problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
5. Let the focus of the parabola $y^{2}=4 x$ be $F$, the vertex be $O$, and $M$ be a moving point on the parabola. Then the maximum value of $\frac{|M O|}{|M F|}$ is $(\quad)$.
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{2 \sqrt{3}}{3}$
(C) $\frac{4}{3}$
(D) $\sqrt{3}$ | 5. B.
Given $F(1,0), O(0,0)$.
Let point $M(x, y)$. Clearly, $x \geqslant 0$.
Then $\left(\frac{|M O|}{|M F|}\right)^{2}=\frac{x^{2}+y^{2}}{(x-1)^{2}+y^{2}}$ $=\frac{x^{2}+4 x}{x^{2}+2 x+1}=\frac{4}{3}-3\left(\frac{1}{x+1}-\frac{1}{3}\right)^{2} \leqslant \frac{4}{3}$. When $x=2$, the equality holds.
Therefore, the ma... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,470 |
Example 4 As shown in Figure 4, in $\triangle A B C$, $A B=A C$, and $F$ is a point on side $B C$. The perpendicular bisectors of $B F$ and $C F$ intersect $A B$ and $A C$ at points $D$ and $E$, respectively. The point symmetric to $F$ with respect to $D E$ is $P$. If $P$ lies on the perpendicular bisector of $A B$, pr... | Proof As shown in Figure 4, connect $EF$ and $PC$ intersecting at point $O$, and connect $PA, PD, PE, PF, DF$.
Obviously, $PB=PA, DB=DF, EF=EC$.
Since $AB=AC$, quadrilateral $ADFE$ is a parallelogram.
Therefore, $\angle BAC=\angle DFE=\angle FEC$.
From the symmetry of points $F, P$ about $DE$, we get
$DP=DF=DB, EP=EF=E... | CP=BP+DE | Geometry | proof | Yes | Yes | cn_contest | false | 725,471 |
6. Suppose the axial section of an inverted conical container is an equilateral triangle. Water is poured into this container, and a solid sphere with radius $r$ is placed inside. If the sphere is submerged in the water and is tangent to the container walls and the water surface, then the height of the water level afte... | 6. D.
Figure 3 shows the axial section $P A B$ of a cone, with the center of the sphere being O. It is easy to see that,
$$
\begin{array}{c}
P C=O C+P O \\
=r+2 r=3 r, \\
A C=\sqrt{3} r, \\
V_{P A B}=\frac{1}{3} \pi(\sqrt{3} r)^{2} \cdot 3 r=3 \pi r^{3} .
\end{array}
$$
Let the water surface after removing the sphere... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,472 |
7. As shown in Figure 1, given that the side length of square $A B C D$ is $3, E$ is the midpoint of $D C$, and $A E$ intersects $B D$ at point $F$. Then $\overrightarrow{F D} \cdot \overrightarrow{D E}=$ $\qquad$ | $=7 .-\frac{3}{2}$.
From the given, we know $\frac{D F}{F B}=\frac{D E}{A B}=\frac{1}{2}$.
Therefore, $\overrightarrow{F D} \cdot \overrightarrow{D E}=-D F \cdot D E \cos \angle F D E$
$=-\frac{1}{3} \times 3 \sqrt{2} \times \frac{3}{2} \times \frac{\sqrt{2}}{2}=-\frac{3}{2}$. | -\frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,473 |
8. The constant term in the expansion of $\left(x^{2}+x-\frac{1}{x}\right)^{6}$ is $\qquad$ (answer with a specific number). | 8. -5 .
From the conditions, the constant term is
$$
C_{6}^{2}(-1)^{4}+C_{6}^{3}(-1)^{3}=-5 .
$$ | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,474 |
9. Let the sequence $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, satisfying $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$. Then the value of $S_{20}$ is $\qquad$. | 9. 0 .
From the condition, we know $a_{1}=\frac{\left(a_{1}+1\right)^{2}}{4}$, solving this gives $a_{1}=1$.
When $n \geqslant 2$, from $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$, we know
$$
S_{n-1}=\frac{\left(a_{n-1}+1\right)^{2}}{4} \text {. }
$$
Then $a_{n}=\frac{1}{4}\left(a_{n}+1\right)^{2}-\frac{1}{4}\left(a_{... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,475 |
10. The number of positive integers not exceeding 2012 and having exactly three positive divisors is $\qquad$ . | 10. 14.
Let $1 \leqslant a \leqslant 2012$ and $a$ has only three positive divisors. Then $a$ is the square of a prime number, i.e., $a=p^{2} \leqslant 2012$.
Thus, $2 \leqslant p \leqslant 43$.
$$
\begin{array}{c}
\text { Hence } p=2,3,5,7,11,13,17,19,23, \\
29,31,37,41,43 .
\end{array}
$$ | 14 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,476 |
11. Given acute angles $\angle A, \angle B$ satisfy $\tan (A+B)=2 \tan A$.
Then the maximum value of $\tan B$ is $\qquad$ | 11. $\frac{\sqrt{2}}{4}$.
Notice,
$$
\begin{array}{l}
\tan B=\tan [(A+B)-A] \\
=\frac{\tan (A+B)-\tan A}{1+\tan (A+B) \cdot \tan A} \\
=\frac{2 \tan A-\tan A}{1+2 \tan ^{2} A}=\frac{\tan A}{1+2 \tan ^{2} A} \\
\leqslant \frac{\tan A}{2 \sqrt{2} \tan A}=\frac{\sqrt{2}}{4} .
\end{array}
$$
Equality holds if and only if... | \frac{\sqrt{2}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,477 |
13. Let the function $f(x)=\sin x+\sqrt{3} \cos x+1$.
(1) Find the maximum and minimum values of the function $f(x)$ on $\left[0, \frac{\pi}{2}\right]$;
(2) If real numbers $a$, $b$, and $c$ satisfy
$$
a f(x)+b f(x-c)=1
$$
for any $x \in \mathbf{R}$, find the value of $\frac{b \cos c}{a}$. | Three, 13. (1) From the given conditions,
$$
f(x)=2 \sin \left(x+\frac{\pi}{3}\right)+1 \text {. }
$$
From $0 \leqslant x \leqslant \frac{\pi}{2} \Rightarrow \frac{\pi}{3} \leqslant x+\frac{\pi}{3} \leqslant \frac{5 \pi}{6}$.
Thus, $\frac{1}{2} \leqslant \sin \left(x+\frac{\pi}{3}\right) \leqslant 1$.
Therefore, when ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,479 |
14. Given $a, b, c \in \mathbf{R}_{+}$, satisfying $a b c(a+b+c)=1$.
(1) Find the minimum value of $S=(a+c)(b+c)$;
(2) When $S$ takes the minimum value, find the maximum value of $c$.
保留了源文本的换行和格式。 | 14. (1) Since
$$
\begin{array}{l}
(a+c)(b+c)=a b+a c+b c+c^{2} \\
=a b+(a+b+c) c \\
=a b+\frac{1}{a b} \geqslant 2 \sqrt{a b \cdot \frac{1}{a b}}=2,
\end{array}
$$
the equality holds when $a b=1$, so when $a=b=1$, $c=\sqrt{2}-1$, $S$ can take the minimum value 2.
(2) When $S$ takes the minimum value, $a b=1$, thus,
$$... | \sqrt{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,480 |
15. The line $y=k x+1$ intersects the left branch of the hyperbola $x^{2}-y^{2}=1$ at points $A$ and $B$. The line $l$ passes through the point $(-2,0)$ and the midpoint of $A B$. Find the range of the y-intercept $b$ of the line $l$.
15. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. By combining the eq... | $$
\left\{\begin{array}{l}
\Delta=4 k^{2}-8\left(k^{2}-1\right)>0 \\
x_{1}+x_{2}=-\frac{2 k}{k^{2}-1}0 .
\end{array}\right.
$$
Solving, we get $1<k<\sqrt{2}$.
Then $y_{1}+y_{2}=k\left(x_{1}+x_{2}\right)+2$
$$
=-\frac{2 k^{2}}{k^{2}-1}+2=-\frac{2}{k^{2}-1} \text {. }
$$
Therefore, the midpoint of $A B$ is $\left(-\fra... | (-\infty,-2-\sqrt{2}) \cup(2,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,481 |
Example 5 As shown in Figure 5, given that the semicircle $\odot O$ is tangent to the sides $AD$, $DC$, and $CB$ of the convex quadrilateral $ABCD$, and the center $O$ is on $AB$. If points $A$, $B$, $C$, and $D$ are concyclic, prove:
$$
AB = AD + BC.
$$ | Proof As shown in Figure 5, let the intersection of lines $A D$ and $B C$ be point $M$, and let $E$ and $F$ be the points of tangency of the semicircle with $A D$ and $B C$, respectively. Draw a tangent line to the semicircle $\odot O$ parallel to $A B$, intersecting $M A$ and $M B$ at points $P$ and $Q$, respectively,... | AB = AD + BC | Geometry | proof | Yes | Yes | cn_contest | false | 725,482 |
16. Let the function $f_{n}(x)=x^{n}(1-x)^{2}$ have its maximum value $a_{n}(n=1,2, \cdots)$ on the interval $\left[\frac{1}{2}, 1\right]$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Prove that for any positive integer $n(n \geqslant 2)$, $a_{n} \leqslant \frac{1}{(n+2)^{2}}$ holds;
... | 16. (1) Notice,
$$
\begin{array}{l}
f_{n}^{\prime}(x)=n x^{n-1}(1-x)^{2}-2 x^{n}(1-x) \\
=x^{n-1}(1-x)[n(1-x)-2 x] .
\end{array}
$$
When $x \in\left[\frac{1}{2}, 1\right]$, from $f_{n}^{\prime}(x)=0$, we know $x=1$ or $x=\frac{n}{n+2}$.
