problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Example 4: Do there exist pairwise distinct real numbers $a$, $b$, $c$ such that the three lines $y=ax+b$, $y=bx+c$, $y=cx+a$ in the Cartesian coordinate system intersect at a common point?
(2013, Peking University Admission Examination for Recommended Students) | No solution.
Proof by contradiction. Suppose there exists a point $\left(x_{0}, y_{0}\right)$ where the three lines intersect. Then
$$
\begin{array}{l}
y_{0}=a x_{0}+b, \\
y_{0}=b x_{0}+c, \\
y_{0}=c x_{0}+a .
\end{array}
$$
Without loss of generality, assume $a>b, a>c$. Then from equations (1) and (3), we get
$$
x_{0... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,871 |
Example 5 What is the minimum degree of the highest term of a polynomial with rational coefficients that has $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots?
(2013, Joint Autonomous Admission Examination of Peking University and Other Universities) | Notice that the polynomial
$$
f(x)=\left(x^{2}-2\right)\left[(x-1)^{3}-2\right]
$$
has roots $\sqrt{2}$ and $1-\sqrt[3]{2}$, and its degree is 5. Therefore, the degree of the highest term of a rational-coefficient polynomial with $\sqrt{2}$ and $1-\sqrt[3]{2}$ as roots cannot be less than 5.
If there exists a rationa... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,872 |
Example 6 Proof: The equation $2 x^{3}+5 x-2=0$ has exactly one real root $r$, and there exists a unique strictly increasing sequence of positive integers $\left\{a_{n}\right\}$, such that $\frac{2}{5}=r^{a_{1}}+r^{a_{2}}+\cdots .{ }^{[3]}$
$(2010$, National High School Mathematics Joint Competition) | Let $f(x)=2 x^{3}+5 x-2$. Then $f^{\prime}(x)=6 x^{2}+5>0$.
Therefore, $f(x)$ is monotonically increasing on $\mathbf{R}$.
Thus, $f(x)=0$ has at most one real root.
Since $f(0)=-20$, it follows that $f(x)=0$ has exactly one real root $r \in\left(0, \frac{1}{2}\right)$.
Then $2 r^{3}+5 r-2=0$, which means
$$
\frac{2}{5}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,873 |
Example 7 Given $m n$ real numbers, arranged in an $m \times n$ matrix (denoted as $\left\{a_{i j}\right\}_{m \times n}$), such that each row is strictly increasing from left to right, i.e., for any $i(i=1,2, \cdots, 2013)$, when $j_{1}<j_{2}$, we have $a_{i j_{1}}<a_{i j_{2}}$; now, rearrange the numbers in each colum... | In the sequence $\left\{a_{i j}^{\prime}\right\}_{m \times n}$, the $n$ numbers in each row are increasing from left to right.
Proof by contradiction.
If there exists a pair $a_{p q}^{\prime}>a_{p(q+1)}^{\prime}$, let $a_{k(q+1)}^{\prime}=$ $a_{i_{k}(q+1)}\left(k=1,2, \cdots, m,\left\{i_{1}, i_{2}, \cdots, i_{m}\right\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,874 |
Example 8 Given $a_{i}(i=1,2, \cdots, 10)$ satisfy
$$
\sum_{i=1}^{10} a_{i}=30, \prod_{i=1}^{10} a_{i}<21 \text {, }
$$
Prove: There exists $a_{i}$, such that $a_{i}<1$.
(2012, Peking University Admissions Exam for Exceptional Students) | Proof by contradiction.
Assume the conclusion does not hold, then for any $i(i=1,2$, $\cdots, 10)$, we have $a_{i} \geqslant 1$.
Let $a_{i}=1+b_{i}(i=1,2, \cdots, 10)$.
Then $b_{i} \geqslant 0$, and $b_{1}+b_{2}+\cdots+b_{10}=20$.
Thus, $a_{1} a_{2} \cdots a_{10}$
$$
\begin{array}{l}
=\left(b_{1}+1\right)\left(b_{2}+1\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,875 |
Example 5 Let quadrilateral $ABCD$ be a rectangle, and $K$ be a point on the plane of the rectangle. Lines $KA$ and $KD$ both intersect side $BC$. Perpendiculars are drawn from points $B$ and $C$ to lines $DK$ and $AK$, respectively, and these perpendiculars intersect at point $M$. Prove that $MK \perp AD$.
(17th All-R... | Proof As shown in Figure 5, let $B E$ and $C F$ be the perpendiculars drawn from points $B$ and $C$ to lines $D K$ and $A K$, with feet of the perpendiculars at $E$ and $F$, respectively. Translate $A K$ and $D K$ so that point $A$ moves to $B$ and point $D$ moves to $C$. At this time, point $K$ moves to $P$.
Since $P ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,876 |
1. Given the function $f(x)=a x^{2}+b x+c(a \neq 0)$, and $f(x)=x$ has no real roots. Is it true that $f(f(x))=x$ has real roots? Prove your conclusion.
(2008, Shanghai Jiao Tong University Independent Admission Examination) | Assume there exists $x_{0}$ such that $f\left(f\left(x_{0}\right)\right)=x_{0}$.
Let $f\left(x_{0}\right)=t$. Then $t \neq x_{0}$.
Otherwise, $x_{0}$ would be a real root of $f(x)=x$.
Since $f(t)=f\left(f\left(x_{0}\right)\right)=x_{0}$, the point $\left(t, x_{0}\right)$ lies on the graph of the function $y=f(x)$.
Als... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,877 |
2. Do there exist four positive real numbers, the pairwise products of which are $2$, $3$, $5$, $6$, $10$, and $16$?
$(2011$, Peking University and other universities' joint independent admission examination) | Assume there exist four positive real numbers $a, b, c, d$ whose pairwise products are $2, 3, 5, 6, 10, 16$. Then
$$
\begin{array}{l}
(a b c d)^{3}=2 \times 3 \times 5 \times 6 \times 10 \times 16 \\
=2^{7} \times 3^{2} \times 5^{2} .
\end{array}
$$
Therefore, $a b c d=4 \sqrt{2 \times 3^{2} \times 5^{2}}$ is an irrat... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,878 |
3. Given that $\sin x, \sin y, \sin z$ form a strictly increasing arithmetic sequence. Prove: $\cos x, \cos y, \cos z$ do not form an arithmetic sequence.
(2011, Peking University Mathematics Summer Camp) | Assume $\cos x, \cos y, \cos z$ form an arithmetic sequence. Then
$$
2 \cos y=\cos x+\cos z \text{.}
$$
Also, from the condition, we have $2 \sin y=\sin x+\sin z$.
Squaring both equations and adding them yields
$$
\begin{array}{l}
4=2+2 \cos (z-x) \Rightarrow \cos (z-x)=1 \\
\Rightarrow z-x=2 k \pi \Rightarrow \sin x=... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,879 |
4. Define a real-valued function $f(x)$ over the real numbers. Does there exist $f(n)$ such that for any integer $n$,
$$
f\left(-n^{2}+3 n+1\right)=f^{2}(n)+2 \text { ? }
$$
(2011, Peking University Mathematics Summer Camp) | Prompt: Does not exist.
If there exists $f(n)$, for any integer $n$ we have
$$
f\left(-n^{2}+3 n+1\right)=f^{2}(n)+2 \text {, }
$$
then when $n=1, n=3$ respectively, we have
$$
\begin{array}{l}
f(3)=f^{2}(1)+2 ; \\
f(1)=f^{2}(3)+2 .
\end{array}
$$
Thus, $f(3) \geqslant 2$, and $f(1) \geqslant 2$.
(1) - (2) gives
$$
\... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,880 |
Example 1 As shown in Figure 1, let the hexagon $A C' B A' C B'$ have equal opposite sides. Point $A_{1}$ is such that $\triangle A_{1} B C' \cong \triangle A_{1} B' C$, and points $B_{1}$ and $C_{1}$ are defined similarly. Prove: $A A_{1}$, $B B_{1}$, and $C C_{1}$ are concurrent. | Notice that, $A_{1}$ is the intersection of the perpendicular bisectors of $B B^{\prime}$ and $C C^{\prime}$. Let the perpendicular bisector of $A A^{\prime}$ intersect $B C$ at point $A_{2}$. Similarly, define points $B_{2}$ and $C_{2}$.
Then, by the converse of Desargues' theorem, it suffices to show that $A_{2}$, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,881 |
Given an integer $n$, find all functions $f: \mathbf{Z} \rightarrow \mathbf{Z}$, such that for any integers $x, y$ we have
$$
f(x+y+f(y))=f(x)+n y \text{. }
$$
(2012, China National Training Team Test)
The original solution used Cauchy's equation. This article introduces a solution using pure algebraic manipulation.
T... | Let $S=\{y+f(y) \mid y \in \mathbf{Z}\}$.
First, consider the case when $n=0$.
If $S=\{0\}$, then $f(x)=-x$ for all $x \in \mathbf{Z}$.
It is easy to verify that this satisfies the requirements.
