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 |
|---|---|---|---|---|---|---|---|---|---|
2. As shown in Figure 1, line $k \parallel l$,
$$
\begin{array}{l}
\angle 4-\angle 3=\angle 3-\angle 2 \\
=\angle 2-\angle 1=d>0, \text { where } \\
\text { among, } \angle 3<90^{\circ}, \angle 1=50^{\circ} .
\end{array}
$$
Then the maximum possible integer value of $\angle 4$ is ( ).
(A) $107^{\circ}$
(B) $108^{\circ... | 2. C.
$$
\begin{array}{l}
\text { Given } \angle 3=50^{\circ}+2 d<90^{\circ} \Rightarrow d < 20^{\circ} \\
\Rightarrow \angle 4=50^{\circ}+3 d<50^{\circ}+3 \times 20^{\circ}=110^{\circ}
\end{array}
$$
$\Rightarrow \angle 4$ The maximum possible integer value is $109^{\circ}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,511 |
3. Let $p$ be a prime number. Then the number of integer pairs $(a, b)$ that satisfy
$$
|a+b|+(a-b)^{2}=p
$$
is ( ) pairs.
(A) 3
(B) 4
(C) 5
(D) 6 | 3. D.
Since $a+b$ and $a-b$ have the same parity, we have
$$
p=2 \text {. }
$$
Thus, the integer pairs $(a, b)$ are
$$
\begin{array}{l}
(1,1),(-1,-1),(0,1), \\
(1,0),(-1,0),(0,-1) .
\end{array}
$$
There are 6 pairs in total. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,512 |
4. Let the three sides of $\triangle A B C$ be $B C=2, C A$ $=3, A B=4, h_{a}, h_{b}, h_{c}$ represent the altitudes to sides $B C, C A, A B$ respectively. Then
$$
\left(h_{a}+h_{b}+h_{c}\right)\left(\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right)=(\quad) .
$$
(A) $\frac{41}{6}$
(B) $\frac{39}{4}$
(C) $\frac{38}... | 4. B.
Let the area of $\triangle A B C$ be $S$. Then
$$
\begin{array}{l}
h_{a}=\frac{2 S}{a}, h_{b}=\frac{2 S}{b}, h_{c}=\frac{2 S}{c} . \\
\text { Hence }\left(h_{a}+h_{b}+h_{c}\right)\left(\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right) \\
=\left(\frac{2 S}{a}+\frac{2 S}{b}+\frac{2 S}{c}\right)\left(\frac{a}{... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,513 |
5. As shown in Figure 2, the square $A B C D$ is divided into two equal areas by the line $O E$. Given that the lengths of segments $O D$ and $A D$ are positive integers, and $\frac{C E}{B E}=20$. Then the minimum value of the area of the square $A B C D$ that satisfies the above conditions is ( ).
(A) 324
(B) 331
(C) ... | 5. D.
As shown in Figure 6, since the square $ABCD$ is divided into two equal areas by the line $OE$, the line $OE$ passes through the center $P$ of the square.
Therefore, $BE = GD$.
Let $OD = n$, $AD = m$. Then
$$
\frac{CE}{BE} = \frac{CE}{GD} = \frac{OC}{OD} = \frac{n + m}{n} = 20 \text{.}
$$
Thus, $m = 19n \geq 1... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,514 |
1. As shown in Figure 3, it is known that
$$
\begin{array}{l}
A B=2, B C=A E=6, \\
C E=C F=7, B F=8 .
\end{array}
$$
Then the ratio of the area of quadrilateral $A B D E$ to the area of $\triangle C D F$ is $\qquad$ . | $=1.1$.
Since $A C=B F=8, C E=C F=7, B C=A E=6$, therefore, $\triangle A E C \cong \triangle B C F$. $=S_{\triangle A C F}-S_{\triangle B D C}=S_{\triangle C D F}$. | 1.1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,515 |
Example 5 Given
$\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$, and $a \neq 0$.
Then $\frac{b+c}{a}=$ $\qquad$ | From the given equation, we have
$$
(b-c)^{2}-4(a-b)(c-a)=0 \text {. }
$$
When $a \neq b$, by the discriminant of a quadratic equation, we know that the quadratic equation in $x$
$$
(a-b) x^{2}+(b-c) x+(c-a)=0
$$
has two equal real roots.
$$
\text { Also, }(a-b)+(b-c)+(c-a)=0 \text {, thus, }
$$
$x=1$ is a root of eq... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,516 |
3. As shown in Figure 4, in quadrilateral $ABCD$, let $\angle BAD + \angle ADC = 270^{\circ}$, and $E, F$ are the midpoints of $AD, BC$ respectively, $EF=4$, the shaded parts are semicircles with diameters $AB, CD$ respectively. Then the sum of the areas of these two semicircles is $\qquad$ (the value of pi is $\pi$). | $3.8 \pi$.
As shown in Figure 7, extend $B A$ and $C D$ to intersect at point $M$.
From $\angle B A D + \angle A D C = 270^{\circ}$, we get
$$
\begin{array}{l}
\angle B M C \\
= \angle A M D = 90^{\circ}.
\end{array}
$$
Connect $B D$ and take the midpoint $P$ of $B D$, then connect $P E$ and $P F$.
By the midline th... | 8 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,518 |
$$
\begin{array}{l}
\frac{1}{1 \times 2010}+\frac{1}{2 \times 2009}+\cdots+\frac{1}{2010 \times 1}- \\
\frac{2010}{2011}\left(\frac{1}{1 \times 2009}+\frac{1}{2 \times 2008}+\cdots+\frac{1}{2009 \times 1}\right) \\
=
\end{array}
$$ | 4. $\frac{1}{2021055}$.
For $k=2,3, \cdots, 2010$, we have
$$
\begin{array}{l}
\frac{1}{k(2011-k)}-\frac{2010}{2011} \cdot \frac{1}{(k-1)(2011-k)} \\
=\frac{1}{2011}\left[\left(\frac{1}{k}+\frac{1}{2011-k}\right)-\left(\frac{1}{k-1}+\frac{1}{2011-k}\right)\right] \\
=\frac{1}{2011}\left(\frac{1}{k}-\frac{1}{k-1}\right... | \frac{1}{2021055} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,519 |
5. As shown in Figure 5, in a square $ABCD$ with a side length of 10, six identical smaller squares are inscribed, with vertices $P, Q, M, N$ of the smaller squares lying on the sides of the larger square. Then the sum of the areas of these six smaller squares is | 5.32. 64 .
As shown in Figure 8, passing through
each vertex of the small squares
draw parallel lines to each side,
forming a "string diagram",
where the longer leg of the small right
triangle is $a$, and the shorter leg is $b$. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
2 a+5 b=10, \\
5 a=10
\end{array}\right. ... | 32.64 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,520 |
Three, (10 points) In the convex pentagon $A B C D E$,
$$
\begin{array}{l}
A B=B C=C D=D E=E A, \\
\angle A B C=2 \angle D B E .
\end{array}
$$
Prove: $\angle A B C=60^{\circ}$. | Three, because $\angle A B C=2 \angle D B E$, so,
$$
\angle D B E=\angle A B E+\angle C B D \text {. }
$$
As shown in Figure 9, draw $B P / / A E$, intersecting
$D E$ at point $P$.
Given $A B=A E$,
we know $\angle P B E=\angle A E B$
$=\angle A B E$.
Therefore, $\angle C B D=\angle D B P$.
But from $B C=C D$, we get ... | 60^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 723,521 |
Four, (15 points) Can 2010 be written as the sum of squares of $k$ distinct prime numbers? If so, try to find the maximum value of $k$; if not, please briefly explain the reason. | If 2010 can be written as the sum of squares of $k$ prime numbers, taking the sum of the squares of the smallest 10 distinct primes, then
$$
\begin{array}{l}
4+9+25+49+121+169+ \\
289+361+529+841 \\
=2397>2010 .
\end{array}
$$
Therefore, $k \leqslant 9$.
It is easy to see that there is only one even prime number 2, an... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,522 |
Five. (15 points) In a math competition for second-year junior high school students, a total of 99 middle schools registered to participate, with both male and female participants from each school. Prove: there exist 50 of these schools whose total number of male participants is not less than half of the total number o... | Five, number the participating middle schools as $1,2, \cdots, 99$, and let $x_{i}$ $(i=1,2, \cdots, 99)$ represent the number of male participants from the $i$-th school. Assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{99}$.
Divide the schools numbered $2 \sim 99$ into two groups.
The first group consist... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,523 |
1. Given $\left\{a_{n}\right\}$ is an arithmetic sequence, $S_{n}$ is the sum of its first $n$ terms. Then $-a_{m}<0$, $S_{m+1}<0$ is a ( ) condition.
(A) Sufficient and necessary
(B) Sufficient but not necessary
(C) Necessary but not sufficient
(D) Neither sufficient nor necessary | -,1. A.
In fact,
$$
\begin{array}{l}
\left\{\begin{array}{l}
S_{m}=\frac{m}{2}\left(a_{1}+a_{m}\right)>0, \\
S_{m+1}=\frac{m+1}{2}\left(a_{1}+a_{m+1}\right)<0
\end{array}\right. \\
\Leftrightarrow a_{1}+a_{m+1}<0<a_{1}+a_{m}
\end{array} \Leftrightarrow
$$
Therefore, it is a sufficient and necessary condition. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,524 |
2. Given the function
$$
f(x)=x^{3}+(a+1) x^{2}+(a+1) x+a
$$
has both a maximum and a minimum value in its domain. Then the range of the real number $a$ is ( ).
