problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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2. In a country, some cities have direct two-way flights between them. It is known that one can fly from any city to any other city with no more than 100 flights, and also one can fly from any city to any other city with an even number of flights. Find the smallest positive integer $d$ such that it is guaranteed that f... | Prompt Answer: $d=200$.
First, construct an example, then prove that $d=200$ is feasible. For any two cities $A$ and $B$, consider the shortest path connecting them that has an even length, let its length be $2k$. Use proof by contradiction combined with the extremal principle to show that $k \leqslant 100$. | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,241 |
Example 2 Given $a+b+c=1$,
$$
b^{2}+c^{2}-4 a c+6 c+1=0 \text{. }
$$
Find the value of $a b c$. | Solve: From equation (1), we get $a=1-b-c$.
Substitute into equation (2) and rearrange to get
$$
\begin{array}{l}
b^{2}+5 c^{2}+4 b c+2 c+1=0 \\
\Rightarrow(b+2 c)^{2}+(c+1)^{2}=0 \\
\Rightarrow b+2 c=c+1=0 \Rightarrow b=2, c=-1 \\
\Rightarrow a=1-b-c=0 \Rightarrow a b c=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,242 |
2. Given that there are six circles on a plane with radii no less than 1, and they do not intersect with each other. Prove: if a certain circle intersects with these six circles, then the radius of that circle is no less than 1. | 2. Let the centers of the six circles be $O_{1}, O_{2}, \cdots, O_{6}$, with radii $R_{1}, R_{2}, \cdots, R_{6}$, respectively.
Let the radius of $\odot O$ be $R$, and it intersects with these six circles. Clearly, there exist indices $i, j$ such that $\angle O_{i} O O_{j} \leqslant 60^{\circ}$. By the given condition... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,243 |
1. Define the operation " $*$ ":
$$
a * b=\log _{2}\left(2^{a}+2^{b}\right)(a 、 b \in \mathbf{R}) \text {. }
$$
Let $A=(1 * 3) * 5, B=(2 * 4) * 6$. Then the value of $1 *(A * B)$ is $\qquad$ | 1.7 .
From the definition of $*$, we know that $2^{a * b}=2^{a}+2^{b}$. Therefore, $2^{A}=2^{1 * 3}+2^{5}=2^{1}+2^{3}+2^{5}$, $2^{B}=2^{2}+2^{4}+2^{6}$.
Thus, $2^{1 *(A * B)}=2^{1}+2^{A}+2^{B}$ $=2+2^{1}+2^{2}+\cdots+2^{6}=2^{7}$.
Hence, $1 *(A * B)=7$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,244 |
2. Given the integer coefficient equation about $x$
$$
x^{2}+(k+3) x+(2 k+3)=0
$$
has one positive root and one negative root, and the absolute value of the positive root is less than the absolute value of the negative root. Then the positive root of the equation is $\qquad$ | 2. $\frac{-1+\sqrt{5}}{2}$.
From the conditions, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
(k+3)^{2}-4(2 k+3)>0, \\
2 k+3=x_{1} x_{2}0
\end{array}\right. \\
\Rightarrow k \in\left(-3,-\frac{3}{2}\right) .
\end{array}
$$
Thus, the integer value of $k$ is -2. At this point, the equation is $x^{2}+x-1=0$, and t... | \frac{-1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,245 |
3. In the Cartesian coordinate system $x O y$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{4}=1(a>0)$, respectively. The line $l$ passes through the right vertex $A$ of the hyperbola and the point $B(0,2)$. If the sum of the distances from points $F_{1}$ and $F_{2}... | 3. $(2 \sqrt{2}, 0)$.
From the given, $c=\sqrt{a^{2}+4}$.
It is easy to know, $\sin \angle O A B=\frac{2}{c},\left|B F_{2}\right|=\sqrt{c^{2}+4}$.
The sum of the distances from points $F_{1}$ and $F_{2}$ to line $l$ is
$$
\begin{array}{l}
\left|A F_{1}\right| \sin \angle O A B+\left|A F_{2}\right| \sin \angle O A B \\... | (2 \sqrt{2}, 0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,246 |
4. Given a regular quadrilateral pyramid $P-A B C D$ with all edges of equal length. Taking $A B C D$ as one face, construct a cube $A B C D-E F G H$ on the other side of the pyramid. Then, the cosine value of the angle formed by the skew lines $P A$ and $C F$ is $\qquad$ | 4. $\frac{2+\sqrt{2}}{4}$.
Assume the edge lengths of the quadrilateral pyramid $P-ABCD$ are all 2. Taking the center $O$ of the square $ABCD$ as the origin, and the directions of $\overrightarrow{DA}$, $\overrightarrow{DC}$, and $\overrightarrow{OP}$ as the positive directions of the $x$-axis, $y$-axis, and $z$-axis,... | \frac{2+\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,247 |
5. Let $z$ be a complex number with modulus 1. Then the maximum value of $\left|\frac{z+\mathrm{i}}{z+2}\right|$ is $\qquad$ .
| 5. $\frac{2 \sqrt{5}}{3}$.
Let $z=\cos \theta+\mathrm{i} \sin \theta$.
$$
\begin{array}{l}
\text { Then } p=\left|\frac{z+\mathrm{i}}{z+2}\right|^{2}=\left|\frac{\cos \theta+(\sin \theta+1) \mathrm{i}}{(\cos \theta+2)+\mathrm{i} \sin \theta}\right|^{2} \\
=\frac{2+2 \sin \theta}{5+4 \cos \theta} \\
\Rightarrow 5 p-2=2... | \frac{2 \sqrt{5}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,248 |
6. In a regular $(n+2)$-gon, two vertices are randomly selected, and a line passing through them is drawn. Let $p_{n}$ denote the probability that the difference in the number of vertices on either side of this line does not exceed 1. Then the value of $\sum_{i=1}^{2018} p_{i} p_{i+1}$ is $\qquad$ | 6. $\frac{1009}{1010}$.
By symmetry, without loss of generality, consider a fixed vertex $A$.
When $n=2 k$, among the remaining $2 k+1$ vertices, only the vertex symmetric to $A$ about the center can be chosen, so $p_{2 k}=\frac{1}{2 k+1}$.
When $n=2 k-1$, among the remaining $2 k$ vertices, one of the two farthest p... | \frac{1009}{1010} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,249 |
3. Let real numbers $a$ and $b$ satisfy
$$
\begin{array}{l}
a^{2}\left(b^{2}+1\right)+b(b+2 a)=40, \\
a(b+1)+b=8 .
\end{array}
$$
Find the value of $\frac{1}{a^{2}}+\frac{1}{b^{2}}$.
(2014, National Junior High School Mathematics League) | Hint: Transform the two equations in the conditions to get $a^{2} b^{2}+(a+b)^{2}=40, a b+(a+b)=8$. Let $x=a+b, y=a b$.
Thus, $x^{2}+y^{2}=40, x+y=8$.
Solving yields $(x, y)=(2,6)$ (discard) or $(6,2)$. Therefore, $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{(a+b)^{2}-2 a b}{a^{2} b^{2}}=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,250 |
8. Figure 1 is a road map of a city, where $A$, $B, \cdots, I$ represent nine bus stops. A bus departs from station $A$, travels along the roads, reaches each bus stop exactly once, and finally returns to station $A$. The number of different sequences of stops it can pass through is
| 8. 32.
The bus route can be considered as a circle $\Gamma$ (direction is not considered for now), and each station on $\Gamma$ has exactly two adjacent stations.
Let the two stations adjacent to $I$ on $\Gamma$ be $\alpha$ and $\beta$. There are two cases:
(1) $\{\alpha, \beta\} \cap \{E, F, G, H\} \neq \varnothing$... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,251 |
11. (20 points) In the Cartesian coordinate system $x O y$, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $P$ is any point on the ellipse $C$ that is not on the major axis, and $Q$ is a point on the line segment $F_{1} F_{2}$ such that the inradii ... | 11. As shown in Figure 3.
Let \( P\left(x_{0}, y_{0}\right), Q(m, 0), |P Q|=t, \) and the inradii of \( \triangle P Q F_{1} \) and \( \triangle P Q F_{2} \) are both \( r \). Therefore,
\[
\begin{array}{l}
\frac{\left|F_{1} Q\right|}{\left|F_{2} Q\right|}=\frac{S_{\triangle P Q F_{1}}}{S_{\triangle P Q F_{2}}}=\frac{r... | t=b | Geometry | proof | Yes | Yes | cn_contest | false | 730,252 |
Given integers $x$, $y$, $z$ are not multiples of 7, and $7^{3} \mid \left(x^{7}+y^{7}+z^{7}\right)$. Prove: $7^{2} \mid (x+y+z)$.
untranslated text is preserved in the translation result. | Proof: By Fermat's Little Theorem, we have
$$
7\left|\left(x^{7}-x\right), 7\right|\left(y^{7}-y\right), 7 \mid\left(z^{7}-z\right) \text {. }
$$
Thus, \(7 \mid (x+y+z)\).
Let \(x+y+z=7k\).
By the binomial expansion, we get
$$
\begin{array}{l}
0 \equiv x^{7}+y^{7}+z^{7} \equiv (7k-y-z)^{7}+y^{7}+z^{7} \\
\equiv -(y+z)... | 7^{2} \mid (x+y+z) | Number Theory | proof | Yes | Yes | cn_contest | false | 730,253 |
5. Given real numbers $x, y$ satisfy
$$
\frac{4}{x^{4}}-\frac{2}{x^{2}}=3, y^{4}+y^{2}=3 \text {. }
$$
Then the value of $\frac{4}{x^{4}}+y^{4}$ is $\qquad$
(2008, "Mathematics Weekly Cup" National Junior High School Mathematics Competition) | By observation, we know that $-\frac{2}{x^{2}}$ and $y^{2}$ are two different solutions of the equation $m^{2}+m=3$.
