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 |
|---|---|---|---|---|---|---|---|---|---|
8. A mapping $f$ from set $A$ to set $B$ is called a surjection if for every element $y$ in set $B$, there is at least one $x \in A$ such that $f(x)=y$. Given a five-element set $A=\{1,2,3,4,5\}$ and a three-element set $B=$ $\{\alpha, \beta, \gamma\}$. Then the number of surjections from set $A$ to set $B$ is. | 8. 150 .
The total number of mappings from a five-element set to a three-element set is $3^{5}$. Among these, the mappings that are not surjective can be divided into two categories for calculation: one category is where only one element in $B$ is the image of all elements in $A$, and there are $\mathrm{C}_{3}^{1}=3$ ... | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,909 |
Example 1 Given that $P$ is a point inside $\triangle ABC$. Prove that: $\angle PAB$, $\angle PBC$, $\angle PCA$ include at least one that is less than or equal to $30^{\circ}$.
(32nd IMO) | Proof 1: By the Law of Sines, we have
$$
\begin{array}{l}
\frac{P B}{P C}=\frac{\sin \angle P C B}{\sin \angle P B C}, \frac{P C}{P A}=\frac{\sin }{\sin } \angle P A C \\
\frac{P A}{P B}=\frac{\sin \angle P B A}{\sin \angle P A B} .
\end{array}
$$
Then $\sin \angle P C B \cdot \sin \angle P A C \cdot \sin \angle P B A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,910 |
9. If $x=\frac{1+\sqrt{5} \mathrm{i}}{2}$, then the value of $2 x^{3}+(x+1)^{2}$ is $\qquad$
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 9. $\sqrt{5} \mathrm{i}-\frac{5}{2}$.
$\frac{1+\sqrt{5}}{2}$ is a root of the equation $x^{2}-x+\frac{3}{2}=0$, then
$$
\begin{array}{l}
2 x^{3}+(x+1)^{2}=2 x^{3}+x^{2}+2 x+1 \\
=(2 x+3)\left(x^{2}-x+\frac{3}{2}\right)+2 x-\frac{7}{2}=\sqrt{5} \text { i }-\frac{5}{2}
\end{array}
$$ | \sqrt{5} \mathrm{i}-\frac{5}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,911 |
10. Let $S$ be the area of a triangle inscribed in a circle of radius 1. Then the minimum value of $4 S+\frac{9}{S}$ is $\qquad$ . | $10.7 \sqrt{3}$.
Let $y=4 S+\frac{9}{S}$, it is easy to know that it is a monotonically decreasing function on $\left(0, \frac{3}{2}\right]$. Also, because among the triangles inscribed in a unit circle, the equilateral triangle has the maximum area, which is equal to $\frac{3 \sqrt{3}}{4}$, hence $0<S \leqslant \frac{... | 7 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,912 |
11. Let the set $M=\{1,2, \cdots, 2005\}$. If $X \subseteq M$, $X \neq \varnothing$, and $a_{X}$ is the sum of the largest and smallest numbers in $X$ (if the set $X$ contains only one element, then this element is both the largest and the smallest number), then, for all non-empty subsets $X$ of $M$, the average value ... | 11.2006.
For any non-empty subset $X \subseteq M$, define $X^{\prime}=\{2006-x \mid x \in X \}$. It is clear that $X^{\prime} \subseteq M$. Thus, $f: X \rightarrow X^{\prime}$ provides a one-to-one correspondence on the family of all non-empty subsets of $M$. Note that
$$
\begin{array}{l}
\max (X)+\min \left(X^{\prime... | 2006 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,913 |
12. $A, B, C, D$ are the 4 vertices of a regular tetrahedron, with each edge length being $1 \, \text{m}$. A gecko starts from point $A$ and crawls along the edges, following these rules: it does not change direction midway along any edge, and when it reaches each vertex, it has an equal probability of choosing any of ... | 12. $\frac{182}{729}$.
Let the number of ways the gecko can crawl $n \mathrm{~m}$ and return to vertex $A$ be $a_{n}$, and the number of ways to crawl $7 \mathrm{~m}$ and return to vertex $A$ be $a_{1}$. We can derive the recursive relation $a_{n}=3^{n-1}-a_{n-1}, n \geqslant 2$, and it is easy to see that $a_{1}=0$. ... | \frac{182}{729} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,914 |
13. Given that $n$ and $k$ are natural numbers, and satisfy the inequality $\frac{9}{17}<\frac{n}{n+k}<\frac{8}{15}$. If for a given natural number $n$, there is only one natural number $k$ that makes the inequality true, find the maximum and minimum values of the natural number $n$ that meet the condition. | When $n=10, 11, \cdots, 16$, respectively, we have
$$
\begin{array}{l}
8 \frac{3}{4}<k<8 \frac{8}{9}, 9 \frac{5}{8}<k<9 \frac{7}{9}, \\
10 \frac{1}{2}<k<10 \frac{2}{3}, 11 \frac{3}{8}<k<11 \frac{5}{9}, \\
12 \frac{1}{4}<k<12 \frac{4}{9}, 13 \frac{1}{8}<k<13 \frac{1}{3}, \\
14<k<14 \frac{2}{9},
\end{array}
$$
None of t... | 17, 144 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,915 |
14. Let a line parallel to the major axis of an ellipse $C$ through a point $P$ on the ellipse intersect the two directrices of the ellipse $C$ at points $T_{1}$ and $T_{2}$. The lines connecting these points to the foci $F_{1}$ and $F_{2}$ on the same side intersect at point $Q$. Prove that the points $P, F_{1}, Q, F_... | 14. As shown in Figure 2, let
the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$
$$
=1(a>b>0) \text { have }
$$
foci at
$$
\begin{array}{l}
F_{1}(-c, 0), \\
F_{2}(c, 0) .