Also, $f_{n}(1)=0$, so when $n=1$, $\frac{n}{n+2}=\frac{1}{3} \notin\left[\frac{1}... | S_{n}<\frac{7}{16} | Algebra | proof | Yes | Yes | cn_contest | false | 725,483 |
1. Given the set $M=\{1,3,5,7,9\}$. If the non-empty set $A$ satisfies: the elements of $A$ each increased by 4 form a subset of $M$, and the elements of $A$ each decreased by 4 also form a subset of $M$, then $A=$ $\qquad$ | $-1 .\{5\}$.
$$
\begin{array}{l}
\text { Let } M_{1}=\{x \mid x=m-4, m \in M\} \\
=\{-3,-1,1,3,5\} , \\
M_{2}=\{x \mid x=m+4, m \in M\} \\
=\{5,7,9,11,13\} .
\end{array}
$$
From the problem, we know that $A \subseteq M_{1} \cap M_{2}=\{5\}$.
Since $A$ is a non-empty set, therefore, $A=\{5\}$. | \{5\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,484 |
2. Given two lines $l_{1}: y=2, l_{2}: y=4$, let the graph of the function $y=3^{x}$ intersect $l_{1} 、 l_{2}$ at points $A 、 B$, and the graph of the function $y=5^{x}$ intersect $l_{1} 、 l_{2}$ at points $C 、 D$. Then the coordinates of the intersection point of lines $A B$ and $C D$ are $\qquad$ | 2. $(0,0)$.
$$
\begin{array}{l}
\text { It is easy to see that } A\left(\log _{3} 2,2\right), B\left(\log _{3} 4,4\right), \\
C\left(\log _{5} 2,2\right), D\left(\log _{5} 4,4\right) . \\
\text { Then } k_{A B}=\frac{4-2}{\log _{3} 4-\log _{3} 2}=\frac{2-0}{\log _{3} 2-0}=k_{O A}, \\
k_{C D}=\frac{4-2}{\log _{5} 4-\log... | (0,0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,485 |
3. For a positive integer $n$, if $n=p q\left(p \geqslant q, p, q \in \mathbf{N}_{+}\right)$, when $p-q$ is the smallest, then $p q$ is called the "best factorization" of $n$, and it is defined that $f(n)=\frac{q}{p}$ (for example, the factorizations of 12 are $12 \times 1$, $6 \times 2$, $4 \times 3$, among which, $4 ... | 3. (2) 、(4).
Notice,
$$
\begin{array}{l}
4=2 \times 2,7=7 \times 1,24=6 \times 4, \\
2012=503 \times 4 .
\end{array}
$$
Then $f(4)=1, f(7)=\frac{1}{7}$,
$$
f(24)=\frac{2}{3}, f(2012)=\frac{4}{503} \text {. }
$$
Therefore, (1) and (3) are incorrect, (2) and (4) are correct. | (2) 、(4) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,486 |
4. Given $\triangle A B C$ is an isosceles right triangle, $\angle A$ $=90^{\circ}$, and $\overrightarrow{A B}=a+b, \overrightarrow{A C}=a-b$.
If $a=(\cos \theta, \sin \theta)(\theta \in \mathbf{R})$, then $S_{\triangle A B C}$ $=$ . $\qquad$ | 4. 1.
From the problem, we know that $A B \perp A C,|A B|=|A C|$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
(a+b) \cdot(a-b)=0, \\
|a+b|=|a-b|
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
|a|=|b|, \\
a \cdot b=0 .
\end{array}\right.
\end{array}
$$
Since $|a|=1$, we have $|b|=1$.
According ... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,487 |
5. In a regular tetrahedron $ABCD$, $AO \perp$ plane $BCD$, with the foot of the perpendicular being $O$. Let $M$ be a point on the line segment $AO$ such that $\angle BMC=90^{\circ}$. Then $\frac{AM}{MO}=$ $\qquad$ | 5. 1.
As shown in Figure 3, connect $O B$. Let the edge length of the regular tetrahedron $A B C D$ be $a$. Then
$$
\begin{array}{c}
O B=\frac{\sqrt{3}}{3} a, \\
M B=\frac{\sqrt{2}}{2} a . \\
\text { Therefore, } M O=\sqrt{M B^{2}-O B^{2}} \\
=\frac{\sqrt{6}}{6} a=\frac{1}{2} A O=A M . \\
\text { Hence, } \frac{A M}{M... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,488 |
6. As shown in Figure 1, the three vertices of the right triangle $\triangle ABC$ are all on the given parabola $x^{2}=2 p y(p>0)$, and the hypotenuse $AB \parallel x$-axis. Then the altitude from $C$ to the hypotenuse $|CD|=$ $\qquad$ | 6. $2 p$.
Let $A\left(2 p t_{1}, 2 p t_{1}^{2}\right), C\left(2 p t_{2}, 2 p t_{2}^{2}\right)$. Then $B\left(-2 p t_{1}, 2 p t_{1}^{2}\right)$.
Since $A C \perp B C$, we have $\overrightarrow{A C} \cdot \overrightarrow{B C}=0$, which means $4 p^{2}\left(t_{2}-t_{1}\right)\left(t_{2}+t_{1}\right)+4 p^{2}\left(t_{2}^{2}... | 2p | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,489 |
7. For a certain game activity,
the rewards are divided into first, second, and third prizes (all participants in the game activity will receive a prize), and the corresponding winning probabilities form a geometric sequence with the first term $a$ and a common ratio of 2. The corresponding prize money forms an arithm... | 7.500 .
From the problem, we know the probabilities of winning the first, second, and third prizes are
$$
P_{1}=a, P_{2}=2 a, P_{3}=4 a .
$$
From $P_{1}+P_{2}+P_{3}=1$, we get $a=\frac{1}{7}$.
Thus, $P_{1}=\frac{1}{7}, P_{2}=\frac{2}{7}, P_{3}=\frac{4}{7}$.
The prizes for winning the first, second, and third prizes a... | 500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,490 |
8. Let $p$ and $q$ be two different prime numbers. Then the remainder when $p^{q-1}+q^{p-1}$ is divided by $p q$ is $\qquad$ | 8. 1 .
Since $p$ and $q$ are different prime numbers, by Fermat's Little Theorem, we have
$$
p^{q-1} \equiv 1(\bmod q) \text {. }
$$
Also, $q^{p-1} \equiv 0(\bmod q)$, thus
$$
p^{q-1}+q^{p-1} \equiv(\bmod q) \text {. }
$$
Similarly, $p^{q-1}+q^{p-1} \equiv 1(\bmod p)$.
Therefore, $p^{q-1}+q^{p-1} \equiv 1(\bmod p q)... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,491 |
Example 6 Let quadrilateral $A B C D$ be a cyclic quadrilateral, the angle bisectors of $\angle A$ and $\angle D$ intersect at point $E$, and a line through $E$ parallel to $B C$ intersects sides $A B$ and $D C$ at points $M$ and $N$ respectively. Prove: $A M+D N=M N$.
Translate the above text into English, please kee... | ```
\begin{array}{l}
\text { Given } \angle D F N=\frac{1}{2}\left(180^{\circ}-\angle F N D\right) \\
= \frac{1}{2}\left(180^{\circ}-\angle C\right)=\frac{1}{2} \angle B A D=\angle E A D \\
\Rightarrow A 、 E 、 F 、 D \text { are concyclic } \\
\Rightarrow \angle A F M=\angle A D E \\
\Rightarrow \angle F A M=\angle M A... | null | Geometry | proof | Yes | Yes | cn_contest | false | 725,493 |
10. Transporting utility poles from a construction site by the roadside along a straight road in the same direction to plant them 500 m away on the roadside, plant one at the 500 m mark, and then plant one every 50 m along the roadside. Knowing that the transport vehicle can carry a maximum of 3 poles at a time, to com... | 10. 14000 .
Assuming the completion of transporting and planting 21 utility poles, 3 poles each time, let the round trip distance for the $k(k=1,2, \cdots, 7)$-th time be $a_{k}$. Then
$$
a_{1}=2 \times 600=1200,
$$
and $a_{k+1}=a_{k}+2 \times 150(k=1,2, \cdots, 6)$.
Therefore, $\left\{a_{n}\right\}$ is an arithmetic... | 14000 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 725,494 |
In $\triangle A B C$, it is known that $A B=2$, $A C=1$, and
$$
\cos 2 A+2 \sin ^{2} \frac{B+C}{2}=1 \text {. }
$$
(1) Find the size of $\angle A$ and the length of side $B C$;
(2) If point $P$ moves within $\triangle A B C$ (including the boundary), and the sum of the distances from $P$ to the three sides is $d$, let ... | (1) From
$$
\begin{array}{l}
\cos 2 A+2 \sin ^{2} \frac{B+C}{2}=1 \\
\Rightarrow \cos 2 A-\cos (B+C)=0 \\
\Rightarrow 2 \cos ^{2} A+\cos A-1=0 \\
\Rightarrow \cos A=\frac{1}{2} \text { or } \cos A=-1 .
\end{array}
$$
Since $\angle A \in(0, \pi)$, then $\angle A=\frac{\pi}{3}$.
By the cosine rule,
$$
B C=\sqrt{A B^{2}+... | \left[\frac{\sqrt{3}}{2}, \sqrt{3}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,495 |
(20 points) In the Cartesian coordinate system, a circle with center $C\left(t, \frac{2}{t}\right)$ passes through the origin $O$, and intersects the $x$-axis and $y$-axis at points $A$ and $B$ (different from the origin $O$).
(1) Prove: The area $S$ of $\triangle A O B$ is a constant;
(2) Suppose the line $l: y=-2 x+4... | (1) From the given, the equation of $\odot C$ is
$$
(x-t)^{2}+\left(y-\frac{2}{t}\right)^{2}=t^{2}+\frac{4}{t^{2}} \text {, }
$$
which simplifies to $x^{2}+y^{2}-2 t x-\frac{4}{t} y=0$.
