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
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
5. We perform operations on the cards placed at points $A_{1}, A_{2}, \cdots, A_{n}(n \geqslant 3)$
and point $O$. An operation is defined as one of the following:
(1) If the number of cards at some point $A_{i}$ is no less than 3, then 3 cards can be taken out, and one card can be placed at points $A_{i-1}, A_{i+1}$,... | 5. Consider only the case where the total number of cards equals $n^{2}+3 n+1$.
Adopt the following strategy.
If there is a point $A_{i}$ with no less than 3 cards, then perform operation (1) on the cards at point $A_{i}$. Such an operation increases the number of cards at point $O$ by 1, so after a finite number of op... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,181 |
1. As shown in Figure 1, it is known that rectangle $A B C D$ can be exactly divided into seven small rectangles of the same shape and size. If the area of the small rectangle is 3, then the perimeter of rectangle $A B C D$ is | $-1.19$.
Let the length of the small rectangle be $a$ and the width be $b$. Then $3a = 4b \Rightarrow a = \frac{4}{3}b \Rightarrow ab = \frac{4}{3}b^2 = 3$. Solving this, we get $b = \frac{3}{2}, a = 2$.
Therefore, the perimeter of rectangle $ABCD$ is $2(4a + b) = 8a + 2b = 16 + 3 = 19$. | 19 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,183 |
2. In the Cartesian coordinate system, a point whose both horizontal and vertical coordinates are integers is called an "integer point". The number of integer points on the graph of the function $y=\frac{x+12}{2 x-1}$ is $\qquad$ | 2. 6 .
Notice that $2 y=1+\frac{25}{2 x-1}$.
Since $y$ is an integer, $\frac{25}{2 x-1}$ must also be an integer. Therefore, $2 x-1= \pm 1, \pm 5, \pm 25$.
Solving for $x$ gives $x=-12,-2,0,1,3,13$.
Correspondingly, $y=0,-2,-12,13,3,1$.
Thus, there are 6 integer points. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,184 |
3. If $\triangle A B C$ has exactly one solution when $\angle A B C=60^{\circ}, A C=12, B C=$ $k$, then the range of values for $k$ is $\qquad$ | $3.0<k \leqslant 12$ or $k=8 \sqrt{3}$. As shown in Figure 4, we can obtain
$$
0<k \leqslant 12
$$
or $k=8 \sqrt{3}$. | 3.0<k \leqslant 12 \text{ or } k=8 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,185 |
4. Given that $a$ and $b$ are integers. Then the number of ordered pairs $(a, b)$ that satisfy $a + b + ab = 2008$ is $\qquad$ groups. | 4. 12 .
$$
\begin{array}{r}
\text { Since }(a+1)(b+1) \\
=2009=41 \times 7^{2} \text {, and given }
\end{array}
$$
that $a, b$ take integer values, so, $a+1, b+1$ have 12 factor combinations, i.e., $1 \times 2009, 7 \times 287, 41 \times 49$ and their negatives as well as swaps. | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,186 |
5. The positive integer solutions of the equation $2 x^{2}+y^{2}-2 x y-4 x-30=0$ are $\qquad$ | 5. $(5,10),(7,4),(7,10)$.
Since $y^{2}-2 x y+\left(2 x^{2}-4 x-30\right)=0$ is a quadratic equation in $y$ with positive integer solutions, we have:
$$
34-(x-2)^{2}=k^{2}
$$
is a perfect square.
Thus, $k^{2}$ can only be $25,16,9,4,1,0$.
From $34-(x-2)^{2}=25 \Rightarrow(x-2)^{2}=9$, we get $x=5, y=10$;
From $34-(x-2... | (5,10),(7,4),(7,10) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,187 |
6. In $\triangle A B C$, $A C=B C$, and $A C=\frac{1+\sqrt{5}}{2} A B$, $A E$ and $B D$ are two angle bisectors, $E F / / A C$ intersects $B D$ at point $F$. Prove: $B E=D F$. | 6. Since $A C=\frac{1+\sqrt{5}}{2} A B$, then,
$$
A C-A B=\frac{\sqrt{5}-1}{2} A B,
$$
i.e., $A C(A C-A B)=A B^{2}$.
As shown in Figure 5, take a point $D_{1}$ on segment $A C$ such that $C D_{1}=A B$. From equation (1), we get
$$
\begin{array}{l}
A B^{2}=A D_{1} \cdot A C \\
\Rightarrow \frac{A D_{1}}{A B}=\frac{A B}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,188 |
Example 6 Solve the equation $\frac{13 x-x^{2}}{x+1}\left(x+\frac{13-x}{x+1}\right)=42$. (1998, Changchun City, Jilin Province Mathematics Competition (Grade 9)) | Solve the original equation by transforming it as follows:
$$
\begin{array}{l}
\frac{13 x-x^{2}}{x+1}\left(\frac{x^{2}+13}{x+1}\right)=42 . \\
\text { Let } \frac{13 x-x^{2}}{x+1}=a+b, \\
\frac{x^{2}+13}{x+1}=a-b .
\end{array}
$$
Multiplying (1) and (2), and adding (1) and (2) respectively, we get
$$
a^{2}-b^{2}=42 \te... | x_{1}=1, x_{2}=6, x_{3}=3+\sqrt{2}, x_{4}=3-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,189 |
7. If the equation $\left(x^{2}-1\right)\left(x^{2}-4\right)=k$ has four non-zero real roots, and the four points corresponding to them on the number line are equally spaced, find the value of $k$. | 7. From $\left(x^{2}-1\right)\left(x^{2}-4\right)=k$ having four non-zero real roots $x_{1}, x_{2}, x_{3}, x_{4}\left(x_{1}<x_{2}<x_{3}<x_{4}\right)$, we know that the equation
$$
(t-1)(t-4)=k \Rightarrow t^{2}-5 t+4-k=0
$$
has two distinct positive real roots, and $x_{1}, x_{2}, x_{3}, x_{4}$ correspond to four point... | \frac{7}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,190 |
8. Arrange the numbers $1,2, \cdots, 10$ in a row in some order, such that the sum of any three consecutive numbers does not exceed $n$. Answer the following:
(1) When $n=10$, can it be arranged? Please explain your reasoning;
(2) When it can be arranged, what is the minimum value of $n$? | 8. (1) Suppose $n=10$ is already arranged. Then the sum of the last nine numbers is less than or equal to 30. Thus, the first number is not less than 25, which is a contradiction. Therefore, it cannot be arranged.
(2) With the same consideration as (1).
When $n=11,12,13,14$, none can be arranged.
When $n=15$, the sum o... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,191 |
9. Let $x_{1}, x_{2}, \cdots, x_{n}$ take values 7 or -7, and satisfy
(1) $x_{1}+x_{2}+\cdots+x_{n}=0$;
(2) $x_{1}+2 x_{2}+\cdots+n x_{n}=2009$.
Determine the minimum value of $n$. | 9. First, when $n=34$,
$$
\begin{array}{l}
x_{1}=x_{2}=\cdots=x_{16}=x_{18}=-7, \\
x_{17}=x_{19}=x_{20}=\cdots=x_{34}=7
\end{array}
$$
satisfies the conditions of the problem.
Second, divide both sides of condition (2) by 7, and let $y_{i}=\frac{x_{i}}{7}(i=1,2, \cdots, n)$. Then
$$
\begin{array}{l}
y_{i}= \pm 1, \\
y... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,192 |
1. (15 points) As shown in Figure 2, $M$, $N$, and $P$ are the midpoints of the sides $AB$, $BC$, and $CA$ of $\triangle ABC$, respectively. $BP$ intersects $MN$ and $AN$ at points $E$ and $F$.
(1) Prove:
$$
BF = 2FP \text{; }
$$
(2) Let the area of $\triangle ABC$ be $S$. Find the area of $\triangle NEF$. | 1. (1) As shown in Figure 6, connect $P N$. Then $P N \parallel A B$, and $P N = \frac{1}{2} A B$.
Therefore, $\triangle A B F \sim \triangle N P F$. Hence $\frac{B F}{F P} = \frac{A F}{F N} = \frac{A B}{P N} = 2$.
Thus, $B F = 2 F P$.
(2) As shown in Figure 7, take the midpoint $G$ of $A F$, and connect $M G$. Then
$$... | \frac{1}{24} S | Geometry | proof | Yes | Yes | cn_contest | false | 722,193 |
2. (15 points) Let $x_{1}, x_{2}, \cdots, x_{2008}$ be integers, and satisfy the following conditions:
(1) $-1 \leqslant x_{n} \leqslant 2 \quad (n=1,2, \cdots, 2008)$;
(2) $x_{1}+x_{2}+\cdots+x_{2008}=200$;
(3) $x_{1}^{2}+x_{2}^{2}+\cdots+x_{2008}^{2}=2008$.
Find the minimum and maximum values of $x_{1}^{3}+x_{2}^{3}+... | 2. Let $x_{1}, x_{2}, \cdots, x_{2008}$ have $q$ zeros, $r$ negative ones, $s$ ones, and $t$ twos. Then
$$
\left\{\begin{array}{l}
-r+s+2 t=200, \\
r+s+4 t=2008 .