\end{array}
$$
Let $P(a \cos \alpha, b \sin \alpha)$, then $T_{2}\left(\frac{a^{2}}{c}, b \sin \alpha\right)$.
By symmetry, point $Q$ lies o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,916 |
15. A and B take turns tossing a fair coin, and whoever tosses heads first wins, at which point the game ends, and the loser gets to toss first in the next game.
(1) Find the probability that the first person to toss wins in any given game;
(2) Suppose they play a total of 10 games, and A tosses first in the first game... | 15. (1) In any given match, the probability of the first person to throw winning is $\frac{1}{2}+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{5}+\cdots=\frac{2}{3}$.
(2) From (1), the probability of the second person to throw winning is $1-\frac{2}{3}=\frac{1}{3}$.
Given $P_{1}=\frac{2}{3}$, for $2 \leqslant... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,917 |
16. Let the set $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}, 0<a_{i}<a_{i+1}(i=1,2,3)$. Determine whether there must exist two elements $x, y$ in the set $A$ such that the inequality
$$
(2+\sqrt{3})|x-y|<(x+1)(y+1)+x y
$$
holds. If such elements exist, provide a proof; if not, explain why. | 16. There exist such two numbers.
Assume $x>y$, then the original inequality can be transformed into
$$
\begin{array}{l}
\frac{x-y}{(x+1)(y+1)+x y} < 0,
\end{array}
$$
From $0 < x - y$, we know that the denominator $(x+1)(y+1) + xy > 0$, which is always true for positive $x$ and $y$. Therefore, the inequality holds.
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,918 |
17. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1} a_{n}-1=a_{n}^{2} \text {. }
$$
(1) Prove: $\sqrt{2 n-1} \leqslant a_{n} \leqslant \sqrt{3 n-2}$;
(2) Find the integer $m$, such that $\left|a_{2 \cos }-m\right|$ is minimized. | 17. (1) From the given information, it is easy to see that $\{a_n\}$ is an increasing sequence, and all its terms are positive. Since $a_{n+1} a_{n} - 1 = a_{n}^2$, we have, for $k \geq 2$,
$$
a_{k} = a_{k-1} + \frac{1}{a_{k-1}}.
$$
From this, we get
$$
a_{k}^2 = \left(a_{k-1} + \frac{1}{a_{k-1}}\right)^2 = a_{k-1}^2 ... | 63 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,919 |
1. Given that $\overline{A B C A}$ represents a four-digit number. If the two-digit number $\overline{A B}$ is a prime number, $\overline{B C}$ is a perfect square not equal to 1, and $\overline{C A}$ is the product of a prime number and a perfect square not equal to 1, then the number of four-digit numbers that satisf... | -,1.C.
Since $\overline{A B}$ is a prime number, $B$ can only be $1,3,5, 7,9$. Also, since $\overline{B C}$ is a perfect square not equal to 1, $\overline{B C}$ can only be 16 or 36. Since $\overline{C A}$ is the product of a prime number and a perfect square not equal to 1, $A$ must be 3 or 8.
Therefore, the four-digi... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,920 |
Example 2 In $\triangle ABC$, $I$ is its incenter, $AI$, $BI$, and $CI$ intersect the opposite sides at $A'$, $B'$, and $C'$ respectively. Prove:
$$
\frac{1}{4}<\frac{AI \cdot BI \cdot CI}{AA' \cdot BB' \cdot CC'} \leqslant \frac{8}{27} \text{. }
$$
(32nd IMO) | Let $B C=a, A C=b, A B=c$.
By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
\frac{A^{\prime} B}{A^{\prime} C}=\frac{A B}{A C}=\frac{c}{b} . \\
\text { Then } \frac{A^{\prime} B}{A^{\prime} C+A^{\prime} B}=\frac{c}{b+c} .
\end{array}
$$
Thus, $A^{\prime} B=\frac{a c}{b+c}$.
Also, $\frac{A I}{A^{\prime} l}=\fr... | \frac{1}{4}<\frac{AI \cdot BI \cdot CI}{AA' \cdot BB' \cdot CC'} \leqslant \frac{8}{27} | Geometry | proof | Yes | Yes | cn_contest | false | 716,921 |
2. Given that $a$ and $b$ are rational numbers, and among $a+b$, $a-b$, $a b$, and $\frac{a}{b}$, exactly three of these numbers are equal. Then the value of $(2 a)^{\prime \prime}$ should be ( ).
(A) 0
(B) 1
(C) -1
(D) $\pm 1$ | 2.D.
Since $b \neq 0$, we have $a+b \neq a-b$.
Thus, $ab = \frac{a}{b}$. Solving this, we get $a=0$ or $b= \pm 1$.
If $a=0$, then $b=0$ must hold, which is a contradiction;
If $b=1$, then among $ab$, $\frac{a}{b}$, $a+b$, and $a-b$, it is impossible for three of them to be equal;
When $b=-1$, we have $ab = \frac{a}{b... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,922 |
3. As shown in Figure 1, $EF$ is the midline of $\triangle ABC$, $O$ is a point on $EF$, and satisfies $OE=2OF$. Then
the ratio of the area of $\triangle ABC$ to the area of $\triangle AOC$ is ( ).
(A) 2
(B) $\frac{3}{2}$
(C) $\frac{5}{3}$
(D) 3 | 3.D.
Since $EF$ is the midline of $\triangle ABC$, i.e., $EF \parallel AC$, therefore, $S_{\triangle COB}=S_{\triangle CPB}=\frac{1}{2} S_{\triangle ABC}$.
Thus, $S_{\triangle MOC}+S_{\triangle AOB}=\frac{1}{2} S_{\triangle ABC}$.
