problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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9. Let $f(x)$ be a function defined on $\mathbf{R}$, if $f(0)$ $=1008$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+4)-f(x) \leqslant 2(x+1), \\
f(x+12)-f(x) \geqslant 6(x+5) .
\end{array}
$$
Then $\frac{f(2016)}{2016}=$ $\qquad$ . | 9. 504 .
From the conditions, we have
$$
\begin{array}{l}
f(x+12)-f(x) \\
=(f(x+12)-f(x+8))+ \\
\quad(f(x+8)-f(x+4))+(f(x+4)-f(x)) \\
\leqslant 2((x+8)+1)+2((x+4)+1)+2(x+1) \\
=6 x+30=6(x+5) . \\
\text { Also, } f(x+12)-f(x) \geqslant 6(x+5), \text { thus, } \\
f(x+12)-f(x)=6(x+5) .
\end{array}
$$
Then, $f(2016)$
$$
... | 504 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,813 |
10. When $x$, $y$, $z$ are positive numbers, the maximum value of $\frac{4 x z+y z}{x^{2}+y^{2}+z^{2}}$ is $\qquad$ . | 10. $\frac{\sqrt{17}}{2}$.
Notice,
(1) $x^{2}+\frac{16}{17} z^{2} \geqslant 2 \sqrt{\frac{16}{17}} x z$, equality holds if and only if $x=\frac{4}{\sqrt{17}} z$;
(2) $y^{2}+\frac{1}{17} z^{2} \geqslant 2 \sqrt{\frac{1}{17}} y z$, equality holds if and only if $y=\frac{1}{\sqrt{17}} z$.
$$
\begin{array}{l}
\text { Then... | \frac{\sqrt{17}}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,814 |
Example 1 As shown in Figure $4, \triangle A B C$ has three excircles that are tangent to its circumcircle at points $D, E, F$. Then $A D, B E$, and $C F$ are concurrent. ${ }^{[1]}$ (2007, Japan Mathematical Olympiad) | Prove that according to property $2$, the isogonal conjugate of the Nagel point of $\triangle ABC$ is the intersection point of $AD$, $BE$, and $CF$.
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly. | null | Geometry | proof | Yes | Yes | cn_contest | false | 729,815 |
2. If the complex number $z=\frac{\sqrt{3}}{2}+\frac{1}{2} \mathrm{i}$, then $z^{2016}=(\quad)$.
(A) -1
(B) $-\mathrm{i}$
(C) $\mathrm{i}$
(D) 1 | 2. D.
It is easy to get $z^{3}=\mathrm{i}$. Then
$$
z^{2016}=\left(z^{3}\right)^{672}=i^{672}=\left(i^{4}\right)^{168}=1 .
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,816 |
3. Given the function $f(x)=x^{2}-2 t x+t$. When $x \in$ $[-1,1]$, let the minimum value of $f(x)$ be $m$. Then the maximum value of $m$ is ( ).
(A) -2
(B) 0
(C) $\frac{1}{4}$
(D) 1 | 3. C.
When $t \leqslant -1$, $m=f(-1)=3t+1$, the maximum value of $m$ is -2;
When $-1<t<1$,
$$
m=f(t)=t-t^{2}=-\left(t-\frac{1}{2}\right)^{2}+\frac{1}{4} \text{, }
$$
the maximum value of $m$ is $\frac{1}{4}$;
When $t \geqslant 1$, $m=f(1)=1-t$, the maximum value of $m$ is 0.
In summary, the maximum value of $m$ is $\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,817 |
4. For any positive integers $n$ and $k(k \leqslant n)$, $f(n, k)$ represents the number of positive integers not exceeding $\left[\frac{n}{k}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$) that are coprime with $n$. Then $f(100,3)=(\quad)$.
(A) 11
(B) 13
(C) 14
(D) 19 | 4. C.
From $\left[\frac{100}{3}\right]=33$, we know that the required number is the count of numbers from 1 to 33 that are coprime with 100.
First, remove all even numbers, leaving 17 odd numbers. Then, remove multiples of 5 (a total of three), thus, the required number is 14. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,818 |
Example 2 As shown in Figure $5, I$ is the incenter of $\triangle A B C$. A line through point $A$ parallel to $BC$ intersects the circumcircle $\odot O$ of $\triangle A B C$ at point $A_{1}$. $AI$ intersects $BC$ at point $D$, and $E$ is the point where the incircle $\odot I$ of $\triangle A B C$ touches $BC$. $A_{1} ... | Proof As shown in Figure 5, let $EI$ intersect $\odot I$ at point $F$, connect $AF$, intersecting $BC$ and $\odot O$ at points $S$ and $G$ respectively; draw $JK \parallel BC$ through point $F$, intersecting $AB$ and $AC$ at points $J$ and $K$ respectively.
Then $\triangle JFI \sim \triangle IEB \Rightarrow JF \cdot BE... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,819 |
9. Given a regular quadrilateral pyramid $S-A B C D$ with lateral edge length $4, \angle A S B=30^{\circ}$, take points $E, F, G$ on the lateral edges $S B, S C, S D$ respectively. Then the minimum perimeter of the spatial quadrilateral $A E F G$ is . $\qquad$ | $9.4 \sqrt{3}$.
Unfold the side of the quadrilateral pyramid along $SA$, as shown in Figure 1. When points $A, E, F, G$ are collinear, the perimeter of the original spatial quadrilateral $A E F G$ takes the minimum value $2 S A \sin 60^{\circ}=4 \sqrt{3}$. | 4 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,820 |
10. Given that the circumcenter of $\triangle A B C$ is $O$, and
$$
2 \overrightarrow{O A}+3 \overrightarrow{O B}+4 \overrightarrow{O C}=\mathbf{0} \text {. }
$$
Then $\cos \angle B A C=$ $\qquad$ | 10. $\frac{1}{4}$.
Assume the circumradius of $\triangle A B C$ is 1.
$$
\begin{array}{l}
\text { From } 2 \overrightarrow{O A}+3 \overrightarrow{O B}+4 \overrightarrow{O C}=\mathbf{0} \\
\Rightarrow 2 \overrightarrow{O A}=-3 \overrightarrow{O B}-4 \overrightarrow{O C} \\
\Rightarrow 4=9+16+24 \overrightarrow{O B} \cd... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,821 |
13. Let $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms $S_{n}=2^{n}+r$ (where $r$ is a constant).
Let $b_{n}=2\left(1+\log _{2} a_{n}\right)\left(n \in \mathbf{Z}_{+}\right)$.
(1) Find the sum of the first $n$ terms $T_{n}$ of the sequence $\left\{a_{n} b_{n}\right\}$;
(2) If for any... | Three, 13. (1) From the conditions, it is easy to know
$$
\begin{array}{l}
a_{1}=2+r, a_{2}=S_{2}-S_{1}=2, \\
a_{3}=S_{3}-S_{2}=4 .
\end{array}
$$
Substituting into $a_{2}^{2}=a_{1} a_{3}$, we get $r=-1$.
Thus, $S_{n}=2^{n}-1$.
Then $a_{n}=2^{n-1}, b_{n}=2\left(1+\log _{2} a_{n}\right)=2 n$, $a_{n} b_{n}=n \times 2^{n... | \frac{3 \sqrt{2}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,822 |
Example 3 As shown in Figure 6, in $\triangle A B C$, the external angle bisector of $\angle A$ intersects line $B C$ at point $D, I$ is the incenter of $\triangle A B C$, and $I_{a}$ is the excenter of $\triangle A B C$ opposite to $\angle A$. A perpendicular line is drawn from point $I$ to $D I_{a}$, intersecting the... | Proof of the lemma first.
Lemma As shown in Figure 7, $I, J$ are the incenter and the excenter opposite to $\angle A$ of $\triangle ABC$, respectively. $F$ is the intersection of the external angle bisector of $\angle BAC$ with $BC$. A perpendicular line $IG$ is drawn from $I$ to $JF$, with the foot of the perpendicul... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,823 |
2. Given that $f(x)$ is a periodic function on $\mathbf{R}$ with the smallest positive period of 2, and when $0 \leqslant x<2$, $f(x)=x^{3}-x$. Then the number of intersections between the graph of the function $y=f(x)$ and the $x$-axis in the interval $[0,6]$ is $\qquad$ . | 2.7.
When $0 \leqslant x<2$, let $f(x)=x^{3}-x=0$, we get $x=0$ or 1.
According to the properties of periodic functions, given that the smallest period of $f(x)$ is 2, we know that $y=f(x)$ has six zeros in the interval $[0,6)$.
Also, $f(6)=f(3 \times 2)=f(0)=0$, so $f(x)$ has 7 intersection points with the $x$-axis ... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,824 |
3. Given the function
$$
f(x)=\sin \omega x+\cos \omega x(\omega>0)(x \in \mathbf{R}) \text {. }
$$
If the function $f(x)$ is monotonically increasing in the interval $(-\omega, \omega)$, and the graph of the function $y=f(x)$ is symmetric about the line $x=\omega$, then $\omega=$ . $\qquad$ | 3. $\frac{\sqrt{\pi}}{2}$.
