
\sqrt{a^{2}+a b+b^{2}}+\sqrt{a^{2}+a c+c^{2}} \\
\geqslant 4 \sqrt{\left(\frac{a b}{a+b}\right)^{2}+\left(\frac{a b}{a+b}\right)\left(\frac{a c}{a+c}\right)+\left(\frac{a c}{a+c}\right)^{2}} .
\end{array}
$$ | Prove that for $z_{1}=\left(a+\frac{b}{2}\right)+\frac{\sqrt{3}}{2} b \mathrm{i}$, $z_{2}=\left(\frac{a}{2}+c\right)+\frac{\sqrt{3}}{2} a \mathrm{i}$, we have
$$
\begin{array}{l}
\left|z_{1}\right|+\left|z_{2}\right| \\
=\sqrt{\left(a+\frac{b}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2} b\right)^{2}}+\sqrt{\left(\frac{a}{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,110 |
1. Given $A B$ is the diameter of $\odot O$, $A B=1$, extend $A B$ to point $C$, such that $B C=1$, $C D$ is the tangent of $\odot O$, $D$ is the point of tangency. Then the area of $\triangle A B D$ is $\qquad$ . | ,$- 1 . \frac{\sqrt{2}}{6}$.
By the secant-tangent theorem, we have
$$
C D^{2}=C B \cdot C A=2 \Rightarrow C D=\sqrt{2} \text {. }
$$
As shown in Figure 5, draw $D E \perp O C$ at point $E$.
Then $D E \cdot O C=2 S_{\triangle O D C}=O D \cdot C D=\frac{\sqrt{2}}{2}$
$$
\Rightarrow D E=\frac{\sqrt{2}}{3} \Rightarrow S_... | \frac{\sqrt{2}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,111 |
2. There are seven small balls of the same size, numbered $1,2, \cdots, 7$. Three balls are randomly selected from them. The probability that the sum of the numbers on the three selected balls is odd is $\qquad$. | 2. $\frac{16}{35}$.
Notice,
the sum of the numbers on the three balls is odd
$\Leftrightarrow$ the three numbers are either all odd or one odd and two even.
Thus, the required probability is $\frac{\mathrm{C}_{4}^{3}+\mathrm{C}_{4}^{1} \mathrm{C}_{3}^{2}}{\mathrm{C}_{7}^{3}}=\frac{16}{35}$. | \frac{16}{35} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,112 |
3. Let real numbers $x, y$ satisfy
$$
x^{2}+\sqrt{3} y=4, y^{2}+\sqrt{3} x=4, x \neq y \text {. }
$$
Then the value of $\frac{y}{x}+\frac{x}{y}$ is $\qquad$ | 3. -5 .
From the conditions, we have
$$
\left\{\begin{array}{l}
x^{2}-y^{2}+\sqrt{3} y-\sqrt{3} x=0, \\
x^{2}+y^{2}+\sqrt{3}(x+y)=8 .
\end{array}\right.
$$
From equation (1) and $x \neq y$, we know $x+y=\sqrt{3}$.
Substituting into equation (2) gives $x^{2}+y^{2}=5$.
$$
\begin{array}{l}
\text { Also, } 2 x y=(x+y)^{2... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,113 |
4. If the product of three prime numbers is exactly 23 times their sum, then these three prime numbers are $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. $23,13,3$ or $23,7,5$.
Let the three prime numbers be $p, q, r$.
Thus, $p q r=23(p+q+r)$.
From this, we know that one of $p, q, r$ must be 23.
Without loss of generality, let $p=23$, and $q \geqslant r$.
Then $23 q r=23(23+q+r)$
$$
\begin{array}{l}
\Rightarrow(q-1)(r-1)=24 \\
\Rightarrow(q-1, r-1) \\
\quad=(24,1),(... | 23,13,3 \text{ or } 23,7,5 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 729,114 |
Example 4 In $\triangle A B C$, $D$ and $E$ are the midpoints of sides $A B$ and $A C$ respectively, $B E$ and $C D$ intersect at point $G$, the circumcircle of $\triangle A B E$ and the circumcircle of $\triangle A C D$ intersect at point $P(P$ is not coincident with point $A)$, the extension of $A G$ intersects the c... | Prove: As shown in Figure 12, connect $D E$.