By Vieta's formulas, we have
$$
\begin{array}{l}
\frac{4}{x^{4}}+y^{4}=m_{1}^{2}+m_{2}^{2}=\left(m_{1}+m_{2}\right)^{2}-2 m_{1} m_{2} \\
=(-1)^{2}-2(-3)=7 .
\end{array}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,254 |
6. If in the real number range there is
$$
x^{3}+p x+q=(x-a)(x-b)(x-c),
$$
and $q \neq 0$, then $\frac{a^{3}+b^{3}+c^{3}}{a b c}=$ $\qquad$ | Obviously, $a$, $b$, $c$ are the roots of the equation $x^{3} + p x + q = 0$.
By Vieta's formulas, we have $a + b + c = 0$.
By formula (2), we get
$$
\frac{a^{3} + b^{3} + c^{3}}{a b c} = \frac{a^{3} + b^{3} + c^{3} - 3 a b c}{a b c} + 3 = 3.
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,255 |
Example 1 Given that $\odot O_{2}$ passes through the center of $\odot O_{1}$, the two circles intersect at points $A$ and $B$, $\odot O_{1}$ intersects $O_{1} O_{2}$ at point $C$, line $B C$ intersects $\odot O_{2}$ at another point $D$, line $A D$ intersects $\odot O_{1}$ at another point $E$, and $F$ is the midpoint... | Proof As shown in Figure 5.
By Corollary 6, we know that points $A$ and $C$ are symmetric with respect to $O_{1} D$, and points $E$ and $C$ are symmetric with respect to $O_{1} A$.
Therefore, $\angle E O_{1} A=\angle A O_{1} C=2 \angle A O_{1} D$.
Since $F$ is the midpoint of $A E$, then
$$
\angle A O_{1} F=\frac{1}{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,256 |
Example 2 Given that $A, B$ are two fixed points on circle $\Gamma_{1}$, the center of circle $\Gamma_{2}$ is on circle $\Gamma_{1}$, and it is tangent to $AB$ at point $B$. A line through point $A$ intersects circle $\Gamma_{2}$ at points $D, E$, and line $BD$ intersects circle $\Gamma_{1}$ at another point $F$ differ... | Prove as shown in Figure 6, let $DE$ intersect circle $\Gamma_{1}$ at point $K$, and connect $FK, BK$.
By $\triangle KFD \backsim \triangle BAD \backsim \triangle EAB$
$$
\Rightarrow \frac{FD}{DK}=\frac{AB}{BE} \text{. }
$$
By Corollary 2, we know
$$
DK=KE \text{. }
$$
Notice that, $\angle BEK=\angle ABD$.
Then $BE$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,257 |
Example 3 In an acute $\triangle ABC$, $AB < AC < BC$, the perpendicular bisector of the angle bisector $AD$ of $\angle A$ intersects the circumcircle $\odot O$ of $\triangle ABC$ at points $K$ and $L$ ($K$ on the minor arc $\overparen{AB}$). A circle $\odot K$ is drawn with $K$ as the center and $KA$ as the radius, in... | Prove as shown in Figure 7, connect $K A$ and $K D$.
Since point $K$ is on the perpendicular bisector of $A D$
$\Rightarrow K A=K D \Rightarrow$ point $D$ is on circle $\odot K$.
Similarly, point $D$ is on circle $\odot L$.
Thus, $D$ is the intersection point of $\odot K$ and $\odot L$.
By property 4, we know that $K, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,258 |
Example 4 As shown in Figure 8, in $\triangle ABC$, $AB=AC$, and $I$ is the incenter of $\triangle ABC$. A circle $\odot A$ is drawn with $AB$ as the radius, and a circle $\odot I$ is drawn with $IB$ as the radius. A circle $\Gamma$ passing through points $B$ and $I$ intersects $\odot A$ and $\odot I$ at points $P$ and... | $$
\begin{array}{l}
\text { As shown in Figure 8, let the line } R I \text { intersect } \odot I \text { at points } E \text { and } F, \text { with } F \text { between } P \text { and } R. \\
By Property 3, we know that } E, P, F, R \text { form a harmonic range.} \\
By Corollary 7, we know that } B F \text { bisects ... | 90^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 730,259 |
Example 5 Let the circumcircle of $\triangle ABC$ be $\odot O$, and point $D$ lies on the segment $BC$, not coinciding with the endpoints or the midpoint of $BC$. Given that the circumcircle of $\triangle BOD$ intersects $\odot O$ again at $K$, and intersects line $AB$ again at $E$; the circumcircle of $\triangle COD$ ... | Prove as shown in Figure 9.
By Miquel's theorem for $\triangle AEF$, the circumcircle of $\triangle AEF$ passes through point $O$.
For $\odot O$ and the circumcircles of $\triangle BOD$ and $\triangle COD$, using property 5, since points $B, D, C$ are collinear, it follows that points $L, D, K$ are collinear.
Similarl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,260 |
Example 6 Given that circle $\Gamma$ with diameter $AB$ intersects circle $\odot A$ at points $C$ and $D$, and point $E$ is on $\odot A$ such that $E$ is outside circle $\Gamma$ and $E, C$ are on the same side of line $AB$. Line $BE$ intersects $\odot A$ again at point $F$, and $K$ is a point on circle $\Gamma$ such th... | Proof As shown in Figure 10, let $B E$ intersect the circle $\Gamma$ at another point $X$.
From $\angle A X B=90^{\circ}$, i.e., $A X \perp E F$, we know that $X$ is the midpoint of $E F$.
Connecting $C B$. By Property 1, we know that $C B$ is tangent to $\odot A$ at point $C$, and $\overparen{B D}=\overparen{B C}$.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,261 |
Example 7 The circle $\Gamma$ passing through the vertex $A$ of $\triangle ABC$ intersects sides $AC$ and $AB$ at points $E$ and $F$, respectively, and intersects the circumcircle of $\triangle ABC$ at another point $P$. Prove that the necessary and sufficient condition for the reflection of point $P$ over line $EF$ to... | Proof As shown in Figure 11, connect $F P$.
Sufficiency.
Assume circle $\Gamma$ passes through the circumcenter $O$ of $\triangle A B C$.
By Corollary 4, we know $F P = F B$.
Take a point $D$ on $B C$ such that $\angle F D B = \angle C B A$.
$$
\begin{array}{l}
\text { Thus, } F D = F B = F P . \\
\text { Also, } \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,262 |
6. Given that $D$ is the intersection of the tangents to the circumcircle of $\triangle A B C$ at $A$ and $B$, the circumcircle of $\triangle A B D$ intersects the line $A C$ and the segment $B C$ at another point $E$ and $F$ respectively, and $C D$ intersects $B E$ at point $G$. If $\frac{B C}{B F}=2$, find $\frac{B G... | Using Property 1, let the circumcenter of $\triangle ABC$ be $O$. Then $O, A, D, B$ are concyclic. If point $O$ is not on side $BC$, then $FO \perp BC \Rightarrow \angle OAB=90^{\circ}$, which is a contradiction. Therefore, point $O$ is on $BC$.
Let $CD$ intersect $AB$ at point $M$. Then
$$
\begin{array}{l}
\frac{AM}{M... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,263 |
7. Given that $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, a line through point $B$ intersects $\odot O_{1}$ at another point $C$ and $\odot O_{2}$ at another point $D$. The tangents to $\odot O_{1}$ and $\odot O_{2}$ at points $C$ and $D$ intersect at point $E$, and the line $A E$ intersects the ci... | Let the line $C O_{1}$ and $D O_{2}$ intersect at point $P$. Then, by Corollary 11, point $P$ lies on the circle $\Gamma$.
Assume without loss of generality that $O_{1}$ and $O_{2}$ lie on the segments $C P$ and $D P$, respectively.
Since $\angle P C E = \angle P D E = 90^{\circ}$, points $P, C, E, D$ are concyclic.
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,264 |
Example 3 Let $f(k)$ be the number of non-negative integers $n$ that satisfy the following conditions:
(1) $n$ has exactly $k$ digits in decimal notation, and the leading digit can be 0, i.e., $0 \leqslant n<10^{k}$;
(2) If the digits of $n$ are rearranged in some way, the resulting number can be divisible by 11.
Prove... | 【Analysis】Using the mapping method, for a given positive integer $m$, define sets $M_{a}$ and $M_{b}$.
(a) $M_{a}$ is the set of all integers $n$ with the following properties:
(i) $0 \leqslant n < 10^{2 m}$, i.e., $n$ has $2 m$ digits;
(ii) Fix the last digit of $n$, and rearrange the $2 m-1$ digits to the left of $n$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,265 |
Problem: There are $n(n>3)$ distinct positive integers on the blackboard, all of which are less than $(n-1)!$. For any pair of numbers $a, b (a>b)$, Peter writes $\left[\frac{a}{b}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$) in his notebook. Prove that there must be two equal nu... | Proof by contradiction.
Assume the numbers written by Peter are all distinct.
Let the $n$ numbers on the blackboard be
$$
\begin{array}{l}
1 \leqslant a_{1}(n-1)!+(n-2)!,
\end{array}
$$
which contradicts the problem statement.
