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In the integer sequence $\left\{a_{n}\right\}$, each term takes an integer between $0 \sim 2013$, and from the fourth term onwards, each term is equal to the remainder of the sum of its previous three terms divided by 2014. Find all possible values of the smallest positive period of the sequence $\left\{a_{n}\right\}$. | It is known that the characteristic equation of the sequence $\{a_n\}$ is
$$
\lambda^3 - \lambda^2 - \lambda - 1 = 0 \quad (\text{mod } 2014),
$$
with 13 being one of its roots.
Factoring, we get
$$
(\lambda - 13)(\lambda^2 + 12\lambda + 155) = 0.
$$
Consider the cases modulo 2, 19, and 53 separately.
(1) Clearly, the... | 1, 2, 4, 13, 18, 26, 36, 52, 234, 360, 468, 4680 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,383 |
412 Given that $x, y, z$ are positive real numbers. Prove:
$$
\frac{x y}{x^{2}+y^{2}+2 z^{2}}+\frac{y z}{2 x^{2}+y^{2}+z^{2}}+\frac{z x}{x^{2}+2 y^{2}+z^{2}} \leqslant \frac{3}{4} .
$$ | Proof: Let $u=x^{2}, v=y^{2}, w=z^{2}$. Then
$$
\begin{array}{l}
\frac{x y}{x^{2}+y^{2}+2 z^{2}}+\frac{y z}{2 x^{2}+y^{2}+z^{2}}+\frac{z x}{x^{2}+2 y^{2}+z^{2}} \\
=\frac{x y}{y^{2}+z^{2}+z^{2}+x^{2}}+ \\
\frac{y z}{z^{2}+x^{2}+x^{2}+y^{2}}+ \\
\frac{z x}{x^{2}+y^{2}+y^{2}+z^{2}} \\
\leqslant \frac{x y}{2 \sqrt{\left(y... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,385 |
6. As shown in Figure $12, E$ and $F$ are on the legs $AC$ and $AB$ of the right triangle $\triangle ABC$, respectively. $BE$ intersects $CF$ at point $D$. The circumcircle $\odot M$ of $\triangle AEF$ intersects the circumcircle $\odot O$ of $\triangle ABC$ at point $P$. Prove: $AP \perp PD$. | Connect $AO$ and extend it to intersect $\odot O$ at point $Q$, connect $OM$ and extend it to intersect $AD, AP$ at points $N, L$ respectively. Then $L$ is the midpoint of $AP$, and according to Newton's line theorem, $N$ is the midpoint of $AD$.
Thus, $P, D, Q$ are collinear.
Therefore, $AP \perp PD$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,386 |
Example 2 For any positive integer $n$, prove:
$$
\sum_{k=0}^{n} 2^{k} \mathrm{C}_{n}^{k} \mathrm{C}_{n-k}^{\left[\frac{n-k}{2}\right]}=\mathrm{C}_{2 n+1}^{n},
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
(1994, China Mathematical Olympiad) | Prove the construction of the polynomial function
$$
f(x)=(1+x)^{2 n+1} \text {. }
$$
Observe that, the right side of the equation to be proved is obviously the coefficient of the $n$-th term of $f(x)$.
$$
\begin{array}{l}
\text { Also, } f(x)=(1+x)\left(1+2 x+x^{2}\right)^{n} \\
=(1+x) \sum_{k=0}^{n} \mathrm{C}_{n}^{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,387 |
Example 3 Prove: For any positive integer $n$, among the binomial coefficients $\mathrm{C}_{n}^{m}(0 \leqslant m \leqslant n)$, the number of odd ones is a power of 2. | Prove that representing $n$ in binary as
$$
\begin{array}{l}
n=2^{\alpha_{1}}+2^{\alpha_{2}}+\cdots+2^{\alpha_{k}}\left(\alpha_{1} \geqslant \alpha_{2} \geqslant \cdots \geqslant \alpha_{k}\right) \\
\Rightarrow(1+x)^{n}=(1+x)^{2 \alpha_{1}}(1+x)^{2 \alpha_{2}} \cdots(1+x)^{2 \alpha_{k}} \\
\Rightarrow(1+x)^{n} \\
\qua... | 2^{k} | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,388 |
Example 3 As shown in Figure 5, in $\triangle A B C$, it is given that $A B \neq A C$, $A T$ is the angle bisector, $M$ is the midpoint of $B C$, $H$ is the orthocenter, $H M$ intersects $A T$ at point $D$, and $D E \perp A B$, $D F \perp A C$. Prove: $E 、 H 、 F$ are collinear. | Prove as shown in Figure 6, extend $M H$, intersecting the circumcircle of $\triangle A B C$ at points $P$ and $Q$.
From $\triangle P E B \backsim \triangle P F C \Rightarrow \frac{P B}{P C}=\frac{B E}{C F}$.
From $S_{\triangle B P Q}=S_{\triangle C P Q} \Rightarrow \frac{P B}{P C}=\frac{Q C}{Q B}$.
Since quadrilateral... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,389 |
Find pairs of positive integers $(x, y)$, satisfying
$$
x>y \text {, and }(x-y)^{x y}=x^{y} y^{x} \text {. }
$$
(2013, Taiwan Mathematical Olympiad Training Camp) | Solution 1 Let
$$
(x, y)=d, x=d v, y=d t,(v, t)=1 .
$$
Then $d^{x y-x-y}(v-t)^{x y}=v^{y} t^{x}$
$$
\Rightarrow d^{(x-1)(y-1)-1}(v-t)^{x y}=v^{y} t^{x} \text {. }
$$
If $(x-1)(y-1)-1=-1$
$\Rightarrow x=1$ (discard) or $y=1$
$$
\Rightarrow(x-1)^{x}=x
$$
$\Rightarrow$ no positive integer solution.
If $(x-1)(y-1)-1 \geq... | (x, y) = (4, 2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,390 |
In $\triangle A B C$, prove:
$$
\begin{array}{l}
\frac{1}{1+\cos ^{2} B+\cos ^{2} C}+\frac{1}{1+\cos ^{2} C+\cos ^{2} A}+ \\
\frac{1}{1+\cos ^{2} A+\cos ^{2} B} \leqslant 2 \text {. }
\end{array}
$$ | Promotion In $\triangle A B C$, if $0<\lambda \leqslant 1$, then
$$
\begin{array}{l}
\frac{1}{\lambda+\cos ^{2} B+\cos ^{2} C}+\frac{1}{\lambda+\cos ^{2} C+\cos ^{2} A}+ \\
\frac{1}{\lambda+\cos ^{2} A+\cos ^{2} B} \leqslant \frac{2}{\lambda} .
\end{array}
$$
This article extends the constant $\lambda$ in the above fo... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,391 |
For a non-empty integer set $A$, if it satisfies $a \in A, a-1 \notin A, a+1 \notin A$, then $a$ is called an isolated element of set $A$. Question: For the set
$$
M=\{1,2, \cdots, n\}(n \geqslant 3)
$$
how many $k(k \geqslant 3)$-element subsets of $M$ have no isolated elements? | Let the number of $k$-element subsets of set $M$ without isolated elements be denoted as $P_{k}$.
Let $r=\min \left\{\left[\frac{k}{2}\right], n-k+1\right\}([x]$ represents the greatest integer not exceeding the real number $x$). Then,
$$
P_{k}=\sum_{l=1}^{r} \mathrm{C}_{k-l-1}^{l-1} \mathrm{C}_{n-k+1}^{l} \text {. }
... | P_{k}=\sum_{l=1}^{r} \mathrm{C}_{k-l-1}^{l-1} \mathrm{C}_{n-k+1}^{l} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,392 |
1. Given that $x, y$ are integers, and satisfy $\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}\right)=-\frac{2}{3}\left(\frac{1}{x^{4}}-\frac{1}{y^{4}}\right)$. Then the possible values of $x+y$ are ( ) .
(A) 1
(B) 2
(C) 3
(D) 4 | -1. .
From the given equation, we have
$$
\frac{x+y}{x y} \cdot \frac{x^{2}+y^{2}}{x^{2} y^{2}}=\frac{2}{3} \cdot \frac{x^{4}-y^{4}}{x^{4} y^{4}} \text {. }
$$
Clearly, $x$ and $y$ are not equal to 0.
Thus, $x+y=0$ or $3 x y=2(x-y)$.
If $3 x y=2(x-y)$, then
$$
(3 x+2)(3 y-2)=-4 \text {. }
$$
Since $x$ and $y$ are int... | 1, -1 | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,393 |
2. Given non-negative real numbers $x, y, z$ satisfy $x+y+z=1$. Then the maximum value of $t=2xy+yz+2zx$ is ( ).
(A) $\frac{4}{7}$
(B) $\frac{5}{9}$
(C) $\frac{9}{16}$
(D) $\frac{12}{25}$ | 2. A.
Notice that,
$$
\begin{aligned}
t & =2 x y+y z+2 z x=2 x(y+z)+y z \\
& \leqslant 2 x(y+z)+\frac{1}{4}(y+z)^{2} \\
& =2 x(1-x)+\frac{1}{4}(1-x)^{2} \\
& =-\frac{7}{4}\left(x-\frac{3}{7}\right)^{2}+\frac{4}{7} .
\end{aligned}
$$
When $x=\frac{3}{7}, y=z=\frac{2}{7}$, the equality holds.
At this point, $t=2 x y+y ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,394 |
4. Six different cards are marked with the numbers 2, $2$, 4, $4$, 6, $6$, respectively. Three cards are drawn from them. The probability that the numbers on these three cards can serve as the lengths of the sides of a triangle is ( ).
(A) $\frac{1}{2}$
(B) $\frac{2}{5}$
(C) $\frac{2}{3}$
(D) $\frac{3}{4}$ | 4. B.
If the numbers on the three cards drawn are all different, there are $2 \times 2 \times 2=8$ ways to draw them; if the numbers on the three cards drawn are the same, there are $3 \times 4=12$ ways to draw them. Therefore, from six different cards, there are $8+12=20$ ways to draw three cards.