Notice,
$$
f(x)=\sin \omega x+\cos \omega x=\sqrt{2} \sin \left(\omega x+\frac{\pi}{4}\right),
$$
and the graph of the function $y=f(x)$ is symmetric about the line $x=\omega$.
$$
\begin{array}{l}
\text { Then } f(\omega)=\sqrt{2} \sin \left(\omega^{2}+\frac{\pi}{4}\right)= \pm \sqrt{2} \\
... | \frac{\sqrt{\pi}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,825 |
4. As shown in Figure $1, a$ is the result output in the program flowchart. Then, the coefficient of the term containing $x^{2}$ in the expansion of the binomial $\left(a \sqrt{x}-\frac{1}{\sqrt{x}}\right)^{6}$ is . $\qquad$ | 4. $-\frac{3}{16}$.
When $i=2016$, $a=\frac{1}{2}$.
Notice that,
$$
\begin{array}{l}
T_{r+1}=\mathrm{C}_{6}^{r}(a \sqrt{x})^{r}\left(-\frac{1}{\sqrt{x}}\right)^{6-r} \\
=(-1)^{6-r} a^{r} \mathrm{C}_{6}^{r} x^{r-3} .
\end{array}
$$
Thus, the coefficient of the $x^{2}$ term is
$$
(-1)^{6-5} a^{5} C_{6}^{5}=-\left(\frac... | -\frac{3}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,826 |
6. Let the odd function $f(x)$ have a domain of $[-2,2]$,
and be decreasing in the interval $[-2,0]$, satisfying
$$
f(1-m)+f\left(1-m^{2}\right)<0 \text {. }
$$
Then the range of real number $m$ is $\qquad$ | 6. $[-1,1)$.
Given that the domain of $f(x)$ is $[-2,2]$, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
-2 \leqslant 1-m \leqslant 2, \\
-2 \leqslant 1-m^{2} \leqslant 2
\end{array}\right. \\
\Rightarrow-1 \leqslant m \leqslant \sqrt{3} .
\end{array}
$$
Since $f(x)$ is an odd function and is decreasing in the i... | [-1,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,827 |
7. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>1, b>0)$ with focal distance $2 c$, the line $l$ passes through the points $(a, 0)$ and $(0, b)$, and the sum of the distances from the point $(1,0)$ to the line $l$ and from the point $(-1,0)$ to the line $l$ is $s \geqslant \frac{4}{5} c$. Then the r... | 7. $\left[\frac{\sqrt{5}}{2}, \sqrt{5}\right]$.
Let the line $l: \frac{x}{a}+\frac{y}{b}=1$, i.e.,
$$
b x+a y-a b=0 \text {. }
$$
By the distance formula from a point to a line, and given $a>1$, the distance $d_{1}$ from the point $(1,0)$ to the line $l$ is $\frac{b(a-1)}{\sqrt{a^{2}+b^{2}}}$; the distance $d_{2}$ fr... | \left[\frac{\sqrt{5}}{2}, \sqrt{5}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,828 |
8. It is known that Team A and Team B each have several people. If 90 people are transferred from Team A to Team B, then the total number of people in Team B will be twice that of Team A; if some people are transferred from Team B to Team A, then the total number of people in Team A will be 6 times that of Team B. Then... | 8. 153.
Let the original number of people in team A and team B be $a$ and $b$ respectively.
Then $2(a-90)=b+90$.
Suppose $c$ people are transferred from team B to team A. Then $a+c=6(b-c)$.
From equations (1) and (2), eliminating $b$ and simplifying, we get
$$
\begin{array}{l}
11 a-7 c=1620 \\
\Rightarrow c=\frac{11 a... | 153 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,829 |
10. The sequence $a_{0}, a_{1}, \cdots, a_{n}$ satisfies
$$
a_{0}=\sqrt{3}, a_{n+1}=\left[a_{n}\right]+\frac{1}{\left\{a_{n}\right\}} \text {, }
$$
where, $[x]$ denotes the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then $a_{2016}=$ $\qquad$ | 10. $3024+\sqrt{3}$.
From the given, we have
$$
\begin{array}{l}
a_{0}=1+(\sqrt{3}-1), \\
a_{1}=1+\frac{1}{\sqrt{3}-1}=1+\frac{\sqrt{3}+1}{2}=2+\frac{\sqrt{3}-1}{2}, \\
a_{2}=2+\frac{2}{\sqrt{3}-1}=2+(\sqrt{3}+1)=4+(\sqrt{3}-1), \\
a_{3}=4+\frac{1}{\sqrt{3}-1}=4+\frac{\sqrt{3}+1}{2}=5+\frac{\sqrt{3}-1}{2} .
\end{array... | 3024+\sqrt{3} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,830 |
12. (13 points) In a game activity of a TV entertainment program, each participant needs to complete three tasks: $A$, $B$, and $C$. It is known that the probabilities for participant Jia to complete tasks $A$, $B$, and $C$ are $\frac{3}{4}$, $\frac{3}{4}$, and $\frac{2}{3}$, respectively. Each task is independent of t... | 12. (1) Let the events of contestant A completing projects $A$, $B$, and $C$ be denoted as events $A$, $B$, and $C$, respectively, and assume they are mutually independent. The event of completing at least one project is denoted as event $D$.
$$
\begin{array}{l}
\text { Then } P(D)=1-P(\bar{A} \bar{B} \bar{C}) \\
=1-\f... | \frac{243}{128} a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,831 |
13. (13 points) As shown in Figure 2, in the pyramid $P-ABCD$, $PA \perp$ base $ABCD$, $BC=CD=2$, $AC=4$, $\angle ACB = \angle ACD = \frac{\pi}{3}$, $F$ is the midpoint of $PC$, and $AF \perp PB$. Find:
(1) the length of $PA$;
(2) the sine of the dihedral angle $B-AF-D$.
| 13. (1) As shown in Figure 3, connect $B D$, intersecting $A C$ at point $O$. Taking $O$ as the origin, the directions of $\overrightarrow{O B}$, $\overrightarrow{O C}$, and $\overrightarrow{A P}$ are the positive directions of the $x$-axis, $y$-axis, and $z$-axis, respectively, to establish the spatial rectangular coo... | 2 \sqrt{3}, \frac{3 \sqrt{7}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,832 |
14. (13 points) Let the sequence $\left\{a_{n}\right\}\left(n \in \mathbf{Z}_{+}\right)$ have the sum of the first $n$ terms as $S_{n}$, and the point $\left(a_{n}, S_{n}\right)$ lies on the graph of $y=\frac{1}{6}-\frac{1}{3}-x$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) If $c_{1}=... | 14. (1) It is easy to know that $S_{n}=\frac{1}{6}-\frac{1}{3} a_{n}$.
When $n \geqslant 2$,
$$
\begin{array}{l}
a_{n}=S_{n}-S_{n-1}=\frac{1}{3} a_{n-1}-\frac{1}{3} a_{n} \\
\Rightarrow a_{n}=\frac{1}{4} a_{n-1} . \\
\text { Also, } S_{1}=\frac{1}{6}-\frac{1}{3} a_{1} \Rightarrow a_{1}=\frac{1}{8} \\
\Rightarrow a_{n}... | \frac{1}{3} \leqslant \frac{1}{c_{2}}+\frac{1}{c_{3}}+\cdots+\frac{1}{c_{n}}<\frac{3}{4} | Algebra | proof | Yes | Yes | cn_contest | false | 729,833 |
1. Given points $A(3,1), B\left(\frac{5}{3}, 2\right)$, and the four vertices of $\square A B C D$ are all on the graph of the function $f(x)=\log _{2} \frac{a x+b}{x-1}$. Then the area of $\square A B C D$ is $\qquad$. | $-1 . \frac{26}{3}$.
$$
\text { Given } \begin{aligned}
f(3) & =1, f\left(\frac{5}{3}\right)=2 \\
\Rightarrow a & =1, b=1 .
\end{aligned}
$$
Thus, $f(x)=\log _{2} \frac{x+1}{x-1}$.
Clearly, $f(x)$ is an odd function, and the graph of $f(x)$ is symmetric about the origin $O$.
From the given, $l_{A B}: 3 x+4 y-13=0$, t... | \frac{26}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,834 |
2. Let the set
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{5}\right\}\left(a_{i} \in \mathbf{R}_{+}, i=1,2, \cdots, 5\right) .
$$
If the set of the products of the four elements in all four-element subsets of set $A$ is $B=\{2,3,4,6,9\}$, then the sum of the elements in set $A$ is $\qquad$ | 2. $\frac{49}{6}$.
Obviously, in all the four-element subsets of set $A$, each element appears 4 times.
Then $\left(a_{1} a_{2} \cdots a_{5}\right)^{4}=1296 \Rightarrow a_{1} a_{2} \cdots a_{5}=6$.
Thus, the products of the four elements of set $A$ are 2, 3, $\frac{2}{3}$, 1, $\frac{3}{2}$, and their sum is $\frac{49}... | \frac{49}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,835 |
1. As shown in Figure $9, \triangle A B C$ has a pseudo-incircle $\odot O_{1}$ at $A$ that intersects $B C$ at points $E$ and $F$, and is tangent to the circumcircle of $\triangle A B C$ at point $D$. $I$ is the incenter of $\triangle A B C$. Prove: $ID$ bisects $\angle E D F$.
untranslated text:
如图 $9, \triangle A B... | Prompt: Extend $D E$ and $D F$, intersecting the circumcircle of $\triangle A B C$ at points $G$ and $P$, respectively, and it can be proven that $G P \parallel E F$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,836 |
7. In tetrahedron $ABCD$, $\angle ADB = \angle BDC = \angle CDA = 60^{\circ}$, $AD = BD = 3$, $CD = 2$. Then the volume of the circumscribed sphere of tetrahedron $ABCD$ is $\qquad$ | $7.4 \sqrt{3} \pi$
It is known that $\triangle A D B$ is an equilateral triangle, and $C A = C B$. As shown in Figure 2, let $P$ and $M$ be the midpoints of $A B$ and $C D$ respectively, and connect $P D$ and $P C$. Then the plane $A B D \perp$ plane $P D C$.
