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 |
|---|---|---|---|---|---|---|---|---|---|
7. Let $A(0, b)$ be a vertex of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. It is known that the line $y=x-3$ intersects the ellipse at points $P$ and $Q$, and the centroid of $\triangle A P Q$ is the right focus of the ellipse. Then the coordinates of the right focus of the ellipse are $\qquad$ | 7. $\left(\frac{27-3 \sqrt{5}}{19}, 0\right)$.
As shown in Figure 7, let $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$, and the right focus of the ellipse $F(c, 0)$.
Then, by the centroid formula of $\triangle A P Q$, we have
$$
\begin{array}{l}
c=\frac{x_{1}+x_{2}}{3}, \\
0=\frac{y_{1}+y_{2}+b}{3} .
\... | \left(\frac{27-3 \sqrt{5}}{19}, 0\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,827 |
8. Two players, A and B, are playing a table tennis match, using the 11-point system, which means that for each ball, the winner gets 1 point and the loser gets 0 points. When one side accumulates 11 points, if the other side has less than 10 points, the one who reaches 11 points wins; if the other side has 10 points, ... | $8.1-\frac{\mathrm{C}_{20}^{10}}{2^{20}}$.
We only need to find the probability $P$ that one side reaches 11 points but cannot win yet. At this point, the score in the first 20 balls is 10:10. It can be seen that, in the first 20 balls, arranging 10 positions for A to win, there are $\mathrm{C}_{20}^{10}$ possibilities... | 1-\frac{\mathrm{C}_{20}^{10}}{2^{20}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,828 |
9. (16 points) Prove that for any four non-coplanar points $A, B, C, D$, there exists a unique rhombus $P Q M N$ such that $P \in A B, Q \in B C, M \in C D, N \in D A$; If points $A, B, C, D$ are coplanar, does the conclusion hold? If it holds, provide a proof; if not, give a counterexample.
---
The translation prese... | 9. Uniqueness.
As shown in Figure 8, let quadrilateral $P Q M N$ be a rhombus that meets the conditions. Then
$P N / / Q M$
$\Rightarrow P N / /$ plane $B C D$
$\Rightarrow P N / / B D$.
Therefore,
$P N / / B D / / Q M$.
Similarly, $P Q / / A C / / M N$.
Thus, $\frac{P N}{B D}+\frac{P Q}{A C}=\frac{A P}{A B}+\frac{B P}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,829 |
10. (20 points) An ellipse is constructed with the side $AB$ of the equilateral $\triangle ABC$ as its major axis. If the ellipse intersects the circumcircle of $\triangle ABC$ at exactly two points, find the range of the ellipse's eccentricity $e$.
保留源文本的换行和格式，直接输出翻译结果。 | 10. As shown in Figure 9, let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with the semi-focal distance being $c, A(-a, 0)$, and $B(a, 0)$.
Assume the ellipse and the circumcircle of $\triangle A B C$ have a common point $P(x, y) (y > 0)$ other than points $A$ and $B$.
Since $A,... | \left(0, \frac{\sqrt{6}}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,830 |
11. (20 points) Let the sequence of rational numbers $\left\{a_{n}\right\}$ be defined as follows:
$a_{k}=\frac{x_{k}}{y_{k}}$, where $x_{1}=y_{1}=1$, and
if $y_{k}=1$, then $x_{k+1}=1, y_{k+1}=x_{k}+1$;
if $y_{k} \neq 1$, then $x_{k+1}=x_{k}+1, y_{k+1}=y_{k}-1$.
How many terms in the first 2011 terms of this sequence ... | 11. The sequence is
$$
\frac{1}{1}, \frac{1}{2}, \frac{2}{1}, \frac{1}{3}, \frac{2}{2}, \frac{3}{1}, \cdots, \frac{1}{k}, \frac{2}{k-1}, \cdots, \frac{k}{1}, \cdots \text {. }
$$
Group it as follows:
$$
\begin{array}{l}
\left(\frac{1}{1}\right),\left(\frac{1}{2}, \frac{2}{1}\right),\left(\frac{1}{3}, \frac{2}{2}, \fra... | 213 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,831 |
One. (40 points) As shown in Figure 3, let $\odot O_{1}$ and $\odot O_{2}$ intersect at two distinct points $P$ and $Q$. The common tangent closer to point $Q$ touches $\odot O_{1}$ and $\odot O_{2}$ at points $A$ and $B$, respectively. Extend $P Q$ to intersect $A B$ at point $R$. Draw a line through point $P$ interse... | As shown in Figure 10, let the midpoints of $CP$ and $DP$ be $E$ and $F$ respectively, and connect $O_{1} O_{2}$, $O_{1} A$, $O_{1} E$, $O_{2} B$, and $O_{2} F$. Then
$$
\begin{array}{l}
O_{1} A \perp A B, O_{2} B \perp A B, \\
O_{1} E \perp C P, O_{2} F \perp P D .
\end{array}
$$
Draw $O_{1} S \perp O_{2} F$ at point... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,832 |
$$
\begin{array}{l}
\text { II. (40 points) Let } \\
X=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}\left(a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}\right)
\end{array}
$$
be a set of positive integers. Prove: For any integer $p$ $(1 \leqslant p \leqslant S(X))$, there exists a subset $A$ of $X$ such that $S(A... | When $n=1$, by $a_{1} \leqslant S_{0}+1=1$, we get $a_{1}=1$.
Therefore, $X=\{1\}$. The conclusion is obviously true.
Assume that when $n=k$, the conclusion holds.
When $n=k+1$,
$X=\left\{a_{1}, a_{2}, \cdots, a_{k+1}\right\}$,
where $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{k+1}, a_{i} \leqslant S_{i-1}+1(i... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,833 |
Three, (50 points) A dance troupe has $n(n \geqslant 5)$ actors, and they have arranged some performances, each of which is performed by four actors on stage. In one performance, they found that: it is possible to appropriately arrange several performances so that every two actors in the troupe perform on stage togethe... | Three, use $n$ points to represent $n$ actors.
If two actors have performed on the same stage once, then connect the corresponding points with an edge. Thus, the condition of this problem is equivalent to:
Being able to partition the complete graph $K_{n}$ of order $n$ into several complete graphs $K_{4}$ of order 4, ... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,834 |
$$
\begin{array}{l}
\text { Four. (50 points) Let } \\
a_{1}=2, a_{2}=34, \\
a_{n}=34 a_{n-1}-225 \quad a_{n-2}(n \geqslant 3) .
\end{array}
$$
Question: Does there exist a positive integer $k$ such that $2011 \mid a_{k}$? If it exists, find the smallest positive integer $k$; if it does not exist, explain the reason. | There does not exist such a positive integer $k$.
It is easy to see that $a_{n}=9^{n-1}+25^{n-1}=\left(3^{n-1}\right)^{2}+\left(5^{n-1}\right)^{2}$.
Assume there exists a positive integer $k$ such that $2011 \mid a_{k}$, i.e.,
$2011 \mid\left[\left(3^{k-1}\right)^{2}+\left(5^{k-1}\right)^{2}\right]$.
Note that $\left(3... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,835 |
In $\triangle A B C$, it is known that $A C=7, B C=4$, and a line $C G \perp A B$ intersects $A B$ at point $G$. Also, $\angle G E A=\frac{1}{2} \angle A C B, \frac{A G}{G B}=\frac{5}{2}$. Find the length of $C E$ when $\angle A B C$ is respectively an acute angle and an obtuse angle. | (1) As shown in Figure 2, draw $B F / / C N / / E G$, and let $B F$ intersect the extension of $A E$ at point $F$. Draw $B M \perp$ $N C$ intersecting $C N$ at point $O$.
Given $\angle G E A=\frac{1}{2} \angle A C B$, and $C N / / E G$, we have
$$
\begin{array}{l}
\angle G E A=\frac{1}{2} \angle A C B=\angle A C N, \\
... | CE=\frac{6}{7} \text{ or } CE=\frac{34}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,836 |
Example 3 If the product of 1990 positive integers has exactly 1989 distinct prime factors, prove: among these 1990 positive integers, either one of them is a perfect square, or the product of some of them is a perfect square.
(1986, Moscow Mathematical Olympiad) | Proof: Let the set of these 1990 positive integers be $A$. Then $A$ has $2^{1990}-1$ non-empty subsets.
Next, find the product of all elements in each subset, and express the product in the form of the largest possible perfect square multiplied by some primes (for example, $M=$ $2^{6} \times 13^{10}=\left(2^{3} \times... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,837 |
Can a rectangular paper sheet of length $m$ and width $n$ be cut and reassembled into a square paper sheet of equal area (ignoring any loss, $m, n \in \mathbf{N}_{+}$)? If so, provide a method for cutting and reassembling; if not, explain the reason. | Let's solve it. Without loss of generality, assume $m>n$. One method of the front splicing is as follows.