Therefore, $A(2 t, 0), B\left(0, \frac{4}{t}\right)$.
Thus, $S=\frac{1}{2}|O A||O B|$
$=\frac{1}{2}|2 t|\left|\frac{4}{t}\right|=4$ (... | (x-2)^{2}+(y-1)^{2}=5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,496 |
Three. (20 points) As shown in Figure 2, the acute triangle $\triangle ABC$ is inscribed in $\odot O$. A line passing through the center $O$ and perpendicular to the radius $OA$ intersects sides $AB$ and $AC$ at points $E$ and $F$, respectively. Let the tangents to $\odot O$ at points $B$ and $C$ intersect at point $P$... | Three, as shown in Figure 5, draw a line through point $P$ parallel to $E F$, intersecting the extensions of $A B$ and $A C$ at points $M$ and $N$. Then
$$
\begin{array}{l}
\angle P M B \\
=\angle A E O \\
=90^{\circ}-\angle O A E \\
=\angle A C B .
\end{array}
$$
Since $P B$ is a tangent to $\odot O$, we have,
$$
\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,497 |
Four. (30 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{2}, a_{n}=2 a_{n} a_{n+1}+3 a_{n+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If the sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=1+\frac{1}{a_{n... | (1) From
$$
\begin{array}{l}
a_{n}=2 a_{n} a_{n+1}+3 a_{n+1} \\
\Rightarrow a_{n+1}=\frac{a_{n}}{2 a_{n}+3} \Rightarrow a_{n}>0 . \\
\text { Hence } \frac{1}{a_{n+1}}-\frac{3}{a_{n}}=2 \Rightarrow \frac{1}{a_{n+1}}+1=3\left(\frac{1}{a_{n}}+1\right) .
\end{array}
$$
Therefore, $\left\{\frac{1}{a_{n}}+1\right\}$ is a ge... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,498 |
Five, (30 points) For any positive integer $n$, prove:
$$
\frac{1}{3-2}+\frac{1}{3^{2}+2^{2}}+\frac{1}{3^{3}-2^{3}}+\cdots+\frac{1}{3^{n}+(-2)^{n}}<\frac{7}{6} .
$$ | Let $a_{k}=\frac{1}{3^{k}+(-2)^{k}}, S_{n}=\sum_{k=1}^{n} a_{k}$. First, prove that for any $m \in \mathbf{N}_{+}$, we have
$$
a_{2 m}+a_{2 m+1}<\frac{4}{3^{2 m+1}}.
$$
Since $3^{2 m+1}+(-2)^{2 m+1}>0$, we have
$$
a_{2 m}+a_{2 m+1} = \frac{1}{3^{2 m}+4^{2 m}} + \frac{1}{3^{2 m+1}-2^{2 m+1}} < \frac{1}{3^{2 m}} + \frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,499 |
1. Let $a_{1}, a_{2}, \cdots, a_{11}$ be distinct positive integers no less than 2, satisfying $a_{1}+a_{2}+\cdots+a_{11}=407$. Question: Does there exist a positive integer $n$, such that when $n$ is divided by $a_{1}, a_{2}, \cdots, a_{11}, 4 a_{1}, 4 a_{2}, \cdots, 4 a_{11}$ these 22 numbers, the sum of the remainde... | 1. Does not exist.
Assume such an $n$ exists.
Notice that the maximum possible remainder when a positive integer is divided by $m$ is $m-1$. Therefore, the sum of the remainders when divided by $a_{1}, a_{2}, \cdots, a_{11}$ does not exceed $407-11=396$, and the sum of the remainders when divided by $4 a_{1}$, $4 a_{... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,500 |
2. Given that among the vertices of a regular 2012-gon, there exist $k$ vertices such that the convex $k$-gon formed by these $k$ vertices has no two parallel sides. Find the maximum value of $k$.
| 2. The maximum value of $k$ is 1509.
Let $A_{1}, A_{2}, \cdots, A_{2012}$ be the set of vertices of a regular polygon.
Consider the set of four vertices
$$
\begin{array}{l}
\left(A_{1}, A_{2}, A_{1000}, A_{1008}\right),\left(A_{3}, A_{4}, A_{100}, A_{1010}\right), \\
\cdots,\left(A_{1005}, A_{1000}, A_{2011}, A_{2012}... | 1509 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,501 |
3. Let quadrilateral $ABCD$ be a parallelogram, $\angle A$ is an obtuse angle, $H$ is the orthogonal projection of point $A$ on $BC$, and the extension of the median from vertex $C$ of $\triangle ABC$ intersects its circumcircle at point $K$. Prove: $K, H, C, D$ are concyclic. | 3. Let $E$ be the projection of point $B$ onto $A D$. Then quadrilateral $A H B E$ is a rectangle.
Therefore, $\angle H E D = \angle A B C = 180^{\circ} - \angle B C D$.
Thus, $D$, $C$, $H$, and $E$ lie on a circle $\omega$.
Let $M$ be the intersection of the diagonals of rectangle $A H B E$. Then
$$
M A = M B = M H = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,502 |
Example 2 Let $n \geqslant 2$,
$$
A_{n}=\left\{x \in \mathbf{R} \left\lvert\, x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right]\right.\right\} .
$$ | Prove: $A=\underset{n \geqslant 2}{\cup} A_{n}$ is a finite set, and find the maximum and minimum elements of $A$.
(2010, Romanian Mathematical Olympiad (Final)) [Analysis] Start with simple cases in mathematical experiments.
Elements in $A_{2}$ satisfy $x=\left[\frac{x}{2}\right]$, and $x \in \mathbf{Z}$. Hence, $\fra... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,504 |
5. 101 wise men sit in a circle for a roundtable discussion, debating whether Earth orbits Jupiter or Jupiter orbits Earth. At the beginning and at each subsequent moment, each wise man holds one of the two views: Earth orbits Jupiter or Jupiter orbits Earth. The wise men simultaneously announce their views every minut... | 5. If a wise person and at least one of the two adjacent people share the same opinion, then this wise person is called "stable"; otherwise, they are called "unstable".
Obviously, once a wise person becomes stable, they will not change their opinion and will always remain stable.
Since there is an odd number of wise ... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 725,505 |
6. Given $A_{1} 、 B_{1} 、 C_{1}$ are points on the sides $B C$ 、 $C A 、 A B$ of $\triangle A B C$, respectively, satisfying
$$
A B_{1}-A C_{1}=C A_{1}-C B_{1}=B C_{1}-B A_{1},
$$
$I_{A} 、 I_{B} 、 I_{C}$ are the incenters of $\triangle A B_{1} C_{1} 、 \triangle A_{1} B C_{1} 、 \triangle A_{1} B_{1} C$, respectively. Pro... | 6. Let $I$ be the incenter of $\triangle ABC$, and $A_{0}, B_{0}, C_{0}$ be the points where the incircle touches sides $BC, CA, AB$ respectively. Without loss of generality, assume $A_{1}$ lies on segment $A_{0}B$ (other cases can be discussed similarly).
Notice,
$$
\begin{array}{l}
C A_{0} + A C_{0} = C B_{0} + A B_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,506 |
7. At the beginning, there are 10 consecutive positive integers written on the blackboard. The following operation is performed on the numbers on the blackboard: take any two numbers $a$ and $b$, and replace them with the numbers $a^{2}-2011 b^{2}$ and $a b$. After several such operations, all 10 numbers originally on ... | 7. Impossible.
Assume after several operations, we get 10 consecutive positive integers, and all the original 10 numbers have been replaced.
First, we give a lemma.
Lemma: For any given positive integer $k$, the number of multiples of $k$ among the 10 numbers will not decrease after the operation.
This lemma is obviou... | Impossible | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,507 |
8. There are several bus lines in the city. It is known that any two bus lines have exactly one common station, and any bus line has at least 4 stations. Prove: All the stations can be divided into two disjoint groups, such that any bus line contains at least one station from each group. | 8. Take any two bus routes $l_{1}, l_{2}$. Let $A$ be
their common station.
If all other routes pass through station $A$, then group station $A$ into one group, and all other stations into another group. It is clear that this satisfies the requirement.
Assume there exists $l_{3}$ that does not pass through station $A... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,508 |
1. Let $P(x)$ be a polynomial with integer coefficients, and satisfy
$$
P(0)=0, \quad (P(0), P(1), \cdots)=1 \text{. }
$$
Prove: There exist infinitely many positive integers $n$, such that
$$
(P(n)-P(0), P(n+1)-P(1), \cdots)=n \text{. }
$$ | 1. Let $P(x)=a_{r} x^{r}+a_{r-1} x^{r-1}+\cdots+a_{1} x$.
Then $Q(x)=P^{\prime}(x)$
$$
=r a_{r} x^{r-1}+(r-1) a_{r-1} x^{r-2}+\cdots+a_{1} \text {. }
$$
Since $P(x)$ is not always a constant, $Q(x)$ is not always 0. We can choose some positive integer $m$ such that $Q(m) \neq 0$. Let $n$ be any prime number that does... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,509 |
2. Given positive real numbers $a, b, c$ satisfying $abc=1$. Prove:
$$
\frac{1}{a^{5}(b+2 c)^{2}}+\frac{1}{b^{5}(c+2 a)^{2}}+\frac{1}{c^{5}(a+2 b)^{2}} \geqslant \frac{1}{3} \text {. }
$$ | 2. Let
$$
\begin{array}{l}
S_{1}=\frac{1}{a^{5}(b+2 c)^{2}}+\frac{1}{b^{5}(c+2 a)^{2}}+\frac{1}{c^{5}(a+2 b)^{2}}, \\
S_{2}=\frac{1}{a^{3}(b+2 c)}+\frac{1}{b^{3}(c+2 a)}+\frac{1}{c^{3}(a+2 b)} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
S_{2}[a(b+2 c)+b(c+2 a)+c(a+2 b)] \\
\geqslant\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,510 |
3. Given an acute triangle $\triangle A B C$ with the altitudes from $A, B, C$ to the opposite sides $B C, C A, A B$ being $h_{a}, h_{b}, h_{c}$ respectively. If $P$ is a point inside $\triangle A B C$, prove:
$$
\frac{P A}{h_{b}+h_{c}}+\frac{P B}{h_{c}+h_{a}}+\frac{P C}{h_{a}+h_{b}} \geqslant 1 .
$$ | 3. First, prove a lemma.
