problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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Example 4 As shown in Figure $5, \triangle A B C$ is inscribed in $\odot O$, the incircle $\odot I$ touches $A B, A C$ at points $J, K$ respectively, and $A O$ intersects $\odot O$ again at point $D$. Connect $D I$, extend $C A$ to point $F$ such that $A F=B J$. Draw a perpendicular from $F$ to $D I$, intersecting the ... | Prove that as shown in Figure 5, extending $DI$ intersects $\odot O$ at point $P$ and the extension of $FG$ at point $E$.
Then $FE \perp DE$, and $AP \perp IP$.
Since $\angle AJI = \angle AKI = \angle API = 90^{\circ}$
$\Rightarrow A, J, I, K, P$ are concyclic
$\Rightarrow \angle JPK = \angle JAK = \angle BPC$
$\Righta... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,568 |
Proposition 1 For a chord $EF$ of $\odot O$, take a point $P$ on it, and take two points $M, N$ on $EF$. Draw chords $AD, BC$ through $M, N$ respectively, and let $CD, AB$ intersect $EF$ at points $X, Y$. If
$$
\frac{1}{\overline{P M}}+\frac{1}{P N}=\frac{1}{P E}+\frac{1}{P F},
$$
then $\frac{1}{P X}+\frac{1}{P Y}=\fr... | Proposition 1 Proof Note that,
$$
\begin{array}{l}
(E Y, M F)=A(E Y, M F)=A(E B, D F) \\
=C(E D, B F)=C(E N, X F)=(E N, X F) . \\
\text { Hence } \frac{\overline{\overline{M Y}}}{\frac{\overline{E F}}{\overline{ }}}=\frac{\overline{\overline{X N}}}{\frac{\overline{E F}}{F N}} \Rightarrow \overline{E M} \cdot \overline{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,569 |
Proposition 2 For a real-coefficient linear equation $l(x, y)=0$, a real-coefficient quadratic equation $\omega(x, y)=0$, take points $A_{i}\left(x_{i}, y_{i}\right)$ $(i=1,2,3,4)$ satisfying $\omega\left(x_{i}, y_{i}\right)=0$.
If there exists $P\left(x_{0}, y_{0}\right)$ satisfying
$$
l\left(x_{0}, y_{0}\right)=0, H(... | For the proof of Proposition 2, without loss of generality, let
$$
P(0,0), l(x, y)=y \text{, }
$$
otherwise, appropriate translation and rotation can be applied.
First, consider the real-coefficient quadratic equation
$$
\Omega(x, y)=A x^{2}+B y^{2}+C x y+D x+E y+F=0 \text{, }
$$
and find the value of \( H(P, l, \Ome... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,570 |
Let $O, G, K, S_{1}, S_{2}$ be the circumcenter, centroid, symmedian point, first isogonic center, and second isogonic center of $\triangle ABC$, respectively, and let $R$ be the circumradius. Then,
$$
\frac{G S_{1}}{G S_{2}}=\left(\frac{R}{O K}-\sqrt{\frac{R^{2}}{O K^{2}}-1}\right)^{3} .
$$
It is well known that the ... | 【Note】1. With the three sides of $\triangle ABC$ as sides, construct equilateral $\triangle BCD$, equilateral $\triangle CAE$, and equilateral $\triangle ABF$ outside the shape, then $AD$, $BE$, and $CF$ are concurrent, and the point of concurrency is called the first Fermat point; constructing three similar equilatera... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,571 |
1. Given that $\odot O_{1}$ and $\odot O_{2}$ are disjoint, with radii $r_{1}$ and $r_{2}$ respectively ($r_{1} < r_{2}$), $AB$ and $XY$ are two internal common tangents of the two circles, with points $A$ and $X$ on $\odot O_{1}$, and points $B$ and $Y$ on $\odot O_{2}$. The circle with $AB$ as its diameter intersects... | 1. As shown in Figure 1.
Let $A B$ intersect $X Y$ at point $S$. Then $S$ is the internal homothetic center of $\odot O_{1}$ and $\odot O_{2}$, and $\frac{A S}{B S}=\frac{r_{1}}{r_{2}}$.
Let $M$ be the midpoint of $A B$.
Since $\angle A P B=\angle A Q B=90^{\circ}$, we have $M A=M P=M B=M Q$, which implies that $M P$ ... | \sqrt{\frac{r_{1}}{r_{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,573 |
2. Let $G$ be a simple graph of order 100. It is known that for any vertex $u$, there exists another vertex $v$ such that $u$ and $v$ are adjacent, and there is no vertex adjacent to both $u$ and $v$. Find the maximum possible number of edges in graph $G$. (Provided by Yunhao Fu) | 2. Let $G=(V, E)$.
For $u v \in E$, if there is no other vertex adjacent to both $u$ and $v$, then $u v$ is called a "good edge".
Let $E_{0}$ be the set of all good edges, and $G_{0}=\left(V, E_{0}\right)$.
From the problem statement, we know that each vertex in graph $G$ has at least one good edge, i.e., $G_{0}$ has ... | 3822 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,574 |
3. Prove: There exists a constant $C>0$, such that for any positive integer $m$, and any $m$ positive integers $a_{1}, a_{2}, \cdots, a_{m}$, we have
$$
\sum_{i=1}^{m} H\left(a_{i}\right) \leqslant C\left(\sum_{i=1}^{m} i a_{i}\right)^{\frac{1}{2}},
$$
where, $H(n)=\sum_{k=1}^{n} \frac{1}{k}$.
(Supplied by Yu Hongbing... | 3. By the rearrangement inequality, without loss of generality, assume
$$
a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{m} \text {. }
$$
For any positive integer $j$, let $s_{j}$ be the number of $a_{i}$ in $a_{1}, a_{2}, \cdots, a_{m}$ that are not less than $j$. Then when $j > a_{1}$, $s_{j} = 0$, and
$$
\begi... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,575 |
4. Let $A_{1}, A_{2}, \cdots, A_{n}$ be binary subsets of the set $\{1,2, \cdots, 2018\}$, such that the sets $A_{i}+A_{j}(1 \leqslant i \leqslant j \leqslant n)$ are all distinct, where,
$$
A+B=\{a+b \mid a \in A, b \in B\} \text {. }
$$
Find the maximum possible value of $n$.
(Qiu Zhenhua) | 4. The maximum value of $n$ is 4033.
If there exists $A_{i}=A_{j}(i \neq j)$, then $A_{i}+A_{i}=A_{j}+A_{j}$, which contradicts the condition.
Thus, $A_{1}, A_{2}, \cdots, A_{n}$ are all distinct.
For a binary set $A=\{a, b\}(a<b)$, let $S_{i}=\{x-y \mid x, y \in M_{i}, x>y\}$.
If there exists $i \neq j$ such that $S_... | 4033 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,576 |
5. In $\triangle A B C$, $\angle B A C>90^{\circ}$, and $\odot O$ is its circumcircle. The tangent to $\odot O$ at point $A$ intersects the tangents at points $B$ and $C$ at points $P$ and $Q$, respectively. Perpendiculars are drawn from $P$ and $Q$ to $B C$, with feet at $D$ and $E$, respectively. $F$ and $G$ are poin... | 5. As shown in Figure 2.
From the fact that points $A, F, B, E$ are concyclic and the tangent-chord theorem, we have
$$
\angle B E F = \angle B A F = \angle A C B.
$$
Thus, $A C \parallel E F$.
Similarly, $A B \parallel D G$.
Take a point $C'$ on $B C$ such that $C' E = C E$. Then
$$
\begin{array}{l}
\angle Q C' C = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,577 |
6. Find all pairs of positive integers $(x, y)$ such that
$$
(x y+1)(x y+x+2)
$$
is a perfect square.
(Proposed by Nian Xiaosheng) | 6. There do not exist positive integers $x, y$ that satisfy the requirement.
Proof by contradiction.
Assume there exist positive integers $x, y$ such that
$$
(x y+1)(x y+x+2)
$$
is a perfect square.
Then $x y+1=d u^{2}, x y+x+2=d v^{2}$, where $d, u, v \in \mathbf{Z}_{+}$, and $d$ is square-free.
Thus, $d v^{2} y - d... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,578 |
2. In a convex quadrilateral $ABCD$, $AB>AD$, $BC=CD$, $AC$ bisects $\angle BAD$, $\angle BAD=25^{\circ}$. Then the degree measure of $\angle BCD$ is ( ).
(A) $160^{\circ}$
(B) $155^{\circ}$
(C) $150^{\circ}$
(D) $145^{\circ}$ | 2. B.
As shown in Figure 3, take point $D_{1}$ on $AB$ such that $A D_{1}=A D$, and connect $C D_{1}$.
Then point $D$ and $D_{1}$ are symmetric with respect to $A C$.
$$
\begin{array}{l}
\Rightarrow \angle A D C=\angle A D_{1} C, B C=C D=C D_{1} \\
\Rightarrow \angle A D C+\angle A B C=180^{\circ} .
\end{array}
$$
Fr... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,579 |
Conclusion 3 The diameter $M N$ of the semicircle $\odot O$ lies on the side $B C$ of $\triangle A B C$, the sides $A B$ and $A C$ of $\triangle A B C$ are tangent to the semicircle arc at points $F$ and $E$, respectively. Draw $A D \perp B C$ at point $D$, then
(1) $A, F, D, E$ are concyclic;
(2) $A D$ bisects $\angle... | Conclusion 3 Proof: When point $D$ coincides with $O$, the two conclusions are obviously true.
When point $D$ does not coincide with $O$, as shown in Figure 3.
(1) Connect $F O, A O, E O$.
Since $O F \perp A B, O E \perp A C$, it follows that points $A, F, O, E$ are concyclic, with $A O$ as the diameter.
