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. Let $a, b$ be positive integers, and $24a^2 = b^2 - 1$. Prove that exactly one of $a, b$ is a multiple of 5. | (提示: if $a \equiv 0(\bmod 5)$, and $b \equiv 0(\bmod 5)$, then $24 a^{2}-b^{2}+1 \equiv 1(\bmod 5)$, which contradicts $24 a^{2}-b^{2}+1=0$. If $a$ and $b$ are not multiples of 5, then $a \equiv \pm 1, \pm 2(\bmod 5)$, and $b \equiv \pm 1, \pm 2(\bmod 5)$. Therefore, $a^{2} \equiv \pm 1(\bmod 5), b^{2} \equiv \pm 1(\bm... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,217 |
4. In Rt $\triangle A B C$, $\angle A C B=90^{\circ}$, on the hypotenuse $A B$ respectively intercept $A D=$
$$
A C, B E=B C, D E=6 \text{, }
$$
$O$ is the circumcenter of $\triangle C D E$, as shown in Figure 2. Then the sum of the distances from $O$ to the three sides of $\triangle A B C$ is . $\qquad$ | 4.9.
As shown in Figure 8, connect
$$
O A, O B, O C, O D \text {, and }
$$
$O E$.
Since $O C=O D$,
we know that point $O$ lies on the perpendicular bisector of segment $C D$.
Also, $A D=A C$, so $O A$ bisects $\angle C A D$.
Similarly, $O B$ bisects $\angle C B D$.
Therefore, $O$ is the incenter of $\triangle A B C$.... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,218 |
II. (25 points) As shown in Figure 3, the angle bisector $AD$ of $\triangle ABC$ intersects its circumcircle $\odot O$ at point $D$, $I$ is a point on $AD$ such that $AI = BD$, and $AB + AC = 2BC$. Prove that $I$ is the incenter of $\triangle ABC$. | I. Let $y_{1}=\left|x^{2}-2 x-3\right|$, $y_{2}=\left|y_{1}+4 a\right|, y_{3}=a^{2}$.
Obviously, $a=0$ does not satisfy the condition.
The graph of the function $y_{1}=\left|x^{2}-2 x-3\right|$ is shown in Figure 9.
II. As shown in Figure 14, connect $O D$ intersecting $B C$ at point
$M$, and draw $I E$
$\perp A B$ at ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,220 |
2. Given the sum of $2 n+1\left(n \in \mathbf{N}_{+}\right)$ consecutive positive integers is $a$, and the difference between the sum of the squares of the last $n$ numbers and the sum of the squares of the first $n$ numbers is $b$. If $\frac{a}{b}=\frac{11}{60}$, then the value of $n$ is | 2.5
Let the $(n+1)$-th number be $m$. Then
$$
\begin{array}{l}
a=(2 n+1) m, \\
b=\sum_{i=1}^{n}(m+i)^{2}-\sum_{i=1}^{n}(m-i)^{2} \\
=2 m \sum_{i=1}^{n} 2 i=2 m n(n+1) . \\
\text { Hence } \frac{a}{b}=\frac{2 n+1}{2 n(n+1)}=\frac{11}{60} .
\end{array}
$$
Solving gives $n=5$ or $-\frac{6}{11}$ (discard).
| 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,223 |
3. Given a circle centered at the origin with radius $R$ intersects the sides of $\triangle A B C$, where $A(4,0)$, $B(6,8)$, and $C(2,4)$. Then the range of values for $R$ is $\qquad$ | 3. $\left[\frac{8 \sqrt{5}}{5}, 10\right]$.
Draw a perpendicular from the origin $O$ to $AC$, with the foot of the perpendicular being $D$, so point $D$ lies on side $AC$. It is easy to see that
$$
l_{A C}: y=-2 x+8 \Rightarrow 2 x+y-8=0 \text {. }
$$
Then $O D=\frac{|2 \times 0+0-8|}{\sqrt{2^{2}+1}}=\frac{8 \sqrt{5}... | \left[\frac{8 \sqrt{5}}{5}, 10\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,224 |
4. The trajectory equation of the moving point in the complex plane
$$
\begin{aligned}
& Z\left(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}, \frac{2}{\cos \theta+\sin \theta}\right)(\theta \in \mathbf{R}, \theta \\
\neq & \left.k \pi+\frac{3 \pi}{4}, k \in \mathbf{Z}\right)
\end{aligned}
$$
is | 4. $\frac{y^{2}}{2}-x^{2}=1$.
Notice that $x^{2}=\frac{1-\sin 2 \theta}{1+\sin 2 \theta}, y^{2}=\frac{4}{1+\sin 2 \theta}$. Then $\frac{y^{2}}{2}-x^{2}=1$. | \frac{y^{2}}{2}-x^{2}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,225 |
5. Divide the positive integers $1,2, \cdots, 10$ into two groups:
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}, B=\left\{b_{1}, b_{2}, \cdots, b_{n}\right\} \text {. }
$$
Take one number $a$ from $A$ and one number $b$ from $B$ ($a \in A, b \in B$), divide them to get a number $\frac{a}{b}$, which is called the "q... | 5. $\frac{(10!)^{4}}{(4!)^{10}}$.
The product of all quotients is
$$
\begin{array}{l}
S=\prod_{i=1}^{m} \prod_{j=1}^{n} \frac{a_{i}}{b_{j}}=\prod_{i=1}^{m}\left[a_{i}^{n}\left(\prod_{j=1}^{n} b_{j}\right)^{-1}\right] \\
=\left(\prod_{i=1}^{m} a_{i}\right)^{n}\left(\prod_{j=1}^{n} b_{j}\right)^{-m} .
\end{array}
$$
Cl... | \frac{(10!)^{4}}{(4!)^{10}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,226 |
6. In a Cartesian coordinate system, the "traffic distance" between points $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$ is defined as
$$
d(P, Q)=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right| \text {. }
$$
A certain country has 3 castles, marked on the map as $A(2,3)$, $B(-6,9)$, and $C(-3,-8)$. Th... | 6. $(-5,0)$.
Let the new castle be $D(x, y)$.
If $x \geqslant 2$, as shown in Figure 2,
when $y \geqslant 3$, $d(A, D) < d(C, D)$; when $y < 3$,
$d(A, D) < d(B, D)$. This is a contradiction, so $x < 2$.
Similarly, we know
$-8 < y < 9$.
Thus, $d(A, D) = |x-2| + |y-3| = 2-x + |y-3|$,
$d(B, D) = |x+6| + |y-9| = |x+6| + 9... | (-5,0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,227 |
1. For a right-angled triangle with a hypotenuse of 2009, if the two legs are also integers, then its area is $\qquad$ . | $-.1 .432180$.
Let the two legs be $x, y$. Then $x^{2}+y^{2}=2009^{2}$. According to Pythagorean triples and the symmetry of $x, y$, there exist positive integers $m, n, k (m>n)$, such that
$$
\begin{array}{l}
x=2 m n k, y=\left(m^{2}-n^{2}\right) k, \\
2009=\left(m^{2}+n^{2}\right) k .
\end{array}
$$
Since $2009=41 \... | 432180 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,228 |
2. For a regular triangular prism $A B C-A^{\prime} B^{\prime} C^{\prime}$, the side edges and the base edges are all 1. Then the volume of the common part of the tetrahedra $A^{\prime} A B C$, $B^{\prime} A B C$, and $C^{\prime} A B C$ is $\qquad$ . | 2. $\frac{\sqrt{3}}{36}$.
As shown in Figure 2, by symmetry, the intersection point $O$ of the planes $C_{1} A B$, $A_{1} B C$, and $B A C$ projects onto the center $H$ of the equilateral $\triangle A B C$ on the base $A B C$. Let the midpoints of $A B$ and $B C$ be $D$ and $E$, respectively. Then $O H$ is the interse... | \frac{\sqrt{3}}{36} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,229 |
11. (15 points) The sequence of positive integers $\left\{a_{n}\right\}$ satisfies:
$$
a_{4}=16,-\frac{1}{a_{n+1}^{2}}<\frac{1}{a_{n}}+\frac{1}{a_{n+1}}-\frac{2}{n(n+1)}<\frac{1}{a_{n}^{2}} \text {. }
$$
Find the general term formula $a_{n}$. | 11. According to the condition $\frac{1}{a_{n+1}^{2}}\frac{1}{16^{2}}+\frac{1}{6}-\frac{1}{16}=\frac{83}{768}$.
Then $a_{3}0$, and
$a_{3} \geqslant 3$, so
$a_{3}>\frac{48+8 \sqrt{21}}{10}>\frac{48+8 \times 4.5}{10}=8.4$. (3)
According to equations (2) and (3), we get the integer $a_{3}=9$.
Again, taking $n=2$ in the co... | a_{n} = (n+1)^2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,230 |
In $\triangle A B C$, $A B>A C$, point $E$ is on side $A B$, and $A E=\frac{A B-A C}{2}$. A perpendicular line to $A B$ through $E$ intersects the perpendicular bisector of side $B C$ at point $F$. Prove that the distances from point $F$ to lines $A B$ and $A C$ are equal. | Proof: As shown in Figure 4, extend $CA$ to point $K$, on $BE$.
Take a point $G$ such that
$$
EG = EA = \frac{AB - AC}{2}
$$
(at this point, $AG = AB - AC < AB$, so point $G$ is indeed on segment $BE$). Then
$$
\begin{array}{l}
BG = AB - AG \\
= AB - (AB - AC) = AC.
\end{array}
$$
Additionally, by $\mathrm{Rt} \triang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,231 |
249 Find the range of the real number $\lambda$ such that the inequality
$$
\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}} \leqslant \frac{3}{\sqrt{1+\lambda}}
$$
holds for any positive real numbers $x, y, z$ satisfying $x y z=\lambda^{3}$. | Solution: In equation (1), let $x=y=t, z=\frac{1}{t^{2}}$.