\end{array}\right.
$$
Adding the two equations gives $s+3 t=1104$.
Hence $0 \leqslant t \leqslant 368$.
$$
\begin{array}{l}
\text { By } x_{1}^{3}+x_{2}^{3}... | 200 \leqslant x_{1}^{3}+x_{2}^{3}+\cdots+x_{2008}^{3} \leqslant 2408 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,194 |
3. (15 points) Fill seven different perfect squares into seven consecutive boxes so that the sum of any three adjacent boxes is greater than 100. Find the minimum possible value of the sum of these seven perfect squares.
| 3. Since the three perfect squares at positions $1, 4$, and $7$ are all different, considering the smallest possible values, one of these numbers must be at least 9. After positioning 9, the remaining six positions can be divided into two groups, each with three consecutive numbers whose sum is at least 101. Therefore,... | 211 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,195 |
4. (15 points) If there exists a right-angled triangle with all sides of integer length, and the ratio $k$ of the square of its perimeter to its area is an integer, then this integer $k$ is called a "triangle number". Find all such triangle numbers.
Translate the above text into English, please retain the original tex... | 4. Let the three sides of a right-angled triangle be \(m^{2}-n^{2}\), \(2 m n\), and \(m^{2}+n^{2}\) \((m, n \in \mathbf{N}_{+}, m > n)\).
Assume \(m\) and \(n\) are coprime.
Then the perimeter \(l = 2 m^{2} + 2 m n\),
and the area \(S = m n (m^{2} - n^{2})\).
Thus, \(k = \frac{l^{2}}{S} = \frac{4 m (m + n)}{n (m - n)}... | 24, 30, 45 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 722,196 |
5. (15 points) Prove: In any 18 consecutive three-digit numbers, there must be a three-digit number that is divisible by the sum of its three digits. | 5. If 999 is contained in 18 consecutive three-digit numbers, then 999 satisfies the condition.
If 999 is not contained in 18 consecutive three-digit numbers, then there must be a number that is a multiple of 18, and thus this number can be divided by 9. Therefore, the sum of its digits can be divided by 9. So, the su... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,197 |
6. (15 points) On a piece of acute triangle scrap, process a square part so that the four vertices of the square are all on the edges of the triangle. If the three sides of the triangle are $a$, $b$, and $c$ ($a > b > c$), try to find: On which side should the two vertices of the square be placed to maximize the area o... | 6. Let the side lengths of the squares on $a$, $b$, and $c$ be $x_{a}$, $x_{b}$, and $x_{c}$ respectively (Figure 8 shows a scenario where two vertices of the square lie on the edge $a$), and the heights on $a$, $b$, and $c$ are denoted as $h_{a}$, $h_{b}$, and $h_{c}$. Then $h_{a}<h_{b}<h_{c}$. Hence | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,198 |
Example 7 The system of equations $\left\{\begin{array}{l}x+x y+y=1, \\ x^{2}+x^{2} y^{2}+y^{2}=17\end{array}\right.$ has the real solution $(x, y)=$ $\qquad$
(1996, Eastern Airlines Cup Shanghai Junior High School Mathematics Competition) | Let $x=a+b, y=a-b$. Then $x+y=2a, xy=a^{2}-b^{2}$.
Thus, the original system of equations becomes
$$
\left\{\begin{array} { l }
{ a ^ { 2 } - b ^ { 2 } + 2 a = 1 , } \\
{ 2 a ^ { 2 } + 2 b ^ { 2 } + ( a ^ { 2 } - b ^ { 2 } ) ^ { 2 } = 1 7 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=\frac{3}{2}, \\
b= \pm \frac... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,200 |
8. (15 points) A and B are playing a game of writing numbers on the blackboard: each time they write a number, it must be a positive integer not exceeding 10, and no repetition is allowed; starting from the third number, each number must also be the difference of two previously written numbers. The player who cannot wr... | 8. Player B has a winning strategy.
Consider the first number written by Player A.
If Player A writes an even number, Player B writes half of this number, then Player A cannot write again.
If Player A writes an odd number:
(1) If Player A writes 1, then Player B writes 2; if Player A writes 3, then Player B writes 6; ... | Player B has a winning strategy. | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 722,201 |
1. If the set $M=\left\{x \left\lvert\, \log _{\frac{1}{2}}(x-1)>-1\right.\right\}$, $N=\left\{x \mid 1<2^{x}<4\right\}$, then $M \cap N=(\quad)$.
(A) $\{x \mid 1<x<3\}$
(B) $\{x \mid 1<x<2\}$
(C) $\{x \mid 0<x<3\}$
(D) $\{x \mid 0<x<2\}$ | -.1. B.
Notice
$$
\begin{array}{l}
x \in M \Leftrightarrow \log _{\frac{1}{2}}(x-1)>-1 \\
\Leftrightarrow\left\{\begin{array}{l}
x-1>0, \\
x-1<\left(\frac{1}{2}\right)^{-1}=2
\end{array} \Leftrightarrow 1<x<3 ;\right.
\end{array}
$$
$$
x \in N \Leftrightarrow 1<2^{x}<4 \Leftrightarrow 0<x<2 \text {. }
$$
Therefore, $M... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,202 |
2. The increasing interval of the function $f(x)=\log _{\frac{1}{2}}\left(-x^{2}-2 x+3\right)$ is ( ).
(A) $(-3,-1)$
(B) $[-1,1)$
(C) $(-3,-1]$
(D) $(-1,1)$ | 2. B.
From $-x^{2}-2 x+3>0$, we get $-3<x<1$.
Therefore, the vertex of the parabola $y=-x^{2}-2 x+3$ is at $(-1,4)$. Thus, the function $y=-x^{2}-2 x+3$ is decreasing in the interval $[-1,1)$.
$$
\begin{array}{l}
\text { Also, } \frac{1}{2}<1, \text { hence, the function } \\
f(x)=\log _{\frac{1}{2}}\left(-x^{2}-2 x+3... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,203 |
3. If for any $x \in \mathbf{R}$, the function $f(x)$ satisfies
$$
f(x+2.009)=-f(x+2008) \text {, }
$$
and $f(2009)=-2009$,
then $f(-1)=(\quad)$.
(A) 1 (B) -1
(C) 2009
(D) -2009 | 3. D.
Notice
$$
\begin{array}{l}
f(x+2009)=-f(x+2008) \\
=-f((x-1)+2009) \\
=f(x-1+2008)=f(x+2007) .
\end{array}
$$
Then $f(x)$ is a periodic function with a period of 2.
Thus $f(-1)=f(2009-2 \times 1005)$
$$
=f(2009)=-2009 .
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,204 |
4. In $\triangle A B C$, it is known that
$$
\begin{array}{l}
(\sqrt{3} \sin B-\cos B)(\sqrt{3} \sin C-\cos C) \\
=4 \cos B \cdot \cos C,
\end{array}
$$
and $A B+A C=4$. Then the range of $B C$ is ( ).
(A) $(2,4)$
(B) $(2,4]$
(C) $[2,4)$
(D) $[2,4]$ | 4. C.
From the given, we have
$$
\begin{array}{l}
\sqrt{3}(\sin B \cdot \cos C+\cos B \cdot \sin C) \\
=-3(\cos B \cdot \cos C-\sin A \cdot \sin C) \\
\Rightarrow \sqrt{3} \sin (B+C)=-3 \cos (B+C) \\
\Rightarrow \tan (B+C)=-\sqrt{3} .
\end{array}
$$
Since $0<\angle B+\angle C<\pi$, we have
$$
\angle B+\angle C=\frac{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,205 |
5. For any $x \in \mathbf{R}, [x]$ represents the greatest integer not exceeding the real number $x$. Then the set of $x$ that satisfies $\left[\left|x^{2}-1\right|\right]=10$ is $(\quad)$.
(A) $(-2 \sqrt{3},-\sqrt{11})$
(B) $[\sqrt{11}, 2 \sqrt{3}]$
(C) $(-2 \sqrt{3},-\sqrt{11}] \cup[\sqrt{11}, 2 \sqrt{3})$
(D) $[-2 \... | 5. C.
It is known that when $[x] \leqslant x1$, we have
$$
\begin{array}{l}
10 \leqslant x^{2}-1<11 \Rightarrow 11 \leqslant x^{2}<12 \\
\Rightarrow \sqrt{11} \leqslant|x|<2 \sqrt{3} \\
\Rightarrow \sqrt{11} \leqslant x<2 \sqrt{3} \text { or }-2 \sqrt{3}<x \leqslant-\sqrt{11} .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,206 |
6. Given two quadratic equations
$$
a x^{2}+b x+c=0, u x^{2}+v x+w=0(a \neq u)
$$
both have real roots. If the coefficients of the quadratic terms of these two equations are swapped, then $w c>0$ is the ( ) condition for at least one of the two quadratic equations obtained after the swap to have real roots.
(A) Suffic... | 6. A.