Since $A D \pe... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,580 |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If
$$
\begin{array}{l}
a=\frac{\sqrt{6+\sqrt{6+\cdots+\sqrt{6}}},}{2016 \text { nested radicals }}, \\
b=\frac{\sqrt[3]{6+\sqrt[3]{6+\cdots+\sqrt[3]{6}}},}{2 \text { 2017 nested radicals }},
\end{array}
$$
then $[a+b]=$ . $\qquad$ | 3. 4 .
Notice that,
$2.4<\sqrt{6}<a<\frac{\sqrt{6+\sqrt{6+\cdots+\sqrt{9}}}}{2016 \text{ levels }}=3$,
1. $8<\sqrt[3]{6}<b<\frac{\sqrt[3]{6+\sqrt[3]{6+\cdots+\sqrt[3]{8}}}}{2017 \text{ layers }}=2$.
Therefore, $[a+b]=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,581 |
Three, (10 points) If $\sum_{i=1}^{1344}(-1)^{i+1} \frac{1}{i}=\frac{m}{n}$, where $m, n$ are natural numbers, and $(m, n)=1$, prove: $2017 | m$.
| $$
\begin{array}{l}
\frac{m}{n}=\sum_{i=1}^{1344} \frac{1}{i}-2 \sum_{i=1}^{672} \frac{1}{2 i}=\sum_{i=673}^{1344} \frac{1}{i} \\
=\frac{1}{2} \sum_{i=673}^{1344}\left(\frac{1}{i}+\frac{1}{2017-i}\right) \\
=\frac{2017}{2} \sum_{i=673}^{1344} \frac{1}{i(2017-i)} .
\end{array}
$$
Noting that 2017 is a prime number, and... | 2017 \mid m | Number Theory | proof | Yes | Yes | cn_contest | false | 730,582 |
Four. (15 points) As shown in Figure 2, in $\triangle A B C$, $\angle A C B=45^{\circ}$. Through a point $D$ on $B C$, draw a perpendicular to $A B$, with the foot of the perpendicular at $H$. $H D$ intersects the extension of $A C$ at point $E$. If $A B = H D$, prove:
$$
A E^{2}=2 D H^{2}+2 D E^{2} .
$$ | $$
\begin{array}{l}
\text {Rt } \triangle B H D \\
\cong \operatorname{Rt} \triangle F A B \\
\Rightarrow B F=B D, \\
\quad \angle F B A=\angle B D H . \\
\text { Then } \angle F B D \\
=\angle F B A+\angle A B D \\
=\angle B D H+\angle H B D=90^{\circ} .
\end{array}
$$
Therefore, $\triangle F B D$ is an isosceles rig... | A E^{2}=2 D H^{2}+2 D E^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 730,583 |
5. If the cube of a three-digit positive integer is an eight-digit number of the form $\overline{A B C D C D A B}$, then such a three-digit number is | 5. 303 | 303 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,588 |
9. As shown in Figure 1, in $\triangle ABC$, $AM$ is a median, intersecting the incircle $\Gamma$ of $\triangle ABC$ at points $K$ and $L$. If
$$
A K: K L: L M=1: 2: 3,
$$
find $B C: C A: A B$. | 9. Let $A K=t, K L=2 t, L M=3 t$, and circle $\Gamma$ is tangent to $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$ respectively. Then
$$
C D=C E=x .
$$
Thus, $M D=\sqrt{15} t, A E=A F=\sqrt{3} t$.
It is easy to see that $M$ is the midpoint of $B C$, so
$$
\begin{array}{l}
B M=M C=x+\sqrt{15} t, \\
B F=B D=x+2 \sq... | B C: C A: A B=6:(4-\sqrt{5}):(4+\sqrt{5}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,591 |
1. Simplify:
$$
\sum_{k=1}^{2016}(k \sqrt{k+1}+(k+1) \sqrt{k})^{-1}=
$$
$\qquad$ | $$
-1.1-\frac{1}{\sqrt{2017}} .
$$
Notice,
$$
\begin{array}{l}
(k \sqrt{k+1}+(k+1) \sqrt{k})^{-1} \\
=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}} .
\end{array}
$$
Therefore, the required result is $1-\frac{1}{\sqrt{2017}}$. | 1-\frac{1}{\sqrt{2017}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,593 |
3. Place a regular tetrahedron with a volume of 1 inside a cube, then the minimum volume of this cube is $\qquad$ | 3. 3 .
Considering in reverse, for a cube with edge length $a$ (volume $a^{3}$), its largest inscribed regular tetrahedron has vertices formed by the cube's vertices that do not share an edge, and its volume is $\frac{a^{3}}{3}$.
$$
\text { Let } \frac{a^{3}}{3}=1 \text {, then } a^{3}=3 \text {. }
$$ | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,594 |
4. If a vertex of an ellipse is symmetric to one of its foci with respect to the directrix, then the eccentricity of the ellipse is $e=$ . $\qquad$ | 4. $\frac{1}{2}$ or $\frac{\sqrt{2}}{2}$.
Establish a Cartesian coordinate system, and let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
Then the vertices are $A_{1,2}=( \pm a, 0), B_{1,2}=(0, \pm b)$, and the foci are $F_{1,2}=( \pm c, 0)$,
with the directrix equations $l_{1,2}: x... | \frac{1}{2} \text{ or } \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,595 |
8. Arrange all positive integers whose sum of digits is 10 in ascending order to form the sequence $\left\{a_{n}\right\}$. If $a_{n}=2017$, then $n=$ . $\qquad$ | 8. 120.
It is easy to know that there are 9 two-digit numbers $\overline{a b}$ whose digits sum to 10.
For three-digit numbers $\overline{a b c}$ whose digits sum to 10, the first digit $a$ can take any value in $\{1,2, \cdots, 9\}$. Once $a$ is determined, $b$ can take any value in $\{0,1, \cdots, 10-a\}$, which give... | 120 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,596 |
12. (20 points) As shown in Figure 1, $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$. $CD$ is a line segment passing through point $A$, where points $C$ and $D$ lie on $\odot O_{1}$ and $\odot O_{2}$, respectively. Through any point $K$ on line segment $CD$ (excluding the endpoints), draw $KM \parallel... | 12. Let the radii of $\odot O_{1}$ and $\odot O_{2}$ be $r_{1}$ and $r_{2}$, respectively. Then $B C=2 r_{1} \sin \angle B A C, B D=2 r_{2} \sin \angle B A D$.
Since $\angle B A C + \angle B A D = 180^{\circ}$, we have $\frac{B C}{B D} = \frac{r_{1}}{r_{2}}$. Therefore, $\triangle B O_{1} C \backsim \triangle B O_{2} D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,597 |
2. Given the set $M=\{1,99,-1,0,25,-36, -91,19,-2,11\}$, let the non-empty subsets of $M$ be $M_{i}(i=1,2, \cdots, 1023)$. If the product of all elements in each $M_{i}$ is $m_{i}$, then $\sum_{i=1}^{1023} m_{i}=$ $\qquad$ . | 2. -1 .
Let the set $M=\left\{a_{i} \mid i=1,2, \cdots, n\right\}$. Then
$$
\begin{array}{l}
\sum_{i=1}^{2^{n}-1} m_{i}=\prod_{i=1}^{n}\left(a_{i}+1\right)-1 . \\
\text { Given }-1 \in M, \text { we know } \sum_{i=1}^{1023} m_{i}=-1 .
\end{array}
$$ | -1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,599 |
4. Let the three-digit number $n=\overline{a b c}$, where the lengths $a, b, c$ can form an isosceles (including equilateral) triangle. Then the number of such three-digit numbers $n$ is. $\qquad$ | 4. 165.
When $a=b=c$, there are 9 cases in total.
When $a=b \neq c, a=c \neq b, b=c \neq a$, the number of cases for each situation is the same. Taking $a=b \neq c$ as an example:
(1) When $a=b \geqslant 5$, $a$ and $b$ have 5 cases, and $c$ has 8 cases, making a total of 40 cases;
(2) When $a=b=4$, $c$ can be any one... | 165 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,600 |
5. Let $a, b, c$ be non-negative real numbers, satisfying
$$
a+b+c=8, \quad ab+bc+ca=16 \text{. }
$$
Let $m=\min \{ab, bc, ca\}$. Then the maximum possible value of $m$ is . $\qquad$ | 5. $\frac{16}{9}$.
Assume $c \geqslant b \geqslant a$. Then
$$
3 c \geqslant a+b+c=8 \Rightarrow c \geqslant \frac{8}{3} \text {. }
$$
$$
\begin{array}{l}
\text { By } 0 \leqslant(a-b)^{2}=(a+b)^{2}-4 a b \\
=(8-c)^{2}-4(16-c(a+b)) \\
=(8-c)^{2}-4(16-c(8-c)) \\
=c(16-3 c) \\
\Rightarrow 0 \leqslant c \leqslant \frac{1... | \frac{16}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,601 |
6. Given $p(x)=a x^{3}+b x^{2}+c x+d$ is a cubic polynomial, satisfying
$$
p\left(\frac{1}{2}\right)+p\left(-\frac{1}{2}\right)=1000 p(0) \text {. }
$$
Let $x_{1} 、 x_{2} 、 x_{3}$ be the three roots of $p(x)=0$. Then the value of $\frac{1}{x_{1} x_{2}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{1} x_{3}}$ is $\qquad$ . | 6. 1996.