Lemma Let $a=BC, b=CA, c=AB$, and the distances from point $P$ to sides $BC, CA, AB$ be denoted as $p_{a}, p_{b}, p_{c}$, respectively. Then
$$
2 a P A \geqslant (b+c)(p_{b}+p_{c}).
$$
Proof Since $\frac{p_{a}}{h_{a}}=\frac{S_{\triangle P B C}}{S_{\triangle A B C}}$,
$$
\frac{p_{b}}{h_{b}}=\f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,511 |
4. In $\triangle A B C$, $M$ and $N$ are points on sides $A C$ and $B C$ respectively, such that $M N \parallel A B$. Points $P$ and $Q$ are on sides $A B$ and $C B$ respectively, such that $P Q \parallel A C$. The incircle of $\triangle C M N$ touches $A C$ at point $E$, and the incircle of $\triangle B P Q$ touches $... | 4. As shown in Figure 3, let the incircles of $\triangle A B C$, $\triangle M N C$, $\triangle P Q B$, and $\triangle A R S$ be circles $\omega_{1}$, $\omega_{C}$, $\omega_{B}$, and $\omega$ respectively. Let circle $\omega_{1}$ touch $A B$ and $A C$ at points $R_{1}$ and $S_{1}$, and let the incenters of $\triangle A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,512 |
5. Define the sequence $a_{1}, a_{2}, \cdots$ as follows: $a_{1}=1$, for positive integers $n>1$,
$$
a_{n}=a_{\left[\frac{n}{2}\right]}+a_{\left[\frac{n}{3}\right]}+\cdots+a_{\left[\frac{n}{n}\right]}+1 .
$$
Prove: There exist infinitely many positive integers $n$, such that
$$
a_{n} \equiv n\left(\bmod 2^{2010}\right... | 5. First, prove a lemma.
Lemma Let $p$ be a prime. If $p^{s} \mid n$, then $2^{s-1} \mid\left(a_{n}-a_{n-1}\right)$, where $s$ is a positive integer, and $a_{0}=0$.
Proof by induction on $s$.
When $s=1$, the conclusion is obviously true.
Assume the conclusion holds for $s-1$.
For $s \geqslant 2$, prove by contradictio... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,513 |
6. Let $T$ be a finite set composed of positive integers greater than 1. If a subset $S$ of $T$ satisfies that for every $t \in T$, there exists some $s \in S$ such that $(s, t)>1$, then $S$ is called "good". Prove: The number of good subsets of $T$ is odd. | 6. Let the set of all ordered subset pairs $(X, Y)$ that satisfy the following conditions be $\mathscr{A}$, where $X, Y \subseteq T$, and for all $x \in X, y \in Y, (x, y)=1$. Then for any $(X, Y) \in \mathscr{A}, X$ and $Y$ have an empty intersection.
(1) If $X^{\prime}$ is good, then the number of subsets of the form... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,514 |
Example 3 Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1} \in(1,2) \text {, and } a_{n+1}=a_{n}+\frac{n}{a_{n}} \text {. }
$$
Prove: There exists at most one pair $(i, j)$ such that $a_{i}+a_{i} \in \mathbf{Z}$.
(35th Russian Mathematical Olympiad) | 【Analysis】The first term of the sequence $a_{1} \in(1,2)$ is uncertain. First, conduct a mathematical experiment by selecting three relatively special values in $(1,2)$ as the first term of the sequence,
$$
a_{1}=\frac{5}{4}=1 \frac{1}{4}, a_{1}=\frac{3}{2}=1 \frac{1}{2}, a_{1}=\frac{7}{4}=1 \frac{3}{4} .
$$
Observe t... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,515 |
7. Given $P, Q$ are two points inside the acute $\triangle A B C$, and satisfy $\angle P A B=\angle Q A C, \angle P B A=\angle Q B C, D$ is a point on side $B C$. Prove: $\angle D P C+\angle A P B=180^{\circ}$ if and only if $\angle D Q B+\angle A Q C=180^{\circ}$. | 7. If $\angle D P C+\angle A P B=180^{\circ}$, as shown in Figure 4, extend $B P$ to point $P_{1}$ such that $\angle P A P_{1}=\angle P C D$. Then
$$
\angle D P C=180^{\circ}-\angle A P B=\angle A P P_{1} \text {. }
$$
Thus, $\triangle P A P_{1} \backsim \triangle P C D$, and there exists a homothetic rotation transfo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,516 |
8. Let \( m, n (m \geqslant n) \) be positive integers, and \( S \) be the set of all ordered \( n \)-tuples \( \left(a_{1}, a_{2}, \cdots, a_{n}\right) \) of positive integers \( a_{1}, a_{2}, \cdots, a_{n} \) such that \( a_{1} + a_{2} + \cdots + a_{n} = m \). Prove:
\[
\begin{array}{l}
\sum_{S} 1^{a_{1}} 2^{a_{2}} \... | 8. Let $m=k+n, T$ be the set of all ordered $n$-tuples $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ of non-negative integers $b_{1}, b_{2}, \cdots, b_{n}$ that satisfy
$$
b_{1}+b_{2}+\cdots+b_{n}=k
$$
Then it suffices to prove that
$$
\begin{aligned}
n! & \sum_{T} 1^{b_{1}} 2^{b_{2}} \cdots n^{b_{n}} \\
= & \mathrm{C}_{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,517 |
9. Does there exist a positive integer $k$ such that $p=6 k+1$ is a prime, and
$$
\mathrm{C}_{3 k}^{k} \equiv 1(\bmod p) ?
$$ | 9. Does not exist.
Assume there exist $k, p$ satisfying the conditions, set
$$
A=\sum_{i=0}^{p-1}\left(i^{3}-1\right)^{3 k} \text {. }
$$
Since $p-1=6 k$ is a multiple of 3, the equation $x^{3} \equiv 1(\bmod p)$ has three roots, and in the sum $A$, these two terms are 0.
Also, $x^{\frac{p-1}{2}} \equiv \pm 1(\bmod... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,518 |
1. Simplify
$$
\sqrt{a+2+3 \sqrt{2 a-5}}-\sqrt{a-2+\sqrt{2 a-5}}
$$
The result is ( ).
(A) 1
(B) $\sqrt{2}$
(C) $\sqrt{3}$
(D) Depends on $a$ | $-1 . \mathrm{B}$.
Let $\sqrt{2 a-5}=t(t \geqslant 0)$. Then $a=\frac{t^{2}+5}{2}$
Therefore, the original expression $=\sqrt{\frac{(t+3)^{2}}{2}}-\sqrt{\frac{(t+1)^{2}}{2}}$
$$
=\frac{(t+3)-(t+1)}{\sqrt{2}}=\sqrt{2} \text {. }
$$ | \sqrt{2} | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,519 |
2. Given the equation in terms of $x$
$$
\frac{x}{x-2}+\frac{x-2}{x}=\frac{a-2 x}{x^{2}-2 x}
$$
has only one real root. Then the number of real values of $a$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 2. C .
The given equation can be transformed into $2 x^{2}-2 x+4-a=0$.
When $\Delta=4(2 a-7)=0$, i.e., $a=3.5$, it meets the requirement;
When $a>3.5$, the two distinct real roots of $2 x^{2}-2 x+4-a=0$ must include one extraneous root of the original equation, which can only be 0 or 2, thus, $a=4$ or 8, meeting the ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,520 |
3. As shown in Figure 1, in the right triangle $\triangle ABC$, $AB = BC = 8$, points $E$ and $F$ are on sides $AB$ and $BC$ respectively, with $AE = 3$, $CF = 1$, and $P$ is a moving point on the hypotenuse $AC$. Then the minimum value of the perimeter of $\triangle PEF$ is ( ).
(A) $4+\sqrt{74}$
(B) $4 \sqrt{2}+\sqrt... | 3. C.
As shown in Figure 7, construct the symmetric point $E^{\prime}$ of point $E$ with respect to $A C$, and connect $E^{\prime} F$. Then the minimum value of $P E + P F$ is exactly the length of segment $E^{\prime} F$.
Construct $E^{\prime} G \perp B C$ at point
G. Then
$$
E^{\prime} G = A B = 8, F G = 4 .
$$
By t... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,521 |
4. The largest integer not exceeding $(3 \sqrt{2}+2 \sqrt{3})^{6}$ is ( ).
(A) 209520
(B)209 519
(C)209 518
(D) 209517 | 4. B.
Let $3 \sqrt{2}+2 \sqrt{3}=a, 3 \sqrt{2}-2 \sqrt{3}=b$.
Then $a+b=6 \sqrt{2}, ab=6$
$$
\begin{array}{l}
\Rightarrow a^{2}+b^{2}=(a+b)^{2}-2ab=60 \\
\Rightarrow a^{6}+b^{6}=\left(a^{2}+b^{2}\right)^{3}-3(ab)^{2}\left(a^{2}+b^{2}\right) \\
\quad=209520 .