When $t \rightarrow 0$, we have
$$
\begin{array}{l}
\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}} \\
=\frac{2}{\sqrt{1+t}}+\frac{t}{\sqrt{t^{2}+1}} \rightarrow 2
\end{array}
$$
Then $2 \leqslant \frac{3}{\sqrt{1+\lambda}} \Leftrightarrow 00, r... | \left(0, \frac{5}{4}\right] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,232 |
A circle with its center at the center of a regular polygon is called the "central circle" of the regular polygon. Let $\odot O$ be the central circle of the regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, and let $P$ be any point on $\odot O$. The projection of $P$ onto $A_{i} \Lambda_{i+1}\left(i=1,2, \cdots, n... | Solution: Establish a rectangular coordinate system with $O$ as the origin and the line $O A_{n}$ as the $x$-axis.
Without loss of generality, let $\left|O A_{n}\right|=1$. Then
$$
A_{i}\left(\cos \frac{2 i \pi}{n}, \sin \frac{2 i \pi}{n}\right)(i=1,2, \cdots, n),
$$
The equation of $\odot O$ is $x^{2}+y^{2}=R^{2}(R>0)... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,233 |
One. (50 points) As shown in Figure 1, in $\triangle ABC$, $D$ is any point on the angle bisector of $\angle A$, and $E$, $F$ are points on the extensions of $AB$, $AC$ respectively, such that $CE \parallel BD$, $BF \parallel CD$. If $M$, $N$ are the midpoints of $CE$, $BF$ respectively, prove: $AD \perp MN$.
保留源文本的换行... | Extend $BD$ and $CD$, intersecting $AC$ and $AB$ at points $H$ and $G$ respectively. Note the equal heights and equal angles of $\triangle DBG$ and $\triangle DCH$ with respect to vertex $D$. By the area ratio theorem, we have
$$
\frac{BG}{CH}=\frac{S_{\triangle DBC}}{S_{\triangle DCH}}=\frac{DB \cdot DG}{DC \cdot DH}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,234 |
Three, (50 points) If the irreducible fraction $\frac{n}{m}$ satisfies: $m n \leqslant$ 2009 ( $m, n$ are positive integers), then $\frac{n}{m}$ is called a “cow fraction”.
Now, arrange all cow fractions in increasing order to form a sequence $\frac{n_{1}}{m_{1}}, \frac{n_{2}}{m_{2}}, \cdots$, called the “cow sequence... | For any positive integer $n$, the subsequence of the cow sequence with denominators not greater than $n$ is denoted as $T_{n}$.
When $n=1$, the sequence
$$
T_{1}=\left(\frac{1}{1}, \frac{2}{1}; \frac{3}{1}, \cdots, \frac{2009}{1}\right)
$$
is obviously satisfied.
We proceed by induction on $n$.
From the sequence $T_{1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,236 |
Four. (50 points) From the nine equal divisions on the circumference of a circle, any five points are colored red. Prove: There exist six different triangles $\triangle_{1}, \triangle_{2}, \cdots, \triangle_{6}$ with red points as vertices, satisfying
$$
\triangle_{1} \cong \triangle_{2}, \triangle_{3} \cong \triangle_... | (1) In an isosceles trapezoid $ABCD$ with bases $AD$ and $BC$, there exist two pairs of congruent triangles:
$\triangle ABC \cong \triangle DCB, \triangle BAD \cong \triangle CDA$,
and each vertex of the trapezoid appears twice in one pair of congruent triangles.
(2) If $M$ is any point on the perpendicular bisectors ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,237 |
1. Given the function $f(x)=x|1-x|(x \in \mathbf{R})$. Then the solution set of the inequality $f(x)>\frac{1}{4}$ is $\qquad$ . | $$
-1 \cdot\left(\frac{1+\sqrt{2}}{2},+\infty\right) \text {. }
$$
Notice that
$$
f(x)=x|1-x|=\left\{\begin{array}{ll}
-x^{2}+x, & x \leqslant 1 ; \\
x^{2}-x, & x>1 .
\end{array}\right.
$$
As shown in Figure 4, it is known that there exists $x_{0}$ in the interval $(1,+\infty)$ such that $f\left(x_{0}\right)=\frac{1}... | \left(\frac{1+\sqrt{2}}{2},+\infty\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,238 |
2. From the arithmetic sequence $2,5,8,11, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$ | 2.8 .
First, take $2,5,8,11,20,41,110,1640$, it is easy to see that the sum of their reciprocals is 1, i.e., $k=8$ satisfies the requirement.
Second, suppose we take $x_{1}, x_{2}, \cdots, x_{k}$ from the sequence, such that $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{k}}=1$.
Let $y_{i}=\frac{x_{1} x_{2} \cdot... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,239 |
3. In a Cartesian coordinate system, two circles have a common point $(9,6)$ and are both tangent to the $x$-axis. The product of their radii is 68. If their other external common tangent also passes through the origin, then its slope is $\qquad$ . | 3. $\frac{12 \sqrt{221}}{49}$.
Since the line connecting the centers of the two circles passes through the origin, we can set the centers of the two circles as $(a, k a),(b, k b)$. Thus,
$$
\begin{array}{l}
(a-9)^{2}+(k a-6)^{2}=(k a)^{2} \\
\Rightarrow a^{2}-6(2 k+3) a+117=0 .
\end{array}
$$
Similarly, $b^{2}-6(2 k+... | \frac{12 \sqrt{221}}{49} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,240 |
4. Given $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$ are any two points (which can coincide) on the graph of the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{2 x}{1-2 x}, & x \neq \frac{1}{2} \\
-1, & x=\frac{1}{2}
\end{array}\right.
$$
Point $M$ lies on the line $x=\frac{1}{2}$, and $\overrightarrow{... | 4. -2 .
Given that point $M$ is on the line $x=\frac{1}{2}$, let $M\left(\frac{1}{2}, y_{M}\right)$.
Also, $\overrightarrow{A M}=\overrightarrow{M B}$, that is,
$$
\begin{array}{l}
\overrightarrow{A M}=\left(\frac{1}{2}-x_{1}, y_{M}-y_{1}\right), \\
\overrightarrow{M B}=\left(x_{2}-\frac{1}{2}, y_{2}-y_{M}\right) .
\e... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,241 |
5. Each leg of the tripod is 5 units long, and the angles between each pair of legs are equal and fixed. When it is placed on the ground, the height from the top to the ground is 4 units. Later, one leg is damaged, and 1 unit is cut off from the bottom. When it is placed on the ground again, the height from the top to ... | 5. $\frac{144}{\sqrt{1585}}$.
As shown in Figure 5, the original tripod, when erected, forms a regular tetrahedron $P-ABC$, with side edges of length 5 and height $PO=4$. Therefore, the radius of the circumcircle of the base equilateral triangle is
$$
\begin{array}{l}
\frac{1}{\sqrt{3}} AB=\sqrt{5^{2}-4^{2}} \\
=3 .
\... | \frac{144}{\sqrt{1585}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,242 |
3. Let $M=\{1,2, \cdots, 2009\}$. If $n \in M$, such that $S_{n}=\frac{1}{n}\left(1^{3}+2^{3}+\cdots+n^{3}\right)$ is a perfect square, then the number of such $n$ is $\qquad$.
| 3. 44 .
Notice that
$$
S_{n}=\frac{1}{n}\left[\frac{n(n+1)}{2}\right]^{2}=\frac{n(n+1)^{2}}{4} \text {. }
$$
Since $n$ and $n+1$ are coprime, if $S_{n}$ is a perfect square, then $n$ must be a perfect square. When $n$ is an even perfect square, $\frac{n}{4}$ is a perfect square integer; when $n$ is an odd perfect squ... | 44 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,243 |
6. Given $S_{n}$ as the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, $\boldsymbol{a}=\left(S_{n}, 1\right), \boldsymbol{b}=\left(-1,2 a_{n}+2^{n+1}\right), \boldsymbol{a} \perp \boldsymbol{b}$. If $b_{n}=\frac{n-2011}{n+1} a_{n}$, and there exists $n_{0}$ such that for any $k\left(k \in \mathbf{N}... | 6.2009 or 2010.
From $\boldsymbol{a} \perp \boldsymbol{b} \Rightarrow-S_{n}+2 a_{n}+2^{n+1}=0$
$$
\begin{array}{l}
\Rightarrow-S_{n+1}+2 a_{n+1}+2^{n+2}=0 \\
\Rightarrow a_{n+1}=2 a_{n}-2^{n+1} \Rightarrow \frac{a_{n+1}}{2^{n+1}}=\frac{a_{n}}{2^{n}}-1
\end{array}
$$
$\Rightarrow\left\{\frac{a_{n}}{2^{n}}\right\}$ is a... | 2009 \text{ or } 2010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,244 |
7. If $a$, $b$, $c$ are all integers $(0<c<90)$, and satisfy $\sqrt{9-8 \sin 50^{\circ}}=a+b \sin c^{\circ}$. Then the value of $\frac{a+b}{c}$ is | 7. $\frac{1}{2}$.
Notice that
$$
\begin{array}{l}
9-8 \sin 50^{\circ} \\
=9+8 \sin 10^{\circ}-8 \sin 10^{\circ}-8 \sin 50^{\circ} \\
=9+8 \sin 10^{\circ}-8\left[\sin \left(30^{\circ}-20^{\circ}\right)+\sin \left(30^{\circ}+20^{\circ}\right)\right] \\
=9+8 \sin 10^{\circ}-8 \cos 20^{\circ} \\
=9+8 \sin 10^{\circ}-8\lef... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,245 |
8. Arrange 4 identical red balls and 4 identical blue balls in a row, with the positions numbered $1,2, \cdots$, 8 from left to right. If balls of the same color are indistinguishable, the number of arrangements where the sum of the positions of the 4 red balls is less than the sum of the positions of the 4 blue balls ... | 8.31.