Also, $EO=2 FO$, so,
$\frac{S_{\triangle BOF}}{S_{\triangle COE}}=\frac{S_{\triangle MOF... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,923 |
4. Let the three-digit number $\bar{b} c$ be divisible by 3, and let $a$, $b$, and $c$ be the lengths of the sides of an isosceles triangle (including equilateral triangles). Then the number of such three-digit numbers is ( ) .
(A) 21
(B) 36
(C) 45
(D) 63 | 4.C.
From the problem, we know that $a+b+c$ is a multiple of 3, and $a, b, c$ are positive integers from 1 to 9.
When $a=b=c$, there are 9 three-digit numbers that satisfy the condition;
When two of $a, b, c$ are equal, let's assume $a=b$, then there are $441, 447, 552, 558, 663, 669, 771, 774, 882, 885, 993$, 996, a ... | 45 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,924 |
5. In $\triangle A B C$, medians $A D$ and $C F$ intersect at point $G$. If $\angle A F G=45^{\circ}, \angle A G C=60^{\circ}$, then the degree measure of $\angle A C F$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$ | 5.D.
As shown in Figure 5, draw $C M \perp A G$ at point $M$, and connect $M F$. Let $G M = 1$, then $C G = 2, F G = 1$. Therefore,
$$
F G = G M .
$$
Thus,
$$
\begin{array}{l}
\angle M F C \\
= \angle G C M \\
= 30^{\circ},
\end{array}
$$
$\angle M F A = \angle M A F = 15^{\circ}$.
Hence, $A M = F M = C M$.
Therefor... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,925 |
6. In the Cartesian coordinate system, line $l$ intersects the positive semi-axes of the $x$-axis and $y$-axis at points $N$ and $M$, respectively. A square $ABCD$ is inscribed in the right triangle $\triangle MON$, with points $A$ and $B$ on segments $MO$ and $NO$, respectively, and points $C$ and $D$ on segment $MN$.... | 6.A.
As shown in Figure 6, draw $D E \perp y$-axis at point $E$. Then $E(0,7)$. It is easy to see that
$$
\begin{array}{l}
\triangle A E D \cong \triangle B O A . \\
\text { Then } A O=E D=3, \\
O B=A E=7-3=4, \\
A D=5 .
\end{array}
$$
In the right triangle $\triangle A D M$, $A D^{2}=A E \cdot A M$, we can get
$$
A ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,926 |
1. If $\left[-\frac{1}{a}, a\right](a>0)$ contains exactly 10 integers, then the minimum value of $118 a-21+110 a-11$ is | $$
=1 . \frac{1}{9} \text {. }
$$
If $9 \leqslant a<10$, then
$$
|18 a-2|+|10 a-1|=28 a-3 \geqslant 249 ;
$$
If $\frac{1}{10}<a \leqslant \frac{1}{9}$, then
$$
|18 a-2|+|10 a-1|=1-8 a \geqslant \frac{1}{9} \text {. }
$$ | \frac{1}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,927 |
3. Given the polynomial
$$
\begin{aligned}
a_{0}+ & \left(a_{1}+4\right) x+ \\
& \left(a_{2}-10\right) x^{2}+\left(a_{3}+6\right) x^{3}+\left(a_{4}-1\right) x^{4}+ \\
& \left(a_{5}-1\right) x^{5}+a_{6} x^{6}+\cdots+a_{2 \alpha 5} x^{2 \omega 5}
\end{aligned}
$$
can be divided by $x^{2}+3 x-2$, and $\alpha^{2}+3 \alpha... | 3.0 .
Since $\alpha^{2}+3 \alpha-2=0$, $\alpha$ is a root of the equation $x^{2}+3 x-2=0$. Therefore,
$$
\begin{array}{l}
a_{0}+\left(a_{1}+4\right) \alpha+\left(a_{2}-10\right) \alpha^{2}+\left(a_{3}+6\right) \alpha^{3}+ \\
\left(a_{4}-1\right) \alpha^{4}+\left(a_{5}-1\right) \alpha^{5}+a_{6} \alpha^{6}+\cdots+a_{2 \... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,929 |
4. Given the quadratic function $y=a x^{2}+13 x$ passes through two lattice points (points with integer coordinates) in the first quadrant, and their y-coordinates are both prime numbers. Then $a=$ $\qquad$ - | 4. -6 .
Let the lattice point satisfying the given conditions be $(m, n)$.
Then $n=a m^{2}+13 m=m(a m+13)$.
Since $n$ is a prime number, we have $m_{1}=1$ and $a m_{1}+13$ is a prime number, or $a m_{2}+13=1$ and $m_{2}$ is a prime number.
Therefore, $m_{1}=1$ and $a+13$ is a prime number, and $a m_{2}=-12=-2 \times ... | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,930 |
-、(20 points) As shown in Figure 3, in $\triangle A B C$, the angle bisector of $\angle A C B$ intersects the external angle bisector of $\angle A B C$ at point $E$, $P E \perp A C$ at point $P$, $E Q \perp$ $C B$ at point $Q$, and $P Q$ intersects $A E$ and $B E$ at points $S$ and $T$ respectively. Prove: A triangle c... | As shown in Figure 8, rotate $\triangle E Q B$ around point $E$ in a clockwise direction to $\triangle E P D$, such that $E Q$ coincides with $E P$, and point $T$ coincides with point $F$. Then, points $C, P, D$ are collinear, and
$$
\begin{array}{l}
P F=Q T, \\
\angle E P F=\angle E Q T, \\
E F=E T .
\end{array}
$$
D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,931 |
1. Let $f(x)=a x^{2}+b x+c,|f(0)| \leqslant 1$, $|f(1)| \leqslant 1,|f(-1)| \leqslant 1$. Then, the maximum value of $|f(2)|$ is $(\quad)$.