First, cut out a square with side length $n$, then cut out another square with an integer side length from the remaining rectangle, and continue this process to obtain a series of squares (since $m, n \in \mathbf{N... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,838 |
297 Let $\alpha, \beta, \gamma$ be the angles formed by the body diagonal of a rectangular parallelepiped with the three edges meeting at the same vertex. Prove:
$$
\sqrt{\left(\cos ^{2} \alpha\right)^{1-\cos ^{2} \alpha} \cdot\left(\cos ^{2} \beta\right)^{1-\cos ^{2} \beta} \cdot\left(\cos ^{2} \gamma\right)^{1-\cos ^... | Proof Given the known conditions:
$$
\begin{array}{l}
\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1, \\
\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=2 . \\
\quad \text { Then } \sqrt{\left(\cos ^{2} \alpha\right)^{1-\cos ^{2} \alpha}\left(\cos ^{2} \beta\right)^{1-\cos ^{2} \beta}\left(\cos ^{2} \gamma\right... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,839 |
298 Proof: For any real number $x$, $\sin \sin \sin \sin x < \cos \cos \cos \cos x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove that $|\sin x \pm \cos x| \leqslant \sqrt{2}\cos \cos \cos x$
$\Rightarrow$ sinsinsin $x-\cos \cos \cos x$
$>$ sinsinsin $x-\operatorname{cossinsin} x>-\frac{\pi}{2}$
$\Rightarrow \operatorname{sinsinsin} x-\operatorname{cossinsin} x+\frac{\pi}{2}>0$.
Also, $\operatorname{sinsinsin} x-\cos \cos \cos x<1<\frac{\pi... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,840 |
Example 4 In three-dimensional Euclidean space, given any 37 integer points, no three of which are collinear. Prove: one can find three points such that the centroid of the triangle formed by them is also an integer point.
(19th IMO Shortlist) | Proof: Let $\left(x_{i}, y_{i}, z_{i}\right)(i=1,2,3)$ be three non-collinear integer points in space. The coordinates of the centroid $(x, y, z)$ of the triangle formed by these points are
$$
\begin{array}{l}
x=\frac{x_{1}+x_{2}+x_{3}}{3}, y=\frac{y_{1}+y_{2}+y_{3}}{3}, \\
z=\frac{z_{1}+z_{2}+z_{3}}{3} .
\end{array}
$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,841 |
Example 5 The sequence $\left\{G_{n}\right\}$ satisfies
$$
G_{0}=0, G_{1}=1, G_{n}=G_{n-1}+G_{n-2}+1(n \geqslant 2) \text {. }
$$
Prove: For any positive integer $m$, there exist two consecutive terms in $\left\{G_{n}\right\}$ that are both divisible by $m$. [3]
(2008, Estonian National Team Selection Exam) | Prove that if $G_{-1}=0$, then $G_{n}=G_{n-1}+G_{n-2}+1$ holds for $n=1$.
If $m=1$, the conclusion is obvious.
If $m=2$, then by the problem statement and definition, the parity of the sequence is
even, odd, even, even, odd, even, even, …
Thus, there are consecutive terms divisible by $m=2$.
If $m \geqslant 3$, conside... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,842 |
Example 6: Prove that in an arithmetic sequence composed of 40 different positive integers, there is at least one term that cannot be expressed in the form $2^{k}+3^{l}(k, l \in \mathbf{N})$. ${ }^{[4]}$
(2009, IMO China National Team Selection Exam) | Prove: Assume there exists a 40-term arithmetic sequence with all terms different and can be expressed in the form $2^{k}+3^{l}(k, l \in \mathbf{N})$.
Let this arithmetic sequence be $a, a+d, \cdots, a+39 d$, where $a, d \in \mathbf{N}_{+}$.
$$
\begin{array}{l}
\text { Let } a+39 d=2^{p}+3^{q}, \\
m=\left[\log _{2}(a+... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,843 |
Example 7 Let $T$ be the set of all divisors of $2004^{100}$, and $S$ be a subset of $T$ in which no number is a multiple of another. What is the maximum number of elements that $S$ can contain? ${ }^{[s]}$
$(2004$, Canadian Mathematical Olympiad) | Notice that $2004=2^{2} \times 3 \times 167$. Then
$$
T=\left\{2^{a} 3^{b} 167^{c} \mid 10 \leqslant a \leqslant 200, 0 \leqslant b, c \leqslant 100\right\} \text {. }
$$
Let $S=\left\{2^{200-b-c} 3^{b} 167^{c} \mid 0 \leqslant b, c \leqslant 100\right\}$.
For $0 \leqslant b, c \leqslant 100$, we have
$$
0 \leqslant 2... | 101^2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,844 |
1. From the numbers $1,2, \cdots, 2014$, what is the maximum number of numbers that can be selected such that none of the selected numbers is 19 times another? | According to the problem, if $k$ and $19 k$ cannot both appear in $1,2, \cdots, 2014$, and since $2014=19 \times 106$, and $106=5 \times 19+11$, then choose
$$
1,2,3,4,5,106,107, \cdots, 2013 \text {, }
$$
These 1913 numbers satisfy the requirement.
The numbers not chosen are $6,7, \cdots, 105,2014$, a total of 101 nu... | 1913 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,845 |
2. Let $T \subseteq\{1,2, \cdots, 25\}$. If for any two distinct elements $a, b (a \neq b)$ in $T$, their product $ab$ is not a perfect square, find the maximum number of elements in $T$, and the number of subsets $T$ that satisfy this condition. | Prompt: Divide the set $\{1,2, \cdots, 25\}$ into several subsets such that the product of any two elements in the same subset is a perfect square, and the product of any two elements in different subsets is not a perfect square.
To maximize the number of elements in $T$, let
$$
\begin{array}{l}
A_{1}=\{1,4,9,16,25\}, ... | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,846 |
3. Let $p$ (where $p$ is a prime number) prime numbers $a_{1}, a_{2}, \cdots, a_{p}$ form an arithmetic sequence with a common difference of $d(d>0)$, and $a_{1}>p$. Prove: $p \mid d$.
| Because $a_{1}>p$, and $p$ is a prime number, so each $a_{i}(i=1,2, \cdots, p)$ cannot be divisible by $p$.
Since the remainders when $p$ numbers $a_{1}, a_{2}, \cdots, a_{p}$ are divided by $p$ can only be one of the $p-1$ numbers $1,2, \cdots, p-1$, by the pigeonhole principle, at least two of these numbers have the... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,847 |
4. Find the smallest positive integer $n$ such that for any $n$ integers, there exist at least two numbers whose sum or difference is divisible by 1991.
(1991, Australian Mathematical Olympiad) | Let $M=\left\{a_{i} \mid a_{i}=0,1, \cdots, 995\right\}$.
Since $a_{i}+a_{j} \leqslant 995+994=1989<1991$, $0<\left|a_{i}-a_{j}\right| \leqslant 995$,
then the sum and difference of any two numbers in $M$ are not multiples of 1991.
Therefore, $n \geqslant 997$.
Let $a_{1}, a_{2}, \cdots, a_{997}$ be any 997 integers.
I... | 997 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,848 |
Example 3 As shown in Figure 5, given that the radius of the circumcircle $\odot 0$ of quadrilateral $A B C D$ is
2, the diagonals $A C$ and $B D$
intersect at point $E, A E=E C$,
$A B=\sqrt{2} A E$, and $B D=$
$2 \sqrt{3}$. Find the area of quadrilateral $A B C D$.
(2000, National Junior High School Mathematics Compet... | 【Analysis】This problem essentially involves the combination of figures 6 and 7.
Solution From figure 6 and the problem statement, we have
$$
\begin{array}{l}
A B^{2}=A E \cdot A C \Rightarrow \triangle A B E \backsim \triangle A C B \\
\Rightarrow \angle A B E=\angle A C B \Rightarrow A B=A D .
\end{array}
$$
From fig... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,849 |
5. Let $a_{1}, a_{2}, \cdots$ be any infinite sequence of positive integers with the property $a_{k}<a_{k+1}(k \geqslant 1)$. Prove that infinitely many $a_{m}$ in this sequence can be expressed as
$$
a_{m}=x a_{p}+y a_{q}(p \neq q) .
$$ | Divide the sequence $\left\{a_{n}\right\}$ into several subsequences modulo $a_{2}$.
Since the sequence $\left\{a_{n}\right\}$ is an infinite sequence, and the number of subsequences is finite (no more than $a_{2}$), by the pigeonhole principle, at least one subsequence must have infinitely many terms.
Next, consider... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,850 |
Example 1 Find all non-right triangles $ABC$ such that
$$
\begin{array}{l}
\tan A \cdot \tan B \cdot \tan C \\
\leqslant[\tan A]+[\tan B]+[\tan C],
\end{array}
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. (2009, Nanjing University Independent Admission Examination) | Given that in a non-right triangle $ABC$, we have
$$
\tan A \cdot \tan B \cdot \tan C
$$
$$
= \tan A + \tan B + \tan C.
$$
Therefore,
$$
\tan A + \tan B + \tan C
$$
$$
\leq [\tan A] + [\tan B] + [\tan C].
$$
Thus, $\tan A, \tan B, \tan C$ are all integers, and at least two of them are positive integers. Without loss ... | 1, 2, 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,851 |
Example 2 Find a set $S$ composed of positive integers with at least two elements, such that the sum of all elements in $S$ equals the product of all elements in $S$.
(2006, Tsinghua University Independent Admission Examination) | If $S$ contains only two elements, let them be $a, b$, then
$$
\begin{array}{l}
a b=a+b \\
\Rightarrow(a-1)(b-1)=1 .
\end{array}
$$
Also, $a-1 \geqslant 0, b-1 \geqslant 0$, so
$$
\begin{array}{l}
a-1=b-1=1 \\
\Rightarrow a=b=2 .
\end{array}
$$
Contradiction.
