problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
6. Given circle $\Gamma:(x-1)^{2}+y^{2}=r^{2}(r>0)$, parabola $G: y^{2}=4 x$. A line $l$ passing through the point $(1,0)$ intersects the circle $\Gamma$ at points $C$ and $D$ (point $C$ is above the $x$-axis), and intersects the parabola $G$ at points $A$ and $B$ (point $A$ is above the $x$-axis). If there are three l... | 6. $(2,+\infty)$.
(1) When the slope of line $l$ does not exist, $|A C|=$ $|B D|$ holds.
(2) When the slope of line $l$ exists, let
$$
\begin{array}{l}
l: y=k(x-1)(k \neq 0) . \\
\text { Let } A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right) . \\
\text { From }\left\{\begin{array}{l}
y=k(x-1), \\
y^{2}=4 x
\end{... | (2,+\infty) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,584 |
7. Given a line passing through the origin (excluding the $x$-axis) intersects the circle $x^{2}+y^{2}=1$ at points $A$ and $B$, and a fixed point $C(2,0)$, then the minimum value of the area of the circumcircle of $\triangle A B C$ is $\qquad$ . | 7. $\frac{25 \pi}{16}$.
Let $A(\cos \theta, \sin \theta)$. Then
$$
B(-\cos \theta,-\sin \theta)(\theta \neq k \pi, k \in \mathbf{Z}) \text {. }
$$
Let the circumcenter of $\triangle A B C$ be $(x, y)$. Then
$$
\left\{\begin{array}{l}
y=-x \cot \theta, \\
y-\frac{\sin \theta}{2}=\frac{2-\cos \theta}{\sin \theta}\left(... | \frac{25 \pi}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,585 |
8. Let the function be
$$
\begin{array}{l}
f(x)=\frac{1}{2}+\ln \frac{x}{1-x}\left(n \geqslant 2, \text { and } n \in \mathbf{N}_{+}\right) . \\
\text {Then } \sum_{i=1}^{n-1} f\left(\frac{i}{n}\right)=
\end{array}
$$ | 8. $\frac{n-1}{2}$.
Obviously, the domain of $f(x)$ is $(0,1)$, and
$$
f(x)+f(1-x)=1 \text {. }
$$
Then $2 \sum_{i=1}^{n-1} f\left(\frac{i}{n}\right)=\sum_{i=1}^{n-1}\left(f\left(\frac{i}{n}\right)+f\left(\frac{n-i}{n}\right)\right)$ $=n-1$.
Therefore, $\sum_{i=1}^{n-1} f\left(\frac{i}{n}\right)=\frac{n-1}{2}$. | \frac{n-1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,586 |
9. (16 points) Given that $F$ is the focus of the parabola $y^{2}=4 x$, $Q$ is the intersection point of its directrix with the $x$-axis, and line $l$ passes through point $Q$. Let line $l$ intersect the parabola at points $A$ and $B$.
(1) Let the slopes of lines $A F$ and $B F$ be $k_{1}$ and $k_{2}$, respectively. Fi... | 9. (1) Let the line $l: y=k(x+1)$,
$$
A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), F(1,0).
$$
Then $k_{1}+k_{2}=\frac{y_{1}}{x_{1}-1}+\frac{y_{2}}{x_{2}-1}$
$$
\begin{array}{l}
=\frac{k\left(x_{1}+1\right)}{x_{1}-1}+\frac{k\left(x_{2}+1\right)}{x_{2}-1} \\
=\frac{2 k\left(x_{1} x_{2}-1\right)}{\left(x_{1}-1... | x=1(-2<y<2, y \neq 0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,587 |
10. (20 points) Given the sequence $A: a_{1}, a_{2}, \cdots, a_{n}$, satisfying $a_{i} \in\{0,1\}(i=1,2, \cdots, n)$. Define the transformation $T$: $T$ changes each 1 in the sequence $A$ to 0,1, and each 0 to 1,0.
If $A_{0}$ is 0,1,
$$
A_{k}=T\left(A_{k-1}\right)(k=1,2, \cdots),
$$
and let $a_{k}$ be the number of pa... | 10. Let $A_{k}$ contain $b_{k}$ pairs of $(0,1)$.
Since the $(0,0)$ pairs in $A_{k+1}$ can only be derived from the $(0,1)$ pairs in $A_{k}$, we have
$$
a_{k+1}=b_{k} \text {. }
$$
Thus, the $(0,1)$ pairs in $A_{k+1}$ can be generated through two ways:
(1) from the 1s in $A_{k}$;
(2) from the $(0,0)$ pairs in $A_{k}$... | a_{k}=\left\{\begin{array}{ll}\frac{1}{3}\left(2^{k}+1\right), & k \text { is odd; } \\ \frac{1}{3}\left(2^{k}-1\right), & k \text { is even. }\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,588 |
11. (20 points) Let the function
$$
f(x)=\frac{1+\ln (x+1)}{x} \text {, }
$$
$k$ is a positive integer. When $x>0$, $f(x)>\frac{k}{x+1}$ always holds. Find the maximum value of $k$.
untranslated part:
$k$ is a positive integer. When $x>0$, $f(x)>\frac{k}{x+1}$ always holds. Find the maximum value of $k$.
(Note: The... | 11. Since $x>0$, we have
$$
k0)
$$
always holds.
$$
\text{Let } g(x)=\frac{(x+1)[1+\ln (x+1)]}{x}(x>0) \text{.}
$$
Then $g^{\prime}(x)=\frac{x-1-\ln (x+1)}{x^{2}}$.
Let $h(x)=x-1-\ln (x+1)$.
Obviously, $h(2)0$.
If $h\left(x_{0}\right)=0$, then $20$ (since $x>0$), then $h(x)$ is monotonically increasing in $(0,+\infty... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,589 |
一、(40 points) Given a circle is tangent to sides $AB$ and $BC$ of $\triangle ABC$, and also tangent to the circumcircle of $\triangle ABC$ at point $T$. If $I$ is the incenter of $\triangle ABC$, prove:
$$
\angle A T I=\angle C T I .
$$ | Let $\triangle A B C$ have an circumcircle with center $O$ and radius $R$, and a smaller circle with center $O^{\prime}$ and radius $r$. Let $\odot O^{\prime}$ be tangent to $A B$ and $B C$ at points $D$ and $E$, respectively.
Connect $D E$ and $B O^{\prime}$, intersecting at point $I^{\prime}$.
We will prove that poin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,590 |
In the numbers
$$
2010^{1}-1,2010^{2}-1, \cdots, 2010^{n-1}-1
$$
at least one number is divisible by $n$, where, $(n, 2010)$ $=1, n \in \mathbf{N}_{+}$. | Consider the numbers
$$
2010^{0}, 2010^{1}, \cdots, 2010^{n-1} \text {. }
$$
The remainders when these $n$ numbers are divided by $n$ are set as:
$$
r_{0}, r_{1}, \cdots, r_{n-1} \text {. }
$$
Since $n \times 2010^{i} (i=0,1, \cdots, n-1)$, the $n$ remainders $r_{0}, r_{1}, \cdots, r_{n-1}$ can only be $1,2, \cdots, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,591 |
Example 1 For a prime $p$, if for infinitely many positive integers $k$, there exists a sequence of positive integers $n_{1}, n_{2}, \cdots, n_{k}$ satisfying:
(1) When $i=1,2, \cdots, k$, $n_{i} \geqslant \frac{p+1}{2}$;
(2) When $i=1,2, \cdots, k$, $p^{n_{i}}-1$ is a multiple of $n_{i+1}$, and $\frac{p_{i}-1}{n_{i+1}... | Proof: Clearly, $n_{i} \geqslant 3$ and all are odd.
Let $q(q \geqslant 3)$ be the smallest prime factor of $n_{1} n_{2} \cdots n_{k}$, and assume without loss of generality that $q \mid n_{2}$.
By Corollary 1, we know that ord $(2) \mid (q-1)$, hence $\operatorname{ord}_{q}(2)<q$.
But from $q \mid\left(2^{n_{1}}-1\rig... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,592 |
Three. (50 points) Let the function $y=f(x)\left(x, y \in \mathbf{N}_{+}\right)$ satisfy:
(1) For any $a, b \in \mathbf{N}_{+}, a>b$, we have $a f(a)+b f(b)>a f(b)+b f(a)$;
(2) For any $n \in \mathbf{N}_{+}$, we have $f(f(n))=3 n$.
Let $a_{n}=f\left(3^{n}\right)\left(n \in \mathbf{N}_{+}\right)$. Prove:
$$
\frac{1}{a_{... | Three, take $a=n+1, b=n\left(n \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
(n+1) f(n+1)+n f(n) \\
>(n+1) f(n)+n f(n+1) \\
\Rightarrow f(n+1)>f(n) .
\end{array}
$$
Hence, $f(x)$ is an increasing function on $\mathbf{N}_{+}$.
Let $f(1)=m\left(m \in \mathbf{N}_{+}\right)$. Then
$$
f(m)=f(f(1))=3 \text {. }
$$
O... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,593 |
Four. (50 points) In an infinite grid composed of unit squares, each unit square contains an integer. If the integer in each square is equal to the sum of the integers in the two adjacent squares directly above and to the left, and there exists a row $M_{0}$, where all the squares contain positive integers. Let the row... | For the convenience of narration, any column taken in the grid is called
the 0th column, to the left of the 0th column are the -1st column, the -2nd column, …, and to the right of the 0th column are the 1st column, the 2nd column, $\cdots$....