Therefore, $a_{n}=1$.
$$
\begin{array}{l}
\text { If } \max _{1 \leq j \leq n-1} q_{j} \geqs... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,266 |
Question 3 Given $n$ points $p_{1}, p_{2}, \cdots, p_{n}$ in the plane, with no three points collinear. Each point $p_{i}(i=1,2, \cdots, n)$ is arbitrarily colored red or blue. Let $S$ be a set of some triangles with vertex set $\left\{p_{1}, p_{2}, \cdots, p_{n}\right\}$, and has the property: for any two line segment... | When $n \in\{1,2, \cdots, 5\}$, it is clearly impossible to satisfy the conditions of the problem.
When $n=6$, color $p_{1} 、 p_{2} 、 p_{3}$ red, and $p_{4}$ 、 $p_{5} 、 p_{6}$ blue, and consider the triangles:
$$
\begin{array}{l}
\triangle p_{1} p_{2} p_{3} 、 \triangle p_{1} p_{3} p_{4} 、 \triangle p_{1} p_{4} p_{5} 、... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,267 |
3. Given that $A$, $B$, and $C$ are points on $\odot O$, and $O A \perp O B, A B^{2}+O B^{2}=B C^{2}$.
Then the size of $\angle O A C$ is ( ).
(A) $15^{\circ}$
(B) $15^{\circ}$ or $75^{\circ}$
(C) $30^{\circ}$
(D) $30^{\circ}$ or $60^{\circ}$ | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,269 |
4. A class has a total of 50 students, and in the three sports of playing football, basketball, and volleyball, each student can do at least one of them. Some can do two of them, and no one can do all three. If 24 students in the class cannot play football, 25 students cannot play basketball, and 30 students cannot pla... | 4. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 730,270 |
Example 5 Given real numbers $x, y$ satisfy $x+y=3, \frac{1}{x+y^{2}}+\frac{1}{x^{2}+y}=\frac{1}{2}$.
Find the value of $x^{5}+y^{5}$. [3]
(2017, National Junior High School Mathematics League) | From the given equation, we have
$$
\begin{array}{l}
2 x^{2}+2 y+2 x+2 y^{2}=x^{3}+x y+x^{2} y^{2}+y^{3} \\
\Rightarrow 2(x+y)^{2}-4 x y+2(x+y) \\
\quad=(x+y)\left((x+y)^{2}-3 x y\right)+x y+x^{2} y^{2} .
\end{array}
$$
Substituting $x+y=3$ into the above equation, we get
$$
\begin{array}{l}
(x y)^{2}-4 x y+3=0 \\
\Ri... | 123 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,274 |
9. As shown in Figure 5, in $\triangle A B C$, $\angle A C B=90^{\circ}, A C$ $=B C=4, D$ and $E$ are the midpoints of $A C$ and $B C$ respectively. The quadrilateral $A B E D$ is translated to the right along the ray $A B$ to form quadrilateral $P Q R S$. Let $A P=m(0<m<4 \sqrt{2})$. Denote the area of the overlapping... | 9. $y=\left\{\begin{array}{ll}6-\sqrt{2} m, & 0<m \leqslant 2 \sqrt{2} ; \\ \frac{1}{4} m^{2}-2 \sqrt{2} m+8, & 2 \sqrt{2}<m<4 \sqrt{2}\end{array}\right.$ | y=\left\{\begin{array}{ll}6-\sqrt{2} m, & 0<m \leqslant 2 \sqrt{2} ; \\ \frac{1}{4} m^{2}-2 \sqrt{2} m+8, & 2 \sqrt{2}<m<4 \sqrt{2}\end{array}\right.} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,275 |
10. On a $4 \times 4$ chessboard with 16 squares, a piece $M$ starts moving from one of the squares, with the rule that “$M$ completes one move” is: moving three squares continuously, either horizontally first then vertically (or vertically first then horizontally), and $M$ only stops on the third square. For example, ... | 10. 14
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the text "10. 14" is already in a numerical format that is universal and does not require translation, the output remains the same:
10. 14 | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,276 |
11. Given the quadratic function $y=a x^{2}+b x+c(b$ is an integer) whose graph does not lie below the $x$-axis at any point, the graph intersects the $y$-axis at point $C$, and the vertex is at point $E$. The axis of symmetry is to the right of the line $x=c-\frac{1}{12}$. Let $t=\frac{a+2 b+12 c}{a}$.
(1) Find the mi... | 11. (1) According to the problem, we have
$$
a>0, b^{2}-4 a c \leqslant 0 \Rightarrow c \geqslant \frac{b^{2}}{4 a} \text {. }
$$
Therefore, $t=\frac{a+2 b+12 c}{a} \geqslant \frac{a+2 b+\frac{3 b^{2}}{a}}{a}$
$$
=1+2\left(\frac{b}{a}\right)+3\left(\frac{b}{a}\right)^{2}=3\left(\frac{b}{a}+\frac{1}{3}\right)^{2}+\frac... | y=6 x^{2}-2 x+\frac{1}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,277 |
2. Let $n$ be a positive integer, among the following pairs of numbers $a, b$, the one that makes $\frac{b}{a}$ a reduced fraction is ( ).
(A) $a=n+1, b=2 n-1$
(B) $a=2 n-1, b=5 n+2$
(C) $a=n+1, b=3 n+1$
(D) $a=3 n+1, b=5 n+2$ | 2. D.
Notice,
$$
\begin{array}{l}
(3 n+1,5 n+2)=(3 n+1,2 n+1) \\
=(n, 2 n+1)=(n, n+1)=1 .
\end{array}
$$
Therefore, $\frac{5 n+2}{3 n+1}$ is a reduced fraction. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 730,278 |
4. As shown in Figure 1,
Ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ $=1(a>b>0)$, - $O: x^{2}+y^{2}=a^{2}$ intersects the positive $y$-axis at point $B$, a line passing through point $B$ is tangent to ellipse $E$, and intersects circle $\odot O$ at another point $A$. If $\angle A O B=60^{\circ}$, then the ecc... | 4. D.
Since $|O A|=|O B|, \angle A O B=60^{\circ}$, therefore, $\triangle A O B$ is an equilateral triangle.
Thus, the slope of $l_{A B}$ is $\pm \frac{\sqrt{3}}{3}$.
Hence, $l_{A B}: y= \pm \frac{\sqrt{3}}{3} x+a$.
Substituting into the ellipse equation, we get
$$
\begin{array}{l}
\left(a^{2}+3 b^{2}\right) x^{2} \pm... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,279 |
Example 6 Given $a+b+c=5$,
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}=15, a^{3}+b^{3}+c^{3}=47 . \\
\text { Find }\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)
\end{array}
$$
the value. ${ }^{[4]}$
$(2016$, National Junior High School Mathematics League (B Volume)) | From formula (1), we know
$$
\begin{array}{l}
2(a b+b c+c a) \\
=(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=10 \\
\Rightarrow a b+b c+c a=5 .
\end{array}
$$
From formula (2), we know
$$
47-3 a b c=5(15-5) \Rightarrow a b c=-1 \text {. }
$$
And $a^{2}+a b+b^{2}$
$$
\begin{array}{l}
=(a+b)(a+b+c)-(a b+b c+c a) \\
=5(5-... | 625 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,280 |
6. As shown in Figure 2, in the cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with edge length 1, $P$, $Q$, and $R$ are the midpoints of edges $A B$, $A D$, and $A A_{1}$, respectively. A right triangular prism is constructed with $\triangle P Q R$ as its base, such that the three vertices of its other base are also on the... | 6. C.
It is known that the three vertices $P_{1}, Q_{1}, R_{1}$ of the other base of the right triangular prism are the centers of the squares $B C C_{1} B_{1}$, $C D D_{1} C_{1}$, and $A_{1} B_{1} C_{1} D_{1}$, respectively. Therefore, the lateral edge of the right triangular prism $P Q R-P_{1} Q_{1} R_{1}$ is $P P_{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,281 |
7. Let the set $A=\left\{n \left\lvert\, \frac{n}{3} \in \mathbf{N}_{+}\right.\right\}$,
$$
B=\left\{y \mid y=x+4+\sqrt{5-x^{2}}\right\} .
$$
Then the number of elements in the set $A \cap B$ is ().
(A) 1
(B) 2
(C) 3
(D) 4 | 7. B.
Let $x=\sqrt{5} \sin \theta\left(-\frac{\pi}{2} \leqslant \theta \leqslant \frac{\pi}{2}\right)$.
Then $y=\sqrt{5} \sin \theta+4+\sqrt{5} \cos \theta$
$=4+\sqrt{10} \sin \left(\theta+\frac{\pi}{4}\right)$.
By $-\frac{\pi}{4} \leqslant \theta+\frac{\pi}{4} \leqslant \frac{3 \pi}{4}$
$\Rightarrow-\frac{\sqrt{2}}{2... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,282 |
8. Let $x \geqslant y>0$. If there exist real numbers $a, b$ such that $0 \leqslant a \leqslant x, 0 \leqslant b \leqslant y$,
and $(x-a)^{2}+(y-b)^{2}=x^{2}+b^{2}=y^{2}+a^{2}$.
Then the maximum value of $\frac{x}{y}$ is $($.
(A) $\frac{2 \sqrt{3}}{3}$
(B) $\sqrt{2}$
(C) $\frac{\sqrt{6}}{2}$
(D) 1 | 8. A.