(A) $a<-12$
(C) $a>2$ or $a<-1$ | 2. D.
Let $f(x)=x^{3}+(a+1) x^{2}+(a+1) x+a$, then $f^{\prime}(x)=3 x^{2}+2(a+1) x+(a+1)$.
From the given condition, we know that $f^{\prime}(x)=0$ must have two distinct real roots.
Therefore, $\Delta=4(a+1)^{2}-12(a+1)>0$.
Solving this, we get $a>2$ or $a<-1$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,525 |
3. If the set $M=\left\{x \left\lvert\, \frac{|3-x|}{|5-x|} \leqslant \frac{1}{2}\right.\right\}$ and the set $N=\left\{x \mid x^{2}-2 x+c \leqslant 0\right\}$ satisfy $M \cap N=M$, then the range of the real number $c$ is ( ).
(A) $c \leqslant-\frac{44}{9}$
(B) $c \leqslant-\frac{55}{9}$
(C) $c \leqslant-\frac{66}{9}$... | 3. B.
$$
\begin{array}{l}
\text { Given } M=\left\{x \neq 5 \mid 4(3-x)^{2} \leqslant(5-x)^{2}\right\} \\
=\left\{x \neq 5 \mid 3 x^{2}-14 x+11 \leqslant 0\right\} \\
=\left\{x \left\lvert\, 1 \leqslant x \leqslant \frac{11}{3}\right.\right\}, \\
N=\{x \mid 1-\sqrt{1-c} \leqslant x \leqslant 1+\sqrt{1-c}\},
\end{array}... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 723,526 |
Example 6 Given that the three numbers $a$, $b$, and $c$ satisfy the system of equations
$$
\left\{\begin{array}{l}
a+b=8, \\
a b-c^{2}+8 \sqrt{2} c=48 .
\end{array}\right.
$$
Try to find the roots of the equation $b x^{2}+c x-a=0$. | 【Analysis】First, find the values of $ab$ and $a+b$ (expressed as algebraic expressions involving $c$), then construct a quadratic equation based on the relationship between roots and coefficients. Use the discriminant $\Delta \geqslant 0$ to determine the value of $c$, and subsequently find the values of $a$ and $b$.
S... | x_{1}=\frac{-\sqrt{2}+\sqrt{6}}{2}, \quad x_{2}=\frac{-\sqrt{2}-\sqrt{6}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,527 |
4. Given $-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$,
$$
2 \tan \beta=\tan 2 \alpha, \tan (\beta-\alpha)=-2 \sqrt{2} \text {. }
$$
Then $\cos \alpha=(\quad)$.
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{3}}{3}$
(D) $\frac{\sqrt{2}}{3}$ | 4. C.
Let $\tan \alpha=u$.
From $\tan \beta=\frac{1}{2} \tan 2 \alpha=\frac{\tan \alpha}{1-\tan ^{2} \alpha}=\frac{u}{1-u^{2}}$, we get
$$
\begin{array}{l}
\tan (\beta-\alpha)=\frac{\tan \beta-\tan \alpha}{1+\tan \alpha \cdot \tan \beta} \\
=\frac{\frac{u}{1-u^{2}}-u}{1+u \cdot \frac{u}{1-u^{2}}}=u^{3} .
\end{array}
$... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,528 |
5. Given the set of integers
$M=\left\{m \mid x^{2}+m x-36=0\right.$ has integer solutions $\}$, set $A$ satisfies the conditions:
(1) $\varnothing \subset A \subseteq M$;
(2) If $a \in A$, then $-a \in A$.
Then the number of all such sets $A$ is ( ).
(A) 15
(B) 16
(C) 31
(D) 32 | 5. C.
Let $\alpha, \beta$ be the roots of the equation $x^{2}+m x-36=0$. Then $\alpha \beta=-36$. Therefore,
when $|\alpha|=1,|\beta|=36$, $m= \pm 35$;
when $|\alpha|=2,|\beta|=18$, $m= \pm 16$;
when $|\alpha|=3,|\beta|=12$, $m= \pm 9$;
when $|\alpha|=4,|\beta|=9$, $m= \pm 5$;
when $|\alpha|=6,|\beta|=6$, $m=0$.
Thus,... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,529 |
7. Let $z$ be a complex number, $\mathrm{i}$ be the imaginary unit. If $|z|=1$, $|\bar{z}+\mathrm{i}|=1$, then when $(z+\mathrm{i})^{n}\left(n \in \mathbf{N}_{+}\right)$ is a real number, the minimum value of $|z+i|^{n}$ is ( ).
(A) $\sqrt{3}$
(B) 3
(C) $2 \sqrt{3}$
(D) $3 \sqrt{3}$ | 7. D.
Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$.
From $|z|=1,|\bar{z}+\mathrm{i}|=1$, we get
$$
\left\{\begin{array} { l }
{ x ^ { 2 } + y ^ { 2 } = 1 , } \\
{ x ^ { 2 } + ( 1 - y ) ^ { 2 } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x= \pm \frac{\sqrt{3}}{2}, \\
y=\frac{1}{2} .
\end{array}\right.\right.
$... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,531 |
8. In the expansion of the polynomial $(a+b+c+d)^{8}$, the number of terms where the exponent of each variable is not zero is ( ) terms.
(A) 35
(B) 42
(C) 45
(D) 50 | 8. A.
Let $(a+b+c+d)^{8}$ have an arbitrary term $p a^{x_{1}} b^{x_{2}} c^{x_{3}} d^{x_{4}}$ in its expansion.
If the exponent of each letter is not zero, then
$$
\begin{array}{l}
x_{i} \geqslant 1(i=1,2,3,4), \text { and } \\
x_{1}+x_{2}+x_{3}+x_{4}=8 . \\
\text { Let } u_{i}=x_{i}-1(i=1,2,3,4) . \text { Then } \\
u_... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 723,532 |
9. As shown in Figure 1, in the tetrahedron $P-A B C$, the lateral face $P A C \perp$ the base $A B C$, the base $A B C$ is an equilateral triangle with a side length of 1, $P A=P C$, $\angle A P C=90^{\circ}$, and $M$ is the midpoint of edge $B C$. Then the distance between $A B$ and $P M$ is ( ).
(A) $\frac{\sqrt{3}}... | 9. A.
As shown in Figure 3, draw $PO \perp AC$ at point $O$, then $O$ is the midpoint of $AC$. Connect $OM$.
Since $M$ is the midpoint of $BC$, we know that $OM \parallel AB$.
Draw $MN \perp AB$ at point $N$.
It is easy to see that $MN = \frac{\sqrt{3}}{4}$.
Since $OM \parallel AB$, therefore, $MN \perp OM$.
Also, ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,533 |
10. A person needs to climb 10 steps to go from the first floor to the second floor. He can take one step at a time, called a one-step, or two steps at a time, called a two-step, or three steps at a time, called a three-step. He takes a total of 6 steps to go from the first floor to the second floor, and no two consecu... | 10. C.
According to the requirements of the problem, it is not difficult to verify that among these 6 steps, there cannot be no third-order steps, nor can there be more than 1 second-order step. Therefore, it can only be 1 third-order step, 2 second-order steps, and 3 first-order steps.
For the sake of vividness, we ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 723,534 |
11. Given the function
$$
f(x)=a x^{2}-\frac{1}{2} x-\frac{3}{4}(a>0) \text {. }
$$
If on any closed interval of length 2, there always exist two points $x_{1} 、 x_{2}$, such that $\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \geqslant \frac{1}{4}$, then the minimum value of $a$ is $\qquad$. | II. $11 \cdot \frac{1}{4}$.
On the closed interval of length 2 $\left[\frac{1}{4 a}-1, \frac{1}{4 a}+1\right]$,
$$
\begin{array}{l}
f_{\max }(x)=f\left(\frac{1}{4 a}-1\right) \\
=f\left(\frac{1}{4 a}+1\right)=a-\frac{1}{16 a}-\frac{3}{4}, \\
f_{\text {min }}(x)=f\left(\frac{1}{4 a}\right)=-\frac{1}{16 a}-\frac{3}{4} .
... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,535 |
12. Given that the orthocenter of $\triangle A B C$ is $H$. If $B(0,0)$, $C(2,0)$, and point $H$ moves on the circle
$$
(x-1)^{2}+(y+1)^{2}=2
$$
then the locus of the moving point $A$ is $\qquad$ . | 12. The circle $(x-1)^{2}+(y-1)^{2}=2(y \neq 0)$ and the lines $x=0(y \neq 0)$ or $x=2(y \neq 0)$.
Let $A(x, y), H\left(x, y_{1}\right)$.
When $x \neq 0, x \neq 2$,
$$
\frac{y_{1}}{x} \cdot \frac{y}{x-2}=-1, y_{1}=-\frac{x^{2}-2 x}{y} \text {. }
$$
Since point $H$ is on the circle $(x-1)^{2}+(y+1)^{2}=2$, we have
$$
(... | (x-1)^{2}+(y-1)^{2}=2(y \neq 0) \text{ and the lines } x=0(y \neq 0) \text{ or } x=2(y \neq 0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,536 |
13. If the function $f(x)=\ln \frac{\mathrm{e} x}{\mathrm{e}-x}$, then $\sum_{k=1}^{2010} f\left(\frac{k e}{2011}\right)=$ $\qquad$ . | 13.2010.