Let the number in the $x(x \in \mathbf{Z})$ column of the $M_{N}(N \geqslan... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 724,594 |
In trapezoid $A B C D$ as shown in Figure 3,
$$
\begin{array}{l}
A D / / B C, A B=A D, \\
\angle B A D=100^{\circ}, \\
\angle B C D=30^{\circ} .
\end{array}
$$
Prove: $A C=B C$. | Prove: As shown in Figure 3, draw $A E / / D C$ intersecting $B C$ at point $E$. Then
$$
E C=A D=A B, \angle A E B=\angle B C D=30^{\circ} \text {. }
$$
Take the circumcenter $O$ of $\triangle A B E$, and connect $O A, O B, O E$.
It is easy to see that $\angle A O B=2 \angle A E B=60^{\circ}$.
Thus, $\triangle O A B$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,595 |
In the cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$ with edge length $a\left(a \in \mathbf{R}_{+}\right)$ as shown in Figure 4, $E$ is the center of the square $A B B_{1} A_{1}$, and $F, G$ are points moving along the edges $B_{1} C_{1}$ and $D D_{1}$, respectively. Question: What is the maximum value of the area of the pr... | The projection area of the spatial quadrilateral $D E F G$ on the bottom surface is equal, and its orthogonal projection on the bottom surface $A B C D$ is the $\triangle D E_{1} F_{1}$ shown in Figure 5.
Since point $F_{1}$ slides along $B C$, the maximum area is $S_{\triangle D E_{1} C}=\frac{1}{2} a^{2}$, and the m... | \frac{3}{4} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,598 |
Example 2 Does there exist $n \in \mathbf{N}_{+}$, such that
$$
103 \mid n, 2^{2 n+1} \equiv 2(\bmod n)
$$
holds? ${ }^{[2]}$
(2010, Indian National Team Selection Exam) | No solution exists. If it did, by Fermat's Little Theorem, we would have
$$
2^{102} \equiv 1 \pmod{103}.
$$
From the problem and Corollary 2, we have
$$
2^{2n} \equiv 1 \pmod{103},
$$
$$
2^{(102, 2n)} \equiv 1 \pmod{103}.
$$
Notice that $102 = 2 \times 3 \times 17$, and when
$$
(102, 2n) = 2, 3, 6,
$$
it is found to... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,599 |
Example 3 Given $x \in \mathbf{N}_{+}$, if $x^{7} \equiv 1(\bmod 29)$, find the value of $x$ modulo 29. | Solve and verify that 2 is a primitive root modulo 29.
Since $x^{7}=1(\bmod 29)$ has at most 7 solutions, and it is verified that $2^{4} 、 2^{8} 、 2^{12} 、 2^{16} 、 2^{20} 、 2^{24} 、 2^{28}$ are exactly the different solutions of the equation, the equation has exactly these 7 solutions. Therefore, the required are the ... | 2, 16, 25, 28, 24, 20, 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,600 |
Example 4 Let $n(n \geqslant 2)$ be a positive integer. Prove: $n \mid\left(\sum_{k=1}^{n-1} k^{n-1}+1\right) \Leftrightarrow$ for any prime factor $p$ of $n$, $p\left|\left(\frac{n}{p}-1\right),(p-1)\right|\left(\frac{n}{p}-1\right)$.
(2005, Silk Road Mathematical Competition) | Proof: Let $n = Ap$ ($p$ is a prime).
If $(p-1) \mid (n-1)$, by Fermat's Little Theorem, it is easy to see that
$$
\sum_{k=1}^{n-1} k^{n-1} \equiv (n-1) - (A-1) \equiv -A \pmod{p}.
$$
If $(p-1) \nmid (n-1)$, by Theorem 2, we can take a primitive root $r$ of $p$.
Since $(r, p) = 1$, we know that $\{r, 2r, \cdots, (p-1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,601 |
Example 5 Find the number of ordered integer pairs $(a, b)$ such that
$$
x^{2}+a x+b=167 y
$$
has integer solutions $(x, y)$, where $1 \leqslant a, b \leqslant 2004 .^{[4]}$ (2004, Singapore Mathematical Olympiad) | If $x^{2}+a x+b \equiv 0(\bmod 167)$, then completing the square gives
$$
a^{2}-4 b \equiv(2 x+a)^{2}(\bmod 167) .
$$
Therefore, for a fixed $a$, $a^{2}-4 b$ is a quadratic residue modulo 167.
Hence, by the lemma, $b$ can take $\frac{167-1}{2}+1=84$ values modulo 167.
And $\frac{2004}{167}=12$, so each $a$ correspon... | 2020032 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,602 |
Example 6 Let $p$ be an odd prime satisfying $p \equiv 1(\bmod 4)$. Calculate the value of $\sum_{k=1}^{p-1}\left\{\frac{k^{2}}{p}\right\}$ where $(\{x\}=x-[x],[x]$ denotes the greatest integer not exceeding the real number $x$). ${ }^{[5]}$
(5th Hong Kong Mathematical Olympiad (2002)) | Since $p \equiv 1(\bmod 4)$, by Theorem 3 we have
$$
\left(\frac{-1}{p}\right)=(-1)^{\frac{2-1}{2}} \equiv 1(\bmod p) \text {, }
$$
Therefore, -1 is a quadratic residue modulo $p$.
By $\left(\frac{-b}{p}\right)=\left(\frac{b}{p}\right)\left(\frac{-1}{p}\right)$, we know that $b$ is a quadratic residue if and only if $... | \frac{p-1}{4} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,603 |
Example 7 Find the solutions in the set of integers for
$$
x^{2010}-2006=4 y^{2009}+4 y^{2008}+2007 y
$$
[6]
$(2009$, Macedonian Mathematical Olympiad) | Prove that the original equation is equivalent to
$$
x^{2010}+1=\left(4 y^{2008}+2007\right)(y+1) \text {. }
$$
Notice that, the left side of the above equation is a sum of squares, so by Corollary 3, it has only prime factors of the form $4k+1$.
However, $4 y^{2008}+2007 \equiv 3(\bmod 4)$, hence the original equati... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,604 |
Example 8 Proof: There are infinitely many positive integers $n$ such that $\left(n^{2}+1\right) \mid n!$; there are also infinitely many positive integers $n$ such that $\left(n^{2}+1\right) \nmid n!$. ${ }^{[7]}$
(2008, Romanian Mathematical Olympiad) | For the first proposition:
When $n=2 k^{2}$,
$$
\begin{array}{l}
n^{2}+1=\left(2 k^{2}-2 k+1\right)\left(2 k^{2}+2 k+1\right) . \\
\text { Since }\left(2 k^{2}-2 k+1,2 k^{2}+2 k+1\right) \\
=\left(2 k^{2}-2 k+1,4 k\right)=1,
\end{array}
$$
and $2 k^{2}-2 k+1<2 k^{2}=n$, so $2 k^{2}-2 k+1$ is a factor of $n!$.
Taking $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,605 |
Example 3 As shown in Figure 3, in rectangle $A B C D$, it is known that the diagonal length is 2, and $\angle 1=\angle 2=\angle 3=\angle 4$. Then the perimeter of quadrilateral $E F G H$ is ().
(A) $2 \sqrt{2}$
(B) 4
(C) $4 \sqrt{2}$
(D) 6
(2010, Sichuan Province Junior High School Mathematics League (Grade 8)) | As shown in Figure 3, according to the properties of axial symmetry, we know that the hypotenuse of $\triangle I J K$ is the perimeter of quadrilateral $E F G H$. The legs of the right triangle are twice the lengths of the sides of the rectangle. The diagonal of the rectangle and the two sides of the rectangle form a r... | 4 | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,606 |
3. Let $q=2p+1$ ($p, q>0$ and both are prime numbers). Prove: there exists a multiple of $q$ whose sum of digits in decimal representation does not exceed 3.
(2009, Brazilian Mathematical Olympiad) | First, prove that the order of 10 modulo $q$ is $p$.
By Theorem 3, we know that 10 is a quadratic residue modulo $q$. Then consider the set of quadratic residues modulo $q$:
$$
\begin{array}{l}
A=\{0\} \cup\left\{10^{k}(\bmod q), 0 \leqslant k<p\right\}, \\
B=\{q-1\} \cup\left\{-1-10^{k}(\bmod q), 0 \leqslant k<p\right... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,609 |
4. (1) Find all prime numbers $p$ such that $\frac{7^{p-1}-1}{p}$ is a perfect square;
(2) Find all prime numbers $p$ such that $\frac{11^{p-1}-1}{p}$ is a perfect square.
(2009, Turkish Mathematical Olympiad) | If there exist positive integers $x$ and prime $p$ satisfying $p x^{2}=q^{p-1}-1$, then there always exist integers $y, z$ satisfying one of the following two cases:
(1) $q^{\frac{p-1}{2}}-1=2 p y^{2}, q^{\frac{p-1}{2}}+1=2 z^{2}$;
(2) $q^{\frac{p-1}{2}}-1=2 y^{2}, q^{\frac{p-1}{2}}+1=2 p z^{2}$.