(A) 5
(B) 6
(C) 7
(D) 8 | 1.C.
$$
\begin{array}{l}
|f(2)|=|4 a+2 b+c| \\
=|3(a+b+c)+(a-b+c)-3 c| \\
=|3 f(1)+f(-1)-3 f(0)| \\
\leqslant 3|f(1)|+|f(-1)|+3|f(0)| \\
\leqslant 3+1+3=7 .
\end{array}
$$
When $f(x)=-2 x^{2}+1$, $|f(2)|=7$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,932 |
2. There are 4 cards each of red and blue, and each color of cards is marked with the numbers $1,2,3,4$. Arrange all 8 cards into a 2-row, 4-column matrix, requiring that cards with the same number are in the same column. The number of different arrangements is $(\quad)$.
(A) $\left(\mathrm{A}_{4}^{4} \mathrm{~A}_{4}^{... | 2. B.
First, merge 2 cards with the same number into 1 card, resulting in 4 "large" cards. Arrange all of them, there are $A_{4}^{4}$ ways. Then, exchange the positions of the 2 cards in the same column, there are $\left(\mathrm{A}_{2}^{2}\right)^{4}$ ways. Therefore, there are $\mathrm{A}_{4}^{4}\left(\mathrm{~A}_{2}... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,933 |
5. Let vectors $a$ and $b$ satisfy $|a|=1,|b|=2$, and the angle between $a$ and $b$ is $60^{\circ}$. If the angle between the vectors $7 a+2 t b$ and $t a+b$ is obtuse, then the range of the real number $t$ is $\qquad$ | 5. $-7<t<-\frac{1}{2}$, and $t \neq-\frac{\sqrt{14}}{2}$.
From $(7 a+2 t b) \cdot(t a+b)<0$, we get $-7<t<-\frac{1}{2}$.
Moreover, the angle between $7 a+2 t b$ and $t a+b$ must not be a straight angle, hence, $t \neq-\frac{\sqrt{14}}{2}$. | -7<t<-\frac{1}{2}, \text{ and } t \neq-\frac{\sqrt{14}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,934 |
11. On the positive direction of the $x$-axis, take a sequence of points $\left\{A_{j}\right\}, j=1,2, \cdots$, and in the first quadrant on the parabola $y^{2}=\frac{3}{2} x$, take a sequence of points $\left\{B_{k}\right\}, k=1,2, \cdots$, such that $\triangle A_{k-1} B_{k} A_{k}(k=1,2, \cdots)$ are all equilateral t... | 11.2005.
Let the side length of the $n$-th equilateral triangle be $a_{n}$. Then the coordinates of the vertex $B_{n}$ of the $n$-th equilateral triangle on the parabola are
$$
\left(a_{1}+a_{2}+\cdots+a_{n-1}+\frac{a_{n}}{2}, \sqrt{\frac{3}{2}\left(a_{1}+a_{2}+\cdots+a_{n-1}+\frac{a_{n}}{2}\right)}\right) .
$$
From ... | 2005 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,935 |
12. According to the instruction, the robot can complete the following actions on a plane: As shown in Figure 1, it first walks from the origin $O$ in a direction that is $\alpha\left(0 \leqslant \alpha \leqslant \frac{\pi}{2}\right)$ northeast for a period of time, then walks north for a period of time, but the time t... | 12. $(100 \pi-200) \mathrm{m}^{2}$.
As shown in Figure 1, let the position of the robot after walking for $2 \mathrm{~min}$ be point $P(x, y)$. Let the point where the robot changes direction be $A$, with $|O A|=a$, and $|A P|=b$. From the given conditions, we have $a+b=2 \times 10=20$, and
$$
\left\{\begin{array}{l}
... | (100 \pi-200) \mathrm{m}^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,936 |
13. Let the left and right foci of the hyperbola $x^{2}-y^{2}=1$ be $F_{1}$ and $F_{2}$, respectively. If the vertex $P$ of $\triangle P F_{1} F_{2}$ moves on the hyperbola in the first quadrant, find the locus of the incenter of $\triangle P F_{1} F_{2}$ and the locus of the point of tangency of the incircle with side... | Three, 13. As shown in Figure 2, let the two vertices of the hyperbola on the $x$-axis be
$A(1,0), B(-1, 0)$, and $G, H, K$ be the points where the incircle of $\triangle P F_{1} F_{2}$ touches the sides $F_{1} F_{2}$, $P F_{2}$, and $P F_{1}$, respectively. Then we have
$$
\begin{array}{l}
\left|G F_{1}\right|-\left|G... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,937 |
14. Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}_{+}$, define
$$
S_{n}=\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2} \text {. }
$$
(1) Find the minimum value of $S_{n}$;
(2) Under the condition $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, find the minimum value of $S_{n}$;
(3) Under the c... | 14. (1) $S_{n} \geqslant \sum_{i=1}^{n}\left(2 \sqrt{\frac{n-1}{n^{2}}}\right)^{2}$
$$
=4 \sum_{i=1}^{n} \frac{n-1}{n^{2}}=4 \times \frac{n-1}{n} \text {. }
$$
When $x_{1}=x_{2}=\cdots=x_{n}=\frac{\sqrt{n-1}}{n}$, the minimum value $4 \times \frac{n-1}{n}$ is achieved.
$$
\begin{array}{l}
\text { (2) } S_{n}=\sum_{i=1... | n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,938 |
15. In a live military exercise, the Red side has set up 20 positions along a straight line. To test 5 different new weapons, it is planned to equip 5 positions with these new weapons, with the requirement that the first and last positions are not equipped with new weapons, and that among every 5 consecutive positions,... | 15. Let 20 positions be sequentially numbered as $1,2, \cdots, 20$, and let the sequence number of the $k$-th new weapon be $a_{k}(k=1,2,3,4,5)$.
$$
\begin{array}{l}
\text { Let } x_{1}=a_{1}, x_{2}=a_{2}-a_{1}, x_{3}=a_{3}-a_{2}, \\
x_{4}=a_{4}-a_{3}, x_{5}=a_{5}-a_{4}, x_{6}=20-a_{5} .