If $S$ contains three elements, by the conclusion of Examp... | S=\{1,2,3\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,852 |
Example 3 Find the units digit of $\left(7^{2004}+36\right)^{818}$.
$(2004$, Shanghai Jiao Tong University Independent Recruitment Examination) | $$
\begin{array}{l}
\text { Solution: Since } 7^{4} \equiv 1(\bmod 10) \\
\Rightarrow 7^{2004} \equiv 1(\bmod 10) \\
\Rightarrow 7^{2004}+36 \equiv 7(\bmod 10) .
\end{array}
$$
$$
\text { Therefore, the original expression } \equiv 7^{818} \equiv 7^{2} \equiv 9(\bmod 10) \text {. }
$$
Thus, the unit digit is 9. | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,853 |
Example 4 Write down all three-term arithmetic sequences of prime numbers with a common difference of 8.
(2009, Tsinghua University Independent Admission Examination) | Let these three numbers be $a$, $a+8$, $a+16$.
Since the remainders when $a$, $a+8$, $a+16$ are divided by 3 are all different, one of these three numbers must be 3. And since $a+8$, $a+16$ are prime numbers greater than 3, it follows that $a=3$.
Therefore, the sequence of numbers is $3,11,19$. | 3,11,19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,854 |
Example 5 When $p$ and $q$ are both odd numbers, does the equation
$$
x^{2}+2 p x+2 q=0
$$
have rational roots? Prove it.
(2009, Tsinghua University Independent Admissions Exam) | No rational roots.
Proof as follows: Suppose there is a rational root, let it be $\frac{r}{s}$ (where $r, s$ are coprime integers). Then
$$
\frac{r^{2}}{s^{2}}+2 p \cdot \frac{r}{s}+2 q=0,
$$
which means $r^{2}+2 p r s+2 q s^{2}=0$.
If $r$ is odd, then the left side of equation (1) is odd, while the right side is even... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,855 |
Example 6 If the three roots of $x^{3}+a x^{2}+b x+c=0$ are $a, b, c$, and $a, b, c$ are rational numbers not all zero. Find $a, b, c$.
(2005, Shanghai Jiao Tong University Independent Recruitment Examination) | Solve by Vieta's formulas:
$$
\left\{\begin{array}{l}
a+b+c=-a, \\
a b+b c+c a=b, \\
a b c=-c .
\end{array}\right.
$$
From equation (1), we know $c=0$ or $a b=-1$.
(1) If $c=0$, then
$$
\left\{\begin{array}{l}
2 a + b = 0, \\
a b = b
\end{array} \Rightarrow \left\{\begin{array}{l}
a=1, \\
b=-2 .
\end{array}\right.\rig... | (1,-2,0),(1,-1,-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,856 |
Example 7 Does there exist a real number $x$, such that $\tan x + \sqrt{3}$ and $\cot x + \sqrt{3}$ are both rational numbers?
(2009, Peking University Independent Admission Examination) | No solution exists.
Assume they are all rational numbers, let
$$
\tan x+\sqrt{3}=a, \cot x+\sqrt{3}=b,
$$
where $a, b$ are both rational numbers. Then
$$
(a-\sqrt{3})(b-\sqrt{3})=\tan x \cdot \cot x=1,
$$
i.e., $\sqrt{3}(a+b)=a b+2$.
If $a+b=0$, then from equation (1) we get $a= \pm \sqrt{2}$, which is a contradictio... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,857 |
Example 8 Let $a, b, c$ be rational numbers, and $\sqrt{a}+\sqrt{b}+$ $\sqrt{c}$ is also a rational number. Prove: $\sqrt{a}, \sqrt{b}, \sqrt{c}$ are all rational numbers. (2008, Tsinghua University Independent Admission Examination) | Prove that if $a, b, c$ have at least two zeros, then $\sqrt{a}, \sqrt{b}, \sqrt{c}$ are all rational numbers.
If $a, b, c$ have at most one zero, let $\sqrt{a} + \sqrt{b} + \sqrt{c} = p$ ($p$ is a rational number). Then
$$
\begin{array}{l}
\sqrt{a} + \sqrt{b} = p - \sqrt{c} \\
\Rightarrow a + b + 2 \sqrt{a b} = p^{2}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,858 |
Example 9 Let the set $A=\left\{n+n! \mid n \in \mathbf{N}_{+}\right\}, B$ be the complement of $A$ in the set of positive integers.
(1) Prove that it is impossible to obtain an arithmetic sequence with infinitely many terms and a common difference not equal to 0 in the set $B$;
(2) Can an infinite geometric sequence b... | (1) Proof Assume we can find an arithmetic sequence with a non-zero common difference in set $B$, let the first term of this sequence be $a_{1}$, and the common difference be $d\left(a_{1} 、 d \in \mathbf{N}_{+}\right)$. Then let
$$
\begin{array}{l}
m=\frac{\left(a_{1}+d\right)!}{d}+2 . \\
\text { Hence } a_{m}=a_{1}+(... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,859 |
Example 4 As shown in Figure 8, given that $AB$ is the diameter of $\odot O$, $BC \perp AB$, $OC$ is parallel to chord $AD$, a line $DE \perp AB$ is drawn through point $D$ at point $E$, connect $AC$, intersecting $DE$ at point $P$. Question: Are $EP$ and $PD$ equal? Prove your conclusion.
(2003, National Junior High S... | 【Analysis】It is obvious that the basic figure of Figure 3 is hidden in this problem. Therefore, we can extend $AD$ and $BC$ to intersect at point $F$. We have the conclusion: $\frac{EP}{BC}=\frac{PD}{CF}$. We only need to prove that $BC=CF$. From $OC \parallel AF$ and $OB=OA$, we can derive that $BC=CF$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,860 |
Let $a, b, c \in \mathbf{R}_{+}$, and $a+b+c=1$. Prove:
$$
\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\left(c+\frac{1}{c}\right) \geqslant \frac{1000}{27} .
$$
(2008, Nanjing University Independent Admission Examination) | The proofs I have seen are mostly quite complicated and not easy to ponder. Below is a generalization of the problem, which leads to a simple proof of equation (1).
Proposition If $a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers satisfying $\sum_{i=1}^{n} a_{i}=S$ and $\alpha>0, \lambda \geqslant\left(\frac{S}{n}\ri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,861 |
Let $u$ be a root of the equation
$$
x^{3}-3 x+10=0
$$
Let $f(x)$ be a quadratic polynomial with rational coefficients, and
$$
\alpha=\frac{1}{2}\left(u^{2}+u-2\right), f(\alpha)=u .
$$
Find $f(0)$.
(2010, Five Schools Joint Examination for Independent Enrollment) | From the given, we have $u^{3}-3 u+10=0$. Then
$$
\begin{aligned}
\alpha^{2} &=\frac{1}{4}\left(u^{2}+u-2\right)^{2} \\
&=\frac{1}{4}\left(u^{4}+2 u^{3}-3 u^{2}-4 u+4\right) \\
&=\frac{1}{4}\left[u(3 u-10)+2(3 u-10)-3 u^{2}-4 u+4\right] \\
&=-4-2 u .
\end{aligned}
$$
Let $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{Q}, a \... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,862 |
Example 1 Let $\left\{a_{1}, a_{2}, \cdots, a_{\varphi(n)}\right\}$ be a reduced residue system modulo $n$ (i.e., each number is coprime with $n$, and distinct modulo $n$; $\varphi(n)$ is the Euler's function, representing the number of positive integers less than or equal to $n$ and coprime with $n$), and let $N$ deno... | To prove that because when $x$ is a solution to the equation, $-x$ is also a solution, $N$ is even.
Therefore, we can find $\frac{N}{2}$ distinct solutions $a_{1}, a_{2}, \cdots, a_{\frac{N}{2}}$, such that $-a_{1}, -a_{2}, \cdots, -a_{\frac{N}{2}}$ are also solutions to the equation, and these $N$ solutions are all d... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,863 |
Example 2 Let $p=3n+1\left(n \in \mathbf{N}_{+}\right)$ be a prime number. Prove: $\sum_{i=1}^{2n} \mathrm{C}_{p}^{i}$ is divisible by $p^{2}$.
(Adapted from the 57th Putnam Mathematical Competition) | $$
\begin{array}{l}
\frac{1}{p} \mathrm{C}_{p}^{i}=\frac{1}{i} \prod_{k=1}^{i-1} \frac{p-k}{k} \\
\equiv(-1)^{i-1} \frac{1}{i}(\bmod p) . \\
\text { Therefore } \frac{1}{p} \sum_{i=1}^{2 n} \mathrm{C}_{p}^{i} \equiv \sum_{i=1}^{2 n}(-1)^{i-1} \frac{1}{i} \\
\equiv \sum_{i=1}^{2 n} \frac{1}{i}-2 \sum_{i=1}^{n} \frac{1}{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,864 |
Find all functions \( f \) from the set of positive integers to the set of positive integers that satisfy the following condition: for all positive integers \( a \) and \( b \), there exists a non-degenerate triangle with side lengths
\[
a, f(b), f(b + f(a) - 1)
\]
(A non-degenerate triangle is one whose three vertice... | Given the conditions
$$
a, f(b), f(b+f(a)-1)
$$
are positive integers, according to the triangle inequality and the discrete nature of integers, we have
$$
\begin{array}{l}
a + f(b) \geqslant f(b + f(a) - 1) + 1, \\
a + f(b + f(a) - 1) \geqslant f(b) + 1, \\
f(b) + f(b + f(a) - 1) \geqslant a + 1,
\end{array}
$$
wher... | f(n) = n | Number Theory | proof | Yes | Yes | cn_contest | false | 723,866 |
Given real numbers $x, y, z$ satisfy
$$
x y z=32, x+y+z=4 \text {. }
$$
Then the minimum value of $|x|+|y|+|z|$ is $\qquad$ .