By Theorem 3, it is kn... | p=3 \text{ for } q=7, \text{ no } p \text{ for } q=11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,610 |
5. There are infinitely many primes of the form $8 k-1$, $8 k+5$.
| Given a finite number of $p_{1}, p_{2}, \cdots, p_{k}$, imitating the proof of Corollary 4, for the form $8 k-1$ prove that
$$
p=\left(p_{1} p_{2} \ldots p_{k}\right)^{2}-2
$$
has a satisfying prime factor, and for the form $8 k+5$ prove that
$$
p=\left(p_{1} p_{2} \cdots, p_{k}\right)^{2}+4
$$
has a satisfying prime... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,611 |
Given a positive integer $d$. Prove: there exist infinitely many positive integers $n$, such that $d \cdot n!-1$ is a composite number. | Proof First, we prove a lemma.
Lemma Let $p$ be a prime factor of $S=\frac{(2 l)!}{d}+1(l \geqslant d \geqslant 2)$. Then
$$
p \geqslant 2 l+3 \text {, and } p \mid[d \cdot(p-2 l-1)!-1] \text {. }
$$
Proof It is easy to see that $p \geqslant 2 l+1$.
If $p=2 l+1$, then $p \mid d S$, i.e.,
$$
p \mid[(2 l)!+d]=[(p-1)!+d]... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,612 |
Given $\boldsymbol{a}=(1,0), \boldsymbol{b}=\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$, $\boldsymbol{c}=\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right), x \boldsymbol{a}+y \boldsymbol{b}+z \boldsymbol{c}=(1,1)$. Find the minimum value of $x^{2}+y^{2}+z^{2}$.
(2011, Tsinghua University Independent Admission Examination... | Proof: Let $a, b > 0$.
From the problem, we have
\[
\left\{
\begin{array}{l}
a_{1} x + b_{1} y + c_{1} z = a, \\
a_{2} x + b_{2} y + c_{2} z = b.
\end{array}
\right.
\]
Recalling the formula for the dot product of vectors, let
\[
\begin{array}{l}
\boldsymbol{m} = \overrightarrow{O M} = (x, y, z), \boldsymbol{n} = \over... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,613 |
1. The postal postage standard for domestic and foreign printed matter in our country is as follows: 0.7 yuan for up to 100 g, and an additional 0.4 yuan for each additional 100 g (any weight less than 100 g is counted as 100 g). A person is mailing a book from Chengdu to Shanghai, and the book weighs 470 g. The postag... | 1. A. Since $470=100+3 \times 100+70$, the postage is $0.7+3 \times 0.4+0.4=2.3$ (yuan). | A | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 724,614 |
2. Let the fractional equation about $x$ be $\frac{a-2}{2-x}=\frac{a-2}{x-2}$ with infinitely many solutions. Then the number of values for $a$ is ( ).
(A) 0
(B) 1
(C) 2
(D) Infinitely many | 2. B.
Since the fractional equation has a solution, we have
$$
a-2=2-a \Rightarrow a=2 \text {. }
$$
Therefore, $a$ has only 1 value. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,615 |
3. Real numbers $a, b, c$ satisfy $a+b+c=0$, and $abc>0$. Then the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is ( ).
(A) Positive
(B) Negative
(C) Zero
(D) Cannot be determined | 3. B.
From the problem, we know that $a, b, c$ must be two negative and one positive (let's assume $a<0, b<0, c>0$).
Then $c>-a>0 \Rightarrow -\frac{1}{a}>\frac{1}{c}$.
And $\frac{1}{b}<0$, so $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<0$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,616 |
Example 4 As shown in Figure 4, on the sides $AB$ and $AC$ of $\triangle ABC$, take points $Q$ and $P$ respectively, such that
$$
\begin{array}{l}
\angle P B C \\
=\angle Q C B \\
=\frac{1}{2} \angle A .
\end{array}
$$
Prove: $B Q = C P$. | Proof From the given,
$$
\text { we know } \begin{aligned}
& \angle B Q C+\angle C P B \\
= & \angle A+\angle A C B-\frac{1}{2} \angle A+ \\
& \angle A+\angle A C B-\frac{1}{2} \angle A \\
= & \angle A+\angle B+\angle C \\
= & 180^{\circ} .
\end{aligned}
$$
Construct the symmetric point $P^{\prime}$ of point $P$ with ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,617 |
4. If $a$, $b$, and $c$ are the lengths of the sides of a triangle, and satisfy $\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=\frac{1}{a+b-c}$, then it must be true that ( ).
(A) $a=b=c$
(B) $a=b$
(C) $a=c$ or $b=c$
(D) $a^{2}+b^{2}=c^{2}$ | 4. C.
From the fractional simplification, we get
$$
(a+b)(a-c)(b-c)=0 \text {. }
$$
Therefore, $a=c$ or $b=c$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,618 |
5. As shown in Figure 1, given rectangle $A B C D, A B=$ $8, B C=6$. If the rectangle is folded so that vertices $A$ and $C$ coincide, then the length of the fold line $P Q$ is ( ).
(A) $\frac{15}{2}$
(B) 7
(C) 8
(D) $\frac{17}{2}$ | 5. A.
Obviously, $A C$ and $P Q$ bisect each other perpendicularly. Therefore,
$$
\triangle P O C \backsim \triangle A D C \Rightarrow \frac{P O}{A D}=\frac{O C}{D C} \text {, }
$$
we get $P Q=\frac{15}{4} \Rightarrow P Q=\frac{15}{2}$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,619 |
6. Using three 2s, the largest number that can be written is ( ).
(A) $2^{2^{2}}$
(B) 222
(C) 242
(D) greater than 1000 | 6. D.
The largest number is $2^{22}>1000$. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 724,620 |
2. Given $a=\sqrt{3}-1$. Then $a^{2012}+2 a^{2011}-2 a^{2010}=$ | 2. 0 .
Notice that $a^{2}=(\sqrt{3}-1)^{2}=4-2 \sqrt{3}$. Then $a^{2}+2 a-2=0$.
Therefore, $a^{2012}+2 a^{2011}-2 a^{2010}$ $=a^{2010}\left(a^{2}+2 a-2\right)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,622 |
3. Figure 2 is a rectangle composed of 6 squares. If the area of the smallest square is 1, then the area of this rectangle is $\qquad$ .
| 3. 143 .
Let the side lengths of the six squares, from smallest to largest, be
$$
1, x, x, x+1, x+2, x+3 \text{.}
$$
Then, by the equality of the top and bottom sides of the rectangle, we have
$$
\begin{array}{l}
x+x+(x+1)=(x+2)+(x+3) \\
\Rightarrow x=4 .
\end{array}
$$
Thus, the length and width of the rectangle ar... | 143 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,623 |
4. Given $\triangle A B C$
the lengths of the three medians are 3,
4, 5. Then $S_{\triangle M B C}$ is $\qquad$ | 4. 8 .
As shown in Figure 4, extend GD to point $D^{\prime}$, making it twice as long.
Then the side lengths of $\triangle G D^{\prime} C$ are $\frac{2}{3}$ times the lengths of the three medians of $\triangle A B C$.
Therefore, it is a right triangle, and its area is $\frac{8}{3}$.
Additionally, the area of $\triangl... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,624 |
Three. (20 points) There are $m$ regular $n$-sided polygons, and the sum of the interior angles of these $m$ regular polygons can be divided by 8. Find the minimum value of $m+n$.
| Three, from the problem, we know that the total sum of the interior angles of these $m$ regular polygons is $m(n-2) \times 180$.
From $81[180 m(n-2)]$
$\Rightarrow 21 \mathrm{~m}(n-2) \Rightarrow 21 \mathrm{mn}$.
Thus, at least one of $m$ and $n$ is even.
Also, $m \geqslant 1, n \geqslant 3$, and both are integers.
To ... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,625 |
Four. (25 points) There are several (more than enough) socks in red, yellow, blue, and white. If any two socks of the same color can make 1 pair, the question is: What is the minimum number of socks needed to ensure that 10 pairs of socks can be formed? | Solution 1 Since there are 4 colors, among 5 socks, there must be 1 pair.
After taking out 1 pair, 3 socks remain. By adding 2 more socks, another pair can be formed.
Following this logic, the number of pairs of socks $(x)$ and the number of socks needed $(y)$ have the following relationship:
$$
y=2 x+3 \text {. }
$$... | 23 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,626 |
Five. (25 points) As shown in Figure 3, given square $ABCD, BE=BD, CE$ $/ / BD, BE$ intersects $CD$ at point $F$. Prove:
$$
DE=DF \text {. }
$$ | Five, as shown in Figure 5, construct the symmetric point $E^{\prime}$ of $E$ with respect to $BC$, and connect $D E^{\prime}$, $C E^{\prime}$, and $B E^{\prime}$. According to the properties of symmetry, we have
$$
\begin{array}{l}
B E^{\prime}=B E=B D, \\
C E^{\prime}=C E, \\
\angle E C E^{\prime}=90^{\circ} . \\
\te... | DE=DF | Geometry | proof | Yes | Yes | cn_contest | false | 724,627 |
Example 5 As shown in Figure 5, in $\triangle A B C$, $\angle C=90^{\circ}$, $I$ is the intersection of the angle bisectors $A D$ and $B E$ of $\angle A$ and $\angle B$. Given that the area of $\triangle A B I$ is 12. Then the area of quadrilateral $A B D E$ is $\qquad$
(2004, Beijing Middle School Mathematics Competit... | Solve As shown in Figure 5, construct the symmetric points $F, G$ of points $E, D$ with respect to $AD, BE$ respectively. Then $F, G$ lie on $AB$. Connect $IF, IG$. It is easy to know that
$$
\angle AIB=90^{\circ}+\frac{1}{2} \angle C=135^{\circ}.
$$
By the properties of axial symmetry, we have
$$
\begin{array}{l}
IF=... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,628 |
1. Given $0<x, y<1$. Find the maximum value of
$$
\frac{x y(1-x-y)}{(x+y)(1-x)(1-y)}
$$
(Liu Shixiong) | 1. When $x=y=\frac{1}{3}$, the value of the algebraic expression is $\frac{1}{8}$.