Given the discriminants of two quadratic equations are
$$
\Delta_{1}=b^{2}-4 a c, \Delta_{2}=v^{2}-4 u w \text {. }
$$
After swapping the quadratic coefficients, the discriminants of the equations become $\Delta_{1}^{\prime}=b^{2}-4 u c, \Delta_{2}^{\prime}=v^{2}-4 a w$.
Then $\left(\Delta_{1}-\Delta_{1}^{\prim... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,207 |
7. Given that $|a|, |b|$ are integers, and satisfy
$$
\begin{array}{l}
(|a|+|b|)(|a|+3|b|)=105, \\
(a+b) \cdot(a+3 b)=33 .
\end{array}
$$
Then the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$ is ( ).
(A) $30^{\circ}$
(B) $60^{\circ}$
(C) $120^{\circ}$
(D) $150^{\circ}$ | 7. C.
Let $|\boldsymbol{a}|+|\boldsymbol{b}|=m$. Then $m(m+2|\boldsymbol{b}|)=105$.
Given $|b|<m \leqslant 10$, we have
$|b|=\frac{105}{2 m}-\frac{m}{2}<m \Rightarrow m \geqslant 6$.
Also, $105=3 \times 5 \times 7$, so $m=7$.
Thus, $|b|=\frac{105-49}{14}=4,|a|=7-4=3$.
Therefore, $(a+b) \cdot(a+3 b)$
$$
=|a|^{2}+3|b|^{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,208 |
8. In the complex plane, the point corresponding to the complex number $z_{1}$ moves on the line segment connecting 1 and $\mathrm{i}$, and the point corresponding to the complex number $z_{2}$ moves on the circle centered at the origin with a radius of 1. Then the area of the region where the point corresponding to th... | 8. B.
From the given, $z_{1}=t+(1-t) \mathrm{i}(0 \leqslant t \leqslant 1)$.
From $z_{2}=\cos \theta+\mathrm{i} \sin \theta$, we get $z_{1}+z_{2}=t+\cos \theta+(1-t+\sin \theta) \mathrm{i}$.
Let $z_{1}+z_{2}=x+y \mathrm{i}$. Then $\left\{\begin{array}{l}x=t+\cos \theta, \\ y=1-t+\sin \theta \text {. }\end{array}\right... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,209 |
9. The number of spatial quadrilaterals with vertices at the vertices and midpoints of the edges of a tetrahedron is ( ).
(A) 141
(B) 144
(C) 423
(D) 432 | 9. C.
A tetrahedron has four vertices, six edges, and each edge has a midpoint, making a total of ten points. The number of ways to choose four non-coplanar points from these ten points is as follows:
(1) The number of ways to choose four points that are all vertices is only 1;
(2) The number of ways to choose four po... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 722,210 |
Example 8 Given $x+y=a+b, x^{2}+y^{2}=a^{2}+b^{2}$. Prove: $x^{2001}+y^{2001}=a^{2001}+b^{2001}$. $(2001$, Jiangsu Province Taizhou City Mathematics Competition (Grade 8)) | Proof: Let $x=m+n, y=m-n$. Then
$$
\begin{array}{l}
m=\frac{x+y}{2}=\frac{a+b}{2} . \\
\text { Also } x^{2}+y^{2}=(m+n)^{2}+(m-n)^{2} \\
=2\left(m^{2}+n^{2}\right),
\end{array}
$$
Substituting equation (1) into the above equation and combining with $x^{2}+y^{2}=a^{2}+b^{2}$, we get
$$
n^{2}=\frac{(a-b)^{2}}{4} .
$$
T... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,211 |
10. In a regular tetrahedron $P-ABC$, there is a hemisphere whose base coincides with the base of the regular tetrahedron, and the three lateral faces of the regular tetrahedron are all tangent to the hemisphere. If the radius of the hemisphere is 1, then when the volume of the regular tetrahedron is minimized, the hei... | 10. B.
As shown in Figure 3, $O$ is the center of the base of the regular triangular pyramid and also the center of the hemisphere. $CD$ is the height of the base of the regular triangular pyramid. The side face $PAB$ is tangent to the hemisphere at point $E$, and $OE$ is connected. Then $OE \perp PD, OE=1, PO=h$. Let... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,212 |
11. If the line $3 x \sin ^{2} \alpha+y \cos ^{2} \alpha-3=0$ intersects the hyperbola $x^{2}-y^{2}=1$ at only one point, then the coordinates of this point are $\qquad$ . | $=11 \cdot(1,0),\left(\frac{17}{8}, \frac{15}{8}\right)$.
For any $\alpha \in \mathbf{R}$, the line always passes through the fixed point $(1,3)$.
When $\cos ^{2} \alpha=0$, $\sin ^{2} \alpha=1$, the equation of the line is $x=1$, substituting into $x^{2}-y^{2}=1$ yields that the line intersects the hyperbola at only o... | (1,0),\left(\frac{17}{8}, \frac{15}{8}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,213 |
12. For any $x>0$, there is always
$$
f(x)=a-x-|\lg x| \leqslant 0 \text {. }
$$
Then the range of values for $\boldsymbol{a}$ is $\qquad$ | 12. $(-\infty, \lg \mathrm{e}-\lg g \mathrm{e})$.
When $x \geqslant 1$, $|\lg x|=\lg x$.
From $x+\lg x \geqslant 1$, we know $a \leqslant 1$;
When $0 < x < 1$, $|\lg x|=-\lg x > 0$;
When $x > \lg \mathrm{e}$, $f^{\prime}(x) > 0$;
When $x < \lg \mathrm{e}$, $f^{\prime}(x) < 0$;
Since $\ln 10 > 2.3$, $\ln \mathrm{e} = 1... | (-\infty, \lg \mathrm{e} - \lg \lg \mathrm{e}] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,214 |
13. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=1+a_{n}+\sqrt{1+4 a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the general term of the sequence $a_{n}=$ $\qquad$ . | 13. $1+(n-1)(n+\sqrt{5}-1)$.
Let $b_{n}=\sqrt{1+4 a_{n}}$. Then $a_{n}=\frac{b_{n}^{2}-1}{4}$.
Thus, $\frac{b_{n+1}^{2}-1}{4}=1+\frac{b_{n}^{2}-1}{4}+b_{n}$, which means $b_{n+1}^{2}=b_{n}^{2}+4 b_{n}+4=\left(b_{n}+2\right)^{2}$.
Given $b_{1}=\sqrt{1+4 a_{1}}=\sqrt{5}$ and $b_{n}>0$, we know $b_{n+1}=b_{n}+2$, i.e., ... | 1+(n-1)(n+\sqrt{5}-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,215 |
14. Randomly throwing four dice, the probability that the sum of the numbers shown on two of the dice is 9 is $\qquad$ . | 14. $\frac{145}{324}$.
Let the result of each roll of four dice be a basic event, and let $X$ represent the set of all basic events. Clearly, $|X|=6^{4} (|X|$ represents the number of elements in set $X$).
For each $i (i=1,2, \cdots, 6)$, define the event $A_{i}$:
$A_{i}=\{$ all basic events where the number shown on ... | \frac{145}{324} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,216 |
15. (12 points) As shown in Figure 1, plane $m \parallel$ plane $n$, line segment $AD$ intersects planes $m$ and $n$ at points $B$ and $C$ respectively. Another line through point $A$ intersects planes $m$ and $n$ at points $M$ and $P$ respectively, and another line through point $D$ intersects planes $m$ and $n$ at po... | Three, 15. From the fact that plane $m //$ plane $n$, we know $B M // C P, B N // C Q$.
Therefore, $\sin \angle M B N = \sin \angle P C Q$, and
$\frac{B M}{C P} = \frac{A B}{A C}, \frac{B N}{C Q} = \frac{B D}{C D}$.
Also, $S_{\triangle B M N} = \frac{1}{2} B M \cdot B N \sin \angle M B N$,
$S_{\triangle C P Q} = \frac{... | 3 + \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,217 |
16. (12 points) For any positive integer $k(1 \leqslant k \leqslant 9)$, does there exist a corresponding quadratic function $f_{k}(x)$ such that for any positive integer $p$, $f_{k}(\underset{p \uparrow}{\overline{k k \cdots k}})=\underset{2 \rho \uparrow}{\overline{k k \cdots k}}$ (for example, $k=$ $3, p=2, \overlin... | For any positive integer $m$, we have
When $m=2 p$, we have
$$
\begin{array}{l}
\overline{k k \cdots k}=\frac{k}{9 p \uparrow}\left(10^{2 p}-1\right) \\
=\frac{k}{9}\left(10^{p}-1\right)\left(10^{p}+1\right) \\
=\frac{9}{k}\left[\frac{k}{9}\left(10^{p}-1\right)\right]^{2}+2 \cdot \frac{k}{9}\left(10^{p}-1\right) \\
=\... | f_{k}(x)=\frac{9}{k} x^{2}+2 x | Algebra | proof | Yes | Yes | cn_contest | false | 722,218 |
17. (12 points) Given the ellipse $T: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$ and the hyperbola $S: \frac{x^{2}}{m^{2}}-\frac{y^{2}}{n^{2}}=1(m>0, n>0)$ have the same focus $F(2,0)$. Let the asymptote of the hyperbola $S$ in the first quadrant be $l$. If the focus $F$ and the upper vertex $B$ of the ellips... | 17. From the given, we have $a^{2}-b^{2}=m^{2}+n^{2}=4$, the equation of the asymptote $l$ is $y=\frac{n}{m} x$, and the equation of the line $l^{\prime}$, which passes through the focus $F$ and is perpendicular to the line $l$, is $y=-\frac{m}{n}(x-2)$.