From the given, $D E // B C$.
Then $P$ is the Miquel point of the complete quadrilateral $A D B G C E$.
By Property 6, we know that $A P$ is the inner conjugate median of $\triangle A B C$.
By Corollary 6(3), we get $P L // C D$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,115 |
5. As shown in Figure $1, \odot O_{1}$ is externally tangent to $\odot O_{2}$ at point $P$. From point $A$ on $\odot O_{1}$, a tangent line $A B$ is drawn to $\odot O_{2}$, with $B$ as the point of tangency. Line $A P$ is extended to intersect $\odot O_{2}$ at point $C$. Given that the radii of $\odot O_{1}$ and $\odot... | 5. $\frac{\sqrt{6}}{2}$.
Connect $O_{1} O_{2}$. Then $O_{1} O_{2}$ passes through the point of tangency $P$.
Connect $O_{1} A$, $O_{1} P$, $O_{2} P$, $O_{2} C$.
Then isosceles $\triangle O_{1} A P \backsim$ isosceles $\triangle O_{2} P C$
$$
\Rightarrow \frac{A P}{P C}=\frac{O_{1} P}{O_{2} P}=2 \text {. }
$$
Assume $... | \frac{\sqrt{6}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,116 |
7. The integer solutions $(x, y)$ of the indeterminate equation $x^{2}+y^{2}=x y+2 x+2 y$ are in total groups.
The integer solutions $(x, y)$ of the indeterminate equation $x^{2}+y^{2}=x y+2 x+2 y$ are in total groups. | 7.6 .
Rewrite the given equation as
$$
\begin{array}{l}
x^{2}-x(2+y)+y^{2}-2 y=0 \\
\Rightarrow \Delta=(2+y)^{2}-4\left(y^{2}-2 y\right)=-3 y^{2}+12 y+4 \\
\quad=-3(y-2)^{2}+16 \geqslant 0 \\
\Rightarrow|y-2| \leqslant \frac{4}{\sqrt{3}}<3 .
\end{array}
$$
Since $y$ is an integer, thus, $y=0,1,2,3,4$.
Upon calculatio... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,118 |
9. (15 points) As shown in Figure 3, in $\triangle A B C$, $B C=a, C A=b$, $\angle A C B=60^{\circ}$, $\triangle A B D$ is an equilateral triangle, and $P$ is its center. Find the length of $C P$. | II. 9. Connect $P A$ and $P B$. Then $\angle A P B=120^{\circ}$. Also, $\angle A C B=60^{\circ}$, so points $C$, $B$, $P$, and $A$ are concyclic. Note that, $P A=P B=\frac{2}{3} \times \frac{\sqrt{3}}{2} A B=\frac{\sqrt{3}}{3} A B$.
By Ptolemy's theorem, we have
$$
\begin{array}{l}
P A \cdot C B+P B \cdot A C=A B \cdot... | \frac{\sqrt{3}}{3}(a+b) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,119 |
10. (15 points) From the 2015 positive integers 1, 2, $\cdots$, 2015, select $k$ numbers such that the sum of any two different numbers is not a multiple of 50. Find the maximum value of $k$.
untranslated part:
在 1,2 , $\cdots, 2015$ 这 2015 个正整数中选出 $k$ 个数,使得其中任意两个不同的数之和均不为 50 的倍数. 求 $k$ 的最大值.
translated part:
From ... | 10. Classify $1 \sim 2015$ by their remainders when divided by 50.