Notice
$$
\begin{array}{l}
f(x)+f(\mathrm{e}-x) \\
=\ln \left[\frac{\mathrm{e} x}{\mathrm{e}-x} \cdot \frac{\mathrm{e}(\mathrm{e}-x)}{\mathrm{e}-(\mathrm{e}-x)}\right]=2 . \\
\text { Therefore } \sum_{k=1}^{2010} f\left(\frac{k \mathrm{e}}{2011}\right) \\
=\sum_{k=1}^{1005}\left(f\left(\frac{k \mathrm{e}}{201... | 2010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,537 |
Example 7 The equation in terms of $x, y$
$$
x^{2}+x y+2 y^{2}=29
$$
has ( ) sets of integer solutions $(x, y)$.
(A)2
(B) 3
(C) 4
(D) infinitely many | We can consider the original equation as a quadratic equation in $x$
$$
x^{2}+y x+\left(2 y^{2}-29\right)=0 \text {. }
$$
Since the equation has integer roots, the discriminant $\Delta \geqslant 0$, and it must be a perfect square.
By $\Delta=y^{2}-4\left(2 y^{2}-29\right)$
$$
=-7 y^{2}+116 \geqslant 0 \text {, }
$$
... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,538 |
14. The center of symmetry of the graph of the function $f(x)=x^{3}+2 x^{2}+3 x+4$ is . $\qquad$ | 14. $\left(-\frac{2}{3}, \frac{70}{27}\right)$.
Let the point $(a, b)$ be the center of symmetry of the graph of the function $f(x)$.
Then $f(a+x)+f(a-x)=2 b$, that is,
$$
\begin{aligned}
2 b= & {\left[(a+x)^{3}+(a-x)^{3}\right]+} \\
& 2\left[(a+x)^{2}+(a-x)^{2}\right]+6 a+8, \\
b= & a^{3}+3 a x^{2}+2 a^{2}+2 x^{2}+3 ... | \left(-\frac{2}{3}, \frac{70}{27}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,539 |
16. (12 points) Given $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
the chord $AB$ passes through the point $F_{2}$, and
$$
\left|A F_{2}\right|=2\left|F_{2} B\right|, \tan \angle A F_{1} B=\frac{3}{4} .
$$
(1) Find the eccentricity $e$ of the ell... | 16. (1) Let $\left|A F_{2}\right|=2\left|F_{2} B\right|=2 k$.
From $\left|A F_{1}\right|+\left|A F_{2}\right|=2 a$,
$\left|B F_{1}\right|+\left|B F_{2}\right|=2 a$,
we get $\left|A F_{1}\right|=2 a-2 k,\left|B F_{1}\right|=2 a-k$.
Since $\tan \angle A F_{1} B=\frac{3}{4}$, we have
$\cos \angle A F_{1} B=\frac{4}{5}$.
... | \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,541 |
17. (12 points) Given the function $f(x)$ for any real numbers $x, y$, it satisfies $f(x+y)=f(x)+f(y)-3$, and when $x>0$, $f(x)<3$.
(1) Is $f(x)$ a monotonic function on the set of real numbers $\mathbf{R}$? Explain your reasoning;
(2) If $f(6)=-9$, find $f\left(\left(\frac{1}{2}\right)^{2010}\right)$. | 17. (1) For any $x_{1}, x_{2} \in \mathbf{R}$, and $x_{1} < x_{2}$, we have $x_{2} - x_{1} > 0$, so $f\left(x_{2}-x_{1}\right) < 3$, which means $f\left(x_{2}\right) < f\left(x_{1}\right)$.
Thus, $f(x)$ is a monotonically decreasing function on $\mathbf{R}$.
$$
\begin{array}{l}
\text { (2) From } f(6)=f(2)+f(4)-3 \\
=f... | 3-\left(\frac{1}{2}\right)^{2009} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,542 |
18. (15 points) Clothing retailers A and B wish to deal in clothing produced by a certain brand's manufacturing company. The design department of the company, without any information about A and B's sales, randomly provides them with $n$ different designs. A and B independently select the designs they approve of. What ... | 18. Let the set of $n$ styles be denoted as $V$, and the choices of Party A and Party B be denoted as
$$
P_{\text {A }} \subseteq V, P_{\text{B}} \subseteq V .
$$
The combination of Party A's and Party B's choices is referred to as a selection plan, denoted as $\left(P_{\text {A }}, P_{\text {B }}\right)$.
First, we ... | 1-\left(\frac{3}{4}\right)^{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,543 |
19. (15 points) In a certain grade, $n$ students participate in Chinese and
Mathematics exams, with scores ranging from 0 to 100 for each subject. Suppose no two students have exactly the same scores (i.e., at least one subject score is different). Additionally, “A is better than B” means that student A’s scores in bo... | 19. Establish a Cartesian coordinate system $x O y$.
If a student's Chinese score is $i$ points, and mathematics score is $j$ points, let it correspond to the integer point $(i, j)$ on the plane, called a "score point". Thus, the examination results of $n$ students are mapped to $n$ score points within the range
$$
0 ... | 401 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,544 |
1. Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
a_{n}=2^{\frac{2 n+3}{5}}, b_{n}=\frac{1}{n} \log _{2}\left(a_{1} a_{2} \cdots a_{n}\right)\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the general term formula of the sequence $\left\{b_{n}\right\}$ is | 1. $b_{n}=\frac{n+4}{5}$.
From $a_{n}=2^{\frac{2 n+3}{5}}$, we get
$a_{1} a_{2} \cdots a_{n}=2^{\frac{2(1+2+\cdots+n)+3 n}{s}}=2^{\frac{n(n+4)}{s}}$.
Therefore, $b_{n}=\frac{1}{n} \cdot \frac{n(n+4)}{5}=\frac{n+4}{5}$. | b_{n}=\frac{n+4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,545 |
2. Given points $M(0,2)$ and $N(-3,6)$, the distances from these points to line $l$ are $1$ and $4$, respectively. The number of lines $l$ that satisfy these conditions is . $\qquad$ | 2.3.
It is easy to get $M N=5$.
Then the circle $\odot M$ with radius 1 is externally tangent to the circle $\odot N$ with radius 4. Therefore, there are 3 lines $l$ that satisfy the condition. | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,546 |
3. Let the function $f(x)=a x^{2}+x$, it is known that
$$
f(3)<f(4) \text {, }
$$
then the range of the real number $a$ is $\qquad$ . | 3. $\left(-\frac{1}{7},-\frac{1}{17}\right)$.
Since when $n \geqslant 8$, $f(n)>f(n+1)$ always holds, thus, $a\frac{7}{2}$.
Solving, we get $a>-\frac{1}{7}$.
Therefore, $a \in\left(-\frac{1}{7},-\frac{1}{17}\right)$. | \left(-\frac{1}{7},-\frac{1}{17}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,547 |
4. As shown in Figure 1, it is known that $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a cube with a side length of 3, and $P, Q, R$ are points on the edges $A B, A D, A A_{1}$ respectively, with $A P=A Q=A R=1$. Then the volume of the tetrahedron $C_{1} P Q R$ is | 4. $\frac{4}{3}$.
Obviously, $A C_{1} \perp$ plane $P Q R$.
Since $A P=A Q=A R=1$, we have,
$$
\begin{array}{l}
P Q=Q R=R P=\sqrt{2} . \\
\text { Also, } A C_{1}=3 \sqrt{3}, \\
V_{A-P Q R}=\frac{1}{3} \times \frac{1}{2} \times 1^{2} \times 1=\frac{1}{6}, \\
V_{C_{1}-P Q R} \\
=\frac{1}{3} \times \frac{\sqrt{3}}{4} \ti... | \frac{4}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,548 |
Example 8 Given real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
x+y+z=5, \\
x y+y z+z x=3 .
\end{array}
$$
Then the maximum value of $z$ is | Given $x+y+z=5$, we have
$$
y=5-x-z \text {. }
$$
Substituting into $x y+y z+z x=3$ yields a quadratic equation in $x$:
$$
x^{2}+(z-5) x+z^{2}-5 z+3=0 \text {. }
$$
Since $x$ and $z$ are real numbers, we have
$$
\begin{array}{l}
\Delta=(z-5)^{2}-4\left(z^{2}-5 z+3\right) \\
=-3 z^{2}+10 z+13 \geqslant 0 .
\end{array}... | \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,549 |
5. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{n+1}=\frac{1+a_{n}}{1-a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Let $T_{n}=a_{1} a_{2} \cdots a_{n}$. Then $T_{2010}=$ $\qquad$ | 5. -6 .
It is easy to get $a_{1}=2, a_{2}=-3, a_{3}=-\frac{1}{2}, a_{4}=\frac{1}{3}$, $a_{1} a_{2} a_{3} a_{4}=1$.
Also, $a_{5}=2=a_{1}$, by induction it is easy to know $a_{n+4}=a_{n}\left(n \in \mathbf{N}_{+}\right)$.
Therefore, $T_{2010}=T_{4 \times 502+2}=a_{1} a_{2}=-6$. | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,550 |
6. A die is a cube, with the six faces engraved with $1$, $2$, $3$, $4$, $5$, $6$. There are 10 uniform dice. Throwing 4 dice and 3 dice at once, the ratio of the probabilities of getting the sum of the points on the faces of each die as 6 is | 6. 6: 1 .