Let $N$ be the circumcenter of $\triangle A B D$, and $O$ b... | 4 \sqrt{3} \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,837 |
11. (20 points) 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}=1+\frac{1}{a_{n}}+\ln a_{n} \text {. }
$$
Let $S_{n}=\left[a_{1}\right]+\left[a_{2}\right]+\cdots+\left[a_{n}\right]$. Find $S_{n}$. | 11. Combining the known information, from $a_{1}=1$, we get $a_{2}=2$.
Thus, $a_{3}=\frac{3}{2}+\ln 2$.
Since $\frac{1}{2}1$, we have $f^{\prime}(x)>0$, hence, $f(x)$ is monotonically increasing in the interval $\left(2, \frac{5}{2}\right)$, that is,
$$
\begin{array}{l}
f(2)2, \\
f\left(\frac{5}{2}\right)=1+\frac{2}{5... | 2n-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,838 |
Three. (50 points) Nine small balls numbered $1,2, \cdots, 9$ are randomly placed on nine equally spaced points on a circle, with one ball on each point. Let $S$ be the sum of the absolute differences of the numbers on all adjacent pairs of balls. Find the probability of the arrangement that minimizes $S$.
Note: If on... | Three, nine differently numbered small balls are placed on nine equally divided points on a circle, with one ball on each point, equivalent to a circular permutation of nine different elements, with a total of $8!$ ways of placement. Considering the factor of flipping, the number of essentially different ways is $\frac... | \frac{1}{315} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,840 |
518 On a circle, initially write 1 and 2 at opposite positions. Each operation involves writing the sum of two adjacent numbers between them, for example, the first operation writes two 3s, the second operation writes two 4s and two 5s. After each operation, the sum of all numbers becomes three times the previous sum. ... | Observe the pattern, and conjecture that after a sufficient number of operations, the $n$ numbers written equal $\varphi(n)$. After each operation, the property that adjacent numbers are coprime remains unchanged, and the new number written each time is the sum of its two neighbors. Therefore, the number we are looking... | 2016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,841 |
Given 519 As shown in Figure $3, \triangle A B C$ is an isosceles triangle, $\angle C A B=\angle C B A=\alpha$, points $P$ and $Q$ are located on opposite sides of line segment $A B$, $\angle C A P=\angle A B Q=\beta, \angle C B P=$ $\angle B A Q=\gamma$. Prove: $P, C, Q$ are collinear. | Prove that, as shown in Figure 3, connect $P C$ and extend it to intersect $A Q$, $B Q$, and $A B$ at points $Q_{2}$, $Q_{1}$, and $D$ respectively.
$$
\begin{array}{c}
\text { Then } \frac{C D \cdot P Q_{2}}{P C \cdot D Q_{2}}=\frac{S_{\triangle A C D} S_{\triangle A P Q_{2}}}{S_{\triangle A P C} S_{\triangle A D Q_{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,842 |
3. As shown in Figure $11, \odot O_{1}, \odot O_{2}, \odot O_{3}$ are the three pseudo-incircles of $\triangle ABC$, which are internally tangent to the circumcircle $\odot O$ of $\triangle ABC$ at points $D, E, F$ respectively. $\odot O_{1}$ is tangent to $AB, AC$ at points $S, T$ respectively, and the midpoint of arc... | Hint: First, by Mannheim's theorem, it can be proven that the lines $Q D$, $S T$, $O_{2} O_{3}$, and $B C$ concur at $R$. Then prove that $E F$ passes through point $R$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,843 |
Example 1 Find all integer solutions to the indeterminate equation $x^{2}=y^{2}+2 y+13$.
untranslated text remains in its original format and lines. | The original equation can be transformed into
$$
x^{2}-(y+1)^{2}=12 \text {. }
$$
By the difference of squares formula, we get
$$
(x+y+1)(x-y-1)=12 \text {. }
$$
Since \(x+y+1\) and \(x-y-1\) have the same parity, and 12 is an even number, we have
$$
\begin{array}{l}
(x+y+1, x-y-1) \\
=(2,6),(6,2),(-2,-6),(-6,-2) \\
... | (x, y)=(4,-3),(4,1),(-4,1),(-4,-3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,844 |
5. Find all triples $(m, n, p)$ such that $p^{n}+144=m^{2}$ (where $m, n \in \mathbf{Z}_{+}$, and $p$ is a prime).
---
The translation is provided as requested, maintaining the original formatting and structure. | Hint: Convert the original equation to
$$
p^{n}=(m+12)(m-12) \text {. }
$$
The solutions are
$$
(m, n, p)=(13,2,5),(20,8,2),(15,4,3) .
$$ | (13,2,5),(20,8,2),(15,4,3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,846 |
5. As shown in Figure 1, the side length of square $ABCD$ is $1, E$ and $F$ are the midpoints of sides $BC$ and $AD$ respectively. Fold $\triangle ABF$ along the line $BF$, and fold $\triangle CDE$ along the line $DE$. Then, during the folding process, the maximum distance between points $A$ and $C$ is $\qquad$ | 5. $\sqrt{2}$.
In square $ABCD$, connect $AC$, intersecting $BF$ and $DE$ at points $G$ and $H$ respectively.
Then, during the folding process, there is always
$$
AC \leqslant AG + GH + HC = \sqrt{2} \text{, }
$$
When $\triangle ABF$ and $\triangle CDE$ are coplanar, the equality holds. | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,847 |
6. Given the ellipse $C: \frac{x^{2}}{9}+\frac{y^{2}}{8}=1$ with left and right foci $F_{1}$ and $F_{2}$, and left and right vertices $A$ and $B$, the line $l: x=m y+1$ passing through the right focus $F_{2}$ intersects the ellipse $C$ at points $M\left(x_{1}, y_{1}\right)$ and $N\left(x_{2}, y_{2}\right)\left(y_{1}>0,... | 6. $\frac{\sqrt{3}}{12}$.
Let $l_{M N}: x=m y+1$, substituting into the ellipse equation yields $\left(8 m^{2}+9\right) y^{2}+16 m y-64=0$.
Then $y_{1}+y_{2}=-\frac{16 m}{8 m^{2}+9}, y_{1} y_{2}=-\frac{64}{8 m^{2}+9}$
$\Rightarrow y_{1}=\frac{-8 m+24 \sqrt{m^{2}+1}}{8 m^{2}+9}$,
$$
y_{2}=\frac{-8 m-24 \sqrt{m^{2}+1}}{... | \frac{\sqrt{3}}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,848 |
7. Given the function $f(x)=x^{2}+2 x+a \ln x$, for any $t \geqslant 1$ it always holds that $f(2 t-1) \geqslant 2 f(t)-3$. Then the range of the real number $a$ is $\qquad$ . | 7. $a \leqslant 2$.
According to the problem, the original condition is equivalent to: for any $t \geqslant 1$, it always holds that
$$
\begin{array}{l}
2(t-1)^{2}+a \ln (2 t-1)-2 a \ln t \geqslant 0 . \\
\text { Let } u(t)=2(t-1)^{2}+a \ln (2 t-1)-2 a \ln t . \\
\text { Then } u^{\prime}(t)=4(t-1)+\frac{2 a}{2 t-1}-\... | a \leqslant 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,849 |
8. If $f(x)=\sum_{k=0}^{4034} a_{k} x^{k}$ is the expansion of $\left(x^{2}+x+2\right)^{2017}$, then $\sum_{k=0}^{1344}\left(2 a_{3 k}-a_{3 k+1}-a_{3 k+2}\right)=$ $\qquad$ | 8. 2 .
Let $x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$. Then
$$
\begin{array}{l}
x^{2}+x+2=1 \\
\Rightarrow \sum_{k=0}^{1344}\left(a_{3 k}+a_{3 k+1} \omega+a_{3 k+2} \omega^{2}\right)=1
\end{array}
$$
Taking the conjugate of the above equation, we get
$$
\sum_{k=0}^{1344}\left(a_{3 k}+a_{... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,850 |
9. (16 points) In the sequence $\left\{a_{n}\right\}$, $a_{1}=2, a_{n+1}=\frac{a_{n}^{2}}{a_{n}+2}$.
Prove: $\sum_{k=1}^{n} \frac{2 k a_{k}}{a_{k}+2}<4$. | Sure, here is the translated text:
```
9. It is easy to see that $a_{n}>0$.
Thus, $a_{n+1}-a_{n}=\frac{-2 a_{n}}{a_{n}+2}<0$.
Therefore, $\left\{a_{n}\right\}$ is a decreasing sequence.