From $\left\{\begin{array}{l}y=\frac{n}{m} x, \\ y=-\frac{m}{n}(x... | \frac{11 x^{2}}{60}+\frac{11 y^{2}}{16}=1 \text{ and } \frac{5 x^{2}}{4}-\frac{5 y^{2}}{16}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,219 |
18. (15 points) On a circle, there are $k(k \geqslant 3)$ numbers. If any three consecutive numbers, in clockwise order, are denoted as $a, b, c$, it always holds that
$$
b=\alpha a+\beta c(\alpha \geqslant 0, \beta \geqslant 0, \alpha+\beta=1) .
$$
Prove: These $k$ numbers must be equal. | 18. Take any one of the numbers and denote it as $a_{0}$, and the remaining numbers in clockwise order are denoted as $a_{1}, a_{2}, \cdots, a_{k-1}$.
For any positive integer $n \geqslant k, n=p k+i, p$ is a positive integer, $i$ is an integer, and $0 \leqslant i \leqslant k-1$, let $a_{n}=a_{i}$. According to the pr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,220 |
19. (15 points) Prove: A function $f(x)$ defined on $\mathbf{R}$ that is odd can be expressed as the sum of a periodic function and a linear function if and only if the graph of $f(x)$ has a center of symmetry $(a, b)$ different from the point $(0,0)$.
Note: A linear function refers to a function of the form $y=k x+h$... | 19. Points $\left(a_{1}, b_{1}\right)$ and $\left(a_{2}, b_{2}\right)$ are symmetric about $(a, b)$
$$
\Leftrightarrow a=\frac{a_{1}+a_{2}}{2}, b=\frac{b_{1}+b_{2}}{2} \text {. }
$$
Therefore, the graph of function $f(x)$ is symmetric about point $(a, b)$
$\Leftrightarrow$ for any two points $\left(x_{1}, f\left(x_{1}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,221 |
2. Let convex quadrilateral $ABCD$ satisfy $AB=AD=1$, $\angle A=160^{\circ}, \angle C=100^{\circ}$. Then the range of the length of diagonal $AC$ is $\qquad$ . | 2. $\{1\}$.
Since $\angle C=180^{\circ}-\frac{\angle A}{2}$, therefore, points $B$, $C$, and $D$ lie on a circle with $A$ as the center and a radius of 1. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,224 |
3. If the three-digit decimal number $n=abc$ satisfies that $a$, $b$, $c$ form an arithmetic sequence, then the maximum possible value of a prime factor of $n$ is $\qquad$ | 3. 317 .
$$
\begin{array}{l}
31 n, 999=27 \times 37, 987=3 \times 7 \times 47, \\
963=9 \times 107, 951=3 \times 317 .
\end{array}
$$ | 317 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,225 |
4. When the real number $a \in$ $\qquad$, there does not exist a real number $x$ such that $|x+a+1|+\left|x+a^{2}-2\right|<3$. | 4. $(-\infty,-2] \cup[0,1] \cup[3, \infty)$.
No solution $\Leftrightarrow\left|a^{2}-a-3\right| \geqslant 3$
$\Leftrightarrow a^{2}-a \leqslant 0$ or $a^{2}-a-6 \geqslant 0$. | (-\infty,-2] \cup[0,1] \cup[3, \infty) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,226 |
5. The sum of the radii of all circles passing through point $A(1505,1008)$ and tangent to the lines $l_{1}: y=0$ and $l_{2}: y=\frac{4}{3} x$ is $\qquad$ | 5. 2009 .
From $\tan 2 \theta=\frac{4}{3} \Rightarrow \tan \theta=\frac{1}{2}$, the center of the circle lies on the line $y=\frac{x}{2}$, and the center is $(2 r, r)$ (where $r$ is the radius of the circle).
$$
\begin{array}{l}
\text { Hence }(1505-2 r)^{2}+(1008-r)^{2}=r^{2} \\
\Rightarrow 4 r^{2}-8036 r+1505^{2}+10... | 2009 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,227 |
6. Let the quadratic function $f(x)=a x^{2}+b x+c(a>0)$ satisfy $f(5-x)=f(5+x)$. Then the order from smallest to largest of $f(\sqrt{40})$, $f(2 \pi)$, and $f\left(5 \sin 45^{\circ}\right)$ is $\qquad$. | $\begin{array}{l}\text { 6. } f(2 \pi)<f(\sqrt{40})<f\left(5 \sin 45^{\circ}\right) . \\ |\sqrt{40}-5| \approx 1.32,|2 \pi-5| \approx 1.28, \\ \left|\frac{5}{\sqrt{2}}-5\right| \approx 1.46 .\end{array}$ | f(2 \pi)<f(\sqrt{40})<f\left(5 \sin 45^{\circ}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,228 |
7. Let a fixed point $P$ outside the plane $\alpha$ be at a distance $h$ from $\alpha$, and let three moving points $A, B, C$ on $\alpha$ be at distances $a, b, c$ from $P$, respectively, with $\angle PBA = 90^{\circ}$. Then the maximum area of $\triangle ABC$ is $\qquad$ (expressed in terms of $a, b, c, h$). | 7. $\frac{1}{2} \sqrt{a^{2}-b^{2}}\left(\sqrt{b^{2}-h^{2}}+\sqrt{c^{2}-h^{2}}\right)$.
Notice that $|A B|=\sqrt{a^{2}-b^{2}}$.
Let the projection of $P$ on plane $\alpha$ be $Q$. Then $A B \perp B Q$.
Thus, $|B Q|=\sqrt{b^{2}-h^{2}},|C Q|=\sqrt{c^{2}-h^{2}}$.
Therefore, the distance from point $C$ to $A B$ is less tha... | \frac{1}{2} \sqrt{a^{2}-b^{2}}\left(\sqrt{b^{2}-h^{2}}+\sqrt{c^{2}-h^{2}}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,229 |
8. Let the lines $l_{1} / / l_{2}$, and take 10 points $A_{1}, A_{2}, \cdots, A_{10}$ and $B_{1}, B_{2}, \cdots, B_{10}$ on $l_{1}$ and $l_{2}$ respectively. Then the line segments $A_{1} B_{1}, A_{2} B_{2}, \cdots, A_{10} B_{10}$ can divide the strip region between $l_{1}$ and $l_{2}$ into at most $\qquad$ non-overlap... | 8. 56 .
A line segment divides the original region into two parts. The $k$-th line segment can be divided into at most $k$ segments by the previous $k-1$ line segments, and it can add at most $k$ parts. Therefore, $k$ line segments can divide the region into at most $2+2+3+\cdots+k=\frac{k^{2}+k+2}{2}$ parts. | 56 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,230 |
1. (21 points) As shown in Figure 1, given quadrilateral $\square A B C D$ satisfies $\angle B A D>90^{\circ}$. Construct $\triangle D C E$ and $\triangle B C F$ outside the quadrilateral such that $\angle E D C=\angle C B F$, $\angle D C E=\angle B F C$. Connect $E F$, and construct $\triangle E F G$ outside $\triangl... | From the problem, we know that
$\triangle E D C \backsim \triangle C B F \backsim \triangle E G F$.
Therefore, $\frac{A B}{D E}=\frac{D C}{D E}=\frac{B F}{B C}=\frac{B F}{A D}$.
Since $\angle E D A=\angle A B F$, we have
$\triangle E D A \backsim \triangle A B F$.