\end{array}
$$
Then we have $x... | 69600 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,939 |
2. If the circle $x^{2}+y^{2}=n^{2}$ covers at least one maximum point and one minimum point of the function $f(x)=\sqrt{3} \sin \frac{\pi x}{n}$, then the minimum value of the positive integer $n$ is ( ).
(A) 1
(-B) 2
(C) 3
(D) 4 | 2. B.
Since $f(x)=\sqrt{3} \sin \frac{\pi x}{n}$ is an odd function, its graph is symmetric about the origin. Therefore, the circle $x^{2}+y^{2}=n^{2}$ only needs to cover one of the extreme points of $f(x)$. Let $\frac{\pi x}{n}=\frac{\pi}{2}$, solving this gives the nearest maximum point $P\left(\frac{n}{2}, \sqrt{3... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,941 |
3. If the focal distance, minor axis length, and major axis length of an ellipse form an arithmetic sequence, then its eccentricity is ( ).
(A) $\frac{3}{4}$
(B) $\frac{\sqrt{2}}{3}$
(C) $\frac{3}{5}$
(D) $\frac{9}{10}$ | 3.C. From $2 a+2 c=4 b, a^{2}=b^{2}+c^{2}$. Eliminating $b$ yields $3 a=5 c$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
3.C. From $2 a+2 c=4 b, a^{2}=b^{2}+c^{2}$. Eliminating $b$ yields $3 a=5 c$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,942 |
4. For $x \in \mathbf{R}$, the function $f(x+2)+f(x-2)$ $=f(x)$, then it is a periodic function. The smallest positive period of such a function is ( ).
(A) 4
(B) 6
(C) 8
(D) 12 | 4.D.
Let $x-2$ replace $x$ in the equation, then we have
$$
f(x)+f(x-4)=f(x-2) \text {. }
$$
Thus, $f(x+2)=-f(x-4)$.
It follows that $f(x+6)=-f(x)$, hence $f(x+12)=f(x)$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,943 |
5. The graph of the function $y=f(x)$ is $C$, and the graph symmetric to $C$ with respect to the line $x=1$ is $C_{1}$. After shifting $C_{1}$ to the left by 1 unit, the resulting graph is $C_{2}$. Then the function corresponding to $C_{2}$ is ( ).
(A) $y=f(-x)$
(B) $y=f(1-x)$
(C) $y=f(2-x)$
(D) $y=f(3-x)$ | 5. B.
$$
\begin{array}{l}
C_{1}: y=f(2-x), \\
C_{2}: y=f(2-(x+1))=f(1-x) .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,944 |
6. The smallest natural number $n$ that satisfies $n \sin 1 > 1 + 5 \cos 1$ is $\qquad$ . | 6.5 .
Since $\frac{\pi}{4}n \sin 1>1+5 \cos 1>1+5 \cos \frac{\pi}{3}=\frac{7}{2} \text {. }$
Thus, $n>\frac{7}{\sqrt{3}}>4$.
When $n=5$, it is easy to prove that $5 \sin 1>1+5 \cos 1$, which means
$$
5(\sin 1-\cos 1)>1 \text {, }
$$
Squaring it gives $\sin 2<\frac{24}{25}$, which is obviously true.
In fact, because $... | 5 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,945 |
6. When $a, b$ are two distinct positive numbers, which of the following inequalities does not hold? ( ).
(A) $\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)>\left(\sqrt{a b}+\frac{1}{\sqrt{a b}}\right)^{2}$
(B) $\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)>\left(\frac{a+b}{2}+\frac{2}{a+b}\right)^{2}$
(C... | 6. B.
When $a=1, b=5$, we have $\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)=10.4$. And $\left(\frac{a+b}{2}+\frac{2}{a+b}\right)^{2}=\frac{100}{9}>10.4$. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 716,946 |
7. Let $A_{x y}=\frac{\left(1-x^{2}\right)\left(1-y^{2}\right)}{x y}$. If $a+b+c$ $=a b c$, then the value of $A=A_{b x}+A_{a c}+A_{a b}$ is ( ).
(A) 3
(B) -3
(C) 4
(D) -4 | 7.C.
Let $a=\tan \alpha, b=\tan \beta, c=\tan \gamma, \alpha+\beta+\gamma=\pi$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,947 |
8. A certain cargo yard has 2005 vehicles waiting in line to be loaded, with the requirement that the first vehicle must load 9 boxes of cargo, and every 4 adjacent vehicles must load a total of 34 boxes. To meet the above requirements, the minimum number of boxes of cargo should be ( ).
(A) 17043
(B) 17044
(C) 17045
(... | 8. A.
Except for the first vehicle which carries 9 boxes, among the remaining 2004 vehicles, every 4 vehicles carry 34 boxes, so there are $\frac{2004}{4} \times 34+9=17043$ boxes. | 17043 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,948 |
9. Several rectangular prisms with edge lengths of $2, 3, 5$ are combined in the same direction to form a cube with an edge length of 90. The number of small rectangular prisms that a diagonal of the cube passes through is ( ).
(A) 64
(B) 66
(C) 68
(D) 70 | 9. B.
Since the least common multiple of $2,3,5$ is 30, and the number of small rectangular prisms with edge lengths of $2,3,5$ that a diagonal of a cube with edge length 30 passes through is $\left[\frac{30}{2}\right]+\left[\frac{30}{3}\right]+\left[\frac{30}{5}\right]-\left[\frac{30}{2 \times 3}\right]-\left[\frac{3... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,949 |
10. A set of important documents is locked in a safe, and now $n$ keys are given to $n$ students in sequence to open the cabinet, but only one of them can actually open the door. On average, the number of attempts needed to open the door is ( ).