Notice that,
$$
\begin{array}{l}
\frac{a_{n+1}}{a_{n}}=\frac{a_{n}}{a_{n}+2}=1-\frac{2}{a_{n}+2} \\
\leqslant 1-\frac{2}{a_{1}+2}=\fr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,851 |
Example 1 Given a circle $\odot O$ and a point $P$ not on $\odot O$, two moving lines $l$ and $l^{\prime}$ through $P$ intersect $\odot O$ at points $X$ and $Y$, $X^{\prime}$ and $Y^{\prime}$ respectively. Prove: the line connecting the centers of the circumcircles of $\triangle P X Y^{\prime}$ and $\triangle P X^{\pri... | Proof As shown in Figure 1, let the circumcircles of $\triangle P X Y^{\prime}$ and $\triangle P X^{\prime} Y$ be $\odot O_{1}$ and $\odot O_{2}$, respectively. The second intersection point of $\odot O_{1}$ and $\odot O_{2}$ is $Q$. The radius of $\odot O$ is $r$. The line $P Q$ intersects $\odot O$ at points $P_{1}$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,852 |
Given $\left\{y_{n}\right\}$ is a sequence of real numbers, and satisfies $\sum_{i=1}^{n} y_{i}^{3}=0$. Prove: there must exist $k \in\{1,2, \cdots, n\}$, such that
$$
\left|y_{k}\right| \geqslant \sqrt[4]{\frac{27}{4}} \cdot \frac{\left|y_{1}+y_{2}+\cdots+y_{n}\right|}{n} .
$$ | Let $I_{m}=\sum_{k=1}^{n} y_{k}^{m}$.
By the Cauchy-Schwarz inequality, we have
$$
\sum_{k=1}^{n} y_{k}^{2} \geqslant \frac{1}{n}\left(\sum_{k=1}^{n} y_{k}\right)^{2} \Rightarrow n I_{2} \geqslant I_{1}^{2} \text {. }
$$
Again, by the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(\sum_{k=1}^{n}\left(r+y... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,853 |
Given $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\begin{array}{l}
\sum\left(\frac{a^{2}}{c}+\frac{c^{2}}{a}\right)+7 \sum a \\
\geqslant \frac{\left(\sum a\right)^{3}}{\sum a b}+\frac{2\left(\sum a b\right)^{2}}{a b c},
\end{array}
$$
where, “ $\sum ”$ denotes the cyclic sum. | $$
\begin{array}{l}
\sum\left(\frac{a^{2}}{c}+\frac{b^{2}}{c}\right)+7 \sum a \\
\geqslant \frac{\left(\sum a\right)^{3}}{\sum a b}+\frac{2\left(\sum a b\right)^{2}}{a b c} \\
\Leftrightarrow \sum \frac{(a-b)^{2}}{c} \\
\geqslant \frac{\left(\sum a\right)^{3}-3\left(\sum a\right)\left(\sum a b\right)}{\sum a b}+ \\
\fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,854 |
High $\mathbf{5 2 3}$ As shown in Figure 2, line $l_{1}$ intersects circles $\Gamma_{1}$ and $\Gamma_{2}$ at points $A, B_{1}, C_{1}, D$ in sequence, and line $l_{2}$ intersects circles $\Gamma_{1}$ and $\Gamma_{2}$ at points $A_{1}, B, C, D_{1}$ in sequence. If $\angle A B C = \angle A_{1} B_{1} C_{1}, \angle B A C = ... | Proof From the problem, we know that the positions of points $A, B, C, D$ on circle $\Gamma_{1}$ are similar to the positions of points $A_{1}, B_{1}, C_{1}, D_{1}$ on circle $\Gamma_{2}$.
Let lines $l_{1}$ and $l_{2}$ intersect at point $P$. Then
$$
\begin{array}{l}
\triangle P A B \backsim \triangle P A_{1} B_{1}, \t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,855 |
Given 524 As shown in Figure $3, \triangle A B C$ has three altitudes $A D, B E, C F$, with orthocenter $H$. Ray $E F$ intersects the circumcircle $\Gamma_{0}$ of $\triangle A B C$ at point $M, D F$ intersects $B E$ at point $P$, and $D E$ intersects $C F$ at point $Q$. Ray $P Q$ intersects the circumcircle of $\triang... | Let the circumcircles of $\triangle D E F$ and $\triangle B H C$ be $\Gamma$ and $\Gamma_{1}$, respectively.
Since $B, F, H, D$ are concyclic, we have
$$
P B \cdot P H = P D \cdot P F.
$$
Thus, point $P$ lies on the radical axis of circles $\Gamma$ and $\Gamma_{1}$.
Similarly, since $C, E, H, D$ are concyclic, point $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,856 |
Example 2 In the acute triangle $\triangle ABC$, draw $CD \perp AB$ at point $D$, and let $E$ be any point on $CD$. Draw perpendiculars from $D$ to $AC$, $AE$, $BE$, and $BC$, and let the feet of these perpendiculars be $P$, $Q$, $R$, and $S$ respectively. Prove that points $P$, $Q$, $R$, and $S$ are concyclic or colli... | Proof as shown in Figure 2.
From $A P \perp P D, A Q \perp Q D$, we know that points $A, P, Q, D$ are concyclic, and the circle has $A D$ as its diameter, denoted as $\Gamma_{A D}$.
Similarly, points $D, R, S, B$ are concyclic, and the circle has $B D$ as its diameter, denoted as $\Gamma_{B D}$; points $C, P, D, S$ ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,857 |
Example 3 Given that the incenter of a non-isosceles $\triangle ABC$ is $I$, a line perpendicular to $AI$ through $I$ intersects $AC$ and $AB$ at points $B'$ and $C'$, respectively. Points $B_1$ and $C_1$ are on the rays $BC$ and $CB$, respectively, such that $AB = BB_1$ and $AC = CC_1$. If the second intersection poin... | Let $\triangle ABC$ have three interior angles $\angle A$, $\angle B$, and $\angle C$. As shown in Figure 3, without loss of generality, assume $\angle A > \angle C$.
Let the composite transformation $h$ be: first perform an inversion with center $A$ and power $AB \cdot AC$, then reflect over the axis $AI$.
Then $B \x... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,858 |
Example 4 Given that $D$ is a moving point on side $BC$ of $\triangle ABC$, and the incenters of $\triangle ABC$, $\triangle ABD$, and $\triangle ACD$ are $I$, $I_1$, and $I_2$ respectively. The circumcircles of $\triangle IAI_1$ and $\triangle IAI_2$ intersect the circumcircle of $\triangle ABC$ at points $M$ and $N$ ... | Let $I_{A}$ be the center of the excircle opposite to vertex $A$ of $\triangle ABC$. Let the composite transformation $h$: first perform an inversion with $A$ as the inversion center and $AB \cdot AC$ as the inversion power, then perform a reflection with $AI$ as the axis of symmetry.
Then $B \xrightarrow{h} C, C \xrig... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,859 |
1. Given $D, E$ are points on the line $AB$ of $\triangle ABC$, satisfying $AD = AC, BE = BC, \angle A$ and $\angle B$'s bisectors intersect the opposite sides at points $P, Q$, and intersect the circumcircle of $\triangle ABC$ at points $M, N$. Let the circumcenters of $\triangle MBE$ and $\triangle AND$ be $U, V$, re... | Hint: First, perform an inversion with $A$ as the inversion center and $A B \cdot A C$ as the inversion power; then, derive the images of each point under the inversion by using similar and congruent triangles; finally, conclude by noting that the orthocenter and circumcenter lie on two isogonal lines of the triangle.
... | null | Geometry | proof | Yes | Yes | cn_contest | false | 729,860 |
2. Given a quadrilateral $ABCD$ that has both a circumcircle (with center $O$) and an incircle, $DA$ and $CB$ intersect at point $E$, $BA$ and $CD$ intersect at point $F$, $AC$ and $BD$ intersect at point $S$. Points $E'$ and $F'$ are on sides $AB$ and $AD$ respectively, satisfying
$$
\angle BEE' = \angle AEE', \angle ... | Given $O$ as the center of inversion and the radius of the circumcircle of quadrilateral $A B C D$ as the power of inversion, the inversion transformation can yield the image by the power of a point theorem. At this point, to prove that three circles share a common radical axis, it is only necessary to prove that three... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,861 |
3. As shown in Figure 7, circle $\Gamma$ intersects circle $\Gamma_{1}$ at points $A$ and $B$. The tangent to circle $\Gamma_{1}$ at point $A$ intersects circle $\Gamma$ at point $C$ (different from point $A$), and the tangent to circle $\Gamma$ at point $A$ intersects circle $\Gamma_{1}$ at point $C_{1}$ (different fr... | Hint: Taking $A$ as the inversion center for inversion transformation, the conclusion can be proven by the power of a point theorem and the properties of the nine-point circle.
The original text's line breaks and formatting have been preserved. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,862 |
Example 3 Proof: The indeterminate equation
$$
x^{2}+y^{2}-z^{2}=1997
$$
has infinitely many integer solutions. | Proof From equation (1) we get
$$
y^{2}-z^{2}=1997-x^{2} \text {. }
$$
By the proposition, we know that when $1997-x^{2}$ is odd, equation (2) always has integer solutions, at this time, $x$ is even.
$$
\text { Let } x=2 t(t \in \mathbf{Z}) \text {. }
$$
Then equation (2) becomes
$$
\begin{array}{l}
(y+z)(y-z)=1997-4... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,864 |
Example 1 Given a quadrilateral $A_{0} A_{1} A_{2} A_{3}$, through the vertex $A_{0}$, draw perpendiculars $l_{i}$ to $A_{0} A_{i}(i=1,2,3)$. Let $l_{1}$ intersect $A_{2} A_{3}$ at point $P_{1}$, $l_{2}$ intersect $A_{3} A_{1}$ at point $P_{2}$, and $l_{3}$ intersect $A_{1} A_{2}$ at point $P_{3}$. Prove: $P_{1}, P_{2}... | Prove as shown in Figure 1.