The situation of rolling a 6 with three dice is
$$
1,1,4 ; 1,2,3 ; 2,2,2 \text {. }
$$
There are a total of $3+3!+1=10$ ways, with a probability of $\frac{10}{6^{3}}$.
The situation of rolling a 6 with four dice is $1,1,1,3 ; 1,1,2,2$.
There are a total of $4+\mathrm{C}_{4}^{2}=10$ ways, with a probability ... | 6: 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,551 |
7. In $\triangle A B C$, it is known that
$$
B C=5, A C=4, \cos (A-B)=\frac{7}{8} \text {. }
$$
Then $\cos C=$ | 7. $\frac{11}{16}$.
From $B C>A C$, we get $\angle A>\angle B$.
As shown in Figure 3, construct $A D$
such that $\angle B A D=\angle B$. Then
$\angle D A C$
$=\angle A-\angle B$.
Let $A D=B D=x$.
Then $D C=5-x$.
In $\triangle A D C$, by the Law of Cosines, we get $x=3$.
Again, by the Law of Cosines, we get $\cos C=\f... | \frac{11}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,552 |
8. In the Cartesian coordinate system $x O y$, the focus of the parabola $y^{2}=2 x$ is $F$. Let $M$ be a moving point on the parabola. Then the maximum value of $\frac{M O}{M F}$ is $\qquad$ . | 8. $\frac{2 \sqrt{3}}{3}$.
Let point $M(x, y)$. Then
$$
\begin{array}{l}
\left(\frac{M O}{M F}\right)^{2}=\frac{x^{2}+y^{2}}{\left(x+\frac{1}{2}\right)^{2}} \\
=\frac{4 x^{2}+8 x}{4 x^{2}+4 x+1} \\
=1+\frac{4 x-1}{4 x^{2}+4 x+1} .
\end{array}
$$
$$
\text { Let } 4 x-1=t \text {. }
$$
When $t \leqslant 0$, $\frac{M O}... | \frac{2 \sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,553 |
II. (16 points) As shown in Figure 2, $P$ is any point on the semicircle $\odot O$: $x^{2}+y^{2}=1(y \geqslant 0)$ above the $x$-axis, $A$ and $B$ are the two endpoints of the diameter. A square $ABCD$ is constructed with $AB$ as one side, and $PC$, $PD$ intersect $AB$ at points $E$ and $F$ respectively. Prove that $BE... | Let $P(\cos \alpha, \sin \alpha), C(-1,-2)$, $D(1,-2), E\left(x_{1}, 0\right), F\left(x_{2}, 0\right)$.
From the collinearity of points $P, E, C$, we have
$$
\begin{array}{l}
\frac{\sin \alpha+2}{\cos \alpha+1}=\frac{2}{x_{1}+1} . \\
\text { Thus, } x_{1}=\frac{2(\cos \alpha+1)}{\sin \alpha+2}-1 .
\end{array}
$$
Simil... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,554 |
Three. (20 points) Let real numbers $a, m$ satisfy
$$
a \leqslant 1, 0 < m \leqslant 2 \sqrt{3} \text{, }
$$
The function $f(x)=\frac{a m x - m x^{2}}{a + a(1-a)^{2} m^{2}}(x \in (0, a))$. If there exist $a, m, x$ such that $f(x) \geqslant \frac{\sqrt{3}}{2}$, find all real values of $x$. | $$
\begin{array}{l}
\text { Three, when } x \in(0, a) \text {, } \\
a m x - m x^{2} \\
= -m\left(x - \frac{a}{2}\right)^{2} + \frac{m a^{2}}{4} \leqslant \frac{m a^{2}}{4} .
\end{array}
$$
The equality holds if and only if $x = \frac{a}{2}$.
$$
\begin{array}{l}
\text { Therefore, } \frac{\sqrt{3}}{2} \leqslant \frac{a... | x = \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,555 |
$$
\begin{array}{l}
a_{1} \in(1,2), \\
a_{n+1}=a_{n}^{3}-3 a_{n}^{2}+3 a_{n}\left(n \in \mathbf{N}_{+}\right) . \\
\text {Prove: } \sum_{k=1}^{n}\left(a_{k}-a_{k+1}\right)\left(a_{k+2}-1\right)<\frac{1}{4} .
\end{array}
$$ | $$
\begin{array}{l}
a_{n+1}-1=\left(a_{n}-1\right)^{3}. \\
\text { Let } b_{n}=a_{n}-1 \text {. Then } \\
0<b_{1}<1, b_{n+1}=b_{n}^{3}<b_{n}, 0<b_{n}<1. \\
\text { Therefore, }\left(a_{k}-a_{k+1}\right)\left(a_{k+2}-1\right) \\
=\left(b_{k}-b_{k+1}\right) b_{k+2} \\
=\left(b_{k}-b_{k+1}\right) b_{k+1}^{3} \\
<\frac{1}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,556 |
One. (40 points) Given that the incircle $\odot 1$ of $\triangle A B C$ touches sides $A C$ and $A B$ at points $E$ and $F$, respectively, and $M$ is a point on segment $E F$. Prove: The necessary and sufficient condition for the areas of $\triangle M A B$ and $\triangle M A C$ to be equal is $M I \perp B C$. | As shown in Figure 4, draw $M P \perp A C$ and $M Q \perp A B$, with the feet of the perpendiculars being $P$ and $Q$ respectively. Let $\odot I$ be tangent to side $B C$ at point $D$. Then,
$I D \perp B C$,
$I F \perp A B$,
$I E \perp A C$.
Clearly, Rt $\triangle Q F M \backsim \mathrm{Rt} \triangle P E M$.
Thus, $\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,557 |
II. (40 points) Color the sides and diagonals of a convex $n$-gon $A_{1} A_{2} \cdots A_{n}$ with either red or blue, such that no triangle has all three sides blue. For $k=1,2, \cdots, n$, let $b_{k}$ be the number of blue edges emanating from vertex $A_{k}$. Prove that:
$$
b_{1}+b_{2}+\cdots+b_{n} \leqslant \frac{n^{... | Let's assume $b=\max \left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$, and from point $A$ to $A_{1}, A_{2}, \cdots, A_{b}$, $b$ blue edges are drawn. Then, there are no blue edges between $A_{1}, A_{2}, \cdots, A_{6}$, and for the $n-b$ points outside of $A_{1}, A_{2}, \cdots, A_{6}$, each point can have at most $b$ blue ed... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,558 |
Three. (50 points) Let the infinite sequence of positive integers $\left\{a_{n}\right\}$ $\left(n \in \mathbf{N}_{+}\right)$ satisfy
$$
a_{4}=4, a_{n}^{2}-a_{n-1} a_{n+1}=1(n \geqslant 2) \text {. }
$$
Find the general term formula for $\left\{a_{n}\right\}$. | Three, from the known we get $\frac{a_{n}}{a_{n+1}}>\frac{a_{n-1}}{a_{n}}$.
If there is some $n$ such that $\frac{a_{n-1}}{a_{n}} \geqslant 1$, then $a_{n}>a_{n+1}$.
Thus, $a_{n-1} \geqslant a_{n}>a_{n+1}>a_{n+2}>\cdots$, which is clearly impossible (since $\left\{a_{n}\right\}\left(n \in \mathbf{N}_{+}\right)$ is an i... | a_{n}=n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,559 |
Four. (50 points) Let $p$ be a prime number, $p \equiv 3(\bmod 4)$; let $x, y$ be integers, satisfying
$$
p \left\lvert\,\left(x^{2}-x y+\frac{p+1}{4} y^{2}\right)\right. \text {. }
$$
Prove: There exist integers $u, v$, such that
$$
x^{2}-x y+\frac{p+1}{4} y^{2}=p\left(u^{2}-u v+\frac{p+1}{4} v^{2}\right) .
$$ | $$
\begin{array}{l}
\text{From the condition, we know that } p \mid \left[(2 x-y)^{2}+p y^{2}\right]. \text{ Then } p \mid (2 x-y)^{2}. \\
\text{Since } p \text{ is a prime, we have } p \mid (2 x-y). \\
\text{Let } 2 x-y=p k. \text{ Then} \\
x^{2}-x y+\frac{p+1}{4} y^{2} \\
=\frac{1}{4}\left[p y^{2}+(2 x-y)^{2}\right] ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,560 |
1. There is a sequence of numbers: $1,1,5,13,25,41, \cdots$, arranged according to this pattern, the 101st number is ( ).
(A) 19801
(B) 19802
(C) 19901
(D) 19902 | - 1. A.
Calculating the difference between two consecutive numbers, we know the $n$-th number is
$$
\begin{array}{l}
1+4 \times 1+4 \times 2+\cdots+4(n-2) \\
=1+4[1+2+\cdots+(n-2)] \\
=1+4 \times \frac{(n-1)(n-2)}{2} \\
=1+2(n-1)(n-2) .
\end{array}
$$
Therefore, the 101st number is
$$
1+2(101-1)(101-2)=19801 .
$$ | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,561 |
2. Given that $A D$ is the altitude of $\triangle A B C$ (inside the shape). The following four conditions are given:
(1) $A B+B D=A C+C D$;
(2) $A B-B D=A C-C D$;
(3) $A B \cdot B D=A C \cdot C D$;
(4) $A B: B D=A C: C D$.
The number of conditions that can definitely lead to $A B=A C$ is ( ) .
(A) 1
(B) 2
(C.) 3
(D) 4 | 2. D.
As shown in Figure 2, let
$$
\begin{array}{l}
A B=c, A C=b, \\
B D=x, C D=y .