Below is the proof: For any $0 < x, y, z < 1$ such that $x + y + z = 1$. Then
$$
\begin{aligned}
& \frac{x y(1-x-y)}{(x+y)(1-x)(1-y)} \\
= & \frac{x y z}{(x+y)(y+z)(z+x)} \\
\leqslant & \frac{x y z}{2 \sqrt{x y} \cdot 2 \sqrt{y z} \cdot... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,629 |
2. Let the set $M \subseteq\{1,2, \cdots, 2011\}$ satisfy: in any three elements of $M$, there can be found two elements $a, b$, such that $a \mid b$ or $b \mid a$. Find the maximum value of $|M|$ (where $|M|$ denotes the number of elements in the set $M$). (Supplied by Feng Zhigang) | 2. When
$$
M=\left\{1,2,2^{2}, \cdots, 2^{10}, 3,3 \times 2,3 \times 2^{2}, \cdots, 3 \times 2^{9}\right\}
$$
it satisfies the condition, at this time, $|M|=21$.
Assume $|M| \geqslant 22$, let the elements of $M$ be
$$
a_{1}2011,
\end{array}
$$
contradiction.
In summary, the maximum value of $|M|$ is 21. | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,630 |
3. Given an integer $n \geqslant 2$.
(1) Prove: The set $\{1,2, \cdots, n\}$ can be properly arranged into subsets $A_{1}, A_{2}, \cdots, A_{2^{n}}$, such that the number of elements in $A_{i}$ and $A_{i+1}\left(i=1,2, \cdots, 2^{n}\right.$, and $\left.A_{2^{n+1}}=A_{1}\right)$ differ by exactly 1.
(2) For subsets $A_{... | 3. (1) Prove by mathematical induction: there exists a subsequence $A_{1}, A_{2}, \cdots, A_{2^{n}}$ that meets the requirements, and $A_{1}=\{1\}$, $A_{2^{n}}=\varnothing$.
When $n=2$, the sequence $\{1\}, \{1,2\}, \{2\}, \varnothing$ satisfies the requirements.
Assume that when $n=k$, there exists a subsequence $B_... | 0 | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,631 |
4. As shown in Figure $1, A B$ and $C D$ are two chords of different lengths in $\odot O$, intersecting at point $E$. $\odot I$ is internally tangent to $\odot O$ at point $F$, and is tangent to chords $A B$ and $C D$ at points $G$ and $H$, respectively. A line $l$ through point $O$ intersects $A B$ and $C D$ at points... | 4. As shown in Figure 2, draw a tangent line to $\odot O$ parallel to $AB$, and let the point of tangency be $L$. Draw the common tangents of the two circles intersecting at point $S$, and let the intersection of $FS$ and the extension of $BA$ be point $R$. Connect $LF$ and $GF$.
First, prove that points $L$, $G$, and... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,632 |
5. Does there exist an odd number $n(n \geqslant 3)$ and $n$ distinct prime numbers $p_{1}, p_{2}, \cdots, p_{n}$, such that $p_{i}+p_{i+1}(i=1$, $2, \cdots, n, p_{n+1}=p_{1})$ are all perfect squares? Please prove your conclusion.
(Tao Pingsheng) | 5. Suppose there exists an odd number $n(n \geqslant 3)$ and $n$ prime numbers $p_{1}, p_{2}, \cdots, p_{n}$ that meet the requirements.
If $p_{1}, p_{2}, \cdots, p_{n}$ are all odd, then $p_{i}+p_{i+1}$ $(i=1,2, \cdots, n)$ must all be multiples of 4. Thus, the remainders of $p_{1}, p_{2}, \cdots, p_{n}$ when divided... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,633 |
6. Let $a, b, c > 0$. Prove:
$$
\begin{array}{l}
\frac{(a-b)^{2}}{(c+a)(c+b)}+\frac{(b-c)^{2}}{(a+b)(a+c)}+\frac{(c-a)^{2}}{(b+c)(b+a)} \\
\geqslant \frac{(a-b)^{2}}{a^{2}+b^{2}+c^{2}} . \quad \text { (Li Shenghong) }
\end{array}
$$
(Li Shenghong) | 6. Proof 1
$$
\begin{array}{l}
\frac{1}{2}(a-2 b)^{2}+\frac{1}{2}(a-2 c)^{2}+(b-c)^{2} \geqslant 0 \\
\Rightarrow 3\left(a^{2}+b^{2}+c^{2}\right) \geqslant 2 a^{2}+2 a b+2 b c+2 a c \\
\quad=2(a+b)(a+c) \\
\Rightarrow(a+b)(a+c) \leqslant \frac{3}{2}\left(a^{2}+b^{2}+c^{2}\right) .
\end{array}
$$
Similarly, $(b+a)(b+c)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,634 |
7. In $\triangle A B C$, $A B>A C$, the incircle $\odot I$ touches sides $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$ respectively, $M$ is the midpoint of side $B C$, $A H \perp B C$ at point $H$, and the angle bisector $A I$ of $\angle B A C$ intersects lines $D E$ and $D F$ at points $K$ and $L$ respectively. ... | 7. As shown in Figure 4, connect $B I$, $D I$, $B K$, $M L$, $K H$, and extend $C L$ to intersect side $A B$ at point $N$.
Since $C D$ and $C E$ are both tangents to $\odot I$, we have,
$$
\begin{array}{l}
C D=C E . \\
\text { By } \angle B I K=\angle B A I+\angle A B I \\
=\frac{1}{2}(\angle B A C+\angle A B C) \\
=\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,635 |
8. Find all integer pairs $(a, b)$ such that for any positive integer $n$ we have $n \mid\left(a^{n}+b^{n+1}\right)$.
(Chen Yonggao) | 8. The integer pairs $(a, b)$ sought are only $(0,0)$ and $(-1,-1)$.
If one of $a, b$ is 0, obviously, the other must also be 0.
Assume $a b \neq 0$. Take a prime $p$ such that $p>\left|a+b^{2}\right|$.
By Fermat's Little Theorem, we have
$$
a^{p}+b^{p+1}=a+b^{2}(\bmod p) \text {. }
$$
Since $p \mid\left(a^{p}+b^{p+1... | (0,0) \text{ and } (-1,-1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,636 |
2. Given the function $y=f(x)$ has an inverse function $y=$ $f^{-1}(x)$. If the graph of $y=f(x)$ is rotated $90^{\circ}$ counterclockwise around $(1,-1)$, the equation of the resulting curve is ( ).
(A) $y=f^{-1}(-x)-2$
(B) $y=-f^{-1}(-x)-2$
(C) $y=f^{-1}(-x+1)-1$
(D) $y=f^{-1}(-x-1)+1$ | 2. A.
The point $(t, f(t))$ is rotated $90^{\circ}$ counterclockwise around $(1, -1)$, resulting in $(-f(t), t-2)$.
Let $x = -f(t)$. Then
$$
y = t-2 = f^{-1}(-x) - 2 \text{.}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,638 |
3. Let $n$ be a positive integer, and
$$
x=\left(1+\frac{1}{n}\right)^{n}, y=\left(1+\frac{1}{n}\right)^{n+1} \text {. }
$$
Then ( ).
(A) $x^{y}>y^{x}$
(B) $x^{y}=y^{x}$
(C) $x^{y}<y^{x}$
(D) Any of the above is possible | 3. B.
Let $x=\frac{(n+1)^{n}}{n^{n}}, y=\frac{(n+1)^{n+1}}{n^{n+1}}$, taking logarithms easily yields $x^{y}=y^{x}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,640 |
4. If the line $y=x-3$ is tangent to the curve $y=\mathrm{e}^{x+a}$, then the value of the real number $a$ is ( ).
(A) -4
(B) -2
(C) 2
(D) 4 | 4. A.
Let the x-coordinate of the tangent point be $x_{0}$. At $x=x_{0}$, the slope of the curve $y=\mathrm{e}^{\mathrm{x}+a}$ is $\mathrm{e}^{x_{0}+a}$, and the slope of the line $y=x-3$ is 1.
Thus, $\mathrm{e}^{x_{0}+a}=1 \Rightarrow x_{0}=-a$.
Therefore, the y-coordinate of the tangent point is
$$
\begin{array}{l}
... | A | Calculus | MCQ | Yes | Yes | cn_contest | false | 724,641 |
5. On a circle $\odot O$ with radius 1, take a fixed point $A$ and a moving point $B$. Let point $P$ satisfy $A P / / O B$, and $\overrightarrow{A P} \cdot \overrightarrow{A B}=1$. Then the locus of point $P$ is ().
(A) Ellipse
(B) Parabola
(C) Hyperbola
(D) Any of the above is possible | 5. B.
Let's assume $O(0,0)$, $A(1,0)$, and $P(x, y)$.
Since $A P / / O B$, we can set $B(k(x-1), k y)$.