The sum of 8 numbers is 36. Dividing the numbers $1,2, \cdots, 8$ into two groups, there are $\frac{C_{8}^{4}}{2}=35$ ways to do so. When the sums of the two groups are not equal, the smaller sum of 4 numbers is considered the sequence of red balls, and the larger sum of 4 numbers is considered the sequence of bl... | 31 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,246 |
9. (14 points) As shown in Figure 1, given the parabola $x^{2}=2 p y$ $(p>0)$ and the line $y=b$ $(b<0)$, point $P(t, b)$
moves on the line. Two tangent lines are drawn from $P$ to the parabola, touching the parabola at points $A$ and $B$, respectively, and the midpoint of segment $A B$ is $M$.
(1) Find the trajectory ... | (1) From the problem, we have $y=\frac{1}{2 p} x^{2}$, then $y^{\prime}=\frac{1}{p} x$. Let $A\left(x_{1}, \frac{1}{2 p} x_{1}^{2}\right)$ and $B\left(x_{2}, \frac{1}{2 p} x_{2}^{2}\right)$. Thus, $k_{P A}=\frac{x_{1}}{p}, k_{P B}=\frac{x_{2}}{p}$.
Therefore, $\frac{x_{1}}{p}=\frac{\frac{x_{1}^{2}}{2 p}-b}{x_{1}-t}$, w... | 2 \sqrt{-2 p b} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,247 |
10. (15 points) Construct a number table using some natural numbers as shown in Figure 2: Use $a_{i j}$ $(i \geqslant j)$ to represent the $j$-th number in the $i$-th row $(i, j \in \mathbf{N}_{+})$, such that $a_{i 1} = a_{i i} = i$. The other numbers in each row are equal to the sum of the two numbers on their "shoul... | 10. From the problem, we have
$$
\begin{array}{l}
b_{n+1}=\sum_{i=1}^{n+1} a_{(n+1) i} \\
=\left(1+a_{n 1}\right)+\sum_{i=1}^{n-1}\left[a_{n i}+a_{n(i+1)}\right]+\left(a_{n n}+1\right) \\
=2+2 \sum_{i=1}^{n} a_{n i}=2+2 b_{n} .
\end{array}
$$
Thus, $b_{n+1}=2 b_{n}+2$, which implies $\frac{b_{n+1}+2}{b_{n}+2}=2$.
Ther... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,248 |
11. (15 points) Let $f(x)$ be a function defined on the domain $D$. If for any real number $\alpha \in(0,1)$ and any two numbers $x_{1}, x_{2}$ in $D$, it always holds that
$$
f\left(\alpha x_{1}+(1-\alpha) x_{2}\right) \leqslant \alpha f\left(x_{1}\right)+(1-\alpha) f\left(x_{2}\right),
$$
then $f(x)$ is called a $C$ ... | 11. (1) For any $n(0 \leqslant n \leqslant m)$, take
$$
x_{1}=m, x_{2}=0, \alpha=\frac{n}{m} \in[0,1] \text {. }
$$
Since $f(x)$ is a $C$ function on $\mathbf{R}$, $a_{n}=f(n)$, and $a_{0}=0, a_{m}=2 m$, we have
$$
\begin{array}{l}
a_{n}=f(n)=f\left(\alpha x_{1}+(1-\alpha) x_{2}\right) \\
\leqslant \alpha f\left(x_{1}... | m^2 + m | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,249 |
One, (50 points) As shown in Figure 3, $A D$ is the angle bisector of $\triangle A B C$, $I_{1}$ and $I_{2}$ are the incenter of $\triangle A B D$ and $\triangle A C D$ respectively. Construct an isosceles $\triangle I_{1} I_{2} E$ with $I_{1} I_{2}$ as the base, such that $\angle I_{1} E I_{2}=\frac{1}{2} \angle B A C... | As shown in Figure 6, let the incenter of $\triangle ABC$ be $I$, and connect $BI$, $CI$, $DI_1$, and $DI_2$. Take a point $F$ on $I_1E$ such that
$$
\begin{array}{l}
\angle I_1 D F \\
=\angle I_1 I I_2. \text{ (1) }
\end{array}
$$
Connect $I_2F$.
$$
\begin{array}{l}
\text{By } \angle D I_1 I \\
=180^{\circ}-\angle D ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,250 |
II. (50 points) Let $p$ be any given prime number. Prove: there must exist a prime $q$, such that for any integer $n$, the number $n^{p}-p$ cannot be divisible by $q$.
保留源文本的换行和格式，翻译结果如下：
II. (50 points) Let $p$ be any given prime number. Prove: there must exist a prime $q$, such that for any integer $n$, the number ... | The prime number $q$ we are looking for is only related to $p$, not to $n$. Therefore, for any positive integer $k$, if $q \mid\left(n^{p}-p\right)$, then $q \mid\left(n^{\ell}-p^{k}\right)$. That is, if $q \chi\left(n^{k p}-p^{k}\right)$, then $q \chi\left(n^{p}-p\right)$. Thus, the problem is reduced to selecting an ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,251 |
Three, (50 points) Let the natural number $k$ satisfy $1 < k$, such that $a_{m}$ is less than at least $k-1$ of the numbers $a_{1}, a_{2}, \cdots, a_{k}$. It is known that the number of sequences satisfying $a_{m}=1$ is $\frac{100!}{4}$. Find the value of $k$. | Three, rearrange $a_{1}, a_{2}, \cdots, a_{k}$ into $b_{1}t(i=k+1, k+2, \cdots, m-1) \text {. }$
$$
\text{When } t \text{ is fixed, by } b_{1}t, \text{ there are } \mathrm{C}_{100-\text{ ways to choose. }}^{k-2}
$$
Arranging $b_{1}, b_{2}, \cdots, b_{k}$ has $k!$ ways. Thus, determining $a_{1}, a_{2}, \cdots, a_{k}$ ha... | 45 \text{ or } 55 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,252 |
Four. (50 points) Let $a_{1}, a_{2}, \cdots, a_{20}$ be 20 distinct positive integers, and the set $\left\{a_{i}+a_{j} \| \leqslant i, j \leqslant 20\right\}$ contains 201 distinct elements. Find the minimum possible number of distinct elements in the set
$$
\left.|| a_{i}-a_{j}|| 1 \leqslant i, j \leqslant 20\right\}.... | The minimum number of elements in the given set is
100.
Example: Let
$$
\begin{array}{l}
a_{i}=10^{11}+10^{i}, \\
a_{10+i}=10^{11}-10^{i}(i=1,2, \cdots, 10) .
\end{array}
$$
Then $\left\{a_{i}+a_{j} \mid 1 \leqslant i \leqslant j \leqslant 20\right\}$ contains
$$
(20+19+\cdots+1)-10+1=201
$$
different elements.
$$
\b... | 100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,253 |
4. Let $f(x)=x^{4}-\frac{3}{4} x^{2}-\frac{1}{8} x+1$. Then $f\left(\cos \frac{\pi}{7}\right)=$ $\qquad$
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 4. $\frac{15}{16}$.
Let $x=\cos \frac{\pi}{7}$. Then
$$
\begin{array}{l}
x^{4}=\left(\cos ^{2} \frac{\pi}{7}\right)^{2}=\left(\cdot \frac{1+\cos \frac{2 \pi}{7}}{2}\right)^{2} \\
=\frac{1+2 \cos \frac{2 \pi}{7}+\cos ^{2} \frac{2 \pi}{7}}{4} \\
=\frac{3+4 \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}}{8}, \\
-\frac{3}{4} ... | \frac{15}{16} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 721,254 |
$$
\begin{array}{l}
\text { 1. Let } A=\left\{x \in \mathbf{Z} \left\lvert\, \frac{x^{2}}{2009}+\frac{y^{2}}{2008}=1\right.\right\}, \\
B=\{x \mid x=2 n+1, n \in \mathbf{Z}\},
\end{array}
$$
Set $M$ is a subset of $A$, but not a subset of $B$. Then the number of all such sets $M$ is $\qquad$ | $$
\begin{array}{l}
-1.2^{41}\left(2^{45}-1\right) . \\
\text { From } \frac{x^{2}}{2009}+\frac{y^{2}}{2008}=1(x \in \mathbf{Z}) \text { we get }|x| \leqslant 44 \text {. }
\end{array}
$$
Therefore, $A=\{-44,-43, \cdots,-1,0,1, \cdots, 44\}$.
Obviously, $|A|=89$, and the subsets of $A$ total $2^{89}$, among which the ... | 2^{44}\left(2^{45}-1\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,255 |
2. Let $P$ be a point on the plane of $\triangle A B C$, satisfying $\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}=2 \overrightarrow{A B}$.
If $S_{\triangle A B C}=1$, then $S_{\triangle P A B}=$ $\qquad$ | 2. $\frac{1}{3}$.
Let $O$ be the origin. Then
$$
\begin{array}{l}
(\overrightarrow{O A}-\overrightarrow{O P})+(\overrightarrow{O B}-\overrightarrow{O P})+(\overrightarrow{O C}-\overrightarrow{O P}) \\
=\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}=2 \overrightarrow{A B}=2(\overrightarrow{O B}-\overrig... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,256 |
3. In the pyramid $P-ABCD$, the base $ABCD$ is a square with side length $a$, $PA \perp$ plane $ABCD$, $PA = a$, and $M, N$ are the midpoints of $PD, PB$ respectively. Then the cosine value of the angle formed by the skew lines $AM$ and $CN$ is $\qquad$ | 3. $\frac{\sqrt{3}}{6}$.