Thus, $\frac{A F}{A E}=\frac{A B}{D E}=\frac{D C}{D E}=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,231 |
2. (21 points) Given the positive sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{2}=2, a_{n}=\frac{a_{n-2}}{a_{n-1}}(n \geqslant 3) \text {. }
$$
Find the general term formula for $a_{n}$. | 2. Let $b_{n}=\log _{2} a_{n}$. Then
$$
b_{1}=0, b_{2}=1, b_{n}=-b_{n-1}+b_{n-2}(n \geqslant 3) \text {. }
$$
Let $\lambda, \mu$ be the roots of the equation $x^{2}=-x+1$, i.e.,
$$
\begin{array}{l}
\lambda, \mu=-\frac{1 \pm \sqrt{5}}{2} \\
\Rightarrow\left\{\begin{array}{l}
b_{n}-\lambda b_{n-1}=\mu\left(b_{n-1}-\lamb... | a_{n}=2^{\frac{\lambda^{n-1}-\mu^{n-1}}{\lambda-\mu}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,232 |
Example 10 Simplify:
$$
\frac{(\sqrt{x}-\sqrt{y})^{3}+2 x \sqrt{x}+y \sqrt{y}}{x \sqrt{x}+y \sqrt{y}}+\frac{3 \sqrt{x y}-3 y}{x-y} .
$$ | Let $\sqrt{x}=a+b, \sqrt{y}=a-b$. Then
$$
\sqrt{x y}=a^{2}-b^{2}, \sqrt{x}+\sqrt{y}=2 a, \sqrt{x}-\sqrt{y}=2 b \text {. }
$$
Therefore, the original expression is
$$
\begin{aligned}
= & \frac{(2 b)^{3}+2(a+b)^{3}+(a-b)^{3}}{(a+b)^{3}+(a-b)^{3}}+ \\
& \frac{3\left(a^{2}-b^{2}\right)-3(a-b)^{2}}{4 a b} \\
= & \frac{3\le... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,233 |
3. (22 points) Find all solutions to the equation $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{7}\right]=x$ ( $[a]$ denotes the greatest integer not exceeding the real number $a$).
| 3. From the equation, we know that the solution $x$ is an integer.
Let $x=42 p+q(p \in \mathbf{Z}, q \in\{0,1, \cdots, 41\})$.
Then $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{7}\right]=x$
$$
\Leftrightarrow p=\left[\frac{q}{2}\right]+\left[\frac{q}{3}\right]+\left[\frac{q}{7}\right]-q .
$$
Ther... | 0,-6,-12,-14,-18,-20,-21,-24,-26,-27,-28,-30,-32,-33,-34,-35,-36,-38,-39,-40,-41,-44,-45,-46,-47,-49,-50,-51,-52,-53,-55,-57,-58,- | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,234 |
4. (22 points) Suppose a set of planar points $S$ has the following properties:
(1) No three points are collinear;
(2) The distance between any two points is unique.
For two points $A$ and $B$ in $S$, if there exists a point $C \in S$ such that $|A C|<|A B|<|B C|$, then $A B$ is called a "middle edge" of $S$. For thre... | 4. Color all middle edges of $S$ red, and color other edges blue.
When $n \geqslant 6$, according to Ramsey's theorem, there must exist a monochromatic triangle, which must have a middle edge, and it must be a middle edge triangle.
The following set of five points with properties (1) and (2) does not have a middle ed... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,235 |
一、Fill in the Blanks (8 points per question, total 64 points)
1. In an acute triangle $\triangle A B C$, the angle bisector $A D$, median $B M$, and altitude $C H$ intersect at a point $P$. Then $C H$ $\qquad$ $A H$ (fill in “ $>$ ”, “ $<$ ”, or “ $=$ ”). | - 1. $>$.
From Ceva's Theorem, as shown in Figure 2, we have
$$
\begin{array}{l}
\frac{A M}{M C} \cdot \frac{C D}{D B} \cdot \frac{B H}{H A}=1 . \\
\text { Since } A M=M C, \\
\frac{C D}{D B}=\frac{A C}{A B}=\frac{b}{c}, \\
\frac{B H}{H A}=\frac{B C \cos B}{A C \cos A} \\
=\frac{a \cos B}{b \cos A},
\end{array}
$$
Th... | C H > A H | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,236 |
2. Given the sequence of real numbers $\left\{a_{n}\right\}$ defined as
$$
a_{0}=\frac{1}{2}, a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{1}{5 a_{n}}\right)(n \in \mathbf{N}) \text {. }
$$
Let $A_{n}=\frac{5}{5 a_{n}^{2}-1}$. Then $\left\{A_{n}\right\}$ contains a perfect square $\qquad$.
| 2. Infinite.
Let $a_{n}=\frac{p_{n}}{q_{n}}\left(p_{n}, q_{n} \in \mathbf{N}_{+},\left(p_{n}, q_{n}\right)=1\right)$. Then $a_{n+1}=\frac{p_{n+1}}{q_{n+1}}=\frac{1}{2}\left(\frac{p_{n}}{q_{n}}+\frac{q_{n}}{5 p_{n}}\right)=\frac{5 p_{n}^{2}+q_{n}^{2}}{10 p_{n} q_{n}}$
Given $a_{0}=\frac{1}{2}$, we get $a_{1}=\frac{9}{2... | Infinite | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,237 |
3. Non-negative real numbers $a_{1}, a_{2}, \cdots, a_{2010}$ satisfy $\sum_{i=1}^{2010} a_{i}=1$, then the maximum value of $\sum_{i=1}^{2010} \prod_{j=0}^{1004} a_{i+j}$ is $\quad\left(a_{2010+i}=a_{i}\right)$. | 3. $1005^{-1005}$.
Let $A=\sum_{i=1}^{2010} \prod_{j=0}^{1004} a_{i+j}$.
Fix $a_{2}, a_{3}, \cdots, a_{1005}, a_{1007}, \cdots, a_{2010}$, then $A=B a_{1}+C a_{1006}$, where $B, C$ are non-negative constants (determined by the fixed $a_{i}$).
It is easy to see that when one of $a_{1}, a_{1006}$ is 0, $A$ is maximized... | 1005^{-1005} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,238 |
4. For the positive integer $n$, define $a_{n}$ as the unit digit of $n^{(n+1)^{n+2}}$. Then $\sum_{n=1}^{2010} a_{n}=$ $\qquad$ . | 4.5 829 .
When $n \equiv 0,1,5,6(\bmod 10)$, $a_{n} \equiv n^{(n+1)^{n+2}} \equiv n(\bmod 10)$;
When $n \equiv 2,4,8(\bmod 10)$,
$$
\begin{array}{l}
(n+1)^{n+2} \equiv 1(\bmod 4) \\
\Rightarrow a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k+1} \equiv n(\bmod 10) ;
\end{array}
$$
When $n \equiv 3,7,9(\bmod 10)$,
$$
\begin{array}... | 5829 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,239 |
5. For a set, the difference between the maximum and minimum elements is called the "capacity" of the set. Let $2 \leqslant r \leqslant n$, and let $F(n, r)$ denote the arithmetic mean of the capacities of all $r$-element subsets of the set $M=\{1,2, \cdots, n\}$. Then $F(n, r)=$ | 5. $\frac{(r-1)(n+1)}{r+1}$.
The number of $r$-element subsets of set $M$ is $\mathrm{C}_{n}^{r}$. Among these, the number of $r$-element subsets with the smallest element $i$ and the largest element $j (1 \leqslant i < j \leqslant n)$ is $\mathrm{C}_{j-i-1}^{r-2}$. Therefore,
$$
\begin{array}{l}
F(n, r)=\frac{1}{\mat... | \frac{(r-1)(n+1)}{r+1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,240 |
6. There are three points $A$, $B$, and $C$ in sequence on a straight line. A hunting dog at point $A$ discovers three rabbits, one large and two small, moving from point $B$ towards their burrow (point $C$), and immediately chases after them. When the rabbits notice the hunting dog, they hurry towards the burrow. The ... | 6. $\frac{27 a-17 b}{64}$.
Obviously, the hound needs $\frac{a+b}{2} \mathrm{~s}$ to reach the rabbit's den.
Assume the rabbit is $x \mathrm{~m}$ away from point $B$ when it spots the hound, having already used $2 x \mathrm{~s}$. Then, for the remaining $(b-x) \mathrm{m}$, the large rabbit must lead the small rabbit t... | \frac{27 a-17 b}{64} | Other | math-word-problem | Yes | Yes | cn_contest | false | 722,241 |
7. The line $l: x+y=t$ intersects the circle $\odot O: x^{2}+y^{2}=20$ at points $A$ and $B$, and $S_{\triangle O A B}$ is an integer. Then the number of all positive integer values of $t$ that satisfy the condition is $\qquad$ . | 7.2.
Let $\angle A O B=2 \alpha, O C \perp A B$, with the foot of the perpendicular being $C$. Then $O C=\sqrt{20} \cos \alpha=\frac{t}{\sqrt{1+1}}=\frac{t}{\sqrt{2}}$ $\Rightarrow \cos \alpha=\frac{t}{2 \sqrt{10}}$.
Also, $S_{\triangle O A B}=10 \sin 2 \alpha \leqslant 10$, so $\sin 2 \alpha=\frac{k}{10}(k \in\{1,2, ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,242 |
8. Given the function $f:\{0,1, \cdots, 2010\} \rightarrow \mathbf{N}$. If for all possible integers $x$, we have
$$
\begin{array}{l}
f(4 x+2)=f(4 x+1), \\
f(5 x+3)=f(5 x+2), \\
f(7 x+5)=f(7 x+4),
\end{array}
$$
then $f(x)$ can take at most $\qquad$ different values. | 8. 1033 .
For the function $f$, construct the function $g$, defined as:
$$
\begin{array}{l}
g(0)=1, \\
g(i+1)=\left\{\begin{array}{l}
1, f(i+1) \neq f(i) ; \\
0, f(i+1)=f(i),
\end{array}\right.
\end{array}
$$
where $i=0,1, \cdots, 2009$.