(2010, Hubei Province High School Mathematics Competition) | Answer: 12.
The text above has been translated into English, maintaining the original text's line breaks and format. Here is the direct output of the translation result. | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,867 |
Extended Question Given real numbers $x, y \sim z$ satisfy
$x y z=m, x+y+z=n$, constants $m, n>0$.
If $m>\frac{1}{27} n^{3}$, then the minimum value of $|x|+|y|+|z|$ is $2 \xi-n$, where $\xi (\xi>n)$ is the only positive real root of the equation
$$
z^{3}-2 n z^{2}+n^{2} z-4 m=0
$$
If $m \leqslant \frac{1}{27} n^{3}$,... | Proof: It is evident that $x, y, z$ cannot all be negative. Without loss of generality, assume $z>0$. Then $x, y$ have the same sign, and are the roots of the equation
$$
t^{2}-(n-z) t+\frac{m}{z}=0
$$
Thus,
$$
\begin{array}{l}
\text { Hence } \Delta=(n-z)^{2}-\frac{4 m}{z} \geqslant 0(z>0) \\
\Leftrightarrow f(z)=z(n... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,868 |
1. The sides of a triangle are all integers, and the perimeter is 9. The number of non-congruent triangles is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | - 1. C.
$$
9=4+4+1=4+3+2=3+3+3 \text {. }
$$
- 1. C.
$$
9=4+4+1=4+3+2=3+3+3 \text {. }
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,869 |
2. Given $a=\frac{1}{2+\sqrt{3}}$. Then the value of $\sqrt{a^{2}-2 a+1}$ is ( ).
(A) $\sqrt{3}-1$
(B) $1-\sqrt{3}$
(C) $\frac{1}{1+\sqrt{3}}$
(D) $\frac{1}{1-\sqrt{3}}$ | 2. A.
Notice that $a=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}<1$.
Then $\sqrt{a^{2}-2 a+1}=\sqrt{(a-1)^{2}}$ $=1-a=\sqrt{3}-1$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,870 |
Example 5 As shown in Figure 9, let the diagonals $AC$ and $BD$ of convex quadrilateral $ABCD$ intersect at point $M$. Draw a line through $M$ parallel to $AD$, intersecting $AB$ and $CD$ at points $E$ and $F$ respectively, and intersecting the extension of $BC$ at point $O$. $P$ is a point on the circle with center $O... | 【Analysis】It is easy to think of proving $\angle O P F=\angle O E P$, it is only necessary to prove $\triangle O F P \backsim \triangle O P E$, that is, it is necessary to prove $\frac{O F}{O P}=\frac{O P}{O E}$.
And $O P=O M$, then it is necessary to prove $\frac{O F}{O M}=\frac{O M}{O E}$.
Noticing $O M / / A D$, fro... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,871 |
3. Let $a=\sqrt{2}-1, b=3-2 \sqrt{2}, c=\sqrt{3}-\sqrt{2}$. Then the size relationship of $a, b, c$ is ( ).
(A) $a>b>c$
(B) $c>b>a$
(C) $c>a>b$
(D) $a>c>b$ | 3. D.
Notice that $a-c=2 \sqrt{2}-(\sqrt{3}+1)$.
And $(2 \sqrt{2})^{2}=8>(\sqrt{3}+1)^{2}=4+2 \sqrt{3}$, so $a>c$.
Also, $c-b=\sqrt{2}+\sqrt{3}-3$, and $(\sqrt{2}+\sqrt{3})^{2}=5+2 \sqrt{6}>3^{2}=9$, so $c>b$.
Therefore, $a>c>b$. | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 723,872 |
4. Let real numbers $x, y, z$ satisfy
$$
x+\frac{1}{y}=1, y+\frac{1}{z}=1 \text {. }
$$
Then the value of $x y z$ is ( ).
(A) 1
(B) 2
(C) -1
(D) -2 | 4. C.
From $x+\frac{1}{y}=1$, we get $x y=y-1$; from $y+\frac{1}{z}=1$, we get $y z=z-1$.
Multiplying the two equations, we get
$$
\begin{array}{l}
x y z=\frac{y z-(y+z)+1}{y} \\
=z-1+\frac{1-z}{y}=z-1+\frac{-y z}{y}=-1 .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,873 |
5. As shown in Figure 1, in $\triangle A B C$, it is given that $A B=A C$, $\angle A B C=40^{\circ}$, $B D$ is the angle bisector of $\angle A B C$, and $B D$ is extended to point $E$ such that $D E=A D$. Then the degree measure of $\angle E C A$ is ( ).
(A) $30^{\circ}$
(B) $40^{\circ}$
(C) $50^{\circ}$
(D) $60^{\circ... | 5. B.
Take a point $F$ on side $B C$ such that $B F=A B$. It is easy to prove that
$$
\triangle A B D \cong \triangle F B D, \triangle E C D \cong \triangle F C D .
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,874 |
6. As shown in Figure 2, in rectangle $A B C D$, it is known that the diagonal length is 2, and $\angle 1=\angle 2=\angle 3=\angle 4$. Then the perimeter of quadrilateral $E F G H$ is ( ).
(A) $2 \sqrt{2}$
(B) 4
(C) $4 \sqrt{2}$
(D) 6 | 6. B.
According to the reflection relationship, Figure 6 can be drawn. The hypotenuse of $\mathrm{Rt} \triangle I J K$ is the perimeter of quadrilateral $E F G H$.
The legs of the right triangle are twice the lengths of the rectangle's sides, and since the diagonal of the rectangle forms a right triangle with the two... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,875 |
1. Given positive integers $a, b$ satisfy
$$
|b-2|+b-2=0,|a-b|+a-b=0,
$$
and $a \neq b$. Then the value of $a b$ is $\qquad$ . | 2.1.2.
From the given conditions, we know that $a>0, b-2 \leqslant 0, a-b \leqslant 0$. Therefore, $a<b \leqslant 2$. Hence, $a=1, b=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,876 |
2. If $\sqrt{x}-\frac{1}{\sqrt{x}}=1$, then the value of $x^{2}-3 x+1$ is
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. | 2.0. From $\sqrt{x}-\frac{1}{\sqrt{x}}=1$, we get $x+\frac{1}{x}=3$. Then $x^{2}=3 x-1 \Rightarrow x^{2}-3 x+1=0$. | null | Algebra | proof | Yes | Yes | cn_contest | false | 723,877 |
3. Given the side lengths of a trapezoid are $3,4,5,6$. Then the area of this trapezoid is $\qquad$ . | 3. 18.
First, determine the lengths of the two bases. As shown in Figure 7, let trapezoid $ABCD$ have $AD$ and $BC$ as the upper and lower bases, respectively. Draw $AE \parallel CD$. Then $BE$ is the difference between the upper and lower bases.
In $\triangle ABE$, $AB - AE = AB - CD < BE = BC - AD$. Therefore, $AB ... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,878 |
4. As shown in Figure 3, in $\triangle ABC$, it is given that $D$ is a point on side $BC$ such that $AD = AC$, and $E$ is the midpoint of side $AD$ such that $\angle BAD = \angle ACE$. If $S_{\triangle BDE} = 1$, then $S_{\triangle ABC}$ is $\qquad$. | 4.4.
Notice that
$$
\begin{array}{l}
\angle E C D=\angle A C D-\angle A C E \\
=\angle A D C-\angle B A D=\angle A B C .
\end{array}
$$
Therefore, $\triangle E C D \backsim \triangle A B C$.
$$
\text { Then } \frac{C D}{B C}=\frac{E D}{A C}=\frac{E D}{A D}=\frac{1}{2} \text {. }
$$
Thus, $D$ is the midpoint of side ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,879 |
Three, (20 points) For a certain project, Team A alone needs 12 days to complete, and Team B alone needs 9 days to complete. If the two teams are scheduled to work in whole days, how many schemes are there to ensure that the project is completed within 8 days? | Three, let teams A and B work for $x, y$ days respectively to just meet the requirements. Then we have
$$
\left\{\begin{array}{l}
\frac{x}{12}+\frac{y}{9}=1, \\
0 \leqslant x \leqslant 8, \\
0 \leqslant y \leqslant 8,
\end{array}\right.
$$
and $x, y$ are both integers.
From $\frac{x}{12}+\frac{y}{9}=1$, we get $x=12-y... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,880 |
Four. (25 points)
As shown in Figure 4, in the Cartesian coordinate system, points $A$ and $B$ are two points on the graph of a certain linear function, satisfying $\angle A O B=90^{\circ}$, and $A O=B O=2$. If the angle between $A O$ and the $y$-axis is $60^{\circ}$, find this linear function. | Four, draw perpendiculars from points $A$ and $B$ to the $x$-axis, intersecting the $x$-axis at points $M$ and $N$ respectively. Then
$$
\angle A O M=30^{\circ}, \angle B O N=60^{\circ} \text {. }
$$
Therefore, $A M=\frac{1}{2} A O=1$,
$$
O M=\sqrt{A O^{2}-A M^{2}}=\sqrt{3} \text {. }
$$
Similarly, $B N=\sqrt{3}, O N... | y=(2-\sqrt{3}) x+2 \sqrt{3}-2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,881 |
Example 6 Let $a, b, c$ be the lengths of the sides of $\triangle ABC$, and $\frac{a}{b}=\frac{a+b}{a+b+c}$. Then the relationship between the interior angles $\angle A, \angle B$ is ( ).