As shown in Figure 5, in the Cartesian coordinate system $x O y$, construct a rectangle $O A B C$ such that $A(x, 0)$, $B(x, y)$, and $C(0, y)$.
Take points $P$ and $Q$ on sides $O A$ and $A B$ respectively, such that $P(a, 0)$ and $Q(x, y-b)$.
From equation (1), we have
$|P Q|=|Q C|=|C P| \Rightarrow \triangle... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,283 |
11. Let $a, b, c$ be distinct positive integers. Then the minimum value of $\frac{a b c}{a+b+c}$ is . $\qquad$ | 11.1.
Assume $a>b>c$. Then $a \geqslant 3, b \geqslant 2, c \geqslant 1$.
Thus, $a b \geqslant 6, b c \geqslant 2, c a \geqslant 3$.
Therefore, $\frac{a+b+c}{a b c}=\frac{1}{b c}+\frac{1}{c a}+\frac{1}{a b}$
$\leqslant \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$
$\Rightarrow \frac{a b c}{a+b+c} \geqslant 1$.
When $a=3, b=2... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,284 |
Let $\triangle A B C$ have internal angles $\angle A, \angle B, \angle C$ with opposite sides $a, b, c$ respectively,
vector $\boldsymbol{m}=(\sin A, b+c)$,
$$
\boldsymbol{n}=(\sin C-\sin B, a-b),
$$
and there exists a real number $\lambda$ such that $\boldsymbol{m}=\lambda \boldsymbol{n}$.
(1) Find the size of $\angl... | (1) From $\boldsymbol{m}=\lambda \boldsymbol{n}$, we get
$$
\left\{\begin{array}{l}
\sin A=\lambda(\sin C-\sin B), \\
b+c=\lambda(a-b) .
\end{array}\right.
$$
Eliminating $\lambda$ yields
$$
(a-b) \sin A=(b+c)(\sin C-\sin B).
$$
By the Law of Sines, we have
$$
(a-b) a=(b+c)(c-b),
$$
which simplifies to $a^{2}+b^{2}-... | (1,2] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,285 |
Example 7 Given $a+b+c=1$,
$$
\frac{1}{a+1}+\frac{1}{b+3}+\frac{1}{c+5}=0 \text {. }
$$
Find the value of $(a+1)^{2}+(b+3)^{2}+(c+5)^{2}$. (2017, National Junior High School Mathematics League (Grade 8)) | Let $x=a+1, y=b+3, z=c+5$.
Then the given equations can be transformed into
$$
\begin{array}{l}
x+y+z=10, \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \Rightarrow x y+y z+z x=0 .
\end{array}
$$
From formula (1), we get
$$
\begin{array}{l}
(a+1)^{2}+(b+3)^{2}+(c+5)^{2} \\
=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x y+y z+z x) \\
=1... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,286 |
2. Let the complex number $z$ satisfy
$$
\frac{1017 z-25}{z-2017}=3+4 \text { i. }
$$
Then $|z|=$ $\qquad$ | 2.5.
Let $w=3+4 \mathrm{i}$. Then
$$
\begin{aligned}
z & =\frac{2017 w-25}{w-2017}=\frac{2017 w-\bar{w} w}{w-2017} \\
& =\frac{2017-\bar{w}}{2017-w} w .
\end{aligned}
$$
Therefore, $|z|=|w|=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,287 |
5. Let planar vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ satisfy $|\boldsymbol{\alpha}+2 \boldsymbol{\beta}|=3,|2 \boldsymbol{\alpha}+3 \boldsymbol{\beta}|=4$. Then the minimum value of $\boldsymbol{\alpha} \cdot \boldsymbol{\beta}$ is $\qquad$ . | 5. -170 .
Let $\alpha+2 \beta=u, 2 \alpha+3 \beta=v,|u|=3,|v|=4$.
From $\alpha=2 v-3 u, \beta=2 u-v$, we get
$$
\boldsymbol{\alpha} \cdot \boldsymbol{\beta}=-6|u|^{2}-2|\boldsymbol{v}|^{2}+7 u \cdot v \text {. }
$$
When $\boldsymbol{u} \cdot \boldsymbol{v}=-12$,
$\boldsymbol{\alpha} \cdot \boldsymbol{\beta}=-54-32-84... | -170 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,288 |
6. Given that a line with slope $k$ is drawn through a focus of the ellipse $x^{2}+2 y^{2}=3$, intersecting the ellipse at points $A$ and $B$. If $AB=2$, then $|k|=$ $\qquad$ | 6. $\sqrt{1+\sqrt{3}}$.
It is known from the calculation that the parameters of the ellipse are $a=\sqrt{3}, b=c=\sqrt{\frac{3}{2}}$. Let the equation of the line be $y=k(x-c)$. Then the x-coordinates $x_{1}, x_{2}$ of points $A, B$ satisfy
$$
x^{2}+2 k^{2}(x-c)^{2}=3 \Rightarrow x_{1}+x_{2}=\frac{4 k^{2} c}{1+2 k^{2}... | \sqrt{1+\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,289 |
7. Let $\theta \in[0,2 \pi]$ satisfy for $x \in[0,1]$,
$$
f(x)=2 x^{2} \sin \theta-4 x(1-x) \cos \theta+3(1-x)^{2}>0 .
$$
Then the range of $\theta$ is . $\qquad$ | 7. $\left(\frac{\pi}{6}, \pi\right)$.
Notice that, $2 \sin \theta=f(1)>0$.
Let $x \in(0,1)$ satisfy
$2 x^{2} \sin \theta=3(1-x)^{2}$.
Then $f(x)=2 x(1-x)(\sqrt{6 \sin \theta}-2 \cos \theta)>0$.
Since $\sin \theta>0$, and $\sqrt{6 \sin \theta}-2 \cos \theta>0$, it follows that for $x \in[0,1]$, we have $f(x)>0$.
Given ... | \left(\frac{\pi}{6}, \pi\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,290 |
9. (21 points) As shown in Figure 1, let $H$ be the orthocenter of acute $\triangle A B C$, and point $D$ lies on line $A C$ such that $H A=H D$. Quadrilateral $A B E H$ is a parallelogram. Prove: $B 、 E 、 C 、 D 、 H$ are concyclic. | 9. From $A H \perp B C, A H / / B E \Rightarrow B E \perp B C$.
From $H C \perp A B, H E / / A B \Rightarrow H E \perp H C$.
Thus, points $B, E, C, H$ are all on the circle with $E C$ as its diameter.
From $B H \perp A D$ and $H A=H D$, we get that $B H$ is the perpendicular bisector of segment $A D$.
$$
\begin{array}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,291 |
1. Given the sets
$$
A=\{x \mid 1 \leqslant x \leqslant 2\}, B=\left\{x \mid x^{2}-a x+4 \geqslant 0\right\} \text {. }
$$
If $A \subseteq B$, then the range of real number $a$ is
$\qquad$ | 1. $a \leqslant 4$.
It is easy to see that the inequality $x^{2}-a x+4 \geqslant 0$ holds for all $x$ in the interval $[1,2]$.
Therefore, $a \leqslant 4$. | a \leqslant 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,292 |
2. The maximum value of the function $f(x)=7 \sin x+\sin 2 x$ is $\qquad$ . | 2. $\frac{15 \sqrt{15}}{8}$.
From the condition, we know
$$
\begin{array}{l}
f^{\prime}(x)=7 \cos x+2 \cos 2 x \\
=4 \cos ^{2} x+7 \cos x-2 \\
=(4 \cos x-1)(\cos x+2) .
\end{array}
$$
Let $\cos \varphi=\frac{1}{4}\left(0<\varphi<\frac{\pi}{2}\right)$. Then
$$
\sin \varphi=\frac{\sqrt{15}}{4} \text {. }
$$
It is easy... | \frac{15 \sqrt{15}}{8} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 730,293 |
3. Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a, b>0)$ with left and right foci $F_{1}$ and $F_{2}$, respectively, and the hyperbola $C$ intersects the circle $x^{2}+y^{2}=r^{2}(r>0)$ at a point $P$. If the maximum value of $\frac{\left|P F_{1}\right|+\left|P F_{2}\right|}{r}$ is $4 \sqrt{2}$, th... | 3. $2 \sqrt{2}$.
Let $P(r \cos \theta, r \sin \theta), \cos \theta \geqslant 0$. Then
$$
\begin{array}{l}
\frac{\left|P F_{1}\right|+\left|P F_{2}\right|}{r}=\frac{a+e r \cos \theta+e r \cos \theta-a}{r} \\
=2 e \cos \theta .
\end{array}
$$
Therefore, by the maximum value of $\frac{\left|P F_{1}\right|+\left|P F_{2}\... | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,294 |
4. Let $p$ be a prime. If there exists a positive integer $n$, such that $p \mid\left(n^{2}+7 n+23\right)$, then the minimum value of $p$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4.11.
$$
\begin{array}{l}
\text { Let } f(n)=n^{2}+7 n+23 . \\
\text { From } f(n) \equiv 1(\bmod 2), \\
f(n) \equiv \pm 1(\bmod 3), \\
f(n) \equiv 1, \pm 2(\bmod 5), \\
f(n) \equiv-1,2, \pm 3(\bmod 7),
\end{array}
$$
and $f(7)=121$, we know that the smallest positive prime satisfying the conditions is 11. | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 730,295 |
5. In $\triangle A B C$, $A B=3, A C=4, B C=5, I$ is the incenter of $\triangle A B C$, and $P$ is a point within $\triangle I B C$ (including the boundary). If $\overrightarrow{A P}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}(\lambda ; \mu \in \mathbf{R})$, then the minimum value of $\lambda+\mu$ is | 5. $\frac{7}{12}$.