Let $A_{i}$ represent the set of numbers from $1 \sim 2015$ that have a remainder of $i$ when divided by 50, where $i=0,1, \cdots, 49$. Then $A_{1}, A_{2}, \cdots, A_{15}$ each contain 41 numbers, and $A_{0}, A_{16}, A_{17}, \cdots, A_{49}$ each contai... | 977 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,120 |
11. (20 points) As shown in Figure 4, the three sides of $\triangle ABC$ are all positive integers, and the perimeter is 35. $G$ and $I$ are the centroid and incenter of $\triangle ABC$, respectively, and $\angle GIC = 90^{\circ}$. Find the length of side $AB$. | 11. Extend $G I$, intersecting $C B$ and $C A$ (or their extensions) at points $P$ and $Q$ respectively.
Since $C I$ is the angle bisector of $\angle C$ and $\angle G I C=90^{\circ}$, we know that $\triangle C P Q$ is an isosceles triangle.
Draw $G E \perp P C$ and $G F \perp C Q$ from point $G$, and draw $I R \perp ... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,121 |
1. Given the function $f: \mathbf{N} \rightarrow \mathbf{N}$ defined as follows:
$$
f(x)=\left\{\begin{array}{ll}
\frac{x}{2}, & x \text { is even; } \\
\frac{x+7}{2}, & x \text { is odd. }
\end{array}\right.
$$
Then the number of elements in the set $A=\{x \in \mathbf{N} \mid f(f(f(x)))=x\}$ is | - 1.8.
On one hand, when $x \in \{0,1, \cdots, 7\}$, it is calculated that $f(x) \in \{0,1, \cdots, 7\}$,
and it can be verified that $\{0,1, \cdots, 7\} \subseteq A$.
On the other hand, when $x \geqslant 8$,
$$
\frac{x}{2}<x \text{, and } \frac{x+7}{2}<\frac{2 x}{2}<x \text{. }
$$
Then $f(f(f(x)))<8$, which does not... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,122 |
2. In the acute $\triangle A B C$, take a point $P$ such that
$$
\angle P A C=60^{\circ}, \angle P C A=\angle P B A=30^{\circ} \text {, }
$$
$M 、 N$ are the midpoints of sides $A C 、 B C$ respectively. Then the degree of $\angle P N M$ is $\qquad$ | 2. $90^{\circ}$.
From the given conditions, it is easy to know that $\angle A P C=90^{\circ}$.
As shown in Figure 4, extend $C P$ to point $D$ such that $P D=C P$, and connect $A D$ and $D B$.
Then $\angle A P D=90^{\circ}, A D=A C$,
$\angle A D P=\angle A C P=30^{\circ}=\angle P B A$.
Thus, points $A, D, B, P$ are co... | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,123 |
Example 5 In the acute triangle $\triangle ABC$, $AB > AC$, $M$ is the midpoint of side $BC$, and $P$ is a point inside $\triangle ABC$ such that $\angle MAB = \angle PAC$. Let the circumcenters of $\triangle ABC$, $\triangle ABP$, and $\triangle ACP$ be $O$, $O_1$, and $O_2$ respectively. Prove that line $AO$ bisects ... | Proof as shown in Figure 13.
From the given conditions, we know that $AP$ is the internal conjugate median of $\triangle ABC$.
By Property 5, we know there exists a circle $\odot O_{1}^{\prime}$ passing through points $A$ and $B$ and tangent to $AC$ at point $A$, and a circle $\odot O_{2}^{\prime}$ passing through poin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,124 |
4. Given that $\triangle A B C$ and $\triangle A D E$ are both isosceles triangles, $\angle A=90^{\circ}, A B=2 A D=4$. As shown in Figure 1, $\triangle A D E$ is rotated counterclockwise around point $A$ by an angle $\alpha$, and the extension of $B D$ intersects line $C E$ at point $P$, as shown in Figure 2.
Then, d... | 4. $\frac{4 \sqrt{2} \pi}{3}$.
Let $\angle B A D=\alpha$. It is easy to know that,
$$
\begin{array}{l}
\triangle A C E \cong \triangle A B D \Rightarrow \angle E C A=\angle D B A \\
\Rightarrow \angle P C B+\angle C B P=90^{\circ} \Rightarrow \angle B P C=90^{\circ} .