Solution 1: As shown in Figure 1, complete the quadrilateral pyramid into a cube $A B C D-P B^{\prime} C^{\prime} D^{\prime}$. Then $M$ and $N$ are the centers of their respective faces. Let the center of face $P B^{\prime} C^{\prime} D^{\prime}$ be $R$, the midpoint of $A B$ be $S$, and the m... | \frac{\sqrt{3}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,257 |
4. Let $x, y$ be complex numbers. Then
$$
\frac{|3 x+4 y|}{\sqrt{|x|^{2}+|y|^{2}+\left|x^{2}+y^{2}\right|}}
$$
the maximum value is . $\qquad$ | 4. $\frac{5 \sqrt{2}}{2}$.
First, prove: $\frac{|3 x+4 y|}{\sqrt{|x|^{2}+|y|^{2}+\left|x^{2}+y^{2}\right|}} \leqslant \frac{5 \sqrt{2}}{2}$.
It suffices to prove:
$$
|3 x+4 y|^{2} \leqslant \frac{25}{2}\left(|x|^{2}+|y|^{2}+\left|x^{2}+y^{2}\right|\right) \text {, }
$$
which is
$$
\begin{array}{l}
2(3 x+4 y)(3 \bar{x}... | \frac{5 \sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,258 |
5. Let $A$ and $B$ be the two intersection points of the line $y=kx+1$ and the hyperbola $3x^2-y^2=1$, and let $O$ be the origin. If $\triangle OAB$ is a right triangle, then $k=$ $\qquad$ . | 5. $\pm 1$.
Let $A\left(x_{1}, k x_{1}+1\right) 、 B\left(x_{2}, k x_{2}+1\right)$.
From $O A \perp O B$ we get
$$
x_{1} x_{2}+\left(k x_{1}+1\right)\left(k x_{2}+1\right)=0 \text {, }
$$
which simplifies to $\left(1+k^{2}\right) x_{1} x_{2}+k\left(x_{1}+x_{2}\right)+1=0$.
Substituting $y=k x+1$ into $3 x^{2}-y^{2}=1$... | \pm 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,259 |
6. Let $f_{1}(x)=\frac{2 x-1}{x+1}$, for $n \geqslant 2$, define $f_{n}(x)=f_{1}\left(f_{n-1}(x)\right)$.
If $f_{29}(x)=\frac{1+x}{2-x}$, then $f_{2009}(x)=$ $\qquad$ | 6. $\frac{1+x}{2-x}$.
Since $f_{30}(x)=f\left(\frac{1+x}{2-x}\right)=x$, we have,
$$
f_{31}(x)=f_{1}(x) \text {. }
$$
And $2009=30 \times 66+29$, thus,
$$
f_{2009}(x)=f_{29}(x)=\frac{1+x}{2-x} .
$$ | \frac{1+x}{2-x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,260 |
7. Two people play a game: they continuously flip a coin several times, and the game ends when the cumulative number of times the head (or tail) side is up reaches 5. When the game ends, if the cumulative number of times the head side is up reaches 5, then $A$ wins; otherwise, $B$ wins. What is the probability that the... | $7: \frac{93}{128}$.
Consider the scenario where the game ends after 9 rotations. In this case, in the first 8 rotations, heads and tails each appear 4 times, with a probability of $\frac{C_{8}^{4}}{2^{8}}=\frac{35}{128}$. Therefore, the probability that the game ends before 9 rotations is $1-\frac{35}{128}=\frac{93}{1... | \frac{93}{128} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,261 |
8. Let $a_{1}=1, a_{2}=2$, for $n \geqslant 2$ we have
$$
a_{n+1}=\frac{2 n}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} .
$$
If for all positive integers $n \geqslant m$, we have $a_{n}>2+$ $\frac{2008}{2009}$, then the smallest positive integer $m$ is $\qquad$ . | 8.4019 .
According to the problem, the recursive relationship becomes
$$
\begin{array}{l}
a_{n+1}-a_{n}=\frac{n-1}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} \\
=\frac{n-1}{n+1}\left(a_{n}-a_{n-1}\right)(a \geqslant 2) .
\end{array}
$$
Therefore, when $n \geqslant 3$,
$$
\begin{array}{l}
a_{n}-a_{n-1}=\frac{n-2}{n}\left(a_{n... | 4019 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,262 |
9. (14 points) Let $a_{i} \in \mathbf{R}$, and $\left|a_{i}\right| \leqslant 1(i=1,2$, $\cdots, n)$. Prove:
$$
a_{1}+a_{2}+\cdots+a_{n}-a_{1} a_{2} \cdots a_{n} \leqslant n-1 \text {. }
$$ | When $n=2$, the inequality is
$$
a_{1}+a_{2}-a_{1} a_{2} \leqslant 1 \text {. }
$$
The above inequality is equivalent to $\left(a_{1}-1\right)\left(a_{2}-1\right) \geqslant 0$. The conclusion holds.
Assume that when $n=k$, the conclusion holds, i.e.,
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{k}-a_{1} a_{2} \cdots a_{k... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,263 |
10. (15 points) If all the coefficients of a polynomial are natural numbers, it is called a "natural polynomial". How many different natural polynomials $P(x)$ are there such that $P(2)=20$? | 10. For a positive integer $n$, let $A(n)$ denote the number of distinct natural polynomials $P(x)$ that satisfy $P(2) = n$.
It is easy to prove: for any positive integer $m$, we have
$$
\begin{array}{l}
A(2 m+1)=A(2 m) \\
=A(2 m-1)+A(m) .
\end{array}
$$
In fact, for any natural polynomial $P(x)$ that satisfies $P(2)=... | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,264 |
5. If the three-digit number $\overline{a b c}$ satisfies $1 \leqslant a \leqslant b \leqslant c \leqslant 9$, then $\overline{a b c}$ is called an "uphill number". Then, the number of uphill numbers is $\qquad$ | 5. 165 .
For a fixed $b$, the number $c$ can take $10-b$ values, namely $b, b+1, \cdots, 9$; and for a fixed number $a$, the number $b$ can take $a, a+1, \cdots, 9$. Therefore, the number of three-digit increasing numbers is
$$
\begin{array}{l}
\sum_{a=1}^{9} \sum_{b=a}^{9}(10-b)=\sum_{a=1}^{9} \sum_{k=1}^{10-a} k \\
... | 165 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,265 |
11. (15 points) Let $f(x)=x^{2}+b x+c(b, c \in$ $\mathbf{R})$. If for all $x \in \mathbf{R}$, we have $f\left(x+\frac{1}{x}\right) \geqslant 0$, and the maximum value of $f\left(\frac{2 x^{2}+3}{x^{2}+1}\right)$ is 1, find the trajectory of the moving point $P(b, c)$. | 11. Since $x$ and $\frac{1}{x}$ have the same sign, we have:
$$
\left|x+\frac{1}{x}\right|=|x|+\left|\frac{1}{x}\right| \geqslant 2 \text {. }
$$
According to the problem, for all real numbers $x$ satisfying $|x| \geqslant 2$, we have $f(x) \geqslant 0$.
(1) When $f(x)=0$ has real roots, the real roots of $f(x)=0$ are... | 3 b+c+8=0(-5 \leqslant b<-4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,266 |
One, (50 points) In a convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $P$, and the opposite sides $AD$ and $BC$ intersect at point $Q$. Prove that the circumcircles of $\triangle APD$, $\triangle BPC$, $\triangle ACQ$, and $\triangle BDQ$ are concurrent.
untranslated text remains the same ... | If the circumcircles of $\triangle A P D$ and $\triangle B P C$ are tangent, then $P$ is the point of tangency. As shown in Figure 2, draw the common tangent line $l$ through point $P$. Clearly, $\angle 1=\angle 2=\angle 3=\angle 4$, so $A D \parallel B C$. This contradicts the fact that $A D$ and $B C$ intersect at po... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,267 |
Sure, here is the translation:
---
II. (50 points) Let $f(x)$ be a function defined on the set of natural numbers $\mathbf{N}$ and taking values in $\mathbf{N}$, satisfying: for any two distinct natural numbers $a$ and $b$,
$$
f(a) + f(b) - f(a + b) = 2009.
$$
(1) Find $f(0)$;
(2) Let $a_{1}, a_{2}, \cdots, a_{100}$ ... | (1) Take $a=0, b=1$, then $a \neq b$.
By the condition, we have
$$
f(0)+f(1)-f(0+1)=2.009 \text {. }
$$
So, $f(0)=2009$.
(2) Proof: For any natural number $n$ greater than 1, and $n$ distinct natural numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$
\begin{array}{l}
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\lef... | 198891 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,268 |
Four, (50 points) In the 21st Century City, all streets run either east-west or north-south. To enhance security management, security booths are placed at some intersections (no two security booths are on the same street), and the rectangle formed by two security booths as two vertices and streets as edges is called a ... | Let the southernmost, easternmost, northernmost, and westernmost security booths be $A, B, C, D$ (some may coincide). The east-west streets passing through $A, C$ and the north-south streets passing through $B, D$ form a rectangle $M$, within which all security booths are located. Let $n=5k+r (0 \leqslant r \leqslant 4... | \left[\frac{n}{5}\right]+2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,270 |
1. If the graph of the function $y=f(x)$ passes through the point $(2,4)$, then the inverse function of $y=f(2-2x)$ must pass through the point | $-1 .(4,0)$.
Since $f(2)=4$, the function $y=f(2-2x)$ passes through the point $(0,4)$, so its inverse function passes through the point $(4,0)$. | (4,0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,271 |
2. The afternoon class schedule needs to arrange 5 classes: Physics, Chemistry, Biology, and two self-study periods. If the first class cannot be Biology, and the last class cannot be Physics, then, the number of different ways to arrange the schedule is $\qquad$ kinds. | 2.39.