From equation (1) we get
$$
\frac{1}{2} b+2 d=1000 d \Rightarrow \frac{b}{d}=1996 \text {. }
$$
By Vieta's formulas we get
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=-\frac{b}{a}, x_{1} x_{2} x_{3}=-\frac{d}{a} . \\
\text { Therefore } \frac{1}{x_{1} x_{2}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{1} x_{3}}=\frac{x_{1}+... | 1996 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,602 |
Example 2 As shown in Figure $5, C$ is the midpoint of the semicircular arc, $P$ is a point on the extension of the diameter $B A$, a tangent line $P D$ is drawn through point $P$ to the semicircle $\odot O$, with $D$ as the point of tangency, the angle bisector of $\angle D P B$ intersects $A C$ and $B C$ at points $E... | Prove that, as shown in Figure 5, connect $D E$.
By Conclusion 2 and Corollary $2(2)$, we know that $P 、 A 、 E 、 D$, and $D$ 、 $E 、 F 、 C$ are respectively concyclic. Therefore,
$$
\angle P D A=\angle P E A=\angle C E F=\angle C D F .
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,603 |
8. Let $I$ be the incenter of $\triangle A B C$, and
$$
3 \overrightarrow{I A}+4 \overrightarrow{I B}+5 \overrightarrow{I C}=\mathbf{0} \text {. }
$$
Then the size of $\angle C$ is | 8. $90^{\circ}$.
As shown in Figure 2, let $a$, $b$, and $c$ be the sides opposite to $\angle A$, $\angle B$, and $\angle C$ respectively. Draw lines through point $I$ parallel to the three sides of $\triangle ABC$, intersecting at points $D$, $N$, $E$, $M$, $F$, and $G$.
Since $I$ is the incenter of $\triangle ABC$,
... | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,604 |
9. Given $x, y, z \in \mathbf{R}_{+}$, and
$$
\begin{array}{l}
x^{2}+y^{2}+x y=1, y^{2}+z^{2}+y z=2, \\
z^{2}+x^{2}+z x=3 .
\end{array}
$$
Then $x+y+z=$ | 9. $\sqrt{3+\sqrt{6}}$.
As shown in Figure 3, construct a right-angled triangle
$$
\begin{array}{l}
\triangle A B C, A B=\sqrt{3}, C A= \\
\sqrt{2}, B C=1 .
\end{array}
$$
Let $F$ be the Fermat point of $\triangle A B C$, which satisfies
$$
\begin{array}{l}
\angle A F B=\angle B F C \\
=\angle C F A=\frac{2 \pi}{3} .... | \sqrt{3+\sqrt{6}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,605 |
13. (20 points) Let the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ pass through the point $P\left(\frac{\sqrt{6}}{2}, \frac{1}{2}\right)$, with an eccentricity of $\frac{\sqrt{2}}{2}$, and let the moving point $M(2, t)$ $(t>0)$.
(1) Find the standard equation of the ellipse.
(2) Find the equation of the... | 13. (1) From the problem, we know $\frac{\frac{6}{4}}{a^{2}}+\frac{\frac{1}{4}}{b^{2}}=1$. Then $\frac{1}{2}=\frac{c^{2}}{a^{2}}=\frac{a^{2}-b^{2}}{a^{2}}$.
Solving simultaneously, we get $a^{2}=2, b^{2}=1$.
Thus, the equation of the ellipse is $\frac{x^{2}}{2}+y^{2}=1$.
(2) It is easy to see that the midpoint of $O M$... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,606 |
2. Given $\alpha, \beta \in(0, \pi), \tan \alpha, \tan \beta$ are the two roots of the equation $x^{2}+3 x+1=0$. Then the value of $\cos (\alpha-\beta)$ is ( ).
(A) $\frac{1}{3}$
(B) $\frac{2}{3}$
(C) $\frac{\sqrt{5}}{3}$
(D) $\frac{\sqrt{5}}{2}$ | 2. B.
From the given information, we have
$$
\tan \alpha + \tan \beta = -3, \tan \alpha \cdot \tan \beta = 1 \text{.}
$$
Thus, $\tan \alpha$ and $\tan \beta$ are both less than 0.
Assume $\alpha \geqslant \beta$. Then $\pi > \alpha \geqslant \beta > \frac{\pi}{2}$.
Therefore, $\tan \alpha \geqslant \tan \beta, 0 \leqs... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,607 |
Example 3 As shown in Figure $6, A B$ is the diameter of semicircle $\odot O$, $C$ is the midpoint of the semicircle arc, $P$ is a point on the extension of $A B$, $P D$ is tangent to semicircle $\odot O$ at point $D$, the angle bisector of $\angle A P D$ intersects $A C$ and $B C$ at points $E$ and $F$ respectively. P... | Proof 1 As shown in Figure 6, connect $E D, D F$.
By Corollary 2, we know
$$
A E=E D, D F=F B, \angle E D F=90^{\circ} \text {. }
$$
Then $E F^{2}=E D^{2}+D F^{2}=A E^{2}+B F^{2}$.
Therefore, $A E, B F, E F$ can form a right-angled triangle.
Proof 2 As shown in Figure 6, draw $F G / / C A$, intersecting $A B$ at point... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,608 |
4. Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$
with points $A$ and $B$ on the ellipse such that $\overrightarrow{F_{1} A}=3 \overrightarrow{F_{2} B}$. Then the range of the eccentricity of the ellipse is ( ).
(A) $\left(0, \frac{1}{2}\right)$
(B)... | 4. C.
As shown in Figure 2.
From $\overrightarrow{F_{1} A}=3 \overrightarrow{F_{2} B}$, we get
$F_{2} B \parallel F_{1} A, \left|F_{2} B\right|=\frac{1}{3}\left|F_{1} A\right|$.
Thus, the line $A B$ intersects the $x$-axis at point $P(2 c, 0)$.
Let $M$ be a moving point on the ellipse, and the point that divides $P M$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,609 |
5. In $\triangle A B C$, $\overrightarrow{A B} \cdot \overrightarrow{B C}=3 \overrightarrow{C A} \cdot \overrightarrow{A B}$. Then the maximum value of $\frac{|\overrightarrow{A C}|+|\overrightarrow{A B}|}{|\overrightarrow{B C}|}$ is ().
(A) $\frac{\sqrt{5}}{2}$
(B) $\sqrt{3}$
(C) 2
(D) $\sqrt{5}$ | 5. B.
From the given conditions,
$$
\begin{array}{l}
a^{2}+c^{2}-b^{2}=3\left(b^{2}+c^{2}-a^{2}\right) \\
\Rightarrow 2 a^{2}=2 b^{2}+c^{2} . \\
\text { Then } (b+c)^{2}=\left(\frac{\sqrt{2}}{2} \cdot \sqrt{2} b+1 \cdot c\right)^{2} \\
\leqslant\left(\left(\frac{\sqrt{2}}{2}\right)^{2}+1^{2}\right)\left(2 b^{2}+c^{2}\... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,610 |
9. If $P(x, y)$ is a point on the hyperbola $\frac{x^{2}}{8}-\frac{y^{2}}{4}=1$, then the minimum value of $|x-y|$ is . $\qquad$ | 9. 2 .
From the condition, we know that $x^{2}-2 y^{2}-8=0$.
By symmetry, without loss of generality, assume
$$
\begin{array}{l}
x>0, y>0, u=x-y>0 . \\
\text { Then }(y+u)^{2}-2 y^{2}-8=0 \\
\Rightarrow y^{2}-2 u y-u^{2}+8=0 \\
\Rightarrow \Delta=(2 u)^{2}-4\left(-u^{2}+8\right) \geqslant 0 \\
\Rightarrow u \geqslant ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,611 |
10. Let the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ be $1, \alpha$ be the plane passing through the line $B D_{1}$. Then the range of the area of the section cut by $\alpha$ on the cube is $\qquad$ . | 10. $\left[\frac{\sqrt{6}}{2}, \sqrt{2}\right]$.
As shown in Figure 3, let the section be $B M D_{1} N$, and the distance from point $M$ to line $B D_{1}$ be $h$. Then the area of the section is
$$
S=2 S_{\triangle B V D_{1}}=B D_{1} h .
$$
Obviously, the minimum value of $h$ is the distance between the skew lines $A... | \left[\frac{\sqrt{6}}{2}, \sqrt{2}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,612 |
11. Given real numbers $x_{1}, x_{2}, x_{3}$ satisfy
$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{1} x_{2}+x_{2} x_{3}=2 \text {. }
$$
Then the maximum value of $\left|x_{2}\right|$ is $\qquad$ | 11. 2 .
From the condition, we have
$$
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}+x_{3}^{2}=4 \text {. }
$$
Notice that,
$$
\begin{array}{l}
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2} \geqslant \frac{x_{2}^{2}}{2}, \\
x_{3}^{2}+\left(x_{2}+x_{3}\right)^{2} \geqslant \frac{x_{2}^{2}}{2} .
\end{arr... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,613 |
12. Let the set $M=\{1,2, \cdots, 10\}$,
$$
\begin{aligned}
A= & \{(x, y, z) \mid x, y, z \in M, \text { and } \\
& \left.9 \mid\left(x^{3}+y^{3}+z^{3}\right)\right\} .
\end{aligned}
$$
Then the number of elements in set $A$ is $\qquad$ . | 12.243.
Notice that,
when $x \equiv 1(\bmod 3)$, $x^{3} \equiv 1(\bmod 9)$;
when $x \equiv 2(\bmod 3)$, $x^{3} \equiv-1(\bmod 9)$;
when $x \equiv 0(\bmod 3)$, $x^{3} \equiv 0(\bmod 9)$.