\end{array}
$$
Therefore, point $P$ moves on the ... | \frac{4 \sqrt{2} \pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,125 |
5. Real numbers $x, y, a$ satisfy $x+y=a+1$ and $xy=a^{2}-7a+16$. Then the maximum value of $x^{2}+y^{2}$ is $\qquad$ | 5. 32 .
Notice that,
$$
\begin{array}{l}
x^{2}+y^{2}=(x+y)^{2}-2 x y \\
=(a+1)^{2}-2\left(a^{2}-7 a+16\right) \\
=-a^{2}+16 a-31=-(a-8)^{2}+33 . \\
\text { Also, }(x-y)^{2}=(a+1)^{2}-4\left(a^{2}-7 a+16\right) \geqslant 0 \\
\Rightarrow-3 a^{2}+30 a-63 \geqslant 0 \Rightarrow 3 \leqslant a \leqslant 7 .
\end{array}
$$... | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,126 |
Four. (10 points) Let $f(n)$ be a function defined on $\mathbf{Z}$, and satisfies
$$
f(0)=1, f(1)=0 \text {. }
$$
For any $m, n \in \mathbf{Z}$, we have
$$
f(m+n)+f(m-n)=2 f(m) f(n) \text {. }
$$
(1) Prove that $f(n)$ is a periodic function, and determine $f(n)$;
(2) Calculate
$$
\frac{f\left(1^{2}+2^{2}+\cdots+2015^... | (1) Let $n=1$, from equation (1) we get
$$
\begin{array}{l}
f(m+1)+f(m-1)=2 f(m) f(1)=0 \\
\Rightarrow f(m+1)=-f(m-1) \\
\Rightarrow f(m+3)=-f(m+1)=f(m-1) .
\end{array}
$$
Thus, $f(m+4)=f(m)$, which means $f(n)$ is a periodic function with a period of 4.
By periodicity, we have
$$
f(n)=\left\{\begin{array}{ll}
0, & n=... | \frac{1}{1007} | Algebra | proof | Yes | Yes | cn_contest | false | 729,127 |
Five. (15 points) Let sets $S_{1}$ and $S_{2}$ be an equipotent bipartition of $S=\{2,3, \cdots, 2015\}$ (i.e., $S_{1} \cup S_{2}=S$, $S_{1} \cap S_{2}=\varnothing$, and $\left|S_{1}\right|=\left|S_{2}\right|=1007$). Denote the product of all numbers in set $S_{1}$ as $a$, and the product of all numbers in set $S_{2}$ ... | (1) The set $S$ contains 2014 numbers, of which 1007 are even and 1007 are odd.
If not all even numbers are in the same set $S_{1}$ or $S_{2}$, then each of $S_{1}$ and $S_{2}$ contains even numbers.
Thus, both $a$ and $b$ are divisible by 2.
Therefore, $p=a+b$ is divisible by 2.
If all even numbers are in one of the ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,128 |
Six. (15 points) As shown in Figure 3, $AL$, $BM$, and $CN$ are the angle bisectors of $\triangle ABC$, and $\angle ANM = \angle ALC$. Prove:
(1) $\angle ACB = 120^{\circ}$;
(2) $NM^2 + NL^2 = ML^2$.
保留源文本的换行和格式，直接输出翻译结果。 | (1) Let $\angle B A C=\alpha, \angle A B C=\beta$, $\angle A C B=\gamma$. Then $\alpha+\beta+\gamma=\pi$.
As shown in Figure 7, draw $M D / / A L$, intersecting $B C$ at point $D$, and connect $D N$.
Thus, $\angle C M D=\frac{\alpha}{2}$.
Since $\angle A N M=\angle A L C=\angle M D C$, we have $\angle B N M=\angle B D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,129 |
1. A contestant's nine scores, after removing the highest and the lowest, have an average of 91 for the remaining seven scores. In the stem-and-leaf plot of the nine scores made on-site, one data point is unclear and is represented by $x$ in Figure 1.