Substituting these coordinates into $\overrightarrow{A P} \cdot \overrightarrow{A B}=1$, we get
$$
k=\frac{x}{(x-1)^{2}+y^{2}} \text {. }
$$
Using the fact that point $B$ lies on $\odot O$, we can derive that the e... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,642 |
6. After expanding and combining like terms of $(a+b+c+d)^{9}$, the number of terms in the resulting polynomial is ( ).
(A) $\mathrm{C}_{9}^{4}$
(B) $\mathrm{C}_{9}^{3}$
(C) $\mathrm{C}_{12}^{4}$
(D) $\mathrm{C}_{12}^{3}$ | 6. D.
Each term in the obtained polynomial is of the form
$$
k a^{x_{1}} b^{x_{2}} c^{x_{3}} d^{x_{4}}(k>0) \text {. }
$$
Thus, $x_{1}+x_{2}+x_{3}+x_{4}=9\left(x_{i} \geqslant 0\right)$.
It is easy to see that the above equation has $\mathrm{C}_{9+4-1}^{4-1}=\mathrm{C}_{12}^{3}$ sets of integer solutions. | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 724,643 |
1. Nine positive real numbers $a_{1}, a_{2}, \cdots, a_{9}$ form a geometric sequence, and
$$
a_{1}+a_{2}=\frac{3}{4}, a_{3}+a_{4}+a_{5}+a_{6}=15 .
$$
Then $a_{7}+a_{8}+a_{9}=$ . $\qquad$ | 1. 112.
Let the common ratio be $q$. Then, from the given conditions, we have
$$
\begin{array}{l}
a_{1}(1+q)=\frac{3}{4} \\
a_{1} q^{2}\left(1+q+q^{2}+q^{3}\right)=15
\end{array}
$$
Dividing the above two equations yields $q^{2}\left(1+q^{2}\right)=20$.
Thus, $q=2, a_{1}=\frac{1}{4}$.
Therefore, $a_{7}+a_{8}+a_{9}=a_{... | 112 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,644 |
3. If the real numbers $x, \alpha, \beta$ satisfy
$$
x=\log _{3} \tan \alpha=-\log _{3} \tan \beta \text {, and } \alpha-\beta=\frac{\pi}{6} \text {, }
$$
then the value of $x$ is $\qquad$ . | 3. $\frac{1}{2}$.
Let $\tan \alpha=y$. Then $\tan \beta=\frac{1}{y}$, $\tan \beta=\tan \left(\alpha-\frac{\pi}{6}\right)=\frac{y-\frac{1}{\sqrt{3}}}{1+\frac{y}{\sqrt{3}}}$. Therefore, $\frac{1}{y}=\frac{y-\frac{1}{\sqrt{3}}}{1+\frac{y}{\sqrt{3}}}$.
Taking the positive root gives $y=\sqrt{3}$.
Thus, $x=\log _{3} y=\fra... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,646 |
4. Let $A$ and $B$ be the two foci of a hyperbola, and point $C$ lies on the hyperbola. It is known that the three sides of $\triangle A B C$ form an arithmetic sequence, and $\angle A C B=120^{\circ}$. Then the eccentricity of the hyperbola is $\qquad$ | 4. $\frac{7}{2}$.
According to the problem, we can assume $|A C|+|A B|=2|B C|$, and
$$
\begin{array}{l}
|A C|^{2}+|B C|^{2}-|A B|^{2} \\
=2|A C||B C| \cos 120^{\circ} .
\end{array}
$$
From this, $|A C|:|A B|:|B C|=3: 7: 5$.
Thus, the eccentricity of the hyperbola is $\frac{7}{2}$. | \frac{7}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,647 |
5. The function $f(x)$ is defined on $(0, \infty)$, and satisfies
$$
f(x)-2 x f\left(\frac{1}{x}\right)+3 x^{2}=0 \text {. }
$$
Then the minimum value of $f(x)$ is $\qquad$ | 5.3.
From $f(x)-2 x f\left(\frac{1}{x}\right)+3 x^{2}=0$, we get $f\left(\frac{1}{x}\right)-\frac{2}{x} f(x)+\frac{3}{x^{2}}=0$.
By solving the above two equations simultaneously, we obtain $f(x)=x^{2}+\frac{2}{x}$.
By the AM-GM inequality,
$$
x^{2}+\frac{2}{x}=x^{2}+\frac{1}{x}+\frac{1}{x} \geqslant 3\left(x^{2} \cdo... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,648 |
6. The complex number $z$ satisfies
$$
|z|(3 z+2 \mathrm{i})=2(\mathrm{i} z-6) \text {. }
$$
Then $|z|$ equals $\qquad$ . | 6. 2 .
Solution 1 Direct calculation shows that
$$
|3 z+2 \mathrm{i}|^{2}-|\mathrm{i} z-6|^{2}=8\left(|z|^{2}-4\right) \text {. }
$$
From this, if $|z|>2$, then
$$
\begin{array}{l}
|3 z+2 \mathrm{i}|>|\mathrm{i} z-6| \\
\Rightarrow|| z|(3 z+2 \mathrm{i})|>|2(\mathrm{i} z-6)|,
\end{array}
$$
which contradicts the giv... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,649 |
Example 7 Two residential areas $A$ and $B$ on the same side of the riverbank $l$ are at distances of $a \, \text{m}$ and $b \, \text{m}$ from the riverbank (i.e., in Figure 7, $A A^{\prime}=a \, \text{m}, B B^{\prime}=b \, \text{m}$), and $A^{\prime} B^{\prime}=c \, \text{m}$. A green belt $C D$ of length $s \, \text{... | (1) As shown in Figure 9, draw a line segment $A P / / l$, such that $A P = s$. Take the symmetric point of $P$ with respect to $l$ as $P^{\prime}$, and connect $B P^{\prime}$, intersecting $l$ at point $D$. On $l$, to the left of point $D$, intercept $D C = s$. Then $C D$ is the position of the required green belt.
I... | \sqrt{(c-s)^{2} + (b+a)^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,650 |
1. In tetrahedron $ABCD$, $AD \perp$ plane $BCD$, $\angle ABD = \angle BDC = \theta < 45^{\circ}$. It is known that $E$ is a point on $BD$ such that $CE \perp BD$, and $BE = AD = 1$.
(1) Prove: $\angle BAC = \theta$;
(2) If the distance from point $D$ to plane $ABC$ is $\frac{4}{13}$, find the value of $\cos \theta$.
| 1. From $A D=B E=1$, we have
$$
\begin{array}{l}
A B=\frac{1}{\sin \theta}, B D=\frac{\cos \theta}{\sin \theta}, D E=\frac{\cos \theta}{\sin \theta}-1, \\
C D=\frac{D E}{\cos \theta}=\frac{1}{\sin \theta}-\frac{1}{\cos \theta} .
\end{array}
$$
Thus, $A C=\sqrt{A D^{2}+C D^{2}}$
$$
\begin{array}{c}
=\sqrt{1+\left(\frac... | \cos \theta = \frac{4}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,651 |
2. Let $a, b, c, d, e, f$ be real numbers, and $a x^{2}+b x+c \geqslant\left|d x^{2}+e x+f\right|$ for any real number $x$. Prove:
$$
4 a c-b^{2} \geqslant\left|4 d f-e^{2}\right| \text {. }
$$ | 2. If $a=0$, then $b=0, d=0, e=0$, the conclusion holds.
When $a \neq 0$, since $a x^{2}+b x+c \geqslant 0$, we have $a>0, b^{2}-4 a c \leqslant 0$.
Furthermore, assume $d>0$.
Then from $a x^{2}+b x+c \geqslant d x^{2}+e x+f$, we know
$a \geqslant d>0$.
Let $g(x)=d x^{2}+e x+f$.
We discuss in two cases:
If $e^{2}-4 d f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,652 |
3. Let the sequence $\left\{a_{n}\right\}$ be defined as
$$
a_{1}=1, a_{n+1}=2 a_{n}+\sqrt{3 a_{n}^{2}+1}(n \geqslant 1) \text {. }
$$
Prove: (1) When $n>1$, $a_{n+1}+a_{n-1}=4 a_{n}$;
(2) $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{1+\sqrt{3}}{2}$. | 3. (1) From the conditions, we know that $\{a_{n}\}$ is an increasing sequence, and $a_{2}=4$. By rearranging and squaring the recursive formula, we get
$$
\left(a_{n+1}-2 a_{n}\right)^{2}=3 a_{n}^{2}+1 \text {, }
$$
which simplifies to $a_{n+1}^{2}-4 a_{n+1} a_{n}+a_{n}^{2}=1$.
Furthermore, $a_{n}^{2}-4 a_{n} a_{n-1}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,653 |
1. Given $x \in \mathbf{R}, y \in \mathbf{R}$. Then “ $|x|<1$, and $|y|<1$ ” is “ $|x+y|+|x-y|<2$ ” ( ).
(A) a sufficient condition but not a necessary condition
(B) a necessary condition but not a sufficient condition
(C) a sufficient and necessary condition
(D) neither a sufficient condition nor a necessary condition | 1. C.
Notice that,
$$
|x+y|+|x-y|=2 \max \{|x|,|y|\} \text{.}
$$
Therefore, it is a sufficient and necessary condition. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 724,654 |
2. The range of the function $f(x)=\sqrt{x-3}+\sqrt{12-3 x}$ is ( .
(A) $[1, \sqrt{2}]$
(B) $\left[1, \frac{3}{2}\right]$
(C) $[1, \sqrt{3}]$
(D) $[1,2]$ | 2. D.
Since $3 \leqslant x \leqslant 4$, we have $0 \leqslant x-3 \leqslant 1$.
Let $x-3=\sin ^{2} \theta\left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)$.