Since points \( P_{1}, P_{2}, P_{3} \) are on the extensions of the sides of \( \triangle A_{1} A_{2} A_{3} \), and
\[
\angle P_{1} A_{0} A_{1} = \angle P_{3} A_{0} A_{3} = 90^{\circ},
\]
therefore, \( \angle P_{1} A_{0} A_{3} = \angle P_{3} A_{0} A_{1} \).
Similarly, from \( \angle P_{2} A_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,865 |
Example 2 As shown in Figure 2, through any point $P$ inside $\triangle A B C$, draw perpendicular lines $l_{A}, l_{B}, l_{C}$ to $P A, P B, P C$ respectively. If $l_{A}$ intersects $B C$ at point $D, l_{B}$ intersects $A C$ at point $E, l_{C}$ intersects $A B$ at point $F$, prove: $D, E, F$ are collinear. | $$
\begin{aligned}
& \text { Hence, by } P F \perp P C, P E \perp P B, P D \perp P A, \text { we get } \\
& \angle B P F=\angle E P C, \angle B P D=\angle E P A, \\
& \angle A P F+\angle C P D=360^{\circ}-\angle C P F-\angle A P D \\
= & 360^{\circ}-90^{\circ}-90^{\circ}=180^{\circ} . \\
& \text { Then } \frac{C E}{E A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,866 |
Example 3 Given that the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $P$, a line is drawn from $P$ intersecting the lines containing the sides $AB$, $BC$, $CD$, and $DA$ of the quadrilateral at points $E$, $M$, $F$, and $N$ respectively, and $PE=PF$. Prove: $PM=PN$.
| Prove as shown in Figure 3.
According to Menelaus' theorem:
By line $AB$ intersecting $\triangle P M C$, we get
$$
\frac{P E}{E M} \cdot \frac{M B}{B C} \cdot \frac{C A}{A P}=1 \text {, }
$$
By line $C D$ intersecting $\triangle P N A$, we get
$$
\frac{P F}{F N} \cdot \frac{N D}{D A} \cdot \frac{A C}{C P}=1 \text {. }... | P M=P N | Geometry | proof | Yes | Yes | cn_contest | false | 729,867 |
Example 4 Let $D$ be a point on side $BC$ of $\triangle ABC$, and point $P$ lies on segment $AD$. A line through $D$ intersects segments $AB$ and $PB$ at points $M$ and $E$, and the extensions of segments $AC$ and $PC$ at points $F$ and $N$. If $DE = DF$, prove: $DM = DN$. | Prove as shown in Figure 4.
According to Menelaus' theorem:
By the line $P B$ cutting $\triangle A M D$, we get
$$
\frac{M E}{E D} \cdot \frac{D P}{P A} \cdot \frac{A B}{B M}=1,
$$
By the line $A F$ cutting $\triangle D N P$, we get
$$
\frac{D F}{F N} \cdot \frac{N C}{C P} \cdot \frac{P A}{A D}=1 \text {. }
$$
Multip... | D M=D N | Geometry | proof | Yes | Yes | cn_contest | false | 729,868 |
Example $\mathbf{1}$ (1) Give an example of a function $f$ : $\mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$ that satisfies
$$
2 f\left(x^{2}\right) \geqslant x f(x)+x \quad (x>0) \text {. }
$$
(2) Prove: For functions satisfying (1), we have
$$
f\left(x^{3}\right) \geqslant x^{2} \quad (x>1) .
$$ | (1) Solution Let $f(x)=\left\{\begin{array}{ll}1, & 0<x \leq 1 \\ x, & x>1 .\end{array}\right.$ It is easy to verify that $f(x)$ satisfies the given conditions.
(2) Proof From (1) we have
$2 f\left(x^{2}\right)>x \Rightarrow f(x)>\frac{\sqrt{x}}{2}$.
Define the sequence $\left\{a_{n}\right\}: a_{0}=\frac{1}{2}, a_{n+1}... | f\left(x^{3}\right) \geqslant x^{2}(x>1) | Inequalities | proof | Yes | Yes | cn_contest | false | 729,869 |
Example 3 If the function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies for all real numbers $x$,
$$
\sqrt{2 f(x)}-\sqrt{2 f(x)-f(2 x)} \geqslant 2
$$
then the function $f$ is said to have property $P$. Find the largest real number $\lambda$ such that if the function $f$ has property $P$, then for all real numbers ... | Solve: From equation (1) we get
$$
\begin{array}{l}
f(x) \geqslant 2, \\
\sqrt{2 f(x)}-2 \geqslant \sqrt{2 f(x)-f(2 x)} .
\end{array}
$$
Squaring both sides of equation (2) we get
$$
\begin{array}{l}
f(2 x) \geqslant 4 \sqrt{2 f(x)}-4 \\
\Rightarrow f(x) \geqslant 4, \text { and } f(x) \geqslant 4 \sqrt{2 f\left(\frac... | 12+8 \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,871 |
Given $A D$ is the angle bisector of $\angle A$ in $\triangle A B C$, the incircle $\odot I$ of $\triangle A B C$ touches side $B C$ at point $E$, point $A_{1}$ lies on the circumcircle of $\triangle A B C$ such that $A A_{1} / / B C, T$ is the intersection of the circumcircle of $\triangle A E D$ and line $E A_{1}$ (p... | Proof As shown in Figure 1, let the external angle bisector of $\angle BAC$ intersect the circumcircle of $\triangle ABC$ at point $K$. Extend $KI$ to intersect the circumcircle of $\triangle ABC$ at point $J$.
It is easy to see that $KB = KC$, $KA = KA_1$.
$$
\begin{array}{l}
\text{Then } \frac{BJ}{CJ} = \frac{S_{\tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,872 |
Given $A D, B E, C F$ are the three altitudes of acute $\triangle A B C$, point $Z$ is on $A D$, points $X, Y$ are on the extensions of $B E, C F$ respectively, and
$$
\angle A Y B=\angle B Z C=\angle C X A=90^{\circ} .
$$
Prove: $X, Y, Z$ are collinear if and only if the length of the tangent from point $A$ to the ni... | Proof As shown in Figure 2, it is easy to see that $B, C, E, F$, $C, A, F, D$, and $A, B, D, E$ are each sets of four concyclic points.
Combining the projection theorem, we get
$$
A X^{2}=A E \cdot A C=A F \cdot A B=A Y^{2} \Rightarrow A X=A Y.
$$
Similarly, $B Y=B Z, C Z=C X$.
Thus, $\angle A X Y=\angle A Y X, \angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,873 |
Example 4 Find all positive integers $k$ such that there exist positive integers $m, n$ satisfying
$$
m(m+k)=n(n+1) \text {. }
$$ | Solution: Clearly, $k=1$ meets the requirement, at this time, $m=n$.
When $k>1$, it is obvious that $m>3$.
When $k \geqslant 4$, from equation (1) we get
$$
\begin{array}{l}
\left(4 m^{2}+4 m k+k^{2}\right)-k^{2}+1=4 n^{2}+4 n+1 \\
\Rightarrow(2 m+k)^{2}-(2 n+1)^{2}=k^{2}-1 .
\end{array}
$$
By the difference of square... | k=1 \text{ or } k \geqslant 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,874 |
2. Two individuals, A and B, independently flip a fair coin. A flips the coin 10 times, and B flips it 11 times. The probability that B gets more heads than A is $\qquad$ . | 2. $\frac{1}{2}$.
Let "the number of times the coin lands heads up for Yi is more than the number of times the coin lands heads up for Jia when each of them independently tosses the coin 10 times" be event $A$. Then the probability of the desired event is
$$
P=P(A)+\frac{1}{2}(1-2 P(A))=\frac{1}{2} .
$$ | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,875 |
3. In $\triangle A B C$, $A B=2, A C=1, B C=\sqrt{7}$, $O$ is the circumcenter of $\triangle A B C$, and $\overrightarrow{A O}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}$. Then $\lambda+\mu=$ $\qquad$ . | 3. $\frac{13}{6}$.
In $\triangle A B C$, by the cosine rule we have
$$
\angle B A C=120^{\circ} \Rightarrow \overrightarrow{A B} \cdot \overrightarrow{A C}=-1 \text {. }
$$
Also, $O$ is the circumcenter of $\triangle A B C$, so
$$
\begin{array}{l}
\left\{\begin{array}{l}
\overrightarrow{A O} \cdot \overrightarrow{A B... | \frac{13}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,876 |
5. The number of non-empty subsets of the set $\{1,2, \cdots, 2016\}$ whose elements sum to an odd number is $\qquad$ | 5. $2^{2015}$
$$
\text { Let } f(x)=(1+x)\left(1+x^{2}\right) \cdots\left(1+x^{2016}\right) \text {. }
$$
Then the desired value is the sum of the coefficients of the odd powers of $x$ in the expansion of $f(x)$. Therefore, the number of non-empty subsets with an odd sum of elements is $\frac{1}{2}(f(1)-f(-1))=2^{2015... | 2^{2015} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,877 |
6. Define the sequence $\left\{a_{n}\right\}: a_{n}$ is the last digit of $1+2+\cdots+n$, and $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then $S_{2016}=$ $\qquad$ . | 6.7 066 .
From the problem, we know
$$
\begin{array}{l}
\frac{(n+20)(n+20+1)}{2}=\frac{n^{2}+41 n+420}{2} \\
=\frac{n(n+1)}{2}+20 n+210 .