Then the variance of the seven remaining scores is ( ).
(A) $\frac{1... | 1. B.
From the conditions, we know that the data set is
$$
87,87,94,90,91,90,9 x, 99,91,
$$
The highest score is 99, and the lowest score is 87. The remaining data are $87, 94, 90, 91, 90, 90+x, 91$. Therefore,
$$
\begin{array}{l}
\frac{87+94+90+91+90+(90+x)+91}{7}=91 \\
\Rightarrow x=4 .
\end{array}
$$
Thus, the va... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,130 |
2. The interior of a sphere with radius $R$ contains four smaller spheres, each with radius $r$. The maximum possible value of $r$ is ( ).
(A) $\frac{\sqrt{3}}{2+\sqrt{3}} R$
(B) $\frac{\sqrt{6}}{3+\sqrt{6}} R$
(C) $\frac{1}{1+\sqrt{3}} R$
(D) $\frac{\sqrt{5}}{2+\sqrt{5}} R$ | 2. B.
From the conditions, we know that the centers of the four small balls form the four vertices of a regular tetrahedron, with the center of the large ball being the center of this tetrahedron.
The inradius of a regular tetrahedron can be found using the volume method, which gives $\frac{\sqrt{6}}{2} r$. Therefore... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,131 |
Example 6 As shown in Figure 14, let the extensions of the two pairs of opposite sides of convex quadrilateral $ABCD$ intersect at points $E$ and $F$, respectively. The circumcircle of $\triangle BEC$ intersects the circumcircle of $\triangle CFD$ at points $C$ and $P$. Prove that $\angle BAP = \angle CAD$ if and only ... | Proof As shown in Figure 14, note that $P$ is the Miquel point of the complete quadrilateral $A B E C F D$.
Let the circle passing through points $A, E, D$ be $\odot O_{1}$, and the circle passing through points $A, B, F$ be $\odot O_{2}$. By the Miquel theorem of the complete quadrilateral, we know that $\odot O_{1}$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,132 |
3. Given the sequences $\left\{a_{n}\right\} 、\left\{b_{n}\right\}$, for any $n \in \mathbf{Z}_{+}$, it holds that $a_{n}>b_{n}$. As $n \rightarrow+\infty$, the limits of the sequences $\left\{a_{n}\right\} 、\left\{b_{n}\right\}$ are $A 、 B$ respectively. Then ( ).
(A) $A>B$
(B) $A \geqslant B$
(C) $A \neq B$
(D) The r... | 3. B.
If we take $a_{n}=\frac{1}{n}, b_{n}=\frac{1}{n+1}$, then $a_{n}>b_{n}$.
And $\lim _{n \rightarrow+\infty} a_{n}=\lim _{n \rightarrow+\infty} b_{n}=0$, so $A>B$ is not necessarily true, but it can ensure that $A \geqslant B$. | B | Calculus | MCQ | Yes | Yes | cn_contest | false | 729,133 |
7. Given a triangle with sides as three consecutive natural numbers, the largest angle is twice the smallest angle. Then the perimeter of the triangle is $\qquad$ | II, 7.15.
Let the three sides of $\triangle ABC$ be $AB=n+1, BC=n, AC=n-1$, and the angle opposite to side $AC$ be $\theta$, and the angle opposite to side $AB$ be $2\theta$.
According to the Law of Sines and the Law of Cosines, we have
$$
\begin{array}{l}
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,134 |
12. Let the family of lines $M$ be defined as $x \cos \theta + (y-2) \sin \theta = 1 (0 \leqslant \theta \leqslant 2 \pi)$.
For the following four statements:
(1) All lines in the family $M$ pass through a fixed point;
(2) There exists a fixed point $P$ that does not lie on any line in the family $M$;
(3) For any integ... | 12. (2)(3).