(A) 1
(B) $n$
(C) $\frac{n+1}{2}$
(D) $\frac{n-1}{2}$ | 10.C.
Given that the probability of each student opening a locker is $\frac{1}{n}$, the average number of times a locker is opened (i.e., the mathematical expectation) is
$$
1 \times \frac{1}{n}+2 \times \frac{1}{n}+\cdots+n \times \frac{1}{n}=\frac{n+1}{2} .
$$ | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,950 |
11. Let $x \in \mathbf{R}$, for the function $f(x)$ satisfying the condition $f\left(x^{2}+1\right)=x^{4}+5 x^{2}-3$. Then, for all $x \in \mathbf{R}, f\left(x^{2}-1\right)=$ $\qquad$ | $$
\text { II. 11. } x^{4}+x^{2}-9 \text {. }
$$
Using the substitution method, we get.
| x^{4}+x^{2}-9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,951 |
12. After a coordinate paper is folded once, point $A(0,4)$ overlaps with point $B(8,0)$. If point $C(6,8)$ overlaps with point $D(m$, $n)$, then $m+n=$ | 12.14.8.
The equation of the axis of symmetry can be solved as $y=2x-6$. From $\frac{n+8}{2}=(6+m)-6, \frac{n-8}{m-6}=-\frac{1}{2}$, we get $m=7.6, n=7.2$. | 14.8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,952 |
13. A sphere is tangent to all six edges of a regular tetrahedron. If the edge length of the tetrahedron is $a$, then the volume of the sphere is $\qquad$ | 13. $\frac{\sqrt{2} \pi a^{3}}{24}$.
We can transform the regular tetrahedron into a regular tetrahedron inscribed in a cube. In this case, the edge length of the cube is $\frac{\sqrt{2} a}{2}$, thus, the radius of the sphere is $\frac{\sqrt{2} a}{4}$. | \frac{\sqrt{2} \pi a^{3}}{24} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,953 |
14. The elements in set $X$ are positive integers, and have the property: if $x \in X$, then $12-x \in X$. The number of such sets $X$ is $\qquad$. | 14.63.
Let $Y=\{(1,11),(2,10),(3,9),(4,8),(5,7), 6\}$. Therefore, the number of sets that satisfy the condition is $2^{6}-1=63$. | 63 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,954 |
15. (12 points) The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{n+1}=a_{n}^{2}+a_{n}(n \in \mathbf{N}), \\
b_{n}=\frac{1}{1+a_{n}}, S_{n}=b_{1}+b_{2}+\cdots+b_{n}, \\
P_{n}=b_{1} b_{2} \cdots b_{n} .
\end{array}
$$
Try to find the value of $2 P_{n}+S_{n}$. | Three, 15. Since $a_{1}=\frac{1}{2}, a_{n+1}=a_{n}^{2}+a_{n}, n \in \mathbf{N}$, therefore, $a_{n+1}=a_{n}\left(a_{n}+1\right)$. Then
$$
\begin{array}{l}
b_{n}=\frac{1}{1+a_{n}}=\frac{a_{n}^{2}}{a_{n} a_{n+1}}=\frac{a_{n+1}-a_{n}}{a_{n} a_{n+1}}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}, \\
P_{n}=b_{1} b_{2} \cdots b_{n}=\frac... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,955 |
Three. (20 points) Let $x, y, z \geqslant 0, x+y+z=3$.
Prove: $\sqrt{x}+\sqrt{y}+\sqrt{z} \geqslant x y+y z+z x$.
| Three, because $2+(\sqrt{x})^{3}=1+1+(\sqrt{x})^{3} \geqslant 3 \sqrt{x}$, so, $2 \sqrt{x}+x^{2} \geqslant 3 x$.
Similarly, $2 \sqrt{y}+y^{2} \geqslant 3 y, 2 \sqrt{z}+z^{2} \geqslant 3 z$.
Therefore, $2(\sqrt{x}+\sqrt{y}+\sqrt{z})+x^{2}+y^{2}+z^{2}$
$\geqslant 3(x+y+z)=(x+y+z)^{2}$ (since $x+y+z=3$ ).
Expanding yields... | \sqrt{x}+\sqrt{y}+\sqrt{z} \geqslant x y+y z+z x | Inequalities | proof | Yes | Yes | cn_contest | false | 716,956 |
16. (12 points) As shown in Figure 1, given that $D$ is any point on side $AB$ of $\triangle ABC$ with an area of 1, $E$ is any point on side $AC$, and $F$ is any point on segment $DE$. Let $\frac{AD}{AB}=x, \frac{AE}{AC}=y, \frac{DF}{DE}=z$, and $y+z-x=\frac{1}{2}$. Try to find the maximum area of $\triangle BDF$. | 16. As shown in Figure 1, connect $B E$. Then
$$
\begin{array}{l}
S_{\triangle \triangle I N}=z S_{\triangle A D E}=z(1-x) S_{\triangle A B E} \\
=z(1-x) y S_{\triangle U C C}=z(1-x) y .