Thus, for $x \in \mathbf{Z}$, we have
$$
x^{3} \equiv 0,1,-1(\bmod 9) \text {. }
$$
To make $9 \mid\left(x^{3}+y^{3}+z^{3}\right)$, ... | 243 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,614 |
13. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=a, a_{n+1}=\frac{5 a_{n}-8}{a_{n}-1}\left(n \in \mathbf{Z}_{+}\right) .
$$
(1) If $a=3$, prove that $\left\{\frac{a_{n}-2}{a_{n}-4}\right\}$ is a geometric sequence, and find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If for any... | Three, 13. (1) From the given information,
$\frac{a_{n+1}-2}{a_{n+1}-4}=\frac{\frac{5 a_{n}-8}{a_{n}-1}-2}{\frac{5 a_{n}-8}{a_{n}-1}-4}=3 \cdot \frac{a_{n}-2}{a_{n}-4}$.
Thus, $\left\{\frac{a_{n}-2}{a_{n}-4}\right\}$ is a geometric sequence.
Hence, $\frac{a_{n}-2}{a_{n}-4}=\frac{a_{1}-2}{a_{1}-4} \cdot 3^{n-1}=-3^{n-1}... | a \in (3, +\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,615 |
Example 1 Given $a, b, c \in \mathbf{R}_{+}, 0 \leqslant k<2$. Prove:
$$
\sum \frac{a^{2}-b c}{b^{2}+c^{2}+k a^{2}} \geqslant 0,
$$
where, $\sum$ denotes the cyclic sum. | $$
\begin{aligned}
& 2 \sum \frac{a^{2}-b c}{b^{2}+c^{2}+k a^{2}} \\
& =\sum \frac{(a+b)(a-c)+(a-b)(a+c)}{b^{2}+c^{2}+k a^{2}} \\
& =\sum\left(\frac{(a+b)(a-c)}{b^{2}+c^{2}+k a^{2}}+\frac{(c-a)(c+b)}{a^{2}+b^{2}+k c^{2}}\right) \\
& =\sum(a-c)\left(\frac{a+b}{b^{2}+c^{2}+k a^{2}}-\frac{c+b}{a^{2}+b^{2}+k c^{2}}\right) ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,616 |
Example 2 Given $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\begin{array}{l}
\sum \sqrt[4]{\frac{\left(a^{2}+b^{2}\right)\left(a^{2}-a b+b^{2}\right)}{2}} \\
\leqslant \frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right) \sum \frac{1}{a+b} .
\end{array}
$$
(2010, Turkey National Team Selection Exam) | Prove that,
$$
\begin{array}{l}
\sum \sqrt[4]{\frac{\left(a^{2}+b^{2}\right)\left(a^{2}-a b+b^{2}\right)}{2}} \\
=\sum \sqrt[4]{\frac{\left(a^{2}+b^{2}\right)\left(a^{3}+b^{3}\right)}{2(a+b)}} \\
=\sum \sqrt[4]{\frac{a^{3}+b^{3}}{a^{2}+b^{2}} \cdot \frac{a^{2}+b^{2}}{a+b} \cdot \frac{a^{2}+b^{2}}{a+b} \cdot \frac{a+b}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,617 |
6. If $\triangle A B C$ satisfies:
$$
\tan \frac{A}{2} \cdot \tan \frac{C}{2}=\frac{1}{3}, b=\frac{4}{3} a,
$$
then $\sin B=$ | $\begin{array}{l}\text { 6. } \frac{4}{5} \text {. } \\ \text { From } \tan \frac{A}{2} \cdot \tan \frac{C}{2}=\frac{1}{3} \\ \Rightarrow \frac{1-\cos A}{\sin A} \cdot \frac{\sin C}{1+\cos C}=\frac{1}{3} \\ \Rightarrow \frac{a^{2}-(b-c)^{2}}{(a+b)^{2}-c^{2}}=\frac{1}{3} \\ \Rightarrow \frac{a-b+c}{a+b+c}=\frac{1}{3} \\... | \frac{4}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,618 |
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Set $a_{n}=\sum_{k=1}^{n}\left[\frac{n}{k}\right]$. Then the number of even numbers in $a_{1}, a_{2}, \cdots, a_{2018}$ is . $\qquad$ | 8. 1028.
It is easy to see that if and only if $k \mid n$,
$$
\left[\frac{n}{k}\right]-\left[\frac{n-1}{k}\right]=1 \text{. }
$$
Otherwise, $\left[\frac{n}{k}\right]-\left[\frac{n-1}{k}\right]=0$.
Thus, $a_{n}-a_{n-1}$
$$
\begin{array}{l}
=\sum_{k=1}^{n}\left[\frac{n}{k}\right]-\sum_{k=1}^{n-1}\left[\frac{n-1}{k}\rig... | 1028 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,619 |
One, (40 points) Given a set $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ of $n$ positive integers that satisfies: for any two different subsets of set $A$, the sums of their respective elements are not equal. Find the minimum value of $\sum_{i=1}^{n} \sqrt{a_{i}}$.
| Let's assume $1 \leqslant a_{1}<a_{2}<\cdots<a_{n}$. First, we prove that for any $i(1 \leqslant i \leqslant n)$, it must hold that
$$
\sum_{j=1}^{i} a_{j} \geqslant 2^{i}-1 \text {. }
$$
Since the $2^{i}-1$ non-empty subsets of $\left\{a_{1}, a_{2}, \cdots, a_{i}\right\}$ have distinct sums, and these sums are all po... | (\sqrt{2}+1)\left(\sqrt{2^{n}}-1\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,620 |
1. Given $a, b, c \in \left[\frac{1}{\sqrt{2}}, \sqrt{2}\right]$. Prove: $\sum \frac{3}{a+2b} \geqslant \sum \frac{2}{a+b}$. | ```
\begin{array}{l}
=\sum \frac{a-b}{(a+2 b)(a+b)} \\
=\sum\left(\frac{a-b}{(a+2 b)(a+b)}-\frac{a-b}{6 a b}\right) \\
=\sum \frac{(2 b-a)(a-b)^{2}}{6 a b(a+2 b)(a+b)} \geqslant 0 .
\end{array}
``` | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,621 |
Given $m \geqslant 3$ and $m \in \mathbf{N}$, let $a_{1}, a_{2}$, $\cdots, a_{m}>0, n \geqslant m$ and $n \in \mathbf{N}$. Prove:
$\sum_{i=1}^{m}\left(\frac{a_{i}}{a_{i}+a_{i+1}}\right)^{n} \geqslant \frac{m}{2^{n}}\left(\right.$ with the convention $a_{m+1}=a_{1}$ ). | Prove that let $b_{i}=\frac{a_{i+1}}{a_{i}}(i=1,2, \cdots, m)$.
Then $\prod_{i=1}^{m} b_{i}=1$.
Thus, the original inequality $\Leftrightarrow \sum_{i=1}^{m}\left(\frac{1}{1+b_{i}}\right)^{n} \geqslant \frac{m}{2^{n}}$.
By the power mean inequality, we have
$\sqrt[n]{\frac{\sum_{i=1}^{m}\left(\frac{1}{1+b_{i}}\right)^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,622 |
Given $a 、 b 、 c > 0, a + b + c \leqslant 3$. Prove:
$$
\frac{\sqrt{a^{3}}}{(b+c)^{2}}+\frac{\sqrt{b^{3}}}{(c+a)^{2}}+\frac{\sqrt{c^{3}}}{(a+b)^{2}} \geqslant \frac{3}{4} .
$$ | Let \( A = \frac{\sqrt{a^{3}}}{(b+c)^{2}} + \frac{\sqrt{b^{3}}}{(c+a)^{2}} + \frac{\sqrt{c^{3}}}{(a+b)^{2}} \), \( B = 2(ab + bc + ca) \).
Applying the AM-GM inequality, we get
\[
\begin{array}{l}
\frac{\frac{\sqrt{a^{3}}}{(b+c)^{2}}}{A} + \frac{a(b+c)}{B} + \frac{a(b+c)}{B} \geqslant \frac{3 a^{\frac{7}{6}}}{\sqrt[3]... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,623 |
For $c$ being a real number, find the minimum value of $c$ such that for any positive real numbers $a_{1}, a_{2}, a_{3}, a_{4}$, the following inequality holds:
$$
\begin{array}{l}
\frac{1}{4} \sum_{i=1}^{4} a_{i}-\sqrt[4]{a_{1} a_{2} a_{3} a_{4}} \\
\leqslant c \max _{1 \leqslant i<j \leqslant 4}\left\{\left(\sqrt{a_{... | Notice that for any $x_{1}, x_{2}, x_{3}, x_{4} > 0$, we have
$$
\begin{array}{l}
\sum_{i=1}^{4} x_{i}^{2}-4 \sqrt{x_{1} x_{2} x_{3} x_{4}} \\
=\left(x_{1}-x_{2}\right)^{2}+2\left(\sqrt{x_{1} x_{2}}-\sqrt{x_{3} x_{4}}\right)^{2}+\left(x_{3}-x_{4}\right)^{2} .
\end{array}
$$
Without loss of generality, assume $x_{1} \g... | \frac{3}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,624 |
Let $x, y, z$ be positive real numbers. Prove:
$$
\sum y z \sqrt{y^{2}+z^{2}+y z} \geqslant \sqrt{\prod\left(x^{2}+y^{2}+x y\right)},
$$
where $\sum, \Pi$ denote cyclic sums and cyclic products, respectively. | Prove formula (1)
$$
\Leftrightarrow \sum \frac{y z}{\sqrt{\left(x^{2}+y^{2}+x y\right)\left(z^{2}+x^{2}+z x\right)}} \geqslant 1 \text {. }
$$
Notice,
$$
\begin{array}{l}
\sqrt{\left(x^{2}+y^{2}+x y\right)\left(z^{2}+x^{2}+z x\right)} \\
\leqslant x^{2}+y z+x \frac{y^{2}+z^{2}}{y+z} \\
\Leftrightarrow \frac{\left(x^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,625 |
2. Given $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\sum \frac{b+c}{a} \geqslant \frac{\left(\sum a^{2}\right)\left(\sum a b\right)}{a b c \sum a}+3 \text {. }
$$
(2006, Romanian Mathematical Olympiad) | To prove the inequality, subtract 6 from both sides, and simplify to transform it into
$$
\sum(c a+b c-a b)(a-b)^{2} \geqslant 0 .
$$
Without loss of generality, assume $a \geqslant b \geqslant c$.