Let $A P$ intersect $B C$ at point $Q$. Then $\overrightarrow{A Q} \triangleq t \overrightarrow{A P}=t \lambda \overrightarrow{A B}+t_{\mu} \overrightarrow{A C}$.
Since $Q, B, C$ are collinear,
$$
\Rightarrow t \lambda+t \mu=1 \Rightarrow \lambda+\mu=\frac{1}{t} \text {. }
$$
When point $P$ coincid... | \frac{7}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,296 |
Example 9 Let real numbers $s, t$ satisfy the equations
$$
\begin{array}{l}
19 s^{2}+99 s+1=0, t^{2}+99 t+19=0, \text { and } s t \neq 1 . \\
\text { Then } \frac{s t+4 s+1}{t}=
\end{array}
$$
(1999, "Mathematics Weekly Cup" National Junior High School Mathematics Competition) | Solution: Clearly, $t \neq 0$.
Thus, $19\left(\frac{1}{t}\right)^{2}+99 \cdot \frac{1}{t}+1=0$.
By comparison, $s$ and $\frac{1}{t}$ are the two distinct roots of the equation $19 x^{2}+99 x+1=0$ (since $s t \neq 1$, i.e., $s \neq \frac{1}{t}$). Therefore, by Vieta's formulas, we have
$$
\begin{array}{l}
s+\frac{1}{t}=... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,297 |
9. (16 points) Given the function
$$
f(x)=3 \lg (x+2)-\log _{2} x,
$$
the set $M=\left\{n \in \mathbf{Z} \mid f\left(n^{3}-3 n^{2}-10 n+32\right) \geqslant 0\right\}$. Find $M$. | 9. It is easy to know from differentiation that the function $f(x)$ is monotonically decreasing,
$$
\begin{array}{l}
f(8)=0 . \\
\text { Then } 00 . \\
\text { Also, } g(3)=2>0 \text {, so } M=\{-3,2,3,4\} .
\end{array}
$$ | M=\{-3,2,3,4\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,298 |
10. (20 points) Given
the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has an eccentricity of $\frac{1}{2}, F$ is the right focus, $A(-a, 0),|A F|$ $=3$. Let $O$ be the origin, $P$ be a point on the ellipse $C$, $M$ be the midpoint of $A P$, the line $O M$ intersects $x=4$ at point $D$, and the line t... | 10. Let the semi-focal distance of the ellipse $C$ be $c$.
According to the problem,
$$
\frac{c}{a}=\frac{1}{2}, a+c=3 \Rightarrow a=2, c=1 .
$$
Then $b^{2}=a^{2}-c^{2}=3$.
Thus, the equation of the ellipse $C$ is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
Therefore, $A(-2,0)$.
Let $P\left(x_{1}, y_{1}\right)\left(x_{1} \n... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,299 |
II. (40 points) As shown in Figure 1, in non-isosceles $\triangle ABC$, $I$ is the incenter, $\odot O$ is the circumcircle of $\triangle ABC$, $M$ is the midpoint of $BC$, $I$'s projection on $BC$ is $D$, $N$ is the midpoint of arc $\overparen{BAC}$, and $AM$ intersects $DN$ at point $K$. Prove: $AD \perp IK$. | II. First, prove a lemma.
Lemma If $\triangle A B C$ is inversely similar to $\triangle A D E$, and $\angle A B C=\angle A D E=90^{\circ}, B E$ intersects $C D$ at point $P$, then $A P \perp B D$.
Proof As shown in Figure 2, draw $A Q \perp B E$ at point $Q$, and let $A Q$ intersect $C D$ at point $R$; draw $A T \perp... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,300 |
1. Given real numbers $x, y$ satisfy $\left(x^{2}+6 x+12\right)\left(5 y^{2}+2 y+1\right)=\frac{12}{5}$. Then the value of $x y$ is $\qquad$ (2011, Huanggang City, Hubei Province Mathematics Competition) | $\begin{array}{l}\left.12=\left((x+3)^{2}+3\right)\right)\left((5 y+1)^{2}+4\right) \geqslant 12 \\ \Rightarrow x=-3, y=-\frac{1}{5} \Rightarrow x y=\frac{3}{5} .\end{array}$
The translation is as follows:
$\begin{array}{l}\left.12=\left((x+3)^{2}+3\right)\right)\left((5 y+1)^{2}+4\right) \geqslant 12 \\ \Rightarrow ... | \frac{3}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,301 |
1. As shown in Figure 1, quadrilateral $ABCD$ is a square with a side length of 4, and there are 3 quartiles on each side. Choose any one of the quartiles on each side of the square, and connect them in sequence to form a quadrilateral.
Ask: What values can the area of such a quadrilateral take?
Please list these value... | 1. To calculate the area of a quadrilateral, one can first calculate the areas of the four surrounding right-angled triangles, and then subtract them from the area of the larger square. The possible values for the area sought are $6, 7, 7.5, 8, 8.5, 9, 10$. | 6, 7, 7.5, 8, 8.5, 9, 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,302 |
4. Given $P_{1}, P_{2}, \cdots, P_{100}$ as 100 points on a plane, satisfying that no three points are collinear. For any three of these points, if their indices are in increasing order and they form a clockwise orientation, then the triangle with these three points as vertices is called "clockwise". Question: Is it po... | 4. Suppose $P_{1}, P_{2}, \cdots, P_{100}$ are arranged counterclockwise on a circle. At this point, the number of clockwise triangles is 0.
Now, move these points (not necessarily along the circumference). When a point $P_{i}$ crosses the line $P_{j} P_{k}$ during the movement, the orientation (clockwise or countercl... | 2017 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,304 |
Example 1 Given a convex quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $P$, the midpoints of sides $AB$, $BC$, $CD$, and $DA$ are $E$, $F$, $G$, and $H$ respectively. The circumcenters of $\triangle PHE$, $\triangle PEF$, $\triangle PFG$, and $\triangle PGH$ are $O_1$, $O_2$, $O_3$, and $O_4$ ... | Prove that by setting $P$ as the origin to establish the complex plane. Without loss of generality, let $A=a, B=b e, C=-c, D=-d e$, where $a, b, c, d \in \mathbf{R}_{+}, e \in \mathbf{C}$.
By property (1),
$A, B, C, D$ are concyclic
$$
\begin{aligned}
& \Leftrightarrow \frac{A-B}{C-B} \cdot \frac{C-D}{A-D} \in \mathbf{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,305 |
Example 2 Let $H$ be the orthocenter of acute $\triangle A B C$, $O^{\prime}$ be the circumcenter of $\triangle B H C$, $N$ be the midpoint of segment $A O^{\prime}$, and $D$ be the reflection of $N$ over side $B C$. Prove that $A, B, D, C$ are concyclic if and only if $b^{2}+c^{2}-a^{2}=3 R^{2}$, where $a=BC$, $b=CA$,... | To prove: It is known that the reflection of the orthocenter $H$ over $BC$, denoted as $H'$, lies on the circumcircle $\odot O$ of $\triangle ABC$.
Therefore, the circumcircle $\odot O'$ of $\triangle HBC$ and the circumcircle $\odot O$ of $\triangle H'BC$ are symmetric with respect to $BC$.
Establish a complex plane ... | b^{2}+c^{2}-a^{2}=3 R^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 730,306 |
5. Let $H$ be the orthocenter of $\triangle ABC$, and let $D, E, F$ be three points on the circumcircle of $\triangle ABC$ such that $AD \parallel BE \parallel CF$. Let $S, T, U$ be the reflections of $D, E, F$ across the sides $BC, CA, AB$ respectively. Prove that $S, T, U, H$ are concyclic.
(2006, China National Team... | Suppose the circumradius of $\triangle A B C$ is 1. Establish a complex plane with the circumcenter $O$ of $\triangle A B C$ as the origin.
By property (2), we know $H=A+B+C$.
By $B E / / A D$
$$
\begin{array}{l}
\Rightarrow \frac{B-E}{A-D}=\frac{\bar{B}-\bar{E}}{\bar{A}-\bar{D}}=\frac{\frac{E-B}{E B}}{\frac{D-A}{D A}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,307 |
High $\mathbf{560}$ Let $k, l, m$ be given positive integers. Prove: there exist infinitely many positive integers $n$, such that $\frac{\mathrm{C}_{n}^{k}}{m^{l}}$ is a positive integer coprime with $m$.
| Prove that for any prime factor $p$ of $m$, there exists a positive integer $n$ such that $v_{p}\left(\mathrm{C}_{n}^{k}\right)=l v_{p}(m)$, where $v_{p}(x)$ denotes the exponent of $p$ in the prime factorization of $x$.
Let $k$ be represented in base $p$ as
$$
k=\left(\overline{a_{f(p)} a_{f(p)-1} \cdots a_{1} a_{0}}\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,308 |
Question 1 In a non-isosceles $\triangle ABC$, $D$, $E$, $F$ are the midpoints of sides $BC$, $CA$, $AB$ respectively. Draw another tangent from point $D$ to the incircle of $\triangle ABC$ (different from line $BC$), intersecting $EF$ at point $X$. Similarly, construct points $Y$, $Z$. Prove that $X$, $Y$, $Z$ are col... | Lemma 1 In $\triangle ABC$, points $D, E, F$ are on lines $BC, CA, AB$ (different from the vertices of $\triangle ABC$), and $AD, BE, CF$ intersect at point $P$. Line $DE$ intersects $AB, EF$ intersects $BC, FD$ intersects $CA$ at points $X, Y, Z$ respectively. Then $X, Y, Z$ are collinear.