\end{array}
$$
By the AM-GM inequality, we have
$$
z(1-x) y \leqslant\left[\frac{z+(1-x)+y}{3}\right]^{3}=\frac{1}{8} \text {. }
$$
... | \frac{1}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,957 |
17. (12 points) Draw a line through point $P(3+2 \sqrt{2}, 4)$ that intersects the $x$-axis and $y$-axis at points $M$ and $N$, respectively. Find the maximum value of $OM + ON - MN$ (where $O$ is the origin). | 17. A circle is drawn through the point $P(3+2 \sqrt{2}, 4)$, tangent to the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and such that point $P$ lies on the major arc $\overparen{A B}$. The equation of the circle is $(x-3)^{2}+(y-3)^{2}=9$. Thus, the tangent line through point $P$ intersects the $x$-axis ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,958 |
18. (16 points) If positive numbers $a$, $b$, $c$ satisfy
$$
\frac{a}{b+c}=\frac{b}{a+c}-\frac{c}{a+b},
$$
prove: $\frac{b}{a+c} \geqslant \frac{\sqrt{17}-1}{4}$. | 18. Given the condition $\frac{b}{a+c}=\frac{c}{a+b}+\frac{a}{b+c}$.
Let $a+b=x, b+c=y, c+a=z$. Then
$$
a=\frac{x+z-y}{2}, b=\frac{x+y-z}{2}, c=\frac{y+z-x}{2} .
$$
Thus, the original condition can be transformed into
$$
\begin{array}{l}
\frac{x+y}{z}=\frac{y+z}{x}+\frac{z+x}{y}-1 \\
\geqslant \frac{z}{x}+\frac{z}{y}... | \frac{\sqrt{17}-1}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,959 |
19. (14 points) The express train from Yueyang to Pengzhou includes the starting station and the terminal station, a total of six stations, which are denoted as $A, B, C, D, E, F$. One day, Zhang Bing and 18 other passengers boarded the same carriage from Yueyang. Among these passengers, some were from Hubei, and other... | 19. From the conditions, 7 people get off at station $B$, meaning 7 of the 19 passengers are from Hunan. At station $E$, 2 people get off, meaning there are 2 Hunan people on the route $D-E$. There are at least 2 Hunan people on the route $C-D$, and at least 2 people get off at station $D$. Therefore, there are at leas... | 7 \text{ Hunan people and 12 Hubei people at the beginning} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 716,960 |
2. Let $\theta$ be the largest interior angle of a triangle, and satisfies $\sin 8 \theta=\sin 2 \theta$. Then the set of possible values of $\theta$ is $\qquad$ (express using the roster method). | 2. $\left\{\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{\pi}{2}, \frac{7 \pi}{10}, \frac{9 \pi}{10}\right\}$.
From $\sin 8 \theta=\sin 2 \theta$, we get
$8 \theta=2 k \pi+2 \theta$ or $8 \theta=(2 k+1) \pi-2 \theta$,
which means $\theta=\frac{1}{3} k \pi$ or $\theta=\frac{1}{10}(2 k+1) \pi$, where $k \in \mathbf{Z}$.
Since $... | \left\{\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{\pi}{2}, \frac{7 \pi}{10}, \frac{9 \pi}{10}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,962 |
3. A 3x3 grid as shown in Figure 1, each small square is filled with a complex number, and the sum of the complex numbers in each row, each column, and each diagonal is equal. Then the complex number represented by $x$ is $\qquad$ .
Preserving the original text's line breaks and format, the translation is as follows... | 3. $\frac{1}{2}+\frac{1}{2} \mathrm{i}$.
Let the numbers in the other two cells of the first row be $y$ and $z$ (as shown in Figure 3). According to the problem, the sum of the three numbers in each row, each column, and each diagonal is $x+y+z$. Therefore, the middle cell is $x+z-1$, the first cell in the middle row ... | \frac{1}{2}+\frac{1}{2} \mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,963 |
4. As shown in Figure 2, the edge length of the regular tetrahedron $ABCD$ is $6 \, \text{cm}$. There are points $E$ and $F$ on edges $AB$ and $CD$, respectively. If $AE = 1 \, \text{cm}$ and $CF = 2 \, \text{cm}$, then the length of line segment $EF$ is $\qquad$ cm. | 4. $\sqrt{23}$.
$$
\begin{array}{l}
\mathbf{F} \mathbf{E}=\mathbf{F} \mathbf{C}+\mathbf{C B}+\mathbf{B E}, \text { note that } C D \perp A B, \text { so } \\
|\mathbf{F E}|^{2}=(\mathbf{F} \mathbf{C}+\mathbf{C B}+\mathbf{B E})^{2} \\
=|\mathbf{F} C|^{2}+|\mathbf{C B}|^{2}+|\mathbf{B E}|^{2}+ \\
2(\mathbf{F C} \cdot \m... | \sqrt{23} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,964 |
5. If the equation with respect to $x$
$$
4^{x}+(a+3) 2^{x}+5=0
$$
has at least one real root in the interval $[1,2]$, then the range of the real number $a$ is $\qquad$ . | 5. $-\frac{33}{4} \leqslant a \leqslant-3-2 \sqrt{5}$.
From $4^{x}+(a+3) 2^{x}+5=0$, we get $a+3=-\frac{4^{x}+5}{2^{x}}$. Let $t=2^{x}$. Since $x \in[1,2]$, hence $t \in[2,4]$, and $a+3=-\left(t+\frac{5}{t}\right)$.
The function $f(t)=t+\frac{5}{t}$ is decreasing on $[2, \sqrt{5}]$ and increasing on $[\sqrt{5}, 4]$; ... | -\frac{33}{4} \leqslant a \leqslant-3-2 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,965 |
6. $a, b, c, d, e$ are 5 elements randomly selected from the set $\{1,2,3,4,5\}$ (repetition allowed). Then the probability that $a b c d + e$ is odd is $\qquad$ | $a b c d+e=$ odd
$\Leftrightarrow a b c d=$ odd, $e=$ even;
or $a b c d=$ even, $e=$ odd.