If the set $S$ contains non-zero numbers, then $f$ is a periodic function. Let $d$ be the smallest positive period of $f$. Th... | f(x)=mx \text{ or } -(m+1)x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,883 |
2. Given a cyclic quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $E$, the extensions of $DA$ and $CB$ intersect at point $F$, point $G$ satisfies that quadrilateral $ECGD$ is a parallelogram, and $H$ is the reflection of point $E$ over the line $AD$. Prove: $D$, $H$, $F$, and $G$ are concyclic. | 2. As shown in Figure 1.
Since quadrilateral $E C G D$ is a parallelogram and points $A$, $B$, $C$, and $D$ are concyclic, we have:
$$
\begin{array}{l}
\angle G D C=\angle D C E=\angle D C A=\angle D B A, \\
\angle C D A=\angle A B F .
\end{array}
$$
Therefore, $\angle G D F=\angle G D C+\angle C D A$
$$
=\angle D B ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,884 |
3. In an acute triangle $\triangle ABC$, it is known that points $D$, $E$, and $F$ are the projections of points $A$, $B$, and $C$ onto sides $BC$, $CA$, and $AB$, respectively. The incenter of $\triangle AEF$ and $\triangle BDF$ are $I_{1}$ and $I_{2}$, respectively. The circumcenters of $\triangle ACI_{1}$ and $\tria... | 3. As shown in Figure 2.
Let $\angle C A B=\alpha, \angle A B C=\beta, \angle B C A=\gamma$, and the extensions of $A I_{1}$ and $B I_{2}$ intersect at point $I$.
Since $A I_{1}$ and $B I_{2}$ are the angle bisectors of $\angle C A B$ and $\angle A B C$ respectively, $I$ is the incenter of $\triangle A B C$.
Because... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,885 |
4. In $\triangle A B C$, it is known that $A B \neq A C, O$ is the circumcenter of $\triangle A B C$, the angle bisector of $\angle B A C$ intersects $B C$ at point $D$, point $E$ is symmetric to $D$ with respect to the midpoint of $B C$, perpendicular lines to $B C$ through points $D$ and $E$ intersect $A O$ and $A D$... | 4. As shown in Figure 3.
Let the angle bisector of $\angle BAC$ intersect the perpendicular bisector of $BC$ at point $P$. Then $P$ is the midpoint of arc $\overparen{BC}$. Let $OP$ intersect $BC$ at point $M$, then $M$ is the midpoint of $BC$. Let the symmetric point of $Y$ with respect to line $OP$ be $Y'$.
Since $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,886 |
Example 6 As shown in Figure 6, given that $A B$ is the diameter of $\odot O$, $C$ and $D$ are two points on the circumference different from points $A$ and $B$ and on the same side of $A B$. Tangents to the circle are drawn through points $C$ and $D$, intersecting at point $E$. The intersection of line segments $A D$ ... | Prove: As shown in Figure 6, connect $AC$, $BD$ and extend them to intersect at point $P$, then connect $PF$.
Since $AB$ is the diameter of $\odot O$, we have
$BC \perp AP$, $AD \perp BP$.
Therefore, $F$ is the orthocenter of $\triangle ABP$, which implies $PF \perp AB$.
Thus, $\angle PFC = 90^{\circ} - \angle CPF = \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,887 |
6. Given that the circumcenter and incenter of $\triangle A B C$ are $O$ and $I$ respectively, points $D, E, F$ lie on sides $B C, C A, A B$ respectively, and satisfy
$$
B D+B F=C A, C D+C E=A B,
$$
the circumcircles of $\triangle B F D$ and $\triangle C D E$ intersect at a point $P$ different from $D$. Prove: $O P=O I... | ```
6. Let the circumcircles of $\triangle A E F$, $\triangle B F D$, and $\triangle C D E$ be circles $\Gamma_{A}$, $\Gamma_{B}$, and $\Gamma_{C}$, respectively.
By Miquel's theorem, circle $\Gamma_{1}$ passes through point $P$.
As shown in Figure 4, let the lines $A I$, $B I$, and $C I$ intersect circles ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,888 |
7. In a convex quadrilateral $A B C D$, it is known that sides $B C$ and $A D$ are not parallel, $E$ is a point on side $B C$, and quadrilaterals $A B E D$ and $A E C D$ both have inscribed circles. Prove: There exists a point $F$ on side $A D$ such that quadrilaterals $A B C F$ and $B C D F$ both have inscribed circle... | 7. As shown in Figure 6, let the incircles of quadrilaterals $A B E D$ and $A E C D$ be $\odot O_{1}$ and $\odot O_{2}$, respectively. Then a point $F$ exists if and only if $\odot O_{1}$ and $\odot O_{2}$ are also the incircles of quadrilaterals $A B C F$ and $B C D F$, respectively.
Let the tangent from point $B$ to... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,889 |
1. If the graph of the inverse proportion function $y=\frac{k}{x}$ intersects the line $y=4 x-1$ at a point $P$ with an x-coordinate of 1, then $k$ $=(\quad)$.
(A) -3
(B) -1
(C) 1
(D) 3 | -.1. D. From the condition $P(1,3) \Rightarrow 3=\frac{k}{1} \Rightarrow k=3$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,891 |
2. The sum of all real solutions of the equation $x^{2}-2012|x|+2013=0$ is ( ).
(A) -2012
(B) 0
(C) 2012
(D) 2013 | 2. B.
If $x_{0}$ is a real solution to the given equation, then $-x_{0}$ is also a real solution to the equation. Therefore, the sum of all real solutions to the equation is 0. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,892 |
3. Given a square $A B C D$ with side length $1, E$ and $F$ are points on sides $B C$ and $C D$ respectively, and satisfy $B E=C F$. Then the minimum value of the area of $\triangle A E F$ is ( ).
(A) $\frac{\sqrt{2}}{8}$
(B) $\frac{\sqrt{3}}{8}$
(C) $\frac{1}{4}$
(D) $\frac{3}{8}$ | 3. D.
Let $B E=C F=x$. Then $C E=D F=1-x$, and
$$
\begin{array}{l}
S_{\triangle A E F}=1-\frac{1}{2} x-\frac{1}{2}(1-x)-\frac{1}{2} x(1-x) \\
=\frac{1}{2}\left[\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}\right] \geqslant \frac{3}{8} .
\end{array}
$$
When $x=\frac{1}{2}$, the equality holds.
Therefore, the minimum valu... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,893 |
4. Given real numbers $a, b$ satisfy
$$
|a-1|+\sqrt{(a-2)(b-2)^{2}}+\left|b^{2}+1\right|=a \text {. }
$$
Then $a^{b}=(\quad$.
(A) $\frac{1}{4}$
(B) $\frac{1}{2}$
(C) 1
(D) 2 | 4. C.
From the given, we know
$$
1=|a-1|-(a-1)+\sqrt{(a-2)(b-2)^{2}}+\left|b^{2}+1\right| \geqslant 1 \text {, }
$$
when $a=2, b=0$, the equality holds. Therefore, $a^{b}=1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,894 |
6. For any two points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right)$ in a Cartesian coordinate system, the expression $\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right|$ is called the Manhattan distance between $P_{1}$ and $P_{2}$, denoted as $d\left(P_{1}, P_{2}\right)$. If $P_{0}\left(x_{0}, y_{0}\r... | 6. A.
From the condition, we know that the required perpendicular distance $d$ is the minimum value of
$$
S=|x-1|+|2 x+5-1|
$$
which is
$$
\begin{array}{l}
\text { Also, } S=|x-(-2)|+|x-1|+|x-(-2)| \\
\geqslant 3+|x-(-2)| \geqslant 3,
\end{array}
$$
with equality holding if and only if $x=-2$.
Therefore, $d=3$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,895 |
1. If $x$ is an integer, and satisfies the inequality system
$$
\left\{\begin{array}{l}
x-1>0, \\
2 x-1<4,
\end{array}\right.
$$
then $x=$ $\qquad$ | $$
=, 1.2 \text {. }
$$
From the given, we know that $1<x<\frac{5}{2}$. Therefore, the integer $x=2$. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,896 |
2. In an equilateral $\triangle A B C$, it is known that $D$ and $E$ are points on sides $A B$ and $A C$ respectively, and satisfy $A D=C E, B E$ intersects $C D$ at point $F$. Then $\angle B F C=$ $\qquad$ . | 2. $120^{\circ}$.
As shown in Figure 3, note that,
$$
\begin{array}{l}
\triangle A D C \cong \triangle C E B \\
\Rightarrow \angle A C D=\angle C B E . \\
\text { Then } \angle E F C \\
=\angle F B C+\angle F C B \\
=\angle E C F+\angle F C B \\
=60^{\circ} .
\end{array}
$$
Therefore, $\angle B F C=120^{\circ}$. | 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,897 |
1. Given that in the acute triangle $\triangle A B C$, the distance from vertex $A$ to the orthocenter $H$ is equal to the radius of its circumcircle. Then the degree of $\angle A$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$
(2007, National Junior High School Mathematics League) | Answer: C.
The text above has been translated into English, maintaining the original text's line breaks and format. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,898 |
3. Let real numbers $x, y, z$ satisfy
$$
x^{2}+y^{2}+z^{2}-x y-y z-z x=27 \text {. }
$$
Then the maximum value of $|y-z|$ is $\qquad$ | 3. 6 .
The original equation is equivalent to a quadratic equation in $x$
$$
\begin{array}{l}
x^{2}-(y+z) x+y^{2}+z^{2}-y z-27=0 . \\
\text { And } \Delta=(y+z)^{2}-4\left(y^{2}+z^{2}-y z-27\right) \geqslant 0 \\
\Rightarrow(y-z)^{2} \leqslant 36 \Rightarrow|y-z| \leqslant 6 .
\end{array}
$$
When $|y-z|=6$, and $x=\f... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,899 |
4. Given $x_{1}, x_{2}, \cdots, x_{15}$ take values of 1 or -1. Let
$$
S=x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{15} x_{1} \text {. }
$$
Then the smallest positive integer that $S$ can take is $\qquad$ | 4.3.
Let $y_{i}=x_{i} x_{i+1}(i=1,2, \cdots, 15)$, with the convention that $x_{16}=$ $x_{1}$. Then $y_{i}=1$ or -1.
In $y_{1}, y_{2}, \cdots, y_{15}$, let there be $a$ values that are 1 and $b$ values that are -1. Clearly, $a+b=15$.
Also, $1^{a}(-1)^{b}=y_{1} y_{2} \cdots y_{15}=\left(x_{1} x_{2} \cdots x_{15}\righ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,900 |
1. (20 points) Given the parabola $y=x^{2}+2 x+m$ intersects the $x$-axis at points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$, where $x_{1}>x_{2}$, and $x_{1}^{2}+x_{2}^{2}=10$.
(1) Find the value of the real number $m$;
(2) Let $M\left(2, y_{0}\right)$ be a point on the parabola $y=x^{2}+2 x+m$. Find a point $P... | Three, 1. (1) From the conditions, we have
$$
x_{1}+x_{2}=-2, x_{1} x_{2}=m \text {. }
$$
Then $10=x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=4-2 m$
$$
\Rightarrow m=-3 \text {. }
$$
Upon verification, $m=-3$ satisfies the conditions.
(2) From (1), the equation of the parabola is
$$
y=x^{2}+2 x-3 ... | P(-1,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,901 |
2. (25 points) As shown in Figure 1, given that $E$ is a point on side $A B$ of square $A B C D$, and the symmetric point of $A$ with respect to $D E$ is $F, \angle B F C = 90^{\circ}$. Find the value of $\frac{A B}{A E}$. | 2. As shown in Figure 4, extend $E F$ to intersect $B C$ at point $M$, connect $D M$, and let it intersect $C F$ at point $G$.
Then, Rt $\triangle D F M \cong$ Rt $\triangle D C M$. Therefore, $\angle F D M = \angle M D C$, and $F M = C M$.
Thus, $M$ is the midpoint of $B C$.
$$
\begin{array}{l}
\text { Also, } \angle ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,902 |
3. (25 points) On a circle, there are $n$ different positive integers $a_{1}$, $a_{2}, \cdots, a_{n}$ placed in a clockwise direction. If for any number $b$ among the ten positive integers $1, 2, \cdots, 10$, there exists a positive integer $i$ such that $a_{i}=b$ or $a_{i}+a_{i+1}=b$, with the convention that $a_{n+1}... | 3. From the conditions, we know that the $2n$ numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{1}+a_{2}, a_{2}+a_{3}, \cdots, a_{n}+a_{1}$ should include the ten positive integers $1,2, \cdots, 10$. Therefore, $2 n \geqslant 10 \Rightarrow n \geqslant 5$.
When $n=5$, the ten numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{1}+a_{2}, ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,903 |
5. In quadrilateral $A B C D$, it is known that $A B=B C=1$, $\angle A B C=100^{\circ}, \angle C D A=130^{\circ}$. Then $B D=$ ( ).
(A) $\frac{\sqrt{2}}{2}$
(B) 1
(C) $\sqrt{2}$
(D) $\sqrt{3}$ | 5. B.
As shown in Figure 2, extend $A B$ to point $E$ such that $B E = B C$.
Connect $C E$. Then
$$
\begin{array}{l}
\angle A E C = \angle B C E \\
= \frac{1}{2} \angle A B C = 50^{\circ}.
\end{array}
$$
Thus, $\angle A E C + \angle A D C = 180^{\circ}$. Therefore, points $A, E, C, D$ are concyclic, and the center of... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,904 |
1. Let $A$ be the closed region in the Cartesian coordinate plane bounded by the three lines $x=1$, $y=0$, and $y=t(2x-t)$ where $0<t<1$. Prove: In region $A$, the area of the triangle with vertices $P\left(t, t^{2}\right)$ and $Q(1,0)$ does not exceed $\frac{1}{4}$.
(Li Shenghong) | 1. It is easy to see that the closed region formed by these three lines is the interior and boundary of a triangle (as shown in Figure 3), with the three vertices of the triangle being
$$
B\left(\frac{t}{2}, 0\right), Q(1,0), C(1, t(2-t)) \text {. }
$$
Taking any point $X$ within $\triangle B Q C$, the area of $\trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,905 |
2. As shown in Figure 1, in trapezoid $ABCD$, $AB \parallel CD$, $\odot O_{1}$ is tangent to sides $DA$, $AB$, and $BC$, and $\odot O_{2}$ is tangent to sides $BC$, $CD$, and $DA$. Let $P$ be the point of tangency of $\odot O_{1}$ with side $AB$, and $Q$ be the point of tangency of $\odot O_{2}$ with side $CD$. Prove t... | 2. As shown in Figure 4, let the intersection of line $AC$ and $BD$ be point $R$. Connect $O_{1}A$, $O_{1}B$, $O_{1}P$, $O_{2}C$, $O_{2}D$, $O_{2}Q$, $PR$, and $QR$.
Since $BA$ and $BC$ are tangents to $\odot O_{1}$, we have
$$
\angle PBO_{1} = \angle CBO_{1} = \frac{1}{2} \angle ABC.
$$
Similarly, $\angle QCO_{2} = \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,906 |
3. In a group composed of $m$ girls and $n$ boys, any two people either know each other or do not know each other. For any two boys and two girls, at least one boy and one girl do not know each other. Prove: the number of boy-girl pairs who know each other does not exceed $m+\frac{n(n-1)}{2}$. | 3. From the given conditions, for any two boys, there is at most one girl they both know. Consider the girls who know exactly $i$ boys, and let the number of such girls be $x_{i}(1 \leqslant i \leqslant n)$.
Thus, $\sum_{i=1}^{n} x_{i}=m$.
To calculate the number of trios consisting of the above two boys and one girl, ... | m+\frac{n(n-1)}{2} | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,907 |
4. Find the number of polynomials $f(x)=a x^{3}+b x$ that satisfy the following two conditions:
(1) $a, b \in\{1,2, \cdots, 2013\}$;
(2) The difference between any two numbers in $f(1), f(2), \cdots, f(2013)$ is not a multiple of 2013. (Wang Bin) | 4. It is known that the prime factorization of $2013=3 \times 11 \times 61$.
Let $p_{1}=3, p_{2}=11, p_{3}=61$.
For $a, b \in\{1,2, \cdots, 2013\}$, let
$a \equiv a_{i}\left(\bmod p_{i}\right), b \equiv b_{i}\left(\bmod p_{i}\right)$,
where $i=1,2,3$.
By the Chinese Remainder Theorem, we know that $(a, b)$ and $\left(... | 7200 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,908 |
5. Given positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$. Prove: There exist positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$, such that $\sum_{i=1}^{n} x_{i}=1$, and for any positive real numbers $y_{1}, y_{2}, \cdots, y_{n}$ satisfying $\sum_{i=1}^{n} y_{i}=1$, we have $\sum_{i=1}^{n} \frac{a_{i} x_{i}}{x_{i}+... | 5. Let $x_{i}=\frac{a_{i}}{\sum_{i=1}^{n} a_{i}}$. Then
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}=1, \\
\sum_{i=1}^{n} \frac{a_{i} x_{i}}{x_{i}+y_{i}}=\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+y_{i}}\right) .
\end{array}
$$
For $\sum_{i=1}^{n} y_{i}=1\left(y_{1}, y_{2}, \cdots, y_{n} ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,910 |
6. Let set $S$ be an $m(m \geqslant 3)$-element subset of $\{0,1, \cdots, 98\}$, satisfying that for any $x, y \in S$, there exists $z \in S$ such that $x+y \equiv 2 z(\bmod 99)$. Find all possible values of $m$.
(Wang Xinmao provided the problem) | 6. Let $S=\left\{s_{1}, s_{2}, \cdots, s_{m}\right\}$.
Since $S^{\prime}=\left\{0, s_{2}-s_{1}, \cdots, s_{m}-s_{1}\right\}$ also satisfies the condition, we may assume without loss of generality that $0 \in S$.
By the condition, for any $x, y \in S$,
$50(x+y) \equiv z(\bmod 99) \in S$.
In particular, for any $x \in S... | 3,9,11,33,99 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,911 |
7. As shown in Figure 2, given that $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at point $T$, quadrilateral $A B C D$ is inscribed in $\odot O_{1}$, lines $D A$ and $C B$ are tangent to $\odot O_{2}$ at points $E$ and $F$ respectively, line $B N$ bisects $\angle A B F$, and intersects segment $E F$ at point ... | 7. As shown in Figure 5, let the extension of $A M$ intersect $E F$ at point $P$. Connect $A T, B M, B P, B T, C M, C T, E T, T P$.
From $B F$ being tangent to $\odot O_{2}$ at point $F$, we get
$\angle B F T = \angle F E T$.
From $\odot O_{1}$ and $\odot O_{2}$ being externally tangent at point $T$, we know
$\angle M ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,912 |
8. Let $n \geqslant 4$ be an even number. Label the $n$ vertices of a regular $n$-gon with $n$ distinct real numbers in any manner. Starting from a certain edge, label the edges in a clockwise direction as $e_{1}, e_{2}, \cdots, e_{n}$. If the numbers at the two endpoints of an edge are in increasing order in a clockwi... | 8. Proof 1 Without loss of generality, assume the $n$ real numbers marked are $1,2, \cdots, n$. Let $A$ be the number of alternating edge pairs, $B$ be the number of positive edges, and $S=A+B$.
Next, we prove: After swapping the numbers $i$ and $i+1$, the parity of $S$ does not change.
We consider two cases.
(1) The ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,913 |
Find the largest positive integer $n(n \geqslant 3)$, such that there exists a convex $n$-gon, where the tangent values of all its interior angles are integers.
(Proposed by the Problem Committee) | On the one hand, since each interior angle of a regular octagon is $135^{\circ}$, and its tangent value is -1, $n=8$ satisfies the condition.
On the other hand, if $n \geqslant 9$, let the exterior angles of the $n$-sided polygon be $\angle A_{1}, \angle A_{2}, \cdots, \angle A_{n}\left(0<\angle A_{1} \leqslant \angle... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,914 |
Let $x_{k} \in[-2,2](k=1,2, \cdots, 2013)$,
and
$x_{1}+x_{2}+\cdots+x_{2013}=0$. Try to find
$$
M=x_{1}^{3}+x_{2}^{3}+\cdots+x_{2013}^{3}
$$
the maximum value.
(Liu Kangning) | Given $x_{i} \in[-2,2](i=1,2, \cdots, 2013)$, we know that $x_{i}^{3}-3 x_{i}=\left(x_{i}-2\right)\left(x_{i}+1\right)^{2}+2 \leqslant 2$.
The equality holds if and only if $x_{i}=2$ or -1.
Noting that, $\sum_{i=1}^{2013} x_{i}=0$.
Thus, $M=\sum_{i=1}^{2013} x_{i}^{3}=\sum_{i=1}^{2013}\left(x_{i}^{3}-3 x_{i}\right)$
$$... | 4026 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,915 |
Three, As shown in Figure 1, given that $A$ and $B$ are two fixed points on $\odot O$, $C$ is the midpoint of the major arc $\overparen{A B}$, and $D$ is any point on the minor arc $\overparen{A B}$. A tangent line is drawn through $D$ to $\odot O$, intersecting the tangents to $\odot O$ at points $A$ and $B$ at points... | Three, as shown in Figure 3, connect $C D$, intersecting $A B$ at point $K$. Draw a line through point $E$ parallel to $A B$, intersecting the extensions of $C A$ and $C D$ at points $P$ and $Q$ respectively. Connect $B C$.
From $\angle E A P=\angle C B A=\angle C A B=\angle E P A$, we know $P E=A E$.
Similarly, $Q E=D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,916 |
For positive integers $n, a, b$, if $n=a^{2}+b^{2}$, and $a, b$ are coprime, then the pair $(a, b)$ is called a "square decomposition" of $n$ (the order of $a, b$ does not matter). Prove: For any positive integer $k$, $13^{k}$ has exactly one square decomposition.
（Lei Yong, Contributed） | First, prove: If $a$, $b$, and $k$ are positive integers, and satisfy $13^{k}=a^{2}+b^{2}$, then
$a \equiv 5 b(\bmod 13)$ or $b \equiv 5 a(\bmod 13)$.
Since $5^{2} \equiv-1(\bmod 13)$, we have
$$
a^{2} \equiv-b^{2} \equiv(5 b)^{2}(\bmod 13),
$$
which implies $(a+5 b)(a-5 b) \equiv 0(\bmod 13)$.
If $a+5 b \equiv 0(\bmo... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,917 |
Find all non-integer solutions of the equation
$$
x+\frac{13}{x}=[x]+\frac{13}{[x]}
$$
where $[x]$ denotes the greatest integer less than or equal to the real number $x$.
(Proposed by the Problem Committee) | The original equation can be transformed into
$$
(x-[x])\left(1-\frac{13}{x[x]}\right)=0 \text {. }
$$
Since $x \notin \mathbf{Z}$, we have $x-[x] \neq 0$. Therefore,
$1-\frac{13}{x[x]}=0 \Rightarrow x[x]=13$.
Let $x=[x]+r(0<r<1)$. Then,
$$
[x]^{2}+r[x]=13 \quad \text{(1)}
$$
When $[x]>0$, we have
$$
[x]^{2}<13<[x]([... | x=-\frac{13}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,918 |
Six, As shown in Figure 2, given that $M$ is the midpoint of side $BC$ of $\triangle ABC$, $\odot O$ passes through points $A$ and $C$ and is tangent to $AM$, the extension of $BA$ intersects $\odot O$ at point $D$, and line $CD$ intersects $MA$ at point $P$. Prove that $PO \perp BC$. | Six, as shown in Figure 4, take the midpoint $N$ of $CD$, and connect $OA$, $AN$, and $MN$.
Then $OA \perp PA$, $ON \perp PN$.
Thus, points $P$, $A$, $N$, and $O$ are concyclic.
Also, $MN \parallel BD$, so
$$
\angle AMN = \angle PAD = \angle ACD \text{.}
$$
Therefore, points $A$, $M$, $C$, and $N$ are concyclic.
Hence... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,919 |
Example 1 Find the remainder when $10^{10}(100$ ones$)$ is divided by 7. | \begin{array}{l}\text { Sol } 10^{10} \equiv(7+3)^{10^{10}} \equiv 3^{10} \\ \equiv(7+2)^{50 \cdots 0} \equiv 2^{5 \times 10^{9}} \equiv 2^{3 \times 106 \cdots 6+2} \\ \equiv 4(7+1)^{166 \cdots 6} \equiv 4(\bmod 7) .\end{array} | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,920 |
Seven, let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=1, \\
a_{n+1}=\left(1+\frac{k}{n}\right) a_{n}+1(n=1,2, \cdots) .
\end{array}
$$
Find all positive integers $k$ such that every term in the sequence $\left\{a_{n}\right\}$ is an integer.
(Zhang Lei) | When $k=1$, $a_{2}=3, a_{3}=\frac{11}{2}$, which does not satisfy the condition.
When $k=2$, by the given condition we have
$$
\frac{a_{n+1}}{(n+1)(n+2)}=\frac{a_{n}}{n(n+1)}+\frac{1}{(n+1)(n+2)} \text {. }
$$
Thus, $\frac{a_{n}}{n(n+1)}=\frac{a_{1}}{1 \times 2}+\sum_{i=2}^{n} \frac{1}{i(i+1)}=1-\frac{1}{n+1}$.
Theref... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,921 |
Suppose two acquaintances have exactly $n$ common acquaintances, and any two non-acquaintances have exactly $2n$ common acquaintances. If three people are mutually acquainted, they are called a “Peach Garden Trio”.
(1) Find the number of all Peach Garden Trios;
(2) Prove: The $3n$ people can be divided into three group... | (1) First, prove: Each person knows exactly $2n$ people.
In fact, by the condition, we know that the number of people each person knows is no less than $2n$.
Assume $A$ knows $B_{1}, B_{2}, \cdots, B_{m} (m \geqslant 2n+1)$. Then, by the condition, we know that each $B_{i} (1 \leqslant i \leqslant m)$ knows exactly $... | n^3 | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,922 |
1. $P(x) 、 Q(x)$ are two 10th degree monic polynomials with real coefficients. It is known that the equation $P(x)=Q(x)$ has no real roots. Prove: the equation $P(x+1)=Q(x-1)$ has a real root. | 1. Let $P(x)=x^{10}+p_{9} x^{9}+\cdots+p_{0}$,
$$
Q(x)=x^{10}+q_{9} x^{9}+\cdots+q_{0} \text {. }
$$
Then $P(x)-Q(x)$
$$
=\left(p_{9}-q_{9}\right) x^{9}+\left(p_{8}-q_{8}\right) x^{8}+\cdots+\left(p_{0}-q_{0}\right)
$$
has no real roots.
Since every odd-degree real polynomial has a real root, hence
$$
\begin{array}{l... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,923 |
2. For tetrahedron $ABCD$, the insphere and an exsphere are tangent to the plane $BCD$ at different points $X$ and $Y$, respectively. Prove: $\triangle AXY$ is an obtuse triangle. | 2. Consider the homothety transformation with $A$ as the center that transforms the exsphere into the insphere. This transformation maps point $Y$ to point $Z$, which is the point symmetric to $X$ with respect to the center of the insphere.
Since $Z X \perp$ plane $B C D$, point $Z$ lies on segment $A Y$, therefore, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,924 |
3. Find all positive integers $k$ such that the product of the first $k$ odd primes minus 1 is a positive integer raised to an integer power greater than 1. | 3. $k$ does not exist.
Let 3 $p_{1}p_{k}$.
Without loss of generality, we can assume $n$ is a prime (if $n=s t$, then we can replace $n$ with $t$ and $a$ with $a^{s}$).
Since $a^{2}+1$ is not a multiple of 3, we have $n>2$.
If there exists $2 \leqslant i \leqslant k, n=p_{i}$, then $p_{i} \mid\left(a^{p_{i}}+1\right)... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,925 |
4. Write 2013 different real numbers on 2013 cards. Place the cards face down on the table. Two players, A and B, play the following game: in each round, A can arbitrarily select ten cards, and B will tell A one of the ten numbers written on these cards (B does not tell A which card the number is written on). Find the ... | 4. The maximum value of $t$ is $1986=2013-27$.
Let $A_{1}, A_{2}, \cdots, A_{2013}$ be these 2013 cards.
First, note that player B has a strategy to prevent player A from determining the number written on any of the cards $A_{1}, A_{2}, \cdots, A_{27}$.
B divides $T=\{1,2, \cdots, 27\}$ into nine groups
$T_{i}=\{3 i-2... | 1986 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 726,926 |
5. $A$ is a set consisting of 101 distinct non-negative integers, each not exceeding 1000. Prove that the set
$$
\{|x-y| \mid x, y \in A, x \neq y\} \cap\{1,2, \cdots, 100\}
$$
contains at least 10 elements. | 5. Let $A$ consist of $a_{0}<a_{1}<\cdots<a_{100}$. The ten numbers
$$
a_{10}-a_{0}, a_{20}-a_{10}, \cdots, a_{100}-a_{90}
$$
sum to $a_{100}-a_{0} \leqslant 1000$, so one of them is no greater than 100.
Suppose $a_{10 i+10}-a_{10 i} \leqslant 100$. Then
$$
\begin{array}{l}
0<a_{10 i+1}-a_{10 i}<a_{10 i+2}-a_{10 i}<\c... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,927 |
7. The central bank governor of a country hopes to issue 12 integer-valued coins to satisfy: using no more than eight coins (multiple coins of each value allowed) can form any integer value from 1 to 6543. Can the governor's wish be realized? | 7. It can be achieved.
The following proof shows that 12 types of coins can be issued, such that using no more than eight coins, any positive integer value not exceeding 6560 can be formed.
In fact, using coins with denominations of $1, 3, 4$ with at most two coins, any integer value from $1 \sim 8$ can be represente... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,929 |
8. Let the incircle $\odot I$ of $\triangle ABC$ intersect the circumcircle $\Gamma$ of $\triangle AIB$ at points $X$ and $Y$, and their common tangents intersect at point $Z$. Prove that the circumcircle of $\triangle ABC$ is tangent to the circumcircle of $\triangle XYZ$.
Translate the above text into English, pleas... | 8. Let $C I$ intersect the circumcircle $\Gamma^{\prime}$ of $\triangle A B C$ at point $S$. Then $S$ is the center of the circumcircle $\Gamma$ of $\triangle A I B$.
By symmetry, point $Z$ lies on $S C$.
Let the two common tangents of $\odot I$ and circle $\Gamma$ touch circle $\Gamma$ at points $M, N$.
Since $S I$ is... | proof | Other | math-word-problem | Yes | Yes | cn_contest | false | 726,930 |
1. Let $S$ be a set of $n(n \geqslant 5)$ points in the plane. If any four points chosen from $S$ have at least one point connected to the other three, then which of the following conclusions is correct? $\qquad$
(1) There is no point in $S$ that is connected to all other points;
(2) There is at least one point in $S$ ... | -、1. (2).
In the point set $S$, all points are connected to each other, which clearly satisfies the problem. Therefore, conclusions (1) and (4) are incorrect.
Suppose $A$, $B$, and $C$ are three points in the point set $S$ that are not connected to each other, but the remaining $n-3$ points are all connected to each o... | 2 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,932 |
2. Given $x>0, y>0, a=x+y, b=$ $\sqrt{x^{2}-x y+y^{2}}, c=\lambda \sqrt{x y}$. If $a, b, c$ can be the lengths of the three sides of a triangle, then the range of the positive real number $\lambda$ is $\qquad$ | $2.1b$.
If $a, b, c$ can be the lengths of the three sides of a triangle, then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+y+\sqrt{x^{2}-x y+y^{2}}>\lambda \sqrt{x y}, \\
\lambda \sqrt{x y}+\sqrt{x^{2}-x y+y^{2}}>x+y
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
\lambda\frac{x+y-\sqrt{x^{2}-x y+y^{2}}}... | 1<\lambda<3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,933 |
3. In a regular quadrilateral pyramid $P-A B C D$, it is known that $A_{1} 、 C_{1}$ are the midpoints of $P A 、 P C$ respectively. Then
$$
\frac{V_{\text {=floor } A_{1}}-B C_{1} D}{V_{\text {regular quadrilateral pyramid } P-A B C D}}=
$$
$\qquad$ | 3. $\frac{1}{4}$.
Notice,
$$
\begin{array}{l}
\text { Also } V_{=B-A_{1} C_{1} P}=\frac{1}{4} V_{=} B-A C P \\
=\frac{1}{4} V_{=}-A B C=\frac{1}{8} V_{=}-A B C C .
\end{array}
$$ | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,934 |
4. Given the inequality about $x$
$$
(2 x-1)^{2}<a x^{2}
$$
has exactly $n$ positive integer solutions. Then the range of the real number $a$ is $\qquad$ . | $$
\begin{array}{l}
\text { 4. }\left(2-\frac{1}{n}\right)^{2}<a \leqslant\left(2-\frac{1}{n+1}\right)^{2} \text {. } \\
\text { Let } f(x)=|2 x-1|, g(x)=\sqrt{a}|x| \text {. }
\end{array}
$$
If the solution set of $f(x)<g(x)$ contains exactly $n$ positive integers, then they must be $1,2, \cdots, n$. Therefore,
$$
\b... | \left(2-\frac{1}{n}\right)^{2}<a \leqslant\left(2-\frac{1}{n+1}\right)^{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,935 |
5. Given that the derivative function $f^{\prime}(x)$ of $f(x)$ is continuous, and $f(0)=0, f^{\prime}(0)=a$. Let the point $Q(s, f(s))$ be the closest point on the curve $y=f(x)$ to $P(t, 0)$. Then $\lim _{t \rightarrow 0} \frac{s}{t}=$ | 5. $\frac{1}{1+a^{2}}$.
Let $y=P Q^{2}=(x-t)^{2}+f^{2}(x)$. Then $y^{\prime}=2(x-t)+2 f(x) f^{\prime}(x)$.
From the given information,
$$
2(s-t)+2 f(s) f^{\prime}(s)=0 \text {. }
$$
Then $\frac{s}{t}=1-\frac{f(s) f^{\prime}(s)}{t}=1-\frac{s}{t} \cdot \frac{f(s)}{s} f^{\prime}(s)$.
Let $\lim _{t \rightarrow 0} \frac{s... | \frac{1}{1+a^{2}} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 726,936 |
6. Given that the three vertices of $\triangle A B C$ are all on the parabola $y^{2}=2 p x(p>0)$, and the centroid of $\triangle A B C$ is exactly the focus of the parabola. If the equation of the line on which side $B C$ lies is $4 x+y$ $-20=0$, then $p=$ $\qquad$ . | 6. $p=8$.
Let $A\left(\frac{y_{1}^{2}}{2 p}, y_{1}\right), B\left(\frac{y_{2}^{2}}{2 p}, y_{2}\right), C\left(\frac{y_{3}^{2}}{2 p}, y_{3}\right)$.
$$
\begin{array}{l}
\text { From }\left\{\begin{array}{l}
y^{2}=2 p x, \\
4 x+y-20=0
\end{array}\right. \\
\Rightarrow 2 y^{2}+p y-20 p=0 .
\end{array}
$$
From the given ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,937 |
7. In a $100 \times 100$ square grid, any rectangle parallel to the grid sides is chosen. The probability that this rectangle is a square is | 7. $\frac{67}{5050}$.
In an $n \times n$ square grid, the number of squares with side length $k$ is $(n+1-k)^{2}$, so there are
$$
\sum_{k=1}^{n}(n+1-k)^{2}=\frac{n(n+1)(2 n+1)}{6}
$$
squares in total.
Therefore, the required probability is
$$
\begin{array}{r}
P=\frac{\frac{n(n+1)(2 n+1)}{6}}{\left(\mathrm{C}_{n+1}^{... | \frac{67}{5050} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,938 |
8. $\sum_{0 \leqslant i<j \leqslant 50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}$ modulo 31 is | 8. 16 .
Notice,
$$
\begin{array}{l}
\sum_{0 \leqslant i<j \leqslant 50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}=\frac{1}{2} \sum_{0 \leqslant i \neq j \leqslant 50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j} \\
=\frac{1}{2}\left[\sum_{i=0}^{50} \sum_{j=0}^{50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}-\sum_{j=0}^{50}\left(... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,939 |
9. (16 points) Given that $P$ is any point on the directrix of the parabola $x^{2}=4 y$, a tangent line is drawn from $P$ to the parabola, with points of tangency being $A$ and $B$.
(1) Prove: The line $A B$ passes through a fixed point;
(2) Find the minimum value of the area enclosed by the line $A B$ and the parabola... | (1) Let points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and $P(m,-1)$. The equations of the tangent lines passing through points $A$ and $B$ are
$$
y=\frac{x_{1}}{2} x-\frac{x_{1}^{2}}{4}, \quad y=\frac{x_{2}}{2} x-\frac{x_{2}^{2}}{4}.
$$
Since the tangent lines pass through point $P$, we have
$$
-1=\... | \frac{8}{3} | Algebra | proof | Yes | Yes | cn_contest | false | 726,940 |
10. (20 points) Given real numbers
$$
x_{i} \in[-6,10](i=1,2, \cdots, 10), \sum_{i=1}^{10} x_{i}=50 \text {. }
$$
Find the maximum value of $\sum_{i=1}^{10} x_{i}^{2}$, and the conditions that should be satisfied when the maximum value is achieved. | 10. Let $a_{i}=x_{i}+6$. Then $a_{i} \in[0,16]$, and
$$
\begin{array}{l}
\sum_{i=1}^{10} a_{i}=110, \\
\sum_{i=1}^{10} a_{i}^{2}=\sum_{i=1}^{10} x_{i}^{2}+12 \sum_{i=1}^{10} x_{i}+360 \\
=\sum_{i=1}^{10} x_{i}^{2}+960 .
\end{array}
$$
When there are at least two numbers in $a_{i}$ that are not both 0 and not both 16, ... | 772 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,941 |
Example 1 Let $n$ be an odd number greater than 1, and let $c_{1}, c_{2}, \cdots, c_{n}$ be integers. For every permutation $a=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ of $\{1,2, \cdots, n\}$, define $s(a)=\sum_{i=1}^{n} c_{i} a_{i}$. Prove that there exist two different permutations $a$ and $b$ of $\{1,2, \cdots, n\}... | Prove that $\sum_{a} s(a)$ represents the sum of $n!$ permutations $s(a)$ of $\{1,2, \cdots, n\}$.
Calculate $\sum_{a} s(a)(\bmod n!)$ in two ways.
First, assume the conclusion is false, so for each $a, s(a)$ modulo $n!$ has different remainders. Therefore, these remainders are the numbers $0,1, \cdots, n!-1$. Then,
$$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,942 |
Example 2 Proof: For each prime $p>100$ and each integer $r$, there exist integers $a, b$ such that
$$
p \mid\left(a^{2}+b^{5}-r\right) \text {. }
$$
(53rd IMO Shortlist Problem) | Prove that for each $r \in \mathbf{Z}_{p}$, let $S(r)=\left\{(a, b) \in \mathbf{Z}_{p}^{2} \mid\left(a^{2}+b^{5}\right) \equiv r(\bmod p)\right\}$.
The operations are under modulo $p$.
We only need to prove: each $S(r)$ is non-empty.
The basic idea to solve the problem is to count the number of quadruples in two differ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,943 |
Example 2 If $\cos ^{5} \theta-\sin ^{5} \theta<7\left(\sin ^{3} \theta-\cos ^{3} \theta\right)$, where $\theta \in[0,2 \pi)$, then the range of values for $\theta$ is $\qquad$ [2]
(2011, National High School Mathematics Joint Competition) | Solve the original inequality, which is equivalent to
$$
\begin{array}{l}
7 \sin ^{3} \theta+\sin ^{5} \theta>7 \cos ^{3} \theta+\cos ^{5} \theta . \\
\text { Let } f(x)=7 x^{3}+x^{5} .
\end{array}
$$
Notice that, $f^{\prime}(x)=21 x^{2}+5 x^{4} \geqslant 0$ for all $x \in(-\infty,+\infty)$, hence $f(x)$ is an increas... | \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,944 |
Example 3 Let $a, b, c$ be the lengths of the three sides of a right-angled triangle, where $c$ is the length of the hypotenuse. Find the maximum value of $k$ such that $\frac{a^{3}+b^{3}+c^{3}}{a b c} \geqslant k$ holds. ${ }^{[3]}$
(Fourth Northern Mathematical Olympiad Invitational Competition) | Given $a^{2}+b^{2}=c^{2}$, let
$$
\begin{array}{l}
a=c \cos \theta, b=c \sin \theta\left(\theta \in\left(0, \frac{\pi}{2}\right)\right) . \\
\text { Then } f=\frac{a^{3}+b^{3}+c^{3}}{a b c} \\
=\frac{(\sin \theta+\cos \theta)^{3}-3 \sin \theta \cdot \cos \theta(\sin \theta+\cos \theta)+1}{\sin \theta \cdot \cos \theta}... | 2+\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,945 |
Example 4 In $\triangle A B C$, prove:
$$
\begin{array}{l}
\frac{1}{1+\cos ^{2} A+\cos ^{2} B}+\frac{1}{1+\cos ^{2} B+\cos ^{2} C}+ \\
\frac{1}{1+\cos ^{2} C+\cos ^{2} A} \leqslant 2 \text {. }
\end{array}
$$
(7th Northern Mathematical Olympiad Invitational Competition) | Prove that by the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
\sin ^{2} C=\sin ^{2}(A+B) \\
=(\sin A \cdot \cos B+\cos A \cdot \sin B)^{2} \\
\leqslant\left(\sin ^{2} A+\sin ^{2} B\right)\left(\cos ^{2} A+\cos ^{2} B\right) .
\end{array}
$$
Then, $\cos ^{2} A+\cos ^{2} B \geqslant \frac{\sin ^{2} C}{\sin ^{2} A+\si... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,946 |
Example 5 Function
$$
y=\sqrt{\sin x}+\sqrt{\cos x}\left(0 \leqslant x \leqslant \frac{\pi}{2}\right)
$$
The range of the function is $\qquad$
(2001, Shanghai Jiao Tong University Independent Admission Examination) | Notice,
$$
\begin{array}{l}
y^{4}=(\sqrt{\sin x}+\sqrt{\cos x})^{4} \\
=[(\sqrt{\sin x}+\sqrt{\cos x})(\sqrt{\sin x}+\sqrt{\cos x})]^{2} \\
\geqslant(\sin x+\cos x)^{2} \\
\geqslant \sin ^{2} x+\cos ^{2} x=1\left(x=0 \text { or } \frac{\pi}{2} \text { when equality holds }\right), \\
y^{4}=(\sqrt{\sin x}+\sqrt{\cos x})... | 1 \leqslant y \leqslant 8^{\frac{1}{4}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,947 |
Example 7 Given $\tan \alpha+\tan \beta+\tan \gamma=\frac{17}{6}$, $\cot \alpha+\cot \beta+\cot \gamma=-\frac{4}{5}$, $\cot \alpha \cdot \cot \beta+\cot \beta \cdot \cot \gamma+\cot \gamma \cdot \cot \alpha=-\frac{17}{5}$.
Then $\tan (\alpha+\beta+\gamma)=$ $\qquad$ [5]
(2012, Xin Zhi Cup Shanghai High School Mathemati... | Let $\tan \alpha=x, \tan \beta=y, \tan \gamma=z$.
Then $x+y+z=\frac{17}{6}$,
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{4}{5}$,
$\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=-\frac{17}{5}$.
(1) $\div$ (3) gives
$x y z=-\frac{5}{6}$.
(2) $\times$ (4) gives $x y+y z+z x=\frac{2}{3}$.
Therefore, $\tan (\alpha+\beta+\gamma)... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,948 |
For example, in $\triangle ABC$, the measures of the three interior angles satisfy
$$
\frac{\angle A}{\angle B}=\frac{\angle B}{\angle C}=\frac{1}{3} \text {. }
$$
Find the value of $T=\cos A+\cos B+\cos C$. ${ }^{[6]}$
(2011, National High School Mathematics League Shanxi Province Preliminary Contest) | Let $\angle A=\theta$. Then $\angle B=3 \theta, \angle C=9 \theta$.
$$
\begin{array}{l}
\text { By } \theta+3 \theta+9 \theta=\pi \Rightarrow \theta=\frac{\pi}{13} \text {. } \\
\text { Hence } T=\cos \theta+\cos 3 \theta+\cos 9 \theta \\
=\cos \theta+\cos 3 \theta-\cos 4 \theta \\
=2 \cos \theta \cdot \cos 2 \theta-2 ... | \frac{1+\sqrt{13}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,949 |
Example 9 Let $\sin \alpha + \cos \alpha = a (0 \leqslant a \leqslant \sqrt{2})$. Find the expression for $\sin ^{n} \alpha + \cos ^{n} \alpha$ in terms of $a$.
(2005, Fudan University Independent Admission Examination) | Given $\sin \alpha + \cos \alpha = a$
$$
\Rightarrow \sin \alpha \cdot \cos \alpha = \frac{a^2 - 1}{2}
$$
$\Rightarrow \sin \alpha, \cos \alpha$ are the roots of the equation $t^2 - a t + \frac{a^2 - 1}{2} = 0$
$$
\Rightarrow \sin \alpha = \frac{a + \sqrt{2 - a^2}}{2}, \cos \alpha = \frac{a - \sqrt{2 - a^2}}{2}
$$
(can... | \left( \frac{a + \sqrt{2 - a^2}}{2} \right)^n + \left( \frac{a - \sqrt{2 - a^2}}{2} \right)^n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,950 |
The third question: An exam consists of $m$ questions, with $n$ students participating, where $m, n \geqslant 2$ are given integers. The scoring rule for each question is: if exactly $x$ students fail to answer the question correctly, then each student who answers the question correctly gets $x$ points, and those who f... | Solution 1 First, use mathematical induction to prove the inequality $p_{1}+p_{n} \leqslant m(n-1)$ for a positive integer $m$.
(1) When $m=1$, if the student who gets $p_{n}$ points answers this question correctly, then the other students should also answer this question correctly.
At this time, $p_{1}=p_{2}=\cdots=p_... | m(n-1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,953 |
Example 3 Find the value of $\sum_{k=1}^{\frac{n(n+1)}{2}}\left[\frac{-1+\sqrt{1+8 k}}{2}\right]$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Consider $f(x)=\frac{x(x+1)}{2}:[1, n] \rightarrow \left[1, \frac{n(n+1)}{2}\right]$
then $f^{-1}(x)=\frac{-1+\sqrt{1+8 x}}{2}$.
Consider the integer points within the rectangular region $x=1, x=n$, $y=1, y=\frac{n(n+1)}{2}$,
as shown in Figure 1.
It is easy to see that the number of integer points within the rectangle... | \frac{n\left(n^{2}+1\right)}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,954 |
Example 1 Suppose
$$
(1+\sqrt{2})^{n}=x_{n}+y_{n} \sqrt{2}\left(x_{n}, y_{n} \in \mathbf{Z}\right) \text {. }
$$
Find $\lim _{n \rightarrow+\infty} \frac{x_{n}}{y_{n}}$.
(2000, Fudan University Admissions Exam for Recommended Students) | Solution 1 Construct the dual form.
Consider the dual form $(1-\sqrt{2})^{n}$, by the binomial theorem we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
(1+\sqrt{2})^{n}=x_{n}+y_{n} \sqrt{2}, \\
(1-\sqrt{2})^{n}=x_{n}-y_{n} \sqrt{2}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x_{n}=\frac{1}{2}\left[(1+\s... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,956 |
Example 2 Proof: For any positive integer $n$, $(1+\sqrt{2})^{n}$ can always be expressed in the form $\sqrt{s}+\sqrt{s-1}\left(s \in \mathbf{Z}_{+}\right)$.
(2012, Joint Autonomous Admission Examination of Peking University and Other Universities) | Prove that
$$
(1+\sqrt{2})^{n}=a_{n}+b_{n} \sqrt{2}\left(a_{n}, b_{n} \in \mathbf{Z}_{+}\right) \text {. }
$$
Then $(1-\sqrt{2})^{n}=a_{n}-b_{n} \sqrt{2}$.
Adding, subtracting, and multiplying the above two equations, we get
$$
\left\{\begin{array}{l}
a_{n}=\frac{1}{2}\left[(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}\right], \\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,957 |
Example 3 Given $a_{n}=(2+\sqrt{7})^{2 n+1}, b_{n}$ is the fractional part of $a_{n}$. Then when $n \in \mathbf{Z}_{+}$, the value of $a_{n} b_{n}$ ( ). ${ }^{[1]}$
(A) must be an irrational number
(B) must be an even number
(C) must be an odd number
(D) can be either an irrational number or a rational number
$(2008$, ... | Notice that,
$$
\begin{aligned}
(\sqrt{7} & +2)^{2 n+1}-(\sqrt{7}-2)^{2 n+1} \\
& =2\left(\mathrm{C}_{2 n+1}^{0} 2^{2 n+1}+\mathrm{C}_{2 n+1}^{2} 2^{2 n-1} \times 7+\cdots+\right. \\
& \left.\mathrm{C}_{2 n+1}^{2 n} 2 \times 7^{n}\right) \in \mathbf{Z}_{+} .
\end{aligned}
$$
While $0<(\sqrt{7}-2)^{2 n+1}<1$, hence $(\... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,958 |
Example 4 Try to find the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$.
[2] | Notice that,
$$
(\sqrt{2}+\sqrt{3})^{2012}=(5+2 \sqrt{6})^{1006} \text {. }
$$
Clearly, $0<(5-2 \sqrt{6})^{1006}<1$,
$$
\begin{array}{l}
(5+2 \sqrt{6})^{1006}+(5-2 \sqrt{6})^{1006} \\
=2\left(C_{1006}^{0} 5^{1006}+C_{1006}^{2} 5^{1004} \times 24+\cdots+\right. \\
\left.\quad C_{1006}^{1006} 24^{503}\right) \in \mathbf... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,959 |
Example 5 Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=1$,
$$
\begin{aligned}
a_{1}=4, a_{2}=49 \text {, and } \\
\left\{\begin{array}{l}
a_{n+1}=7 a_{n}+6 b_{n}-3, \\
b_{n+1}=8 a_{n}+7 b_{n}-4
\end{array}(n=0,1, \cdots) .\right.
\end{aligned}
$$
Prove: $a_{n}(n=0,1, \cdots)$ is a... | Prove that from $b_{n}=\frac{a_{n+1}-7 a_{n}+3}{6}$, we know
$$
b_{n+1}=\frac{a_{n+2}-7 a_{n+1}+3}{6} .
$$
Substituting the condition, we get
$$
\frac{a_{n+2}-7 a_{n+1}+3}{6}=8 a_{n}+7 \cdot \frac{a_{n+1}-7 a_{n}+3}{6}-4 \text {. }
$$
Simplifying, we obtain $a_{n+2}=14 a_{n+1}-a_{n}-6$.
By the method of undetermined ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,960 |
Example 6 The general term formula of the sequence $\left\{f_{n}\right\}$ is
$$
f_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Let $S_{n}=\mathrm{C}_{n}^{1} f_{1}+\mathrm{C}_{n}^{2} f_{2}+\cdots+\mathrm{C}_... | Let $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$. Then
$$
\begin{array}{l}
S_{n}=\frac{1}{\sqrt{5}} \sum_{i=1}^{n} \mathrm{C}_{n}^{i}\left(\alpha^{i}-\beta^{i}\right) \\
=\frac{1}{\sqrt{5}} \sum_{i=0}^{n} \mathrm{C}_{n}^{i}\left(\alpha^{i}-\beta^{i}\right) \\
=\frac{1}{\sqrt{5}}\left(\sum_{i=0}^{n} \mathrm... | 3|n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,961 |
Example 7 Let $x=(15+\sqrt{220})^{19}+(15+\sqrt{200})^{22}$. Find the unit digit of the number $x$.
| Solve the conjugate expression
$$
y=(15-\sqrt{220})^{19}+(15-\sqrt{220})^{82} \text {. }
$$
Then $x+y$
$$
\begin{aligned}
= & (15+\sqrt{220})^{19}+(15-\sqrt{220})^{19}+ \\
& (15+\sqrt{220})^{82}+(15-\sqrt{220})^{82} .
\end{aligned}
$$
By the binomial theorem, for any $n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,962 |
1. Let $A$ be a set of integers. If for any $x, y \in A$ (allowing $x, y$ to be the same), and for every integer $k$, we have $x^{2}+k x y+y^{2} \in A$, then the set $A$ is called "admissible". Find all non-zero integer pairs $(m, n)$ such that the only admissible set containing $m, n$ is the set of all integers. | 1. The integer pair $(m, n)$ satisfies the condition if and only if $(m, n)=1$.
Assume $(m, n)=d>1$. Then the set
$$
A=\{\cdots,-2 d,-d, 0, d, 2 d, \cdots\}
$$
is allowed, because if $d$ divides $x, y$, then for each integer $k$, we have $d \mid\left(x^{2}+k x y+y^{2}\right)$.
Clearly, $m, n \in A, A \neq \mathbf{Z}$.... | (m, n)=1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,963 |
Example 4 Positive integers $a_{1}, a_{2}, \cdots, a_{2006}$ (allowing repetition). What is the minimum number of distinct numbers in $a_{2}, \cdots, a_{2006}$? ${ }^{[1]}$
(2006, China Mathematical Olympiad) | This problem is solved using combinatorial geometry methods.
Let the positive integers $a_{1}, a_{2}, \cdots, a_{n}$ be such that $\frac{a_{i}}{a_{i+1}} (1 \leqslant i \leqslant n-1)$ are all distinct. Then the minimum number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{n}$ is $k=\left\lceil\frac{1+\sqrt{4 n-7}}{2}... | 46 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,964 |
3. Find all integers $m(m \geqslant 2)$, such that every integer $n\left(\frac{m}{3} \leqslant n \leqslant \frac{m}{2}\right)$ divides the binomial coefficient $\mathrm{C}_{n}^{m-2 n}$. | 3. The integer $m$ that satisfies the condition is all prime numbers.
First, verify that all prime numbers satisfy the condition.
A stronger conclusion can be proven:
If $p$ is a prime number, then for each integer $n\left(1 \leqslant n \leqslant \frac{p}{2}\right)$, $n \mid \mathrm{C}_{n}^{p-2 n}$.
If $p=2$, then $n=... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,965 |
4. For an integer $a$, if the equation
$$
\left(m^{2}+n\right)\left(n^{2}+m\right)=a(m-n)^{3}
$$
has positive integer solutions, then the integer $a$ is called "friendly".
(1) Prove: The set $\{1,2, \cdots, 2012\}$ contains at least 500 friendly integers;
(2) Determine whether $a=2$ is friendly. | 4. (1) Integers of the form $a=4 k-3\left(k \geqslant 2, k \in \mathbf{Z}_{+}\right)$ are all friendly.
In fact, $m=2 k-1>0$ and $n=k-1>0$ satisfy the equation (1) when $a=4 k-3$, i.e.,
$$
\begin{array}{l}
\left(m^{2}+n\right)\left(n^{2}+m\right) \\
=\left[(2 k-1)^{2}+(k-1)\right]\left[(k-1)^{2}+(2 k-1)\right] \\
=(4 ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,966 |
5. For every non-negative integer $n$, define $\operatorname{rad}(n)$ as follows: if $n=0$ or 1, then $\operatorname{rad}(n)=1$; if the prime factors of $n$ are $p_{1}, p_{2}, \cdots, p_{k}\left(p_{1}<p_{2}<\cdots<p_{k}\right)$, then $\operatorname{rad}(n)=p_{1} p_{2} \cdots p_{k}$. Find all non-negative integer coeffi... | 5. First, we explain: $f(x)=a x^{m}(a, m \in \mathbf{N})$ satisfies the condition.
If $f(x)$ is the zero polynomial, the conclusion is obviously true.
If $f(x)$ has at least one positive coefficient, then $f(1)>0$.
Let $p$ be a prime. Then
$$
\begin{array}{l}
f(n) \equiv 0(\bmod p) \\
\Rightarrow f\left(n^{\operatornam... | f(x)=a x^{m} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,967 |
6. Let $x, y \in \mathbf{Z}_{+}$. If for every positive integer $n$, we have $\left(2^{n} y+1\right) \mid\left(x^{2 n}-1\right)$, prove: $x=1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 6. First prove: For each positive integer $y$, there exist infinitely many primes $p \equiv 3(\bmod 4)$, such that $p$ divides some integer of the form $2^{n} y+1$.
It suffices to consider the case where $y$ is odd.
Let $2 y+1=p_{1}^{\varepsilon_{1}} p_{2}^{e_{2}} \cdots p_{r}^{e_{r}}$ be the prime factorization of $2 ... | proof | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,968 |
8. Prove: For each prime $p(p>100)$ and each integer $r$, there exist integers $a, b$ such that
$$
p \mid\left(a^{2}+b^{5}-r\right) .
$$ | 8. All congruence relations in the entire solution are modulo $p$.
Fix $p$, and let $\mathscr{P}=\{0,1, \cdots, p-1\}$ be the set of residue classes modulo $p$. For each $r \in \mathscr{P}$, set
$S_{r}=\left\{(a, b) \in \mathscr{P} \times \mathscr{P} \mid a^{2}+b^{5} \equiv r\right\}$,
$s_{r}=|S_{r}|$.
We need to prov... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,969 |
$$
\begin{array}{l}
A=\{2,0,1,3\}, \\
B=\left\{x \mid -x \in A, 2-x^{2} \notin A\right\} .
\end{array}
$$
Then the sum of all elements in set $B$ is | ,$- 1 .-5$.
It is easy to know that $B \subseteq\{-2,0,-1,-3\}$.
When $x=-2,-3$, $2-x^{2}=-2,-7 \notin A$; when $x=0,-1$, $2-x^{2}=2,1 \in A$.
Therefore, the set $B=\{-2,-3\}$.
Thus, the sum of all elements in set $B$ is -5. | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,970 |
2. In the Cartesian coordinate system $x O y$, it is known that points $A$ and $B$ lie on the parabola $y^{2}=4 x$, and satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B}=-4, F$ is the focus of the parabola. Then $S_{\triangle O F A} \cdot S_{\triangle O F B}=$ $\qquad$ . | 2. 2 .
From the problem, we know the point $F(1,0)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Thus, $x_{1}=\frac{y_{1}^{2}}{4}, x_{2}=\frac{y_{2}^{2}}{4}$.
Then $-4=\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}+y_{1} y_{2}$
$$
\begin{array}{l}
=\frac{1}{16}\left(y_{1} y_{2}\right)^{2... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,971 |
3. In $\triangle A B C$, it is known that
$$
\sin A=10 \sin B \cdot \sin C, \cos A=10 \cos B \cdot \cos C \text {. }
$$
Then $\tan A=$ $\qquad$ | 3. 11 .
From $\sin A-\cos A$
$$
\begin{array}{l}
=10(\sin B \cdot \sin C-\cos B \cdot \cos C) \\
=-10 \cos (B+C)=10 \cos A \\
\Rightarrow \sin A=11 \cos A \\
\Rightarrow \tan A=11 .
\end{array}
$$ | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,972 |
4. Given a regular tetrahedron $P-ABC$ with a base edge length of 1 and a height of $\sqrt{2}$. Then its inscribed sphere radius is $\qquad$ . | 4. $\frac{\sqrt{2}}{6}$.
As shown in Figure 2, let the projections of the center of the sphere $O$ onto the planes $ABC$ and $ABP$ be $H$ and $K$ respectively, and let the midpoint of side $AB$ be $M$. The inradius of the sphere is $r$. Then, $P, K, M$ and $P, O, H$ are collinear, $\angle PHM = \angle PKO = \frac{\pi}... | \frac{\sqrt{2}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,973 |
Example 5 For an integer $n(n \geqslant 4)$, find the smallest integer $f(n)$ such that for any positive integer $m$, in any $f(n)$-element subset of the set $\{m$, $m+1, \cdots, m+n-1\}$, there are at least three pairwise coprime elements.
$(2004$, National High School Mathematics Competition) | Let $m=2$.
By the principle of inclusion-exclusion, the number of elements in the set $\{2,3, \cdots, n+1\}$ that are divisible by 2 or 3 is
$$
\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right] \text {. }
$$
Any three of these numbers will either have two that are both divisible by 2, or... | f(n)=\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,974 |
6. From $1,2, \cdots, 20$, choose any five different numbers, the probability that at least two of them are consecutive is $\qquad$ | 6. $\frac{232}{323}$.
Let $a_{1}<a_{2}<\cdots<a_{5}$ be chosen from $1,2, \cdots, 20$.
If $a_{1}, a_{2}, \cdots, a_{5}$ are not adjacent, then
$$
1 \leqslant a_{1}<a_{2}-1<a_{3}-2<a_{4}-3<a_{5}-4 \leqslant 16 \text {. }
$$
This implies that the number of ways to choose five non-adjacent numbers from $1,2, \cdots, 20$... | \frac{232}{323} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,975 |
7. If real numbers $x, y$ satisfy $x-4 \sqrt{y}=2 \sqrt{x-y}$, then the range of values for $x$ is $\qquad$ . | 7. $\{0\} \cup[4,20]$.
Let $\sqrt{y}=a, \sqrt{x-y}=b(a, b \geqslant 0)$, at this point,
$$
x=y+(x-y)=a^{2}+b^{2},
$$
and the given equation becomes
$$
a^{2}+b^{2}-4 a=2 b.
$$
Thus, $a, b$ satisfy the equation
$$
(a-2)^{2}+(b-1)^{2}=5(a, b \geqslant 0).
$$
As shown in Figure 3, in the $aOb$ plane, the trajectory of p... | \{0\} \cup[4,20] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,976 |
8. Given a sequence $\left\{a_{n}\right\}$ with nine terms, where $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in \left\{2,1,-\frac{1}{2}\right\}$. Then the number of such sequences is | 8.491.
Let $b_{i}=\frac{a_{i+1}}{a_{i}}(1 \leqslant i \leqslant 8)$. Then for each sequence $\left\{a_{n}\right\}$ that meets the conditions, it satisfies
$$
\prod_{i=1}^{8} b_{i}=\prod_{i=1}^{8} \frac{a_{i+1}}{a_{i}}=\frac{a_{9}}{a_{1}}=1,
$$
and $b_{i} \in\left\{2,1,-\frac{1}{2}\right\}(1 \leqslant i \leqslant 8)$.... | 491 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,977 |
9. (16 points) Given a sequence of positive numbers $\left\{x_{n}\right\}$ satisfying
$$
S_{n} \geqslant 2 S_{n-1}(n=2,3, \cdots) \text {, }
$$
where, $S_{n}=x_{1}+x_{2}+\cdots+x_{n}$. Prove: There exists a constant $C>0$, such that
$$
x_{n} \geqslant 2^{n} C(n=1,2, \cdots) \text {. }
$$ | When $n \geqslant 2$,
$$
S_{n} \geqslant 2 S_{n-1} \Leftrightarrow x_{n} \geqslant x_{1}+x_{2}+\cdots+x_{n-1} \text {. }
$$
For the constant $C=\frac{1}{4} x_{1}$, we will use mathematical induction to prove:
$$
x_{n} \geqslant 2^{n} C(n=1,2, \cdots) \text {. }
$$
When $n=1$, the conclusion is obviously true.
$$
\tex... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,978 |
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