Proof of Lemma 1 Consider $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,309 |
Question 3 In a non-isosceles $\triangle ABC$, the incircle $\Gamma$ touches the sides $BC$, $CA$, $AB$ at points $R$, $S$, $T$ respectively. Points $D$, $E$, $F$ lie on the lines $BC$, $CA$, $AB$ (different from the vertices of $\triangle ABC$), and $AD$, $BE$, $CF$ intersect at point $P$. Through point $D$, draw anot... | Proof As shown in Figure 2, let lines $X R$ and $X D$ intersect $S T$ at points $U$ and $L$, respectively; lines $Y S$ and $Y E$ intersect $T R$ at points $V$ and $M$, respectively; and lines $Z T$ and $Z F$ intersect $R S$ at points $W$ and $N$, respectively.
It is easy to see that the polars of points $A$, $B$, and ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,310 |
1. For a positive integer $n$, define $A_{n}$ as the set of all prime numbers $p$ with the following property: there exist positive integers $a, b$ such that $\frac{a+b}{p}$ and $\frac{a^{n}+b^{n}}{p^{2}}$ are both integers coprime to $p$. When $A_{n}$ is a finite set (including the empty set), use $f(n)$ to denote the... | 1. (1) When $n=1$, obviously, there does not exist a prime $p$ that satisfies the condition. Hence $A_{1}=\varnothing$.
When $n=2$, for any prime $p$, take $a=p$, $b=p^{2}$, so that $\frac{a+b}{p}=1+p$ and $\frac{a^{n}+b^{n}}{p^{2}}=1+p^{2}$ are both integers coprime to $p$. Therefore, $A_{2}$ contains all primes, whi... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,311 |
3. Let $q$ be a positive integer that is not a perfect cube. Prove: there exists a positive real number $c$, such that for any positive integer $n$, we have
$$
\left\{n q^{\frac{1}{3}}\right\}+\left\{n q^{\frac{2}{3}}\right\} \geqslant c n^{-\frac{1}{2}},
$$
where $\{x\}$ denotes the fractional part of the real number... | 3. $c=\left(13 q^{\frac{4}{5}}\right)^{-1}$ satisfies the requirement.
Proof by contradiction.
Assume there exists a positive integer $n$ such that
$\left\{n q^{\frac{1}{3}}\right\}+\left\{n q^{\frac{2}{3}}\right\}n$.
Then there exist distinct pairs $\left(u_{1}, v_{1}\right)$, $\left(u_{2}, v_{2}\right)$ in the set $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,312 |
5. Given an odd number $n \geqslant 3$, color each cell of an $n \times n$ grid with two colors, black and white. Two cells with the same color that share a vertex are called "adjacent". For any two cells $a$ and $b$, if there exists a sequence of cells $c_{1}, c_{2}, \cdots, c_{k}$ such that $c_{1}=a, c_{k}=b, c_{i}(i... | 5. Consider the problem of promotion.
For an $m \times n$ grid, where $m, n$ are odd numbers not less than 3, color the grid in black and white. Under a given coloring scheme, all cells can be divided into several connected components, such that cells in the same connected component are connected, and cells in differe... | \frac{1}{4}(n+1)^{2}+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,313 |
6. Given that $n, k$ are positive integers, $n>k$. Given real numbers $a_{1}, a_{2}, \cdots, a_{n} \in(k-1, k)$. Let positive real numbers $x_{1}, x_{2}$, $\cdots, x_{n}$ satisfy that for any $k$-element subset $I$ of $\{1,2, \cdots, n\}$, we have $\sum_{i \in I} x_{i} \leqslant \sum_{i \in I} a_{i}$. Find the maximum ... | 6. The maximum value is $a_{1} a_{2} \cdots a_{n}$.
If $x_{i}=a_{i}(1 \leqslant i \leqslant n)$, then $x_{1}, x_{2}, \cdots, x_{n}$ satisfy the conditions, and $x_{1} x_{2} \cdots x_{n}=a_{1} a_{2} \cdots a_{n}$.
Next, we prove: $x_{1} x_{2} \cdots x_{n} \leqslant a_{1} a_{2} \cdots a_{n}$.
When $k=1$, the conclusion ... | a_{1} a_{2} \cdots a_{n} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,314 |
1. Given that $AB$ is a line segment of length 8, and point $P$ is at a distance of 3 from the line containing $AB$. Then the minimum value of $AP \cdot PB$ is $\qquad$. | $-, 1.24$
Draw a perpendicular from point $P$ to the line $AB$, with the foot of the perpendicular being $H$. Then
$$
\begin{array}{l}
\frac{1}{2} AB \cdot PH = S_{\triangle PAB} = \frac{1}{2} AP \cdot PB \sin \angle APB \\
\Rightarrow AP \cdot PB = \frac{24}{\sin \angle APB} \geqslant 24 .
\end{array}
$$
When $\angle... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,315 |
Example 3 Let quadrilateral $A B C D$ be convex, and let $P, Q, R, S$ be points on segments $A B, B C, C D, D A$, respectively. Suppose that the intersecting line segments $P R$ and $Q S$ divide the convex quadrilateral $A B C D$ into four convex quadrilaterals with perpendicular diagonals. Prove that the points $P, Q,... | Prove that by taking the intersection point $O$ of $P R$ and $Q S$ as the origin to establish the complex plane. Without loss of generality, assume $A, B, C, D$ are arranged clockwise.
Let $P=p e_{1}, R=r e_{1}, Q=q e_{2}, S=s e_{2}$, where $p, q, r, s \in \mathbf{R}, e_{1}, e_{2} \in \mathbf{C}, |e_{1}|=|e_{2}|=1$.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,316 |
2. Given the parabola $y=\sqrt{2} x^{2}$ intersects with the lines $y=1, y=2, y=3$ to form three line segments. The area of the triangle formed by these three line segments is $\qquad$ . | 2. 2 .
The intersection points of the lines $y=1, y=2, y=3$ with the parabola $y=\sqrt{2} x^{2}$ are all symmetric about the $y$-axis, so the lengths of the three line segments are $2 x_{1}, 2 x_{2}, 2 x_{3}$, where $x_{1}, x_{2}, x_{3}$ are the positive roots of the equations $\sqrt{2} x_{1}^{2}=1, \sqrt{2} x_{2}^{2}... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,317 |
3. For any $x \in[0,1]$, we have $|a x+b| \leqslant 1$.
Then the maximum value of $|b x+a|$ is $\qquad$ | 3. 2 .
Let $f(x)=a x+b$. Then
$$
\begin{array}{l}
b=f(0), a=f(1)-f(0) . \\
\text { Hence }|b x+a|=|f(0) x+f(1)-f(0)| \\
=|f(0)(x-1)+f(1)| \\
\leqslant|f(0)||x-1|+|f(1)| \\
\leqslant 1+1=2 .
\end{array}
$$
When $a=2, b=-1, x=0$, the maximum value 2 is obtained. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,318 |
4. Let the increasing function $f(x)$ defined on $(0,+\infty)$ satisfy: for any $x \in(0,+\infty)$, we have $f(x)>-\frac{6}{x}$, and $f\left(f(x)+\frac{6}{x}\right)=5$.
Then $f(10)=$ $\qquad$ | 4. $\frac{12}{5}$.
Let $f(a)=5$ where $a>0$.
By the definition of an increasing function, we have
$$
\begin{array}{l}
f(a)+\frac{6}{a}=a \Rightarrow 5+\frac{6}{a}=a \Rightarrow a=3 \\
\Rightarrow f(x)+\frac{6}{x}=3 \Rightarrow f(10)=3-\frac{6}{10}=\frac{12}{5} .
\end{array}
$$ | \frac{12}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,319 |
5. Given the set
$$
A=\{n|n \in \mathbf{N}, 11| S(n), 11 \mid S(n+1)\},
$$
where $S(m)$ denotes the sum of the digits of the natural number $m$. Then the smallest number in set $A$ is $\qquad$ . | 5.2899999.
Let the smallest number in $A$ be $n=\overline{a_{1} a_{2} \cdots a_{t}}$,
$$
S(n)=a_{1}+a_{2}+\cdots+a_{t} \text {. }
$$
If the unit digit of $n$, $a_{t} \neq 9$, then
$$
\begin{array}{l}
n+1=\overline{a_{1} a_{2} \cdots a_{t-1} a_{t}^{\prime}}\left(a_{t}^{\prime}=a_{t}+1\right) . \\
\text { Hence } S(n+1... | 2899999 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,320 |
Five. (15 points) Let $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2016}=\frac{a}{b}$, where $a, b$ are natural numbers, and $(a, b)=1$. Prove:
$$
2017^{2} \mid a .
$$ | $$
\begin{array}{l}
\frac{2 a}{b}=2 \sum_{k=1}^{2016} \frac{1}{k}=\sum_{k=1}^{2016}\left(\frac{1}{k}+\frac{1}{2017-k}\right) \\
=\sum_{k=1}^{2016} \frac{2017}{k(2017-k)} .
\end{array}
$$
Thus, $\frac{2 a}{2017 b}=\sum_{k=1}^{2016} \frac{1}{k(2017-k)}$.