\end{array}
$$
Then $\frac{(n+20)(n+21)}{2}$ and $\frac{n(n+1)}{2}$ have the same last digit, i.e., $a_{n+20}=a_{n}$.
$$
\begin{array}{l}
\text { Therefore, } S_{2016}=S_{16}+100 S... | 7066 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,878 |
Example 5 Proof: The number $97^{97}$ cannot be expressed as the sum of cubes of several consecutive integers. ${ }^{[2]}$ | Prove the sum of cubes formula:
$$
1^{3}+2^{3}+\cdots+(n-1)^{3}+n^{3}=\frac{n^{2}(n+1)^{2}}{4} \text {. }
$$
In fact, it is only necessary to prove that $97^{97}$ cannot be expressed as the sum of cubes of several consecutive positive integers.
Assume there exist $m, n \in \mathbf{N}$, such that
$$
97^{97}=(m+1)^{3}+(... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,879 |
II. (16 points) Given real numbers $x, y$ satisfy $2^{x}+3^{y}=4^{x}+9^{y}$.
Try to find the range of values for $U=8^{x}+27^{y}$. | Let $a=2^{x}, b=3^{y}$.
Then the given condition becomes
$$
\begin{array}{l}
a+b=a^{2}+b^{2} \quad (a, b>0) \\
\Rightarrow\left(a-\frac{1}{2}\right)^{2}+\left(b-\frac{1}{2}\right)^{2}=\frac{1}{2} .
\end{array}
$$
From the image in the $a O b$ plane, we know
$$
t=a+b \in(1,2] \text {. }
$$
Also, $a b=\frac{(a+b)^{2}-\... | (1,2] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,880 |
1. Given the function $f(x)$ satisfies $f(1)=2$, and
$$
f(x+1)=\frac{1+f(x)}{1-f(x)}
$$
for any $x$ in its domain. Then $f(2016)=$
$\qquad$ | $-1 . \frac{1}{3}$.
From $f(x+1)=\frac{1+f(x)}{1-f(x)}$, it is easy to get
$$
f(x+2)=-\frac{1}{f(x)}, f(x+4)=f(x) \text {. }
$$
Thus, $f(x)$ is a periodic function with a period of 4.
Given $f(2)=-3, f(3)=-\frac{1}{2}, f(4)=\frac{1}{3}$
$$
\Rightarrow f(2016)=f(4)=\frac{1}{3} \text {. }
$$ | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,881 |
2. From five positive integers $a, b, c, d, e$, any four are taken to find their sum, resulting in the set of sums $\{44,45,46,47\}$, then $a+b+c+d+e=$ $\qquad$ . | 2. 57 .
From five positive integers, if we take any four to find their sum, there are five possible ways, which should result in five sum values. Since the set $\{44,45, 46,47\}$ contains only four elements, there must be two sum values that are equal. Therefore,
$$
\begin{array}{l}
44+44+45+46+47 \\
\leqslant 4(a+b+c... | 57 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,882 |
3. Given that $\triangle A B C$ is an equilateral triangle, and the ellipse $\Gamma$ has one focus at $A$, and the other focus $F$ lies on the line segment $B C$. If the ellipse $\Gamma$ passes exactly through points $B$ and $C$, then its eccentricity is $\qquad$ | 3. $\frac{\sqrt{3}}{3}$.
Let the side length of the equilateral $\triangle A B C$ be $x$, the semi-major axis of the ellipse $\Gamma$ be $a$, and the semi-focal distance be $c$.
According to the problem, we have
$$
|A C|+|C F|=|A B|+|B F|=2 a \text {. }
$$
Therefore, $F$ is the midpoint of $B C$, so
$$
x+\frac{x}{2}=... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,883 |
4. Given real numbers $x, y$ satisfy $\frac{x^{2}}{3}+y^{2}=1$. Then
$$
P=|2 x+y-4|+|4-x-2 y|
$$
the range of values for $P$ is . $\qquad$ | 4. $[2,14]$.
Let $x=\sqrt{3} \cos \theta, y=\sin \theta(\theta \in[0,2 \pi))$.
Then $2 x+y=2 \sqrt{3} \cos \theta+\sin \theta$
$$
\begin{array}{l}
=\sqrt{13} \sin (\theta+\alpha)<4, \\
x+2 y=\sqrt{3} \cos \theta+2 \sin \theta \\
=\sqrt{7} \sin (\theta+\beta)<4 .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence } P=|2 ... | [2,14] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,884 |
Example 6 Try to find the non-negative integer solutions $x, y, z$ that satisfy the equation $2^{x}+3^{y}=z^{2}$. | If $y=0$, then
$2^{x}=z^{2}-1=(z+1)(z-1)$.
Let $\left\{\begin{array}{l}z+1=2^{s}, \\ z-1=2^{t}\end{array}\right.$. $(s>t, x=s+t)$.
Subtracting the two equations gives $2^{s}-2^{t}=2$.
Solving this, we get $t=1, s=2$.
At this point, $x=3, y=0, z=3$.
If $y>0$, then
$$
2^{x}+3^{y} \equiv(-1)^{x}(\bmod 3) \text {. }
$$
If... | x=3, y=0, z=3; x=0, y=1, z=2; x=4, y=2, z=5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,885 |
8. Let one edge of a tetrahedron be 6, and the other edges all be 5. Then the radius of the circumscribed sphere of this tetrahedron is $\qquad$ | 8. $\frac{20 \sqrt{39}}{39}$.
In the tetrahedron $P A B C$, let
$$
B C=6, P A=P B=P C=A B=A C=5 \text {. }
$$
Let the midpoint of $B C$ be $D$, and the circumcenter of $\triangle A B C$ be $E$.
Then $B D=C D=3$,
$$
A E=B E=C E=\frac{25}{8}, D E=\frac{7}{8} \text {. }
$$
Since $P A=P B=P C=5$, we have
$P E \perp$ pla... | \frac{20 \sqrt{39}}{39} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,886 |
9. Given that $M N$ is a moving chord of the circumcircle of equilateral $\triangle A B C$ with side length $2 \sqrt{6}$, $M N=4$, and $P$ is a moving point on the sides of $\triangle A B C$. Then the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is | 9.4.
Let the circumcenter of $\triangle ABC$ be $O$.
It is easy to find that the radius of $\odot O$ is $r=2 \sqrt{2}$.
Also, $MN=4$, so $\triangle OMN$ is an isosceles right triangle, and
$$
\begin{aligned}
& \overrightarrow{O M} \cdot \overrightarrow{O N}=0,|\overrightarrow{O M}+\overrightarrow{O N}|=4 . \\
& \text ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,887 |
12. Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies
$$
f(1)=\frac{10}{3} \text {, }
$$
and for any real numbers $x, y$, it always holds that
$$
f(x) f(y)=f(x+y)+f(x-y) .
$$
Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{n}=3 f(n)-f(n-1)\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(1) Find the... | 12. (1) In equation (1), let $x=1, y=0$, we get $f(1) f(0)=2 f(1)$.
Also, $f(1)=\frac{10}{3}$, then $f(0)=2$.
In equation (1), let $x=n, y=1$, we get
$$
\begin{array}{l}
f(n) f(1)=f(n+1)+f(n-1) \\
\Rightarrow f(n+1)=\frac{10}{3} f(n)-f(n-1) \\
\Rightarrow a_{n+1}=3 f(n+1)-f(n) \\
\quad=9 f(n)-3 f(n-1) \\
\quad=3(3 f(n)... | S_{n}<1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,888 |
1. Let the complex number $z$ satisfy $|z|=1$. Then the maximum value of $\mid z^{3}+3 z+2 \mathrm{i}$ is ( ).
(A) $4 \sqrt{2}$
(B) $3 \sqrt{3}$
(C) $2 \sqrt{5}$
(D) $4 \sqrt{3}$ | -.1. B.
Let's assume $z=a+bi$ from the given conditions.
Thus, $a^{2}+b^{2}=1$.
Then, $z^{3}+3z+2i=4a^{3}-\left(4b^{3}-6b-2\right)i$.
Therefore, $\left|z^{3}+3z+2i\right|^{2}=\left(4a^{3}\right)^{2}+\left(4b^{3}-6b-2\right)^{2}$ $=-16b^{3}-12b^{2}+24b+20$.
It is easy to see that when $b=\frac{1}{2}$, the above expressi... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,889 |
2. In $\triangle A B C$, the sides $a, b, c$ opposite to $\angle A, \angle B, \angle C$ form a geometric sequence. Then
$$
\frac{\sin A \cdot \cot C+\cos A}{\sin B \cdot \cot C+\cos B}
$$
the range of values is ( ).
(A) $(0,+\infty)$
(B) $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$
(C) $\left(0, \frac{\sq... | 2. B.
Let the common ratio of $a, b, c$ be $q$.
Then $b=a q, c=a q^{2}$.
$$
\begin{array}{l}
\text { Hence } \frac{\sin A \cdot \cot C+\cos A}{\sin B \cdot \cot C+\cos B} \\
=\frac{\sin A \cdot \cos C+\cos A \cdot \sin C}{\sin B \cdot \cos C+\cos B \cdot \sin C} \\
=\frac{\sin (A+C)}{\sin (B+C)}=\frac{\sin B}{\sin A}=... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,890 |
3. Let the function
$$
\begin{array}{l}
f(x, y) \\
=\sqrt{x^{2}+y^{2}-6 y+9}+\sqrt{x^{2}+y^{2}+2 \sqrt{3} x+3}+ \\
\quad \sqrt{x^{2}+y^{2}-2 \sqrt{3} x+3} .