From $x \cos \theta+(y-2) \sin \theta=1$, we know that the distance from point $P(0,2)$ to each line in the line system $M$ is
$$
d=\frac{1}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=1,
$$
which means the line system $M$ consists of all the tangents to the circle $\odot C: x^{2}+(y-2)^{2}=1$. Therefore, t... | (2)(3) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,135 |
Example 1 Find the minimum value of the function
$$
f(x)=|x-1|+|x-2|+\cdots+|x-10|
$$
. ${ }^{[1]}$ | 【Analysis】By the geometric meaning of absolute value, $\sum_{i=1}^{n}\left|x-a_{i}\right|$ represents the sum of distances from the point corresponding to $x$ on the number line to the points corresponding to $a_{i}(i=1,2, \cdots, n)$. It is easy to know that when the point corresponding to $x$ is in the middle of the ... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,136 |
13. (16 points) As shown in Figure 3, given that $AB$ is the hypotenuse of the right triangle $\triangle ABC$, and $I$ is its incenter. If the circumradius of $\triangle IAB$ is $R$, and the inradius of the right triangle $\triangle ABC$ is $r$, prove:
$$
R \geqslant (2 + \sqrt{2}) r.
$$ | In $\triangle I A B$,
$$
\angle A I B=180^{\circ}-\frac{1}{2}(\angle C A B+\angle C B A)=135^{\circ} \text {. }
$$
By the Law of Sines, we get $R=\frac{A B}{2 \sin \angle A I B}=\frac{\sqrt{2}}{2} A B$.
In the right triangle $\triangle A B C$, let $\angle B A C=\theta\left(0^{\circ}<\theta<90^{\circ}\right)$. Then the... | R \geqslant (2 + \sqrt{2}) r | Geometry | proof | Yes | Yes | cn_contest | false | 729,137 |
14. (16 points) As shown in Figure 4, $A$ and $B$ are the common vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. $P$ and $Q$ are moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and sati... | Prove: (1) $O, P, Q$ are collinear;
(2) If the slopes of the lines $AP, BP, AQ, BQ$ are respectively
14. (1) Note that,
$\overrightarrow{A P}+\overrightarrow{B P}=2 \overrightarrow{O P}, \overrightarrow{A Q}+\overrightarrow{B Q}=2 \overrightarrow{O Q}$.
Also, $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrigh... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,138 |
2. Can a straight line divide any convex polygon into two parts of equal perimeter, and with the largest side of both parts being equal? What if the requirement is changed to the smallest side being equal instead? | 2. Maximum edge equality is feasible, minimum edge equality is not.
For a convex polygon and a point $M$ on its boundary, a corresponding point $N$ (related to point $M$, denoted as $N(M))$ can be found such that $MN$ divides the convex polygon into two parts of equal perimeter, $M A_{1} \cdots A_{m} N$ and $M A_{m+1}... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,139 |
6. A country has several cities, each with a number. In the country's flight guide, there is an indicator between the numbers of each pair of cities indicating whether there is a direct flight between these two cities. For any two cities, given the numbers $M$ and $N$, if the city with number $M$ is changed to number $... | 6. Impossible.
Figure 4 shows the original direct flight relationships between cities.
Notice that this relationship graph is symmetrical about any diameter passing through the midpoint of the smallest chord. Therefore, by symmetry, the numbers of two adjacent cities can be swapped, and the original direct flight rel... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 729,141 |
Example 2 Figure 1 is a map of an industrial area, a highway (thick line) runs through the area, and seven factories $A_{1}, A_{2}$, $\cdots, A_{7}$ are distributed on both sides of the highway, connected to the highway by some small roads (thin lines). Now, a long-distance bus station is to be set up on the highway, w... | 【Analysis】The optimal station location is the point that minimizes the sum of the distances from the seven factories to the station.
Since the distances from each factory to the road intersection are fixed, we only need to consider the sum of the distances from the five small intersections $F, E, D, C, B$ to the stati... | [a, b] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,142 |
2. Given that $A D$, $B E$, and $C F$ are the altitudes of acute $\triangle A B C$, and $H$ is the orthocenter of $\triangle A B C$. Then $H$ is the ( ) of $\triangle D E F$.