\end{array}
$$
Since $0<b<1$, we have
$$
209519<a^{6}<20952... | 209519 | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,522 |
5. Square $A B C D$, square $B E F G$, and square $P K R F$ are positioned as shown in Figure 2, with point $G$ on line segment $D K$, and the side length of square $B E F G$ is 2. Then the area of $\triangle D E K$ is $(\quad)$.
(A) 4
(B) 3
(C) 2
(D) $\sqrt{2}$ | 5. A.
As shown in Figure 8, connect $D B$, $G E$, and $F K$.
Then $D B / / G E / / F K$
$$
\begin{array}{l}
\Rightarrow S_{\triangle D E K}=S_{\triangle D G E}+S_{\triangle E G K}=S_{\triangle E B G}+S_{\triangle F E G} \\
\quad=S_{\text {square BEFG }}=4 .
\end{array}
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,523 |
1. The numbers $a_{1}, a_{2}, \cdots, a_{100}$ satisfy the following condition: For $k=1,2, \cdots, 100, a_{k}$ is less than the sum of the other 99 numbers by $k$. Then $a_{50}=$ | $$
\text { II. 1. } \frac{75}{98} \text {. }
$$
Let $S=a_{1}+a_{2}+\cdots+a_{10}$. Then
$$
a_{k}=\left(S-a_{k}\right)-k \Rightarrow k+2 a_{k}=S \text {. }
$$
Summing over $k$ we get
$$
\begin{array}{l}
(1+2+\cdots+100)+2 S=100 S \\
\Rightarrow S=\frac{2525}{49} \Rightarrow a_{50}=\frac{S-50}{2}=\frac{75}{98} .
\end{a... | \frac{75}{98} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,525 |
Example 4 The sequence of positive integers $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1}=a_{n}+b_{n}(n \geqslant 1) \text {, }
$$
where $b_{n}$ is obtained by reversing the digits of $a_{n}$ (the leading digit of $b_{n}$ can be zero), for example, when $a_{1}=170$, we have $a_{2}=241, a_{3}=383, a_{4}=766$, etc. Questi... | 【Analysis】First, we conduct a simple mathematical experiment, taking $a_{1}=1$, then
$$
\begin{array}{l}
a_{2}=2, a_{3}=4, a_{4}=8, a_{5}=16, \\
a_{6}=77, a_{7}=847=7 \times 11^{2} .
\end{array}
$$
Conjecture: $a_{7}$ is not a prime number and contains a certain special prime factor.
The prime number 3, which is unre... | a_{7} \neq 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,526 |
2. The sum of the maximum and minimum values of the algebraic expression $\sqrt{t-2}+\sqrt{4-t}$ is equal to
保留了源文本的换行和格式。 | $2.2+\sqrt{2}$.
Let $y=\sqrt{t-2}+\sqrt{4-t}$.
Obviously, $y \geqslant 0, 2 \leqslant t \leqslant 4$.
$$
\begin{array}{l}
\text { Then } y^{2}=2+2 \sqrt{-(t-3)^{2}+1} \\
\Rightarrow 2 \leqslant y^{2} \leqslant 4 \Rightarrow \sqrt{2} \leqslant y \leqslant 2 \\
\Rightarrow y_{\text {max }}+y_{\text {min }}=2+\sqrt{2} .
\... | 2+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,527 |
3. As shown in Figure 4, in the "dart-shaped" quadrilateral $ABCD$, $AB=4\sqrt{3}$, $BC=8$, $\angle A=\angle B=\angle C=30^{\circ}$. Then the distance from point $D$ to $AB$ is $\qquad$ | 3. 1.
As shown in Figure 9, extend $A D$ to intersect $B C$ at point $E$. Then $A E=B E$.
Draw $E F \perp A B$ at point $F$.
It is easy to see that $A F=B F=2 \sqrt{3}$,
$$
E F=2, A E=4 \text {. }
$$
Thus, $C E=4$.
Also, $\angle A D C=90^{\circ}$
$\Rightarrow D E=2$
$\Rightarrow D$ is the midpoint of $A E$
$\Rightarr... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,528 |
4. As shown in Figure 5, an equilateral $\triangle ABC$ with side length 26 is inscribed in a circle, and chord $DE \parallel BC$. The extension of $DE$ intersects the extensions of $AB$ and $AC$ at points $F$ and $G$, respectively. If the lengths of segments $AF$ and $DF$ are both positive integers, then the chord $DE... | 4. 16 .
Let $A F=x, D F=y$. Then
$$
B F=x-26, D E=x-2 y, E F=x-y \text {. }
$$
By the secant theorem, we have
$$
\begin{array}{l}
A F \cdot B F=D F \cdot E F \\
\Rightarrow x(x-26)=y(x-y) \\
\Rightarrow x^{2}-(26+y) x+y^{2}=0 .
\end{array}
$$
Since $A F$ and $D F$ are both positive integers, we have
$$
\Delta=(26+y)... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,529 |
One. (20 points) As shown in Figure 6, in the right triangle $\triangle ABC$, $\angle BAC=90^{\circ}$, $AD \perp BC$ at point $D$, $P$ is the midpoint of $AD$, the extension of $BP$ intersects $AC$ at point $E$, $EF \perp BC$ at point $F$, and the extension of $FE$ intersects the circumcircle $\odot O$ of $\triangle AB... | As shown in Figure 10, line $F G$ intersects $\odot O$ and the extension of $B A$ at points $N$ and $M$ respectively.
Given $A D \perp B C$,
$E F \perp B C$,
we have $\frac{A P}{E M}=\frac{B P}{B E}$
$$
=\frac{P D}{E F} \text {. }
$$
Since $A P=P D$
$$
\Rightarrow E M=E F \text {. }
$$
It is easy to prove
Rt $\triang... | 6(\sqrt{2}-1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,530 |
II. (25 points) Given the parabola $y=a x^{2}+b x+2$ intersects the $x$-axis at points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)\left(x_{1}<x_{2}\right)$, and $x_{1}, x_{2}$ are the two real roots of the equation $x^{2}-x-6=0$, $C$ is the intersection point of the parabola with the $y$-axis.
(1) Find the equation... | (1) From $x^{2}-x-6=0$, we solve to get $x_{1}=-2, x_{2}=3$.
Thus, $A(-2,0), B(3,0)$.
Therefore, the equation of the axis of symmetry of the parabola is $x=\frac{1}{2}\left(x_{1}+x_{2}\right)=\frac{1}{2}$.
(2) By setting $x=0$ in the equation of the parabola, we get $C(0,2)$.
Then $y_{A C}=x+2, y_{B C}=-\frac{2}{3} x+... | \left(-\frac{4}{7}, 0\right),\left(\frac{6}{7}, 0\right),\left(\frac{2}{9}, 0\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,531 |
Three. (25 points) Write out a sequence of consecutive positive integers starting from 1, then erase one of the numbers, so that the average of the remaining numbers is $43 \frac{14}{17}$. What is the number that was erased? | Three, suppose we have written $n$ consecutive positive integers 1, 2, ..., $n$. If the number $k$ is erased, then
$$
\begin{array}{l}
\frac{(1+2+\cdots+n)-k}{n-1}=43 \frac{14}{17} \\
\Rightarrow \frac{n}{2}+\frac{n-k}{n-1}=43 \frac{14}{17} \\
\Rightarrow \frac{n}{2} \leqslant 43 \frac{14}{17} \leqslant \frac{n}{2}+1 \... | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,532 |
2. Given the function
$$
\begin{array}{l}
f(x)=f_{1}(x) \\
\quad=\left\{\begin{array}{ll}
16(x-0.25)^{2}, & 0 \leqslant x<0.5 ; \\
16(x-0.75)^{2}, & 0.5 \leqslant x \leqslant 1 .
\end{array}\right.
\end{array}
$$
When $n \geqslant 2$,
$$
f_{n}(x)=f\left(f_{n-1}(x)\right)(x \in[0,1]) \text {. }
$$
Then the number of r... | $2.4^{2012}$.
It is easy to know that the domain and range of the function $f(x)$ are both the interval $[0,1]$. From the graph of $f(x)$, we know that the domain of $f(x)$ can be divided into 4 monotonic intervals, meaning that when $x$ runs through $[0,1]$ once, $f(x)$ runs through $[0,1]$ four times.
Further inducti... | 4^{2012} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,534 |
3. Let $\left\{a_{n}\right\}$ be a geometric sequence with real terms, the sum of the first $k$ terms is $S_{k}$, and the common ratio $q$ satisfies $|q| \neq 1$. If $S_{6 n}=$ $2 S_{4 n}+11 S_{2 n}$, then $\frac{S_{10 n}}{S_{8 n}}=$ $\qquad$ | 3. $\frac{341}{85}$.
Let $S_{k}=A\left(1-B^{k}\right)(A \neq 0,|B| \neq 1)$.
Substitute $S_{6 n}=2 S_{4 n}+11 S_{2 n}$, we get
$1-B^{6 n}=2\left(1-B^{4 n}\right)+11\left(1-B^{2 n}\right)$.
Let $B^{2 n}=x>0$. Then
$1-x^{3}=2\left(1-x^{2}\right)+11(1-x)(x \neq 1)$
$\Rightarrow 1+x+x^{2}=2(1+x)+11$
$\Rightarrow x=4$.
Thu... | \frac{341}{85} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,535 |
4. On two concentric circles with radii of 3 and $\sqrt{2}$, respectively, a point $P$ and a point $Q$ are randomly taken. Then the probability that the distance $P Q$ between these two points is no greater than $\sqrt{17}$ is $\qquad$ | 4. $\frac{3}{4}$.
Assume the center of the concentric circles is at the origin of the coordinate system, and let
$$
P(3 \cos x, 3 \sin x), Q(\sqrt{2} \cos y, \sqrt{2} \sin y) \text {, }
$$
where $x, y \in [0, 2\pi]$.
$$
\begin{array}{l}
\text { Then } P Q \leqslant \sqrt{17} \Leftrightarrow P Q^{2} \leqslant 17 \\
\L... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,536 |
Example 5 Find all prime numbers $p$ that satisfy the following condition: for any prime number $q$ less than $p$, the remainder $r$ when $p$ is divided by $q$ has no square factor greater than 1. ${ }^{[3]}$
(1999, IMO China National Training Team Selection Exam) | 【Analysis】Select simple prime numbers for mathematical experiments.
When $p=2$, there is obviously no prime number $q$ less than 2, so the conclusion holds.
When $p=3$, $q=2, r=1$, the conclusion holds.
When $p=5$, if $q=2$, then $r=1$;
if $q=3$, then $r=2$.
The conclusion holds.
When $p=7$, if $q=2$, then $r=1$;
if $q... | 2,3,5,7,13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,537 |
5. Given that $f(x)$ is a function defined on $\mathbf{R}$. If $f(0)=0$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+4)-f(x) \leqslant x^{2}, \\
f(x+16)-f(x) \geqslant 4 x^{2}+48 x+224,
\end{array}
$$
then $f(64)=$ $\qquad$ | 5. 19840.
Notice that,
$$
\begin{array}{l}
f(x+4)-f(x) \\
=(f(x+16)-f(x))-(f(x+16)- \\
f(x+12))-(f(x+12)-f(x+8))- \\
(f(x+8)-f(x+4)) \\
\geqslant\left(4 x^{2}+48 x+224\right)-(x+12)^{2}- \\
(x+8)^{2}-(x+4)^{2} \\
= x^{2} .
\end{array}
$$
Since $f(x+4)-f(x) \leqslant x^{2}$, we have,
$$
\begin{array}{l}
f(x+4)-f(x)=x... | 19840 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,538 |
6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{0}=0, \\
a_{n}=a_{n-1}^{2}+2 n a_{n-1}+n^{2}-n-1\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Then $S_{n}=\sum_{k=1}^{n} a_{k} \mathrm{C}_{n}^{k}=$ $\qquad$ | 6. $-2^{n-1} n$.
From the recurrence relation, we get
$$
\begin{array}{l}
a_{n}+(n+1)=\left(a_{n-1}+n\right)^{2} \\
=\left[a_{n-2}+(n-1)\right]^{2 \times 2}=\cdots \\
=\left(a_{0}+1\right)^{2 n}=1
\end{array}
$$
Thus, $a_{n}=-n$.
Then, $S_{n}=-\sum_{k=1}^{n} k \mathrm{C}_{n}^{k}=-\sum_{k=1}^{n} n \mathrm{C}_{n-1}^{k-... | -2^{n-1} n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,539 |
7. The line $x+y-1=0$ passing through the point $M(1,0)$ intersects the circle $x^{2}+y^{2}=r^{2}$ at points $A$ and $B$, and $A M=2 M B$. Then the radius of the circle $r=$ $\qquad$ | 7. $\sqrt{5}$.
Let the intersection points of the circle with the $x$-axis be
$$
P(-r, 0), Q(r, 0) \text {. }
$$
Then $M A \cdot M B = M P \cdot M Q$.
Since $A M = 2 M B$, we have,
$$
\begin{array}{l}
2 M B^{2} = M P \cdot M Q \\
= (r+1)(r-1) = r^{2} - 1, \\
A B = 3 M B = 3 \sqrt{\frac{r^{2} - 1}{2}} .
\end{array}
$$... | \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,540 |
8. $E$ and $F$ are the midpoints of edges $BD$ and $CD$ of the regular tetrahedron $ABCD$, respectively. The cosine value of the dihedral angle formed by plane $ABC$ and plane $AEF$ is $\qquad$ | 8. $\frac{5 \sqrt{33}}{33}$.
Let the edge length of a regular tetrahedron be 1, and the dihedral angle between plane $ABC$ and plane $AEF$ be $\theta$.
It is easy to know that $EF \parallel BC$.
Draw a line $AP$ through point $A$ parallel to $BC$.
Then $EF \parallel BC \parallel AP$, and $AP$ is the edge of the dihedr... | \frac{5 \sqrt{33}}{33} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,541 |
9. (16 points) Given that $f(x)$ is a function defined on the set of real numbers $\mathbf{R}$, $f(0)=2$, and for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(5+2 x)=f(-5-4 x), \\
f(3 x-2)=f(5-6 x) .
\end{array}
$$
Find the value of $f(2012)$. | In equation (1), let $x=\frac{1}{2} y-\frac{3}{2}(y \in \mathbf{R})$, we get $f(2+y)=f(1-2 y)$.
In equation (2), let $x=\frac{1}{3} y+\frac{2}{3}(y \in \mathbf{R})$, we get $f(y)=f(1-2 y)$.
Thus, $f(2+y)=f(y)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2012)=f(2 \times 1006+0)=f(0)=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,542 |
10. (20 points) Solve the equation in the set of real numbers
$$
\sqrt{3}\left(x^{2}-\frac{1}{x^{2}}\right)-4\left(x+\frac{1}{x}\right)=8
$$ | 10. Let $x=\tan \theta\left(\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right)$. Then the equation can be transformed into
$$
\begin{array}{l}
\sqrt{3}\left(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}-\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\right)-4\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin ... | x_{1}=-1, x_{2}=2+\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,543 |
11. (20 points) Let $A$, $B$, and $C$ be three distinct points on the parabola $4 x^{2} = y + 1$. If the lines $AB$ and $AC$ are both tangent to the parabola $x^{2} = y$, prove that the line $BC$ is also tangent to the parabola $x^{2} = y$. | 11. Let the parabola $4 x^{2}=y+1$ have three distinct points $A\left(p, 4 p^{2}-1\right), B\left(q, 4 q^{2}-1\right)$, $C\left(r, 4 r^{2}-1\right)(p, q, r$ are distinct $)$.
Then the equation of the line $A B$ is
$$
\begin{array}{l}
(x-p)\left(4 p^{2}-4 q^{2}\right)=\left(y-4 p^{2}+1\right)(p-q) \\
\Leftrightarrow y=4... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,544 |
One, (40 points) As shown in Figure 1, let the extensions of the two pairs of opposite sides of the convex quadrilateral $ABCD$ intersect at points $P$ and $Q$, respectively. The angle bisector of $\angle BPC$ intersects $AD$ and $BC$ at points $E$ and $G$, respectively. The angle bisector of $\angle AQB$ intersects $A... | Connect $A C$. Let $Q T$ intersect $G F$, $P C$ intersect $F G$, and $P G$ intersect $Q F$ at points $M$, $W$, and $O$ respectively.
By lines $P F$, $P W$ intersecting $\triangle E M G$ respectively, we get
$$
\begin{array}{l}
\frac{E A}{A M} \cdot \frac{M F}{F G} \cdot \frac{G P}{P E}=1, \\
\frac{E D}{D M} \cdot \frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,545 |
$$
\begin{array}{l}
\text { II. (40 points) Find } \\
f(x, y) \\
=\sqrt{\cos 4 x+7}+\sqrt{\cos 4 y+7}+ \\
\sqrt{\cos 4 x+\cos 4 y-8 \sin ^{2} x \cdot \sin ^{2} y+6}
\end{array}
$$
the minimum value. | II. Notice that,
$$
\begin{array}{l}
\cos 4 x+7=2 \cos ^{2} 2 x+6 \\
=2\left(2 \cos ^{2} x-1\right)^{2}+6 \\
=8\left(\cos ^{4} x-\cos ^{2} x+1\right) .
\end{array}
$$
Similarly, $\cos 4 y+7=8\left(\cos ^{4} y-\cos ^{2} y+1\right)$.
$$
\begin{array}{l}
\text { and } \cos 4 x+\cos 4 y-8 \sin ^{2} x \cdot \sin ^{2} y+6 \... | 4 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,546 |
Three. (50 points) Given an infinite sequence $\left\{a_{n}\right\}$ with all terms being integers, satisfying: $p, q, r \in \mathbf{Z}$,
$$
\begin{array}{l}
a_{1}=p^{2}+q^{2}+r^{2}+1, \\
a_{2}=p^{2}-q^{2}+r^{2}-2, \\
a_{3}=p^{2}-5 q^{2}-6 r^{2}+3, \\
a_{4}=3 p^{2}-11 q^{2}-18 r^{2}+15, \\
a_{n}=3 a_{n-1}-2 a_{n-2}+2 a... | Consider the sequence $\left\{a_{n}\right\}$ and its modulo $m$ residue sequence $\left\{b_{n}\right\}$. Consider the sequence of quadruples formed by the terms of $\left\{b_{n}\right\}$:
$$
\begin{array}{l}
\left(b_{1}, b_{2}, b_{3}, b_{4}\right),\left(b_{2}, b_{3}, b_{4}, b_{5}\right), \cdots, \\
\left(b_{m^{4}+1}, b... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,547 |
Example 1 Given that $x, y, z$ are three non-negative real numbers, satisfying
$$
3 x+2 y+z=5, x+y-z=2 \text{. }
$$
If $S=2 x+y-z$, then the sum of the maximum and minimum values of $S$ is ( ).
(A) 5
(B) 6
(C) 7
(D) 8
(2008, National Junior High School Mathematics League Sichuan Preliminary) | Solve the system of equations $\left\{\begin{array}{l}3 x+2 y+z=5, \\ x+y-z=2 .\end{array}\right.$
The solution is $x=\frac{7-3 y}{4}, z=\frac{y-1}{4}$.