According to the principle of inclusion-exclusion, the number of different ways to schedule classes is $\frac{5!}{2}-2 \times \frac{4!}{2}+\frac{3!}{2}=39$. | 39 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,272 |
3. For a regular octagon $A_{1} A_{2} \cdots A_{8}$ with side length 1, take any two points $A_{i} 、 A_{j}$. Then the maximum value of $\overrightarrow{A_{i} A_{j}} \cdot \overrightarrow{A_{1} \vec{A}_{2}}$ is $\qquad$ . | 3. $\sqrt{2}+1$.
According to the geometric meaning of the dot product of vectors, we only need to look at the projection of the vector in the direction of $\overrightarrow{A_{1} A_{2}}$. Therefore, the maximum value is $\sqrt{2}+1$. | \sqrt{2}+1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,273 |
5. The figure obtained by intersecting a cube with a plane passing through the center of the sphere circumscribing the cube could be $\qquad$
(1) triangle
(2) square
(3) trapezoid
(4) pentagon
(5) hexagon | 5.(2), (5).
By symmetry, the resulting figure should be a centrally symmetric figure, and (2), (5) can be intercepted. | (2), (5) | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,275 |
6.50 The sum of 50 positive numbers is 231, and the sum of their squares is 2009. Then, the maximum value of the largest number among these 50 numbers is $\qquad$ . | 6. 35 .
Let the maximum number be $x$, and the other 49 numbers be $a_{1}, a_{2}$, $\cdots, a_{49}$, and $x \geqslant a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{49}$. Then
$$
\begin{array}{l}
\sum_{i=1}^{49} a_{i}=231-x, \\
2009=x^{2}+\sum_{i=1}^{49} a_{i}^{2} \geqslant x^{2}+\frac{1}{49}\left(\sum_{i=1}^{49}... | 35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,276 |
6. If $n \in \mathbf{N}$, and $\sqrt{n^{2}+24}-\sqrt{n^{2}-9}$ is a positive integer, then $n=$ $\qquad$ | 6.5.
Notice that $\sqrt{n^{2}+24}-\sqrt{n^{2}-9}$
$$
=\frac{33}{\sqrt{n^{2}+24}+\sqrt{n^{2}-9}},
$$
thus $\sqrt{n^{2}+24}+\sqrt{n^{2}-9}$ could be $1,3,11,33$. Therefore, the solution is $n=5$. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,277 |
7. For a real number $x$, $[x]$ denotes the greatest integer not exceeding the real number $x$. For some integer $k$, there are exactly 2008 positive integers $n_{1}, n_{2}, \cdots, n_{2008}$, satisfying
$$
k=\left[\sqrt[3]{n_{1}}\right]=\left[\sqrt[3]{n_{2}}\right]=\cdots=\left[\sqrt[3]{n_{2008}}\right],
$$
and $k \m... | 7.668.
If $\sqrt[3]{n}-1n \text {. }$
Also, if $k \mid n$, then $n$ can take $k^{3}, k^{3}+k, \cdots, k^{3}+$ $3 k^{2}+3 k$, a total of $3 k+4$.
Therefore, $3 k+4=2008 \Rightarrow k=668$. | 668 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,278 |
8. Team $A$ and Team $B$ are having a table tennis team competition. Each team has three players, and each player will play once. The three players of Team $A$ and Team $B$ are $A_{1}, A_{2}, A_{3}, B_{1}, B_{2}, B_{3}$, respectively, and the winning probability of $A_{i}$ against $B_{j}$ is $\frac{i}{i+j}(1 \leqslant ... | 8. $\frac{91}{60}$.
It can be discussed that when $A_{1}: B_{3}, A_{2}: B_{1}, A_{3}: B_{2}$, the maximum expected value is $\frac{91}{60}$. | \frac{91}{60} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,279 |
9. (14 points) Given $f(x)=\sin 2 x+3 \sin x+$ $3 \cos x(0 \leqslant x<2 \pi)$. Find:
(1) the range of $f(x)$;
(2) the intervals of monotonicity of $f(x)$. | II. 9. (1) Let
$$
\sin x + \cos x = t \quad (-\sqrt{2} \leqslant t \leqslant \sqrt{2}).
$$
Then $\sin 2x = t^2 - 1$.
We need to find the range of $g(t) = t^2 + 3t - 1$.
Since $g(t) = \left(t + \frac{3}{2}\right)^2 - \frac{13}{4}$, when $t = \pm \sqrt{2}$,
$g(t)$ attains its extreme values, i.e., the range of $f(x)$ is... | \left[0, \frac{\pi}{4}\right], \left[\frac{5\pi}{4}, 2\pi\right] \text{ 为增区间}, \left[\frac{\pi}{4}, \frac{5\pi}{4}\right] \text{ 为减区间} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,280 |
$\begin{array}{l}\text { 10. (15 points) Given } a>0, b>0, \\ \log _{a} a=\log _{12} b=\log _{16}(a+b) . \\ \text { Find the value of } \frac{b}{a} .\end{array}$ | 10. Let $\log _{9} a=\log _{12} b=\log _{16}(a+b)=k$. Then $a=9^{k}, b=12^{k}, a+b=16^{k}$. Therefore, $(a+b) a=b^{2}$. Solving this, we get $\frac{b}{a}=\frac{1+\sqrt{5}}{2}$. | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,281 |
11. (15 points) Through the point $(2,3)$, draw a moving line $l$ intersecting the ellipse $\frac{x^{2}}{4}+y^{2}=1$ at two distinct points $P$ and $Q$. Draw the tangents to the ellipse at $P$ and $Q$, and let the intersection of these tangents be $M$.
(1) Find the equation of the locus of point $M$;
(2) Let $O$ be the... | 11. (1) According to the problem, let the equation of line $l$ be $y=k(x-2)+3$.
When combined with the ellipse, we get
$$
\left(1+4 k^{2}\right) x^{2}+8 k(3-2 k) x+4\left(4 k^{2}-12 k+8\right)=0 \text {. }
$$
Therefore, $\Delta=64(3 k-2)>0 \Rightarrow k>\frac{2}{3}$.
Let $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2},... | x-y+1=0 \text{ or } 11 x-4 y-10=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,282 |
One, (50 points) As shown in Figure 1, the circumcenter of acute $\triangle ABC$ is $O$. The lines $BO$ and $CO$ intersect the sides $AC$ and $AB$ at points $B'$ and $C'$, respectively. The line $B'C'$ intersects the circumcircle of $\triangle ABC$ at points $P$ and $Q$. If $AP = AQ$, prove that $\triangle ABC$ is an i... | As shown in Figure 2, connect $BP$ and $QC$. Given $AP = AQ$, we have $\angle APQ = \angle AQP$. Therefore,
$$
\begin{array}{l}
\angle PBA \\
=\angle AQP \\
=\angle APC'.
\end{array}
$$
Also, $\angle PAB$ is a common angle, so
$\triangle APC' \sim \triangle ABP$.
Thus, $AC' \cdot AB = AP^2$.
Similarly, $AB' \cdot AC =... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,283 |
II. (50 points) Given the sequence $\left\{a_{n}\right\}$ defined by
$$
a_{1}=\frac{2}{3}, a_{n+1}=a_{n}^{2}+a_{n-1}^{2}+\cdots+a_{1}^{2}\left(n \in \mathbf{N}_{+}\right)
$$
Determine. If for any $n\left(n \in \mathbf{N}_{+}\right)$,
$$
\frac{1}{a_{1}+1}+\frac{1}{a_{2}+1}+\cdots+\frac{1}{a_{n}+1}<M
$$
holds true. Fin... | Given that when $n \geqslant 2$, $a_{n+1}=a_{n}^{2}+a_{n}$. Also, $a_{2}=a_{1}^{2}=\frac{4}{9}=\left(\frac{1}{3}\right)^{2}+\frac{1}{3}$. Let's set $b_{1}=\frac{1}{3}$ and $b_{n}=a_{n}$ for $n \geqslant 2$.
Then, $b_{n+1}=b_{n}^{2}+b_{n}\left(n \in \mathbf{N}_{+}\right)$.
We will prove by mathematical induction that:
$... | \frac{57}{20} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,284 |
Three, (50 points) (1) Prove: There exist infinitely many positive integers $n$, such that $3^{n}+2$ and $5^{n}+2$ are both composite.
(2) Determine whether there exist positive integers $p$ and $q$, such that for any $n (n \geqslant 2007)$, at least one of $p \cdot 3^{n}+2$ and $q \cdot 5^{n}+2$ is a prime number. Pro... | (1) Notice that $3^{n}+2=3\left(3^{n-1}-1\right)+5$.
By Fermat's Little Theorem, $3^{4} \equiv 1(\bmod 5)$.
Let $n-1=4 r$. Then $51\left(3^{n}+2\right)$, i.e., when $n=4 r+1\left(r \in \mathbf{N}_{+}\right)$, $3^{n}+2$ is composite.
Also, $5^{n}+2=5\left(5^{n-1}-1\right)+7$, by Fermat's Little Theorem, $5^{6} \equiv 1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,285 |
Four. (50 points) Given a square with side length $n$ and $(n+1)^{2}$ points inside it, where no three points are collinear. Prove: It is possible to select three of these points such that the area of the triangle formed by these three points does not exceed $\frac{1}{2}$.
untranslated part:
(No additional text to tr... | Let the convex hull of these $(n+1)^{2}$ points be a $k$-sided polygon.
(1) If $k \geqslant 4 n$, then the perimeter of the $k$-sided polygon does not exceed $4 n$ (since the convex hull is inside the square). Therefore, there exist two consecutive sides whose lengths sum to no more than 2. Then the area of the triangl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,286 |
7. The minimum value of the function $f(x)=\sqrt{9(x-3)^{2}+\left(x^{2}-12\right)^{2}}+$ $\sqrt{9 x^{2}+\left(x^{2}-\frac{9}{4}\right)^{2}}$ is $\qquad$ | 7. $\frac{57}{4}$.