\end{array}
$$
Then $c^{2}-x^{2}$
$$
=b^{2}-y^{2} \text {. }
$$
Factoring gives
$$
\begin{array}{l}
(c+x)(c-x) \\
=(b+y)(b-y) .
\end{array}
$$
(1) If $c+x=b+y$, then $c-x=b-y$.
Adding the two equations yields $2 c=2 b \Rightarrow c=... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,562 |
3. The number of positive integer solutions to the equation $a b c=2(a+b+c)$ is ( ).
(A) 3
(B) 6
(C) 12
(D) 15 | 3. D.
Notice that the original equation is symmetric with respect to $a$, $b$, and $c$.
Without loss of generality, assume $a \leqslant b \leqslant c$. Then $a+b \leqslant 2 c$.
Substituting into the original equation, we get $2 < a b \leqslant 6$.
When $a b=3$, we get
$$
(a, b, c)=(1,3,8) \text {; }
$$
When $a b=4$,... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,563 |
4. Given that the equation of line $l$ is $y=\frac{1}{2} x+1$, and a line $P Q \perp l$ is drawn through point $P(1,1)$, with $Q$ being the foot of the perpendicular. Then the equation of line $P Q$ is ( ).
(A) $y=-\frac{1}{2} x+3$
(B) $y=-\frac{1}{2} x+4$
(C) $y=-2 x+3$
(D) $y=-2 x+4$ | 4. C.
Let the equation of line $P Q$ be $y=k x+b$.
Since $P Q \perp l$, we know $k=-2$.
Substituting $P(1,1)$, we get $b=3$.
Therefore, the equation of line $P Q$ is $y=-2 x+3$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,564 |
5. If $n \leqslant 2011$, then the number of positive integers $n$ such that $1+17 n$ is a perfect square is ( ) .
(A) 20
(B) 22
(C) 24
(D) 26 | 5. A.
Let $1+17 n=a^{2}$ (where $a$ is a positive integer). Then $n=\frac{a^{2}-1}{17}=\frac{(a+1)(a-1)}{17}$.
Since $n$ is a positive integer and 17 is a prime number, we have $a+1=17 b$ or $a-1=17 c$ (where $b, c$ are positive integers).
(1) When $a+1=17 b$, $n=\frac{a^{2}-1}{17}=17 b^{2}-2 b \leqslant 2011$.
Since ... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,565 |
6. In $\triangle A B C$, it is known that $\frac{1}{\tan \frac{A}{2}}+\frac{1}{\tan \frac{C}{2}}=\frac{4}{\tan \frac{B}{2}}$, and $b=4$. Then $a+c=(\quad)$.
(A) 5
(B) 6
(C) 7
(D) 8 | 6. B.
In Figure 3, let the incircle of $\triangle ABC$ be $\odot I$, with radius $r$. $\odot I$ is tangent to sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Connect $IA$, $IB$, $IC$, $ID$, $IE$, and $IF$.
By the tangent segment theorem, we have
$$
AF = p - a, \quad BD = p - b, \quad CE = p - c,
... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,566 |
2. In quadrilateral $A B C D$, it is known that $A B=B C=$ $C D$, and $\angle A=80^{\circ}, \angle D=40^{\circ}$. Then the degree measure of $\angle B$ is $\qquad$. | 2. $80^{\circ}$.
As shown in Figure 4, draw $BO \parallel CD$, and connect $AO, DO$.
Then quadrilateral $BCDO$
is a rhombus. Therefore,
$$
\begin{array}{l}
OB=OD \\
=BC=AB. \\
\text{By } \angle ABC + \angle C \\
=360^{\circ} - \angle BAD - \angle ADC = 240^{\circ}, \\
\angle OBD + \angle C = 180^{\circ}, \\
\angle A... | 80^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,568 |
3. Given the function
$$
y=(x-m)(x-n)-1(m<n),
$$
and $a, b(a<b)$ are the roots of the equation
$$
(x-m)(x-n)-1=0.
$$
Then the possible order of the real numbers $a, b, m, n$ is (connect with “<”).
| 3. $a<m<n<b$.
As shown in Figure 5, the function $y=(x-m)(x-n)-1$ is a quadratic function, and its graph is a parabola opening upwards, intersecting the $x$-axis at points $(a, 0)$ and $(b, 0)$.
When $x=m$ or $x=n$, $y=-1<0$. Therefore, $a<m<n<b$. | a<m<n<b | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,569 |
2. Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$ | Given $a=\frac{-1+\sqrt{5}}{2}$, we can obtain the corresponding quadratic equation $a^{2}+a-1=0$.
Thus, the original expression is
$$
\begin{array}{l}
=\frac{\left(a^{2}+a-1\right)\left(a^{3}-a\right)-2(a-1)}{a\left(a^{2}-1\right)} \\
=\frac{-2}{a^{2}+a}=-2 .
\end{array}
$$ | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,570 |
4. In $\triangle A B C$, points $D$ and $E$ are on sides $A B$ and $A C$ respectively, and $B E$ intersects $C D$ at point $O$. Then the size relationship between $\frac{S_{\triangle A D E}}{S_{\triangle A B C}}$ and $\frac{S_{\triangle D O E}}{S_{\triangle B O C}}$ is $\qquad$ (fill in "equal" or "not equal"). | 4. Equality.
As shown in Figure 6, let
$$
\begin{array}{l}
S_{\triangle A D E}=a, S_{\triangle D O E}=b, S_{\triangle B O C}=c, \\
S_{\triangle A O D}=x, S_{\triangle C O E}=y . \\
\text { By } \frac{S_{\triangle B O D}}{S_{\triangle D O E}}=\frac{B O}{O E}=\frac{S_{\triangle B O C}}{S_{\triangle C O E}}, \text { we g... | equality | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,571 |
One, (20 points) Given real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
\bar{x}-y=\frac{x}{y+z}, \\
z^{2}=x(y+z)-y(x-y) .
\end{array}
$$
Find the value of $\frac{y^{2}+z^{2}-x^{2}}{2 y z}$. | $$
\begin{array}{l}
\text { From equation (1), we get } y^{2}-x^{2}=-y(x+z). \\
\text { From equation (2), we get } z^{2}=x z+y^{2}. \\
\text { Then } \frac{y^{2}+z^{2}-x^{2}}{2 y z}=\frac{z^{2}-y(x+z)}{2 y z} \\
=\frac{x z+y^{2}-y(x+z)}{2 y z} \\
=\frac{1}{2} \cdot \frac{x-y}{y} \cdot \frac{z-y}{z} \\
=\frac{1}{2} \cd... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,572 |
II. (25 points) In a cyclic quadrilateral $ABCD$, it is given that $\frac{AB}{AD}=\frac{BC}{CD}>1$. Prove that the tangents to the circle at points $A$ and $C$ and the line $BD$ are concurrent. | As shown in Figure 7, draw tangents to the circle through $A$ and $C$, intersecting line $BD$ at points $P$ and $Q$ respectively. It is easy to see that
$$
\begin{array}{l}
\triangle A B P \backsim \triangle D A P . \\
\text { Then } \frac{A B}{A D}=\frac{P B}{P A}=\frac{P A}{P D} . \\
\text { Hence } \frac{A B^{2}}{A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,573 |
Three, (25 points) Given that $a$ and $b$ are integers, and satisfy $a-b$ is a prime number, $ab$ is a perfect square. If $a \geqslant 2011$, find the minimum value of $a$.
| Three, let $a-b=m$ (where $m$ is a prime number) and $ab=n^2$ (where $n$ is a positive integer).
$$
\begin{array}{l}
\text { From }(a+b)^{2}-4ab=(a-b)^{2} \\
\Rightarrow(2a-m)^{2}-4n^{2}=m^{2} \\
\Rightarrow(2a-m+2n)(2a-m-2n)=m^{2} \times 1 .
\end{array}
$$
Since $2a-m+2n$ and $2a-m-2n$ are both positive integers, and... | 2025 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,574 |
1. Given $f(x)=-\log _{3}\left(x^{2}-a x-a\right)$ is monotonically increasing on $(-\infty, 1-\sqrt{3})$. Then the range of values for $a$ is $\qquad$ . | $$
-1.2-2 \sqrt{3} \leqslant a \leqslant 2 \text {. }
$$
Since $y=\log _{3}\left(x^{2}-a x-a\right)$ is monotonically decreasing on $(-\infty, 1-\sqrt{3})$, we have:
$$
\left\{\begin{array}{l}
x_{0}=\frac{a}{2} \geqslant 1-\sqrt{3}, \\
(1-\sqrt{3})^{2}-a(1-\sqrt{3})-a \geqslant 0 .
\end{array}\right.
$$
Solving this,... | 2-2 \sqrt{3} \leqslant a \leqslant 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,575 |
2. The range of the function $f(x)=\sqrt{x-3}+\sqrt{12-3 x}$ is $\qquad$ . | 2. $[1,2]$.
Since the domain of $f(x)$ is $3 \leqslant x \leqslant 4$, we have $0 \leqslant x-3 \leqslant 1$.
Let $x-3=\sin ^{2} \theta\left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)$. Then
$$
\begin{array}{l}
f(x)=\sqrt{x-3}+\sqrt{3(4-x)} \\
=\sin \theta+\sqrt{3\left(1-\sin ^{2} \theta\right)} \\
=\sin \theta... | [1,2] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,576 |
3. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that there are skew lines on the diagonals of the square faces. If the distance between two such skew lines is 1, then the volume of the cube is $\qquad$ | 3. 1 or $3 \sqrt{3}$.