From the problem, we know
$$
x=\frac{7-3 y}{4} \geqslant 0, y \geqslant 0, z=\frac{y-1}{4} \geqslant 0 \text {. }
$$
Solving, we get $1 \leqslant y \leqslant \frac{7}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,548 |
Example 2 Let $x, y$ be real numbers. Then the minimum value of the algebraic expression
$$
2 x^{2}+4 x y+5 y^{2}-4 x+2 y-5
$$
is $\qquad$ [1]
(2005, National Junior High School Mathematics League Wuhan CASIO Cup Selection Competition) | $$
\begin{array}{l}
\text { Original expression }= x^{2}+4 x y+4 y^{2}+x^{2}-4 x+4+ \\
y^{2}+2 y+1-10 \\
=(x+2 y)^{2}+(x-2)^{2}+(y+1)^{2}-10 .
\end{array}
$$
Therefore, when $x=2, y=-1$, the minimum value of the required algebraic expression is -10. | -10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,549 |
5. Given $m>0$. If the function
$$
f(x)=x+\sqrt{100-m x}
$$
has a maximum value of $g(m)$, find the minimum value of $g(m)$.
(2011, National High School Mathematics League Sichuan Province Preliminary Contest) | First, use the discriminant method to find $g(m)$.
Notice that the original function is $y-x=\sqrt{100-m x}$. Squaring both sides and rearranging, we get
$$
x^{2}+(m-2 y) x+y^{2}-100=0 \text {. }
$$
By $\Delta \geqslant 0$, we have $y \leqslant \frac{m}{4}+\frac{100}{m}$.
Thus, $g(m)=\frac{m}{4}+\frac{100}{m}$
$$
\geq... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,551 |
The proposition that the three chords $A D, B E, C F$ in a circle are concurrent (considering three parallel lines as concurrent at the point at infinity) is that the necessary and sufficient condition is $\frac{A B}{B C} \cdot \frac{C D}{D E} \cdot \frac{E F}{F A}=1$. | To prove that due to the proposition not specifying the arrangement order of the endpoints of the three chords on the circumference, the possible scenarios can be summarized into 8 types. Figures 1, 2, and 3 are three of these scenarios (Figure 2 is obtained by swapping points \( B \) and \( E \) in Figure 1).
Let \( ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,552 |
Example 1 As shown in Figure $4, P$ and $Q$ are the midpoints of the diagonals $A C$ and $B D$ of the cyclic quadrilateral $A B C D$. If $\angle B P A=$ $\angle D P A$, prove: $\angle A Q B=\angle C Q B .{ }^{[1]}$
(2011, National High School Mathematics Competition) | Proof As shown in Figure 4, extend $DP$ and $BP$ to intersect the circle at points $E$ and $F$, respectively, and let $AC$ and $BD$ intersect at point $K$.
Since $\angle CPE = \angle DPA = \angle BPA$, and $P$ is the midpoint of chord $AC$, we know that $AB = EC$.
Similarly, $CF = DA$.
For the cyclic hexagon $ABECFD$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,553 |
Example 2 Given that $AB$ is a chord of $\odot O$, point $P$ is on the extension of $AB$, $C$ is a point on $\odot O$ such that $PC$ is tangent to $\odot O$, and diameter $CD$ intersects $AB$ at a point inside $\odot O$. Let $DB$ intersect $OP$ at point $E$. Prove: $AC \perp CE .^{[2]}$
$(2008$, Singapore National Team... | Prove that in the figure, $PO$ and its extension intersect $\odot O$ at points $G$ and $H$, and the extension of $AO$ intersects $\odot O$ at point $F$.
To prove $AC \perp CE$, i.e., $\angle ACE = 90^{\circ}$, it suffices to prove that $F, E, C$ are collinear, which means proving that $GH, FC, DB$ are concurrent.
For... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,554 |
Example 3 As shown in Figure 6, let $\odot O_{1}$ and $\odot O_{2}$ intersect, and $P$ be one of their intersection points. A common external tangent to $\odot O_{1}$ and $\odot O_{2}$ touches the circles at points $A$ and $B$, respectively. A line through point $A$ and perpendicular to $BP$ intersects the line segment... | Let line $AB$ intersect $O_{1}O_{2}$ at point $S$ (if $AB \parallel O_{1}O_{2}$, then consider $S$ as the point at infinity), and let $SP$ intersect $\odot O_{1}$ and $\odot O_{2}$ at points $H$ and $G$ respectively.
Since the homothetic center of $\odot O_{1}$ and $\odot O_{2}$ is $S$, we have
$AP \parallel BG$.
Thus,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,555 |
Example 4 Let trapezoid $ABCD$ be inscribed in circle $\omega$, satisfying $AB \parallel CD, G$ be a point inside $\triangle BCD$, ray $AG, BG$ intersect circle $\omega$ at points $P, Q$ respectively, the line through $G$ parallel to $AB$ intersects $BD, BC$ at points $R, S$ respectively. Prove that $P, Q, R, S$ are co... | Proof of Necessity.
As shown in Figure 7, let the rays $C G$ and $D G$ intersect the circle $\omega$ at points $E$ and $F$, respectively, and let $Q E$ and $B D$, $Q F$ and $B C$ intersect at points $R^{\prime}$ and $S^{\prime}$, respectively.
For the cyclic hexagon $B D F Q E C$, by Pascal's theorem, the points $R^{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,556 |
Example 5 As shown in Figure $9, \triangle A B C$ has an incircle that touches $B C, C A, A B$ at points $D, E, F$ respectively. $P$ is a point inside the incircle, and line segments $P A, P B, P C$ intersect the incircle at points $X, Y, Z$ respectively. Prove that $X D, Y E, Z F$ are concurrent. ${ }^{[4]}$
(7th Nort... | Proof As shown in Figure 9, connect $F X, X E, E Z, Z D, D Y, Y F$. Let the diameter of the incircle of $\triangle A B C$ be $d$.
From $\frac{F X}{\sin \angle 1}=\frac{A X}{\sin \angle A F X}, \sin \angle A F X=\frac{F X}{d}$
$\Rightarrow \sin \angle 1=\frac{F X^{2}}{d A X}$.
Similarly, $\sin \angle 2=\frac{X E^{2}}{d ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,557 |
Example 6. Given $\triangle A B C$ with the incircle $\odot I$ touching $B C$ at point $D, K$ is any point on $A D, B K 、 C K$ intersect $\odot I$ at points $E 、 F$. Prove: $B F 、 A D 、 C F$ are concurrent. | Proof As shown in Figure 10, draw auxiliary lines.
From Example 5, we know that $M F$, $D P$, and $N E$ are concurrent, denote this point as $Q$.
$$
\begin{array}{l}
\text { Then } \frac{S_{\triangle P E N}}{S_{\triangle D E N}}=\frac{P Q}{D Q}=\frac{S_{\triangle P M F}}{S_{\triangle D M F}} \\
\Rightarrow \frac{P E \c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,558 |
1. Given a quadrilateral $ABCD$ inscribed in $\odot O$, the extensions of sides $AB$ and $DC$ intersect at point $P$, and the extensions of sides $AD$ and $BC$ intersect at point $Q$. Two tangents are drawn from $Q$ to the circle, touching the circle at points $E$ and $F$. Prove that points $P$, $E$, and $F$ are collin... | To prove that $F E$, $C D$, and $A B$ are concurrent, for the cyclic hexagon $F C A E D B$, it is sufficient to show that
$$
\frac{F C}{C B} \cdot \frac{B D}{D E} \cdot \frac{E A}{A C}=1 .
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,559 |
Example 3 If $y=\sqrt{1-x}+\sqrt{x-\frac{1}{2}}$ has a maximum value of $a$ and a minimum value of $b$, then the value of $a^{2}+b^{2}$ is $\qquad$ [2]
(2011, "Mathematics Weekly" Cup National Junior High School Mathematics Competition) | Solve: From $1-x \geqslant 0$, and $x-\frac{1}{2} \geqslant 0$, we get
$$
\frac{1}{2} \leqslant x \leqslant 1 \text {. }
$$
Then $y^{2}=\frac{1}{2}+2 \sqrt{-x^{2}+\frac{3}{2} x-\frac{1}{2}}$
$$
=\frac{1}{2}+2 \sqrt{-\left(x-\frac{3}{4}\right)^{2}+\frac{1}{16}} \text {. }
$$
Since $\frac{1}{2}<\frac{3}{4}<1$, therefor... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,560 |
2. From a point $P$ outside $\odot O$, draw a tangent to $\odot O$, with $A$ as the point of tangency. Draw a secant line $PD$ from point $P$ intersecting $\odot O$ at points $C$ and $D$ ($PC < PD$). Let $E$ be a point on $CD$, and $AE$ intersects $\odot O$ at another point $B$. The line $BC$ intersects $PA$ at point $... | Hint: It is only necessary to prove that $D C$, $T G$, and $B A$ are concurrent. For the cyclic hexagon $D T B C G A$, it is only necessary to prove
$$
\frac{D T}{T B} \cdot \frac{B C}{C G} \cdot \frac{G A}{A D}=1 .
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,561 |
3. In $\triangle A B C$, $\angle A C B=90^{\circ}$, the incircle $\odot I$ of $\triangle A B C$ touches sides $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$, respectively. Connect $A D$, intersecting the incircle $\odot I$ at point $P$, and connect $B P$ and $C P$. If $\angle B P C=90^{\circ}$, prove:
$$
A E+A P=P... | Let $P B$ and $P C$ intersect $\odot I$ at points $H$ and $G$ respectively. From Example 6, we can assume that $B G$, $C H$, and $A D$ are concurrent at point $R$.
Let $C H$ intersect $D G$ at point $S$.
By the harmonic properties of the complete quadrilateral $C G P R B D$, we know that $C$, $S$, $R$, and $H$ form a h... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,562 |
Example 1 Given $n\left(n \in \mathbf{N}_{+}\right)$ is an even number. Prove: $\frac{1}{1!\cdot(n-1)!}+\frac{1}{3!\cdot(n-3)!}+\cdots+$ $\frac{1}{(n-1)!\cdot 1!}=\frac{2^{n-1}}{n!}$.