For each $i \in\{1,2, \cdots, 2016\}$, the indeterminate equation ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,321 |
2. Given an arithmetic sequence $\left\{a_{n}\right\}$ with a non-zero common difference, and $a_{2} 、 a_{3} 、 a_{9}$ form a geometric sequence. Then the value of $\frac{a_{4}+a_{5}+a_{6}}{a_{2}+a_{3}+a_{4}}$ is
(A) $\frac{8}{3}$
(B) $\frac{7}{3}$
(C) 3
(D) $\frac{5}{3}$ | 2. A.
Let the common difference be $d$.
From $a_{3}^{2}=a_{2} a_{9}$, we get $\left(a_{2}+d\right)^{2}=a_{2}\left(a_{2}+7 d\right)$.
Combining $d \neq 0$, we know $d=5 a_{2} \neq 0$.
Thus, $a_{4}+a_{5}+a_{6}=3 a_{5}=3\left(a_{2}+3 d\right)=48 a_{2}$,
$$
a_{2}+a_{3}+a_{4}=3 a_{3}=3\left(a_{2}+d\right)=18 a_{2} \text {,... | \frac{8}{3} | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,322 |
Example 4 In an acute triangle $\triangle ABC$, $\angle A > 60^{\circ}$, $H$ is the orthocenter of $\triangle ABC$, points $M$ and $N$ are on sides $AB$ and $AC$ respectively, such that $\angle HMB = \angle HNC = 60^{\circ}$, $O$ is the circumcenter of $\triangle HMN$, point $D$ is on the same side of line $BC$ as $A$,... | 【Analysis】The complex number method proof in [3] is slightly complicated and involves too much calculation. If a complex plane is established with $H$ as the origin, the process is greatly simplified.
Proof Establish a complex plane with $H$ as the origin and $H A$ as the real axis. To simplify the calculation, assume... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,323 |
3. The curve $y=\log _{2} x$ is moved one unit in the positive direction of the $x$-axis, and then two units in the negative direction of the $y$-axis, resulting in curve $C$. Among the following curves, which one is symmetric to $C$ with respect to the line $x+y=0$? ( ).
(A) $y=2^{x+2}-1$
(B) $y=-2^{x+2}-1$
(C) $y=-2^... | 3. C.
Take any point $\left(2^{t}, t\right)$ on the curve $y=\log _{2} x$, after translation, we get the point $\left(2^{t}+1, t-2\right)$ on curve $C$, and its symmetric point with respect to $x+y=0$ is $\left(2-t,-2^{t}-1\right)$.
Let $x=2-t, y=-2^{t}-1$.
Eliminating $t$ yields $y=-2^{2-x}-1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,324 |
2. Given a regular 2017-gon $A_{1} A_{2} \cdots A_{2017}$ inscribed in the unit circle $\odot O$, take any two different vertices $A_{i} 、 A_{j}$. Then the probability that $\overrightarrow{O A_{i}} \cdot \overrightarrow{O A_{j}}>\frac{1}{2}$ is . $\qquad$ | 2. $\frac{1}{3}$.
Notice that, $\overrightarrow{O A_{i}} \cdot \overrightarrow{O A_{j}}>\frac{1}{2} \Leftrightarrow \angle A_{i} O A_{j}<60^{\circ}$.
For any given point $A_{i}$, the number of ways to choose point $A_{j}$ such that $\angle A_{i} O A_{j}<60^{\circ}$ is $2\left[\frac{2017}{6}\right]=672$ (where [x] deno... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,325 |
4. Let $F$ be the left focus of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and $A$ be a point on the ellipse in the first quadrant. Draw a tangent line from point $A$ to the circle $x^{2}+y^{2}=b^{2}$, with the point of tangency being $P$. Then $|A F|-|A P|=$ $\qquad$ | 4. a.
Let $F(-c, 0), A(a \cos t, b \sin t)\left(t \in\left(0, \frac{\pi}{2}\right)\right)$.
Then $|A F|=\sqrt{(a \cos t+c)^{2}+(b \sin t)^{2}}$
$$
\begin{array}{l}
=\sqrt{a^{2} \cos ^{2} t+2 a c \cos t+c^{2}+\left(a^{2}-c^{2}\right) \sin ^{2} t} \\
=\sqrt{a^{2}+2 a c \cos t+c^{2} \cos ^{2} t}=a+c \cos t .
\end{array}
... | a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,326 |
5. Place a sphere inside a cone, which is tangent to the cone's side and base. Then the maximum ratio of the sphere's surface area to the cone's surface area is $\qquad$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. $\frac{1}{2}$.
Let the radius of the base of the cone be $r$, and the angle between the slant height and the base be $2 \theta$. Then the radius of the inscribed sphere $R=r \tan \theta$.
$$
\begin{array}{l}
\text { By } S_{\text {sphere }}=4 \pi R^{2}=4 \pi r^{2} \tan ^{2} \theta, \\
S_{\text {cone }}=\pi r^{2}\le... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,327 |
6. Given that for all $x \in[-1,1]$, we have $x^{3}-a x+1 \geqslant 0$.
Then the range of the real number $a$ is $\qquad$ | 6. $\left[0, \frac{3 \sqrt[3]{2}}{2}\right]$.
When $x=0$, the inequality obviously holds.
When $x \neq 0$, let $f(x)=x^{2}+\frac{1}{x}$.
If $x \in[-1,0)$, then the original inequality is equivalent to
$$
f(x) \leqslant a \text {. }
$$
Notice that, $f(x)$ is a decreasing function on the interval $[-1,0)$, and its maxi... | \left[0, \frac{3 \sqrt[3]{2}}{2}\right] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,328 |
Example 5 Given $\triangle A B C$ with three unequal sides, its incircle touches the three sides at points $D, E, F$ respectively. Let $L$ be the reflection of point $D$ over line $E F$, $M$ be the reflection of point $E$ over line $F D$, and $N$ be the reflection of point $F$ over line $D E$. The lines $A L$ and $B C$... | Proof: Let the semi-perimeter of $\triangle ABC$ be $s$, the radii of the circumcircle $\odot O$ and the incircle $\odot I$ be $R$ and $r$ respectively, $AI$ intersects $BC$ at point $T$, and $DL$ intersects $EF$ at point $X$.
$$
\begin{array}{l}
\text { Then } D L=2 D X=4 r \sin \angle D E F \cdot \sin \angle D F E \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,329 |
1. Let the three interior angles of $\triangle A B C$ be $\angle A, \angle B$, and $\angle C$. Prove: If the midpoint of $B C$ is $M$, then
$$
\cot \angle B A M=2 \cot A+\cot B \text {. }
$$ | Three, 1. Take the midpoint $N$ of $AB$, and draw $MK \perp AB$ at point $K$.
Then $\cot \angle BAM = \frac{\overline{KA}}{KM}$,
$\cot A = \cot \angle BNM = \frac{\overline{KN}}{KM}$,
$\cot B = \frac{\overline{BK}}{KM}$ (assuming the direction of $\overline{BA}$ is positive).
The conclusion is equivalent to $\overline{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,330 |
3. If the integer $n \geqslant 2$, prove:
$$
\prod_{k=2}^{n}\left(1+\frac{1}{k^{2}}\right)<2 .
$$ | 3. In the familiar inequality $1+x \leqslant \mathrm{e}^{x}$, taking $x=\frac{1}{2^{2}}, \frac{1}{3^{2}}, \cdots, \frac{1}{n^{2}}$, we get
the left side of equation (1) $\leqslant \mathrm{e}^{\sum^{n}_{k=2} \frac{1}{k^{2}}}$.
Notice that,
$$
\begin{array}{l}
\sum_{k=2}^{n} \frac{1}{k^{2}}<\sum_{k=2}^{n} \frac{1}{k^{2}-... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,331 |
In $\triangle A B C$, if
$$
\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=2 \text {, }
$$
then the maximum value of $\cos A+\cos B+2 \cos C$ is ().
(A) $\frac{3 \sqrt{2}}{2}$
(B) $\sqrt{5}$
(C) $\frac{4 \sqrt{3}}{3}$
(D) $\sqrt{6}$ | - 1. B.
From the problem, we know
$$
\begin{array}{l}
4=2 \sin ^{2} A+2 \sin ^{2} B+2 \sin ^{2} C \\
\Rightarrow \cos 2 A+\cos 2 B+2 \cos ^{2} C=0 \\
\Rightarrow 2 \cos (A+B) \cdot \cos (A-B)+2 \cos ^{2} C=0 \\
\Rightarrow-4 \cos A \cdot \cos B \cdot \cos C=0
\end{array}
$$
$\Rightarrow \cos A, \cos B, \cos C$ must ha... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,332 |
2. Given that the surface area of a rectangular prism is $\frac{45}{2}$, and the total length of its edges is 24. Then the maximum value of the angle formed between the body diagonal of the rectangular prism and its edge is ( ).
(A) $\arccos \frac{1}{3}$
(B) $\arccos \frac{\sqrt{2}}{3}$
(C) $\arccos \frac{\sqrt{3}}{9}$... | 2. D.
Let the lengths of the edges passing through the same vertex be $a$, $b$, and $c$ ($a \geqslant b \geqslant c$). Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
2(a b+b c+c a)=\frac{45}{2}, \\
4(a+b+c)=24
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a+b=6-c, \\
a b-c^{2}+6 c=\frac{45}{4} .
\end{arra... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,333 |
5. Given positive numbers $a, b, c, d$ satisfy $a+2b=1, c+2d=1$. Then the minimum value of $\frac{1}{a}+\frac{1}{bcd}$ is ( ).