Obviously, $\frac{1}{3} f(x)$
$$
=\sqrt{(x-3)^{2}+\left(\frac{x^{2}}{3}-4\right)^{2}}+\sqrt{x^{2}+\left(\frac{x^{2}}{3}-\frac{3}{4}\right)^{2}}
$$
Let $y=\frac{x^{2}}{3}$. This is the equation of a parabola, with its focus at $F\left(0, \frac{3}{4}\right)$, and the directrix
equation is $y=-\frac{3... | \frac{57}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,287 |
1. Let $a, b \in \mathbf{N}_{+}$, when $a^{2}+b^{2}$ is divided by $a+b$, the quotient is $q$, and the remainder is $r$. Then the number of pairs $(a, b)$ that satisfy $q^{2}+r=2009$ is $\qquad$ pairs. | -1.0 .
From $2\left(a^{2}+b^{2}\right) \geqslant(a+b)^{2}$, we get $\frac{a^{2}+b^{2}}{a+b} \geqslant \frac{a+b}{2}$.
Then $q+1>q+\frac{r}{a+b}=\frac{a^{2}+b^{2}}{a+b}$ $\geqslant \frac{a+b}{2} \geqslant \frac{r+1}{2}$.
Thus, $r<2 q+1$.
From $q^{2}+r=2009<(q+1)^{2}$, we get $q \geqslant 44$. From $q^{2}+r=2009 \geqslan... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,288 |
2. Four families are sequentially denoted as A, B, C, and D, with four members in each family. Everyone is divided into four groups for a four-person competition. The probability that at least two groups, each with four people from different families, is $\qquad$
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 2. $\frac{85(4!)^{4}}{4 \times 16!}$
Let four groups be numbered as $1, 2, 3, 4$. Let the event that exactly $k$ groups, each with four people from different families, be denoted as $A_{k}(k=0,1,2,3,4)$. Then the number of ways to divide 16 people into four groups is $\frac{16!}{(4!)^{4}}$. It is easy to see that $A_{... | \frac{85(4!)^{8}}{4 \times 16!} | Number Theory | proof | Yes | Yes | cn_contest | false | 721,289 |
3. Given in $\triangle A B C$, $A C \geqslant A B$, side $B C$ is divided into $n$ ($n$ is an odd number) equal parts. Let $\alpha$ represent the angle subtended at point $A$ by the segment containing the midpoint of side $B C$, $h$ be the altitude from $A$ to side $B C$, and $B C=a$. If $\tan \alpha=\frac{4 n h}{\left... | 3. $90^{\circ}$.
Draw a perpendicular from point $A$ to $BC$, with the foot of the perpendicular being $D$. Let the segment containing the midpoint of $BC$ be $B' C'$. Then
$$
B B'=\frac{n-1}{2 n} a, B C'=\frac{n+1}{2 n} a .
$$
Let $\overrightarrow{B D}=x \overrightarrow{B C}(x \in \mathbf{R})$. Then
$$
D B'=\left(\f... | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,290 |
4. Given $a, b, c, d \in \mathbf{Z}$,
$$
f(x)=a x^{3}+b x^{2}+c x+d \text {. }
$$
Then among $A(1,1)$, $B(2009,1)$, $C(-6020,2)$, $D(2,-6020)$, at most $\qquad$ of them can be on the curve $y=f(x)$. | 4.3.
If $A$ and $C$ are both on $y=f(x)$, then $f(1)=1, f(-6020)=2$.
It is easy to see that $\left(x_{1}-x_{2}\right) \mid \left(f\left(x_{1}\right)-f\left(x_{2}\right)\right)$. Therefore, $[1-(-6020)] \mid (f(1)-f(-6020))=-1$, which is a contradiction.
Thus, $A$ and $C$ cannot both be on $y=f(x)$.
Similarly, $B$ and ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,291 |
5. Given a tetrahedron $ABCD$ divided by $n+1$ (where $n$ is odd) planes $\pi_{0}$, $\pi_{1}, \cdots, \pi_{n}$ parallel to $AB$ and $CD$ and equally spaced, into $n$ parts, where $AB \subset \pi_{0}$ and $CD \subset \pi_{n}$. Let the volumes of these $n$ parts be $V_{1}, V_{2}, \cdots, V_{n}$. Then $\left(\sum_{i \text... | 5. $\left(n^{3}+1\right):\left(n^{3}-1\right)$.
As shown in Figure 1, plane $E F G H$ is parallel to edges $A B$ and $C D$. A plane $E H K$ is drawn through $E H$ parallel to plane $B C D$.
Let $A E=t A C(t \in(0,1))$. Then,
$$
\begin{array}{c}
V_{A-E H K} \\
=t^{3} V_{A-B C D}, \\
V_{B F G-N E H}=\frac{3(1-t)}{t} V_... | \left(n^{3}+1\right):\left(n^{3}-1\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,292 |
6. Given a prime number $p$ satisfying $p \equiv-1(\bmod 4), n$ is a positive integer. Then “ $2 \sqrt{1+4 p n^{2}}$ is a positive integer” is a “ $2+$ $2 \sqrt{1+4 p n^{2}}$ is a perfect square” $\qquad$ condition (fill in “sufficient but not necessary”, “necessary and sufficient”, or “necessary but not sufficient”).
... | 6. Necessary and Sufficient.
Necessity is evident.
Sufficiency: From $2 \sqrt{1+4 p n^{2}}=m(m \in \mathbf{N})$, we have $1+4 p n^{2}=\frac{m^{2}}{4} \in \mathbf{N}$.
Also, $1+4 p n^{2}=(2 t+1)^{2}(t \in \mathbf{N})$, so $p n^{2}=t(t+1)$.
Since $(t, t+1)=1$, one of $t$ or $t+1$ must be a perfect square, and the other ... | Necessary and Sufficient | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,293 |
7. The equation $\sum_{k=1}^{2 n+1} \cos ^{2} k x=n\left(n \in \mathbf{N}_{+}\right)$ has $\qquad$ solutions in $[0,2 \pi)$. | $$
\begin{array}{l}
7.8 n+2 . \\
\sum_{k=1}^{2 n+1} \cos ^{2} k x=n \\
\Leftrightarrow \sum_{k=1}^{2 n+1}(\cos 2 k x+1)=2 n \\
\Leftrightarrow \sum_{k=1}^{2 n+1} \cos 2 k x=-1 .
\end{array}
$$
Obviously, $x \neq t \pi(\iota \in \mathbf{Z})$. Then
$$
\begin{array}{l}
\sum_{k=1}^{2 n+1} \cos 2 k x=\operatorname{Re}\left... | 8n+2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,294 |
8. Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms $S_{n}$ satisfying
$$
S_{n}+a_{n-1}=\frac{n^{2}-2}{(n+1)!}(n=2,3, \cdots) \text {, }
$$
and $a_{1}=-\frac{1}{2}$.
Then the general term formula $a_{n}=$ $\qquad$ | 8. $\frac{1}{\sqrt{5}}\left[\left(\frac{-1-\sqrt{5}}{2}\right)^{n}-\left(\frac{-1+\sqrt{5}}{2}\right)^{n}\right]+\frac{n}{(n+1)!}$.
From $S_{n}+a_{n-1}=\frac{n^{2}-2}{(n+1)!}$, we know
$$
\begin{array}{l}
S_{n}+S_{n-1}-S_{n-2} \\
=\frac{n^{2}-2}{(n+1)!}=-\frac{1}{(n+1)!}-\frac{1}{n!}+\frac{1}{(n-1)!} .
\end{array}
$$
... | \frac{1}{\sqrt{5}}\left[\left(\frac{-1-\sqrt{5}}{2}\right)^{n}-\left(\frac{-1+\sqrt{5}}{2}\right)^{n}\right]+\frac{n}{(n+1)!} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,295 |
9. (14 points) A soldier needs to check for landmines in a circular area with a radius of $2 a \mathrm{~m}$. The effective range of the detection instrument he uses is $a \mathrm{~m}$. Find the minimum distance the soldier needs to walk starting from a point $A$ on the boundary of the circle, to complete the inspection... | $=9 \cdot\left(\frac{4 \pi}{3}+2 \sqrt{3}\right) a$.
First, find the shortest path $\Gamma$ for a soldier to start from $A$, inspect all points on the boundary of the circle $D(O, 2a)$, and return to $A$.
Below, we use proof by contradiction to show:
(1) Any line segment connecting two points on $\Gamma$ lies within th... | \left(\frac{4 \pi}{3} + 2 \sqrt{3}\right) a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,296 |
10. (15 points) Given a set of positive integers $M=$ $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, the sum of its elements is 2009. Find the maximum value of the product of its elements. | 10. $\frac{63!}{6}$.
Let 2009 be written as the sum of several different positive integers as
$$
2009=a_{1}+a_{2}+\cdots+a_{n},
$$
where, $a_{1}1$. Otherwise,
$$
\begin{array}{l}
a_{1} a_{2} \cdots a_{n}=a_{2} a_{3} \cdots a_{n} \\
a_{1} \text {, }
$$
a contradiction.
(3) There does not exist $1 \leqslant i \leqslan... | \frac{63!}{6} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,297 |
8. Let $M=\{1,2, \cdots, 17\}$. If there are four distinct numbers $a, b, c, d \in M$, such that $a+b \equiv c+d(\bmod 17)$, then $\{a, b\}$ and $\{c, d\}$ are called a “balanced pair” of set $M$. The number of balanced pairs in set $M$ is $\qquad$. | 8.476.
Divide the circumference into 17 equal parts, and label the points in clockwise order as $A_{1}, A_{2}, \cdots, A_{17}$. Then
$$
\begin{array}{l}
m+n \equiv k+l(\bmod 17) \\
\Leftrightarrow A_{m} A_{n} / / A_{k} A_{l} .