(2012, East China University of Science and Technology Independent Recruitment Examination) | Prove that by the binomial theorem,
$$
\begin{array}{l}
\mathrm{C}_{n}^{0}+\mathrm{C}_{n}^{1}+\cdots+\mathrm{C}_{n}^{n}=(1+1)^{n}=2^{n}, \\
\mathrm{C}_{n}^{0}-\mathrm{C}_{n}^{1}+\cdots+(-1)^{n} \mathrm{C}_{n}^{n}=(1-1)^{n}=0 .
\end{array}
$$
Subtracting the above two equations and rearranging, we get
$$
\mathrm{C}_{n}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,563 |
Example 2 Proof: For any positive integer $n$, $(1+\sqrt{2})^{n}$ can always be expressed in the form $\sqrt{s}+\sqrt{s-1}\left(s \in \mathbf{N}_{+}\right)$ (for example, $\left.(1+\sqrt{2})^{2}=\sqrt{9}+\sqrt{8}\right)$.
(2012, Joint Autonomous Admission Examination of Peking University and Other Universities) | 【Analysis】This problem also appeared in the Tsinghua University Summer Camp exam in 2010. We can consider the conjugate form of $(1+\sqrt{2})^{n}$, which is $(1-\sqrt{2})^{n}$.
Proof By the binomial theorem, we know
$$
(1+\sqrt{2})^{n}=\sqrt{a}+\sqrt{b}\left(a 、 b \in \mathbf{N}_{+}\right) \text {. }
$$
Then $(1-\sqrt... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,564 |
Example 3 There are $n$ students in a class, who are to participate in a table tennis doubles match. The class held a warm-up match first. These $n$ students paired up with each other, and each pair participated in one warm-up match exactly. Please write down all possible values of $n$, and provide a match arrangement ... | Assume that $k$ warm-up matches have been held.
Since two pairs of students appeared in one match, $\mathrm{C}_{n}^{2}=2 k \Rightarrow 4 k=n(n-1)$.
Thus, $n=4 l$ or $n=4 l+1\left(l \in \mathbf{N}_{+}\right)$.
Next, we prove that when $n=4 l$ or $n=4 l+1\left(l \in \mathbf{N}_{+}\right)$, a match schedule that meets the... | 4 l \text{ or } 4 l+1\left(l \in \mathbf{N}_{+}\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,565 |
Example 4: From the numbers $1, 2, \cdots, 2012$, select a set of numbers such that the sum of any two numbers cannot be divisible by their difference. How many such numbers can be selected at most?
(2012, Joint Autonomous Admission Examination of Peking University and Other Universities) | Solve: Divide $1,2, \cdots, 2012$ into
$$
\begin{array}{l}
(1,2,3),(4,5,6), \cdots, \\
(2008,2009,2010),(2011,2012)
\end{array}
$$
these 671 groups.
If at least 672 numbers are taken, then by the pigeonhole principle, there must be two numbers in the same group, let's say $a$ and $b$ ($a>b$).
Thus, $a-b=1$ or 2.
When ... | 671 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,566 |
Example 5 Some scientists work in a research institute. On a certain day during the 8-hour work period, each scientist visited the coffee shop at least once. It is known that for any two scientists, the total time that at least one of them was in the coffee shop is at least $x(x>4)$ hours. Find the maximum possible num... | Let the research institute have $n$ scientists. $t_{ij}$ represents the time when exactly one of the $i$-th and $j$-th scientists is in the coffee room.
Let $S=\sum_{1 \leqslant i<j \leqslant n} t_{i}$. Then $S \geqslant \mathrm{C}_{n}^{2} x$.
On the other hand, divide the 8-hour workday into finite segments $t_{1}, t_... | n \leqslant 2\left[\frac{x}{2 x-8}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,567 |
2.18 A football league consists of 18 teams playing a round-robin tournament, meaning each round the 18 teams are divided into 9 pairs, and each pair plays one match. In the next round, the teams are re-grouped to play again, for a total of 17 rounds, ensuring that each team plays against each of the other 17 teams exa... | First, construct a tournament scheme: number the 18 teams sequentially as $1, 2, \cdots, 18$, and form two subsets according to odd and even numbers:
$$
A=\{1,3, \cdots, 17\}, B=\{2,4, \cdots, 18\},
$$
Then, the teams $(x, y)$ for each match are assigned to rounds based on the remainder when $x+y$ is divided by 9. Fir... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,569 |
3. Consider a complete graph with $n$ vertices. The vertices and edges of this complete graph are colored according to the following rules:
(1) Two edges emanating from the same vertex have different colors;
(2) The color of each vertex is different from the colors of the edges emanating from it.
For each fixed $n$, f... | Obviously, at least $n$ colors are needed, because the $n-1$ edges extending from a single point must all have different colors, and they must also be different from the color of that point.
A coloring scheme that meets the conditions of the problem using exactly $n$ colors does exist.
Let the $n$ points be $v_{0}, v... | n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,570 |
Example 4 Algebraic expression
$$
\sqrt{9 x^{2}+4}+\sqrt{9 x^{2}-12 x y+4 y^{2}+1}+\sqrt{4 y^{2}-16 y+20}
$$
reaches its minimum value when $x$ and $y$ are $\qquad$ .
(2005, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | Solve as shown in Figure 1.
Original expression
\[
\begin{aligned}
= & \sqrt{\left[0-(-2)^{2}\right]^{2}+(3 x-0)^{2}}+ \\
& \sqrt{(1-0)^{2}+(2 y-3 x)^{2}}+ \\
& \sqrt{(3-1)^{2}+(4-2 y)^{2}} \\
= & A B+B C+C D \geqslant A D,
\end{aligned}
\]
where, \( A(-2,0), B(0,3 x), C(1,2 y), D(3,4) \), and when points \( B \) and ... | x=\frac{8}{15}, y=\frac{6}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,571 |
Proposition 2 If $m$ is a $\mathrm{T}$-type number, then $k^{2} m$ and $3 m$ are both $\mathrm{T}$-type numbers; the product of two $\mathrm{T}$-type numbers is still a $\mathrm{T}$-type number. | Proposition 2 Proof If
$$
m=a^{2}+a b+b^{2}(a<b),
$$
then $k^{2} m=(k a)^{2}+(k a)(k b)+(k b)^{2}$,
$$
\begin{array}{l}
3 m=3\left(a^{2}+a b+b^{2}\right) \\
=(2 a+b)^{2}+(2 a+b)(b-a)+(b-a)^{2}.
\end{array}
$$
Let $m=a^{2}+a b+b^{2}, n=x^{2}+x y+y^{2}$.
(1) When $m=n$, we have
$$
\begin{array}{l}
m n=m^{2}=\left(a^{2}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,572 |
Proposition 3 Any positive integer of the form $3 n-1$, is not a $\mathrm{T}$-type number (or referred to as “non-$\mathrm{T}$-type number”).
The translation maintains the original text's line breaks and format. | Proof of Proposition 3 by contradiction.
If $3 n-1=a^{2}+a b+b^{2}$, take both sides modulo 3.
(1) If $a \equiv b(\bmod 3)$, then
$$
a^{2}+a b+b^{2} \equiv 0(\bmod 3),
$$
which is a contradiction;
(2) If $a \equiv \pm 1(\bmod 3), b \equiv 0(\bmod 3)$, or $a \equiv 1(\bmod 3), b \equiv-1(\bmod 3)$, then
$$
a^{2}+a b+b^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,573 |
Proposition 4 If in the factorization of a positive integer $m$, there is an odd power of a prime of the form $3 n-1$, then $m$ is not a $\mathrm{T}$-type number. | Proposition 4 Proof Let $p=3n-1$,
$$
m=p^{2k+1}m_{1}=p^{2k} \cdot pm_{1}\left(\left(p, m_{1}\right)=1\right) \text{. }
$$
Suppose $pm_{1}$ is a $\mathrm{T}$-type number, i.e., $pm_{1}=a^{2}+ab+b^{2}$, then $p \mid 4pm_{1}$.
And $4pm_{1}=(2a+b)^{2}+3b^{2}=x^{2}+3b^{2}$, where $x=2a+b$.
By $p \mid (x^{2}+3b^{2})$, if ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,574 |
Proposition 6 Let $(a, b)=1$, for the $\mathrm{T}$-type number $m=a^{2}+a b+b^{2}$,
there exist a $\mathrm{T}$-type number $n=x^{2}+x y+y^{2}$ and a positive integer $k$, such that $m n=k^{2}+k+1$. | Proposition 6 Proof: If $a=b=1$, then $m=3$. In this case, there exists $n=2^{2}+2+1=7, m n=21=4^{2}+4+1$, i.e., $k=4$ satisfies the condition.
Assume $1 \leqslant a<b$. Since $(a, b)=1$, there exist positive integers $x, y$ such that $b y-a x=1$. Take $n=x^{2}+x y+y^{2}$, then
$$
\begin{array}{l}
m n=\left(a^{2}+a b+... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,576 |
Proposition 7 Let $(a, b)=1$, for positive integer $k$ and $\mathrm{T}$-type number $n=a^{2}+a b+b^{2}$, if there exists a positive integer $m$ satisfying $m n=k^{2}+k+1$, then $m$ is also a $\mathrm{T}$-type number, i.e., there exist positive integers $x, y$ such that $m=x^{2}+x y+y^{2}$. | The proof of Proposition 7 is as follows:
$$
m n^{2}=\left(k^{2}+k+1\right)\left(a^{2}+a b+b^{2}\right)
$$
is a $\mathrm{T}$-type number. Suppose $m=r^{2} m_{1}$ (where $m_{1}$ does not contain any square factor greater than 1), then $m_{1}(r n)^{2}$ is a $\mathrm{T}$-type number.
By Proposition 4, we know that $m_{1}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,577 |
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