(A) orthocenter
(B) centroid
(C) circumcenter
(D) incenter | 2. D.
It is easy to prove that points $A, E, H, F$ are concyclic $\Rightarrow \angle E F H = \angle E A H$. Similarly, $\angle H B D = \angle D F H$. Also, $\angle H B D = \angle E A H$, thus, $\angle E F H = \angle D F H$ $\Rightarrow F H$ bisects $\angle D F E$. Similarly, $E H$ bisects $\angle D E F$. Therefore, $H... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,143 |
5. Given real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\sqrt{a+b+c+d}+\sqrt{a^{2}-2 a+3-b}- \\
\sqrt{b-c^{2}+4 c-8}=3 .
\end{array}
$$
Then the value of $a-b+c-d$ is ( ).
(A) $-7 \quad \square \quad$ -
(B) -8
(C) -4
(D) -6 | 5. A.
$$
\begin{array}{l}
\text { Given } a^{2}-2 a+3-b \geqslant 0, \\
b-c^{2}+4 c-8 \geqslant 0,
\end{array}
$$
we know
$$
\begin{array}{l}
b \leqslant-(a+1)^{2}+4 \leqslant 4, \\
b \geqslant(c-2)^{2}+4 \geqslant 4 .
\end{array}
$$
Thus, $b=4, a=-1, c=2$.
Substituting these into the known equations gives $d=4$.
The... | -7 | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,144 |
6. As shown in Figure 1, given that $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, point $C$ is on $\odot O_{1}$ and outside $\odot O_{2}$, the extensions of $C A$ and $C B$ intersect $\odot O_{2}$ at points $D$ and $E$, respectively, with $A C=3$, $A D=6$, and the radius of $\odot O_{1}$ is 2. Then t... | 6. C.
Connect $C O_{1}$ and extend it to intersect $\odot O_{1}$ at point $F$, and intersect $D E$ at point $G$. Connect $A B$ and $A F$.
Then $\angle A F C=\angle A B C=\angle D$,
$$
\angle A C F=\angle G C D \text {. }
$$
Therefore, $\triangle A C F \backsim \triangle G C D$
$$
\begin{array}{l}
\Rightarrow \angle G... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,145 |
3. Given a positive real number $x$ satisfies
$$
x^{3}+x^{-3}+x^{6}+x^{-6}=2754 \text {. }
$$
then $x+\frac{1}{x}=$ | 3. 4 .
Transform the given equation into
$$
\begin{aligned}
& \left(x^{3}+x^{-3}\right)^{2}+\left(x^{3}+x^{-3}\right)-2756=0 \\
\Rightarrow & \left(x^{3}+x^{-3}+53\right)\left(x^{3}+x^{-3}-52\right)=0 .
\end{aligned}
$$
Notice that, $x^{3}+x^{-3}+53>0$.
Thus, $x^{3}+x^{-3}=52$.
Let $b=x+x^{-1}$.
Then $x^{3}+x^{-3}=\l... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,147 |
Example 3 Given real numbers $x_{1}, x_{2}, \cdots, x_{10}$ satisfy $\sum_{i=1}^{10}\left|x_{i}-1\right| \leqslant 4, \sum_{i=1}^{10}\left|x_{i}-2\right| \leqslant 6$. Find the average value $\bar{x}$ of $x_{1}, x_{2}, \cdots, x_{10}$. ${ }^{[2]}$ | 【Analysis】From the given conditions, we have
$$
\begin{array}{l}
10=\left|\sum_{i=1}^{10}\left[\left(x_{i}-1\right)-\left(x_{i}-2\right)\right]\right| \\
\leqslant \sum_{i=1}^{10}\left|x_{i}-1\right|+\sum_{i=1}^{10}\left|x_{i}-2\right| \leqslant 4+6=10 \\
\Rightarrow \sum_{i=1}^{10}\left|x_{i}-1\right|=4, \sum_{i=1}^{1... | 1.4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,148 |
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