$a b c d$ being odd has $3^{4}=81$ possibilities, $a b c d$ being even has $5^4-3^{4}=544$ possibilities. Therefore, the required probability is
$$
P=\frac{81 \times 2+544 \times 3}{5^{5}}=\frac{1794}{3125} .
$$ | \frac{1794}{3125} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,966 |
7. For any real numbers $x, y$, the function $f(x)$ satisfies
$$
f(x)+f(y)=f(x+y)-xy-1 \text{.}
$$
If $f(1)=1$, then for negative integers $n$, the expression for $f(n)$ is $\qquad$. | 7. $f(n)=\frac{n^{2}+3 n-2}{2}$.
Let $y=1$, by the given condition we have
$$
f(x+1)=f(x)+f(1)+x+1=f(x)+x+2 \text {, }
$$
i.e., $f(x+1)-f(x)=x+2$.
In particular, $f(1)-f(0)=2$, which means $f(0)=-1$.
When $n$ is a negative integer,
$$
\begin{array}{l}
f(n)=[f(n)-f(n+1)]+[f(n+1)- \\
f(n+2)]+\cdots+[f(-1)-f(0)]+f(0) \\... | \frac{n^{2}+3 n-2}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,968 |
8. Real numbers $x, y, z$ satisfy $x+y+z=0$, and $x^{2}+y^{2}+z^{2}=1$. Let $m$ be the maximum of $x^{2}, y^{2}, z^{2}$, then the minimum value of $m$ is $\qquad$ . | 8. $\frac{1}{2}$.
Since substituting $-x, -y, -z$ for $x, y, z$ respectively does not change the conditions or the conclusion, by symmetry we can assume without loss of generality that $x \geqslant 0, y \geqslant 0, z \leqslant 0$.
Also, $z=-(x+y)$, so,
$2 z^{2}=z^{2}+(x+y)^{2}=x^{2}+y^{2}+z^{2}+2 x y \geqslant 1$.
Th... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,969 |
II. (14 points) Let $f(x)=\sqrt{a x^{2}+b x}$. Find the value of the real number $a$ such that: there exists at least one positive number $b$ for which the domain and range of $f(x)$ are the same. | If $a=0$, then for each positive number $b, f(x)=\sqrt{b x}$ has both its domain and range as $[0,+\infty)$, so $a=0$ satisfies the condition.
If $a>0$, then for positive number $b, f(x)=\sqrt{a x^{2}+b x}$ has the domain
$$
\begin{array}{l}
D=\left\{x \mid a x^{2}+b x \geqslant 0\right\} \\
=\left(-\infty,-\frac{b}{a... | 0 \text{ or } -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,970 |
Three, (14 points) Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $\left(a, b \in \mathbf{R}_{+}\right)$ with semi-focal distance $c$, and $b^{2}=a c$. $P, Q$ are any two points on the hyperbola, and $M$ is the midpoint of $P Q$. When the slopes $k_{P Q}$ and $k_{O M}$ of $P Q$ and $O M$ both exist, fi... | Three, since $M$ is the midpoint of $P Q$, we can set $M\left(x_{0}, y_{0}\right)$, $P\left(x_{0}+\alpha, y_{0}+\beta\right), Q\left(x_{0}-\alpha, y_{0}-\beta\right)$.
Thus, $k_{O M}=\frac{y_{0}}{x_{0}}, k_{P Q}=\frac{\beta}{\alpha}$.
Also, points $P$ and $Q$ are on the hyperbola, so,
$$
\begin{array}{l}
b^{2}\left(x_{... | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,971 |
Four, (16 points) Let $[x]$ denote the greatest integer not exceeding the real number $x$. Find the number of elements in the set
$$
\left\{n \left\lvert\, n=\left[\frac{k^{2}}{2005}\right]\right., 1 \leqslant k \leqslant 2004, k \in \mathbf{N}\right\}
$$ | From $\frac{(k+1)^{2}}{2005}-\frac{k^{2}}{2005}=\frac{2 k+1}{2005} \leqslant 1$, we solve to get
$k \leqslant 1002$.
When $k=1,2, \cdots, 1002$, we have
$$
\left[\frac{(k+1)^{2}}{2005}\right]=\left[\frac{k^{2}}{2005}\right]
$$
or $\left[\frac{(k+1)^{2}}{2005}\right]=\left[\frac{k^{2}}{2005}\right]+1$.
Since $\left[\fr... | 1503 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,972 |
Five. (16 points) The general term formula of the sequence $\left\{f_{n}\right\}$ is
$$
f_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right], n \in \mathbf{Z}_{+} \text {. }
$$
Let $S_{n}=\mathrm{C}_{n}^{1} f_{1}+\mathrm{C}_{n}^{2} f_{2}+\cdots+\mathrm{C}_{n}... | Let $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$, then
$$
\begin{array}{l}
S_{n}=\frac{1}{\sqrt{5}} \sum_{i=1}^{n} \mathrm{C}_{n}^{i}\left(\alpha^{i}-\beta^{i}\right)=\frac{1}{\sqrt{5}} \sum_{i=0}^{n} \mathrm{C}_{n}^{i}\left(\alpha^{i}-\beta^{i}\right) \\
=\frac{1}{\sqrt{5}}\left(\sum_{i=0}^{n} \mathrm{C}_... | 3 \mid n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,973 |
2. Point $P\left(\ln \left(2^{x}+2^{-x}-\tan \frac{\pi}{6}\right), \cos 2\right)(x$ $\in \mathbf{R})$ is located in the ( ).
(A) first quadrant
(B) second quadrant
(C) third quadrant
(D) fourth quadrant | 2.D.
Since $2^{x}+2^{-x}-\tan \frac{\pi}{6} \geqslant 2-\frac{\sqrt{3}}{3}>1$, therefore,
$$
\begin{array}{l}
\ln \left(2^{x}+2^{-x}-\tan \frac{\pi}{6}\right)>0 . \\
\text { