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5. A one-to-one correspondence from the set of integers $\{\cdots,-2,-1,0,1,2, \cdots$ to the set of natural numbers $\{0,1,2, \cdots\}$ is $\qquad$ | 5. $f(n)=\left\{\begin{array}{ll}2 n, & n \in \mathbf{N} ; \\ -1-2 n, & n \in \mathbf{Z}_{-} .\end{array}\right.$
Let non-negative integers correspond one-to-one with even numbers in the set of natural numbers, and let negative integers correspond one-to-one with odd numbers in the set of natural numbers. This gives a... | f(n)=\left\{\begin{array}{ll}2 n, & n \in \mathbf{N} ; \\ -1-2 n, & n \in \mathbf{Z}_{-} .\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,878 |
6. Read the proof of the following problem and point out one mistake.
Problem: There are six points on a plane, any three of which are the vertices of a triangle with three unequal sides. Then, among these triangles, there is one whose shortest side is also the longest side of another triangle.
Proof: Step 1, connect... | 6. The third step is incorrect, the reason being: it cannot be deduced that “any three edges can form a triangle” or the second step is incorrect, the reason being: it cannot be deduced that $a_{1}<a_{2}<\cdots<a_{15}$.
The third step is incorrect. Because from “any three points are vertices of a triangle with unequal... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,880 |
Three. (20 points) Through the focus $\boldsymbol{F}(c, 0)$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, draw the line $y=(x-c) \tan \theta\left(\frac{\pi}{2}<\theta<\pi\right)$ intersecting the two directrices of the hyperbola at points $A$ and $B$. If the circle with $A B$ as its diameter passes thro... | Solution 1: The directrices of the hyperbola are $x= \pm \frac{a^{2}}{c}$.
Substituting into the line equation, we get
$$
\begin{array}{l}
A\left(\frac{a^{2}}{c},-\frac{c^{2}-a^{2}}{c} \tan \theta\right), \\
B\left(-\frac{a^{2}}{c},-\frac{c^{2}+a^{2}}{c} \tan \theta\right) .
\end{array}
$$
Since the circle with diamet... | \theta=\pi-\arcsin \frac{a^{2}}{c^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,881 |
(1) On a one-dimensional line, a line segment is a closed centrally symmetric figure, with Proposition 1: Two distinct points determine a line segment;
(2) On a two-dimensional plane, a circle is a closed centrally symmetric figure, with Proposition 2: Three non-collinear points determine a circle;
(3) In three-dimensi... | Four, the answer is affirmative.
From the fact that points $A, B, C, D$ are not coplanar, it follows that points $B, C, D$ are not collinear. According to Proposition 2, points $B, C, D$ can determine a circle, with the center of the circle denoted as $E$, and the plane determined by $B, C, D$ denoted as $\alpha$. Cons... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,882 |
Five. (20 points) Given non-zero real numbers $a, b, c, t$ satisfy
$$
\left\{\begin{array}{l}
a=t b+c, \\
b=c\left(1+t+t^{2}\right) .
\end{array}\right.
$$
(1) Prove that the quadratic equation
$$
c x^{2}+c(b-2 c) x-\left(b^{2}+c^{2}\right)(b-c)=0
$$
must have real roots, and $2 c-b-a$ is one of the real roots;
(2) Wh... | (1) Solution 1: From $b=c\left(1+t+t^{2}\right)$, we have
$$
\begin{array}{l}
4 b c=4 c^{2}\left(1+t+t^{2}\right) \\
=3 c^{2}+c^{2}(1+2 t)^{2} \geqslant 3 c^{2},
\end{array}
$$
Thus, the discriminant of the quadratic equation is
$$
\begin{array}{l}
\Delta=c^{2}(b-2 c)^{2}+4 c\left(b^{2}+c^{2}\right)(b-c) \\
=b^{2}\lef... | c=1, t=2 | Algebra | proof | Yes | Yes | cn_contest | false | 720,883 |
一、(50 points) As shown in Figure 2, line segments $AC$ and $BD$ intersect at point $G$. Take any point $E$ on the extension of $AD$, forming quadrilateral $DBCE$. Prove that the circumcircles of $\triangle ACE$, $\triangle BCG$, and $\triangle BDE$ are concurrent. | Let the circumcircles of $\triangle A C E$ and $\triangle B C G$ be circles $\omega_{1}$ and $\omega_{2}$, respectively. Since the two circles already have a common point $C$, their positions are either tangent or intersecting.
(1) Circles $\omega_{1}$ and $\omega_{2}$ are tangent.
Since point $G$ is inside circle $\om... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,884 |
Three. (50 points) Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms as $S_{n}, a_{1}=1$ and $2 S_{n}=a_{n} a_{n+1}\left(n \in \mathbf{N}_{+}\right)$.
(1) Prove: $a_{n} \neq 0\left(n \in \mathbf{N}_{+}\right)$, and find the general term formula for $\left\{a_{n}\right\}$;
(2) Construct the s... | (iii) From $a_{1}=1$ and $2 S_{1}=a_{1} a_{2}$, we get $a_{2}=2$. From $a_{1}=1, a_{2}=2$ and $2 S_{2}=a_{2} a_{3}$, we get $a_{3}=3$. Below, we use mathematical induction to prove:
$$
a_{n}=n\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
i.e., $\left\{\begin{array}{l}a_{2 k-1}=2 k-1, \\ a_{2 k}=2 k\end{array}\left(k... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,886 |
Find all integers $n$ such that $\frac{n^{3}-n+5}{n^{2}+1}$ is an integer.
Initial 241 | Solution: Notice that
$$
\begin{array}{l}
n^{3}-n+5=n\left(n^{2}+1\right)-(2 n-5) . \\
\text { Then } \frac{n^{3}-n+5}{n^{2}+1} \in \mathbf{Z} \Leftrightarrow \frac{2 n-5}{n^{2}+1} \in \mathbf{Z} \\
\Leftrightarrow \frac{n(2 n-5)}{n^{2}+1} \in \mathbf{Z}\left(n \text { and } n^{2}+1 \text { are coprime }\right) \\
\Lef... | n=0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,887 |
As shown in Figure 2, $\triangle ABC$ is an acute triangle. The circle with $AB$ as its diameter intersects $AC$ at point $D$ and intersects the altitude $CH$ of side $AB$ at points $E$ and $F$. The semicircle with $AC$ as its diameter intersects the extension of $BD$ at point $G$. Connecting $GF$ intersects the circle... | Proof: As shown in Figure 2, connect $A E$, $A F$, $A G$, and $E G$.
By the projection theorem, we have
$$
\begin{array}{l}
A F^{2}=A H \cdot A B, \\
A G^{2}=A D \cdot A C .
\end{array}
$$
It is easy to see that points $B$, $C$, $D$, and $H$ are concyclic. Therefore,
$$
A H \cdot A B=A D \cdot A C \text {. }
$$
From ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,888 |
Given a height of 241, find the minimum value of the function
$$
S_{n}=|n-1|+2|n-2|+\cdots+100|n-100|
$$
$\left(n \in \mathbf{N}_{+}\right)$. | Solution: When $n \geqslant 100$,
$$
\begin{array}{l}
S_{n}=(n-1)+2(n-2)+\cdots+100(n-100) \\
=n(1+2+\cdots+100)-\left(1^{2}+2^{2}+\cdots+10^{2}\right) \\
\geqslant S_{100} .
\end{array}
$$
Therefore, the minimum value of $S_{n}$ can only be obtained when $1 \leqslant n \leqslant 100$.
When $2 \leqslant n \leqslant 10... | 99080 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,889 |
Example 1 Find four distinct natural numbers such that the sum of any two of them can be divided by their difference. If the sum of the largest and smallest of these four numbers is to be minimized, what is the sum of the middle two numbers?
(3rd Hua Luogeng Cup) | Analysis: Let $a_{1}, a_{2}, a_{3}, a_{4}$ be four numbers that meet the conditions, and $a_{1}8$, which contradicts the condition that $a_{1}+a_{4}$ is the smallest.
In summary, $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(2,3,4,6)$, the sum of the middle two numbers is 7. . | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,890 |
Example 2 If a natural number $N$ is appended to the right of any natural number, the resulting number can be divided by $N$ (for example, 2 appended to 35 results in 352, which is divisible by 2), then $N$ is called a "magic number". Among the natural numbers less than 130, how many magic numbers are there? | Analysis: To calculate how many magic numbers there are, we first need to clarify what constitutes a magic number. Although the problem provides the definition of a magic number, it is not convenient to directly use this definition to determine whether a number is a magic number. Therefore, we need to explore a simplif... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,891 |
Example 1 Let $P$ and $Q$ be two moving points on the circumference of the circle $x^{2}+y^{2}=1$, and they satisfy the condition that with a fixed point $A\left(0, \frac{1}{2}\right)$ inside the circle, $\angle P A Q = \frac{\pi}{2}$. Find the equation of the locus of point $M$, the intersection of the two tangents pa... | Solution 1: Let $M\left(x_{0}, y_{0}\right)$. Then the equation of the line containing the chord $P Q$ is $x_{0} x+y_{0} y=1$.
Substituting into $x^{2}+y^{2}=1$ gives
$$
\left(x_{0}^{2}+y_{0}^{2}\right) x^{2}-2 x_{0} x+1-y_{0}^{2}=0 \text {. }
$$
Let $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$. Then
... | 3 x^{2}+3 y^{2}+4 y-8=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,892 |
Three. (50 points) Let $a_{1}, a_{2}, \cdots, a_{n}$ be distinct positive integers, and their least common multiple be $b_{n}$. Prove that:
$$
b_{n} \geqslant \frac{n^{2}}{3} .
$$ | Three, let $b_{n}=p_{1}^{a_{1}} p_{2}^{a} \cdots p_{k}^{\sigma_{k}^{2}}$, where $p_{1}(m+1)^{2}\left(p\right.$ is a prime number greater than 3, $\left.m \in \mathbf{N}_{+}\right)$,
$$
\begin{array}{l}
3^{m} \geqslant(m+1)^{2}\left(m \geqslant 2, m \in \mathbf{N}_{+}\right), \\
3^{m}>2^{m}>\frac{(m+1)^{2}}{3}\left(m \i... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,893 |
Initially 243 Hexagon $A B C D E F$ is inscribed in $\odot O$, $A F // D C, F E // C B, E D // B A, A B+B C=$ $2 C D$. Constructing squares on each of the six sides, the sum of the areas of these six squares is 2008. Find the perimeter of hexagon $A B C D E F$. | Solution: As shown in Figure 3, from $A F$
$/ / D C, F E / / C B, E D / /$
$B A$, we know that $A D, B E, C F$ are
all diameters of $\odot O$, thus
$$
\begin{array}{l}
D E=A B, \\
E F=B C, \\
F A=C D .
\end{array}
$$
Let $C D=x, A B=x-d$. Then
$$
\begin{array}{l}
B C=x+d, D E=x-d, \\
A F=x, E F=x+d .
\end{array}
$$
A... | 108 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,894 |
As shown in Figure 4, from a point $P$ outside circle $\odot O$, two tangents to $\odot O$ are drawn, with $A$ and $B$ as the points of tangency. Another secant line $PCD$ is drawn from $P$ to $\odot O$. A tangent to $\odot O$ is drawn through $C$, intersecting $PA$ and $DA$ at points $F$ and $E$, respectively. Prove t... | Proof: As shown in Figure 4, let $DF$ intersect $\odot O$ at $K_{1}$, and connect $AK_{1}$ and $K_{1}C$. It is easy to see that,
$$
\begin{array}{l}
\triangle FAK_{1} \backsim \triangle FDA, \\
\triangle FCK_{1} \backsim \triangle FDC. \\
\text{Then } \frac{AK_{1}}{AD}=\frac{AF}{FD}=\frac{CF}{FD}=\frac{K_{1}C}{CD}.
\en... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,895 |
In the acute triangle $\triangle A B C$, if $n \in \mathbf{N}$, then
$$
\frac{\cos ^{n} A}{\cos B+\cos C}+\frac{\cos ^{n} B}{\cos C+\cos A}+\frac{\cos ^{n} C}{\cos A+\cos B} \geqslant \frac{3}{2^{n}} \text {. }
$$ | Proof: The proof is divided into two parts using mathematical induction.
(1) $n$ is even.
(i) When $n=0$,
the left side of equation (1)
$$
\begin{array}{l}
=\frac{1}{\cos B+\cos C}+\frac{1}{\cos C+\cos A}+\frac{1}{\cos A+\cos B} \\
\geqslant \frac{3}{\sqrt[3]{(\cos B+\cos C)(\cos C+\cos A)(\cos A+\cos B)}} \\
\geqslan... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,896 |
244 Given real numbers $a, b, c, d$ satisfy
$$
a+b+c+d=ab+ac+ad+bc+bd+cd=3 \text{. }
$$
Find the maximum real number $k$, such that the inequality
$$
a+b+c+2ab+2bc+2ca \geqslant k d
$$
always holds. | Given:
$$
\begin{aligned}
& (a+b+c+d)^{2}=3^{2} \\
\Rightarrow & a^{2}+b^{2}+c^{2}+d^{2}+2 a b+2 a c+2 a d+ \\
& 2 b c+2 b d+2 c d=9 \\
\Rightarrow & a^{2}+b^{2}+c^{2}+d^{2}=3 .
\end{aligned}
$$
Let \( y=(x-a)^{2}+(x-b)^{2}+(x-c)^{2} \).
Then \( y=3 x^{2}-2(a+b+c) x+\left(a^{2}+b^{2}+c^{2}\right) \).
Since \( y \geqsl... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,897 |
Example 2 The tangents at two distinct points $A, B$ on the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ are perpendicular to each other. Prove: the locus of the intersection point $M$ of the two tangents is $x^{2}+y^{2}=a^{2}+b^{2}$. | Proof: Let $M\left(x_{0}, y_{0}\right)$. Then the equation of the line containing the chord of contact $AB$ is
$$
\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1.
$$
Substituting into the equation of the ellipse $C$ and eliminating $y$ gives
$$
\left(1-\frac{x_{0} x}{a^{2}}\right)^{2}=\frac{y_{0}^{2}}{b^{2}}\left(1-\fra... | x^{2}+y^{2}=a^{2}+b^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 720,898 |
Example 3 Construct a line parallel to the known line $l$: $y=\frac{1}{4} x+1$ through the point $Q(-1,-1)$, intersecting the hyperbola $\frac{x^{2}}{4}-y^{2}=1$ at points $M$ and $N$.
(1) Prove: $Q$ is the midpoint of segment $M N$;
(2) Construct the tangents $l_{1}$ and $l_{2}$ to the hyperbola at points $M$ and $N$,... | Prove: (1) $l_{\text {MN }}: y=\frac{1}{4}(x-3)$.
Substitute into the hyperbola $\frac{x^{2}}{4}-y^{2}=1$, we get
$$
3 x^{2}+6 x-25=0 \text {. }
$$
Let $M\left(x_{1}, y_{1}\right) 、 N\left(x_{2}, y_{2}\right)$. Then $x_{1} 、 x_{2}$ are the two roots of equation (1), and we have $x_{1}+x_{2}=-2$.
Thus, $y_{1}+y_{2}=\fr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,899 |
Example 4 As shown in Figure 2, let the equation of the parabola be $x^{2}=2 p y$ $(p>0)$, and let $M$ be any point on the line $y=-2 p$. Draw tangents from $M$ to the parabola, with points of tangency at $A$ and $B$.
(1) Prove that the x-coordinates of points $A$, $M$, and $B$ form an arithmetic sequence.
(2) When poi... | Proof: (1) Let $M\left(x_{0}, y_{0}\right)$. Then the equation of the line containing the chord $AB$ is
$$
x_{0} x=p\left(y+y_{0}\right) \text {. }
$$
Substituting into the parabola $x^{2}=2 p y$ and eliminating $y$ gives
$$
x^{2}-2 x_{0} x+2 p y_{0}=0 \text {. }
$$
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2},... | M(0,-2 p) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,900 |
Example 5 As shown in Figure 3, let $P\left(x_{0}, y_{0}\right)$ be a fixed point, and $x_{0}>0, y_{0}>0, x_{0} y_{0}>$ 1. A moving line through $P$ intersects the curve $C: x y=1(x>0)$ at points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Find the equation of the locus of the intersection point $M$ ... | Solution 1: Let the equation of line $A B$ be
$$
y-y_{0}=k\left(x-x_{0}\right)(k0, y_{M}>0$.
From equation (2), we know $x_{M} \in\left(0, \frac{2}{y_{0}}\right), y_{M} \in\left(0, \frac{2}{x_{0}}\right)$. Therefore, the required trajectory is the open line segment
$$
y_{0} x_{M}+x_{0} y_{M}=2\left(x_{M} \in\left(0, \f... | y_{0} x_{M}+x_{0} y_{M}=2\left(x_{M} \in\left(0, \frac{2}{y_{0}}\right), y_{M} \in\left(0, \frac{2}{x_{0}}\right)\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,901 |
Example 6 Given the function $f(x)=x+\frac{t}{x}(t>0)$ and the point $P(1,0)$, draw two tangent lines $P M$ and $P N$ from $P$ to the curve $y=f(x)$, with the points of tangency being $M$ and $N$.
(1) Let $|M N|=g(t)$, find the expression for the function $g(t)$.
(2) Does there exist a $t$ such that $M$, $N$, and $A(0,... | Solution: (1) The equation of the chord of tangents $MN$ passing through the external point $P(1,0)$ of the curve $y=x+\frac{t}{x}(t>0)$ is
$$
y=2 x+2 t \text {. }
$$
Substituting into $y=x+\frac{t}{x}(t>0)$, we get $x^{2}+2 t x-t=0$.
Thus, $x_{1}+x_{2}=-2 t, x_{1} x_{2}=-t \ldots$
Therefore, $g(t)=\sqrt{\left(x_{1}-x... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,902 |
1. Let the tangent lines $M A$ and $M B$ through the point $M(4,2)$ to the circle $C: x^{2}+y^{2}=10$ touch the circle at points $A$ and $B$. Find the length of the minor arc between points $A$ and $B$. | (提示: 弦 $A B$. 所在直线的 equation is $2 x+y$ $=5$, the distance from the circle center to this line is $\sqrt{5}$, thus $\angle A O B=$ $90^{\circ}$, the length of the minor arc between the tangent points $A 、 B$ is $\frac{\sqrt{10} \pi}{2}$.) | \frac{\sqrt{10} \pi}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,903 |
2. In the Cartesian coordinate system $x O y$, for the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the focal distance is $2 c$. A circle $M$ is drawn with $O$ as the center and $a$ as the radius. If two tangents are drawn from $P\left(\frac{a^{2}}{c}, 0\right)$ to circle $M$ and they are perpendicular t... | (The line equation of chord $AB$ is $x=c$. It is easy to know that $\triangle OAP$ is an isosceles right triangle. Therefore, $e=\frac{c}{a}=$ $\left.\cos 45^{\circ}=\frac{\sqrt{2}}{2}.\right)$ | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,904 |
3. Given the circle $C: x^{2}+y^{2}=r^{2}$ and the line $l: x+$ $2 y=2 r$, there is a point $M$ on $l$. Two tangents $M A$ and $M B$ are drawn from $M$ to the circle. Find the equation of the locus of the midpoint $N$ of the chord $A B$.
Translate the above text into English, please retain the original text's line bre... | (It is known that $N$ is the intersection point of the line $O M$ and $A B$. Let $M\left(x_{0}, y_{0}\right)$. Then the equation of the line $A B$ where the chord of contact lies is $x_{0} x+y_{0} y=r^{2}, l_{o m}: y=\frac{y_{0}}{x_{0}} x$. Solving the system of equations gives $x_{0}=\frac{x r^{2}}{x^{2}+y^{2}}, y_{0}... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,905 |
4. As shown in Figure 4, let point \( P\left(x_{0}, y_{0}\right) \) be on the line \( x=m \) (where \( y \neq \pm m \) and \( 0 < m < 1 \)). Draw two tangents \( PA \) and \( PB \) from \( P \) to the hyperbola \( x^{2} - y^{2} = 1 \), with points of tangency at \( A \) and \( B \). The fixed point \( M \left( \frac{1}... | (Hint: (1) The equation of the line containing the chord $AB$ is $x_{0} x-y_{0} y=1$. Given $x_{0}=m$, then $l_{AB}: mx-y_{0} y=1$. Substitute $M\left(\frac{1}{m}, 0\right)$ into this line equation to check.
(2) Let $A\left(x_{1}, y_{1}\right)$. The equation of the perpendicular line $AN$ is $y-y_{1}=-x+x_{1}$. From $y... | \left(x - \frac{1}{3m}\right)^{2} - y^{2} = \frac{2}{9} | Geometry | proof | Yes | Yes | cn_contest | false | 720,906 |
Example 3 Write a two-digit prime number after another two-digit prime number that is not equal to it, to form a four-digit number. It is known that this four-digit number can be divided by half the sum of these two prime numbers. Find all such four-digit numbers.
(1992, Yingchun Cup Mathematics Competition) | Solution: Let the two two-digit prime numbers be $x$ and $y$ $(x \neq y)$. Then the four-digit number obtained by writing $y$ after $x$ is $100x + y$. According to the problem, there exists an integer $k$ such that
$$
100x + y = k \cdot \frac{x + y}{2}.
$$
Therefore, $(x + y) \mid (200x + 2y)$, which means
$$
(x + y) ... | 1353, 5313, 1947, 4719, 2343, 4323, 2937, 3729 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,907 |
5. As shown in Figure 5, the moving chord $AB$ of the parabola $y=x^{2}$ lies on a line that is tangent to the circle $x^{2}+y^{2}=1$. The two tangent lines to the parabola at points $A$ and $B$ intersect at point $M$. Find the equation of the locus of point $M$.
保留了原文的换行和格式。 | (The equation of the line containing the chord $AB$ is $y+y_{M}=2 x_{M} x$. It is easy to know that $d=\frac{\left|y_{M}\right|}{\sqrt{1+4 x_{M}^{2}}}=1$, i.e., $y_{M}^{2}-4 x_{M}^{2}=1$. Substituting $y+y_{M}=2 x_{M} x$ into $y=x^{2}$, we get $x^{2}-2 x_{M} x+y_{M}=0$. Therefore, $\Delta=4 x_{M}^{2}-4 y_{M}=y_{M}^{2}-... | y^{2}-4 x^{2}=1(y \leqslant-1 \text{ or } y>2+\sqrt{5}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,908 |
6. Given a line $l$ passing through the point $(0,1)$ intersects the curve $C$: $y=x+\frac{1}{x} (x>0)$ at two distinct points $M$ and $N$. Find the locus of the intersection of the tangents to the curve $C$ at points $M$ and $N$. (2007, National High School Mathematics Competition) | (The line equation of the chord $MN$ is $l$: $\left(y_{0}-2 x_{0}\right) x+x_{0} y=2$. Since $l$ passes through the point $(0,1)$, substituting gives $x_{0}=2$. At this point, the line $l:\left(y_{0}-4\right) x+2 y=2$. Substituting into the curve $C: y=x+\frac{1}{x}(x>0)$, we get $\left(y_{0}-2\right) x^{2}-$ $2 x+2=0$... | x=2\left(2<y<\frac{5}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,909 |
110 players participate in a table tennis tournament, with each pair of players playing one match. If player $i$ beats player $j$, player $j$ beats player $k$, and player $k$ beats player $i$, this is called a "triangle." Let $W_{i}$ and $L_{i}$ represent the number of wins and losses, respectively, for the $i$-th play... | Solution: Let the points $-v_{1}, v_{2}, \cdots, v_{10}$ represent the 10 players. If $v_{i}$ beats $v_{j}$, then draw a directed edge between the corresponding points: $v_{i} \rightarrow v_{j}$. Thus, we obtain a 10-vertex directed graph $G$, then
$$
\text { and } \begin{aligned}
d^{+}\left(v_{i}\right) & =W_{i}, d^{-... | 40 | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,910 |
Example 2: Six non-negative integers whose sum is 2003 are written at the vertices of a regular hexagon. Bert performs the following operation: he can select a vertex, erase the number written there, and replace it with the absolute value of the difference of the numbers on the adjacent vertices. Prove that Bert can pe... | $\qquad$Divide the six vertices into a "positive" and "inverted" triangle as shown in Figure 1. Since the sum of the numbers filled at the six vertices is the odd number 2003, the sum of the numbers at the vertices of these two triangles must be one odd and one even (but this situation may change after operations).
(1)... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,911 |
Question 1 Let $f(x)$ be a periodic function, $T$ and 1 are periods of $f(x)$ and $0<a_{n}>a_{n+1}>0(n=$ $1,2, \cdots)$, and each $a_{n}(n=1,2, \cdots)$ is a period of $f(x)$ ${ }^{[1]}$.
(2008, National High School Mathematics Competition) | Proof l: (1) Omitted.
(2) Suppose there does not exist such a sequence $\left\{a_{n}\right\}$, i.e., there are only a finite number of irrational numbers $a_{n}$, each of which is a period of $f(x)$.
Assume there are $k$ irrational numbers, each of which is a period of $f(x)$, and $1>a_{1}>a_{2}>\cdots>a_{k}>0$.
Then,... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,912 |
Question 2 Let $a_{k}>0(k=1,2, \cdots, 2008)$. Prove: there exists a sequence $\left\{x_{i}\right\}$ satisfying the following conditions if and only if $\sum_{k=1}^{2008} a_{k}>1$:
(1) $0=x_{0}<x_{n}<x_{n+1}(n=1,2, \cdots)$;
(2) $\lim _{n \rightarrow \infty} x_{n}$ exists;
$$
\begin{array}{l}
\quad(3) x_{n}-x_{n-1}=\su... | Proof: Sufficiency: as per the standard answer.
Necessity: From condition (3), we can get
$$
\begin{array}{l}
x_{n}-x_{n-1} \\
= \sum_{k=1}^{2008} a_{k} x_{n+k}-\sum_{k=0}^{2007} a_{k+1} x_{n+k} \quad(n=1,2, \cdots) \\
=\left(a_{1} x_{n+1}+a_{2} x_{n+2}+\cdots+a_{2008} x_{n+2008}\right)- \\
\left(a_{1} x_{n}+a_{2} x_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,913 |
Let $p$ be a prime number satisfying $p \equiv 3(\bmod 8)$. Find all integer solutions $(x, y)$ to the equation $y^{2}=x^{3}-p^{2} x$. (2007, Indian National Team Selection Exam) | The original equation is equivalent to
$$
y^{2}=(x-p)(x+p) x \text {. }
$$
We discuss the following cases. All letters in the discussion below are natural numbers.
(1) $p \nmid y$.
Since $(x-p, x)=(x+p, x)=1$, if $x$ is even, then $(x-p, x+p)=1$. In this case, $x$, $x+p$, and $x-p$ are all perfect squares, but
$$
x+p... | (0,0),(p, 0),(-p, 0) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,914 |
1. Given an acute triangle $\triangle P B C, P B \neq P C$. Let $A$ and $D$ be points on sides $P B$ and $P C$ respectively, and connect $A C$ and $B D$ intersecting at point $O$. Draw $O E \perp A B$ at $E$ and $O F \perp C D$ at $F$, and let the midpoints of segments $B C$ and $A D$ be $M$ and $N$ respectively.
(1) I... | 1. (1) As shown in Figure 1, let $Q$ and $R$ be the midpoints of sides $OB$ and $OC$, respectively. Connect $EQ$, $MQ$, $FR$, and $MR$. Then
$$
\begin{array}{l}
EQ=\frac{1}{2} OB=RM, \\
MQ=\frac{1}{2} OC=RF.
\end{array}
$$
Also, quadrilateral $OQMR$ is a parallelogram, so
$$
\angle OQM = \angle ORM.
$$
Given that poi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,915 |
2. Find all prime pairs $(p, q)$ such that $p q \mid\left(5^{p}+5^{q}\right)$.
(Fu Yunhao) | 2. If $21 p q$, let's assume $p=2$. Then
$$
2 q\left|\left(5^{2}+5^{q}\right) \Rightarrow q\right|\left(5^{q}+25\right) \text {. }
$$
By Fermat's Little Theorem, we know $q \mid\left(5^{q}-5\right)$, thus $q \mid 30$, i.e., $q=2,3,5$.
It is easy to verify that the prime pair $(2,2)$ does not meet the requirements, whi... | (2,3),(3,2),(2,5),(5,2),(5,5),(5,313),(313,5) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,916 |
3. Let $m, n (4 < m < n)$ be given integers, $A_{1} A_{2} \cdots A_{2 n+1}$ be a regular $(2 n+1)$-gon, and $P=\left\{A_{1}, A_{2}, \cdots, A_{2 n+1}\right\}$. Find the number of convex $m$-gons with vertices in $P$ and exactly two acute interior angles.
（Leng Gangsong, problem contributor） | 3. First, prove a lemma.
Lemma: A convex $m$-gon with vertices in $P$ has at most two acute angles; and if there are two acute angles, these two angles must be adjacent.
Proof of the lemma: Let the convex $m$-gon be $P_{1} P_{2} \cdots P_{m}$. We only consider the case where there is at least one acute angle. Without... | (2 n+1)\left(\mathrm{C}_{n+1}^{m-1}+\mathrm{C}_{n}^{m-1}\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,917 |
Example 4 If for all integers $m, n$, there exist integers $x, y$ such that $a x+b y=m, c x+d y=n$, prove:
$$
|a d-b c|=1
$$ | Analysis: The conditions of the problem actually tell us that for all integers $m, n$, the corresponding system of equations has integer solutions. Clearly, we cannot exhaust "all integers $m, n$", but we can choose appropriate $m, n$ to solve the corresponding system of equations, and then use the fact that their solu... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,918 |
4. Given an integer $n(n \geqslant 3)$, real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $\min _{1<i<j<n}\left|a_{i}-a_{j}\right|=1$. Find the minimum value of $\sum_{k=1}^{n}\left|a_{k}\right|^{3}$.
(Zhu Huawei) | 4. Let $a_{1}<a_{2}<\cdots<a_{n}$.
Then for $1 \leqslant k \leqslant n$ we have
$$
\begin{array}{l}
\left|a_{k}\right|+\left|a_{n-k+1}\right| \geqslant\left|a_{n-k+1}-a_{k}\right| \\
\geqslant|n+1-2 k| . \\
\text { Hence } \sum_{k=1}^{n}\left|a_{k}\right|^{3}=\frac{1}{2} \sum_{k=1}^{n}\left(\left|a_{k}\right|^{3}+\lef... | \frac{1}{32}\left(n^{2}-1\right)^{2} \text{ (for odd } n\text{)}, \frac{1}{32} n^{2}\left(n^{2}-2\right) \text{ (for even } n\text{)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,919 |
5. In a convex $n$-gon $P$, each side and each diagonal is colored with one of $n$ colors. For which $n$ does there exist a coloring such that for any three different colors among these $n$ colors, there is a triangle whose vertices are vertices of the polygon $P$ and whose three sides are colored with these three colo... | 5. When $n(n \geqslant 3)$ is odd, there exists a coloring method that meets the requirements; when $n$ is even, no such coloring method exists.
Since every three vertices form a triangle, the number of triangles is $\mathrm{C}_{n}^{3}$, and the number of color combinations is also $\mathrm{C}_{n}^{3}$. Therefore, the... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,920 |
6. Given an integer $n(n \geqslant 3)$. Prove: There exists a set $S$ of $n$ distinct positive integers such that for any two different non-empty subsets $A$ and $B$ of $S$, the numbers
$$
\frac{\sum_{x \in A} x}{|A|} \text { and } \frac{\sum_{x \in B} x}{|B|}
$$
are coprime composites (here $\sum_{x \in X} x$ and $|X... | 6. Let $f(X)$ denote the arithmetic mean of the elements in a finite set of numbers $X$.
(1) Prove: There exists a set $S_{1}$ of $n$ distinct positive integers such that for any two different non-empty subsets $A$ and $B$ of $S_{1}$, the numbers $f(A)$ and $f(B)$ are distinct positive integers.
In fact, choose an inte... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,921 |
1. Given $x+y=x^{-1}+y^{-1} \neq 0$. Then the value of $x y$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2 | $-1 . C$.
Notice that $x+y=\frac{1}{x}+\frac{1}{y}=\frac{x+y}{x y}$.
Since $x+y \neq 0$, we have $x y=1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,922 |
3. Four students, A, B, C, and D, participate in the $4 \times 100 \mathrm{~m}$ relay race at the school track and field meet. If the four students are arranged in any order, then the probability that A exactly hands the baton to B is $($ ).
(A) $\frac{1}{4}$
(B) $\frac{1}{6}$
(C) $\frac{1}{8}$
(D) $\frac{1}{12}$ | 3. A.
By listing or drawing a tree diagram, we can know that there are 24 possible orders for the four students' competition, among which, the cases where Jia hands over to Yi are 6. Therefore, the required probability is $\frac{6}{24}=\frac{1}{4}$. | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,924 |
4. As shown in Figure 2, the graphs of two inverse proportion functions $y=\frac{k_{1}}{x}$ and $y=\frac{k_{2}}{x}\left(k_{1}>k_{2}>\right.$
$0)$ in the first quadrant are curves $C_{1}$
and $C_{2}$, respectively. Let point $P$ be on $C_{1}$,
$P E \perp x$-axis at point $E$, intersecting $C_{2}$ at point $A$, and $P D ... | 4. B.
Let $P(m, n)$. Since point $P$ is on $C_{1}$, we have,
$$
\begin{array}{l}
m n=k_{1}, A\left(m, \frac{k_{2}}{m}\right), B\left(\frac{k_{2}}{n}, n\right) . \\
S_{\text {quadrilateral } P A O B}=S_{\text {quadrilateral } P E O D}-S_{\triangle O E 1}-S_{\triangle O B D} \\
=m n-\frac{1}{2}\left(m \cdot \frac{k_{2}}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,925 |
5. In the Cartesian coordinate system, a point whose both horizontal and vertical coordinates are integers is called an "integer point". The image of the quadratic function $y=-x^{2}+6 x-\frac{27}{4}$ and the x-axis enclose a closed figure, which is colored red. The number of integer points inside this red region and o... | 5.C.
The graph of the quadratic function $y=-x^{2}+6 x-\frac{27}{4}$ intersects the $x$-axis at two points $\left(\frac{3}{2}, 0\right)$ and $\left(\frac{9}{2}, 0\right)$. Between $x=\frac{3}{2}$ and $x=\frac{9}{2}$, there are 3 integers: 2, 3, 4.
When $x=2,4$, $y=\frac{5}{4}$, the integers that satisfy $0 \leqslant ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,926 |
6. Xiao Ming divided a cube with an edge length of 4 into 29 smaller cubes with integer edge lengths. Then the number of smaller cubes with an edge length of 1 is ( ).
(A) 22
(B) 23
(C) 24
(D) 25 | 6. C.
If a cube with an edge length of 3 is separated out, then there can only be 1 cube with an edge length of 3, and the remaining ones are all cubes with an edge length of 1, totaling 37, which does not meet the requirement.
Let the number of cubes with an edge length of 2 be $x$, and the number of cubes with an e... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,927 |
7. As shown in Fig. $3, \odot O$
is tangent to the hypotenuse $AB$ of the right triangle $\triangle ABC$ at point $D$, intersects the leg $AC$ at point $E$, and $DE \parallel BC$. Given $AE = 2\sqrt{2}$, $AC = 3\sqrt{2}$, and $BC = 6$. Then the radius of $\odot O$ is ( ).
(A) 3
(B) 4
(C) $4\sqrt{3}$
(D) $2\sqrt{3}$ | 7.D.
Notice $A B=\sqrt{(3 \sqrt{2})^{2}+6^{2}}=3 \sqrt{6}$.
From $D E / / B C \Rightarrow \triangle A E D \backsim \triangle A C B$
$$
\Rightarrow \frac{E D}{C B}=\frac{A E}{A C} \Rightarrow E D=6 \times \frac{2 \sqrt{2}}{3 \sqrt{2}}=4 \text {. }
$$
In Figure 3, draw $O F \perp D E$, with $F$ being the foot of the pe... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,928 |
Example 5 Choose $n$ numbers from $1,2, \cdots, 9$, among which there must be several numbers (at least one, or all), the sum of which can be divisible by 10. Find the minimum value of $n$.
(2008, National Junior High School Mathematics Competition) | Solution: When $n \leqslant 4$, since no combination of $n$ numbers from $1,3,5,8$ can sum to a multiple of 10, we have $n \geqslant 5$.
Below, we prove that $n=5$ meets the requirement, i.e., we prove that from $1,2, \cdots, 9$, any 5 numbers chosen will always include some numbers whose sum is a multiple of 10.
We u... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,929 |
8.7 The lengths of seven line segments $a_{1}, a_{2}, \cdots$, $a_{7}$ are all integers, satisfying $a_{1}<a_{2}<\cdots<a_{7}$, and no three of these seven line segments can form a triangle. If $a_{1}=1, a_{7}=$ 21, then $a_{6}=(\quad)$.
(A) 18
(B) 13
(C) 8
(D) 5 | 8.B.
Only when $a_{2}=2, a_{3}=a_{1}+a_{2}=3, a_{4}=$ $a_{2}+a_{3}=5, a_{5}=a_{3}+a_{4}=8, a_{6}=a_{4}+a_{5}=$ 13, none of the 7 segments can form a triangle with any other two, so $a_{6}=13$. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,930 |
11. As shown in Figure 5, in quadrilateral $A B C D$, $\angle A=$ $\angle B C D=90^{\circ}, B C=$ $C D, E$ is a point on the extension of $A D$. If $D E=A B$ $=3, C E=4 \sqrt{2}$, then the length of $A D$ is | 11.5.
In Figure 5, connect $A C$. It is easy to prove that
$\triangle C D E \cong \triangle C B A, \angle A C E=90^{\circ}$.
Since $C A=C E=4 \sqrt{2}$, therefore, $A E=8$.
Thus, $A D=5$. | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,933 |
12. Given that $\triangle A B C$ is an obtuse triangle, with its longest side $A C$ having a point $P (P$ not coinciding with $A$ or $C)$, a line $l$ is drawn through $P$, such that the triangle formed by $l$ intersecting $\triangle A B C$ is similar to the original triangle. Then, such a line $l$ can be drawn $\qquad$... | 12.2 or 3.
As shown in Figure 11, where
$$
\begin{array}{l}
\angle A B P_{1}=\angle C, \\
\angle C B P_{2}=\angle A .
\end{array}
$$
When point $P$ is located between point $A$ and $P_{1}$ (including point $P_{1}$) or between point $C$ and $P_{2}$ (including point $P_{2}$), there are 3 lines that satisfy the conditio... | 2 \text{ or } 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,934 |
13. Xiao Wang walks along the street at a constant speed, and he notices that a No. 18 bus passes him from behind every $6 \mathrm{~min}$, and a No. 18 bus comes towards him every $3 \mathrm{~min}$. Assuming that each No. 18 bus travels at the same speed, and the No. 18 bus terminal dispatches a bus at fixed intervals,... | 13.4 .
Let the speed of bus No. 18 be $x \mathrm{~m} / \mathrm{min}$, the walking speed of Xiao Wang be $y \mathrm{~m} / \mathrm{min}$, and the distance between two consecutive buses traveling in the same direction be $s \mathrm{~m}$. From the problem, we have
$$
\left\{\begin{array}{l}
6 x-6 y=s, \\
3 x+3 y=s .
\end{... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,935 |
15. (12 points) Two students, Xiao Wang and Xiao Li, are playing the game "Rock, Paper, Scissors." The game rules are: winning a round earns 3 points, drawing a round earns 1 point, and losing a round earns 0 points. The game consists of seven rounds. The player with the higher score at the end of the game wins.
(1) If... | Three, 15. (1) Let Xiao Wang win $x$ times and draw $y$ times $(x, y$ are natural numbers). Then
$$
\left\{\begin{array}{l}
3 x + y = 10, \\
x + y \leqslant 7, \\
03(4-x-y)+y, \\
x+y \leqslant 4, \\
0 < x \leqslant 4.
\end{array}\right.
$$
Solving, we get $(x, y)=(4,0),(3,1),(3,0)$,
$$
(2,2),(2,1),(1,3) \text{.}
$$
T... | (x, y)=(4,0),(3,1),(3,0),(2,2),(2,1),(1,3) | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 720,937 |
16. (12 points) In the Cartesian coordinate system $x O y$, the graph of the linear function $y=k x+b+2(k \neq 0)$ intersects the positive $x$-axis and $y$-axis at points $A$ and $B$, respectively, such that
$$
\mathrm{S}_{\triangle O A B}=|O A|+|O B|+3 \text{. }
$$
(1) Express $k$ in terms of $b$;
(2) Find the minimum... | 16. (1) Let $x=0$, we get $y=b+2(b>-2)$; let $y=0$, we get $x=-\frac{b+2}{k}>0(k>0)$.
(2) From (1), we have
$$
\begin{array}{l}
S_{\triangle O A B}=\frac{(b+2)(b+5)}{b}=\frac{b^{2}+7 b+10}{b} \\
=b+\frac{10}{b}+7=\left(\sqrt{b}-\sqrt{\frac{10}{b}}\right)^{2}+7+2 \sqrt{10} .
\end{array}
$$
When $\sqrt{b}-\sqrt{\frac{10... | 7+2 \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,938 |
17. (12 points) As shown in Figure 7, $AB$, $AC$, and $AD$ are three chords in a circle, and point $E$ is on side $AD$, with $AB=AC=AE$. Please explain the reasons for the following equations.
(1) $\angle CAD=2 \angle DBE$;
(2) $AD^2 - AB^2 = BD \cdot DC$. | 17. (1) As shown in Figure 13, connect $B C$.
Since $A B=A C=A E$, we have:
$$
\begin{array}{c}
\angle 5=\angle 2, \\
\angle 2+\angle 3=\angle 6. \\
Also, \angle 4+\angle 5 \\
=\angle 6=\angle 2+\angle 3,
\end{array}
$$
$$
\text{ then } \angle 4=\angle 3 \text{. }
$$
And $\angle 1=\angle 4+\angle 3$, so $\angle 1=2 \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,939 |
1. Let $a, b, c, d, e, f$ be a permutation of $1, 2, 3, 4, 5, 6$. If $2|\overline{a b}, 3| \overline{a b c}, 4|\overline{a b c d}, 5| \overline{a b c d e}$, $61 \overline{a b c d e f}$, find $\overline{a b c d e f}$.
(1982, Beijing Mathematical Competition (Grade 8)) | (It is easy to know that $b$, $d$, $f$ are a permutation of $2$, $4$, $6$, and thus $a$, $c$, $e$ are a permutation of $1$, $3$, $5$. From $5 \mid \overline{a b c d e}$ we get $e=5$, so, $(a, c)=(1,3)$ or $(3,1)$. By enumeration, we get $\overline{a b c d e f}=123654$ or 321654.) | 123654 \text{ or } 321654 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,940 |
18. (14 points) As shown in Figure 8, in the Cartesian coordinate system $x O y$, the rectangle $O A B C$ has its sides on the $x$-axis and $y$-axis, with $O A=10 \text{ cm}$ and $O C=6 \text{ cm}$. There are two moving points $P$ and $Q$ starting from $O$ and $A$ simultaneously. Point $P$ moves uniformly along $O A$, ... | 18. (1)(i) Notice that
$$
\begin{aligned}
& S_{\triangle C P Q}=S_{\text {quadrilateral OABC }}-S_{\triangle O C P}-S_{\triangle P R Q}-S_{\triangle B C C} \\
= & 60-\frac{1}{2} \times 6 t-\frac{1}{2}(10-t) \cdot \frac{1}{2} t- \\
& \frac{1}{2} \times 10\left(6-\frac{1}{2} t\right) \\
= & \frac{1}{4} t^{2}-3 t+30 \\
= ... | a=\frac{4}{3}, Q\left(10, \frac{8}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,941 |
1. If the equation
$$
x^{2}+(1-2 \mathrm{i}) x+3 m-\mathrm{i}=0(m \in \mathbf{R})
$$
has one real root, then its other root is ( ).
(A) $-\frac{3}{2}+2$ i
(B) $\frac{3}{2}-2 \mathrm{i}$
(C) $-\frac{1}{2}+2$ i.
(D) $\frac{1}{2}-2 \mathrm{i}$ | $-1 . C$.
Let the real root of the equation be $\alpha$, and the other root be $a+b \mathrm{i}$ $(a, b \in \mathbf{R})$. Then
$$
\begin{array}{l}
x^{2}+(1-2 \mathrm{i}) x+3 m-\mathrm{i} \\
=(x-\alpha)[x-(a+b \mathrm{i})] \\
=x^{2}-(\alpha+a+b \mathrm{i}) x+\alpha(a+b \mathrm{i})=0 .
\end{array}
$$
By comparing coeffic... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,942 |
2. If the function $f(x) = \log _{a} x(a>0$, and $a \neq 1)$ satisfies $f\left(\frac{2}{a}\right)>f\left(\frac{3}{a}\right)$, then the solution set for $f\left(1-\frac{1}{x}\right)>1$ is ( ).
(A) $0<x<\frac{1}{a}$
(B) $0<x<\frac{1}{1-a}$
(C) $1<x<\frac{1}{a}$
(D) $1<x<\frac{1}{1-a}$ | 2.D.
From $f\left(\frac{2}{a}\right)=\log _{a} \frac{2}{a}>f\left(\frac{3}{a}\right)=\log _{a} \frac{3}{a}$, we get $0<a<1$.
Therefore, the original inequality is equivalent to $0<1-\frac{1}{x}<a$.
Solving this inequality yields
$$
-1<-\frac{1}{x}<a-1 \Rightarrow 1<x<\frac{1}{1-a} .
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,943 |
3. The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are both arithmetic sequences of positive integers, with common differences $d_{1}$ and $d_{2}$, respectively, and both are greater than zero. If the sets $A=\left\{a_{n} \mid n \in \mathbf{N}_{+}\right\}$ and $B=\left\{b_{n} \mid n \in \mathbf{N}_{+}\r... | 3.B.
Let $\left[d_{1}, d_{2}\right]$ denote the least common multiple of $d_{1}$ and $d_{2}$.
If $a_{1}=b_{1}=a$, then for any natural number $n$, we have $a+n\left[d_{1}, d_{2}\right] \in C$.
Conversely, for any $c_{n} \in C$, then
$$
c_{n}=a+\left(n_{1}-1\right) d_{1}=a+\left(n_{2}-1\right) d_{2} \text {, }
$$
whic... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,944 |
4. For the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{m}=1$, the circle with the eccentricity as the radius and the right focus as the center is tangent to the asymptotes of the hyperbola. Then $m=(\quad)$.
(A) $\frac{3}{2}$
(B) $\frac{4}{3}$
(C) $\frac{5}{4}$
(D) $\frac{6}{5}$ | 4. B.
Notice that $a=2, b=\sqrt{m}$,
$$
c=\sqrt{4+m}, e=\frac{\sqrt{4+m}}{2} \text {. }
$$
The equations of the asymptotes are $\frac{x}{2} \pm \frac{y}{\sqrt{m}}=0$, i.e.,
$$
\sqrt{m} x \pm 2 y=0 \text {. }
$$
The distance from the right focus $(\sqrt{4+m}, 0)$ to the asymptote is $d=b=\sqrt{m}$.
Thus, $\sqrt{m}=\f... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,945 |
5. Given $\boldsymbol{a}=(\cos \alpha, \sin \alpha), \boldsymbol{b}=(\cos \beta, \sin \beta)$, $|a-b|=\frac{2 \sqrt{5}}{5}$. If $0<\alpha<\frac{\pi}{2},-\frac{\pi}{2}<\beta<0$, and $\sin \beta=-\frac{5}{13}$, then $\sin \alpha=(\quad)$.
(A) $\frac{31}{65}$
(B) $\frac{32}{65}$
(C) $\frac{33}{65}$
(D) $\frac{34}{65}$ | 5.C.
$$
\begin{array}{l}
|a-b|^{2}=(\cos \alpha-\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2} \\
=2-2(\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta) \\
=2-2 \cos (\alpha-\beta)
\end{array}
$$
That is, $2-2 \cos (\alpha-\beta)=\frac{4}{5}, \cos (\alpha-\beta)=\frac{3}{5}$.
Given $\alpha=(\alpha-\beta)+\beta$... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,946 |
7. There are 6 seats arranged in a row, and three people are to be seated, with exactly two adjacent empty seats. The number of different seating arrangements is ( ).
(A) 48
(B) 60
(C) 72
(D) 96 | 7.C.
Arrange the seats in order as $1,2, \cdots, 6$.
If the adjacent empty seats are 1 and 2 or 5 and 6, there are $\mathrm{C}_{3}^{1} \cdot \mathrm{A}_{3}^{3}$ arrangements each; if the adjacent empty seats are 2 and 3, 3 and 4, or 4 and 5, there are $\mathrm{C}_{2}^{1} \cdot \mathrm{A}_{3}^{3}$ arrangements each.
T... | 72 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,947 |
8. The real coefficient quadratic equation in one variable
$$
x^{2}+(a+1) x+a+b+1=0
$$
has two real roots $x_{1}, x_{2}$. If $0<x_{1}<1$ and $x_{2}>1$, then the range of $\frac{a}{b}$ is ( ).
(A) $\left(-2,-\frac{1}{2}\right)$
(B) $\left(-2, \frac{1}{2}\right)$
(C) $\left(-1,-\frac{1}{2}\right)$
(D) $\left(-1, \frac{1... | 8. A.
Let $f(x)=x^{2}+(a+1) x+a+b+1$. Then
$$
\left\{\begin{array} { l }
{ 0 1 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ f ( 0 ) > 0 , } \\
{ f ( 1 ) < 0, } \\
{ 2 a+b+3 < 0 . }
\end{array}\right.\right.\right.
$$
Draw the feasible region, the line $a+b+1=0$ and $2 a+b+3=0$ intersect at point $A(-... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,948 |
9. In the quadrilateral pyramid $P-ABCD$,
$$
\begin{array}{l}
\overrightarrow{AB}=(4,-2,3), \overrightarrow{AD}=(-4,1,0), \\
\overrightarrow{AP}=(-6,2,-8) .
\end{array}
$$
Then the height $h$ of this quadrilateral pyramid is $(\quad)$.
(A) 1
(B) 2
(C) 13
(D) 26 | 9. B.
Let the height of the quadrilateral pyramid be $P H, \overrightarrow{P H}=(x, y, z)$. Then $\overrightarrow{P H} \cdot \overrightarrow{A B}=4 x-2 y+3 z=0$,
$\overrightarrow{P H} \cdot \overrightarrow{A D}=-4 x+y=0$.
Solving these, we get $y=4 x, z=\frac{4}{3} x$.
$\overrightarrow{P H}=\left(x, 4 x, \frac{4}{3} x... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,949 |
2. Given 99 | $\overline{62 x y 427}$. Find $950 x+24 y+1$. (1997, Hope Cup Mathematics Invitational)
Find the value of $950 x+24 y+1$ if 99 divides $\overline{62 x y 427}$. (1997, Hope Cup Mathematics Invitational) | (提示: From $91 \overline{62 x y 427}$, we get $91(x+y+21)$. Therefore, $9 \mid(x+y+3)$. Noting that $3 \leqslant x+y+3 \leqslant$ 21, hence $x+y=6$ or 15. From $11 \overline{62 x y 427}$, we get $x-$ $y=-2$ or 9. Also note that $x+y$ and $x-y$ have the same parity, solving gives $(x, y)=(2,4)$. Therefore, $950 x+24 y+1=... | 1997 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,950 |
10. For a real number $x$, $[x]$ denotes the greatest integer not exceeding the real number $x$. It is known that the sequence of positive numbers $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right),
$$
where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\... | 10. B.
Notice
$$
\begin{array}{l}
S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right)=\frac{1}{2}\left[\left(S_{n}-S_{n-1}\right)+\frac{1}{S_{n}-S_{n-1}}\right], \\
S_{n}+S_{n-1}=\frac{1}{S_{n}-S_{n-1}}, \\
S_{n}^{2}=S_{n-1}^{2}+1 .
\end{array}
$$
Since $S_{1}=a_{1}=1$, we have $S_{n}^{2}=n, S_{n}=\sqrt{n}$. Therefor... | 18 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,951 |
11. In a free-throw test, as long as a person makes 3 shots, they are qualified and do not need to shoot anymore. However, each person can shoot at most 5 times. The probability that a player with a shooting accuracy of $\frac{2}{3}$ qualifies in the test is $\qquad$ . | $=, 11 . \frac{64}{81}$.
The player:
The probability of making 3 out of 3 shots is $\left(\frac{2}{3}\right)^{3}=\frac{8}{27}$;
The probability of making 3 out of 4 shots is $\frac{2}{3} \times\left(\frac{2}{3}\right)^{2} \times \frac{1}{3} C_{3}^{2}=\frac{8}{27}$;
The probability of making 3 out of 5 shots is $\frac{2... | \frac{64}{81} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,952 |
12. As shown in Figure 1, goods are transported from a riverside city $A$ to another location $B$. The shortest distance from $B$ to the riverbank at point $C$ is $b \mathrm{~km}$. The distance along the river from $A$ to $C$ is $a \mathrm{~km}$. If the waterway transportation cost is
$\frac{1}{n}$ of the highway tra... | 12. $\frac{6}{\sqrt{n^{2}-1}}$.
Assume the road transportation cost is 1. Then the water transportation cost is $\frac{1}{n}$. Let $D C=x \text{ km}$. Then the total transportation cost from $A$ to $B$ is $f(x)=\frac{1}{n}(a-x)+\sqrt{x^{2}+b^{2}}$. $f^{\prime}(x)=-\frac{1}{n}+\frac{2 x}{2 \sqrt{x^{2}+b^{2}}}=\frac{n x... | \frac{b}{\sqrt{n^{2}-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,953 |
13. As shown in Figure 2, in quadrilateral $ABCD$, $AC = l_{1}, BD = l_{2}$. Then $(\overrightarrow{AB}+\overrightarrow{DC}) \cdot (\overrightarrow{BC}+\overrightarrow{AD})=$ $\qquad$ | 13. $l_{1}^{2}-l_{2}^{2}$.
As shown in Figure 4, let $E, F, G, H$ be the midpoints of $AB, BC, CD, DA$ respectively. Then
$$
\begin{array}{l}
\overrightarrow{EG}=\overrightarrow{EB}+\overrightarrow{BC}+\overrightarrow{CG}. \\
Also, \overrightarrow{EG}=\overrightarrow{EA}+\overrightarrow{AD}+
\end{array}
$$
$\overright... | l_{1}^{2}-l_{2}^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,954 |
15. (12 points) $F(1,0)$ is a fixed point, $P(0, b)$ is a moving point on the $y$-axis, and point $M(a, 0)$ satisfies $\overrightarrow{P M} \cdot \overrightarrow{P F} = 0$. If point $N$ satisfies $2 \overrightarrow{P N} + \overrightarrow{N M} = 0$, find:
(1) The equation of the trajectory curve $C$ of point $N$;
(2) Th... | (1) Notice that $\overrightarrow{P M}=(a,-b), \overrightarrow{P F}=(1,-b)$.
From $\overrightarrow{P M} \cdot \overrightarrow{P F}=0$, we get $a+b^{2}=0$.
Let point $N(x, y)$. Then
$\overrightarrow{P N}=(x, y-b), \overrightarrow{N M}=(a-x,-y)$.
Thus, $2(x, y-b)+(a-x,-y)=0$, which means $a+x=0, y-2 b=0$.
Therefore, $x=-a... | x=-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,956 |
16. (12 points) Let the function
$$
f(x)=x(1+x)^{2}(x \in(-\infty, 0]) \text {. }
$$
(1) Find the critical points of $f(x)$;
(2) For any $a<0$, let $F(a)$ denote the point where $f(x)$ attains its minimum value on $[a, 0]$. Find the minimum value of $k=\frac{F(a)}{a}$. | 16. (1) From
$$
\begin{array}{l}
f^{\prime}(x)=(1+x)^{2}+2 x(1+x) \\
=(1+x)(1+3 x)=0,
\end{array}
$$
we solve for $x_{1}=-1, x_{2}=-\frac{1}{3}$.
When $x-\frac{1}{3}$, $f^{\prime}(x)>0$;
When $-1< x < -\frac{1}{3}$, $f^{\prime}(x)<0$;
When $x > -\frac{1}{3}$, $f^{\prime}(x)>0$.
When $a < -\frac{4}{3}$,
$$
F(a)=f(a)=a... | \frac{1}{9} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 720,957 |
17. (12 points) Given $x>0, y>0, z>0$, and $x y z=1$. Prove:
$$
1<\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}<2 .
$$ | 17. For any $a>0$, let $b=a x, c=b y$.
From $x y z=1$, we get $x=\frac{b}{a}, y=\frac{c}{b}, z=\frac{a}{c}$. Therefore,
$$
\begin{array}{l}
\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c} \\
>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{b}{a+b+c}=1 . \\
\text { Also, } \frac{a}{a+b}+\frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,958 |
18. (15 points) Let $K, L, M, N$ be points on the edges $AB, BC, CD, DA$ of tetrahedron $ABCD$, respectively. If $K, L, M, N$ are coplanar, and $\frac{AN}{AD}=\frac{BL}{BC}$, prove that $\frac{DM}{MC}=\frac{AK}{KB}$. | 18. Let the plane $K L M N$ be $\alpha$, and the line $A C$ be $l$.
(1) If $l / / a$ (as shown in Figure 6).
From $\frac{A N}{A D}=\frac{B L}{B C}$, we have $\frac{D N}{N A}=\frac{C L}{L B}$.
Since $l / / \alpha$, then $M N / / l, K L / / l$.
Thus, $\frac{D M}{M C}=\frac{D N}{N A}, \frac{A K}{K B}=\frac{C L}{L B}$, whi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,959 |
19. (15 points) Given the function
$$
f(x)=a x^{4}+b x^{3}+c x^{2}+d x
$$
satisfies the following conditions:
(1) $a, b, c, d$ are all greater than 0;
(2) For any $x \in\{-2,-1,0,1,2\}$, $f(x)$ is an integer;
(3) $f(1)=1, f(5)=70$.
Determine whether, for every integer $x$, $f(x)$ is an integer, and provide an argument... | 19. Since $f(1)=a+b+c+d$,
$$
\begin{array}{l}
f(-1)=a-b+c-d, \\
f(1)+f(-1)=2 a+2 c, \\
f(1)-f(-1)=2 b+2 d,
\end{array}
$$
Therefore, $2 a+2 c$ and $2 b+2 d$ are both integers.
Since $f(2)=16 a+8 b+4 c+2 d$,
$$
\begin{array}{l}
f(-2)=16 a-8 b+4 c-2 d, \\
f(2)+f(-2)=32 a+8 c, \\
f(2)-f(-2)=16 b+4 d,
\end{array}
$$
Ther... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,960 |
3. For a positive integer $n$, its decimal representation consists only of the digits 0 and 1, and it is divisible by 225. Find the minimum value of $n$.
| (Tip: From $25 \mid n$, we know that the last two digits of $n$ are multiples of 25. But since all the digits of $n$ are 0 and 1, the last two digits of $n$ must be 00. From $91n$, we know that the sum of the digits of $n$ is a multiple of 9, so $n$ must contain at least 9 ones. Therefore, the smallest value of $n$ is ... | 11111111100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,961 |
1. If $x-\frac{1}{x}=\sqrt{5}$, then the value of $\frac{x^{10}+x^{6}+x^{4}+1}{x^{10}+x^{8}+x^{2}+1}$ is ( .
(A) $\frac{47}{42}$
(B) $\frac{42}{47}$
(C) $\frac{45}{49}$
(D) $\frac{49}{45}$ | $$
\begin{array}{l}
\text { Given } x^{2}+\frac{1}{x^{2}}-2=5, \text { i.e., } x^{2}+\frac{1}{x^{2}}=7. \text { Rearranging, we get } x^{4}+1=7 x^{2}. \\
\text { Therefore, the original expression is } \frac{\left(x^{6}+1\right)\left(x^{4}+1\right)}{\left(x^{2}+1\right)\left(x^{8}+1\right)} \\
=\frac{\left(x^{2}+1\righ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,962 |
2. In the Cartesian coordinate system, point $A$ is in the first quadrant, point $B$ is in the second quadrant, and point $C$ is on the coordinate axis, satisfying that $\triangle A B C$ is a right triangle. The maximum number of points $C$ is $a$, and the minimum number of points $C$ is $b$. The value of $a+b$ is ( )
... | 2.D.
Since point $A$ is in the first quadrant and point $B$ is in the second quadrant, the perpendicular lines through $A$ and $B$ to $AB$ must intersect the $x$-axis at two points, and intersect the $y$-axis at most at two points, and at least at 0 points; the circle with $AB$ as its diameter must intersect the $y$-a... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,963 |
3. A bag contains 2000 red balls, 1000 black balls; 10 yellow balls, these balls differ only in color. To ensure that 1000 balls of the same color are drawn, at least ( ) balls should be drawn.
(A) 1010
(B) 2000
(C) 2008
(D) 2009 | 3. D.
If 999 red balls, 999 black balls, and 10 yellow balls are drawn, a total of $999 \times 2+10=2008$ balls are drawn, and there are no 1000 balls of the same color.
According to the pigeonhole principle, at least 2009 balls must be drawn to ensure that 1000 balls of the same color are drawn. | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,964 |
4. In Rt $\triangle A B C$, $\angle A=30^{\circ}, \angle C=$ $90^{\circ}$, equilateral triangles $\triangle A B D$ and $\triangle A C E$ are constructed outward from $A B$ and $A C$ respectively, and $D E$ is connected, intersecting $A C$ and $A B$ at points $F$ and $G$. Then the value of $\frac{A F}{A G}$ is ( ).
(A) ... | 4.C.
As shown in Figure 3, draw $E M \perp A C$ at $M$, extend $E M$ to intersect $A B$ at $N$, and connect $D N$. Then $M$ is the midpoint of $A C$, and $M N \parallel C B$. Therefore, $N$ is the midpoint of $A B$. Thus, $D N \perp A B$.
Also, $\angle E A N = \angle D A M = 60^{\circ} + 30^{\circ} = 90^{\circ}$, whic... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,965 |
5. Given the quadratic function $f(x)$ satisfies $f(-2)=1$, $f(2)=2$, and $x \leqslant f(x) \leqslant \frac{1}{4}\left(x^{2}+4\right)$ for all real numbers $x$. Then the value of $f(4)$ is ( ).
(A) $\frac{17}{4}$
(B) $\frac{9}{2}$
(C) $\frac{19}{4}$
(D) 5 | Let $f(x)=a x^{2}+b x+c(a \neq 0)$.
From the given information, we have
$$
\left\{\begin{array}{l}
f(-2)=4 a-2 b+c=1, \\
f(2)=4 a+2 b+c=2 .
\end{array}\right.
$$
Solving these, we get $b=\frac{1}{4}, c=\frac{3}{2}-4 a$.
Also, $x \leqslant f(x) \leqslant \frac{1}{4}\left(x^{2}+4\right)$ holds for all real numbers $x$, ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,966 |
$1 . \underbrace{66 \cdots}_{100 \uparrow} \underbrace{77 \cdots 7}_{100 \uparrow} \underbrace{88 \cdots 8}_{100 \uparrow} 8$ divided by 72 has a remainder of | $$
\begin{array}{l}
\underbrace{66 \cdots 6}_{100 \uparrow} \underbrace{77 \cdots}_{100 \uparrow} \underbrace{88 \cdots 8}_{100 \uparrow} \\
=\underbrace{66 \cdots 677 \cdots}_{10 \uparrow} \underbrace{77}_{100 \uparrow} \underbrace{88 \cdots 82000}_{96 \uparrow}+72 \times 95+48 . \\
\underbrace{66 \cdots}_{100 \uparro... | 48 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,968 |
2. In trapezoid $A B C D$, $A D / / B C, A B=A C$, $B C=B D=(\sqrt{2}+1) C D$. Then the degree measure of $\angle B A C+\angle B D C$ is $\qquad$ . | 2.180 .
As shown in Figure 5, draw $A E \perp B C$ at $E$ and $D F \perp B C$ at $F$. Then
$$
\begin{array}{l}
B E=E C=\frac{1}{2} B C, \\
\angle C A E=\frac{1}{2} \angle B A C,
\end{array}
$$
and quadrilateral $A E F D$ is a rectangle. | 180 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,969 |
3. Given that $n$ is an integer, and the quadratic equation in $x$
$$
(n-1)^{2} x^{2}-5 n(n-1) x+\left(6 n^{2}-n-1\right)=0
$$
has at least one integer root. Then the sum of all possible values of $n$ is $\qquad$ | $$
[(n-1) x-(2 n-1)][(n-1) x-(3 n+1)]=0 .
$$
Since $n \neq 1$, then
$$
\begin{array}{l}
x_{1}=\frac{2 n-1}{n-1}=2+\frac{1}{n-1}, \\
x_{2}=\frac{3 n+1}{n-1}=3+\frac{4}{n-1} .
\end{array}
$$
Because the original equation has at least one integer root, and $n$ is an integer, so, $n-1= \pm 1, \pm 2, \pm 4$.
Solving gives... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,970 |
4. Given that $\alpha$ is an acute angle, and $3 \sin \alpha+4 \cos \alpha=5$. Then $\tan \alpha=$ $\qquad$ . | 4. $\frac{3}{4}$.
In $\triangle ABC$, $\angle ACB=90^{\circ}$, let $\angle A=\alpha$. Then $\sin \alpha=\frac{a}{c}, \cos \alpha=\frac{b}{c}$. From the given information and the Pythagorean theorem, we have
$$
\left\{\begin{array}{l}
\frac{3 a}{c}+\frac{4 b}{c}=5, \\
a^{2}+b^{2}=c^{2} .
\end{array}\right.
$$
Solving ... | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,971 |
4. Add three digits after 764 to make the resulting six-digit number divisible by 8, 9, and 11. | (Let the six-digit number be $\overline{764 a b c}$. Let $M=7+4$ $+b=11+b, N=c+a+6$. Then $\overline{a b c}$ is divisible by 8, $M+N=17+a+b+c$ is divisible by 9, $M-N=$ $5+b-a-c$ is divisible by 11. It is easy to see that $17 \leqslant M+N \leqslant$ $44,-13 \leqslant M-N \leqslant 14$. Therefore, $M+N=18,27,36$, $M-N=... | 764280 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,972 |
One, (20 points) Given real numbers $a, b, c, x, y, z$ satisfy that at least one of $a^{2}+b^{2}-c^{2}$ and $x^{2}+y^{2}-z^{2}$ is negative. Prove:
$$
(a x+b y-c z)^{2} \geqslant\left(a^{2}+b^{2}-c^{2}\right)\left(x^{2}+y^{2}-z^{2}\right) .
$$ | Let's assume $a^{2}+b^{2}-c^{2}<0$. Then $c \neq 0$. Construct a quadratic function of $t$, denoted as
$$
m=\left(a^{2}+b^{2}-c^{2}\right) t t_{1}^{2}+2(a x+b y-c) t+\left(x^{2}+y^{2}-z^{2}\right) \text {. }
$$
Then the graph of the quadratic function $m$ opens downwards. Note that
$$
m=(a t+x)^{2}+(b t+y)^{2}-(c t+z)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,973 |
II. (25 points) In $\triangle ABC$, the angle bisectors of $\angle A$ and $\angle B$ intersect at point $P$. If $S_{\text{quadrilateral } ABDE} = 2 S_{\triangle APB}$, prove that $\angle ACB = 90^{\circ}$. | As shown in Figure 7, draw $GH \perp AP$, intersecting $AC$ and $AB$ at $G$ and $H$ respectively, and connect $GB$, $GD$, and $CP$. It is easy to see that
$$
\triangle APG \cong \triangle APH,
$$
which implies $GP = HP$.
Thus, $S_{\triangle APG} = S_{\triangle APH}$,
$$
S_{\triangle BPG} = S_{\triangle BPF} \text{ : }... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,974 |
Three, (25 points) In any permutation of $1,2, \cdots, 200$, there always exists a sum of 20 consecutive numbers that is not less than $a$. Find the maximum value of $a$.
| Three, take any permutation, and divide it into 10 groups, each containing 20 numbers. Let $A_{1}=\left(b_{1}, b_{2}, \cdots, b_{20}\right), A_{2}=\left(b_{21}, b_{2}\right.$, $\left.\cdots, b_{40}\right), \cdots, A_{10}=\left(b_{181}, b_{182}, \cdots, b_{200}\right)$.
Take the average of the 10 groups of numbers
$$
\b... | 2010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,975 |
1. As shown in Figure 1, the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ divide it into four parts, with areas $S_{1}, S_{2}, S_{3}, S_{4}$, respectively. Given that $S_{1}>1, S_{2}>1$. Then $S_{3}+S_{4}$ ).
$(\mathrm{A})=2$
(B) $>2$
(C) $<2$
(D) Not necessarily | -,1.B.
It is known that $S_{1} S_{2}=S_{3} S_{4}$.
Therefore, $S_{3}+S_{4} \geqslant 2 \sqrt{S_{3} S_{4}}=2 \sqrt{S_{1} S_{2}}>2$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,976 |
2. Connecting the intersection points of the hyperbola $x y=20$ and the circle $x^{2}+y^{2}=41$ in sequence forms a convex quadrilateral. The area of this quadrilateral is ( ).
(A) 18
(B) 20
(C) 22
(D) 30 | 2.A.
Let $A\left(x_{0}, y_{0}\right)\left(x_{0}>0, y_{0}>0\right)$.
From the fact that the two curves are symmetric with respect to the origin and the line $y=x$, the coordinates of the other three intersection points are $B\left(y_{0}, x_{0}\right)$,
$$
C\left(-x_{0},-y_{0}\right) 、 D\left(-y_{0},-x_{0}\right) \text ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,977 |
3. Given an acute triangle $\triangle A B C$. The following judgments are given:
(1) Three segments of lengths $\sin 2 A 、 \sin 2 B 、 \sin 2 C$ can definitely form a triangle;
(2) Three segments of lengths $\cos A 、 \cos B 、 \cos C$ can definitely form a triangle;
(3) Three segments of lengths $\cos A 、 \cos B 、 \sin C... | 3.C.
$$
\begin{array}{l}
\sin 2 A+\sin 2 B-\sin 2 C \\
=2 \sin (A+B) \cdot \cos (A-B)-2 \sin C \cdot \cos C \\
=2 \sin C[\cos (A-B)+\cos (A+B)] \\
=4 \cos A \cdot \cos B \cdot \sin C>0 .
\end{array}
$$
Similarly, $\sin 2 A+\sin 2 C-\sin 2 B>0$,
$$
\sin 2 B+\sin 2 C-\sin 2 A>0 \text {. }
$$
Therefore, (1) is correct.
... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,978 |
4. Given a spatial quadrilateral $A B C D, A B=a, B C=$ $b, C D=c, D A=d$. Then $\overrightarrow{A C} \cdot \overrightarrow{B D}=(\quad)$.
(A) $\frac{1}{2}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$
(B) $-\frac{1}{2}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$
(C) $\frac{1}{2}\left(a^{2}+c^{2}-b^{2}-d^{2}\right)$
(D) $\frac{1}{2}\l... | 4.D.
$$
\begin{array}{l}
\overrightarrow{A C} \cdot \overrightarrow{B D}=(\overrightarrow{A B}+\overrightarrow{B C}) \cdot(\overrightarrow{B C}+\overrightarrow{C D}) \\
=\overrightarrow{B C}^{2}+\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C D}+\overrightarrow{C D} \cdot \o... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,979 |
5. The range of the function $y=[\sin x \cdot \cos x]+[\sin x+\cos x]$ is $(\quad)([x]$ denotes the greatest integer not exceeding the real number $x$”.
(A) $\{-2,-1,0,1,2\}$
(B) $\{-2,-1,0,1\}$
(C) $\{-1,0,1\}$
(D) $\{-2,-1,1\}$ | 5.D.
$$
y=\left[\frac{1}{2} \sin 2 x\right]+\left[\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\right]
$$
The following discussion assumes $k \in \mathbf{Z}$.
(1) When $2 k \pi \leqslant x \leqslant 2 k \pi+\frac{\pi}{2}$, $y=1$;
(2) When $2 k \pi+\frac{\pi}{2}<x \leqslant 2 k \pi+\frac{3 \pi}{4}$, $y=-1$;
(3) When $2 k ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,980 |
6. Given $S=\{1,2, \cdots, 216\}, A \subseteq S$. If the sum of any two elements in set $A$ cannot be divisible by 6, then the maximum number of elements in set $A$ is ( ).
(A) 36
(B) 52
(C) 74
(D) 90 | $$
\begin{array}{c}
\text { 6.C. } \\
\text { Let } S_{k}=\{x \in S \mid x=6 n+k, n \in \mathbb{N}(k= \\
0,1, \cdots, 5) \text {, and } S=\bigcup_{k=0}^{5} S_{k} .
\end{array}
$$
It is known that $\operatorname{card}\left(S_{k}\right)=36$. Then, in set $A$, it cannot simultaneously contain elements from $S_{1}$ and $S... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,981 |
1. Given the general term of the sequence $\left\{a_{n}\right\}$
$$
a_{n}=\frac{(n+1)^{4}+n^{4}+1}{(n+1)^{2}+n^{2}+1} \text {. }
$$
Then the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, $S_{n}=$ $\qquad$ | $$
\text { II. 1. } \frac{n\left(n^{2}+3 n+5\right)}{3} \text {. }
$$
Simplifying, we get $a_{n}=n^{2}+n+1$.
$$
\begin{array}{l}
\text { Therefore, } S_{n}=\sum_{k=1}^{n}\left(k^{2}+k+1\right) \\
=\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}+n \\
=\frac{1}{3} n\left(n^{2}+3 n+5\right)
\end{array}
$$ | \frac{n\left(n^{2}+3 n+5\right)}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,982 |
5. Use $0,1, \cdots, 9$ to form a ten-digit number without repeated digits that is divisible by 11. Find the largest and smallest numbers among them. | (Prompt: Let the ten-digit number be $A=\overline{x_{1} x_{2} \cdots x_{10}}$. Let $x=x_{1}+x_{3}+x_{5}+x_{7}+x_{9}, y=x_{2}+x_{4}+x_{6}+$ $x_{8}+x_{10}$. Then $11 \mid (x-y)$. Since $10 \leqslant x, y \leqslant 25$, therefore, $|x-y|=0,11,22$. Because $x_{1}, x_{2}, \cdots, x_{10}$ is a permutation of $0,1, \cdots, 9$... | 9876524130, 1024375869 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,983 |
2. Given $f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9}$, $x \in[0,1]$. Provide the following conclusions:
(1) $f(x)>0$;
(2) $f(x)<0$;
(3) There exists $x_{0} \in[0,1)$, such that $f\left(x_{0}\right)=0$;
(4) There exists $x_{0} \in[0,1]$, such that $f\left(x_{0}\right)<0$.
Among them, the correct conclusion numbers are... | 2. (1)
$$
\begin{array}{l}
f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9} \\
=x^{3}(1-x)+3 x^{3}-3 x^{2}+(x-1)^{2}+\frac{4}{9} \\
=x^{3}(1-x)+(x-1)^{2}+\frac{3}{2} x^{3}+\frac{3}{2} x^{3}+\frac{4}{9}-3 x^{2} \\
\geqslant x^{3}(1-x)+(x-1)^{2}+3 \sqrt[3]{\frac{3}{2} x^{3} \cdot \frac{3}{2} x^{3} \cdot \frac{4}{9}}-3 x^{2} ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,984 |
3. As shown in Figure 2, given a regular tetrahedron $A B C D$ with edge length 1, $M$ is the midpoint of $A C$, and $P$ lies on the line segment $D M$. Then the minimum value of $A P+B P$ is | 3. $\sqrt{1+\frac{\sqrt{6}}{3}}$.
Let $\angle B D M=\theta$. In $\triangle B D M$,
$$
B D=1, B M=M D=\frac{\sqrt{3}}{2} \text {. }
$$
Thus, $\cos \theta=\frac{\sqrt{3}}{3}, \sin \theta=\frac{\sqrt{6}}{3}$.
As shown in Figure 5, rotate $\triangle B D M$ around $D M$ so that $\triangle B D M$ lies in the plane $A C D$,... | \sqrt{1+\frac{\sqrt{6}}{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,985 |
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