Given $S_{a} \geqslant 0, S_{b} \geqslant 0, S_{b}+S_{c} \geqslant 0$, we know the original inequality holds. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,626 |
3. Given $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\sum \frac{b+c-2 a}{a^{2}+b c} \geqslant 0 \text {. }
$$ | 提示 化简后变形为
$$
\sum\left(c^{2}+a b\right)(a+b-c)(b-a)^{2} \geqslant 0 \text {. }
$$
不妨设 $a \geqslant b \geqslant c$.
由 $S_{b} \geqslant 0, S_{c} \geqslant 0, b^{2} S_{a}+a^{2} S_{b} \geqslant 0$, 知原不等式成立.
Hint, after simplification, it transforms to
$$
\sum\left(c^{2}+a b\right)(a+b-c)(b-a)^{2} \geqslant 0 \text {. }
$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,627 |
4. Given $x, y, z \in [1,2]$. Prove:
$$
\left(\sum x\right)\left(\sum \frac{1}{x}\right) \geqslant 6 \sum \frac{x}{y+z} \text {. }
$$
(2006, Vietnam Mathematical Olympiad) | To prove the inequality, subtract 9 from both sides, and after simplification and transformation, we get
$$
\sum(x-y)^{2} z(x+y)((z+x)(y+z)-3 x y) \geqslant 0 .
$$
Assume without loss of generality that $2 \geqslant x \geqslant y \geqslant z \geqslant 1$.
We can prove: $S_{x} \geqslant 0, S_{y} \geqslant 0, S_{y}+S_{z... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,628 |
5. Given $a, b, c \geqslant 1, a+b+c=9$. Prove: $\sqrt{\sum ab} \leqslant \sum \sqrt{a}$. | Hint: Simplify and transform to
$$
\sum(\sqrt{a}-\sqrt{b})^{2}\left(\frac{(\sqrt{a}+\sqrt{b})^{2}}{6}-1\right) \geqslant 0 \text {. }
$$
Assume without loss of generality that $a \geqslant b \geqslant c$. Then $a \geqslant 3, b \geqslant 1, c \geqslant 1$.
It can be proven that: $S_{b} \geqslant 0, S_{c} \geqslant 0, ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,629 |
6. Given $x, y, z \in \mathbf{R}_{+}$. Prove:
$$
\left(\sum x y\right)\left(\sum \frac{1}{(x+y)^{2}}\right) \geqslant \frac{9}{4} \text {. }
$$
(1996, Iran Mathematical Olympiad) | Hint: Simplify and transform to
$$
\sum\left(\frac{2}{(z+x)(y+z)}-\frac{1}{(x+y)^{2}}\right)(x-y)^{2} \geqslant 0 \text {. }
$$
Assume without loss of generality that $x \geqslant y \geqslant z$.
We can prove: $S_{y}>0, S_{z}>0$.
It is easy to show that $\frac{x-z}{x+z} \geqslant \frac{y-z}{y+z}$.
Thus $\sum S_{z}(x-y... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,630 |
Question 1 As shown in Figure 1, in the acute triangle $\triangle ABC$, $AB \neq AC$, $K$ is the midpoint of the median $AD$, $DE \perp AB$ at point $E$, $DF \perp AC$ at point $F$, the lines $KE$ and $KF$ intersect $BC$ at points $M$ and $N$ respectively, and the circumcenters of $\triangle DEM$ and $\triangle DFN$ ar... | Proof of a lemma first.
Lemma In $\triangle ABC$, $D$ is the midpoint of $BC$, and the circle $\Gamma$ with diameter $AD$ intersects $AB$ and $AC$ at points $E$ and $F$ respectively. The tangents to circle $\Gamma$ at $E$ and $F$ intersect at point $P$. Prove: $PD \perp BC$.
Proof Draw $AQ \parallel BC$, intersecting ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,632 |
Example 3 Given $a, b, c \in \mathbf{R}_{+}, a+b+c=abc$. Prove:
$$
\sum \frac{\sqrt{1+a^{2}}}{a}-\sum \frac{1}{a} \geqslant \sqrt{3} .
$$ | Let $\frac{1}{a}=x, \frac{1}{b}=y, \frac{1}{c}=z$.
Then the given equation is equivalent to $x y+y z+z x=1$.
$$
\begin{array}{l}
\text { Therefore, equation (3) } \Leftrightarrow \sqrt{1+x^{2}}-\sum x \geqslant \sqrt{3} \\
\Leftrightarrow \sum \sqrt{(x+y)(x+z)}-\sum x \geqslant \sqrt{3 \sum x y} \\
\Leftrightarrow \su... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,633 |
Proposition 1 Let $A O_{2}$ be a diameter of circle $\Gamma_{1}$ with center $O_{1}$. Let $\Gamma_{2}$ be a circle with center $O_{2}$ and radius not exceeding the length of $AO_{2}$. The line $l$ is the line through the two intersection points of circles $\Gamma_{1}$ and $\Gamma_{2}$. Take a point $B$ on circle $\Gamm... | The proof of Proposition 1 utilizes the theorem of power of a point with respect to a circle ${ }^{[1]}$, which states that the difference of the powers of a point with respect to two non-concentric circles is equal to twice the product of the distance between the centers of the circles and the distance from the point ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,635 |
Inference 1 For $\triangle A B C, D$ is the tangency point of the A-excircle and the circumcircle, $E$ is the tangency point of the incircle and side $B C$, the line $D E$ intersects the circumcircle at a second point $F$, then $A F / / B C$. | Proof of Inference 1 As shown in Figure 1, let $I$ be the incenter of $\triangle A B C$, and $G, H$ be the midpoints of arcs $\overparen{B A C}$ and $\overparen{B C}$, respectively.
It is easy to see that $G, I, D$ are collinear ${ }^{[2]}, A, I, H$ are collinear.
By the properties of the incenter, it is known that th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,636 |
Inference 2 For $\triangle A B C, D$ is the tangency point of the A-excircle and the circumcircle, $E$ is a point on the circumcircle such that $A E \perp I E, T$ is the intersection of the angle bisector of $\angle B A C$ and $B C$, the second intersection of line $D T$ with the circumcircle is $F$, then $E F / / B C$... | For the proof of Corollary 2, as shown in Figure 2, let $L$ be the antipode of $A$ on the circumcircle, and $G, H$ be the midpoints of arcs $\overparen{B A C}$ and $\overparen{B C}$, respectively. Then quadrilateral $A H L G$ is a rectangle, and $G L / / I T$. Since $A E \perp I E$, points $E, I, L$ are collinear. Comb... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,637 |
Proposition 2 For $\triangle A B C, D$ is the tangency point of the $A$-mixtilinear incircle with the circumcircle, $J$ is the tangency point of the $A$-excircle with side $B C$. Then $\angle B A J=\angle C A D$.
| For the proof of Proposition 2, as shown in Figure 3, let $G$ and $H$ be the midpoints of arcs $\overparen{B A C}$ and $\overparen{B C}$, respectively. Let $E$ be the point where the incircle touches side $B C$, and $F$ be the point symmetric to $A$ with respect to the diameter $G H$. The second intersection point of l... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,638 |
Question 2 As shown in Figure 5, in $\triangle ABC$, $I$ is the incenter, $AI$ intersects $BC$ at point $T$, and the incircle touches side $BC$ at point $E$. A line $AF$ is drawn through point $A$ parallel to $BC$, intersecting the circumcircle at the second point $F$. Take a point $K$ on $EF$ such that points $A$, $E$... | Proof: Let $D$ be the point of tangency of the A-excircle of $\triangle ABC$ with the circumcircle, and $G$ be the midpoint of arc $\overparen{BAC}$.
By Corollary 1 and homothety, we know that $D, E, F$ and $G, I, D$ are collinear respectively.
Thus, $\angle IDK = \angle IDA$.
Since $A, K, T, E$ are concyclic, then
\[
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,639 |
Example 4 Given $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\sum a \leqslant \sum \frac{a b}{a+b}+\frac{1}{2} \sum \frac{a b}{c} \text {. }
$$
(2009, Oliforum Mathematical Olympiad) | Proof Note that,
$$
\begin{array}{l}
\text { Equation (4) } \Leftrightarrow 2 \sum a \leqslant \sum \frac{2 a b}{a+b}+\sum \frac{a b}{c} \\
\Leftrightarrow \sum\left(\frac{a+b}{2}-\frac{2 a b}{a+b}\right) \leqslant \frac{1}{2} \sum\left(\frac{a b}{c}+\frac{b c}{a}-2 b\right) \\
\Leftrightarrow \sum \frac{(a-b)^{2}}{2(a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,640 |
3. Anna and Bob are playing a "directing" game on the edges of an infinite square grid. Define each step of the directing operation as follows: choose any undirected unit edge, and determine a direction (there are two choices) along the orientation of the edge as the direction of the edge. The game starts with Anna, an... | 3. Bob does not have a winning strategy, Anna has a way to prevent Bob from winning.
Definition: For two unit edges with a common vertex, if the common vertex is exactly at the bottom and leftmost of its three vertices (including one common vertex), then these two edges form an "LL corner"; if the common vertex is exa... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,641 |
6. Given an increasing sequence composed of powers of 3 or the sum of several different powers of 3: $1,3,4,9,10,12,13, \cdots$. Then the 100th term of this sequence is $\qquad$ | 6. 981.