\end{array}
$$
Note that none of the chords connecting any pair of the 17 equally divided p... | 476 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,298 |
11. (15 points) Let $n \in \mathbf{N}_{+}, n \geqslant 4$. For a permutation $\left\{x_{1}, x_{2}, \cdots, x_{2 n}\right\}$ of $\{1,2, \cdots, 2 n\}$, if there exists $1 \leqslant i \leqslant 2 n-1$ such that $x_{i}+x_{i+1}=2 n+1$, then the permutation is said to have property $P$. Let the number of permutations with p... | 11. Let $\left(x_{1}, x_{2}, \cdots, x_{2 n}\right)$ be a permutation where $k$ is adjacent to $2 n+1-k$ in the set $N_{k}(1 \leqslant k \leqslant n)$. We need to find
$$
\begin{array}{l}
\left|\bigcap_{i=1}^{t} N_{k_{i}}\right|\left(1 \leqslant k_{1}\mathrm{C}_{n}^{1} \times 2 \times(2 n-1)!-\mathrm{C}_{n}^{2} \times ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,299 |
In the scalene $\triangle ABC$, $AD$, $BE$, and $CF$ are the angle bisectors of the three interior angles $(D \in BC, E \in AC, F \in AB)$, and $K_{a}$, $K_{b}$, $K_{c}$ are three points on the incircle $\odot I$ of $\triangle ABC$ such that $DK_{a}$, $EK_{b}$, and $FK_{c} \left(K_{a} \notin BC, K_{b} \notin CA, K_{c} ... | (1) Let the incircle $\odot I$ of $\triangle ABC$ touch $BC$, $CA$, and $AB$ at points $T_{a}$, $T_{b}$, and $T_{c}$, respectively.
It is easy to see that $K_{a}$ is symmetric to $T_{a}$ with respect to $AD$.
Since $T_{b}$ and $T_{c}$ are symmetric with respect to $AD$ and $I \in AD$, it follows that on $\odot I$, $\ov... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,300 |
Let $n \in \mathbf{N}_{+}$. Find the positive real number $T(n)$ such that the set of real numbers $x$ satisfying the inequality $\sum_{k=1}^{n} \frac{k}{x-k} \geqslant T(n)$ is the union of disjoint intervals, and the total length of these intervals is 2009. | Consider the function $f(x)=\sum_{k=1}^{n} \frac{k}{x-k}-T(n)$.
It is easy to see that when $k=1,2, \cdots, n$,
$$
\lim _{x \rightarrow k^{+}} f(x)=+\infty, \lim _{x \rightarrow k^{-}} f(x)=-\infty \text {. }
$$
And $f(x)$ is a strictly decreasing continuous function in $(k, k+1)$, therefore, $f(x)=0$ has exactly one ... | T(n)=\frac{n(n+1)}{4018} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,301 |
Three, (50 points) Find the number of positive integers $t$ not exceeding 2009, such that for all natural numbers $n$, $\sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1} t^{k}$ is coprime with 2009.
---
The translation maintains the original text's line breaks and formatting. | $$
\begin{array}{l}
\text { Three, let } \\
\begin{array}{l}
x= \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1} t^{k}=\frac{1}{\sqrt{t}} \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1}(\sqrt{t})^{2 k+1}, \\
\begin{array}{l}
y= \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k} t^{k}=\sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k}(\sqrt{t})^{2 k} .... | 980 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,302 |
Four. (50 points) Let $M$ be a set composed of $n\left(n \in \mathbf{N}_{+}\right)$ distinct positive integers, where each element has prime factors no greater than 100, and there do not exist four distinct elements in $M$ such that the product of these four numbers is a fourth power. Find the maximum value of $n$.
u... | Prime numbers not greater than 100 are
$2,3,5,7,11,13,17,19,23,29,31,37,41$, $43,47,53,59,61,67,71,73,79,83,89,97$.
Denoted as $p_{i}(i=1,2, \cdots, 25)$. $\alpha_{i} \in \mathbf{N}(i=1,2, \cdots, 25)$.
The tuple $\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{25}\right)$, classified by parity, has $2^{25}$ types. Thu... | 3 \times 2^{25} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,303 |
1. Given the identity
$$
\begin{array}{l}
x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4} \\
=(x+1)^{4}+b_{1}(x+1)^{3}+b_{2}(x+1)^{2}+ \\
\quad b_{3}(x+1)+b_{4} .
\end{array}
$$
Express $b_{3}$ in terms of $a_{1}, a_{2}, a_{3}, a_{4}$, then $b_{3}=$
$\qquad$ | $$
-1 .-4+3 a_{1}-2 a_{2}+a_{3} \text {. }
$$
Substituting $x=-1$ into the known identity, we get
$$
b_{4}=1-a_{1}+a_{2}-a_{3}+a_{4} .
$$
Rearranging and simplifying, we have
$$
\begin{array}{l}
\left(x^{4}-1\right)+a_{1}\left(x^{3}+1\right)+a_{2}\left(x^{2}-1\right)+a_{3}(x+1) \\
=(x+1)^{4}+b_{1}(x+1)^{3}+b_{2}(x+1)... | -4+3 a_{1}-2 a_{2}+a_{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,304 |
2. There is a $19 \times 19$ square chessboard. If two horizontal lines and two vertical lines are randomly selected, the probability that the enclosed shape is a square is $\qquad$ . | 2. $\frac{13}{190}$.
Squares with side length 1 have $19^{2}$, squares with side length 2 have $18^{2}$, ... squares with side length 19 have $1^{2}$. Therefore, the total number of squares is
$$
1^{2}+2^{2}+\cdots+19^{2}=\frac{19 \times 20 \times 39}{6} \text { (squares) } .
$$
The required probability is
$$
P=\frac... | \frac{13}{190} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,305 |
3. A line segment $AB$ of length 4 moves along the positive x-axis, and another line segment $CD$ of length 2 moves along the positive y-axis. If the four endpoints $A$, $B$, $C$, and $D$ are concyclic, then the locus of the center of this circle is $\qquad$ | 3. $x^{2}-y^{2}=3(x>2, y>1)$.
As shown in Figure 2, let the center of the circle be $M(x, y)(x>2, y>1)$. Then $A(x-2,0)$,
$B(x+2,0)$.
$C(0, y-1)$,
$D(0, y+1)$.
From $|M A|^{2}=|M C|^{2}$,
we get $2^{2}+y^{2}=x^{2}+1^{2}$.
Therefore, the required locus is a segment of a hyperbola, whose equation is $x^{2}-y^{2}=3(x>2, ... | x^{2}-y^{2}=3(x>2, y>1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,306 |
4. Given that $a, b$ are positive real numbers, $n \in \mathbf{N}_{+}$. Then the function
$$
f(x)=\frac{\left(x^{2 n}-a\right)\left(b-x^{2 n}\right)}{\left(x^{2 n}+a\right)\left(b+x^{2 n}\right)}
$$
has a maximum value of $\qquad$ . | 4. $\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)^{2}$.
Let $t=x^{2 n}$. Then $t \geqslant 0$, H
$$
\begin{aligned}
f & =\frac{(t-a)(b-t)}{(t+a)(b+t)} \\
& =\frac{-t^{2}+(a+b) t-a b}{t^{2}+(a+b) t+a b} \\
& =-1+\frac{2(a+b) t}{t^{2}+(a+b) t+a b} .
\end{aligned}
$$
When $t=0$, $f=-1$;
When $t>0$,
$$
\begin{... | \left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,307 |
5. As shown in Figure 1, the plane containing
square $A B C D$ and the plane containing
square $A B E F$ form a $45^{\circ}$ dihedral angle.
Then the angle formed by the skew lines $A C$ and
$B F$ is $\qquad$ | 5. $\arccos \frac{2-\sqrt{2}}{4}$.
Let's assume the side length of the square is 1. Then
$$
\begin{array}{l}
\cos \langle\overrightarrow{A C}, \overrightarrow{B F}\rangle=\frac{\overrightarrow{A C} \cdot \overrightarrow{B F}}{\sqrt{2} \cdot \sqrt{2}} \\
=\frac{(\overrightarrow{A D}+\overrightarrow{D C}) \cdot(\overrig... | \arccos \frac{2-\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,308 |
9. (14 points) Given three non-collinear points $A, B, C$ in the plane, construct an ellipse with segment $AB$ as one of its axes (major or minor) such that the ellipse does not pass through point $C$, and intersects the lines $AC, BC$ at points $E, F$ respectively. Draw the tangents to the ellipse at $E, F$, and let t... | 9. First, consider an ellipse with side $AB$ as one of its axes, as shown in Figure 4, establishing a Cartesian coordinate system. Let the equation of the ellipse be
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a \neq
$$
$b)$, which intersects the lines $AC$ and $BC$ at points $E$ and $F$, respectively. Tangents to the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,309 |
6. The sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, \\
a_{n+2}=2 a_{n+1}-a_{n}+2(n=1,2, \cdots) .
\end{array}
$$
Then the sum of its first $n$ terms is | 6. $\frac{1}{3} n\left(n^{2}+2\right)$.
From the given, we have $a_{n+2}-a_{n+1}=\left(a_{n+1}-a_{n}\right)+2$. Therefore, the sequence $\left\{a_{n+1}-a_{n}\right\}$ is an arithmetic sequence with a common difference of 2, and the first term is $a_{2}-a_{1}=2$. Hence, $a_{n+1}-a_{n}=2 n$. Thus,
$$
\begin{array}{l}
a_... | \frac{1}{3} n\left(n^{2}+2\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,310 |
7. In the Cartesian coordinate plane, the 4 points $A(1,2)$, $B(3,1)$, $C(2,3)$, $D(4,0)$ have a sum of the squares of their distances to the line $y=kx$ denoted as $S$. When $k$ varies, the minimum value of $S$ is $\qquad$ | $7.22-\sqrt{185}$.