Let the edge length of the cube be $x$.
If the distance between the diagonal lines $A C$ and $B_{1} D_{1}$ is 1, then $x=1$.
Thus, the volume of the cube is 1.
If the distance between the diagonal lines $A C$ and $B C$ is 1, then $\frac{\sqrt{3}}{3} x=1, x=\sqrt{3}$.
Therefore, the volume of the ... | 1 \text{ or } 3 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,577 |
4. Given the sequence $\left\{x_{n}\right\}$ :
$$
1,3,3,3,5,5,5,5,5, \cdots
$$
formed by all positive odd numbers arranged from smallest to largest, and each odd number $k(k=1,3,5, \cdots)$ appears consecutively $k$ times. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$, then $a+b+c+d=$ $\qqu... | 4.3.
For $k^{2}+1 \leqslant n \leqslant(k+1)^{2}$, $x_{n}=2 k+1, k=[\sqrt{n-1}]$,
therefore, $x_{n}=2[\sqrt{n-1}]+1$.
Thus, $(a, b, c, d)=(2,1,-1,1)$.
So $a+b+c+d=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,578 |
5. Given the equation $x+8 y+8 z=n(n \in \mathbf{N})$ has 666 sets of positive integer solutions $(x, y, z)$. Then the maximum value of $n$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 5. 304 .
Prove that when $m>1(m \in \mathbf{N})$, $y+z=m$ has $m-1$ sets of positive integer solutions. Therefore, the original equation also has $m-1$ sets of positive integer solutions.
From $1+2+\cdots+(m-1)=666$, solving gives $m=37$ or -36 (discard).
Thus, $1+8 \times 37 \leqslant n \leqslant 8+8 \times 37=304$. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,579 |
6. $\sum_{i=0}^{50} \sum_{j=0}^{50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}$ modulo 31 is
$\qquad$ . | 6.1.
Original expression $=\sum_{i=0}^{50} C_{50}^{i} \cdot \sum_{j=0}^{50} C_{50}^{j}=\left(2^{50}\right)^{2}=2^{100}$.
And $2^{5} \equiv 1(\bmod 31)$, so
Original expression $=2^{100} \equiv 1(\bmod 31)$.
Therefore, the remainder is 1. | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,580 |
Example 1 Suppose there are $n(n \geqslant 2)$ players participating in a tournament, and every two players play one match against each other, with each match having a clear winner and loser. Prove that there exists a player $A$, such that any other player either loses to $A$, or loses to a player who was defeated by $... | To prove that the player $A$ we are looking for, intuitively, should be the "strongest" player. Therefore, among these $n$ players, let the player with the most wins be $A$ (considering the extreme case).
Below is the proof that player $A$ meets the requirements of the problem.
For any other player $B$, if $B$ does not... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,581 |
Example 2 Distribute 1600 peanuts among 100 monkeys. Prove: no matter how they are distributed, at least 4 monkeys will get the same number of peanuts, and design a distribution method so that no 5 monkeys get the same number of peanuts. | To ensure that no 4 monkeys receive the same number of peanuts, we can consider the extreme scenario.
The most economical (i.e., using the fewest peanuts) way to distribute them is: 3 monkeys get 0 peanuts, 3 monkeys get 1 peanut, ..., 3 monkeys get 32 peanuts, and one monkey gets 33 peanuts. The total number of peanu... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,582 |
5. There are 20 teams participating in the national league. Question: What is the minimum number of matches that must be played so that in any group of three teams, at least two teams have played against each other? | Let team $A$ have the minimum number of matches (which is $k$ matches). Then
(1) There are $k$ teams that have played against team $A$, and each of these teams has played at least $k$ matches;
(2) There are $19-k$ teams that have not played against team $A$, and these teams must all have played against each other, othe... | 90 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,583 |
6. Given points $P$, $Q$, $R$ are on the sides $AB$, $BC$, $CA$ of $\triangle ABC$ respectively, and $\quad BP=PQ=QR=RC=1$.
Then the maximum area of $\triangle ABC$ is ( ).
(A) $\sqrt{3}$
(B) 2
(C) $\sqrt{5}$
(D) 3 | 6. B.
Notice that
$S_{\triangle B P Q} \leqslant \frac{1}{2} \times 1 \times 1 \times \sin \angle B P Q \leqslant \frac{1}{2}$.
Similarly, $S_{\triangle P Q R} \leqslant \frac{1}{2}, S_{\triangle Q R C} \leqslant \frac{1}{2}$.
Let $\triangle D P R$ be the reflection of $\triangle Q P R$ about $P R$.
If point $D$ coinc... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,584 |
1. Among the integers $1,2, \cdots, 2011$, the number of integers that can be expressed in the form $[x[x]]$ is $\qquad$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. | 2. 1.990.
Let $x=k+a, k=[x], 0 \leqslant a<1$. Then $[x[x]]=[k(k+a)]=k^{2}+[k a]$.
Therefore, $k^{2}, k^{2}+1, \cdots, k^{2}+k-1(k=1,2$, $\cdots, 44)$ can all be expressed in the form of $[x[x]]$. Hence, the number of integers that meet the requirement is
$$
1+2+\cdots+44=990 \text { (numbers). }
$$ | 990 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,585 |
2. In a certain kingdom, there are 32 knights, some of whom are servants to other knights. Each servant can have at most one master, and each master must be richer than any of his servants. If a knight has at least four servants, he is ennobled as a noble. If it is stipulated that a servant of $A$'s servant is not a se... | 2.7.
According to the problem, the richest knight is not a servant of any other knight, so at most 31 knights can be servants of other knights.
Since each noble has at least four servants, there can be at most 7 nobles.
Number the 32 knights as $1,2, \cdots, 32$, with wealth decreasing as the number increases, then
... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,586 |
3. Given $x_{1}=x_{2011}=1$, $\left|x_{n+1}\right|=\left|x_{n}+1\right|(n=1,2, \cdots, 2010)$.
Then $x_{1}+x_{2}+\cdots+x_{2010}=$ $\qquad$ | 3. -1005 .
From the given, it is easy to obtain
$$
x_{n+1}^{2}=x_{n}^{2}+2 x_{n}+1(n=1,2, \cdots, 2010) \text {. }
$$
By summing up and organizing these 2010 equations, we get
$$
\begin{array}{l}
x_{2011}^{2}=2\left(x_{1}+x_{2}+\cdots+x_{2010}\right)+2011 . \\
\text { Therefore, } x_{1}+x_{2}+\cdots+x_{2010}=-1005 .
... | -1005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,587 |
4. Among the positive integers not greater than 100, the ordered integer pairs $(m, n)$ that satisfy
$$
\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n}
$$
are $\qquad$ pairs. | 4. 170 .
Notice that $\sqrt{2} n-1<m<\sqrt{2}(n+1)$.
For each $n$, the number of $m$ is given by
$$
\begin{array}{l}
{[\sqrt{2}(n+1)]-[\sqrt{2} n-1]} \\
=[\sqrt{2}(n+1)]-[\sqrt{2} n]+1
\end{array}
$$
Since $100<71 \sqrt{2}<101<72 \sqrt{2}$, we have $n \leqslant 71$.
But when $n=71$, $m=100$.
Therefore, the number of ... | 170 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,588 |
One, (20 points) When $1<x<5$, find the minimum value of the algebraic expression
$$
\frac{\sqrt{(7 x+1)(5-x)(x-1)(7 x+5)}}{x^{2}}
$$ | $$
\begin{array}{l}
\frac{\sqrt{(7 x+1)(5-x)(x-1)(7 x+5)}}{x^{2}} \\
=\sqrt{\frac{\left(7 x^{2}-2 x-5\right)\left(-7 x^{2}+34 x+5\right)}{x^{4}}} \\
=\sqrt{\left(7 x-\frac{5}{x}-2\right)\left(-7 x+\frac{5}{x}+34\right)} \text {. } \\
\end{array}
$$
Let $t=7 x-\frac{5}{x}$. Then
$$
\begin{array}{l}
\text { Equation (1)... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,589 |
II. (25 points) Given 15 quadratic equations $x^{2}-p_{i} x+q_{i}=0(i=1,2, \cdots, 15)$ with coefficients $p_{i} 、 q_{i}$ taking values from $1,2, \cdots, 30$, and these coefficients are all distinct. If an equation has a root greater than 20, it is called a "good equation." Find the maximum number of good equations. | Second, if there exists an equation $x^{2}-p_{i} x+q_{i}=0$ with two roots $x_{1}, x_{2}$, then from $x_{1}+x_{2}=p_{i}>0, x_{1} x_{2}=q_{i}>0$, we know that both roots are positive.
Next, if a certain equation has a root greater than 20, then $p_{i}=x_{1}+x_{2}>20$. However, among the numbers $1,2, \cdots, 30$, there... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,590 |
Three. (25 points) As shown in Figure 2, given points $A$ and $B$ are two distinct points outside circle $\odot O$, point $P$ is on $\odot O$, and $PA$, $PB$ intersect $\odot O$ at points $D$ and $C$ respectively, different from point $P$, and $AD \cdot AP = BC \cdot BP$.
(1) Prove: $\triangle OAB$ is an isosceles tria... | (1) Draw $A T$ tangent to $\odot O$ at point $T$, and connect $O T$.