The terms of the sequence are given by $\sum_{i=0}^{n} 3^{i} a_{i}$, where,
$$
a_{i} \in\{0,1\}(i=1,2, \cdots, n) \text {. }
$$
When $n=5$, there are $2^{6}-1=63$ numbers that can be formed, and the 64th term is $3^{6}=729$.
Starting from the 65th term, there are $2^{5}-1=31$ terms that do not contain $3^{5}... | 981 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,642 |
II. (16 points) Let $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$, and the equation
$$
\cos 2 x-4 a \cos x-a+2=0
$$
has two distinct solutions. Find the range of values for $a$. | $$
2 \cos ^{2} x-4 a \cos x-a+1=0 \text {. }
$$
Let $t=\cos x$.
From $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$, we know $0 \leqslant t \leqslant 1$.
Then the problem is transformed into the equation
$$
f(t)=2 t^{2}-4 a t-a+1=0
$$
in $0 \leqslant t \leqslant 1$.
(1) The equation has a unique solution in $0 ... | \frac{3}{5}<a \leqslant 1 \text{ or } a=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,643 |
Three. (20 points) Suppose a circle intersects a rectangular hyperbola at points $A_{1}, A_{2}, A_{3}, A_{4}$, where $A_{1}$ and $A_{2}$ are the endpoints of a diameter of the circle.
(1) Prove: The midpoint of segment $A_{3} A_{4}$ is the center of the hyperbola;
(2) Find the angle between the tangents to the hyperbol... | (1) Let's assume:
The equation of the hyperbola is $x y=a(a \neq 0)$,
The equation of the circle is $x^{2}+y^{2}+2 D x+2 E y+F=0$,
The intersection points are $A_{i}\left(x_{i}, y_{i}\right)(i=1,2,3,4)$.
By eliminating $y$, we get
$$
x^{4}+2 D x^{3}+F x^{2}+2 a E x+a^{2}=0 \text {. }
$$
Then $x_{i}$ are the roots of e... | 90^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 730,644 |
$$
\begin{array}{l}
A=\left\{x \mid \log _{2}(x-1)<1\right\}, \\
B=\{x|| x-a \mid<2\} .
\end{array}
$$
If $A \cap B \neq \varnothing$, then the range of real number $a$ is
$\qquad$ | -1. $(-1,5)$.
From $\log _{2}(x-1)<1 \Rightarrow A=(1,3)$,
$$
|x-a|<2 \Rightarrow B=(a-2, a+2) \text {. }
$$
If $A \cap B=\varnothing$, then
$a+2 \leqslant 1$ or $a-2 \geqslant 3$
$\Rightarrow a \leqslant-1$ or $a \geqslant 5$.
Therefore, when $A \cap B \neq \varnothing$, the range of values for $a$ is $(-1,5)$. | (-1,5) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,645 |
2. Given that $f(x)$ is an odd function defined on $\mathbf{R}$, and the function $y=f(x+1)$ is an even function. When $-1 \leqslant x \leqslant 0$, $f(x)=x^{3}$. Then $f\left(\frac{9}{2}\right)=$ $\qquad$ . | 2. $\frac{1}{8}$.
From the function $y=f(x+1)$ being an even function, we know
$$
f(-x+1)=f(x+1) \text {. }
$$
Since $f(x)$ is an odd function, then
$$
\begin{array}{l}
f(x+2)=f(-x)=-f(x), \\
f(x+4)=-f(x+2)=f(x) .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence } f\left(\frac{9}{2}\right)=f\left(\frac{1}{2}\right)=-... | \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,646 |
Example 6 Given that $x, y, z$ are non-negative real numbers, and $x^{2}+y^{2}+z^{2}=2(xy+yz+zx)$. Prove: $\frac{x+y+z}{3} \geqslant \sqrt[3]{2xyz}$.
(2015, Moldova Mathematical Olympiad) | To prove that when $x, y, z$ have one of them being 0, the inequality obviously holds.
It suffices to prove the case where $x, y, z$ are all positive real numbers.
Without loss of generality, assume $x \geqslant y \geqslant z$. From
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=2(x y+y z+z x) \\
\Leftrightarrow(\sqrt{x}+\sqrt{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,647 |
5. In the tetrahedron $P-ABC$, $\triangle ABC$ is an equilateral triangle with side length $2 \sqrt{3}$, $PB=PC=\sqrt{5}$, and the dihedral angle $P-BC-A$ is $45^{\circ}$. Then the surface area of the circumscribed sphere of the tetrahedron $P-$ $ABC$ is $\qquad$ | 5. $25 \pi$.
As shown in Figure 2, take the midpoint $D$ of $BC$, and connect $AD$, $PD$.
From the problem, we know that
$AD \perp BC, PD \perp BC, PD=\sqrt{2}$.
Thus, $\angle PDA$ is the plane angle of the dihedral angle $P-BC-A$, $\angle PDA=45^{\circ}, BC \perp$ plane $PAD$, plane $PAD \perp$ plane $ABC$.
Construct... | 25 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,648 |
7. In $\triangle A B C$, the side lengths opposite to $\angle A 、 \angle B 、 \angle C$ are $a 、 b 、 c$, respectively, and
$$
\begin{array}{l}
\sin C \cdot \cos \frac{A}{2}=(2-\cos C) \sin \frac{A}{2}, \\
\cos A=\frac{3}{5}, a=4 .
\end{array}
$$
Then the area of $\triangle A B C$ is . $\qquad$ | 7.6 .
From equation (1) we know
$$
\begin{array}{l}
2 \sin \frac{A}{2}=\sin \left(C+\frac{A}{2}\right) \\
\Rightarrow 2 \sin A=2 \sin \left(C+\frac{A}{2}\right) \cdot \cos \frac{A}{2} \\
=\sin C+\sin B .
\end{array}
$$
Thus, $c+b=2a$.
Also, $a^{2}=b^{2}+c^{2}-2 b c \cos A$, which means
$$
4^{2}=b^{2}+(8-b)^{2}-2 b(8-... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,649 |
8. If the equation with respect to $x$
$$
x^{2}+a x+b-3=0(a, b \in \mathbf{R})
$$
has real roots in the interval $[1,2]$, then the minimum value of $a^{2}+(b-4)^{2}$ is . $\qquad$ | 8. 2 .
From the problem, we know that $b=-x^{2}-a x+3$.
Then $a^{2}+(b-4)^{2}$
\[
\begin{array}{l}
=a^{2}+\left(-x^{2}-a x-1\right)^{2} \\
=x^{2}(x+a)^{2}+2\left(x^{2}+a x\right)+a^{2}+1 \\
=\left(x^{2}+1\right)(x+a)^{2}+x^{2}+1 .
\end{array}
\]
Also, $x \in[1,2]$, so,
\[
a^{2}+(b-4)^{2} \geqslant x^{2}+1 \geqslant 2... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,650 |
9. The function
$$
f(x)=\sqrt{2 x-7}+\sqrt{12-x}+\sqrt{44-x}
$$
has a maximum value of $\qquad$ | 9.11.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(\sqrt{2 x-7}+\sqrt{12-x}+\sqrt{44-x})^{2} \\
\leqslant(3+2+6)\left(\frac{2 x-7}{3}+\frac{12-x}{2}+\frac{44-x}{6}\right) \\
=11^{2},
\end{array}
$$
The equality holds if and only if $\frac{9}{2 x-7}=\frac{4}{12-x}=\frac{36}{44-x}$, which is when $x=8$... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,651 |
10. Let $A$, $B$, and $C$ be three distinct points on $\odot O$, and $\angle A O B=120^{\circ}$, point $C$ lies on the minor arc $\overparen{A B}$ (point $C$ does not coincide with $A$ or $B$). If $\overrightarrow{O C}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}(\lambda, \mu \in \mathbf{R})$, then the range o... | 10. $(1,2]$.
Connect $O C$, intersecting $A B$ at point $D$. Let $\overrightarrow{O D}=m \overrightarrow{O C}$. Then, by the problem, $\overrightarrow{O D}=m \lambda \overrightarrow{O A}+m \mu \overrightarrow{O B}$, and $m \lambda+m \mu=1 \Rightarrow \lambda+\mu=\frac{1}{m}$.
Assume the radius of the circle is 1, and... | 1<\lambda+\mu \leqslant 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,652 |
12. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ passes through the point $P(-2,1)$, and the eccentricity is $\frac{\sqrt{2}}{2}$. Two perpendicular lines are drawn through point $P$ intersecting the ellipse at points $A$ and $B$ ($A$ and $B$ do not coincide with point $P$). Prove: The line $... | 12. According to the problem, we have
$$
\frac{4}{a^{2}}+\frac{1}{b^{2}}=1 \text {, and } \frac{c}{a}=\frac{\sqrt{a^{2}-b^{2}}}{a}=\frac{\sqrt{2}}{2} \text {. }
$$
Solving, we get $a^{2}=6, b^{2}=3$.
Thus, the ellipse $C: \frac{x^{2}}{6}+\frac{y^{2}}{3}=1$.
It is clear that the line $l_{A B}$ has a slope, let
$$
l_{A ... | \left(-\frac{2}{3},-\frac{1}{3}\right) | Geometry | proof | Yes | Yes | cn_contest | false | 730,653 |
$$
\begin{array}{l}
M=\left\{x \left\lvert\, x=\sin \frac{(2 m-3) \pi}{6}\right., m \in \mathbf{Z}\right\}, \\
N=\left\{y \left\lvert\, y=\cos \frac{n \pi}{3}\right., n \in \mathbf{Z}\right\} .