Then $g$ is a mapping from $\{0,1, \cdots, 2010\}$ to $\{0,1\}$, and for all pos... | 1033 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,243 |
Example 2 Let positive numbers $x, y, z$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{2}+x y+\frac{y^{2}}{3}=25, \\
\frac{y^{2}}{3}+z^{2}=9, \\
z^{2}+z x+x^{2}=16 .
\end{array}\right.
$$
Find the value of $x y+2 y z+3 z x$. | If we solve the system of three quadratic equations in the conventional way, first finding the values of $x$, $y$, and $z$, and then calculating the value of the algebraic expression, we will inevitably fall into tedious calculations.
In fact, the original system of equations can be transformed into
$$
\left\{\begin{ar... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,244 |
Example 3 The system of equations $\left\{\begin{array}{l}x+y=2, \\ x y-z^{2}=1\end{array}\right.$ has how many sets of real solutions? | Given
\[
\left\{\begin{array}{l}
x+y=2, \\
x y=z^{2}+1,
\end{array}\right.
\]
we are reminded of the inverse of Vieta's formulas. By the inverse of Vieta's formulas, $x$ and $y$ are the two real roots of the equation
\[
t^{2}-2 t+\left(z^{2}+1\right)=0.
\]
Then,
\[
\Delta=(-2)^{2}-4\left(z^{2}+1\right) \geqslant 0 \Rig... | (x, y, z) = (1, 1, 0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,245 |
2. Prove: When $n$ is odd or a multiple of 4, the equation $x^{2}-y^{2}=n$ has positive integer solutions. | Hint: Construct the equations $(k+1)^{2}-k^{2}=2 k+1$ and $(k+1)^{2}-(k-1)^{2}=4 k$. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,246 |
Three. (25 points) Given that $n$ is an integer. Prove:
$$
(n+1)^{6}+n^{6}+(n-1)^{6}-3 n^{2}-2
$$
is divisible by 60. | Three, notice that
$$
\begin{array}{l}
(n+1)^{6}+n^{6}+(n-1)^{6}-3 n^{2}-2 \\
=\left[(n+1)^{6}-(n+1)^{2}\right]+\left(n^{6}-n^{2}\right)+ \\
\quad\left[(n-1)^{6}-(n-1)^{2}\right] .
\end{array}
$$
Next, we prove that when $m$ is an integer, $m^{6}-m^{2}$ is divisible by 60.
Notice that
$$
\begin{array}{l}
m^{6}-m^{2}=... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,247 |
1. Let $[x]$ be the greatest integer not exceeding the real number $x$. For any $x \in \mathbf{R}$, we have $[x]+[x+a]=[2 x]$. Then the set of all possible values of the positive real number $a$ is | -1. $\left\{\frac{1}{2}\right\}$.
Let $x=\frac{1}{2}$. Then
$$
\left[\frac{1}{2}+a\right]=1 \Rightarrow a \in\left[\frac{1}{2}, \frac{3}{2}\right) \text {. }
$$
For any $x \in\left(0, \frac{1}{2}\right)$, we always have
$$
[x+a]=0 \Rightarrow x+a<1,
$$
which means $a<1-x$ always holds.
Therefore, $a \leqslant \frac{1... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,248 |
2. If the equation with respect to $x$
$$
\sqrt{1+a-x}-a+\sqrt{x}=0
$$
has real solutions, then the range of positive real number $a$ is $\qquad$ | 2. $\left[\frac{1+\sqrt{5}}{2}, 1+\sqrt{3}\right]$.
Let $t=\sqrt{x}$. Then $0 \leqslant t \leqslant \sqrt{1+a}$.
The original equation becomes
$\sqrt{1+a-t^{2}}=a-t$.
Squaring both sides of equation (1) and rearranging, we get
$2 t^{2}-2 a t+a^{2}-a-1=0$.
By $\Delta=4 a^{2}-8\left(a^{2}-a-1\right) \geqslant 0$, we get... | \left[\frac{1+\sqrt{5}}{2}, 1+\sqrt{3}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,249 |
3. Let $\left(2+x-2 x^{2}\right)^{1005}=\sum_{k=0}^{2010} a_{k} x^{k}$. Then
$$
\begin{array}{l}
a_{1}+3 a_{3}+5 a_{5}+\cdots+2009 a_{2009} \\
=
\end{array}
$$ | 3. 1005.
Let $f(x)=\left(2+x-2 x^{2}\right)^{1005}$. Then $f^{\prime}(x)=1005\left(2+x-2 x^{2}\right)^{1004}(1-4 x)$. Therefore, $f^{\prime}(1)=1005 \times(-3)$, $f^{\prime}(-1)=1005 \times 5$.
Given $f(x)=\sum_{k=0}^{2010} a_{k} x^{k}$, we have
$$
\begin{array}{l}
f^{\prime}(x)=\sum_{k=1}^{2010} k a_{k} x^{k-1} . \\
... | 1005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,250 |
4. In $\triangle A B C$, if $\qquad$
$$
\frac{\overrightarrow{A B} \cdot \overrightarrow{B C}}{3}=\frac{\overrightarrow{B C} \cdot \overrightarrow{C A}}{2}=\frac{\overrightarrow{C A} \cdot \overrightarrow{A B}}{1}
$$
then $\tan A=$ $\qquad$ | 4. $\sqrt{11}$.
From the given information,
$$
\begin{array}{l}
\frac{c^{2}+a^{2}-b^{2}}{3}=\frac{a^{2}+b^{2}-c^{2}}{2} \\
=\frac{b^{2}+c^{2}-a^{2}}{1}=\frac{a^{2}+b^{2}+c^{2}}{6} .
\end{array}
$$
Therefore, $a^{2}: b^{2}: c^{2}=5: 3: 4$.
By the cosine rule,
$$
\cos A=\frac{1}{2 \sqrt{3}} \Rightarrow \tan A=\sqrt{11}... | \sqrt{11} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,251 |
5. If the three views (front view, side view, top view) of a tetrahedron are all isosceles triangles with a base of 2 and a height of 2, then the surface area of this geometric body is | 5. $2+\sqrt{5}+\sqrt{21}$.
The base $ABC$ of the tetrahedron $P-ABC$ is an isosceles triangle with a base of 2 and a height of 2; the front side $PBC$ is an isosceles triangle with a base of 2 and a height of $\sqrt{5}$; the other two sides are non-isosceles triangles with a base of $\sqrt{5}$ and a height of $\sqrt{\... | 2+\sqrt{5}+\sqrt{21} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,252 |
6. If the polynomial $f(x)=x^{3}-6 x^{2}+a x+a$ has three roots $x_{1}, x_{2}, x_{3}$ that satisfy
$$
\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0 \text {, }
$$
then the value of the real number $a$ is .. $\qquad$ | 6. -9 .
From the problem, we have
$$
\begin{array}{l}
g(t)=f(t+3) \\
=(t+3)^{3}-6(t+3)^{2}+a(t+3)+a \\
=t^{3}+3 t^{2}+(a-9) t+4 a-27
\end{array}
$$
The three roots \( t_{1}, t_{2}, t_{3} \) satisfy \( t_{1}^{3}+t_{2}^{3}+t_{3}^{3}=0 \).
By the relationship between roots and coefficients, we get
$$
\begin{array}{l}
\l... | -9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,253 |
7. Let $a_{1}, a_{2}, \cdots, a_{10}$ be a permutation of $2000, 2001, \cdots$, 2009, and let the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ be $S_{n}$. Then the probability that the permutation $a_{1}, a_{2}, \cdots, a_{10}$ satisfies “ $S_{i}(1 \leqslant i \leqslant 10)$ are not multiples of 3"... | 7. $\frac{1}{50}$.
Let a permutation of $2000, 2001, \cdots, 2009$ be a basic event $M$. Then the total number of basic events is
$$
N=\mathrm{A}_{10}^{10} \text {. }
$$
Next, we calculate the number of basic events contained in the event $M$.
(1) The first term cannot be a multiple of 3, and the other terms can be m... | \frac{1}{50} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,254 |
8. Let $a_{i}\left(i \in \mathbf{N}_{+}, i \leqslant 2009\right)$ be non-negative real numbers, and $\sum_{i=1}^{2009} a_{i}=1$. Then the maximum value of $\sum_{i \neq j, i l j} a_{i} a_{j}$ is | 8. $\frac{5}{11}$.
If $a_{i}$ and $a_{j} (i < j, i \nmid j)$ are both not 0, then adjust $\left(a_{i}, a_{j}\right)$ to $\left(a_{i}+a_{j}, 0\right)$ or $\left(0, a_{i}+a_{j}\right)$ (this is determined by the sum of the numbers whose indices have a divisibility relationship with $a_{i}$ or $a_{j}$), at this point, th... | \frac{5}{11} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,255 |
9. (16 points) Let $x \in(0,1]$. Find the range of the function $y=$ $\frac{3 x^{6}+15 x^{2}+2}{2 x^{6}+15 x^{4}+3}$. | Let $x=\tan \frac{\alpha}{2}\left(\alpha \in\left(0, \frac{\pi}{2}\right]\right)$. Then
$$
3 \alpha \in\left(0, \frac{3 \pi}{2}\right], \cos 3 \alpha \in[-1,1) \text {, }
$$
and $\cos \alpha=\frac{1-x^{2}}{1+x^{2}}$.