(A) $\frac{1}{16}$
(B) 8
(C) $\frac{1}{5}$
(D) 25 | 5. D.
By the AM-GM inequality and the given conditions, we have
$$
\begin{array}{l}
2 c d \leqslant\left(\frac{c+2 d}{2}\right)^{2}=\frac{1}{4} \\
\Rightarrow c d \leqslant \frac{1}{8} .
\end{array}
$$
Equality in (1) holds if and only if $c=\frac{1}{2}, d=\frac{1}{4}$.
From (1), we get
$\frac{1}{a}+\frac{1}{b c d} \... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,334 |
6. Let $f(x)$ be a function defined on $\mathbf{R}$, satisfying
$$
\left|f(x)+\cos ^{2} x\right| \leqslant \frac{3}{4},\left|f(x)-\sin ^{2} x\right| \leqslant \frac{1}{4} .
$$
Then the analytical expression of the function $f(x)$ is . $\qquad$ | 2,6. $f(x)=\sin ^{2} x-\frac{1}{4}$.
From the given and
$$
\begin{array}{l}
1=\sin ^{2} x+\cos ^{2} x \\
\leqslant\left|f(x)+\cos ^{2} x\right|+\left|f(x)-\sin ^{2} x\right| \\
\leqslant \frac{3}{4}+\frac{1}{4}=1,
\end{array}
$$
we get $\left|f(x)+\cos ^{2} x\right|=\frac{3}{4},\left|f(x)-\sin ^{2} x\right|=\frac{1}{4... | f(x)=\sin ^{2} x-\frac{1}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,335 |
Example 6 Given that $H$ is the orthocenter of a non-isosceles $\triangle ABC$, $E$ is the midpoint of $AH$, and the incircle of $\triangle ABC$ touches sides $AB$ and $AC$ at points $C'$ and $B'$, respectively. $F$ is the reflection of $E$ over the line $B'C'$. Prove that $F$, the incenter $I$, and the circumcenter $O... | Proof: Let $\angle A=2 \alpha, \angle B=2 \beta, \angle C=2 \gamma$.
Establish a complex plane with $A$ as the origin and $A I$ as the real axis. Let the symmetric points of $O$ and $I$ with respect to the line $B^{\prime} C^{\prime}$ be $O_{1}$ and $I_{1}$, respectively. Denote the foot of the perpendicular from $B^{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,336 |
10. In $\triangle A B C$, $A B=\sqrt{2}, A C=\sqrt{3}$, $\angle B A C=30^{\circ}$, and $P$ is any point in the plane of $\triangle A B C$. Then the minimum value of $\mu=\overrightarrow{P A} \cdot \overrightarrow{P B}+\overrightarrow{P B} \cdot \overrightarrow{P C}+\overrightarrow{P C} \cdot \overrightarrow{P A}$ is $\... | 10. $\frac{\sqrt{2}}{2}-\frac{5}{3}$.
Notice,
$$
\begin{aligned}
\mu= & \overrightarrow{P A} \cdot \overrightarrow{P B}+\overrightarrow{P B} \cdot \overrightarrow{P C}+\overrightarrow{P C} \cdot \overrightarrow{P A} \\
= & \overrightarrow{P A} \cdot(\overrightarrow{P A}+\overrightarrow{A B}+\overrightarrow{P A}+\overr... | \frac{\sqrt{2}}{2}-\frac{5}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,337 |
13. (25 points) Given the ellipse $C: \frac{x^{2}}{4}+y^{2}=1$ with its upper vertex $M$ and lower vertex $N$, and $T(t, 2)(t \neq 0)$ is a point on the line $y=2$. The lines $T M$ and $T N$ intersect the ellipse $C$ at points $E$ and $F$ respectively. If the area of $\triangle T M N$ is $k$ times the area of $\triangl... | 13. It is known that $l_{T M}: y=\frac{x}{t}+1$.
Substituting into the ellipse equation, we get $E\left(\frac{-8 t}{t^{2}+4}, \frac{t^{2}-4}{t^{2}+4}\right)$. Similarly,
$$
l_{T N}: y=\frac{3 x}{t}-1, F\left(\frac{24 t}{t^{2}+36}, \frac{-t^{2}+36}{t^{2}+36}\right) .
$$
Then $l_{E F}: y-\frac{t^{2}-4}{t^{2}+4}=-\frac{... | \frac{4}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,338 |
2. The range of the function $y=\frac{\sin 2 x-3}{\sin x+\cos x-2}$ is | 2. $[2-\sqrt{2}, 2+\sqrt{2}]$.
Notice that,
$$
\begin{aligned}
y & =\frac{\sin 2 x-3}{\sin x+\cos x-2} \\
& =\frac{(\sin x+\cos x)^{2}-4}{\sin x+\cos x-2} \\
& =\sin x+\cos x+2 .
\end{aligned}
$$
Since $-\sqrt{2} \leqslant \sin x+\cos x \leqslant \sqrt{2}$, therefore,
$$
2-\sqrt{2} \leqslant y \leqslant 2+\sqrt{2} \t... | [2-\sqrt{2}, 2+\sqrt{2}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,339 |
1. Given a cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $P$. Point $Q$ lies on line $BD$ (not coinciding with $B$) such that $PQ = PB$. There are two points $R$ and $S$ in the plane such that quadrilaterals $CAQR$ and $DBCS$ are parallelograms. Prove that points $C$, $Q$, $R$, and $S$ are... | Take $P$ as the origin to establish the complex plane.
Let $A=a \mathrm{e}^{\mathrm{i} \alpha}, B=b \mathrm{e}^{\mathrm{i} \beta}, C=-c \mathrm{e}^{\mathrm{i} \alpha}, D=-d \mathrm{e}^{\mathrm{i} \beta}$.
Then $Q=-B=-b \mathrm{e}^{\mathrm{i} \beta}, a c=b d$,
$$
R=C+Q-A=C-B-A, S=C+D-B \text {. }
$$
Thus, $C, Q, R, S$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,340 |
4. Arrange all positive odd numbers in ascending order, take the first number as $a_{1}$, take the sum of the next two numbers as $a_{2}$, then take the sum of the next three numbers as $a_{3}$, and so on, to get the sequence $\left\{a_{n}\right\}$, that is,
$$
a_{1}=1, a_{2}=3+5, a_{3}=7+9+11, \cdots \cdots
$$
Then $... | 4. 44100 .
Notice that, $\sum_{i=1}^{20} a_{i}$ is the sum of the first $\sum_{i=1}^{20} i=210$ odd numbers. Therefore, $210^{2}=44100$. | 44100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,341 |
5. Let the line $y=k x-2$ intersect the parabola $y^{2}=8 x$ at points $A$ and $B$. If the x-coordinate of the midpoint of segment $A B$ is 2, then the length of segment $A B$ is $|A B|=$ $\qquad$ . | 5. $2 \sqrt{15}$.
According to $k x=y+2$, we get
$$
\begin{array}{l}
k y^{2}=8 k x=8 y+16, \\
y_{1,2}=\frac{4}{k} \pm \frac{4}{k} \sqrt{k+1} .
\end{array}
$$
Thus, $x_{1,2}=\frac{2}{k}+\frac{4}{k^{2}} \pm \frac{4}{k^{2}} \sqrt{k+1}$.
According to the condition, we have
$$
2=\frac{x_{1}+x_{2}}{2}=\frac{2}{k}+\frac{4}{... | 2 \sqrt{15} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,342 |
7. The last digit of $\sum_{k=0}^{201}(10 k+7)^{k+1}$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 7.6.
It is easy to see that $\sum_{k=0}^{201}(10 k+7)^{k+1} \equiv \sum_{k=0}^{201} 7^{k+1}(\bmod 10)$.
Notice that, for any natural number $n$, the last digits of $7^{4 n+1}$, $7^{4 n+2}$, $7^{4 n+3}$, and $7^{4 n+4}$ are sequentially $7$, $9$, $3$, and $1$, and the last digit of their sum is 0.
$$
\text{Therefore, }... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,343 |
8. Let $M=\{1,2, \cdots, 2017\}$ be the set of the first 2017 positive integers. If one element is removed from the set $M$, and the sum of the remaining elements is exactly a perfect square, then the removed element is $\qquad$ . | 8. 1677 .
Notice,
$$
\begin{array}{l}
1+2+\cdots+2017=\frac{2017 \times 2018}{2} \\
>[\sqrt{2017 \times 1009}]^{2}=1426^{2} \\
=2033476 . \\
\text { Then } \frac{2017 \times 2018}{2}-2033476=1677 .
\end{array}
$$ | 1677 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,344 |
2. The solution set of the inequality $\sqrt{x-x^{2}}+\sqrt{4 x-x^{2}} \geqslant 2$ is $\qquad$ . | 2. $\left\{\frac{4}{5}\right\}$.
From the domain, we get $0 \leqslant x \leqslant 1$.
$$
\begin{array}{l}
\text { By } \sqrt{x-x^{2}}+\sqrt{4 x-x^{2}} \geqslant 2 \\
\Rightarrow \sqrt{4 x-x^{2}} \geqslant 2-\sqrt{x-x^{2}} \\
\Rightarrow 4 \sqrt{x-x^{2}} \geqslant 4-3 x \\
\Rightarrow 25 x^{2}-48 x+16 \leqslant 0 \\
\R... | \left\{\frac{4}{5}\right\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 730,345 |
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