\end{array}
$$
Then the relationship between $M$ and $N$ is ( ).
(A) $M \subsetneq N$
(B) $N \varsubsetneqq M$
(C) $M=N$
(D) ... | 一, 1. B.
By periodicity, $M=N=\left\{ \pm 1, \pm \frac{1}{2}\right\}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,654 |
2. Given $a=\log _{8} 5, b=\log _{4} 3, c=\frac{2}{3}$. Then the size relationship of $a$, $b$, and $c$ is ( ).
(A) $a>b>c$
(B) $b>c>a$
(C) $b>a>c$
(D) $c>b>a$ | 2. C.
Notice,
$$
a=\log _{2} \sqrt[3]{5}>1, b=\log _{2} \sqrt{3}>1, c=\frac{2}{3}<1 \text {. }
$$
From $5^{2}<3^{3} \Rightarrow \sqrt[3]{5}<\sqrt{3}$. Therefore, $c<a<b$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,655 |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=1, a_{n+1}=n+1+a_{n}\left(n \in \mathbf{Z}_{+}\right) .
$$
Then the value of $\left[\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2018}}\right]$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 2... | 3. B.
$$
\begin{array}{l}
a_{n}-a_{1}=\sum_{k=1}^{n-1}\left(a_{k+1}-a_{k}\right)=\sum_{k=1}^{n-1}(k+1) \\
=\frac{n(n+1)}{2}-1 \\
\Rightarrow \frac{1}{a_{n}}=\frac{2}{n(n+1)} .
\end{array}
$$
Therefore, the required value is $\left[2\left(1-\frac{1}{2019}\right)\right]=2$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,656 |
7. If the irreducible fraction $\frac{p}{q}\left(p, q \in \mathbf{Z}_{+}\right)$ is converted to a decimal as $0.18 \cdots$, then when $q$ is the smallest, the value of $p$ is ().
(A) 9
(B) 7
(C) 5
(D) 2 | 7. D.
From the problem, we have
$0.18<\frac{p}{q}<0.19 \Leftrightarrow \frac{100 p}{19}<q<\frac{100 p}{18}$.
It is evident that as $p$ increases, the lower bound of $q$ also increases.
When $p=1$, there is no integer $q$ that satisfies the condition;
When $p=2$, $q=11$ satisfies the condition.
Therefore, $q_{\text {mi... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,657 |
Example $\mathbf{8}$ Given $a, b, c \in \mathbf{R}_{+}, abc=1$. Prove:
$$
\sum a^{3}+\sum \frac{a b}{a^{2}+b^{2}} \geqslant \frac{9}{2} \text {. }
$$
(2015, Moldova Mathematical Olympiad) | Proof Note that,
Equation (6)
$$
\begin{aligned}
\Leftrightarrow & \sum a^{3}-3 a b c+\sum\left(\frac{a b}{a^{2}+b^{2}}-\frac{1}{2}\right) \geqslant 0 \\
\Leftrightarrow & \frac{1}{2}\left(\sum a\right)\left(\sum(a-b)^{2}\right)- \\
& \sum \frac{(a-b)^{2}}{2\left(a^{2}+b^{2}\right)} \geqslant 0 \\
\Leftrightarrow & \su... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,658 |
9. Let $\triangle A B C$ have internal angles $\angle A, \angle B, \angle C$ with opposite sides $a, b, c$ respectively, and $\angle A - \angle C = \frac{\pi}{2}, a, b, c$ form an arithmetic sequence. Then the value of $\cos B$ is . $\qquad$ | II. $\frac{3}{4}$.
It is easy to know,
$$
\begin{array}{l}
2 b=a+c \\
\Leftrightarrow 2 \sin (A+C)=\sin A+\sin C \\
\Leftrightarrow 2 \sin \left(2 C+\frac{\pi}{2}\right)=\sin \left(C+\frac{\pi}{2}\right)+\sin C \\
\Leftrightarrow 2 \cos 2 C=\cos C+\sin C \\
\Rightarrow \cos C-\sin C=\frac{1}{2} .
\end{array}
$$
Theref... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,659 |
10. As shown in Figure 1, the parabola $y^{2}=2 p x(p>0)$ has a focus at $F$ and a directrix at $l$. A line passing through point $F$ intersects the parabola at points $A$ and $B$, and $|A B|=3 p$. Let the projections of points $A$ and $B$ on $l$ be $A^{\prime}$ and $B^{\prime}$, respectively. A point $M$ is randomly t... | 10. $\frac{1}{3}$.
The required probability is
$$
\begin{array}{c}
\frac{S_{\triangle A^{\prime} B^{\prime} F}}{S_{\text {trapezoid } A A^{\prime} B^{\prime} B}}=\frac{\left|A^{\prime} B^{\prime}\right| p}{\left|A^{\prime} B^{\prime}\right|\left(\left|A A^{\prime}\right|+\left|B B^{\prime}\right|\right)} \\
=\frac{p}{... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,660 |
One. (20 points) Given the function
$$
f(x)=2 \cos x(\cos x+\sqrt{3} \sin x)-1(x \in \mathbf{R}) \text {. }
$$
(1) Find the intervals where the function $f(x)$ is monotonically increasing;
(2) Let points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots$, $P_{n}\left(x_{n}, y_{n}\right), \cdots$ a... | (1) From the problem, we have
$$
f(x)=\cos 2 x+\sqrt{3} \sin 2 x=2 \sin \left(2 x+\frac{\pi}{6}\right) \text {. }
$$
Therefore, the monotonic increasing interval of $f(x)$ is
$$
\left[-\frac{\pi}{3}+k \pi, \frac{\pi}{6}+k \pi\right](k \in \mathbf{Z}) \text {. }
$$
(2) Let $t_{n}=2 x_{n}+\frac{\pi}{6}, t_{1}=2 x_{1}+\f... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,661 |
Three. (20 points) As shown in Figure 3, $M$ is the midpoint of side $BC$ of acute $\triangle ABC$, $\odot O$ passes through point $A$ and is tangent to line $BC$ at point $C$, line $AM$ intersects $\odot O$ at a second point $D$, and line $BD$ intersects $\odot O$ at a second point $E$. Prove:
$$
\angle EAC = \angle B... | Three, by the power of a point theorem
$$
\begin{array}{l}
M C^{2}=M D \cdot M A \\
\Rightarrow B M^{2}=M D \cdot M A \\
\Rightarrow \triangle A B M \backsim \triangle B D M \\
\Rightarrow \angle A B C=\angle B D M=\angle A D E=\angle A C E .
\end{array}
$$
Also, the angle between the chord and the tangent $\angle A C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,662 |
2. Let the sequence $\left\{a_{n}\right\}$ satisfy:
$$
\begin{array}{l}
a_{1}=1, a_{n+1}=5 a_{n}+1(n=1,2, \cdots) \\
\text { then } \sum_{n=1}^{2018} a_{n}=
\end{array}
$$ | $\begin{array}{l} \text { 2. } \frac{5^{2019}}{16}-\frac{8077}{16} . \\ \text { Given } a_{n+1}=5 a_{n}+1 \\ \Rightarrow a_{n+1}+\frac{1}{4}=5\left(a_{n}+\frac{1}{4}\right) \\ \Rightarrow a_{n}=\frac{5^{n}}{4}-\frac{1}{4} . \\ \text { Then } \sum_{n=1}^{2018} a_{n}=\frac{1}{4} \sum_{n=1}^{2018} 5^{n}-\frac{2018}{4} \\ ... | \frac{5^{2019}}{16}-\frac{8077}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,664 |
3. Given $\alpha, \beta \in\left(\frac{3 \pi}{4}, \pi\right)$,
$$
\cos (\alpha+\beta)=\frac{4}{5}, \sin \left(\alpha-\frac{\pi}{4}\right)=\frac{12}{13} \text {. }
$$
Then $\cos \left(\beta+\frac{\pi}{4}\right)=$ $\qquad$ | 3. $-\frac{56}{65}$.
From the problem, we have
$$
\sin (\alpha+\beta)=-\frac{3}{5}, \cos \left(\alpha-\frac{\pi}{4}\right)=-\frac{5}{13} .
$$
Then $\cos \left(\beta+\frac{\pi}{4}\right)$
$$
\begin{array}{l}
=\cos (\alpha+\beta) \cdot \cos \left(\alpha-\frac{\pi}{4}\right)+\sin (\alpha+\beta) \cdot \sin \left(\alpha-\... | -\frac{56}{65} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,665 |
4. Choose any two numbers from $2, 4, 6, 7, 8, 11, 12, 13$ to form a fraction. Then, there are $\qquad$ irreducible fractions among these fractions. | 4. 36 .
Among 7, 11, 13, choose one number and among $2, 4, 6, 8, 12$, choose one number to form a reduced fraction, there are $2 \mathrm{C}_{3}^{1} \mathrm{C}_{5}^{1}$ $=30$ kinds; among $7, 11, 13$, choose two numbers to form a reduced fraction, there are $\mathrm{A}_{3}^{2}=6$ kinds.
There are a total of 36 differe... | 36 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,666 |
7. In $\triangle A B C$, $A B+A C=7$, and the area of the triangle is 4. Then the minimum value of $\sin A$ is $\qquad$ . | 7. $\frac{32}{49}$.
From $A B + A C = 7 \Rightarrow A B \cdot A C \leqslant \frac{49}{4}$, and from $\frac{1}{2} A B \cdot A C \sin A = 4 \Rightarrow \sin A \geqslant \frac{32}{49}$. When $A B = A C = \frac{7}{2}$, the equality holds in both expressions. Therefore, the minimum value of $\sin A$ is $\frac{32}{49}$. | \frac{32}{49} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,667 |
8. Let $f(x)=|x+1|+|x|-|x-2|$. Then $f(f(x))+1=0$ has $\qquad$ distinct solutions.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 8. 3 .