(A) $\angle B>2 \angle A$
(B) $\angle B=2 \angle A$
(C) $\angle B<2 \angle A$
(D) Uncertain
(2000, National Junior High School Mathe... | The given proportional equation can be simplified as
$$
b^{2}=a(a+c) \Rightarrow \frac{a}{b}=\frac{b}{a+c} \text {. }
$$
Therefore, as shown in Figure 11, we can
construct similar triangles with $a, b$ and $b, a+c$
as corresponding sides. It is easy to see that
$$
\angle B=2 \angle A \text {. }
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,882 |
Five. (25 points) As shown in Figure 5, in the right triangle $\triangle ABC$, it is known that $\angle C=90^{\circ}$, $AM$ and $AN$ are the median and the angle bisector of $\angle BAC$ on side $BC$, respectively. A perpendicular line $CD$ is drawn from point $C$ to $AN$ at point $D$. Prove: (1) $DM=\frac{1}{2}(AB-AC)... | (1) As shown in Figure 8, extend $C D$ to intersect $A B$ at point $C^{\prime}$.
$$
\begin{array}{l}
\text { Since } \angle C^{\prime} A D \\
=\angle C A D \\
\Rightarrow C D=D C^{\prime} .
\end{array}
$$
Since $\angle A D C^{\prime}=\angle A D C=90^{\circ} \Rightarrow A C^{\prime}=A C$. Because $A M$ is the median on... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,883 |
1. Given an integer $n(n \geqslant 3)$. Find the largest real number $M$, such that for any sequence of positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$, there exists a permutation $y_{1}, y_{2}, \cdots, y_{n}$, satisfying
$$
\sum_{i=1}^{n} \frac{y_{i}^{2}}{y_{i+1}^{2}-y_{i+1} y_{i+2}+y_{i+2}^{2}} \geqslant M,
$$
w... | 1. Let
$$
\begin{array}{l}
F\left(x_{1}, x_{2}, \cdots, x_{n}\right) \\
=\sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i+1}^{2}-x_{i+1} x_{i+2}+x_{i+2}^{2}} .
\end{array}
$$
First, take $x_{1}=x_{2}=\cdots=x_{n-1}=1, x_{n}=\varepsilon$. In this case, all permutations are the same under cyclic permutation, then
$$
\begin{array}{l... | n-1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,884 |
2. Given $n(n>1)$ is an integer, $k$ is the number of distinct prime factors of $n$. Prove: There exists an integer $a$ $\left(1<a<\frac{n}{k}+1\right)$, such that $n \mid\left(a^{2}-a\right)$. | Due to $p_{1}^{\alpha_{k}}, p_{2}^{\alpha_{k}}, \cdots, p_{k}^{\alpha_{k}}$ being pairwise coprime, by the Chinese Remainder Theorem, for every $i(1 \leqslant i \leqslant k)$, the system of congruences
$$
\left\{\begin{array}{l}
x \equiv 1\left(\bmod p_{i}^{\alpha_{i}}\right), \\
x \equiv 0\left(\bmod p_{j}^{\alpha_{j}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,885 |
3. Let the number of vertices of a simple graph $G$ be $3 n^{2}(n \geqslant 2$, $n \in \mathbf{Z})$. It is known that the degree of each vertex in $G$ does not exceed $4 n$, at least one vertex has a degree of 1, and there is a path of length no more than 3 between any two different vertices. Prove: The minimum number ... | 3. For any two different vertices $u, v$, if the length of the shortest path between them is $k$, then the distance between them is called $k$.
Consider graph $G$, whose vertex set is
$\left\{x_{1}, x_{2}, \cdots, x_{3 n^{2}-n}, y_{1}, y_{2}, \cdots, y_{n}\right\}$,
where, $y_{i}$ and $y_{j}(1 \leqslant i0$ (by proper... | \frac{7}{2} n^{2}-\frac{3}{2} n | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,886 |
4. As shown in Figure 1, let $H$ be the orthocenter of acute $\triangle A B C$, and $P$ be a point on the circumcircle arc $\overparen{B C}$. Connect $P H$ to intersect arc $\overparen{A C}$ at point $M$. Let point $K$ be on arc $\overparen{A B}$ such that line $K M$ is parallel to the Simson line of point $P$ with res... | 4. Prove: $J K=J M$.
As shown in Figure 2, draw a perpendicular from point $P$ to $B C$, intersecting the circumcircle at point $S$ and $B C$ at point $L$. Let the projection of $P$ on $A B$ be $N$, and connect $A S$, $N L$, $N P$, and $B P$.
Since $B$, $P$, $L$, and $N$ are concyclic,
$\angle S L N=\angle N B P=\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,887 |
5. Let $a_{1}, a_{2}, \cdots$ be a permutation of all positive integers. Prove: There exist infinitely many positive integers $i$, such that
$$
\left(a_{i}, a_{i+1}\right) \leqslant \frac{3}{4} i .
$$ | 5. Assume the conclusion does not hold. Then there exists $i_{0}$, such that when $i \geqslant i_{0}$, we have
$$
\left(a_{i}, a_{i+1}\right)>\frac{3}{4} i .
$$
Fix a positive integer $M\left(M>i_{0}\right)$.
When $i \geqslant 4 M$, we have
$$
\left(a_{i}, a_{i+1}\right)>\frac{3}{4} i \geqslant 3 M \text {. }
$$
Thus... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,888 |
3. As shown in Figure 16, quadrilateral $ABCD$ is a square with side length $a$. A circular arc centered at $D$ with radius $DA$ intersects a semicircle with diameter $BC$ at another point $P$. Extending $AP$ intersects $BC$ at point $N$. Then $\frac{BN}{NC}=$ $\qquad$
(2004, National Junior High School Mathematics Lea... | Hint: The problem involves the basic figures of overlapping and hidden intersecting circles, common tangents, similarity, and projections. Extend $C P$ and $D A$ to meet at point $F$, and connect $B P$. Answer: $\frac{1}{2}$.
| \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,890 |
4. As shown in Figure 17, if $P A = P B, \angle A P B = 2 \angle A C B, A C$ intersects $P B$ at point $D$, and $P B = 4$, $P D = 3$, then $A D \cdot D C$ equals ( )
(A) 6
(B) 7
(C) 12
(D) 16
(2001, National Junior High School Mathematics Competition) | From the problem, we can construct a circle with $P$ as the center and $P A$ as the radius. Let $B P$ intersect the circle at point $E$. Then, by the intersecting chords theorem, we have
$$
A D \cdot D C=B D \cdot D E=7 \text{. }
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,891 |
1. Let the polynomial $f(x)$ satisfy: for any $x \in \mathbf{R}$, we have
$$
f(x+1)+f(x-1)=2 x^{2}-4 x .
$$
Then the minimum value of $f(x)$ is $\qquad$ | - 1. - -2 .
Since $f(x)$ is a polynomial, we know that $f(x+1)$ and $f(x-1)$ have the same degree as $f(x)$. Therefore, $f(x)$ is a quadratic polynomial (let's assume $f(x) = ax^2 + bx + c$). Then,
$$
\begin{array}{l}
f(x+1) + f(x-1) \\
= 2ax^2 + 2bx + 2(a + c) = 2x^2 - 4x.
\end{array}
$$
Thus, $a = 1, b = -2, c = -1... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,892 |
Example 7 As shown in Figure 12, given that $P$ is a point outside $\odot O$, $PS$ and $PT$ are two tangents to $\odot O$. A secant line $PAB$ through point $P$ intersects $\odot O$ at points $A$ and $B$, and intersects $ST$ at point $C$. Prove:
$$
\frac{1}{PC}=\frac{1}{2}\left(\frac{1}{PA}+\frac{1}{PB}\right) \text {.... | 【Analysis】First consider a special case, where the line $PAB$ passes through the center of the circle (as shown in Figure 13), then we have
$$
\begin{array}{l}
\frac{1}{P A}+\frac{1}{P B}=\frac{P A+P B}{P A \cdot P B} . \\
\text { and } P A \cdot P B=P S^{2}=P C \cdot P O, \\
P A+P B=P A+P O+O B \\
=P A+A O+P O=2 P O,
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,893 |
2. Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
$$
a_{k} b_{k}=1(k=1,2, \cdots) \text {, }
$$
the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is $A_{n}=\frac{n}{n+1}$. Then the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ $B_{n}=$ . $\qquad$ | 2. $\frac{n(n+1)(n+2)}{3}$.
Notice that
$$
\begin{array}{l}
a_{1}=A_{1}=\frac{1}{2}, \\
a_{n}=A_{n}-A_{n-1}=\frac{1}{n(n+1)}, \\
b_{n}=\frac{1}{a_{n}}=n(n+1) .
\end{array}
$$
Thus, $B_{n}=\sum_{k=1}^{n}\left(k^{2}+k\right)=\frac{n(n+1)(n+2)}{3}$. | \frac{n(n+1)(n+2)}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,894 |
3. The range of the function $f(x)=\frac{\sqrt{1-x^{2}}}{x+2}$ is | 3. $\left[0, \frac{\sqrt{3}}{3}\right]$.
Let $y=\sqrt{1-x^{2}}(-1 \leqslant x \leqslant 1)$. Then the image of point $P(x, y)$ is the part of the unit circle above the $x$-axis (as shown in Figure 3).
Rewrite the function as
$$
\begin{array}{l}
f(x) \\
=\frac{y-0}{x-(-2)},
\end{array}
$$
which represents the slope of... | \left[0, \frac{\sqrt{3}}{3}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,895 |
4. A line $l$ with a slope of 2 is drawn through the focus $F$ of the parabola $y^{2}=8 x$. If $l$ intersects the parabola at points $A$ and $B$, then the area of $\triangle O A B$ is $\qquad$ | $4.4 \sqrt{5}$.
The focus of the parabola is $F(2,0), l_{A B}: y=2 x-4$.
Substituting into the parabola equation yields
$$
y^{2}-4 y-16=0 \text {. }
$$
Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$. Then,
$$
\begin{array}{l}
y_{1}+y_{2}=4, y_{1} y_{2}=-16 . \\
\text { Thus }\left(y_{1}-y_{2}\right)^{2}... | 4 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,896 |
5. Given that $\angle A, \angle B$ are acute angles, satisfying
$$
\frac{\sin A}{\sin B}=\cos (A+B) \text{. }
$$
Then the maximum value of $\tan A$ is | 5. $\frac{\sqrt{2}}{4}$.
$$
\begin{array}{l}
\text { Given } \sin [(A+B)-B]=\cos (A+B) \cdot \sin B \\
\Rightarrow \sin (A+B) \cdot \cos B=2 \cos (A+B) \cdot \sin B \\
\Rightarrow \tan (A+B)=2 \tan B .
\end{array}
$$
Thus $\tan A=\tan [(A+B)-B]$
$$
\begin{array}{l}
=\frac{\tan (A+B)-\tan B}{1+\tan (A+B) \cdot \tan B} ... | \frac{\sqrt{2}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,897 |
6. Let the inradius of a regular tetrahedron be 1. Then the minimum value of its volume is $\qquad$ .
| $6.8 \sqrt{3}$.
Let the center of the base of the regular tetrahedron $P-ABC$ be $H$. Then the center of the inscribed sphere $O$ lies on $PH$.
If $AH$ intersects $BC$ at point $D$, then $PD \perp BC$, and the point of tangency $E$ of the sphere $O$ with the plane $PBC$ lies on $PD$.
Considering the section
$\triangle... | 8 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,898 |
8. Divide the elements of the set $M=\{1,2, \cdots, 12\}$ into three disjoint subsets: $M=A \cup B \cup C$, where,
\[
\begin{aligned}
A & =\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}, \\
B & =\left\{b_{1}, b_{2}, b_{3}, b_{4}\right\}, \\
C & =\left\{c_{1}, c_{2}, c_{3}, c_{4}\right\}\left(c_{1}<c_{2}<c_{3}<c_{4}\right),
... | 8. $\{8,9,10,12\},\{7,9,11,12\},\{6,10,11,12\}$.
From $\sum_{i=1}^{4} c_{i}=\sum_{i=1}^{4} a_{i}+\sum_{i=1}^{4} b_{i}$, we get
$2 \sum_{i=1}^{4} c_{i}=\sum_{i=1}^{4} a_{i}+\sum_{i=1}^{4} b_{i}+\sum_{i=1}^{4} c_{i}$
$=1+2+\cdots+12=78$.
Therefore, $\sum_{i=1}^{4} c_{i}=39$.
Without considering the pairing scenarios, le... | \{8,9,10,12\},\{7,9,11,12\},\{6,10,11,12\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,900 |
9. (20 points) As shown in Figure 2, a chord $AB$ of circle $\odot O$ divides the circle into two parts. $M$ and $N$ are the midpoints of the two arcs, respectively. With point $B$ as the center of rotation, the segment $AMB$ is rotated clockwise by an angle to become segment $A_1MB$. If the midpoints of $AA_1$ and $MN... | 9. As shown in Figure 10, let the midpoints of $A B$ and $A_{1} B$ be $E$ and $F$ respectively.
Since in $\odot O$, points $E$ and $F$ coincide and are the intersection of diameter $M N$ and $A B$, we have:
$$
\begin{array}{c}
M F \cdot E N=A E \cdot E B \\
=P E \cdot P F . \\
\text { Therefore, } \frac{M F}{P F}=\fra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,901 |
10. (25 points) Given
Ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$,
$\odot O: x^{2}+y^{2}=b^{2}$.
From any point $P$ on the ellipse, different from its vertices, draw two tangents to $\odot O$, with the points of tangency being $M$ and $N$. If the intercepts of line $MN$ on the $x$-axis and $y$-axis a... | 10. As shown in Figure 11, let \( P\left(x_{0}, y_{0}\right) \).
Since \( M \) and \( N \) are the points of tangency of \( \odot O \),
\( O M \perp M P \),
\( O N \perp N P \).
Therefore, \( O, M, P, N \) are concyclic. The diameter of this circle is \( O P \), so the center is \( P_{0}\left(\frac{x_{0}}{2}, \frac{y_... | \frac{a^{2}}{n^{2}}+\frac{b^{2}}{m^{2}}=\frac{a^{2}}{b^{2}} | Geometry | proof | Yes | Yes | cn_contest | false | 723,902 |
11. (25 points) For a set $M=\left\{p_{1}, p_{2}, \cdots, p_{2_{n}}\right\}$ consisting of $2 n$ prime numbers, its elements can be paired to form $n$ products, resulting in an $n$-element set. If
$$
\begin{aligned}
A & =\left\{a_{1} a_{2}, a_{3} a_{4}, \cdots, a_{2 n-1} a_{2 n}\right\} \\
\text { and } \quad B & =\lef... | 11. Six elements can form fifteen different "slips of paper," listed as follows:
$$
\begin{array}{l}
\{a b, c d, e f\},\{a b, c e, d f\},\{a b, c f, d e\}, \\
\{a c, b d, e f\},\{a c, b e, d f\},\{a c, b f, d e\}, \\
\{a d, b c, e f\},\{a d, b e, c f\},\{a d, b f, c e\}, \\
\{a e, b c, d f\},\{a e, b d, c f\},\{a e, b ... | 60 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,903 |
2. If real numbers $x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
(x-1)^{2011}+(x-1)^{2009}+2010 x=4020, \\
(y-1)^{2011}+(y-1)^{2009}+2010 y=0,
\end{array}\right.
$$
then $x+y=$ $\qquad$ . | 2. 2 .
Transform the original system of equations into
$$
\begin{array}{l}
\left\{\begin{array}{l}
(x-1)^{2011}+(x-1)^{2009}+2010(x-1)=2010, \\
(y-1)^{2011}+(y-1)^{2009}+2010(y-1)=-2010 .
\end{array}\right. \\
\text { Let } f(t)=t^{2011}+t^{2009}+2010 t(t \in \mathbf{R}) .
\end{array}
$$
Then the function $f(t)$ is a... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,906 |
3. Given $a, b, c \in \mathbf{R}$, and
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \text {, }
$$
then there exists an integer $k$, such that the following equations hold for
$\qquad$ number of them.
(1) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1... | 3. 2 .
From the given equation, we have
$$
\begin{array}{l}
\frac{(b c+a c+a b)(a+b+c)-a b c}{a b c(a+b+c)}=0 . \\
\text { Let } P(a, b, c) \\
=(b c+a c+a b)(a+b+c)-a b c \\
=(a+b)(b+c)(c+a) .
\end{array}
$$
From $P(a, b, c)=0$, we get
$$
a=-b \text { or } b=-c \text { or } c=-a \text {. }
$$
Verification shows that... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,907 |
4. Let two fixed points in the plane be $A(-3,0)$ and $B(0,-4)$, and let $P$ be any point on the curve $y=\frac{12}{x}(x>0)$. Draw $PC \perp x$-axis and $PD \perp y$-axis, with the feet of the perpendiculars being $C$ and $D$, respectively. Then the minimum value of $S_{\text{quadrilateral } ACD}$ is | 4. 24.
Notice that
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \\
=2\left(x+\frac{9}{x}\right)+12 \geqslant 24 .
\end{array}
$$
The equality holds if and only if $x=3$. Therefore, the minimum value of $S_{\text {quadrilateral } A B C D}$ is 24. | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,908 |
5. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and satisfy $\sum_{k=1}^{n} a_{k}=1, \sum_{k=1}^{n} \frac{1}{a_{k}}=4$.
Then the value of $\prod_{k=1}^{n} a_{k}$ is $\qquad$ | 5. $\frac{1}{4}$.
By the Cauchy-Schwarz inequality, we know that $\left(\sum_{k=1}^{n} a_{k}\right) \sum_{k=1}^{n} \frac{1}{a_{k}} \geqslant n^{2}$.
Also, since $\left(\sum_{k=1}^{n} a_{k}\right) \sum_{k=1}^{n} \frac{1}{a_{k}}=4$, we have $n^{2} \leqslant 4$, and $n$ is a positive integer. Therefore, $n=1$ or 2.
When ... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,909 |
6. Given $k_{1}, k_{2}, \cdots, k_{n}$ are $n$ distinct positive integers, and satisfy $\sum_{i=1}^{n} k_{i}^{3}=2024$. Then the maximum value of the positive integer $n$ is $\qquad$ . | 6. 8 .
From the problem, we know that when $n \geqslant 9$, we have
$$
\sum_{i=1}^{n} k_{i}^{3} \geqslant \sum_{i=1}^{n} i^{3} \geqslant \sum_{i=1}^{9} i^{3}=2025>2024,
$$
which is a contradiction. Therefore, $n \leqslant 8$.
Also, $2^{3}+3^{3}+\cdots+9^{3}=2024$, so the maximum value of the positive integer $n$ is 8... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,910 |
7. Given the function $f(x)=\frac{x^{2}}{1+x^{2}}$. If $m=\sum_{k=1}^{101} f(k), n=\sum_{k=2}^{101} f\left(\frac{1}{k}\right)$, then $m+n=$ | 7. $\frac{201}{2}$.
When $x \neq 0$,
$$
\begin{array}{l}
f(x)+f\left(\frac{1}{x}\right)=\frac{x^{2}}{1+x^{2}}+\frac{\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}} \\
=\frac{x^{2}}{1+x^{2}}+\frac{1}{1+x^{2}}=1 .
\end{array}
$$
Since $f(1)=\frac{1^{2}}{1+1^{2}}=\frac{1}{2}$, therefore,
$$
\begin{array}{l}
m+n=\sum_{k=1}^{101} f(k... | \frac{201}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,911 |
8. Given a sphere with radius $r$ that is tangent to two other spheres with radius $R$, and all three spheres are tangent to the two half-planes of a dihedral angle of $60^{\circ}$. Then the value of $\frac{R}{r}$ is $\qquad$ | 8. $\frac{5 \pm \sqrt{13}}{4}$.
As shown in Figure 2, let the centers of two spheres with radius $R$ be $O_{1}$ and $O_{2}$, and the center of the sphere with radius $r$ be $O_{3}$.
According to the problem, points $O_{1}$, $O_{2}$, and $O_{3}$ determine a plane that bisects the given dihedral angle, and the distance... | \frac{5 \pm \sqrt{13}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,912 |
9. (15 points) In the tetrahedron $S-ABC$, it is known that $SC \perp$ plane $ABC$, $AB=BC=CA=4\sqrt{2}$, $SC=2$, and $D$, $E$ are the midpoints of $AB$, $BC$ respectively. If point $P$ moves on $SE$, find the minimum value of the area of $\triangle PCD$.
---
The translation preserves the original text's formatting a... | 9. As shown in Figure 3, in the tetrahedron $S-ABC$, the base $ABC$ is an equilateral triangle, and $D$ is the midpoint of $AB$. Then
$$
\begin{array}{l}
|CD|=\frac{\sqrt{3}}{2}|AB| \\
=\frac{\sqrt{3}}{2} \times 4 \sqrt{2} \\
=2 \sqrt{6}.
\end{array}
$$
Thus, to find the minimum value of the area of $\triangle PCD$, w... | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,913 |
10. (20 points) Given a fixed point $M(-3,0)$, and $Q$, $P$ are moving points on the $x$-axis and $y$-axis respectively, such that $M P \perp P Q$. Point $N$ is on the line $P Q$, and $\overrightarrow{P N}=-\frac{3}{2} \overrightarrow{N Q}$.
(1) Find the equation of the trajectory $C$ of the moving point $N$.
(2) Draw ... | 10. (1) Let points $N(x, y)$, $P\left(0, y^{\prime}\right)$, and $Q\left(x^{\prime}, 0\right)$ with $\left(x^{\prime}>0\right)$.
From $\overrightarrow{P N}=-\frac{3}{2} \overrightarrow{N Q}$, we get $x^{\prime}=\frac{x}{3}$, $y^{\prime}=-\frac{y}{2}$.
From $M P \perp P Q$, we get
$$
\begin{array}{l}
\frac{y^{\prime}-0}... | \left(\frac{11}{3}, 0\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,914 |
Dirichlet's Theorem Let $\theta$ be an irrational number. Then for any positive integer $n$, there exist integers $p, q(|q| \leqslant n)$, satisfying
$$
|q \theta-p|<\frac{1}{n} \text {. }
$$ | Prove that dividing the interval $[0,1]$ into $n$ equal parts, each part has a length of $\frac{1}{n}$.
Consider $n+1$ numbers $\{j \theta\}(j=0,1, \cdots, n)$. Here, $\{j \theta\}$ represents the fractional part of $j \theta$, i.e.,
$$
\{j \theta\}=j \theta-[j \theta] \text {. }
$$
Thus, $\{j \theta\} \in(0,1)$.
Sin... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,915 |
11. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{12}{11}, a_{n+1}=\frac{4 a_{n}}{2 a_{n}+1}(n=1,2, \cdots) \text {. }
$$
(1) Find the general term formula of $\left\{a_{n}\right\}$, and prove: for any $x>0$,
$$
a_{n} \geqslant \frac{3}{2+x}-\frac{3}{(2+x)^{2}}\left(\frac{3}{4^{n}}-x\r... | 11. (1) From $a_{n+1}=\frac{4 a_{n}}{2 a_{n}+1}$, we get
$$
\frac{1}{a_{n+1}}=\frac{1}{2}+\frac{1}{4 a_{n}} \text {. }
$$
It can be seen that $\frac{1}{a_{n+1}}-\frac{2}{3}=\frac{1}{4}\left(\frac{1}{a_{n}}-\frac{2}{3}\right)=\cdots=\frac{1}{4^{n+1}}$.
Thus, $\frac{1}{a_{n}}-\frac{2}{3}=\frac{1}{4^{n}} \Rightarrow a_{n... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,916 |
12. (15 points) On a plane, there are $n(n \geqslant 3)$ non-collinear points $A_{1}, A_{2}, \cdots, A_{n}$, and next to each point $A_{i}(i=1,2, \cdots, n)$, a number $a_{i}$ is marked. If a line passes through two or more of these points, then the sum of the numbers marked next to these points is zero. Prove: All the... | 12. Let $S=a_{1}+a_{2}+\cdots+a_{n}$. Draw lines $l_{i j}(j=1,2, \cdots, m)$ through point $A_{i}(i=$ $1,2, \cdots, n)$, such that all other points appear exactly once on these lines except for $A_{i}$.
According to the problem, $2 \leqslant m \leqslant n-1$, and
$(m-1) a_{i}+S=0$.
Summing over $i$ yields $(m-1+n) S=0$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,917 |
$$
\begin{array}{l}
0.123 \times 958958+877 \times 613.613-34.5 \times 1231.23 \\
=(\quad) .
\end{array}
$$
(A) -613613
(B) 613613
(C) 61361.3
(D) None of the above answers is correct | - 1. B.
Original expression
$$
\begin{array}{l}
=1.001(123 \times 958-345 \times 123+877 \times 613) \\
=1.001(123 \times 613+877 \times 613) \\
=1.001 \times 613 \times 1000 \\
=613613
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,918 |
2. Let $x$, $y$, $z$ all be real numbers, satisfying
$$
\begin{array}{l}
a=-x^{2}+y z-\frac{1}{2011}, \\
b=-y^{2}+x z-\frac{1}{2011}, \\
c=-z^{2}+x y-\frac{1}{2011} .
\end{array}
$$
Then among $a$, $b$, $c$, at least one value ( ).
(A) is greater than 0
(B) is equal to 0
(C) is less than 0
(D) is not greater than 0 | 2. C.
Notice
$$
\begin{array}{l}
-2(a+b+c) \\
=2 x^{2}-2 y z+2 y^{2}-2 x z+2 z^{2}-2 x y+\frac{6}{2011} \\
=(x-y)^{2}+(y-z)^{2}+(z-x)^{2}+\frac{6}{2011} \\
>0 .
\end{array}
$$
Therefore, $a+b+c<0$.
Hence, at least one of $a$, $b$, and $c$ is less than 0. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,919 |
3. In $\triangle A B C$, $\angle B=2 \angle C, A C=6$. Then the range of $A B$ is ( ).
(A) $A B>3$
(B) $3<A B<6$
(C) $3 \leqslant A B<6$
(D) None of the above | 3. B.
As shown in Figure 5, extend $C B$ to point $D$ such that $B D = A B$, and connect $A D$. Then,
$$
\begin{array}{c}
\angle D = \angle B A D . \\
\text { Therefore, } \angle C \\
= \frac{1}{2} \angle A B C = \angle D .
\end{array}
$$
Thus, $A D = A C = 6$.
In $\triangle A B D$, since $A B + B D > A D$, we have,
... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,920 |
4. If $a$ and $b$ are both prime numbers, and $a^{4}+13b=107$, then the last digit of $a^{2011}+b^{2012}$ is ( ).
(A) 9 or 3
(B) 8 or 3
(C) 9 or 7
(D) 1 or 3 | 4. A.
From the condition, we know that $a^{4}$ and $13 b$ must be one odd and one even.
If $a^{4}$ is even, then $a=2, b=7$. At this time,
$$
\begin{array}{l}
a^{2011}+b^{2012}=2^{2011}+7^{2012} \\
=\left(2^{4}\right)^{502} \times 2^{3}+\left(7^{4}\right)^{503} .
\end{array}
$$
From $\left(2^{4}\right)^{502}$ having ... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,921 |
5. As shown in Figure 1, in $\square A B C D$, $A B=3, B C=7$, $B E$ is perpendicular to the angle bisector $A E$ of $\angle B A D$ at point $E$, $F$ is the midpoint of $C D$, and $B D$ intersects $E F$ at point $G$. Then the value of $E G: G F$ is ( ).
(A) $3: 4$
(B) $3: 7$
(C) $1: 2$
(D) $4: 7$ | 5. D.
As shown in Figure 6, extend $B E$ to intersect $A D$ at point $H$. It is easy to see that
$$
\begin{array}{c}
\triangle A H E \cong \triangle A B E \\
\Rightarrow E H=B E, \\
A H=A B=3 .
\end{array}
$$
Then $H D=A D-A H$
$$
=7-3=4 \text {. }
$$
Thus, $E F$ is the midline of trapezoid $H B C D$.
Therefore, $E ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,922 |
1. Let integers $a$, $b$, $c$ satisfy
$$
\begin{array}{l}
a^{2}+b^{4}+c^{6} \leqslant \sqrt{5} . \\
\text { Then }(-2)^{a}+(-2)^{b}+(-2)^{c}=
\end{array}
$$ | 2. $\pm 3$ or 0 or $\frac{3}{2}$.
From the given, we have $a^{2}+b^{4}+c^{6}=0$ or 1 or 2.
When $a^{2}+b^{4}+c^{6}=0$, $a=b=c=0$. At this time, $(-2)^{a}+(-2)^{b}+(-2)^{c}=3$.
When $a^{2}+b^{4}+c^{6}=1$, one of $a, b, c$ is +1 or -1, and the other two are 0. At this time,
$$
(-2)^{a}+(-2)^{b}+(-2)^{c}=0 \text{ or } \fr... | \pm 3 \text{ or } 0 \text{ or } \pm \frac{3}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,924 |
2. If $\sqrt{x}-\frac{1}{\sqrt{x}}=-\sqrt{2}$, then $x^{2}-\frac{1}{x^{2}}=$ $\qquad$ | 2. $-8 \sqrt{3}$.
From $\sqrt{x}-\frac{1}{\sqrt{x}}=-\sqrt{2} \Rightarrow x+\frac{1}{x}=4$.
Thus, $\left(x-\frac{1}{x}\right)^{2}=\left(x+\frac{1}{x}\right)^{2}-4=12$.
Therefore, $x-\frac{1}{x}= \pm 2 \sqrt{3}$.
Since $\sqrt{x}-\frac{1}{\sqrt{x}}=-\sqrt{2}<0$, we know $0<x<1$.
Hence, $x-\frac{1}{x}=-2 \sqrt{3}$.
So, $... | -8 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,925 |
Example 1 Proof: There exist infinitely many positive integers, which are multiples of 2005, and each of these numbers, when written in decimal form, has an equal number of occurrences of the digits $0,1, \cdots, 9$ (Note: leading zeros are not counted). [1]
$(2005$, Austrian Mathematical Olympiad) | First, note that $2005=5 \times 401$.
Let $M=1234678905$.
Set $N_{k}=M\left[10^{10(k-1)}+10^{10(k-2)}+\cdots+10^{10}+1\right]$,
Consider the set $T=\left\{N_{1}, N_{2}, \cdots, N_{401}\right\}$.
If there is no element in $T$ that is a multiple of 401, then these 401 elements modulo 401 can have at most 400 different re... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,926 |
3. As shown in Figure 2, in $\triangle A B C$, $A B=A C$, $\angle B A C=78^{\circ}$, $P$ is a point inside $\triangle A B C$, and $\angle B C P=42^{\circ}$, $\angle A P C$ $=162^{\circ}$. Then the degree measure of $\angle P B C$ is | 3. $21^{\circ}$.
As shown in Figure 7, construct an equilateral $\triangle APO$ on side $AP$ at point $B$, and connect $BO$.
It is easy to see that
$$
\begin{array}{l}
\angle ABC = \angle ACB \\
= 51^{\circ}.
\end{array}
$$
Then $\angle PCA = \angle ACB - \angle BCP$ $= 51^{\circ} - 42^{\circ} = 9^{\circ}$.
Thus, $\a... | 21^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,927 |
4. Given the equation in terms of $x$
$$
x^{2}+k x+\frac{3}{4} k^{2}-3 k+\frac{9}{3}=0 \quad (k \text { is a real number })
$$
the two real roots are $x_{1}$ and $x_{2}$. Then the value of $\frac{x_{1}^{2011}}{x_{2}^{2012}}$ is
$\qquad$ | 4. $-\frac{2}{3}$.
According to the problem, we have $\Delta=k^{2}-4\left(\frac{3}{4} k^{2}-3 k+\frac{9}{2}\right) \geqslant 0$. Thus, $(k-3)^{2} \leqslant 0$.
Also, $(k-3)^{2} \geqslant 0$, so $(k-3)^{2}=0$.
Solving this, we get $k=3$.
At this point, the equation becomes $x^{2}+3 x+\frac{9}{4}=0$.
Solving this, we ge... | -\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,928 |
One. (20 points) As shown in Figure 3, in the Cartesian coordinate system, point $A$ is on the $y$-axis, points $B$ and $C$ are on the $x$-axis, $A O=8$, $A B=A C$, $\sin \angle A B C=\frac{4}{5}$, point $D$ is on $A B$, and $C D$ intersects the $y$-axis at point $E$, satisfying $S_{\triangle C O E}=S_{\triangle A D E}... | Given $\sin \angle A B C=\frac{A O}{A B} \Rightarrow A B=10$.
By the Pythagorean theorem,
$$
B O=\sqrt{A B^{2}-A O^{2}}=6 \text {. }
$$
It is easy to see that $\triangle A B O \cong \triangle A C O$.
Therefore, $C O=B O=6$.
Thus, $A(0,-8) 、 B(6,0) 、 C(-6,0)$.
Let point $D(m, n)$.
From $S_{\triangle C O E}=S_{\triangl... | y=\frac{2}{27} x^{2}-\frac{8}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,929 |
```
II. (25 points) As shown in Figure 4, given that $AB$ is the diameter of $\odot O$, $CD$ is tangent to $\odot O$ at point $D$, and $D$ is the midpoint of arc $\overparen{AB}$. $AC$ intersects $\odot O$ at point $E$, the extensions of $OD$ and $BE$ intersect at point $F$, and $BF$ intersects $CD$ at point $G$. Prove... | II. As shown in Figure 8, connect $DE$ and $CF$. It is easy to know that
$\angle AOD = 90^{\circ}$.
Then $\angle AED$
$= \frac{1}{2} \angle AOD$
$= 45^{\circ}$.
Since $AB$ is the diameter of $\odot O$ and $CD$ is the tangent of $\odot O$, we know
$\angle AEB = \angle CDO = 90^{\circ}$.
Thus, $\angle CEF = \angle CDF = ... | (2,3,15,90), (2,15,3,90), (3,2,15,90), (3,15,2,90), (15,2,3,90), (15,3,2,90), (2,5,5,50), (5,2,5,50), (5,5,2,50) | Geometry | proof | Yes | Yes | cn_contest | false | 723,930 |
1. Let $f_{1}(x)=\left\{\begin{array}{ll}x, & x \in \mathbf{Q} \text {; } \\ \frac{1}{x}, & x \notin \mathbf{Q} .\end{array}\right.$ For positive integers $n$ greater than 1, define $f_{n}(x)=f_{1}\left(f_{n-1}(x)\right)$.
Then for all positive integers $n$, we have $f_{n}(x)=$ | $$
\text { -、1. } f_{n}(x)=\left\{\begin{array}{ll}
x, & x \in \mathbf{Q} \\
x^{(-1)^{n}}, & x \notin \mathbf{Q}
\end{array}\right.
$$
When $x \in \mathbf{Q}$,
$$
f_{2}(x)=f_{1}\left(f_{1}(x)\right)=f_{1}(x)=x \text {; }
$$
When $x \notin \mathbf{Q}$,
$$
f_{2}(x)=f_{1}\left(f_{1}(x)\right)=f_{1}\left(\frac{1}{x}\righ... | f_{n}(x)=\left\{\begin{array}{ll} x, & x \in \mathbf{Q} \\ x^{(-1)^{n}}, & x \notin \mathbf{Q} \end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,931 |
5. Let $x, y \geqslant 0$, and $x+y \leqslant \sqrt{\frac{2}{3}}$. Then
$$
\sqrt{2-3 x^{2}}+\sqrt{2-3 y^{2}}
$$
the minimum value is $\qquad$ | 5. $\sqrt{2}$.
Obviously, $0 \leqslant x, y \leqslant \sqrt{\frac{2}{3}}$.
Let $x=\sqrt{\frac{2}{3}} \cos \alpha, y=\sqrt{\frac{2}{3}} \cos \beta$, where $0^{\circ} \leqslant \alpha, \beta \leqslant 90^{\circ}$.
Then $\sqrt{2-3 x^{2}}+\sqrt{2-3 y^{2}}$
$$
=\sqrt{2}(\sin \alpha+\sin \beta),
$$
and $\cos \alpha+\cos \b... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,935 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.