Let the distances from points $A$, $B$, $C$, and $D$ to the line $k x-y=0$ be $d_{1}$, $d_{2}$, $d_{3}$, and $d_{4}$, respectively. Then
$$
\begin{array}{l}
d_{1}=\frac{|k-2|}{\sqrt{k^{2}+1}}, d_{2}=\frac{|3 k-1|}{\sqrt{k^{2}+1}}, \\
d_{3}=\frac{|2 k-3|}{\sqrt{k^{2}+1}}, d_{4}=\frac{|4 k|}{\sqrt{k^{2... | 22-\sqrt{185} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,311 |
8. The positive integer $n$ ensures that in every $n$-element subset of the set $\{1,2, \cdots, 2008\}$, there are 2 elements (which can be the same), whose sum is a positive integer power of 2. Then the minimum value of $n$ is $\qquad$ | 8.1003.
First, if $n=1002$, take
$$
\begin{array}{l}
A=\{5,6,7\}, B=\{17,18, \cdots, 24\}, \\
C=\{33,34, \cdots, 39\}, \\
D=\{1025,1026, \cdots, 2008\} .
\end{array}
$$
Then $S=A \cup B \cup C \cup D$.
Thus, $|S|=3+8+7+984=1002$.
It is easy to verify that the sum of any two (possibly the same) elements in $S$ is not ... | 1003 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,312 |
9. (14 points) Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=1+2+\cdots+n\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
take all multiples of 3 from this sequence to form a new sequence $b_{1}, b_{2}, \cdots, b_{m}, \cdots$. Find the sum of the first $2 m$ terms of the sequence $\left\{b_{m... | Obviously, $a_{n}=\frac{n(n+1)}{2}$.
If $a_{n}=3 t\left(t \in \mathbf{Z}_{+}\right)$, then since $n$ and $n+1$ are coprime, one of $n$ and $n+1$ must be a multiple of 3.
When $n=3 k\left(k \in \mathbf{Z}_{+}\right)$, $a_{3 k}=\frac{3 k(3 k+1)}{2}$;
When $n+1=3 k$, i.e., $n=3 k-1\left(k \in \mathbf{Z}_{+}\right)$,
$$
a_... | \frac{3}{2} m(m+1)(2 m+1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,313 |
10. (14 points) In $\triangle A B C$, $B C=a, C A=$ $b$, a square $A B E F$ is constructed outward with side $A B$, $O$ is the center of the square $A B E F$, $M$ and $N$ are the midpoints of sides $B C$ and $C A$, respectively. When $\angle B C A$ varies, find the maximum value of $O M+O N$.
---
Please note that the... | 10. As shown in Figure 3, in $\triangle O B M$, by the Law of Cosines (let $A B = c$, $\angle A B C = \beta$)
$$
\begin{aligned}
& O M^{2} \\
& =\left(\frac{\sqrt{2}}{2} c\right)^{2}+ \\
& \left(\frac{a}{2}\right)^{2}-2 \cdot \frac{\sqrt{2}}{2} c \cdot \frac{a}{2} \cos \left(\beta+45^{\circ}\right) \\
= & \frac{c^{2}}{... | \frac{\sqrt{2}+1}{2}(a+b) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,314 |
11. (16 points) Let $A(n, k)$ denote the number of $k$-element subsets of the set $\{1,2$, $\cdots, n\}$ that do not contain consecutive integers. Find $A(n, k)$. | 11. Clearly, $A(n, 1)=n$.
When $k \in \mathbf{Z}_{+}, k \geqslant 2$, and $n<2 k-1$,
$$
A(n, k)=0 \text {. }
$$
When $k \in \mathbf{Z}_{+}, k \geqslant 2$, and $n \geqslant 2 k-1$, let $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ be a $k$-element subset of $\{1,2, \cdots, n\}$ that does not contain consecutive intege... | A(n, k)=\mathrm{C}_{n-(k-1)}^{k} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,315 |
12. (16 points) On the Cartesian plane, a point whose both coordinates are rational numbers is called a rational point. Find the smallest positive integer $k$ such that: for every circle that contains $k$ rational points on its circumference, the circle must contain infinitely many rational points on its circumference. | 12. First, prove: If a circle's circumference contains 3 rational points, then the circumference must contain infinitely many rational points.
Let $\odot C_{0}$ in the plane have 3 rational points $P_{i}\left(x_{i}, y_{i}\right)(i=1,2,3)$ on its circumference, with the center $C_{0}\left(x_{0}, y_{0}\right)$.
Since t... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,316 |
1. The real number sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{0} \neq 0,1, a_{1}=1-a_{0}, \\
a_{n+1}=1-a_{n}\left(1-a_{n}\right)(n=1,2, \cdots) .
\end{array}
$$
Prove: For any positive integer $n$, we have
$$
a_{0} a_{1} \cdots a_{n}\left(\frac{1}{a_{0}}+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\rig... | 1. From the given conditions, we have
$$
\begin{array}{l}
1-a_{n+1}=a_{n}\left(1-a_{n}\right)=a_{n} a_{n-1}\left(1-a_{n-1}\right) \\
=\cdots=a_{n} a_{n-1} \cdots a_{1}\left(1-a_{1}\right)=a_{n} \cdots a_{1} a_{0},
\end{array}
$$
which means
$$
a_{n+1}=1-a_{0} a_{1} \cdots a_{n}(n=1,2, \cdots) \text {. }
$$
We will pr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,317 |
2. As in 1, in $\triangle ABC$, $AB=AC$, its incircle $\odot I$ touches $BC, CA, AB$ at points $D, E, F$, respectively. $P$ is a point on the arc $\overparen{EF}$ (excluding point $D$). Let line segment $BP$ intersect $\odot I$ again at point $Q$, and lines $EP, EQ$ intersect line $BC$ at points $M, N$, respectively. P... | 2. (1) Connect $E F$. From the given conditions, we know $E F / / B C$. Therefore,
$$
\begin{array}{l}
\angle A B C=\angle A F E=\angle A F P+\angle P F E \\
=\angle P E F+\angle P F E=180^{\circ}-\angle F P E .
\end{array}
$$
Thus, points $P, F, B, M$ are concyclic.
$$
\begin{array}{l}
\underset{E N}{E M}=\frac{\sin ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,318 |
3. Let integers $m(m \geqslant 2), a_{1}, a_{2}, \cdots, a_{m}$ all be positive integers. Prove: There exist infinitely many positive integers $n$, such that the number
$$
a_{1} \times 1^{n}+a_{2} \times 2^{n}+\cdots+a_{m} \times m^{n}
$$
is always composite.
(Supplied by Chen Yonggao) | 3. Take the prime factor $p$ of $a_{1}+2 a_{2}+\cdots+m a_{m}$. By Fermat's Little Theorem, for any $k(1 \leqslant k \leqslant m)$, we have
$$
h^{\mu} \equiv k(\bmod p) .
$$
Therefore, for any positive integer $n$ we have
$$
\begin{array}{l}
a_{1} \times 1^{p^{n}}+a_{2} \times 2^{p^{n}}+\cdots+a_{m} \times m^{p^{n}} \... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,319 |
5. Points on a straight line where the distance between adjacent points is 1 are called unit points: There is a frog, allowing any frog to jump to the symmetric point of another frog with respect to a third frog. Prove: No matter how many times the frogs jump, the distance between the points where the four frogs are lo... | 5. Place the frogs on a number line for discussion.
Assume the initial positions of the four frogs are 1, $2, 3, 4$. Note that, frogs on odd positions will still be on odd positions after each jump, and frogs on even positions will still be on even positions after each jump. Therefore, after any number of jumps, there... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,322 |
6. Let $x, y, z \in (0,1)$, satisfy
$$
\sqrt{\frac{1-x}{y z}}+\sqrt{\frac{1-y}{z x}}+\sqrt{\frac{1-z}{x y}}=2 \text {. }
$$
Find the maximum value of $x y z$.
(Tang Lihua, provided) | 6. Let $u=\sqrt[5]{x y z}$. Then, by the condition and the AM-GM inequality, we have
$$
\begin{array}{l}
2 u^{3}=2 \sqrt{x y z}=\frac{1}{\sqrt{3}} \sum \sqrt{x(3-3 x)} \\
\leqslant \frac{1}{\sqrt{3}} \sum \frac{x+(3-3 x)}{2}=\frac{3 \sqrt{3}}{2}-\frac{1}{\sqrt{3}}(x+y+z) \\
\leqslant \frac{3 \sqrt{3}}{2}-\sqrt{3} \cdot... | \frac{27}{64} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,323 |
7. Let $n$ be a given positive integer. Find the largest integer $k$ such that there exist sets
$$
\begin{array}{l}
A=\left\{x_{1}, x_{2}, \cdots, x_{k}\right\}, \\
B=\left\{y_{1}, y_{2}, \cdots, y_{k}\right\}, \\
C=\left\{z_{1}, z_{2}, \cdots, z_{k}\right\}
\end{array}
$$
satisfying for any $j(1 \leqslant j \leqslant... | 7. From the conditions, we have
$$
\begin{array}{l}
k n=\sum_{i=1}^{k}\left(x_{i}+y_{i}+z_{i}\right) \\
\geqslant 3 \sum_{i=0}^{k-1} i=\frac{3 k(k-1)}{2} .
\end{array}
$$
Therefore, \( k \leqslant\left[\frac{2 n}{3}\right]+1 \).
Below is an example for \( k=\left[\frac{2 n}{3}\right]+1 \).
If \( n=3 m \), for \( 1 \le... | \left[\frac{2 n}{3}\right]+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,324 |
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