Thus, $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$
$$
\begin{array}{l}
=\frac{4\left(1-x^{2}\right... | \left(\frac{2}{3}, \frac{3}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,256 |
3. If the real numbers $a, b, c, d, e$ satisfy the conditions
$$
\begin{array}{l}
a+b+c+d+e=8, \\
a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16 .
\end{array}
$$
Determine the maximum value of $e$. | $$
\begin{array}{l}
f(x)=4 x^{2}+2(a+b+c+d) x+\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \\
=(x+a)^{2}+(x+b)^{2}+(x+c)^{2}+(x+d)^{2} \\
\geqslant 0 .
\end{array}
$$
Therefore, $\Delta \leqslant 0$. Solving this, we get $0 \leqslant e \leqslant \frac{16}{5}$. | \frac{16}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,257 |
10. (20 points) Given an ellipse centered at the origin, with foci on the $x$-axis, the length of the major axis is twice the length of the minor axis, and it passes through the point $M(2,1)$. A line $l$ parallel to $OM$ has a $y$-intercept of $m (m<0)$, and intersects the ellipse at two distinct points $A$ and $B$. F... | 10. Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
Then $\left\{\begin{array}{l}a=2 b, \\ \frac{4}{a^{2}}+\frac{1}{b^{2}}=1\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=8, \\ b^{2}=2 \text {. }\end{array}\right.\right.$
Therefore, the equation of the ellipse is $\frac{x^{2}... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,258 |
11. (20 points) Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1, a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Try to find $\left[a_{2009}^{2}\right]$. | 11. From $a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a_{n}}$, we get $a_{n+1}^{2}=\frac{a_{n}^{2}}{n^{2}}+\frac{n^{2}}{a_{n}^{2}}+2$. Given $a_{1}=1$, we have $a_{2}^{2}=4, a_{3}^{2}=4, a_{4}^{2}=4+\frac{25}{36}$.
Next, we use mathematical induction to prove:
When $n \geqslant 4$, $n+\frac{2}{n}\frac{n+1}{n^{2}}+\frac{n^{2}}{n+1... | 2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,259 |
Initial 269 The incircle of a triangle trisects one of its medians. Find the ratio of the three sides of this triangle.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve as shown in Figure 2, let $E, F$ be the intersection points of the incircle of $\triangle ABC$ and the median $BD$, satisfying $BF=EF=ED$. Let $P, Q, R$ be the points where the incircle touches the three sides. Clearly, $PA=QA$.
From $BF=FE=ED$, we get $DF=BE$.
From $PB^2=BF \cdot BE=DE \cdot DF=DQ^2$
$\Rightarro... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,261 |
Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}, x y z>0$,
and
$$
\begin{array}{l}
\sqrt[3]{\frac{2010}{x^{2}}+\frac{2011}{y^{2}}+\frac{2012}{z^{2}}} \\
=\sqrt[3]{2010}+\sqrt[3]{2011}+\sqrt[3]{2012} .
\end{array}
$$
Find the value of $x+y+z$. | Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}=k$, obviously $k \neq 0$.
Then $2010=x^{3} k, 2011=y^{3} k, 2012=z^{3} k$.
From the given, we have
$$
\sqrt[3]{x k+y k+z k}=\sqrt[3]{x^{3} k}+\sqrt[3]{y^{3} k}+\sqrt[3]{z^{3} k},
$$
which means $\sqrt[3]{k} \sqrt[3]{x+y+z}=\sqrt[3]{k}(x+y+z)$.
Since $k \neq... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,262 |
In rectangle $ABCD$, $AB=2a, BC=2b, P$ is a moving point, the extensions of $CP$ and $DP$ intersect $AB$ (or its extension) at points $F$ and $E$ respectively. If $\overrightarrow{AE} \cdot \overrightarrow{BF}=-\overrightarrow{EF}^2$, find the equation of the trajectory of point $P$. | Solve: Establish a rectangular coordinate system as shown in Figure 3, with the line $AB$ as the $x$-axis and the midpoint of segment $AB$ as the origin. Then
$$
\begin{array}{l}
A(-a, 0), B(a, 0), \\
C(a, 2b), \\
D(-a, 2b).
\end{array}
$$
Let $P(x, y)$, $E\left(x_{1}, 0\right)$, and $F\left(x_{2}, 0\right)$.
First, f... | \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,264 |
5. Is the statement "If $a$ and $b$ are both irrational numbers, then $a^{b}$ is irrational" correct? If correct, please provide a proof; if not correct, please explain why. | Hint: $\sqrt{2}$ is an irrational number. According to the original proposition, if $\sqrt{2}$ years is an irrational number, then $\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}$ is also an irrational number. However, $\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}$ $=(\sqrt{2})^{\sqrt{2} \cdot \sqrt{2}}=2$ is a rational nu... | not found | Number Theory | proof | Yes | Yes | cn_contest | false | 722,266 |
6. Let $a, b, c, d$ all be positive numbers. Prove that there exists a triangle with side lengths
$$
\sqrt{b^{2}+c^{2}}, \sqrt{a^{2}+c^{2}+d^{2}+2 c d}, \sqrt{a^{2}+b^{2}+d^{2}+2 a b}
$$
and find the area of this triangle. | To construct a rectangle $ABCD$ with side lengths $a+b$ and $c+d$, take points $E$ and $F$ on sides $AB$ and $AD$ respectively, such that $AE=b$ and $AF=c$. Then $\triangle CEF$ meets the requirements. Its area is $\frac{1}{2}(ac+bc+bd)$. | \frac{1}{2}(ac+bc+bd) | Geometry | proof | Yes | Yes | cn_contest | false | 722,267 |
Example 1 Find all integers $x$ such that $1+5 \times 2^{x}$ is the square of a rational number. ${ }^{[1]}$
(2008, Croatia National Training (Grade 2)) | Discuss the following two scenarios:
(1) If $1+5 \times 2^{x}$ is a perfect square, then $x \in \mathbf{N}$. Let $1+5 \times 2^{x}=y^{2}(y \in \mathbf{N})$. Hence, $(y+1)(y-1)=5 \times 2^{x}$.
If $x=0$, then $y^{2}=6$, which is impossible. Therefore, $x \neq 0$.
Since $y+1$ and $y-1$ have the same parity, both must be ... | x=-2 \text{ or } 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,269 |
Example 2 Find the positive integer solutions of the equation $3^{x}-5^{y}=z^{2}$. (2009, Balkan Mathematical Olympiad) | Taking modulo 2, we get
$$
0 \equiv z^{2}(\bmod 2) \Rightarrow 21 z \text {. }
$$
Taking modulo 4, we get
$$
(-1)^{x}-1 \equiv 0(\bmod 4) \Rightarrow 2 \mid x \text {. }
$$
Let $x=2 x_{1}$. Then $\left(3^{x_{1}}+z\right)\left(3^{x_{1}}-z\right)=5^{y}$.
Let $\left\{\begin{array}{l}3^{x_{1}}+z=5^{\alpha}, \\ 3^{x_{1}}-... | (x, y, z)=(2,1,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,270 |
Example 3 Find all non-negative integer solutions to the equation $2^{x}+2009=3^{y} \times 5^{z}$.
(2009, Middle European Mathematical Olympiad) | Obviously, $y, z$ are not both 0. Otherwise, the right side $=10$, then $(-1)^{x}-1 \equiv 0(\bmod 3)$.
Therefore, $2 \mid x$.
If $z>0$, then $2^{x}-1 \equiv 0(\bmod 5)$.
And $2^{x} \equiv 2,-1,-2,1(\bmod 5)$, so $4 \mid x$.
In summary, $x$ is even.
If $x=0$, then
$$
\begin{array}{l}
3^{y} \times 5^{z}=2^{x}+2009=2010 ... | (x, y, z)=(4,4,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,271 |
Example 4 Find all positive integers $n$, such that $2^{n-1} n+1$ is a perfect square.
(2004, Slovenia IMO National Team Selection Test | Let $2^{n-1} n+1=m^{2}\left(m \in \mathbf{N}_{+}\right)$. Then $2^{n-1} n=(m+1)(m-1)$.
When $n=1,2,3,4$, $2^{n-1} n+1$ is not a perfect square.
Therefore, $n \geqslant 5, 16 \mid (m+1)(m-1)$.
Since $m+1$ and $m-1$ have the same parity, both $m+1$ and $m-1$ are even, and $m$ is odd.
Let $m=2 k-1\left(k \in \mathbf{N}_{+... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,272 |
Example 4 Let $\overline{abc}$ be a prime number in decimal notation. Prove: $b^{2}-4ac$ is not a perfect square. | Proof by contradiction.