Notice,
$$
f(x)=\left\{\begin{array}{ll}
-x-3, & x \leqslant-1 ; \\
x-1, & -1 < x \leqslant 2 ; \\
2x-3, & x > 2 .
\end{array}\right.
$$
From $f(f(x))+1=0$
$\Rightarrow f(x)=-2$ or 0
$\Rightarrow x=-1$ or -3 or $\frac{1}{3}$, a total of three solutions. | null | Other | math-word-problem | Yes | Yes | cn_contest | false | 730,668 |
13. (20 points) Let real numbers $x_{1}, x_{2}, \cdots, x_{2018}$ satisfy
$$
\begin{array}{l}
x_{n+1}^{2} \leqslant x_{n} x_{n+2}(n=1,2, \cdots, 2016), \\
\prod_{n=1}^{2018} x_{n}=1 .
\end{array}
$$
Prove: $x_{1009} x_{1010} \leqslant 1$. | 13. From the condition, we know that $x_{n}$ and $x_{n+2}$ have the same sign.
Proof by contradiction. Assume $x_{1009} x_{1010}>1$. From the condition, we know that $x_{1}, x_{2}, \cdots, x_{2018}$ have the same sign.
$$
\begin{array}{l}
\text { From } x_{n+1}^{2} \leqslant x_{n} x_{n+2} \\
\Rightarrow \frac{x_{n+1}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,669 |
1. Given $a>b>2$, and $a+b$, $a-b$, $a b$, $\frac{b}{a}$ form a geometric sequence in a certain order. Then $a=$ $\qquad$ . | $$
-1.7+5 \sqrt{2} .
$$
Since $a>b>2$, we have
$$
\begin{array}{l}
a b-(a+b)=(a-1)(b-1)-1>0, \\
\frac{b}{a}b^{2}=\frac{b}{a} \cdot a b$, then when the common ratio is greater than 1,
it can only be $\frac{b}{a} 、 a-b 、 a+b 、 a b$ forming a geometric sequence, at this time,
$$
\begin{array}{l}
(a-b)(a+b)=a b \cdot \fra... | 7+5 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,671 |
2. Given $f(x)=\lg (x+1)-\frac{1}{2} \log _{3} x$. Then the set
$$
M=\left\{n \mid f\left(n^{2}-8 n-2018\right) \geqslant 0, n \in \mathbf{Z}\right\}
$$
the number of subsets of $M$ is $\qquad$. | 2. 1.
For any $0< x_2 < x_1$, we have
$$
\begin{array}{l}
f\left(x_{1}\right)-f\left(x_{2}\right)=\lg \frac{x_{1}}{x_{2}}-\frac{\lg \frac{x_{1}}{x_{2}}}{\lg 9}>0 .
\end{array}
$$
Thus, $f(x)$ is a decreasing function on the interval $(0,+\infty)$. Note that, $f(9)=0$.
Therefore, when $x>9$, $f(x)f(9)=0$.
Hence, $f(x)... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,672 |
4. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with left and right foci $F_{1}$ and $F_{2}$, upper and lower vertices $B_{1}$ and $B_{2}$, and right vertex $A$, the line $A B_{1}$ intersects $B_{2} F_{1}$ at point D. If $2\left|A B_{1}\right|=3\left|B_{1} D\right|$, then the eccentricity of ... | 4. $\frac{1}{4}$.
Connect $B_{1} F_{2}$. Then $B_{1} F_{2} / / B_{2} D$.
$$
\begin{array}{l}
\text { Hence } \frac{\left|A B_{1}\right|}{\left|B_{1} D\right|}=\frac{\left|A F_{2}\right|}{\left|F_{1} F_{2}\right|} \\
\Rightarrow \frac{3}{2}=\frac{a-c}{2 c} \\
\Rightarrow e=\frac{c}{a}=\frac{1}{4} .
\end{array}
$$ | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,673 |
Example 1 Given that two circles intersect at points $P$ and $Q$, $AC$ and $BD$ are two chords passing through $P$, and $H_{1}$ and $H_{2}$ are the orthocenters of $\triangle PCD$ and $\triangle PAB$. Prove: The point symmetric to $Q$ with respect to $AC$ lies on $H_{1}H_{2}$. ${ }^{[2]}$ | Prove as shown in Figure 4, let $D H_{1}$ intersect the larger circle at point $E$, and $Q E$ intersect the smaller circle at point $F$.
Then $\angle D E Q=\angle D P Q=\angle B A Q=\angle B F E$
$\Rightarrow B F / / D E \Rightarrow B F \perp A C$
$\Rightarrow B F$ passes through point $H_{2}$.
From $H_{1}$ being the o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,674 |
Example 2 As shown in Figure 5, point $C$ is on $BD$, the incenter of $\triangle ABC$ is $X$, the excenter of $\triangle ACD$ opposite vertex $A$ is $Y$, the excenter of $\triangle ABD$ opposite vertex $B$ is $Z$, and the orthocenter of $\triangle XYZ$ is $H$. Prove: point $H$ lies on $CD$.
untranslated text remains ... | Proof From the given, we have
$$
\begin{array}{l}
\angle X A Z=180^{\circ}-\angle X A C-\angle Z A D \\
= 180^{\circ}-\frac{1}{2}\left(180^{\circ}-\angle C A D\right) \\
= 90^{\circ}+\frac{1}{2} \angle C A D, \\
\angle X Y Z=90^{\circ}-\frac{1}{2} \angle C A D .
\end{array}
$$
Therefore, points $A, X, Y, Z$ are concy... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,675 |
In a $100 \times 2018$ grid, fill each cell with 0 or 1, such that each column contains at least 75 ones. Prove: There exists a way to remove 95 rows so that the resulting $5 \times 2018$ grid has at most one column entirely composed of 0s. | Let $[x]$ denote the greatest integer not exceeding the real number $x$.
From the problem, the number of 1s in the original table is at least
$$
75 \times 2018 = 151350.
$$
By $\left[\frac{151350}{100}\right] + 1 = 1514$, we know there must be a row (denoted as the $i_{1}$-th row) with at least 1514 1s. Removing the c... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,676 |
Three, (50 points) Given $2 n$ positive real numbers $a_{1}, a_{2}, \cdots$, $a_{2 n}$ satisfying $a_{1}<a_{2}<\cdots<a_{2 n}$. Denote
the above text in English, please retain the original text's line breaks and format, and output the translation result directly.
Three, (50 points) Given $2 n$ positive real numbers ... | $$
\begin{array}{l}
\text { Three, let } U=\frac{1}{n-1} \sum_{1 \leqslant i(S+T)\left(S\left(T^{2}-V\right)+T\left(S^{2}-U\right)\right) \\
\Leftrightarrow(S+T)(S V+T U)>S T(T-S)^{2} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(S+T)(S V+T U) \geqslant(\sqrt{S T U}+\sqrt{T S V})^{2} ... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,677 |
For positive real numbers $a_{1}, a_{2}, \cdots, a_{n}\left(n \in \mathbf{Z}_{+}\right.$, $n \geqslant 2$ ) satisfying:
$$
a_{1} \leqslant 2, \sum_{i=1}^{k} a_{i} \leqslant 3 \times 2^{k-1}-1(k=2,3, \cdots, n),
$$
prove: When $n \geqslant 2$, $\frac{2}{a_{1}}+\sum_{k=2}^{n} \frac{3 \times 2^{k-2}}{a_{k}} \geqslant n$. | Prove that,
$$
\begin{array}{l}
\frac{2}{a_{1}}+\frac{a_{1}}{2} \geqslant 2, \\
\frac{3 \times 2^{k-2}}{a_{k}}+\frac{a_{k}}{3 \times 2^{k-2}} \geqslant 2(k=2,3, \cdots, n) .
\end{array}
$$
Adding the above $n$ inequalities, we get
$$
\begin{array}{l}
\frac{2}{a_{1}}+\sum_{k=2}^{n} \frac{3 \times 2^{k-2}}{a_{k}} \\
\ge... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,679 |
Given
$$
f(x)=x^{4}+a x^{3}+b x^{2}+c x+d
$$
has four complex roots $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$,
$$
g(x)=x^{3}-b x^{2}+(a c-4 d) x+4 b d-a^{2} d-c^{2}
$$
has three complex roots $\beta_{1}, \beta_{2}, \beta_{3}$. Prove:
$$
\prod_{1 \leqslant i<j \leqslant 4}\left(\alpha_{i}-\alpha_{j}\right)^{2}=... | Prove that without loss of generality,
$$
\left\{\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right\}=\{e, f, g, h\} .
$$
By Vieta's formulas, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
a=-(e+f+g+h), \\
b=e f+e g+e h+f g+f h+g h, \\
c=-(e f g+e f h+f g h+e g h), \\
d=e f g h .
\end{array}\right. \\
\text { ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,680 |
Given an acute triangle $\triangle ABC$, circles $\odot O_{1}$ and $\odot O_{2}$ pass through point $A$ and are tangent to $BC$ at points $B$ and $C$, respectively. The other intersection point of the two circles is $D$. A line through point $A$ different from $AD$ intersects $\odot O_{1}$ and $\odot O_{2}$ at points $... | Prove: As shown in Figure 2, connect $A D$.
Since $B C$ is tangent to both circles, we have
$$
\begin{array}{l}
\angle C B D=\angle B A D, \angle B C D=\angle C A D \\
\Rightarrow \angle C B D+\angle B C D=\angle B A C \\
\Rightarrow \angle B A C+\angle B D C=180^{\circ} .
\end{array}
$$
Also, $\triangle A B C$ is an ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,681 |
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