\end{array}
$$
Then the minimum value of $f(x, y)$ is ( ).
(A) $3+2 \sqrt{3}$
(B) $2 \sqrt{3}+2$
(C) 6
(D) 8 | 3. C.
$$
\begin{array}{l}
\text { Let } A(0,3), B(-\sqrt{3}, 0), C(\sqrt{3}, 0) \text {, } \\
D(0,1), P(x, y) \text {. }
\end{array}
$$
Then \( f(x, y) = |PA| + |PB| + |PC| \).
Notice that,
$$
\angle ADB = \angle BDC = \angle CDA = 120^{\circ} \text {. }
$$
Thus, \( D \) is the Fermat point of \( \triangle ABC \).
Th... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,891 |
4. Arrange the numbers $2, 3, 4, 6, 8, 9, 12, 15$ in a row so that the greatest common divisor of any two adjacent numbers is greater than 1. The total number of possible arrangements is ( ) .
(A) 720
(B) 1014
(C) 576
(D) 1296 | 4. D.
First, divide the eight numbers into three groups:
I $(2,4,8)$, II $(3,9,15)$, III $(6,12)$.
Since the numbers in group I and group II have no common factors, the arrangement that satisfies the condition must be:
(1) After removing the numbers 6 and 12, the remaining numbers are divided into three parts and arra... | 1296 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 729,892 |
7. In the expansion of $(2+\sqrt{x})^{2 n+1}$, the sum of the coefficients of the terms where the power of $x$ is an integer is $\qquad$ . | 7. $\frac{3^{2 n+1}+1}{2}$.
Notice, $T_{r+1}=\mathrm{C}_{2 n+1}^{r} x^{\frac{2 n+1-r}{2}} 2^{r}$.
Since the power index of $x$ is an integer, $r$ must be odd.
$$
\begin{array}{l}
\text { Let } S=\mathrm{C}_{2 n+1}^{1} 2+\mathrm{C}_{2 n+1}^{3} 2^{3}+\cdots+\mathrm{C}_{2 n+1}^{2 n+1} 2^{2 n+1} . \\
\text { Also, }(1+2)^... | \frac{3^{2 n+1}+1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,893 |
9. As shown in Figure 1, in $\triangle A B C$,
$$
\begin{array}{l}
\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}, \\
\overrightarrow{A H} \cdot \overrightarrow{B C}=0, \\
\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0 .
\end{array}
$$
Then the eccentricity of the hyperbola passing through point $C$ an... | 9. 2 .
Given $\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow(\overrightarrow{C B}-\overrightarrow{C A}) \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow A C=B C$.
From $\overrightarrow{A H} \cdot \overrightarrow{B C}=0 \Rightarrow A H \perp B C$.
Since $\cos \frac... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,894 |
11. (20 points) If $\lg a+\lg b+\lg c=0$, prove:
$$
1<\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}<2 .
$$ | Three, 11. Since $\lg a+\lg b+\lg c=0$, hence $a, b, c>0$, and $abc=1$.
Assume without loss of generality that $c \leqslant a \leqslant b$.
When $a=b=c=1$, we have
$$
\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=\frac{3}{2} \text {. }
$$
When $a, b, c$ are not all equal, then
$$
00 .
\end{array}
$$
Since $b>1$, thus, $g... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,895 |
1. Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{n}=\frac{n}{n+2016}$. If $x_{2016}=x_{m} x_{n}$, then a solution for the positive integers $m, n$ is $\{m, n\}$ $=$ $\qquad$ . | 1. $\{4032,6048\}$. (Not unique)
$$
\begin{array}{l}
\text { Given } x_{2016}=x_{m} x_{n} \\
\Rightarrow \frac{1}{2}=\frac{m}{m+2016} \times \frac{n}{n+2016} \\
\Rightarrow(m+2016)(n+2016)=2 m n \\
\Rightarrow(m-2016)(n-2016)=2 \times 2016^{2} \text {. } \\
\text { If } m-2016=2016, n-2016=2 \times 2016 \\
\Rightarrow ... | \{4032,6048\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,897 |
3. If for all positive numbers $x, y$, we have
$$
\sqrt{x}+\sqrt{y} \leqslant a \sqrt{x+y} \text {, }
$$
then the minimum value of the real number $a$ is $\qquad$ | 3. $\sqrt{2}$.
$$
\begin{array}{l}
\text { Given }\left(\frac{\sqrt{x}}{\sqrt{x+y}}\right)^{2}+\left(\frac{\sqrt{y}}{\sqrt{x+y}}\right)^{2}=1 \\
\Rightarrow \frac{\sqrt{x}}{\sqrt{x+y}}+\frac{\sqrt{y}}{\sqrt{x+y}} \leqslant \sqrt{2} .
\end{array}
$$
When $x=y$, the equality holds. | \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,899 |
4. As shown in Figure $2, P$ is a point on the incircle of square $A B C D$, and let $\angle A P C=\alpha$, $\angle B P D=\beta$. Then $\tan ^{2} \alpha+\tan ^{2} \beta$ $=$ | 4. 8 .
As shown in Figure 5, establish a Cartesian coordinate system.
Let the equation of the circle be $x^{2}+y^{2}=r^{2}$.
Then the coordinates of the vertices of the square are
$$
A(-r,-r), B(r,-r), C(r, r), D(-r, r) \text {. }
$$
If $P(r \cos \theta, r \sin \theta)$, then the slopes of the lines $P A, P B, P C$, ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,900 |
6. Let $x$ be an acute angle. Then the maximum value of the function $y=\sin x \cdot \sin 2 x$ is $\qquad$ | 6. $\frac{4 \sqrt{3}}{9}$.
$$
\begin{array}{l}
\text { Given } y=2 \sin ^{2} x \cdot \cos x \\
\Rightarrow y^{2}=4 \sin ^{4} x \cdot \cos ^{2} x \\
\quad \leqslant 2\left(\frac{2 \sin ^{2} x+2 \cos ^{2} x}{3}\right)^{3}=\frac{16}{27} \\
\Rightarrow y \leqslant \frac{4 \sqrt{3}}{9} .
\end{array}
$$
When $\cos ^{2} x=\f... | \frac{4 \sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,901 |
9. As shown in Figure 3, $C D$ is a diameter of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. A line parallel to $C D$ is drawn through the left vertex $A$ of the ellipse's major axis, intersecting the ellipse at another point $N$ and the line of the ellipse's minor axis at point $M$. Prove: $A M \cdot A N=C... | 9. The equation of the ellipse is $\left\{\begin{array}{l}x=a \cos \theta, \\ y=b \sin \theta\end{array}\right.$
It is known that $A(-a, 0)$.
Let $l_{A N}:\left\{\begin{array}{l}x=-a+t \cos \theta, \\ y=t \sin \theta .\end{array}\right.$
Substituting into the ellipse equation, we get
$$
\begin{array}{l}
\left(b^{2} \co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,902 |
3. Consider whether the positive integer solutions to the equation $x^{2}+y^{3}=z^{2}$ are finite or infinite in number. ${ }^{[2]}$ | Hint: Transform the original equation into $z^{2}-x^{2}=y^{3}$, then use the difference of squares formula. There are infinitely many positive integer solutions. | There\ are\ infinitely\ many\ positive\ integer\ solutions. | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,903 |
10. As shown in Figure 4, $D$ is the excenter of $\triangle A B C$, and the point $A$ is symmetric to point $E$ with respect to the line $D C$. Prove:
(1) $B 、 C 、 E$
are collinear;
(2) $A 、 B 、 D 、 E$
are concyclic. | 10. As shown in Figure 6, connect $I B$. Let the point $A$ be symmetric to $B C$'s perpendicular bisector at $A_{1}$.
Then $\triangle A_{1} C B \cong \triangle A B C$.
If the incenters of $\triangle A B C$ and $\triangle A_{1} C B$ are $I$ and $I_{1}$ respectively, then quadrilateral $I I_{1} B C$ is an isosceles trape... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,904 |
2. Represent the integer 2016 as
$$
2016=\frac{\left(a_{1}!\right)\left(a_{2}!\right) \cdots\left(a_{m}!\right)}{\left(b_{1}!\right)\left(b_{2}!\right) \cdots\left(b_{n}!\right)} \text {, }
$$
where, $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{m}, b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}, a_{i} ... | 2. B.
Given $2016=2^{5} \times 7 \times 3^{2}, 7>3>2$, and 7 is in the numerator, we know $a_{1}=7$.
There does not exist a prime $p$ such that $p$ is a prime factor of 2016, then $b_{1}<p<7$.
The largest prime less than 7 must be in the numerator, so $b_{1}=5$.
$$
\text { Therefore, } 2016=\frac{(7!)(4!)(2!)}{5!} \... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,905 |
3. Given that $M$ and $N$ are points on the hypotenuse $BC$ of an isosceles right $\triangle ABC$,
$$
AB = AC = 6 \sqrt{2}, BM = 3, \angle MAN = 45^{\circ}.
$$
Then $NC = (\quad)$.
(A) 3
(B) $\frac{7}{2}$
(C) 4
(D) $\frac{9}{2}$ | 3. C.
As shown in Figure 4, $\triangle C A N$ is rotated $90^{\circ}$ clockwise around point $A$ to $\triangle A B D$, and $M D$ is connected.
$$
\begin{array}{l}
\text { Then } \angle M B D=\angle A B M+\angle A B D \\
=45^{\circ}+45^{\circ}=90^{\circ}, \\
B D=C N, A D=A N, \\
\angle M A D=\angle B A C-\angle M A N \... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,906 |
4. The ten smallest positive odd numbers $1, 3, \cdots, 19$ are arranged on a circle. Let $m$ be the maximum value of the sum of any one of the ten numbers and its two adjacent numbers. Then the minimum value of $m$ is ( ).
(A) 31
(B) 32
(C) 33
(D) 34 | 4. C.
As shown in Figure 5, the ten numbers given on the circumference are divided into four parts: $S_{1}, S_{2}, S_{3}, S_{4}$, where $S_{1}$ contains only the number $1$, and $S_{2}, S_{3}, S_{4}$ each contain three adjacent numbers.
Thus, the sum of the nine numbers in $S_{2}, S_{3}, S_{4}$ is
$$
3+5+\cdots+19=99... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 729,907 |
4. Given that the last two decimal digits of the square of an integer are 09. Prove: The hundreds digit of this number's square is even. ${ }^{[3]}$ | Let the integer be $a$, and $a^{2}=\overline{b 09}$.
Then $a^{2}=100 b+9$.
By the difference of squares formula, we get
$$
(a+3)(a-3)=100 b \text{. }
$$
Thus, both $a+3$ and $a-3$ are even.
It is easy to prove that $8 \mid (a+3)(a-3)$.
Therefore, $8 \mid 100 b$, which implies $2 \mid b$. | 2 \mid b | Number Theory | proof | Yes | Yes | cn_contest | false | 729,908 |
2. From the numbers $1,2, \cdots, 2017$, select $n$ numbers such that the difference between any two of these $n$ numbers is a composite number. The maximum value of $n$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result direct... | 2. 505 .
Notice that 4 is the smallest composite number.
Take all numbers of the form $4k+1$. Clearly, the difference between any two such numbers is a multiple of 4, and thus composite. This is the best possible scenario, because in any sequence of eight consecutive numbers $m, m+1, \cdots, m+7$, the differences betw... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 729,909 |
3. In a certain football league, a double round-robin system (i.e., two teams play each other twice) is used, with $m$ teams participating. At the end of the competition, a total of $9 n^{2}+6 n+32$ matches were played, where $n$ is an integer. Then $m=$ $\qquad$. | 3.8 or 32.
According to the problem, we have
$$
\begin{array}{l}
9 n^{2}+6 n+32=2 \times \frac{1}{2} m(m-1) \\
\Rightarrow(6 n+2)^{2}-(2 m-1)^{2}=-125 \\
\Rightarrow(6 n+2 m+1)(6 n-2 m+3)=-125 .
\end{array}
$$
Obviously, $6 n+2 m+1 \geqslant 6 n-2 m+3$.
Thus, $6 n+2 m+1$ can take the values $1, 5, 25, 125$, and corre... | 8 \text{ or } 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,910 |
2. Given real numbers $x, y$ satisfy $x^{2}+y^{2}-x y=12$. Then the maximum value of $x^{2}-y^{2}$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2. $8 \sqrt{3}$.
Let $m=x-y, n=x+y$.
Then $x=\frac{m+n}{2}, y=\frac{n-m}{2}$.
The original condition becomes $3 m^{2}+n^{2}=48$, hence
$$
x^{2}-y^{2}=m n \leqslant 8 \sqrt{3} \text {. }
$$
When $x=\sqrt{6}+\sqrt{2}, y=\sqrt{6}-\sqrt{2}$, the equality holds. | 8 \sqrt{3} | Number Theory | proof | Yes | Yes | cn_contest | false | 729,911 |
3. If three numbers are taken simultaneously from the 14 integers $1,2, \cdots, 14$, such that the absolute difference between any two numbers is not less than 3, then the number of different ways to choose is $\qquad$ | 3. 120 .
Let the three integers taken out be $x, y, z (x<y<z)$.
$$
\begin{array}{l}
\text { Let } a=x, b=y-x-2, \\
c=z-y-2, d=15-z .
\end{array}
$$
Thus, $a, b, c, d \geqslant 1$.
If $a, b, c, d$ are determined, then $x, y, z$ are uniquely determined.
Since $a+b+c+d=11$, it is equivalent to dividing 11 identical ball... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,912 |
Example 1 Given positive real numbers $a, b$ and integer $n \geqslant 2$, let $f(x)=(x+a)(x+b)$.
For non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$,
$$
\begin{array}{l}
x_{1}, x_{2}, \cdots, x_{n}, \text{ find } \\
F=\sum_{1 \leqslant i<j \leqslant n} \min \left\{f\left... | Solve:
$$
\begin{array}{l}
\min \left\{f\left(x_{i}\right), f\left(x_{j}\right)\right\} \leqslant \sqrt{f\left(x_{i}\right) f\left(x_{j}\right)} \\
=\sqrt{\left(x_{i}+a\right)\left(x_{i}+b\right)\left(x_{j}+a\right)\left(x_{j}+b\right)} \\
\leqslant \frac{\left(x_{i}+a\right)\left(x_{j}+b\right)+\left(x_{j}+a\right)\le... | \frac{n-1}{2 n}\left(1+n(a+b)+n^{2} a b\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,913 |
Example 2 Let $x_{i} \geqslant 0(i=1,2, \cdots, n)$, and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \sqrt{\frac{k}{j}} x_{k} x_{j}=1$.
Find the maximum and minimum values of $f=\sum_{i=1}^{n} x_{i}$.
(2001, National High School Mathematics League Competition) | 【Analysis】From the problem, we have
$$
\begin{array}{l}
f^{2}=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i0$, such that $z_{i}=c(\sqrt{i}-\sqrt{i-1})(1 \leqslant i \leqslant n)$.
Given $\sum_{i=1}^{n} z_{i}^{2}=1$, we find
$$
c=\frac{1}{\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}}},
$$
i.e., $z_{i}=\frac{\sqrt{i}-... | f_{\max }=\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,914 |
Example 2 Given that $BC$ is the diameter of $\odot O$, $A$ is a point on $\odot O$, $0<\angle AOB<120^{\circ}$, $D$ is the midpoint of arc $\overparen{AB}$ (excluding point $C$), a line through the center $O$ parallel to $DA$ intersects $AC$ at point $I$, the perpendicular bisector of $OA$ intersects $\odot O$ at poin... | Proof As shown in Figure 2.
Since $E F$ is the perpendicular to $O A$, therefore, $A$ is the midpoint of arc $\overparen{E F}$, which means $C A$ is the angle bisector of $\angle E C F$.
By a well-known theorem, it suffices to prove that $A I=A F$.
Also, $\angle B O D=\frac{1}{2} \angle B O A=\angle B C A$
$\Rightarrow... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,915 |
Three, (50 points) Given an integer $m \geqslant 3$, let $a_{1}, a_{2}, \cdots, a_{m}>0, n \geqslant m$, and $n \in \mathbf{Z}_{+}$.
Prove: $\sum_{i=1}^{m}\left(\frac{a_{i}}{a_{i}+a_{i+1}}\right)^{n} \geqslant \frac{m}{2^{n}}\left(a_{m+1}=a_{1}\right)$. | Let $b_{i}=\frac{a_{i+1}}{a_{i}}(i=1,2, \cdots, m)$.
Then, $\prod_{i=1}^{m} b_{i}=1$.
The inequality to be proved is equivalent to
$$
\sum_{i=1}^{m}\left(\frac{1}{1+b_{i}}\right)^{n} \geqslant \frac{m}{2^{n}} \text {. }
$$
By the power mean inequality, we have
$$
\sqrt[n]{\frac{\sum_{i=1}^{m}\left(\frac{1}{1+b_{i}}\ri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,916 |
Four. (50 points) On a plane, there are $n$ points, no three of which are collinear. Each pair of points is connected by a line segment, and each line segment is colored either red or blue. A triangle with all three sides of the same color is called a "monochromatic triangle." Let the number of monochromatic triangles ... | For $n=2 k(k \geqslant 3)$, we uniformly prove: the minimum value of $S$ is $\frac{k(k-1)(k-2)}{3}$.
Since there are $\mathrm{C}_{2 k}^{3}$ triangles, the number of non-monochromatic triangles is $\mathrm{C}_{2 k}^{3}-S$.
We call an angle with two adjacent sides (sharing a common point) of the same color a monochroma... | \frac{k(k-1)(k-2)}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,917 |
$$
\begin{array}{l}
\text { Find the largest positive integer } n, \text { such that for positive real } \\
\text { numbers } \alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}, \text { we have } \\
\quad \sum_{i=1}^{n} \frac{\alpha_{i}^{2}-\alpha_{i} \alpha_{i+1}}{\alpha_{i}^{2}+\alpha_{i+1}^{2}} \geqslant 0\left(\alpha_{n+1}... | Let $a_{i}=\frac{\alpha_{i+1}}{\alpha_{i}}(i=1,2, \cdots, n)$.
Then $\prod_{i=1}^{n} a_{i}=1$, and $a_{i}>0$.
The inequality to be proved becomes
$\sum_{i=1}^{n} \frac{1-a_{i}}{1+a_{i}^{2}} \geqslant 0$.
Let $x_{i}=\ln a_{i}(i=1,2, \cdots, n)$.
Then $\sum_{i=1}^{n} \frac{1-\mathrm{e}^{x_{i}}}{1+\left(\mathrm{e}^{x_{i}}... | 5 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,918 |
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