Let the radius of $\odot O$ be $R$. Then
$$
A D \cdot A P=A T^{2}=O A^{2}-R^{2} \text {. }
$$
Similarly, $B C \cdot B P=O B^{2}-R^{2}$.
From the given, it is easy to see that $O A=O B$.
(2) From the proof in (1), we know
$$
p(2 p+1)=(m-1)^{2}-9 \text... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,591 |
1. Given the inequality
$$
3 x+4 \sqrt{x y} \leqslant a(x+y)
$$
holds for all positive numbers $x$ and $y$. Then the minimum value of the real number $a$ is . $\qquad$ | -1.4 .
From the problem, we know that $a \geqslant\left(\frac{3 x+4 \sqrt{x y}}{x+y}\right)_{\text {max }}$.
$$
\text { Also, } \frac{3 x+4 \sqrt{x y}}{x+y} \leqslant \frac{3 x+(x+4 y)}{x+y}=4 \text {, }
$$
with equality holding if and only if $x=4 y>0$.
Therefore, the minimum value of $a$ is 4. | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,592 |
2. In a regular tetrahedron $ABCD$, let
$$
\overrightarrow{AE}=\frac{1}{4} \overrightarrow{AB}, \overrightarrow{CF}=\frac{1}{4} \overrightarrow{CD} \text {, }
$$
$\vec{U} \overrightarrow{DE}$ and $\overrightarrow{BF}$ form an angle $\theta$. Then $\cos \theta=$ | 2. $-\frac{4}{13}$.
Let the edge length of the regular tetrahedron be 4. Then
$$
\begin{array}{l}
\overrightarrow{B F} \cdot \overrightarrow{D E}=(\overrightarrow{B C}+\overrightarrow{C F}) \cdot(\overrightarrow{D A}+\overrightarrow{A E}) \\
=\overrightarrow{C F} \cdot \overrightarrow{D A}+\overrightarrow{B C} \cdot \... | -\frac{4}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,593 |
Example 1 Find the smallest positive integer $n$, such that $n^{3}+2 n^{2}$ is a square of an odd number.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | From $n^{3}+2 n^{2}=n^{2}(n+2)$ being an odd number, we know that $n$ is an odd number.
To make $n^{2}(n+2)$ a perfect square, then $n+2$ must be a perfect square.
Therefore, the smallest positive integer $n=7$. | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,594 |
3. Given the equation in $x$
$$
x^{2}+(a-2010) x+a=0 \quad (a \neq 0)
$$
has two integer roots. Then the value of the real number $a$ is $\qquad$. | 3.4024.
Let the roots of the equation be $x_{1} 、 x_{2}\left(x_{1} \leqslant x_{2}\right)$.
By Vieta's formulas, we have
$$
x_{1}+x_{2}=-(a-2010), x_{1} x_{2}=a \text {. }
$$
Then $x_{1} x_{2}+x_{1}+x_{2}=2010$, which means
$$
\left(x_{1}+1\right)\left(x_{2}+1\right)=2011 \text {. }
$$
Since 2011 is a prime number, ... | 4024 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,595 |
4. $[x]$ is the greatest integer not exceeding the real number $x$. It is known that the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{3}{2}, a_{n+1}=a_{n}^{2}-a_{n}+1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then $m=\left[\sum_{k=1}^{2011} \frac{1}{a_{k}}\right]$ is . $\qquad$ | 4.1.
$$
\begin{array}{l}
\text { Given } a_{n+1}=a_{n}^{2}-a_{n}+1 \\
\Rightarrow a_{n+1}-1=a_{n}\left(a_{n}-1\right) \\
\Rightarrow \frac{1}{a_{n+1}-1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}} . \\
\text { Then } \frac{1}{a_{n}}=\frac{1}{a_{n}-1}-\frac{1}{a_{n+1}-1} . \\
\text { Therefore } m=\left[\frac{1}{a_{1}-1}-\frac{1}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,596 |
5. As shown in Figure 1, in a certain city, there is a neat grid of roads between two places, $M$ and $N$. $A_{1}, A_{2}, A_{3}, A_{4}$ are four intersections located on one diagonal of the road grid. Now, two people, A and B, start from $M$ and $N$ respectively and randomly choose a shortest path along the streets, wa... | 5. $\frac{41}{100}$.
If A and B walk along the shortest path, they can only meet at $A_{1}, A_{2}, A_{3}, A_{4}$. If A and B meet at $A_{2}$, A has $\mathrm{C}_{3}^{1}$ ways to walk from $M$ to $A_{2}$, and $\mathrm{C}_{3}^{1}$ ways to walk from $A_{2}$ to $N$, making a total of $\mathrm{C}_{3}^{1} \mathrm{C}_{3}^{1}$... | \frac{41}{100} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,597 |
6. Given a function with period $T=4$
$$
f(x)=\left\{\begin{array}{ll}
m \sqrt{1-x^{2}}, & x \in(-1,1] ;(m>0) . \\
1-|x-2|, & x \in(1,3]
\end{array}\right.
$$
If the equation $3 f(x)=x$ has exactly five real solutions, then the range of values for $m$ is | 6. $\left(\frac{\sqrt{15}}{3}, \sqrt{7}\right)$.
To make the equation $3 f(x)=x$ have exactly five real solutions, the line $y=\frac{x}{3}$ must intersect with $y=m \sqrt{1-(x-4)^{2}}$ at two points, and not intersect with $y=m \sqrt{1-(x-8)^{2}}$ (as shown in Figure 3).
Figure 3
Solving the system of equations, for t... | \left(\frac{\sqrt{15}}{3}, \sqrt{7}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,598 |
7. Let $f(x)$ be a monotonically increasing function defined on $(0,+\infty)$, and for any positive number $x$, we have
$$
f\left(f(x)+\frac{1}{x}\right)=\frac{1}{f(x)} .
$$
Then $f(1)=$ $\qquad$ | 7. $\frac{1-\sqrt{5}}{2}$.
In equation (1), let $x=1$, we get
$$
f(f(1)+1)=\frac{1}{f(1)} \text {. }
$$
Let $f(1)=a$. Then $f(a+1)=\frac{1}{a}$.
Now let $x=a+1$, substituting in we get
$$
\begin{array}{l}
f\left(f(a+1)+\frac{1}{a+1}\right)=\frac{1}{f(a+1)} \\
\Rightarrow f\left(\frac{1}{a}+\frac{1}{a+1}\right)=a=f(1)... | \frac{1-\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,599 |
8. Let the function be
$$
\begin{array}{l}
f(x)= \sqrt{10-6 \cos x}+\sqrt{\frac{17}{8}-\frac{3 \sqrt{2}}{2} \sin x}+ \\
\sqrt{19-2 \sqrt{2} \cos x-8 \sin x} .
\end{array}
$$
Then the minimum value of $f(x)$ is $\qquad$ | 8. $\frac{21 \sqrt{2}}{4}-1$.
Notice
$$
\begin{aligned}
f(x)= & \sqrt{(\cos x-3)^{2}+\sin ^{2} x}+ \\
& \sqrt{\cos ^{2} x+\left(\sin x-\frac{3 \sqrt{2}}{4}\right)^{2}}+ \\
& \sqrt{(\cos x-\sqrt{2})^{2}+(\sin x-4)^{2}} .
\end{aligned}
$$
Let $A(3,0) 、 B\left(0, \frac{3 \sqrt{2}}{4}\right) 、 C(\sqrt{2}, 4)$ 、 $P(\cos x... | \frac{21 \sqrt{2}}{4}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,600 |
9. (16 points) Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with an eccentricity of $\sqrt{3}$, and the right directrix equation is $x=\frac{\sqrt{3}}{3}$. For the circle $\odot O: x^{2}+y^{2}=2$, a moving point $P\left(x_{0}, y_{0}\right)$ $\left(x_{0} y_{0} \neq 0\right)$ has a tang... | 9. From the given information, we have
$$
a=1, c=\sqrt{3}, b^{2}=c^{2}-a^{2}=2 \text {. }
$$
Thus, the equation of the hyperbola $C$ is
$$
x^{2}-\frac{y^{2}}{2}=1 \text {. }
$$
Also, the equation of the tangent line to the circle $\odot O: x^{2}+y^{2}=2$ at point $P\left(x_{0}, y_{0}\right)$ is
$$
x_{0} x+y_{0} y=2 \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,601 |
10. (20 points) Given the function
$$
f(x)=a(|\sin x|+|\cos x|)-3 \sin 2 x-7 \text {, }
$$
where $a$ is a real parameter. Find all pairs $(a, n)(n \in \mathrm{N}_{+})$ such that the function $y=f(x)$ has exactly 2011 zeros in the interval $(0, n \pi)$. | 10. First, the function $f(x)$ has a period of $\pi$, and is symmetric about $x=\frac{k \pi}{2}+\frac{\pi}{4}(k \in \mathbf{Z})$, i.e.,
$$
\begin{array}{l}
f(x+\pi)=f(x), \\
f\left(k \pi+\frac{\pi}{2}-x\right)=f(x)(k \in \mathbf{Z}) .
\end{array}
$$
Second, $f\left(\frac{k \pi}{2}\right)=a-7$,
$$
\begin{array}{l}
f\le... | (a, n)=(7,503),(5 \sqrt{2}, 2011),(2 \sqrt{2}, 2011) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,602 |
11. (20 points) Let $f(x)$ be a non-decreasing function on $[0,1]$, and satisfy:
(1) $f(0)=0$;
(2) $f\left(\frac{x}{3}\right)=\frac{f(x)}{2}$;
(3) $f(1-x)=1-f(x)$.