To form the sides ... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 728,395 |
5. Let $[t]$ denote the greatest integer not exceeding the real number $t$, and let $\{t\}=t-[t]$. Given that the real number $x$ satisfies $x^{3}+\frac{1}{x^{3}}=$
18. Then $\{x\}+\left\{\frac{1}{x}\right\}=(\quad)$.
(A) $\frac{1}{2}$
(B) $3-\sqrt{5}$
(C) $\frac{3-\sqrt{5}}{2}$
(D) 1 | 5. D.
Let $x+\frac{1}{x}=a$. Then
$$
\begin{array}{l}
x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}-1\right) \\
=\left(x+\frac{1}{x}\right)\left[\left(x+\frac{1}{x}\right)^{2}-3\right]=a\left(a^{2}-3\right) .
\end{array}
$$
Thus $a\left(a^{2}-3\right)=18$
$$
\begin{array}{l}
\Rightarrow(... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,396 |
1. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a+b+c=1, \\
\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}=1 .
\end{array}
$$
Then $a b c=$ | $$
\begin{aligned}
& \frac{1}{1-2 c}+\frac{1}{1-2 a}+\frac{1}{1-2 b}=1 \\
\Rightarrow & (1-2 a)(1-2 b)+(1-2 b)(1-2 c)+(1-2 a)(1-2 c) \\
& =(1-2 a)(1-2 b)(1-2 c) \\
\Rightarrow & 2-2(a+b+c)=8 a b c \\
\Rightarrow & a b c=0 .
\end{aligned}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,397 |
2. The largest positive integer $n$ for which the inequality $\frac{9}{17}<\frac{n}{n+k}<\frac{8}{15}$ holds for a unique integer $k$ is $\qquad$ | 2. 144 .
From the problem, we know that $\frac{7}{8}<\frac{k}{n}<\frac{8}{9}$.
By the uniqueness of $k$, we have
$$
\begin{array}{l}
\frac{k-1}{n} \leqslant \frac{7}{8}, \text { and } \frac{k+1}{n} \geqslant \frac{8}{9} . \\
\text { Therefore, } \frac{2}{n}=\frac{k+1}{n}-\frac{k-1}{n} \geqslant \frac{8}{9}-\frac{7}{8}... | 144 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,398 |
4. Given positive integers $a, b, c$ satisfy
$$
1<a<b<c, a+b+c=111, b^{2}=a c \text {. }
$$
then $b=$ $\qquad$ | 4. 36.
Let $(a, c)=d, a=a_{1} d, c=c_{1} d, a_{1}, c_{1}$ be positive integers, and $\left(a_{1}, c_{1}\right)=1, a_{1}<c_{1}$.
Then $b^{2}=a c=d^{2} a_{1} c_{1} \Rightarrow d^{2}\left|b^{2} \Rightarrow d\right| b$.
Let $b=b_{1} d\left(b_{1} \in \mathbf{Z}_{+}\right)$. Then $b_{1}^{2}=a_{1} c_{1}$.
Since $\left(a_{1},... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,399 |
Three. (25 points) Let $n$ be an integer. If there exist integers $x$, $y$, $z$ satisfying $n=x^{3}+y^{3}+z^{3}-3 x y z$, then $n$ is said to have property $P$.
Among the numbers 1, 5, 2013, 2014, which numbers have property $P$ and which do not? Explain your reasoning. | Three, taking $x=1, y=z=0$, we get
$$
1=1^{3}+0^{3}+0^{3}-3 \times 1 \times 0 \times 0 \text {. }
$$
Thus, 1 has property $P$.
Taking $x=y=2, z=1$, we get
$$
5=2^{3}+2^{3}+1^{3}-3 \times 2 \times 2 \times 1 \text {. }
$$
Therefore, 5 has property $P$.
Next, consider numbers with property $P$.
Let $f(x, y, z)=x^{3}+y^... | 1, 5, 2014 \text{ have property } P; 2013 \text{ does not have property } P | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,400 |
Three. (25 points) Let $n$ be an integer. If there exist integers $x, y, z$ satisfying
$$
n=x^{3}+y^{3}+z^{3}-3 x y z \text {, }
$$
then $n$ is said to have property $P$.
(1) Determine whether $1, 2, 3$ have property $P$;
(2) Among the 2014 consecutive integers $1, 2, \cdots, 2014$, how many do not have property $P$? | (1) Let $x=1, y=z=0$, we get
$$
1=1^{3}+0^{3}+0^{3}-3 \times 1 \times 0 \times 0 \text {. }
$$
Thus, 1 has property $P$.
Let $x=y=1, z=0$, we get
$$
2=1^{3}+1^{3}+0^{3}-3 \times 1 \times 1 \times 0 \text {. }
$$
Thus, 2 has property $P$.
If 3 has property $P$, then there exist integers $x, y, z$ such that
$$
\begin{a... | 448 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,401 |
Example 5 As shown in Figure $8, M$ and $N$ are the midpoints of the minor arcs $\overparen{A C}$ and $\overparen{A B}$ of the circumcircle of acute $\triangle A B C$, respectively. A line parallel to $M N$ is drawn through point $A$, intersecting the circumcircle at point $A^{\prime}$. $I$ is the incenter of $\triangl... | Prove as shown in Figure 8.
(1) By the properties of the incenter, we have
$$
N A=N I=N I_{1}, M A=M I=M I_{2} \text {. }
$$
Thus, $\triangle A M N \cong \triangle I M N$.
Also, $\triangle A M N \cong \triangle A^{\prime} N M$, which indicates that quadrilateral $A^{\prime} M I N$ is a parallelogram.
Therefore, $A^{\p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,402 |
2. $\frac{1-\sin ^{6} \theta-\cos ^{6} \theta}{1-\sin ^{4} \theta-\cos ^{4} \theta}=$ | 2. $\frac{3}{2}$.
Notice,
$$
\begin{array}{l}
\sin ^{4} \theta+\cos ^{4} \theta \\
=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-2 \sin ^{2} \theta \cdot \cos ^{2} \theta \\
=1-2 \sin ^{2} \theta \cdot \cos ^{2} \theta, \\
\sin ^{6} \theta+\cos ^{6} \theta \\
=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\left(... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,403 |
5. Given $P$ as any point on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{15}=1$, and $EF$ as any diameter of the circle $\odot N:(x-1)^{2}+y^{2}=4$. Then the range of $\overrightarrow{P E} \cdot \overrightarrow{P F}$ is $\qquad$ | 5. $[5,21]$.
Let $N$ be the center, and
$$
\begin{array}{l}
\overrightarrow{P E}=\overrightarrow{P N}+\overrightarrow{N E}, \\
\overrightarrow{P F}=\overrightarrow{P N}+\overrightarrow{N F}=\overrightarrow{P N}-\overrightarrow{N E},
\end{array}
$$
we know
$$
\begin{array}{l}
\overrightarrow{P E} \cdot \overrightarrow... | [5,21] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,404 |
7. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If $n$ is a positive integer, then
$$
\sum_{n=1}^{2014}\left(\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]\right)=
$$
$\qquad$ | 7. 2027091.
Let $f(n)=\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]$. Then
$$
\begin{array}{l}
f(0)=0, f(1)=0, f(2)=1, \\
f(3)=2, f(4)=3, f(5)=3 .
\end{array}
$$
For a positive integer $k$, we have
$$
\begin{array}{l}
f(6 k)=\left[\frac{6 k}{2}\right]+\left[\frac{6 k}{3}\right]+\left[\fra... | 2027091 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,406 |
8. In a certain World Cup, the Brazilian team has a probability of $\frac{1}{2}$ to win against each opponent, a probability of $\frac{1}{3}$ to draw, and a probability of $\frac{1}{6}$ to lose. Winning a match earns 3 points, drawing earns 1 point, and losing earns 0 points. It is known that each team plays three matc... | 8. $\frac{1}{12}$.
Let $\xi$ be the points Brazil scores in the group stage. Then $\xi=0,1, \cdots, 7,9$.
From the problem, we have
$$
\begin{array}{l}
P(\xi=4)=C_{3}^{1} \times \frac{1}{2} C_{2}^{1} \times \frac{1}{3} C_{1}^{1} \times \frac{1}{6}=\frac{1}{6}, \\
P(\xi=5)=C_{3}^{2}\left(\frac{1}{3}\right)^{2} \times \... | \frac{1}{12} | Other | math-word-problem | Yes | Yes | cn_contest | false | 728,407 |
II. (16 points) Find all natural numbers $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square. | Let $N$ be the square of the desired natural number.
Below, we discuss different cases.
(1) When $n \leqslant 8$,
$$
N=2^{n}\left(2^{8-n}+2^{11-n}+1\right) \text {. }
$$
Since the result inside the parentheses is odd, for $N$ to be a square number, $n$ must be even.
By verifying $n=2,4,6,8$ one by one, we find that $... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,408 |
2. Let the complex number $z=\frac{1+\mathrm{i}}{1-\mathrm{i}}$ (where $\mathrm{i}$ is the imaginary unit). Then $\mathrm{C}_{2014}^{0}+\mathrm{C}_{2014}^{1} z+\cdots+\mathrm{C}_{2014}^{2014} z^{2014}=(\quad)$.
(A) $2^{1007} \mathrm{i}$
(B) $-2^{1007} \mathrm{i}$
(C) $-2^{1005}$
(D) $-2^{1004}$ | 2. B.
From the problem, we have $z=\mathrm{i}$.
Notice that, $(1+\mathrm{i})^{2}=2 \mathrm{i}$.
Thus, the required expression is
$$
\begin{array}{l}
(1+z)^{2014}=(1+\mathrm{i})^{2014}=(2 \mathrm{i})^{1007} \\
=(2 \mathrm{i})^{2 \times 503} \times 2 \mathrm{i}=-2^{1007} \mathrm{i} .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,409 |
3. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, one of its asymptotes intersects the parabola $y=x^{2}+1$ at only one point. Then the eccentricity of the hyperbola is ( ).
(A) $\frac{5}{4}$
(B) 5
(C) $\frac{\sqrt{5}}{2}$
(D) $\sqrt{5}$ | 3. D.
From the problem, we know that the asymptotes of the hyperbola are $y= \pm \frac{b}{a} x$. By combining and simplifying with the parabola, we get
$$
a x^{2} \pm b x+a=0 \text {. }
$$
Let $\Delta=0$, solving gives $b^{2}=4 a^{2} \Rightarrow b=2 a$.
Thus, the eccentricity of the hyperbola is
$$
e=\frac{c}{a}=\fra... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,410 |
4. In $\triangle A B C$, it is known that the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, and $1+\frac{\tan A}{\tan B}=\frac{2 c}{b}$. Then the size of $\angle A$ is ( ).
(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{3}$
(D) $\frac{2 \pi}{3}$ | 4. C.
Simplifying the equation in the question and combining with the sine formula, we get
$$
\begin{array}{l}
\frac{\sin B \cdot \cos A+\sin A \cdot \cos B}{\sin B \cdot \cos A}=\frac{2 \sin C}{\sin B} \\
\Rightarrow \cos A=\frac{1}{2} \Rightarrow \angle A=60^{\circ} .
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,411 |
6. It is known that there are two people each wearing red and yellow clothes, and one person wearing blue clothes. Now, arrange these five people in a row, requiring that people wearing the same color clothes cannot be adjacent. Then the number of different arrangements is ( ).
(A) 24
(B) 28
(C) 36
(D) 48 | 6. D.
$$
\mathrm{A}_{5}^{5}-3\left(\mathrm{~A}_{2}^{2} \mathrm{~A}_{2}^{2} \cdot 3\right) \mathrm{A}_{2}^{2}=48 \text {. }
$$ | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 728,413 |
Example 7 As shown in Figure 10, let $D$ be any point on the circumcircle $\Gamma$ of $\triangle A B C$. Draw two tangents from $D$ to the incircle, intersecting side $B C$ at points $M$ and $N$. Let $T$ be the midpoint of arc $\overparen{B A C}$, and $I$ be the incenter of $\triangle A B C$. Extend $T I$ to intersect ... | Prove as shown in Figure 10, extend $D N$ and $D M$, intersecting the circle $\Gamma$ at points $E$ and $F$ respectively.
By Poncelet's closure theorem, the line $E F$ must also be tangent to the incircle of $\triangle A B C$.
Extend $E I$ and $F I$, intersecting the circle $\Gamma$ at points $V$ and $U$ respectively.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,414 |
10. Given the line $x+y-k=0(k>0)$ intersects the circle $x^{2}+y^{2}=4$ at two distinct points $A$ and $B$, and $O$ is the origin, with $|\overrightarrow{O A}+\overrightarrow{O B}| \geqslant \frac{\sqrt{3}}{3}|\overrightarrow{A B}|$. Then the range of $k$ is $(\quad)$.
(A) $(\sqrt{3},+\infty)$
(B) $[\sqrt{2},+\infty)$
... | 10. C.
Combining the equations of the line and the circle, and noting that there are two intersection points, we have $\Delta>0$. Solving this, we get $-2 \sqrt{2}<k<2 \sqrt{2}$.
As shown in Figure 7, construct rhombus $A O B C$, and let the diagonals $O C$ and $A B$ intersect at point $M$.
From the inequality in th... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,415 |
12. As shown in Figure 3, in the right trapezoid $A B C D$, it is known that $A B \perp A D, A D=D C$ $=1, A B=3$, and a moving point
$P$ is inside a circle centered at point $C$ and tangent to the line $B D$. If
$$
\overrightarrow{A P}=\alpha \overrightarrow{A D}+\beta \overrightarrow{A B}(\alpha, \beta \in \mathbf{R}... | 12. D.
To find the extremum of $\alpha+\beta$, we might as well assume that point $P$ is on $\odot C$.
It is easy to know that the radius of $\odot C$ is $\frac{1}{\sqrt{10}}$. Therefore,
$$
\begin{aligned}
\overrightarrow{A P}= & \left(1+\frac{1}{\sqrt{10}} \sin \gamma\right) \overrightarrow{A D}+ \\
& \frac{1}{3}\lef... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,417 |
13. Given a regular tetrahedron $P-ABC$, points $P, A, B, C$ are all on a sphere with radius $\sqrt{3}$. If $PA, PB, PC$ are mutually perpendicular, then the distance from the center of the sphere to the plane $ABC$ is $\qquad$. | $=13 \cdot \frac{\sqrt{3}}{3}$.
Let $P A=P B=P C=x$. Then $A B=\sqrt{2} x$.
Let the centroid of $\triangle A B C$ be $M$, and the center of the circumscribed sphere of the regular tetrahedron be $O, O M=y$. Then
$$
\left.\begin{array}{l}
A M=\frac{2 \sqrt{3}}{3} \times \frac{\sqrt{2}}{2} x=\frac{\sqrt{6}}{3} x .
\end{a... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,418 |
14. Given the functions
$$
\begin{array}{l}
f(x)=\left\{\begin{array}{ll}
-x^{2}+x, & x \leqslant 1 ; \\
\log _{\frac{1}{3}} x, & x>1,
\end{array}\right. \\
g(x)=|x-k|+|x-1| .
\end{array}
$$
If for any $x_{1} 、 x_{2} \in \mathbf{R}$, we have $f\left(x_{1}\right) \leqslant$ $g\left(x_{2}\right)$, then the range of the ... | 14. $k \leqslant \frac{3}{4}$ or $k \geqslant \frac{5}{4}$.
From the problem, we know that
for any $x_{1}, x_{2} \in \mathbf{R}, f\left(x_{1}\right) \leqslant g\left(x_{2}\right)$
$\Leftrightarrow$ for any $x_{1}, x_{2} \in \mathbf{R}, f_{\text {max }} \leqslant g_{\text {min }}$.
Notice that, when $x=\frac{1}{2}$, $f... | k \leqslant \frac{3}{4} \text{ or } k \geqslant \frac{5}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,419 |
15. Let $f_{1}(x)=\frac{2}{1+x}, f_{n+1}(x)=f_{1}\left(f_{n}(x)\right)$, and $a_{n}=\frac{f_{n}(0)-1}{f_{n}(0)+2}$. Then $a_{2014}=$ $\qquad$ . | 15. $-\left(\frac{1}{2}\right)^{2015}$.
From the problem, we know
$$
\begin{array}{c}
a_{n+1}=\frac{f_{n+1}(0)-1}{f_{n+1}(0)+2}=\frac{f_{1}\left(f_{n}(0)\right)-1}{f_{1}\left(f_{n}(0)\right)+2} \\
=\frac{\frac{2}{1+f_{n}(0)}-1}{\frac{2}{1+f_{n}(0)}+2}=\frac{1-f_{n}(0)}{4+2 f_{n}(0)}=-\frac{1}{2} a_{n} .
\end{array}
$$... | -\left(\frac{1}{2}\right)^{2015} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,420 |
16. Let the function $f(x)=k x, g(x)=\frac{\ln x}{x}$. If the equation $f(x)=g(x)$ has two real solutions for $x$ in the interval $\left[\frac{1}{\mathrm{e}}, \mathrm{e}\right]$, then the range of the real number $k$ is $\qquad$. | 16. $k \in\left[\frac{1}{\mathrm{e}^{2}}, \frac{1}{2 \mathrm{e}}\right)$.
Notice that, $g(x)=\frac{\ln x}{x}$.
Then $g^{\prime}(x)=\frac{\frac{1}{x} \cdot x-\ln x}{x^{2}}=\frac{1-\ln x}{x^{2}}$.
It is easy to see that $g(x)$ is monotonically increasing on the interval $(0, \mathrm{e})$ and monotonically decreasing on ... | \left[\frac{1}{\mathrm{e}^{2}}, \frac{1}{2 \mathrm{e}}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,421 |
17. (10 points) In $\triangle A B C$, let the sides opposite to $\angle A$, $\angle B$, and $\angle C$ be $a$, $b$, and $c$, respectively, satisfying
$a \cos C = (2 b - c) \cos A$.
(1) Find the size of $\angle A$;
(2) If $a = \sqrt{3}$, and $D$ is the midpoint of side $B C$, find the range of $A D$. | 17. (1) By the sine rule, we have
$$
\begin{array}{l}
\sin A \cdot \cos C=2 \sin B \cdot \cos A-\sin C \cdot \cos A \\
\Rightarrow \sin (A+C)=2 \sin B \cdot \cos A \\
\Rightarrow \sin B=2 \sin B \cdot \cos A .
\end{array}
$$
Since $\sin B \neq 0$, then $\cos A=\frac{1}{2}$.
Given $0<\angle A<\pi$, thus, $\angle A=\fra... | \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,422 |
19. (12 points) In a certain batch of a certain type of light bulb, 200 samples were randomly selected, and their lifespans were tracked and recorded in a frequency distribution table as shown in Table 1. Based on their lifespan, the light bulbs are classified into three grades: premium, standard, and substandard. Ligh... | 19. (1) $a=0.15, b=30$.
(2) From Table 1, we know that there are 50 premium quality light bulbs, 100 standard quality light bulbs, and 50 defective light bulbs in the sample.
Thus, the ratio of premium quality, standard quality, and defective light bulbs is $50: 100: 50=1: 2: 1$.
Therefore, according to stratified samp... | E(X)=\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,423 |
22. (12 points) Given
$$
f(x)=x \ln x-a x, g(x)=-x^{2}-2 \text {. }
$$
(1) For all $x \in(0,+\infty)$, $f(x) \geqslant g(x)$, find the range of the real number $a$;
(2) When $a=-1$, find the maximum and minimum values of the function $f(x)$ on the interval $[m, m+3](m>0)$;
(3) Prove that for all $x \in(0,+\infty)$, $\l... | 22. (1) From the problem, for all $x \in(0,+\infty)$,
$$
\begin{array}{l}
f(x) \geqslant g(x) \Leftrightarrow x \ln x-a x \geqslant-x^{2}-2 \\
\Leftrightarrow a \leqslant \ln x+x+\frac{2}{x} .
\end{array}
$$
Let $F(x)=\ln x+x+\frac{2}{x}$. Then
$$
\begin{array}{l}
F^{\prime}(x)=\frac{1}{x}+1-\frac{2}{x^{2}} \\
=\frac{... | proof | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 728,424 |
3. Given that for any $m \in\left[\frac{1}{2}, 3\right]$, we have $x^{2}+m x+4>2 m+4 x$.
Then the range of values for $x$ is | $$
\begin{array}{l}
x>2 \text { or } x<2 m+4 x \\
\Leftrightarrow m(x-2)+(x-2)^{2}>0 . \\
\text { Let } g(m)=m(x-2)+(x-2)^{2}, \text { where, }
\end{array}
$$
$$
m \in\left[\frac{1}{2}, 3\right] \text {. }
$$
The problem is converted to $g(m)$ being always greater than 0 for $m \in\left[\frac{1}{2}, 3\right]$.
$$
\tex... | x>2 \text{ or } x<-1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,426 |
4. Given the real number $a \geqslant-2$, and
$$
\begin{array}{l}
A=\{x \mid-2 \leqslant x \leqslant a\}, \\
B=\{y \mid y=2 x+3, x \in A\}, \\
C=\left\{z \mid z=x^{2}, x \in A\right\} .
\end{array}
$$
If $C \subseteq B$, then the range of values for $a$ is $\qquad$ | 4. $\left[\frac{1}{2}, 3\right]$.
From the problem, we know $B=[-1,2 a+3]$.
To make $C \subseteq B$, it is only necessary that the maximum element of set $C$ is in set $B$.
$$
\text { Hence }\left\{\begin{array}{l}
(-2)^{2} \leqslant 2 a+3, \\
a^{2} \leqslant 2 a+3
\end{array} \Rightarrow \frac{1}{2} \leqslant a \leqs... | \left[\frac{1}{2}, 3\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,427 |
6. Let $m>1$, under the constraint conditions
$$
\left\{\begin{array}{l}
y \geqslant x, \\
y \leqslant m x, \\
x+y \leqslant 1
\end{array}\right.
$$
the maximum value of the objective function $z=x+m y$ is less than 2. Then the range of values for $m$ is . $\qquad$ | 6. $(1,1+\sqrt{2})$.
The objective function is transformed into $y=-\frac{1}{m} x+\frac{z}{m}$.
Given $m>1$, we know $-1<-\frac{1}{m}<0$.
The feasible region represented by the system of inequalities is the shaded area in Figure 2.
According to the geometric meaning of the objective function, the maximum value of the... | (1,1+\sqrt{2}) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,428 |
7. The axial section $S A B$ of the circular cone is an equilateral triangle with a side length of 2, $O$ is the center of the base, $M$ is the midpoint of $S O$, and the moving point $P$ is within the base of the cone (including the circumference). If $A M \perp M P$, then the length of the trajectory formed by point ... | 7. $\frac{\sqrt{7}}{2}$.
Establish a rectangular coordinate system in space.
Let $A(0,-1,0), B(0,1,0)$,
$S(0,0, \sqrt{3}), M\left(0,0, \frac{\sqrt{3}}{2}\right), P(x, y, 0)$.
Thus, $\overrightarrow{A M}=\left(0,1, \frac{\sqrt{3}}{2}\right), \overrightarrow{M P}=\left(x, y,-\frac{\sqrt{3}}{2}\right)$.
Since $A M \perp M... | \frac{\sqrt{7}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,429 |
8. In the right trapezoid $A B C D$, it is known that $A D \perp A B$, $A B / / D C, A B=4, A D=D C=2$. Let $N$ be the midpoint of side $D C$, and $M$ be a moving point within or on the boundary of trapezoid $A B C D$. Then the maximum value of $\overrightarrow{A M} \cdot \overrightarrow{A N}$ is | 8. 6 . | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,430 |
10. Given that the asymptotes of a hyperbola with foci on the $x$-axis pass through the intersection points of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{16}=1$ and $\frac{a x^{2}}{16}+\frac{y^{2}}{4}=1(0<a \leqslant 1)$. Then the range of the eccentricity of the hyperbola is $\qquad$ | 10. $\left[\sqrt{2}, \frac{\sqrt{21}}{3}\right)$.
By solving the system of two ellipse equations, we get $y^{2}=\frac{4-a}{3} x^{2}$. Therefore, $e^{2}=1+\frac{4-a}{3}=\frac{7-a}{3} \in\left[2, \frac{7}{3}\right)$. Hence, $e \in\left[\sqrt{2}, \frac{\sqrt{21}}{3}\right)$. | \left[\sqrt{2}, \frac{\sqrt{21}}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,431 |
12. (14 points) In $\triangle A B C$, let the sides opposite to $\angle A$, $\angle B$, and $\angle C$ be $a$, $b$, and $c$ respectively, satisfying
$\sqrt{3} \tan A \cdot \tan B - \tan A - \tan B = \sqrt{3}$.
(1) Find the size of $\angle C$;
(2) If $c=2$, and $\triangle A B C$ is an acute triangle, find the range of $... | 12. (1) From the given equation, we have
$$
\begin{array}{l}
\frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} = -\sqrt{3} \\
\Rightarrow \tan (A + B) = -\sqrt{3} \\
\Rightarrow \tan C = \sqrt{3} \Rightarrow \angle C = \frac{\pi}{3}.
\end{array}
$$
(2) From (1), we know $\frac{\pi}{6} < \angle A < \frac{\pi}{2}$.
By the ... | \frac{20}{3} < a^{2} + b^{2} \leqslant 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,432 |
13. (16 points) Given the hyperbola \( C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \) \((a>0, b>0)\) with an eccentricity of 2, a line \( l \) passing through the point \( P(0, m) \) \((m>0)\) with a slope of 1 intersects the hyperbola \( C \) at points \( A \) and \( B \), and
\[
\overrightarrow{A P}=3 \overrightarro... | 13. (1) From the eccentricity of the hyperbola being 2, we know $c=2a, b=\sqrt{3}a$.
Thus, the equation of the hyperbola can be written as $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{3a^{2}}=1$.
Also, the line $l: y=x+m$, when combined with the hyperbola equation, gives
$$
2x^{2}-2mx-m^{2}-3a^{2}=0.
$$
Let points $A(x_{1}, y_{1... | x^{2}-\frac{y^{2}}{3}=1, M(-1,0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,433 |
2. Given the equation in terms of $x$
$$
x^{2}-4 x \log _{2} a+3\left(\log _{2} a\right)^{2}=0
$$
the two roots belong to the intervals $(1,3)$ and $(6,+\infty)$, respectively. Then the range of the real number $a$ is $\qquad$ . | 2. $a \in(4,8)$.
The original equation is
$$
\begin{array}{l}
\left(x-\log _{2} a\right)\left(x-3 \log _{2} a\right)=0 \\
\Rightarrow x_{1}=\log _{2} a, x_{2}=3 \log _{2} a=3 x_{1} .
\end{array}
$$
Since both roots are positive, it must be that
$$
\begin{array}{l}
x_{1}=\log _{2} a \in(1,3), \\
x_{2}=3 \log _{2} a \i... | a \in(4,8) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,434 |
5. Let $A$ be a set of $m$ real numbers, with the sum of its elements being $a$; $B$ be a set of $n$ real numbers, with the sum of its elements being $b$. Define
$$
C=\{\boldsymbol{x} \mid \boldsymbol{x}=(s, t), s \in A, t \in B\} \text {. }
$$
Let $\boldsymbol{x} y$ each take all vectors in $C$ (allowing $\boldsymbol... | 5. $n^{2} a^{2}+m^{2} b^{2}$.
From the coordinate form of vectors, we get
$$
\begin{array}{l}
\sum_{\boldsymbol{x} \in C} \boldsymbol{x}=\sum_{s \in A} \sum_{t \in B}(s, t)=\sum_{s \in A}\left(\sum_{t \in B} s, \sum_{t \in B} t\right) \\
=\sum_{s \in A}(n s, b)=\left(n \sum_{s \in A} s, \sum_{s \in A} b\right) \\
=(n ... | n^{2} a^{2}+m^{2} b^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,435 |
6. Given that the hyperbola has asymptotes $2 x \pm y=0$, and it passes through the intersection point of the lines $x+y-3=0$ and $2 x-y+3 t=0$, where $-2 \leqslant t \leqslant 5$. Then the maximum possible value of the real axis length of the hyperbola is $\qquad$ . | 6. $4 \sqrt{3}$.
It is easy to know that the intersection point of the two lines is $(1-t, 2+t)$.
First, consider the case where the foci of the hyperbola are on the $x$-axis.
$$
\text { Let } \frac{x^{2}}{a^{2}}-\frac{y^{2}}{4 a^{2}}=1(a>0) \text {. }
$$
Notice that, $-2 \leqslant t \leqslant 5$. Then
$$
\begin{arra... | 4 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,436 |
7. As shown in Figure 1, under the rules of Chinese chess, the "pawn" at point $A$ can reach point $B$ through a certain path (before crossing the river, the pawn can only move to the adjacent intersection directly in front of it each step; after crossing the river, it can move to the adjacent intersections in front, t... | 7. 6561.
Assume the chessboard has 10 horizontal lines from bottom to top, sequentially labeled as the 1st, 2nd, ..., 10th rows, and 9 vertical lines from left to right, sequentially labeled as the 1st, 2nd, ..., 9th columns. For example, point $A$ is located at the 4th row and 5th column.
Note that, during the movem... | 6561 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,437 |
8. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, and
$$
a=5, b=4, \cos (A-B)=\frac{31}{32} \text {. }
$$
Then the area of $\triangle A B C$ is . $\qquad$ | 8. $\frac{15 \sqrt{7}}{4}$.
By the Law of Sines, we have
$$
\begin{array}{l}
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{a+b}{\sin A+\sin B}=\frac{a-b}{\sin A-\sin B} \\
\Rightarrow 9(\sin A-\sin B)=\sin A+\sin B \\
\Rightarrow 18 \sin \frac{A-B}{2} \cdot \cos \frac{A+B}{2}=2 \sin \frac{A+B}{2} \cdot \cos \frac{A-B}{2} \\... | \frac{15 \sqrt{7}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,438 |
9. (16 points) Given a tetrahedron $P-ABC$ with each edge of the base being 1 and the height being $\sqrt{2}$, the center of its inscribed sphere is $O$, and the radius is $r$. Find the area of the planar region formed by points within the base $ABC$ that are no more than $2r$ away from point $O$. | 9. First, find the radius $r$ of the inscribed sphere.
As shown in Figure 3, let the projections of the sphere center $O$ on the plane $ABC$ and the plane $ABP$ be $H$ and $K$, respectively, and let the midpoint of $AB$ be $M$. Then, $P$, $K$, $M$ and $P$, $O$, $H$ are collinear, respectively. Thus,
$$
\begin{array}{l}... | \frac{1}{4} + \frac{\pi}{24} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,439 |
11. (20 points) Find all cubic polynomials $P(x)$ such that for all $x, y \geqslant 0$, we have
$$
P(x+y) \geqslant P(x)+P(y) .
$$ | 11. Let $P(x)=a x^{3}+b x^{2}+c x+d(a \neq 0)$. Then the original inequality is equivalent to
$$
3 a x y(x+y)+2 b x y \geqslant d(\text { for any } x, y \geqslant 0) \text {. }
$$
Let $x, y$ be sufficiently large, we get $a>0$.
Let $x=y=0$, we get $d \leqslant 0$.
Under these conditions, equation (1) can be rewritten a... | 8 b^{3} \geqslant 243 a^{2} d | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,440 |
Example 1 As shown in Figure 1, in $\square A B C D$, it is known that $A E \perp B C$, $A F \perp C D$, and $H$ is the orthocenter of $\triangle A E F$. Prove:
$$
A C^{2}=A H^{2}+E F^{2} .[2]
$$
(2012, Sichuan Province Junior High School Mathematics League Final (Grade 8)) | Prove as follows:
Connect $H E$ and
$H F$.
From the problem, we know
$$
\begin{array}{l}
H E \perp A F, \\
H F \perp A E .
\end{array}
$$
Also, $A E \perp B C, A F \perp C D$, so
$H E \parallel F C, H F \parallel C E$.
Thus, quadrilateral $C E H F$ is a parallelogram.
Construct $E G \parallel A H$, intersecting $A B$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,441 |
Example 2 As shown in Figure 3, it is known that $A A^{\prime} 、 B B^{\prime} 、 C C^{\prime}$ intersect at point $O$, and
$$
\begin{array}{l}
A A^{\prime}=B B^{\prime}=C C^{\prime}=1, \\
\angle A O C^{\prime}=\angle B O A^{\prime} \\
=\angle C O B^{\prime}=60^{\circ} .
\end{array}
$$
Prove: $S_{\triangle A O C^{\prime... | Prove as shown in Figure 4, translate $\triangle A O C^{\prime}$ 1 unit to the right to $\triangle E C D$, extend $O B^{\prime}$ and $D E$, intersecting at point $F$.
From $\angle F D O=\angle A O C^{\prime}=\angle C O B^{\prime}=60^{\circ}$, we know that $\triangle O D F$ is an equilateral triangle, and $O D=1$.
From... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,442 |
3. In a convex pentagon, it is known that
$$
\begin{array}{l}
A B=B C=C D=D E=E A, \\
\angle A B C=2 \angle D B E .
\end{array}
$$
Prove: $\angle A B C=60^{\circ}$.
$(2010$, Beijing Middle School Mathematics Competition (Grade 8)) | Proof 1 As shown in Figure 13, rotate $\triangle B C D$ counterclockwise around point $B$ to $\triangle B A D^{\prime}$, and prove that $\triangle A D E^{\prime}$ is an equilateral triangle.
Proof 2 Construct $\angle F B E$
$$
=\angle A B E \text {, intersecting } D E \text { at }
$$
point $F$, prove that $A E / / B F... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,443 |
Let $[x]$ denote the greatest integer not exceeding the real number $x$. Suppose $m, n \in \mathbf{Z}_{+}, m < n$. Prove:
$$
\begin{array}{l}
\mathrm{C}_{n}^{j}+\mathrm{C}_{n}^{m+j}+\mathrm{C}_{n}^{2 m+j}+\cdots \\
=\frac{2^{n}}{m}\left[1+2 \sum_{k=1}^{\left[\frac{m-1}{2}\right]} \cos \frac{k \pi}{m} \cdot \cos \frac{(... | Let $\omega=\cos \frac{2 \pi}{m}+\mathrm{i} \sin \frac{2 \pi}{m}$.
$$
\begin{array}{l}
\text { Then } \omega^{k}=\cos \frac{2 k \pi}{m}+\mathrm{i} \sin \frac{2 k \pi}{m} \\
\Rightarrow 1+\omega^{k}=1+\cos \frac{2 k \pi}{m}+\mathrm{i} \sin \frac{2 k \pi}{m} \\
=2 \cos \frac{k \pi}{m}\left(\cos \frac{k \pi}{m}+\mathrm{i}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,444 |
5. Let $P$ be a point inside $\triangle ABC$, $\angle PBA=10^{\circ}$, $\angle BAP=20^{\circ}$, $\angle PCB=30^{\circ}$, $\angle CBP=40^{\circ}$. Prove that $\triangle ABC$ is an isosceles triangle.
(25th United States of America Mathematical Olympiad) | As shown in the figure,
15, considering $\angle P C B$
$=30^{\circ}$, construct the reflection point
$B^{\prime}$ of point $B$ about $P C$,
resulting in $\triangle B C B^{\prime}$ being
an equilateral triangle.
Thus, $\angle P B B^{\prime}$
$=20^{\circ}$ and $P B=P B^{\prime}$, we get
$\angle P B^{\prime} B=\angle P A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,446 |
Example 2: Two people, A and B, play the following game: They take turns writing a natural number on the blackboard, with the requirement that the newly written number cannot be expressed as a non-negative integer linear combination of the numbers already on the blackboard. That is, after writing numbers \(a_{1}, a_{2}... | First, we prove a lemma.
Lemma: Let $a, b$ be coprime positive integers. Then $ab - a - b$ is the largest integer that cannot be expressed as $ax + by$ where $x, y \in \mathbb{N}$.
Proof: By the coprimality of $a, b$ and Bézout's identity, for any integer $n$, there exist integers $x, y$ such that $n = ax + by$.
$$
\b... | A | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,447 |
Example 4 Color the elements of the set $M=\{1,2, \cdots, n\}$ with three colors, each element being colored with one color. Let
$$
A=\{(x, y, z) \mid x, y, z \in M, x+y+z \equiv
$$
$0(\bmod n), x, y, z$ have the same color $\}$,
$$
B=\{(x, y, z) \mid x, y, z \in M, x+y+z \equiv
$$
$0(\bmod n), x, y, z$ have different ... | Let the sets of numbers colored with three different colors in set $M$ be $R$, $S$, and $T$ respectively $\left(\sum|R|=n\right)$.
$$
\begin{array}{l}
\text { Let } f(x)=\left(\sum_{r \in R} x^{r}\right)\left(\sum_{s \in S} x^{s}\right)\left(\sum_{t \in T} x^{t}\right), \\
g(x)=\left(\sum_{r \in R} x^{t}\right)^{3}+\le... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,448 |
Example 5 Let $p$ be a prime, $a, k \in \mathbf{Z}_{+}$, and $p^{a}<k<2 p^{a}$. Prove: There exists a positive integer $n<p^{2 a}$, satisfying
$$
\mathrm{C}_{n}^{k} \equiv n \equiv k\left(\bmod p^{a}\right) .
$$ | To prove that $n \equiv k\left(\bmod p^{a}\right)$, let $n=k+t p^{a}$.
From $1 \leqslant n=k+t p^{a}<p^{2 a}$
$\Rightarrow-\frac{k}{p^{a}} \leqslant t<p^{a}-\frac{k}{p^{a}}$.
Given $p^{a}<k<2 p^{a} \Rightarrow 1<\frac{k}{p^{a}}<2$
$\Rightarrow$ the integer $t$ satisfies $-1 \leqslant t \leqslant p^{a}-2$.
Therefore, to... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,449 |
Example 6 For an integer $k>1$, let $f(k)$ be the number of ways to decompose $k$ into a product of positive integers greater than 1 (for example, $12=12=2 \times 6=3 \times 4=2 \times 2 \times 3$, so $f(12)=$ $4$). Prove: If $n$ is an integer greater than 1, and $p$ is any prime factor of $n$, then $f(n) \leqslant \fr... | First, prove: $f(n) \leqslant \sum_{d \mid \frac{n}{p}} f(d)$.
Let $n > 1$ be an integer with any factorization $n = n_1 n_2 \cdots n_k$. Since $p \mid n$, it follows that $p \mid n_i (1 \leqslant i \leqslant k)$, and without loss of generality, assume $p \mid n_1$.
This factorization corresponds to $d = \frac{n}{n_1} ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,450 |
Example 3 Given that $P$ is a point inside an acute $\triangle A B C$ and minimizes $P A+P B+P C$. Determine the position of point $P$ and prove your conclusion. ${ }^{[3]}$
(2009, National Junior High School Mathematics Competition, Tianjin Preliminary) | Solve: Construct equilateral triangles $\triangle A B A^{\prime}$ and $\triangle A C A^{\prime \prime}$ outwardly on sides $AB$ and $AC$, respectively. Then the intersection of $A^{\prime} C$ and $A^{\prime \prime} B$ is point $P$.
In fact, as shown in Figure 5, rotate $\triangle A P B$ counterclockwise by $60^{\circ}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,451 |
1. Prove: For any integer $n>16$, there exists a set $S$ of $n$ positive integers, with the property: if a subset $A$ of $S$ satisfies that for any $a, a^{\prime} \in A\left(a \neq a^{\prime}\right)$, we have $a+a^{\prime} \notin S$, then $|A| \leqslant 2\lfloor\sqrt{3 n}\rfloor$. | Let $k \in \mathbf{Z}(1<k<n)$,
$$
\begin{array}{l}
n=k q+r(0 \leqslant r<k) . \\
\text { For } i=1,2, \cdots, k, \text { let } \\
S_{i}=\left\{2^{i-1} m \mid q \leqslant m \leqslant 2 q-1\right\} .
\end{array}
$$
Then $\left|S_{i}\right|=q$, and for any $1 \leqslant i<j \leqslant k$, we have
$$
S_{i} \cap S_{j}=\varno... | 2\lfloor\sqrt{3 n}\rfloor | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,452 |
2. Let $k>3$ be a given odd number. Prove: there exist infinitely many positive odd numbers $n$, such that there are two positive integers $d_{1} 、 d_{2}$, satisfying $d_{1} 、 d_{2}$ both divide $\frac{n^{2}+1}{2}$, and $d_{1}+d_{2}=n+k$. | $$
\begin{array}{l}
2 d_{1} d_{2} \mid\left(n^{2}+1\right) \\
\Rightarrow 2 d_{1} d_{2} \mid\left[\left(d_{1}+d_{2}-k\right)^{2}+1\right] \\
\Rightarrow\left(d_{1}+d_{2}-k\right)^{2}+1=2 d \cdot d_{1} d_{2} .
\end{array}
$$
Let $d_{1}=d_{2}=1$, then $2 d=(k-2)^{2}+1$.
It remains to prove that the equation
$$
\left[(k-... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,453 |
3. Prove: There exists an unbounded sequence of positive integers $\left\{a_{n}\right\}\left(a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \leqslant \cdots\right)$, such that there exists a positive integer $M$, for all positive integers $n \geqslant M$, if $n+1$ is not a prime, then all prime factors of $n!+1... | Let $p$ be a prime, $p \mid(n!+1)$ and $n+1$ is not a prime. Then $p>n+1$.
Let $p-n-1=t\left(t \in \mathbf{Z}_{+}\right)$.
Then by Wilson's theorem we have
$(p-1)!\equiv-1(\bmod p)$.
Thus, $(p-t-1)!\cdot t!\equiv(-1)^{t+1}(\bmod p)$.
Given $p-t-1=n$ and $p \mid (n!+1)$, we know $p \mid\left[t!+(-1)^{t+1}\right]$.
Hence... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,454 |
5. Let $a>b>1, b$ be an odd number, and $n$ be a positive integer. If $b^{n} \mid\left(a^{n}-1\right)$, prove: $a^{b}>\frac{3^{n}}{n}$. | Lemma: Let $p$ be an odd prime, $a \in \mathbf{Z}, (p, a)=1$, $\delta_{p^{m}}(a)=d_{m}, p^{\lambda} \|\left(a^{d_{1}}-1\right)$. Then
$$
d_{m}=\left\{\begin{array}{ll}
d_{1}, & m \leqslant \lambda ; \\
d_{1} p^{m-\lambda}, & m \geqslant \lambda .
\end{array} \quad a^{b}>p^{\lambda} \geqslant 3^{n}>\frac{3^{n}}{n},\righ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,455 |
Example 2 Given six points in a plane, any three of which form the vertices of a scalene triangle. Prove: Among these triangles, there must be a triangle whose shortest side is also the longest side of another triangle.
保留了源文本的换行和格式。 | Proof: Let the six given points be $A_{1}, A_{2}, \cdots, A_{6}$. For each triangle formed by these points, color the shortest side red.
Thus, each triangle must have one red side.
When the shortest side of all triangles is colored red, the sides that are not colored can also be considered to have a color (such as whit... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,456 |
Example 3 Consider the sky as a plane, on which there are $n$ stars. At point $P$ in the sky, there is a sounding observation instrument, and $k$ radio telescopes are fixedly installed on it, as shown in Figure 1.
Assume that the angle that the $j$-th telescope can illuminate is $\alpha_{j}$, and $\sum_{j=1}^{k} \alph... | Prove that, as shown in Figure 2, with a fixed point $P$ as the center, draw a circle $\odot P$. For $n$ stars $M_{1}, M_{2}, \cdots, M_{n}$, let the line segment $P M_{i} (i=1,2, \cdots, n)$ intersect $\odot P$ at point $A_{i}$. Then $A_{1}, A_{2}, \cdots, A_{n}$ are $n$ fixed points on the circle $\odot P$.
Consider ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,457 |
Example 4 Color the space in five colors in any way (each point belongs to one color, and each color has points).
(1) Prove: There exists a plane that contains at least four different colored points;
(2) Does there necessarily exist a plane with all five colors?
This is originally a classification problem of spatial po... | (1) Proof If there exists a four-color line $l$, then the plane containing $l$ is the one sought; if there exists a three-color line $l$, then a fourth-color point $M$ can be taken outside the line $l$, and the plane through $M$ and the line $l$ is the one sought.
Suppose that no line contains more than two colors. Fo... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,458 |
Example 6 If the three vertices of any inscribed equilateral triangle in a circle are all colored with different colors, then such a colored circle is called a "good circle". Let the circle with center $P$ and radius $r$ be denoted as $\odot(P, r)$.
(1) Is it possible to color each point in $\odot(O, 2)$ and its interi... | (1) There does not exist a coloring method that satisfies the conditions.
Proof by contradiction.
Assume there exists a coloring method that satisfies the conditions, and let the three colors be $s_{0}, s_{1}, s_{2}$.
As shown in Figure 4, take any point $M$ on $\odot(O, 2)$, and construct a circle with $OM$ as the dia... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,459 |
Example 4 Given that $P$ and $Q$ are two points inside the square $ABCD$, and $\angle PAQ = \angle PCQ = 45^{\circ}$. Prove:
$$
S_{\triangle PAB} + S_{\triangle PCQ} + S_{\triangle QAD} = S_{\triangle QCD} + S_{\triangle PAQ} + S_{\triangle PBC} \text{. }
$$
(Adapted from the 2011 "Mathematics Weekly Cup" National Juni... | Proof As shown in Figure 6, rotate $\triangle A Q D$ 90° clockwise around point $A$ to $\triangle A Q^{\prime} B$, and rotate $\triangle C Q D$ 90° counterclockwise around point $C$ to $\triangle C Q^{\prime \prime} B$. Connect $P Q^{\prime}$ and $P Q^{\prime \prime}$. It is easy to see that
$$
\begin{array}{l}
\triang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,460 |
Example 1 On the Cartesian plane, given the parabola $y=1-x^{2}$ and the line $y=x+a(-1<a<1)$ intersect at points $A$ and $B$, and point $C(1,0)$. Question: For what value of $a$ is the area of $\triangle ABC$ maximized? Find the maximum area of $\triangle ABC$. ${ }^{[1]}$
$(2010$, Shanghai $\mathrm{TI}$ Cup High Scho... | Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
$$
\begin{array}{l}
\text { From }\left\{\begin{array}{l}
y=1-x^{2}, \\
y=x+a
\end{array}\right. \\
\Rightarrow x^{2}+x+a-1=0 \\
\Rightarrow x_{1}+x_{2}=-1, x_{1} x_{2}=a-1 \\
\Rightarrow|A B|^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=2\l... | \frac{3 \sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,461 |
Example 2 Given a positive number $p$ and a parabola $C: y^{2}=2 p x$ $(p>0)$, $A\left(\frac{p}{6}, 0\right)$ is a point on the axis of symmetry of the parabola $C$, and $O$ is the vertex of the parabola $C$. $M$ is any point on the parabola $C$. Find the maximum value of $\frac{|O M|}{|A M|}$. ${ }^{[2]}$
(2013, Xin Z... | Let $M(x, y)$. Then
$$
\begin{array}{l}
x \geqslant 0, \text { and } y^{2}=2 p x(p>0). \\
\text { Hence }\left(\frac{|O M|}{|A M|}\right)^{2}=\frac{x^{2}+y^{2}}{\left(x-\frac{p}{6}\right)^{2}+y^{2}} \\
=\frac{x^{2}+2 p x}{x^{2}-\frac{p x}{3}+\frac{p^{2}}{36}+2 p x}=\frac{x^{2}+2 p x}{x^{2}+\frac{5 p x}{3}+\frac{p^{2}}{... | \frac{3 \sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,462 |
Example 3 Given the parabola $C: y^{2}=2 p x(p>0)$ intersects with line $l$ at points $A$ and $B$, the line connecting point $A$ and the vertex $O$ of the parabola intersects the directrix at point $B^{\prime}$, and the line connecting point $B$ and the vertex $O$ of the parabola intersects the directrix at point $A^{\... | (1) Proof: Let $l_{A B}: x=c+m y$.
Then point $A\left(c+m y_{1}, y_{1}\right), B\left(c+m y_{2}, y_{2}\right)$.
Since $A A^{\prime} / / B B^{\prime} / / x$ axis, we have,
$$
A^{\prime}\left(-\frac{p}{2}, y_{1}\right), B^{\prime}\left(-\frac{p}{2}, y_{2}\right) \text {. }
$$
From the collinearity of points $A 、 O 、 B^... | 2 p^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 728,463 |
Example 4 Given that quadrilateral $ABCD$ is an inscribed parallelogram in the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
and is circumscribed about a circle centered at the origin. Find
(1) the maximum and minimum values of $AB$;
(2) the maximum and minimum values of the area of quadrilateral $ABC... | Given that quadrilateral $ABCD$ is a rhombus, $OA \perp OB$.
Without loss of generality, let point $A\left(r_{1} \cos \theta, r_{1} \sin \theta\right)$,
$B\left(r_{2} \cos \left(\theta+\frac{\pi}{2}\right), r_{2} \sin \left(\theta+\frac{\pi}{2}\right)\right)$.
Then $B\left(-r_{2} \sin \theta, r_{2} \cos \theta\right), ... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,464 |
Example 5 Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and an inscribed parallelogram with one pair of opposite sides passing through the foci $F_{1}$ and $F_{2}$ of the ellipse. Find the maximum area of the parallelogram. ${ }^{[4]}$
(2013, National High School Mathematics League Shandong Province Preliminary... | Solve the general ellipse:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
Corresponding problems.
With $F_{1}$ as the pole and $F_{1} x$ as the polar axis, the equation of the ellipse is transformed into
$$
\rho=\frac{e p}{1+e \cos \theta}=\frac{b^{2}}{a-c \cos \theta} .
$$
Therefore, the length of the chord ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,465 |
Example 6: Through the right focus $F$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, two perpendicular chords $A B$ and $C D$ are drawn. Let the midpoints of $A B$ and $C D$ be $M$ and $N$ respectively.
(1) Prove that the line $M N$ must pass through a fixed point, and find this fixed point;
(2) If... | $$
\begin{array}{l}
\text { Then } l_{M N}: y-\frac{b^{2} c k}{b^{2} k^{2}+a^{2}} \\
=-\frac{\left(a^{2}+b^{2}\right) k}{\left(k^{2}-1\right) a^{2}}\left(x-\frac{a^{2} c}{b^{2} k^{2}+a^{2}}\right) \\
\Rightarrow y=-\frac{\left(a^{2}+b^{2}\right) k}{\left(k^{2}-1\right) a^{2}}\left(x-\frac{a^{2} c}{a^{2}+b^{2}}\right) \... | \frac{b^{3}}{4 a} | Geometry | proof | Yes | Yes | cn_contest | false | 728,466 |
Example 7 Given that $A$ and $B$ are two moving points on the parabola $C: y^{2}=4 x$, point $A$ is in the first quadrant, and point $B$ is in the fourth quadrant. Lines $l_{1}$ and $l_{2}$ pass through points $A$ and $B$ respectively and are tangent to the parabola $C$, and $P$ is the intersection point of $l_{1}$ and... | (1) Proof: Let
$$
A\left(\frac{y_{1}^{2}}{4}, y_{1}\right), B\left(\frac{y_{2}^{2}}{4}, y_{2}\right)\left(y_{1}>0>y_{2}\right) \text {. }
$$
Then $l_{1}: y_{1} y=2\left(x+\frac{y_{1}^{2}}{4}\right)$,
$$
l_{2}: y_{2} y=2\left(x+\frac{y_{2}^{2}}{4}\right) \text {. }
$$
Let $P\left(x_{0}, y_{0}\right)$. Then
$$
y_{1} y_... | \frac{128 \sqrt{3}}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,467 |
2. If $x^{2}+y^{2}=1$, then
$$
\sqrt{x^{2}-2 x+1}+\sqrt{4 y^{2}+4 y+1}+\sqrt{x y-2 x+y-2}
$$
is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 2. D.
From the conditions, we know that $-1 \leqslant x \leqslant 1, -1 \leqslant y \leqslant 1$, and
$$
\begin{array}{l}
x y-2 x+y-2 \geqslant 0 \\
\Rightarrow(x+1)(y-2) \geqslant 0 .
\end{array}
$$
Since $y-2<0, x+1 \geqslant 0$, therefore,
$$
x+1=0 \Rightarrow x=-1 \text {. }
$$
Thus, $y=0$.
Therefore, the origin... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,468 |
3. Given $\triangle A B C$ with the three interior angles $\angle A, \angle B, \angle C$ and the corresponding side lengths $a, b, c$, where $a, b, c$ are rational numbers, and they satisfy
$$
(a+\sqrt{2})^{2}=(b+\sqrt{2})(c+\sqrt{2}) \text {. }
$$
Then $\angle A=$ ( ).
(A) $60^{\circ}$
(B) $75^{\circ}$
(C) $90^{\circ... | 3. A.
From the given conditions,
$$
\begin{array}{l}
a^{2}+2+2 \sqrt{2} a=(b c+2)+(b+c) \sqrt{2} \\
\Rightarrow\left\{\begin{array}{l}
a^{2}+2=b c+2, \\
2 a=b+c
\end{array}\right. \\
\Rightarrow(b+c)^{2}=4 a^{2}=4 b c \\
\Rightarrow(b-c)^{2}=0 \Rightarrow b=c .
\end{array}
$$
Thus, $a=b=c$. Therefore, $\angle A=60^{\... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,469 |
Example 5 Given $A B=A C, \angle B A C=\angle C D E=$ $90^{\circ}, D C=D E, F$ is the midpoint of $B E$. Prove:
$$
F A=F D \text {, and } F A \perp F D .{ }^{[4]}
$$
(2013, Sichuan Province Junior High School Mathematics League Final (Grade 8)) | Proof As shown in Figure 7, extend $AF$ to point $G$, such that $FG = AF$. Connect $DG$ and $EG$.
From $BF = EF, \angle AFB = \angle GFE, FA = FG$, we get $\triangle AFB \cong \triangle GFE$.
Thus, $AB = GE, \angle B = \angle FEG$.
Notice,
$$
\begin{array}{l}
\angle B + \angle BAC + \angle CAD + \angle CDA + \\
\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,470 |
6. Let $[x]$ denote the greatest integer not exceeding the real number $x$ (for example, $[3.14]=3, [0]=0, [-3.14]=-4$). If
a positive integer $x$ satisfies $x-[\sqrt{x}]^{2}=9$, then $x$ is called a "good number". Then, among the 2014 positive integers from $1$ to $2014$, the number of good numbers is ( ).
(A) 36
(B)... | 6. B.
Let $[\sqrt{x}]=n$. Then $x=n^{2}+9$.
Thus $n \leqslant \sqrt{n^{2}+9}<n+1 \Rightarrow n \geqslant 5$.
Also, $n^{2}+9 \leqslant 2014 \Rightarrow n \leqslant 44$.
Therefore, $5 \leqslant n \leqslant 44$.
Hence, there are 40 values of $n$ that satisfy the condition, which means there are 40 good numbers. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 728,471 |
2. Given $n+2$ real numbers
$$
a_{1}, a_{2}, \cdots, a_{n}, 16, a_{n+2} \text {, }
$$
where the average of the first $n$ numbers is 8, the average of the first $n+1$ numbers is 9, and the average of these $n+2$ numbers is 10. Then the value of $a_{n+2}$ is $\qquad$ | 2. 18 .
From the condition, $\frac{8 n+16}{n+1}=9 \Rightarrow n=7$.
Also, $\frac{8 \times 7+16+a_{n+2}}{7+1+1}=10 \Rightarrow a_{n+2}=18$. | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,472 |
3. (25 points) In the Cartesian coordinate system $x O y$, it is known that points $A\left(-1, y_{0}\right)$ and $B(2,0)$ are both on the parabola $y=a x^{2}+b x(a>0)$, and the area of $\triangle A O B$ is 1.
(1) Find the equation of the parabola;
(2) The equation of line $l$ is $y=x-\frac{25}{4}$, and $P$ is any point... | 3. (1) From the condition, we know $y_{0}>0$, and
$$
\frac{1}{2} \times 2 y_{0}=1 \Rightarrow y_{0}=1 \text {. }
$$
Therefore, the equation of the required parabola is
$$
y=\frac{1}{3} x^{2}-\frac{2}{3} x \text {. }
$$
(2) As shown in Figure 4, let the line $l_{1} / / l$, and suppose it is tangent to the parabola at p... | \left(\frac{5}{2}, \frac{5}{12}\right), \frac{25 \sqrt{2}}{12} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,473 |
Example 6 In $\triangle A B C$, it is known that $\angle A B C=$ $\angle B A C=70^{\circ}, P$ is a point inside the triangle, $\angle P A B=40^{\circ}$, $\angle P B A=20^{\circ}$. Prove: $P A+P B=P C .^{[5]}$
(2012, Beijing Middle School Mathematics Competition (Grade 8)) | Prove as shown in Figure 8, construct the reflection point \(P'\) of point \(P\) about \(AC\), and connect \(P'A\), \(P'P\), and \(P'C\).
Then \(\angle PAC = 30^\circ\).
Thus, \(\angle PAP' = 60^\circ\).
Since \(PA = P'A\), \(\triangle PAP'\) is an equilateral triangle.
Therefore, \(PA = PP'\),
\(\angle APP' = 60^\circ... | PA + PB = PC | Geometry | proof | Yes | Yes | cn_contest | false | 728,474 |
2. Function
$$
y=\sin ^{2} x+\sin x \cdot \cos x-2 \cos ^{2} x
$$
The range of the function is . $\qquad$ | 2. $\left[-\frac{\sqrt{10}+1}{2}, \frac{\sqrt{10}-1}{2}\right]$.
From the problem, we have
$$
\begin{aligned}
y & =\frac{1-\cos 2 x}{2}+\frac{1}{2} \sin 2 x-(\cos 2 x+1) \\
& =\frac{1}{2}(\sin 2 x-3 \cos 2 x)-\frac{1}{2} \\
& =\frac{\sqrt{10}}{2} \sin (2 x-\varphi)-\frac{1}{2},
\end{aligned}
$$
where $\varphi \in\lef... | \left[-\frac{\sqrt{10}+1}{2}, \frac{\sqrt{10}-1}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,475 |
3. In $\triangle A B C$, it is known that
$$
\angle A=30^{\circ}, 2 \overrightarrow{A B} \cdot \overrightarrow{A C}=3 \overrightarrow{B C}^{2} \text {. }
$$
Then the cosine of the largest angle of $\triangle A B C$ is | 3. $-\frac{1}{2}$.
Let $A B=c, A C=b, B C=a$.
From $2 \overrightarrow{A B} \cdot \overrightarrow{A C}=3 \overrightarrow{B C}^{2} \Rightarrow 2 c b \cos A=3 a^{2}$.
Since $\angle A=30^{\circ}$, we have $b c=\sqrt{3} a^{2}$.
Combining with the sine rule, we get
$$
\begin{array}{l}
\sin B \cdot \sin C=\sqrt{3} \sin ^{2} ... | -\frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,476 |
5. If $\sqrt{3-a}-\sqrt{a+1}>\frac{1}{2}$ always holds, then the range of values for $a$ is . $\qquad$ | 5. $\left[-1,1-\frac{\sqrt{31}}{8}\right)$.
From the problem, we know that $3-a \geqslant 0$, $a+1 \geqslant 0$, and
$$
\sqrt{3-a}-\sqrt{a+1}>0 \text {. }
$$
Thus, $-1 \leqslant a1+\frac{\sqrt{31}}{8}$ or $a<1-\frac{\sqrt{31}}{8}$.
In summary, $-1 \leqslant a<1-\frac{\sqrt{31}}{8}$. | \left[-1,1-\frac{\sqrt{31}}{8}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,478 |
7. In tetrahedron $ABCD$, it is known that $AB=AC=3, BD=BC=4, BD \perp$ plane $ABC$.
Then the radius of the circumscribed sphere of tetrahedron $ABCD$ is . $\qquad$ | 7. $\frac{\sqrt{805}}{10}$.
Let the circumcenter of $\triangle ABC$ be $O_{1}$, and the radius be $r$. Let the circumcenter of the tetrahedron $ABCD$ be $O$, and the radius be $R$.
Obviously, $O_{1} O \perp$ plane $ABC$.
As shown in Figure 2, let the midpoint of $BD$ be $E$.
Since $O B=O D$, we have $O E \perp B D$.
... | \frac{\sqrt{805}}{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,479 |
9. If $a \in A$, and $a-1 \notin A, a+1 \notin A$, then $a$ is called an isolated element of set $A$. Therefore, the number of four-element subsets of set $M=\{1,2, \cdots, 9\}$ without isolated elements is $\qquad$ . | 9. 21 .
Consider the smallest element $i$ and the largest element $j$ in a set that satisfies the condition.
Let this set be $A$. Then $i+1 \in A, j-1 \in A$ (otherwise, $i$ or $j$ would be an isolated element).
Thus, $A=\{i, i+1, j-1, j\}$.
And $2 \leqslant i+1 < j-1 \leqslant 8$, so the number of ways to choose $i+... | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,480 |
10. Let the eccentricities of an ellipse and a hyperbola with common foci be $e_{1}$ and $e_{2}$, respectively. If the length of the minor axis of the ellipse is twice the length of the imaginary axis of the hyperbola, then the maximum value of $\frac{1}{e_{1}}+\frac{1}{e_{2}}$ is . $\qquad$ | 10. $\frac{5}{2}$.
Let the imaginary semi-axis length of the hyperbola be $b$. Then the minor semi-axis length of the ellipse is $2b$.
Assume the foci are $(\pm c, 0)(0 < b < c)$. Then
$$
\frac{1}{e_{1}}=\frac{\sqrt{c^{2}+(2 b)^{2}}}{c}, \frac{1}{e_{2}}=\frac{\sqrt{c^{2}-b^{2}}}{c} \text {. }
$$
Thus, $\frac{1}{e_{1}... | \frac{5}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,481 |
Example 7 Given that $B D$ is the angle bisector of $\triangle A B C$, $E$ and $F$ are the feet of the perpendiculars from points $A$ and $C$ to $B D$, and $P$ is the foot of the perpendicular from point $D$ to $B C$. Prove that $\angle D P E=\angle D P F$.
(2002, 17th Latin American Mathematical Olympiad) | Given that points $D$, $P$, $C$, and $F$ are concyclic, and $A E \parallel C F$.
Thus, $\angle D P F = \angle D C F = \angle D A E$.
Next, we will prove: $\angle D A E = \angle D P E$.
As shown in Figure 9, construct the reflection of point $P$ about $B D$, denoted as $P'$, and connect $P' D$ and $P' E$.
Since $B D$ b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,482 |
1. Divide the sequence of positive integers $1,2, \cdots$ from left to right into segments such that the first segment has $1 \times 2$ numbers, the second segment has $2 \times 3$ numbers, $\cdots$, the $n$-th segment has $n \times(n+1)$ numbers, $\cdots$. Then 2014 is in the $\qquad$ segment. | $-, 1.18$.
$$
\begin{array}{l}
\text { Let } S_{n}=1 \times 2+2 \times 3+\cdots+n(n+1) \\
=\left(1^{2}+2^{2}+\cdots+n^{2}\right)+(1+2+\cdots+n) \\
=\frac{n(n+1)(n+2)}{3} .
\end{array}
$$
If 2014 is in the $(n+1)$-th segment, since there are $S_{n}$ numbers before this segment, then $S_{n}<2014 \leqslant S_{n+1}$.
Sinc... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,483 |
2. In a right-angled tetrahedron $ABCD$, if the sum of the six edge lengths is 6, then the maximum value of its volume is $\qquad$ . | 2. $\frac{4}{3(1+\sqrt{2})^{3}}$.
Let edges $D A$, $D B$, and $D C$ be mutually perpendicular.
Let $D A=a$, $D B=b$, $D C=c$.
Then $V=\frac{1}{6} a b c \Rightarrow a b c=6 V$.
And $A B=\sqrt{a^{2}+b^{2}}$, $B C=\sqrt{b^{2}+c^{2}}$,
$C A=\sqrt{c^{2}+a^{2}}$,
Thus $6=a+b+c+\sqrt{a^{2}+b^{2}}+$
$$
\begin{aligned}
& \sqrt... | \frac{4}{3(1+\sqrt{2})^{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,484 |
Three. (20 points) The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy the conditions $a_{1}=b_{1}=1, a_{n+1}=a_{n}+2 b_{n}, b_{n+1}=a_{n}+b_{n}$. Prove: for each positive integer $n$,
(1) $\frac{a_{2 n-1}}{b_{2 n-1}}\sqrt{2}$;
(2) $\left|\frac{a_{n+1}}{b_{n+1}}-\sqrt{2}\right|<\left|\frac{a_{n}}{b_... | Three, (1) From the problem, we know
$$
\begin{array}{l}
a_{n+1}^{2}-2 b_{n+1}^{2}=\left(a_{n}+2 b_{n}\right)^{2}-2\left(a_{n}+b_{n}\right)^{2} \\
=-\left(a_{n}^{2}-2 b_{n}^{2}\right) .
\end{array}
$$
Therefore, \( a_{n}^{2}-2 b_{n}^{2}=-\left(a_{n-1}^{2}-2 b_{n-1}^{2}\right) \)
$$
\begin{array}{l}
=(-1)^{2}\left(a_{n... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,486 |
2. The function
$$
f(x)=\sqrt{3} \sin ^{2} x+\sin x \cdot \cos x-\frac{\sqrt{3}}{2}\left(x \in\left[\frac{\pi}{12}, \frac{\pi}{2}\right]\right)
$$
has the range . $\qquad$ | 2. $\left[-\frac{1}{2}, 1\right]$.
From the problem, we have
$$
\begin{array}{l}
f(x)=\sqrt{3} \times \frac{1-\cos 2 x}{2}+\frac{1}{2} \sin 2 x-\frac{\sqrt{3}}{2} \\
=\frac{1}{2} \sin 2 x-\frac{\sqrt{3}}{2} \cos 2 x=\sin \left(2 x-\frac{\pi}{3}\right) .
\end{array}
$$
Since $x \in\left[\frac{\pi}{12}, \frac{\pi}{2}\r... | \left[-\frac{1}{2}, 1\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,487 |
3. In the tetrahedron $D-ABC$, it is known that $AB=$ $BC=2, AB \perp BC, BC \perp CD, DA \perp AB$, $\angle CDA=60^{\circ}$. Then the volume of the tetrahedron $D-ABC$ is $\qquad$ . | 3. $\frac{4}{3}$.
As shown in Figure 2, draw $DE \perp$ plane $ABC$ at point $E$, and connect $EA$, $EC$.
Since $BC \perp CD$, $DA \perp AB$, therefore,
$EC \perp CB$,
$EA \perp AB$.
Thus, quadrilateral $EABC$ is a rectangle.
Given $AB = BC$, we know that rectangle $EABC$ is a square, and $DA = DC$.
Also, $\angle CD... | \frac{4}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,488 |
4. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: x^{2}-\frac{y^{2}}{24}=1$, respectively, and let $P$ be a point on the hyperbola $C$ in the first quadrant. If $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{4}{3}$, then the radius of the incircle of $\triangle P F_{1} F_{2}$ is . $\q... | 4. 2 .
Let $\left|P F_{1}\right|=4 t$. Then $\left|P F_{2}\right|=3 t$.
Thus $4 t-3 t=\left|P F_{1}\right|-\left|P F_{2}\right|=2$
$$
\Rightarrow t=2,\left|P F_{1}\right|=8,\left|P F_{2}\right|=6 \text {. }
$$
Combining with $\left|F_{1} F_{2}\right|=10$, we know that $\triangle P F_{1} F_{2}$ is a right triangle, $P... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,489 |
6. If the fraction $\frac{p}{q}\left(p, q \in \mathbf{Z}_{+}\right)$ is converted to a decimal as
$$
\frac{p}{q}=0.198 \cdots,
$$
then when $q$ takes the minimum value, $p+q=$ . $\qquad$ | 6. 121 .
Given $\frac{p}{q}=0.198 \cdots5 p$.
Let $q=5 p+m\left(m \in \mathbf{Z}_{+}\right)$. Then
$$
\begin{array}{l}
\frac{p}{5 p+m}=0.198 \cdots \\
\Rightarrow 0.198(5 p+m)<p<0.199(5 p+m) \\
\Rightarrow 19.8 m<p<39.8 m .
\end{array}
$$
When $m=1$, $20 \leqslant p \leqslant 39$, taking $p=20, m=1$, $q$ is minimized... | 121 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,490 |
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