Assume there exists a prime number $\overline{a b c}$ in decimal notation such that $b^{2}-4 a c$ is a perfect square. Considering the form of the result to be proven, we can consider the (auxiliary) quadratic equation
$$
f(x)=a x^{2}+b x+c=0 .
$$
The given condition implies
$$
p=f(10)=a \times... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,273 |
Example 5 Find all pairs of integers $(x, y)$ such that
$$
1+2^{x}+2^{2 x+1}=y^{2} \text {. }
$$
(47th IMO) | When $x=-1$, the left side is 2, which is not a perfect square; when $x \leqslant -2$, the left side is not an integer.
Let $(x, y)$ be a solution to this equation. Then $x \geqslant 0$ and $(x, -y)$ is also a solution.
When $x=0$, $y= \pm 2$, the equation has solutions $(0,2)$, $(0,-2)$.
When $x>0$, let $(x, y)$ be ... | (x, y)=(0,2),(0,-2),(4,23),(4,-23) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,274 |
1. Given that $3^{a}+7^{b}$ is a perfect square. Find all ordered integer pairs $(a, b)$.
(2009, Canadian Mathematical Olympiad) | Hint: Using congruence analysis and prime factor analysis, we classify and discuss to get $(a, b)=(1,0),(2,1)$. | (a, b)=(1,0),(2,1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,275 |
2. Find all triples of positive integers $(a, b, c)$ such that
$$
a^{2}+2^{b+1}=3^{c} \text {. }
$$ | Using congruence analysis and prime factor analysis, we get
$$
(a, b, c)=(1,2,2),(7,4,4) \text {. }
$$ | (1,2,2),(7,4,4) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,276 |
3. Find all positive integer solutions to the equation $3^{x}+4^{y}=5^{x}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Hint: Using congruence analysis, prime factor analysis, and classification discussion, we get $(x, y, z)=(2,2,2)$. | (2,2,2) | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 722,277 |
5. Prove that the positive integer solution $(x, y, z)$ of $8^{x}+15^{y}=17^{x}$ is unique. | Hint: Using congruence analysis and prime factor analysis, the unique positive integer solution is $(x, y, z)=(2,2,2)$. | (x, y, z)=(2,2,2) | Number Theory | proof | Yes | Yes | cn_contest | false | 722,279 |
Example 5 Solve the equation
$$
\sqrt{x^{2}+3 x+7}-\sqrt{x^{2}-x+10}=1 \text {. }
$$ | Solving irrational equations by the usual method (transposition, squaring, simplification) can be quite cumbersome. Observing the characteristics of the given equation, one can think of the difference of squares formula. Construct the identity
$$
\left(x^{2}+3 x+7\right)-\left(x^{2}-x+10\right)=4 x-3 \text {. }
$$
Div... | x=3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,283 |
Question 2 Real numbers $a_{1}, a_{2}, a_{3}, a_{4}$ satisfy
$$
a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant a_{4} \geqslant 0, a_{1}+a_{2}+a_{3}+a_{4}=1 \text {. }
$$
Prove:
$$
\begin{array}{l}
\max \left\{\sum_{i=1}^{4} \sqrt{a_{i}^{2}+a_{i} a_{i-1}+a_{i-1}^{2}+a_{i-1} a_{i-2}},\right. \\
\left.\sum_{i=1}^{4} \sq... | Proof 1 It is sufficient to prove
$$
\begin{array}{l}
\sum_{i=1}^{4} \sqrt{a_{i}^{2}+a_{i} a_{i-1}+a_{i-1}^{2}+a_{i-1} a_{i-2}}+ \\
\sum_{i=1}^{4} \sqrt{a_{i}^{2}+a_{i} a_{i+1}+a_{i+1}^{2}+a_{i+1} a_{i+2}} \geqslant 4
\end{array}
$$
By
$$
\begin{array}{l}
\text { } \sqrt{a_{i}^{2}+a_{i} a_{i-1}+a_{i-1}^{2}+a_{i-1} a_{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,284 |
Question 3 Let $D$ be a point on side $BC$ of $\triangle ABC$, satisfying $\angle CAD = \angle CBA$. Circle $\odot O$ passes through points $B$ and $D$, and intersects line segments $AB$ and $AD$ at points $E$ and $F$, respectively. $BF$ intersects $DE$ at point $G$, and $M$ is the midpoint of $AG$. Prove: $CM \perp AO... | Proof 1 As shown in Figure 1, let the radius of $\odot O$ be $R$. By the power of a point theorem, we have $C O^{2}=C D \cdot C B+R^{2}$.
Since $\angle C A D=\angle C B A$ and $\angle A C D$ is a common angle, $\Rightarrow \triangle A C D \backsim \triangle B C A \Rightarrow \frac{A C}{B C}=\frac{C D}{A C}$ $\Rightarro... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,285 |
Question 4: Circle $\Gamma$ is internally tangent to circle $\omega$ at point $S$, the chord $AB$ of circle $\Gamma$ is tangent to circle $\omega$ at point $T$, the center of circle $\omega$ is $O$, and $P$ is a point on line $AO$. Prove:
$$
P B \perp A B \Leftrightarrow P S \perp T S .
$$ | Proof 1 Necessity.
As shown in Figure 4, connect $OT$, $SB$. Draw $BK \parallel TS$ intersecting line $AS$ at point $K$, connect $PK$, and draw the common tangent $MN$ of the two circles through point $S$.
Then $\angle AST = \angle TSM - \angle ASM$
$$
= \angle STA - \angle ABS = \angle TSB.
$$
Since $BK \parallel TS$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,286 |
Proposition 1 The smallest inscribed equilateral triangle in a regular pentagon has one vertex at the midpoint of one of the pentagon's sides, and the side opposite this vertex is parallel to the side of the pentagon containing the vertex. | Prove that, as shown in Figure 1, in the regular pentagon $ABCDE$, $\triangle RST$ is an equilateral triangle, with $R$ being the midpoint of side $AB$, and points $S, T$ lying on $CD, DE$ respectively, such that $ST \parallel AB$.
Use proof by contradiction.
Assume there exists an equilateral $\triangle UVW$ inscribed... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,287 |
In $\triangle A B C$, the angle bisector of $\angle B C A$ intersects the circumcircle of $\triangle A B C$ at point $R$, intersects the perpendicular bisector of side $B C$ at point $P$, and intersects the perpendicular bisector of side $A C$ at point $Q$. Let $K$ and $L$ be the midpoints of $B C$ and $A C$, respectiv... | Proof (1) As shown in Figure 1, extend $PK$ to intersect $LQ$ at point $O$, then $O$ is the circumcenter of $\triangle ABC$. Connect $OC$, $OR$, and let point $M$ be on the extension of $AC$.
By the external angle bisector theorem, we have
$$
\frac{TB}{TA}=\frac{CB}{CA}.
$$
Thus, $\frac{TB}{TA}=\frac{CK}{CL}$.
Since ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,288 |
1. Given points $A(2,3)$, $B(5,4)$, $C(-4,1)$, connecting these three points in sequence. Then these three points ( ).
(A) form an equilateral triangle
(B) form a right triangle
(C) form an acute triangle
(D) lie on the same straight line | 一、1. D.
It is easy to know that the equation of line $AB$ is $y=\frac{1}{3} x+\frac{7}{3}$. When $x=-4$, $y=1$.
Therefore, point $C$ lies on line $AB$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,289 |
2. The number of triangles with integer side lengths and a perimeter of 20 is ( ).
(A) 4
(B) 6
(C) 8
(D) 12 | 2. C.
Let the three sides of a triangle be $a, b, c(a \geqslant b \geqslant c)$. From $a+b+c=20 \Rightarrow a \geqslant 7$.
Also, $a+b>a, 2a<20 \Rightarrow a<10$, thus $7 \leqslant a \leqslant 9$.
Therefore, we can list
$$
\begin{array}{l}
(a, b, c) \\
=(9,9,2),(9,8,3),(9,7,4),(9,6,5), \\
\quad(8,8,4),(8,7,5),(8,6,6),... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 722,290 |
3. Let $N=3^{1001}+7^{1002}+13^{1003}$. Then the units digit of $N$ is ( ).
(A) 3
(B) 6
(C) 9
(D) 0 | 3. C.
Since the unit digit of $3^{1001}$ is 3, the unit digit of $7^{1002}$ is 9, and the unit digit of $13^{1003}$ is 7, therefore, the unit digit of $N$ is 9. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 722,291 |
4. Given that $P$ is a point inside square $A B C D$. If $P A: P B: P C=1: 2: 3$, then the degree measure of $\angle A P B$ is ().
(A) $120^{\circ}$
(B) $135^{\circ}$
(C) $150^{\circ}$
(D) None of the above | 4. B.
As shown in Figure 3, draw $B P^{\prime} \perp B P$ such that $B P^{\prime} = B P$, and connect $P^{\prime} A$. It is easy to see that
$$
\begin{array}{l}
\triangle P^{\prime} A B \cong \triangle P C B \\
\Rightarrow P^{\prime} A = P C .
\end{array}
$$
Let $P A = k$. Then
$$
P B = 2 k, P C = P^{\prime} A = 3 k ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,292 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.