Find the value of $f\left(\frac{17}{2010}\right)$. | 11. From (3) we have
$$
\begin{array}{l}
f(1)=1-f(0)=1, \\
f\left(\frac{1}{3}\right)=\frac{f(1)}{2}=\frac{1}{2}, \\
f\left(\frac{2}{3}\right)=1-f\left(\frac{1}{3}\right)=\frac{1}{2} . \\
\text { Hence } f(x)=\frac{1}{2}\left(x \in\left[\frac{1}{3}, \frac{2}{3}\right]=\left[\frac{670}{2010}, \frac{1340}{2010}\right]\rig... | \frac{9}{256} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,603 |
One. (40 points) As shown in Figure 2, given a point $P$ outside circle $\odot O$, two tangents $P E$ and $P F$ and two secants $P D A$ and $P C B$ are drawn to $\odot O$. Connect the chord $E F$, which intersects the secants $P D A$ and $P C B$ at points $M$ and $N$ respectively. Prove:
$$
\begin{array}{l}
\text { (1)... | (1) The auxiliary lines are shown in Figure 4.
By the theorem of common sides, we have $\frac{P D}{P A}=\frac{S_{\triangle P E D}}{S_{\triangle P E A}}=\frac{S_{\triangle P F D}}{S_{\triangle P F 1}}$.
From $\triangle P E D \backsim \triangle P A E$ and $\triangle P F D \backsim \triangle P A F$, we get
$$
\begin{array... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,604 |
Example 2 Let $n$ be a positive integer, $A$ be a $2n$-digit number, and each digit of $A$ is $4$; $B$ is an $n$-digit number, and each digit of $B$ is $8$. Prove: $A+2B+4$ is a perfect square.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation r... | Notice that
$$
A=\frac{4}{9}\left(10^{2 n}-1\right), B=\frac{8}{9}\left(10^{n}-1\right) \text {. }
$$
Thus, $A+2 B+4$
$$
\begin{array}{l}
=\frac{4}{9}\left(10^{2 n}-1\right)+\frac{16}{9}\left(10^{n}-1\right)+4 \\
=\frac{4}{9} \times 10^{2 n}+\frac{16}{9} \times 10^{n}+\frac{16}{9} \\
=\left(\frac{2 \times 10^{n}+4}{3}... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 723,605 |
Does there exist a real number $k$, such that
$$
f(x, y, z)=\frac{x^{5}+y^{5}+z^{5}+k\left(x^{2}+y^{2}+z^{2}\right)\left(x^{3}+y^{3}+z^{3}\right)}{x+y+z}
$$
is a polynomial in three variables. | Assume there exists such a real number $k$. Then
$$
x^{5}+y^{5}+z^{5}+k\left(x^{2}+y^{2}+z^{2}\right)\left(x^{3}+y^{3}+z^{3}\right)
$$
has a factor $x+y+z$.
Let $y=z=1$. Then $x+2$ is a factor of
$$
x^{5}+2+k\left(x^{3}+2\right)\left(x^{2}+2\right)
$$
Perform polynomial division.
The remainder of dividing $(1+k) x^{5... | k=-\frac{5}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,606 |
Three. (50 points) In a certain drill activity, the drill leader arranged $n$ students, numbered $1 \sim n (n>3)$, in a circular formation, and they performed a $1-2-3$ cyclic counting. The drill leader recorded the numbers of the students who reported, and required students who reported 1 and 2 to leave the formation,... | Three, the numbers recorded by the leader are $a_{1}$, $a_{2}, \cdots, a_{n}, a_{n+1}, \cdots, a_{m}$, with their sum being $S_{n}$. Thus,
(1) After each 1-2-3 counting cycle, the total number of students decreases by 2, but the sum of the numbers remains unchanged.
(2) If $n=3^{k}$, after $3^{k-1}$ 1-2-3 counting cycl... | 16011388 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,607 |
Four. (50 points) Find all positive integers $n$, such that there exists an integer $m$, making $2^{n}-1$ a divisor of $m^{2}+289$.
保留源文本的换行和格式，翻译结果如下：
Four. (50 points) Find all positive integers $n$, such that there exists an integer $m$, making $2^{n}-1$ a divisor of $m^{2}+289$.
---
Note: The translation preser... | For $n=2^{k}$ (where $k$ is a non-negative integer).
On one hand, if $n$ has an odd factor $s>1$, then $2^{s}-1$ is a factor of $2^{n}-1=\left(2^{s}\right)^{\frac{n}{1}}-1$, i.e.,
$2^{n}-1=\left(2^{2}\right)^{\frac{n}{4}}-1$
$=\left(2^{s}-1\right)\left[\left(2^{s}\right)^{\frac{n}{4}-1}+\left(2^{s}\right)^{\frac{n}{3}... | n=2^{k} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,608 |
Initial 293 Given the equation
$$
x^{3}-(2 a+11) x^{2}+\left(a^{2}+11 a+28\right) x-28 a=0
$$
all of whose roots are positive integers. Find the value of $a$ and the roots of the equation. | It is known that $x=a$ is a root of the original equation.
Thus, $a$ is a positive integer.
Factoring the left side of the equation, we get
$$
(x-a)\left[x^{2}-(a+11) x+28\right]=0 \text {. }
$$
Since all roots of the original equation are positive integers, then
$$
x^{2}-(a+11) x+28=0
$$
has a discriminant
$$
\Delta... | a=18,5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,609 |
Initial 294 Question: Does there exist a positive integer $b$ such that 2011 can be written as $\overline{x y z}$ in base $b$, and
$$
x+y+z=2+0+1+1 \text {. }
$$ | Solution: Since $\overline{x y z}_{(b)}=x b^{2}+y b+z=2011$, $x+y+z=4$,
thus, $x\left(b^{2}-1\right)+y(b-1)=2007$, which means $(b-1)[(b+1) x+y]=2007$.
Therefore, $(b-1) \mid 2007$.
Also, $b^{2} \leqslant 2011<b^{3}$, so $10<\sqrt[3]{2011}<b \leqslant \sqrt{2011}<45$.
Hence, $9<b-1<44$.
Given $2007=3^{2} \times 223$, t... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,610 |
293 Find
$$
c_{n}=\frac{1}{2}\left[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right](n \in \mathbf{N})
$$
the highest power of the prime factor 7 in it. | Solve for the modulo $7,\left\{c_{n}\right\}$ as a modular periodic sequence, with a period of $8, c_{0} \sim c_{7}$ modulo 7 remainders are
$$
1,2,0,5,6,5,0,2 \text {. }
$$
Thus, $7 \mid c_{n} \Leftrightarrow n \equiv 2(\bmod 4)$.
Let $b_{n}=\frac{1}{2 \sqrt{3}}\left[(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}\right]$
be the d... | \alpha+1, n \equiv 2(\bmod 4), 7^{a} \| n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,611 |
Given $a, b, c, m \in \mathbf{R}$, prove:
$$
\begin{array}{l}
\frac{a^{m+2}}{a+b}+\frac{b^{m+2}}{b+c}+\frac{c^{m+2}}{c+a} \\
\geqslant \frac{1}{2}\left(a^{m+1}+b^{m+1}+c^{m+1}\right) .
\end{array}
$$ | $$
\begin{array}{l}
\frac{a^{m+2}}{a+b}=\frac{a^{m+1}(a+b)-a^{m+1} b}{a+b} \\
=a^{m+1}-\frac{a^{m+1} b}{a+b} \geqslant a^{m+1}-\frac{a^{m}(a+b)^{2}}{4(a+b)} \\
=a^{m+1}-\frac{1}{4} a^{m}(a+b)=\frac{3}{4} a^{m+1}-\frac{1}{4} a^{m} b .
\end{array}
$$
Similarly, $\frac{b^{m+2}}{b+c} \geqslant \frac{3}{4} b^{m+1}-\frac{1}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,612 |
Example 3 Let $m, n$ be positive integers, and satisfy
$$
2001 m^{2}+m=2002 n^{2}+n \text{. }
$$
Prove: $m-n$ is a perfect square. | Given that $m>n$.
Let $m=n+k\left(k \in \mathbf{N}_{+}\right)$. Then equation (1) becomes
$$
n^{2}-4002 n k-2001 k^{2}-k=0 \text {, }
$$
which is
$$
\begin{array}{l}
(n-2001 k)^{2} \\
=(2001 k)^{2}+2001 k^{2}+k \\
=k(2001 \times 2002 k+1) .
\end{array}
$$
Since $(k, 2001 \times 2002 k+1)=1$, both $k$ and $2001 \times... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,613 |
Example 4 Find all pairs of positive integers $(m, n)$ such that $m^{2}-4 n$ and $n^{2}-4 m$ are both perfect squares. | By symmetry, assume $m \leqslant n$.
(1) If $m \leqslant n-1$, then $n^{2}-4 m \geqslant n^{2}-4 n+4=(n-2)^{2}$.
Also, $n^{2}-4 m \leqslant(n-1)^{2}=n^{2}-2 n+1$, and the
equality does not hold, so
$$
n^{2}-4 m=(n-2)^{2} \Rightarrow n=m+1 \text {. }
$$
Thus, $m^{2}-4 n=(m-2)^{2}-8=t^{2}\left(t \in \mathbf{N}_{+}\right... | (m, n)=(4,4),(5,6),(6,5) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,614 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.