Then $f(x)=\sqrt{x-3}+\sqrt{3(4-x)}$
$=\sin \theta+\sqrt{3\left(1-\sin ^{2} \theta\right)}$
$=\sin \theta+\sqrt{3} \cos \theta=2 \sin \left(\theta+\frac{\pi... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,655 |
3. A box contains 3 black balls and 4 white balls. Now, a ball is randomly drawn from the box each time, and the ball is not put back after it is drawn. Each ball has an equal chance of being drawn, until all balls of one color are drawn. What is the probability that the last ball drawn is black? ( ).
(A) $\frac{3}{5}$... | 3. B.
There are $\mathrm{C}_{7}^{3}=35$ ways to draw all 7 balls, and there are $\mathrm{C}_{6}^{3}=20$ ways to draw all 3 black balls. Therefore, the required probability is $\frac{20}{35}=\frac{4}{7}$. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 724,656 |
4. $l 、 l^{\prime}$ are skew lines that are perpendicular to each other, $l$ is parallel to plane $\alpha$, and $l^{\prime}$ lies in plane $\alpha$. Then, the locus of points in plane $\alpha$ that are equidistant from $l 、 l^{\prime}$ is ( ).
(A) Line
(B) Ellipse
(C) Parabola
(D) Hyperbola | 4. D.
As shown in Figure 3, in plane $\alpha$, take $l^{\prime}$ as the $x$-axis and the projection of $l$ in plane $\alpha$ as the $y$-axis to establish a Cartesian coordinate system. Then, the distance from a moving point $M(x, y)$ in plane $\alpha$ to $l^{\prime}$ is $|y|$, and the distance to $l$ is $\sqrt{x^{2}+a... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,657 |
5. Let $\{a_{n}\}$ be a sequence of positive numbers, and let the sum of the first $n$ terms be $b_{n}$, the product of the first $n$ terms of the sequence $\{b_{n}\}$ be $c_{n}$, and $b_{n}+c_{n}=1$. Then the number in the sequence $\left\{\frac{1}{a_{n}}\right\}$ that is closest to 2011 is ( ).
(A) 1980
(B) 2010
(C) ... | 5. A.
Given $b_{n}=\frac{c_{n}}{c_{n-1}}(n \geqslant 2), c_{1}=b_{1}=\frac{1}{2}$.
$$
\begin{array}{l}
\text { Then } \frac{c_{n}}{c_{n-1}}+c_{n}=1 \Rightarrow \frac{1}{c_{n}}-\frac{1}{c_{n-1}}=1 \\
\Rightarrow \frac{1}{c_{n}}=2+(n-1) \times 1=n+1 \\
\Rightarrow c_{n}=\frac{1}{n+1} \Rightarrow b_{n}=\frac{n}{n+1} \\
\... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,658 |
6. As shown in Figure 1, the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is 1. On the side diagonal $A_{1} D$, take point $M$, and on $C D_{1}$, take point $N$, such that line segment $M N$ is parallel to the diagonal face $A_{1} A C C_{1}$. Then the minimum length of such $M N$ is $(\quad)$.
(A) $\frac{\... | 6. A.
As shown in Figure 4, draw $M M_{1} \perp A D$ at point $M_{1}$, and $N N_{1} \perp D C$ at point $N_{1}$.
It is easy to prove that
$$
M_{1} N_{1} \parallel A C \text{. }
$$
Let $D M_{1}$
$$
=D N_{1}=x \text{. }
$$
Then $M M_{1}=x$,
$$
N N_{1}=1-x \text{. }
$$
Draw $M H \perp N N_{1}$ at point $H$. Then
$$
N ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,659 |
7. Let $\left(1+x-x^{2}\right)^{10}=a_{0}+a_{1} x+\cdots+a_{20} x^{20}$. Then $a_{0}+a_{1}+2 a_{2}+\cdots+20 a_{20}=$ $\qquad$ | 7. -9 .
Let $x=0$, we get $a_{0}=1$.
Differentiating both sides of the given equation, we get
$$
\begin{array}{l}
10\left(1+x-x^{2}\right)^{9}(1-2 x) \\
=a_{1}+2 a_{2} x+\cdots+20 a_{20} x^{19} . \\
\text { Let } x=1 \text {, we get. } \\
a_{1}+2 a_{2}+\cdots+20 a_{20}=-10 .
\end{array}
$$
Then $a_{0}+a_{1}+2 a_{2}+\... | -9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,660 |
1. As shown in Figure 10, in rectangle $A B C D$, $A B=20$, $B C=10$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value.
(1998, Beijing Junior High School Mathematics Competition) | As shown in Figure 10, construct the symmetric point $B^{\prime}$ of point $B$ with respect to line $A C$, and let $B B^{\prime}$ intersect $A C$ at point $E$. Draw $B^{\prime} N \perp A B$ at point $N$, and let $B^{\prime} N$ intersect $A C$ at point $M$. Then, $M$ and $N$ are the required points. The minimum value so... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,661 |
8. Given the line $\sqrt{3} x-y-\sqrt{3}=0$ intersects the parabola $y^{2}=4 x$ at points $A$ and $B$ (point $A$ is above the $x$-axis), and intersects the $x$-axis at point $F$. If $\overrightarrow{O F}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}$, then $\mu^{2}-\lambda^{2}=$ $\qquad$ . | 8. $\frac{1}{2}$.
It is easy to know that $F(1,0)$ is the focus of the parabola $y^{2}=4 x$.
Substituting $y=\sqrt{3} x-\sqrt{3}$ into $y^{2}=4 x$, we get
$$
3 x^{2}-10 x+3=0 \text {. }
$$
Then $A(3,2 \sqrt{3})$ and $B\left(\frac{1}{3},-\frac{2 \sqrt{3}}{3}\right)$.
By the given condition,
$(1,0)=\lambda(3,2 \sqrt{3}... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,662 |
9. Let the set of positive real numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$, and the set $S=\{(a, b) \mid a \in A, b \in A, a-b \in A\}$. Then the set $S$ can have at most $\qquad$ elements. | 9.4950 .
The number of ordered pairs of real numbers $(a, b)$ formed by the elements of set $A$ is $100^{2}=10000$.
Since $a_{i}-a_{i}=0 \notin A$, we have $\left(a_{i}, a_{i}\right) \notin S(i=1,2, \cdots, 100)$.
When $\left(a_{i}, a_{j}\right) \in S$, then $\left(a_{j}, a_{i}\right) \notin S$.
Therefore, the maximum... | 4950 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,663 |
10. The sum of the maximum and minimum values of the function $y=\sin x+\sqrt{2+\cos ^{2} x}$ is $\qquad$ . | 10. $2 \sqrt{2}$.
From $-1 \leqslant \sin x \leqslant 1, 0 \leqslant \cos ^{2} x \leqslant 1$, we know that when $x=\frac{3 \pi}{2}$, $\sin x+\sqrt{2+\cos ^{2} x}$ takes the minimum value $\sqrt{2}-1$.
Let $\sin x=\sqrt{3} \cos \theta$, where
$\theta \in\left[\arccos \frac{1}{\sqrt{3}}, \arccos \left(-\frac{1}{\sqrt{3... | 2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,664 |
11. Let the function $f(x)=\frac{1}{2}+\log _{2} \frac{x}{1-x}$, and define
$$
S_{n}=f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\cdots+f\left(\frac{n-1}{n}\right),
$$
where, $n \in \mathbf{N}_{+}, n \geqslant 2$. Then $S_{n}=$ | 11. $\frac{n-1}{2}$.
When $x_{1}+x_{2}=1$,
$$
f\left(x_{1}\right)+f\left(x_{2}\right)=1+\log _{2} \frac{x_{1} x_{2}}{\left(1-x_{1}\right)\left(1-x_{2}\right)}=1 \text {. }
$$
Then $S_{a}=\frac{1}{2} \sum_{i=1}^{n-1}\left(f\left(\frac{i}{n}\right)+f\left(\frac{n-i}{n}\right)\right)=\frac{n-1}{2}$. | \frac{n-1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,665 |
12. Each vertex of the convex pentagon $A B C D E$ is colored with one of five colors, such that the two endpoints of each diagonal have different colors. The number of such coloring methods is $\qquad$ (answer with a number), | 12. 1020 .
The number of coloring ways where all vertices have different colors is $A_{5}^{5}=$ 120. The number of coloring ways where two adjacent vertices have the same color and the rest of the vertices have different colors is
$$
A_{5}^{1} A_{5}^{4}=5(5 \times 4 \times 3 \times 2)=600 \text { ways. }
$$
The numbe... | 1020 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,666 |
13. Find the intersection points of the curves $y=2-\sqrt{1-4^{x}}$ and $y=2^{x} a$ $(a \in \mathbf{R})$. | 13. Let $t=2^{x}$.
Then, the problem is transformed into finding the intersection of the line $y=a t$ and the arc of the circle $y=2-\sqrt{1-t^{2}}(00$, and $y \leqslant 2$.
At this point, the intersection point is in the first quadrant, so $02$, the line intersects the arc at one point.
From $a t=2-\sqrt{1-t^{2}}$
$\... | \left(\log _{2} \frac{2 a-\sqrt{a^{2}-3}}{a^{2}+1}, 2-\frac{2+a \sqrt{a^{2}-3}}{a^{2}+1}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,667 |
14. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with an eccentricity of $\frac{1}{2}$, $F_{1}$ and $F_{2}$ are the left and right foci, respectively. A line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$. If the maximum area of $\triangle F_{1} A B$ is 6, find the equation... | 14. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
line $l_{A B}: x=m y+c$.
Substituting it into the ellipse equation yields ...
$$
\left(a^{2}+b^{2} m^{2}\right) y^{2}+2 b^{2} c m y-b^{4}=0 \text {. }
$$
Thus, $S_{\triangle F_{1} A B}=\frac{1}{2}\left|F_{1} F_{2}\right|\left|y_{1}-y_{2}\right|$
$$
=c ... | \frac{x^{2}}{8}+\frac{y^{2}}{6}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,668 |
15. As shown in Figure 2, quadrilateral $ABCD$ is an inscribed quadrilateral of $\odot O$, with opposite sides $BC$ and $AD$ intersecting at point $F$, and $AB$ and $DC$ intersecting at point $E$. The circumcircle of $\triangle ECF$ intersects $\odot O$ at another point $H$, $AH$ intersects $EF$ at point $M$, and $MC$ ... | 15. (1) As
in Figure 5, connect $E H$,
$C H$, and $F H$. Then
$\angle F A M$
$=\angle D A H$
$=\angle D C H$.
From the fact that points $E$, $C$, $H$, and
$\boldsymbol{F}$ are concyclic, we have
$$
\begin{array}{l}
\angle D C H \\
=\angle H F M .
\end{array}
$$
Thus, $\angle F A M=\angle H F M$.
Since $\angle H M F... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,669 |
16. Given a function $f_{n}(x)$ defined on $\mathbf{R}$ with the following properties:
(i) $f_{n}(0)=\frac{1}{2}$;
(ii) $n\left(f_{n}\left(\frac{k+1}{n}\right)-f_{n}\left(\frac{k}{n}\right)\right)$
$$
=\left(f_{n}\left(\frac{k}{n}\right)-1\right) f_{n}\left(\frac{k+1}{n}\right)(k=0,1, \cdots) \text {. }
$$
(1) For a fi... | 16. (1) From the given condition, we have
$$
\begin{array}{l}
n\left(\frac{1}{a_{k+1}}-\frac{1}{a_{k}}\right)=\left(\frac{1}{a_{k}}-1\right) \frac{1}{a_{k+1}} \\
\Rightarrow(n+1) a_{k}-n a_{k+1}=1 \\
\Rightarrow \frac{a_{k+1}-1}{a_{k}-1}=1+\frac{1}{n} .
\end{array}
$$
Thus, the sequence $\left\{a_{k}-1\right\}$ is a g... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,670 |
1. Function
$$
f(x)=\sin ^{4} x+\sin x \cdot \cos x+\cos ^{4} x
$$
The maximum value is $\qquad$. | 1. $\frac{9}{8}$.
Notice that,
$$
\begin{array}{l}
f(x)=\sin ^{4} x+\sin x \cdot \cos x+\cos ^{4} x \\
=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cdot \cos ^{2} x+ \\
\quad \sin x \cdot \cos x \\
=1-\frac{1}{2} \sin ^{2} 2 x+\frac{1}{2} \sin 2 x \\
=-\frac{1}{2}\left(\sin 2 x-\frac{1}{2}\right)^{2}+\frac{... | \frac{9}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,671 |
2. Given that $S_{n}$ and $T_{n}$ are the sums of the first $n$ terms of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively, and
$$
\frac{S_{n}}{T_{n}}=\frac{2 n+1}{4 n-2}(n=1,2, \cdots) \text {. }
$$
then $\frac{a_{10}}{b_{3}+b_{18}}+\frac{a_{11}}{b_{6}+b_{15}}=$ $\qquad$ | 2. $\frac{41}{78}$.
Since $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are both arithmetic sequences, we have,
$$
\begin{array}{l}
\frac{a_{10}}{b_{3}+b_{18}}+\frac{a_{11}}{b_{6}+b_{15}}=\frac{a_{10}}{b_{1}+b_{20}}+\frac{a_{11}}{b_{1}+b_{20}} \\
=\frac{a_{1}+a_{20}}{b_{1}+b_{20}}=\frac{S_{20}}{T_{20}}=\frac{2 \ti... | \frac{41}{78} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,673 |
3. If the function
$$
f(x)=\log _{a}\left(4 x+\frac{a}{x}\right)
$$
is increasing on the interval $[1,2]$, then the range of values for $a$ is $\qquad$ . | 3. $(1,4]$.
Let $g(x)=4 x+\frac{a}{x}$.
According to the problem;
when $0<x<\frac{\sqrt{a}}{2}$, $g(x)$ is a decreasing function, and when $x>\frac{\sqrt{a}}{2}$, $g(x)$ is an increasing function.
Since $g(x)$ is an increasing function on the interval $[1,2]$, we have $\frac{\sqrt{a}}{2} \leqslant 1 \Rightarrow a \leq... | (1,4] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,674 |
4. As shown in Figure 1, in the tetrahedron $D-ABC$, it is known that $DA \perp$ plane $ABC$, and $\triangle ABC$ is an equilateral triangle with a side length of 2. Then, when the tangent value of the dihedral angle $A-BD-C$ is 2, $V_{D-ABC}=$ $\qquad$ | 4.2.
Given $D A \perp$ plane $A B C$, we know
$D A \perp A B$, plane $D A B \cdot \perp$ plane $A B C$.
As shown in Figure 3, take the midpoint $O$ of $A B$.
Then, by the problem statement, $C O \perp A B, C O$
is on plane $D A B$, and
$$
C O=\sqrt{3} \text {. }
$$
Draw $O E \perp D B$ at point $E$, and connect $C E$... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,675 |
5. Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies:
(1) $f(1)=1$;
(2) When $00$;
(3) For any real numbers $x, y$,
$$
f(x+y)-f(x-y)=2 f(1-x) f(y) \text {. }
$$
Then $f\left(\frac{1}{3}\right)=$ . $\qquad$ | 5. $\frac{1}{2}$.
Let $x=\frac{2}{3}, y=\frac{1}{3}$. From (3) we get $f(1)-f\left(\frac{1}{3}\right)=2 f\left(\frac{1}{3}\right) f\left(\frac{1}{3}\right)$. Combining $f(1)=1, f\left(\frac{1}{3}\right)>0$, we solve to get $f\left(\frac{1}{3}\right)=\frac{1}{2}$. | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,676 |
6. Given real numbers $x, y$ satisfy $3 x^{2}+4 y^{2}=48$. Then
$$
\sqrt{x^{2}+y^{2}-4 x+4}+\sqrt{x^{2}+y^{2}-2 x+4 y+5}
$$
The maximum value is $\qquad$. | $$
6.8+\sqrt{13} \text {. }
$$
Let point $P(x, y)$ be on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$, and $A(2,0)$ be the right focus of the ellipse. Point $B(1,-2)$ is inside the ellipse.
Let the left focus of the ellipse be $F_{1}(-2,0)$, and the length of the major axis be $2a$. Then
$$
\begin{array}{l}
\sqr... | 8+\sqrt{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,677 |
7. Given positive integers $x, y, z$ satisfying $x y z=(14-x)(14-y)(14-z)$, and $x+y+z<28$. Then the maximum value of $x^{2}+y^{2}+z^{2}$ is . $\qquad$ | 7. 219 .
From the problem, we know that $x$, $y$, and $z$ are all positive integers less than 14. On the other hand, expanding the given equation, we get
$$
2 x y z=14^{3}-14^{2}(x+y+z)+14(x y+y z+z x) \text {. }
$$
Thus, $71 x y z$.
Since $x$, $y$, and $z$ are all less than 14, at least one of $x$, $y$, or $z$ must ... | 219 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,678 |
8. There are five table tennis balls, three of which are new, and two are old (i.e., used at least once). Each time a match is played, two balls are taken out and used, and then all are put back. Let the number of new balls taken in the second match be $\xi$, then the mathematical expectation $E \xi=$ $\qquad$ . | 8. $\frac{18}{25}$.
Let the probability of drawing $m$ new balls in the $i$-th game be $P_{i}(m)$. It is easy to see that
$$
\begin{array}{l}
P_{1}(0)=\frac{\mathrm{C}_{2}^{2}}{\mathrm{C}_{5}^{2}}=\frac{1}{10}, \\
P_{1}(1) \frac{\mathrm{C}_{3}^{1} \mathrm{C}_{2}^{1}}{\mathrm{C}_{5}^{2}}=\frac{6}{10}, \\
P_{1}(2) \frac... | \frac{18}{25} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,679 |
9. For a positive integer $n$, let $x_{n}$ be the real root of the equation
$$
n x^{3}+2 x-n=0
$$
with respect to $x$, and let
$$
a_{n}=\left[(n+1) x_{n}\right](n=2,3, \cdots),
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then
$$
\frac{1}{1005}\left(a_{2}+a_{3}+\cdots+a_{2011}\right)... | 9.2013.
Let $f(x)=n x^{3}+2 x-n$.
It is easy to see that when $n$ is a positive integer, $f(x)$ is an increasing function.
When $n \geqslant 2$,
$$
\begin{array}{l}
f\left(\frac{n}{n+1}\right)=n\left(\frac{n}{n+1}\right)^{3}+2 \times \frac{n}{n+1}-n \\
=\frac{n}{(n+1)^{3}}\left(-n^{2}+n+1\right)0$.
Therefore, when $n ... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,680 |
10. In the Cartesian coordinate system, given the point set $I=\{(x, y) \mid x, y$ are integers, and $0 \leqslant x, y \leqslant 5\}$. Then the number of different squares with vertices in the set $I$ is . $\qquad$ | 10. 105.
It is easy to know that there are only two types of squares that meet the conditions: squares whose sides lie on lines perpendicular to the coordinate axes, called "standard squares," and squares whose sides lie on lines not perpendicular to the coordinate axes, called "oblique squares."
(1) In standard squar... | 105 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,681 |
11. Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ lie on the ellipse $C$. If $x_{1}+x_{2}=\frac{1}{2}$, and $\overrightarrow{A F_{2}}=\lambda \overrightarrow{F_{2} B}$, find the value... | 11. From $\overrightarrow{F_{2} A}=\lambda \overrightarrow{F_{2} B}$, we know that points $A$, $B$, and $F_{2}$ are collinear.
If line $A B \perp x$-axis, then $x_{1}=x_{2}=1$, which does not meet the requirement.
If line $A B$ has a slope (let it be $k$), then the line $l_{A B}: y=k(x-1)$.
From $\left\{\begin{array}{l... | \frac{3+\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,682 |
3. In $\triangle A B C$, $\angle A=75^{\circ}, \angle B=35^{\circ}$, and $D$ is a point on side $B C$ such that $B D=2 C D$. Prove:
$$
A D^{2}=(A C+B D)(A C-C D) \text {. }
$$
(2008, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | Extend $B C$ to point $E$ such that $C E=A C$.
Given that $\angle C=70^{\circ}$, then
$$
\angle E=35^{\circ}=\angle B \text {, }
$$
which means $\triangle A B E$ is an isosceles triangle.
Draw $A M \perp B E$ at point $M$. Then $M$ is the midpoint of side $B E$. Take the midpoint $F$ of $B D$, then
$$
B F=F D=D C \tex... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,683 |
12. Given the quadratic function
$$
f(x)=a x^{2}+2 b x+c(c>b>a) \text {, }
$$
its graph passes through the point $(1,0)$, and intersects with the line $y=-a$. Prove: $0 \leqslant \frac{b}{a}<1$. | 12. From the given, we have $a+2 b+c=0$.
Combining $c>b>a$, we know $a<0$.
Also, from $b>a$, we know $\frac{b}{a}<1$.
Since the graph of $y=f(x)$ intersects the line $y=-a$, the equation $a x^{2}+2 b x+c+a=0$ has real roots. Therefore,
$$
\begin{array}{l}
\Delta=4 b^{2}-4 a(c+a) \\
=4 b^{2}+4(2 b+c)(-2 b) \\
=-4 b(3 b... | 0 \leqslant \frac{b}{a}<1 | Algebra | proof | Yes | Yes | cn_contest | false | 724,684 |
13. As shown in Figure 2, let the circumcircle of acute triangle $\triangle ABC$ be circle $\Gamma$. The tangents to circle $\Gamma$ at points $B$ and $C$ intersect at point $P$. Connect $AP$ and let it intersect $BC$ at point $D$. Points $E$ and $F$ are on sides $AC$ and $AB$ respectively, such that $DE \parallel BA$ ... | 13. (1) From the given, we know $P B=P C$,
$$
\begin{array}{l}
\angle A B P=180^{\circ}-\angle A C B, \\
\angle A C P=180^{\circ}-\angle A B C . \\
\text { Then } \frac{B D}{C D}=\frac{S_{\triangle A B P}}{S_{\triangle A C P}}=\frac{A B \cdot B P' \sin \angle A B \cdot \cdot C P \sin \angle A C P}{A B} \\
=\frac{A B \s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,685 |
14. Given $f(x)=\mathrm{e}^{x}-x-1$ (where e is the base of the natural logarithm). Prove:
(1) $f(x) \geqslant 0$ always holds;
(2) $\left(\frac{1}{2 n}\right)^{n}+\left(\frac{3}{2 n}\right)^{n}+\cdots+\left(\frac{2 n-1}{2 n}\right)^{n}<\frac{\sqrt{\mathrm{e}}}{\mathrm{e}-1}$
for all positive integers $n$. | 14. (1) From $f^{\prime}(x)=\mathrm{e}^{x}-1$, we know
when $x<0$, $f^{\prime}(x)<0$; when $x>0$, $f^{\prime}(x)>0$.
Therefore, $f(x)$ is a decreasing function on the interval $(-\infty, 0]$ and an increasing function on $[0,+\infty)$.
Thus, $f(x) \geqslant f(0)=0$.
(2) From (1), we know the inequality $1+x \leqslant ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,686 |
15. Given $P_{1}, P_{2}, \cdots, P_{35}$ are the 35 vertices of a convex 35-gon in the plane, and the distance between any two points among $P_{1}, P_{2}, \cdots, P_{35}$ is not less than $\sqrt{3}$. Prove: From these 35 points, five points can be selected such that the distance between any two of these five points is ... | 15. First, prove a lemma.
Lemma Let $P$ be any point among the 35 points $P_{1}, P_{2}, \cdots, P_{35}$. Then, among the remaining 34 points, at most six points are less than 3 units away from point $P$.
Proof By contradiction.
As shown in Figure 4, assume there are 7 points (let's say $P_{1}, P_{2}, \cdots, P_{7}$) t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,687 |
1. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
\frac{2 x+5}{3}-x>5, \\
\frac{x+3}{2}-t<x,
\end{array}\right.
$$
has only five integer solutions. Then the range of $t$ is ( ).
(A) $-6<t<-\frac{11}{2}$
(B) $-6 \leqslant t<-\frac{11}{2}$
(C) $-6<t \leqslant-\frac{11}{2}$
(D) $-6 \leqslant t \leq... | 1. C.
It is known that the solution set of the system of inequalities is
$$
3-2 t<x<20 \text {. }
$$
Since the system of inequalities has only 5 integer solutions, therefore,
$$
14 \leqslant 3-2 t<15 \Rightarrow-6<t \leqslant-\frac{11}{2} \text {. }
$$ | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 724,688 |
2. For the equation in $y$
$$
4 y^{2}+4(2-k) y-k^{2}=0
$$
the two real roots $y_{1}, y_{2}$ satisfy $\left|y_{1}\right|=2+\left|y_{2}\right|$. Then the value of $k$ is ( ).
(A) 1 or 4
(B) 0 or 1
(C) 0 or 4
(D) 0 or 1 or 4 | 2. C.
$$
\begin{array}{l}
\text { Given }\left|y_{1}\right|=2+\left|y_{2}\right| \\
\Rightarrow\left(y_{1}+y_{2}\right)^{2}-2 y_{1} y_{2}-2\left|y_{1} y_{2}\right|=4 \text {. }
\end{array}
$$
By Vieta's formulas, we have
$$
y_{1}+y_{2}=k-2, y_{1} y_{2}=-\frac{k^{2}}{4} \text {. }
$$
Thus, $(k-2)^{2}=4 \Rightarrow k=0... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,689 |
3. As shown in Figure 1, given that $P A$ and $P B$ are tangents to $\odot O$, with $A$ and $B$ being the points of tangency, $C$ and $D$ are points on $A P$ and $B P$ respectively. Connect $O A$, $O B$, $O C$, $O D$, $O P$, and $C D$. Let the perimeter of $\triangle P C D$ be $c$. If $r=2 A P$, then the line $C D$ and... | 3. B.
As shown in Figure 6, draw $O F \perp C D$ at point $F$, extend $C A$ to point $E$ such that $A E=B D$, and connect $O E$.
Then $\triangle A O E \cong \triangle B O D \Rightarrow O E=O D$.
Let $c=2 A P$
$\Rightarrow C D=A C+B D=A C+A E=C E$
$\Rightarrow \triangle E O C \cong \triangle D O C$
$\Rightarrow O A=O F... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,690 |
4. On the side $BC$ of the equilateral $\triangle ABC$, take a point $D$ such that $CD=2BD$. Draw $CH \perp AD$ at point $H$, and connect $BH$. Then $\angle DBH - \angle DAB$ is ( ).
(A) 0
(B) negative
(C) positive
(D) cannot be determined | 4. A.
As shown in Figure 7, draw $A E \perp B C$ at point $E$. Suppose $B C=$
6a. Then
$$
\begin{array}{l}
E C=E B \\
=3 a, \\
B D=2 a, \\
D E=a, \\
D C=4 a .
\end{array}
$$
Therefore, $A, H, E, C$ are concyclic
$$
\Rightarrow D H \cdot D A=D C \cdot D E=B D^{2}
$$
$\Rightarrow B D$ is the tangent to the circumcircle... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,691 |
5. In $\triangle A B C$, it is known that
$$
A B=A C=1, B C=x, \angle A=36^{\circ} \text {. }
$$
Then $\sqrt[3]{x-x^{2}}=(\quad)$.
(A) $\frac{\sqrt{5}}{2}$.
(B) $\frac{3-\sqrt{5}}{2}$
(C) 1
(D) $\frac{\sqrt{5}-1}{2}$ | 5. D.
As shown in Figure 8, construct the angle bisector of $\angle A B C$ intersecting $A C$ at point $D$. Then
$$
\begin{array}{l}
A D=B D=B C=x, \\
C D=1-x, \\
\text { and } \triangle B C D \backsim \triangle A B C \\
\Rightarrow \frac{B C}{A B}=\frac{C D}{B C} \\
\Rightarrow \frac{x}{1}=\frac{1-x}{x} \\
\Rightarr... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,692 |
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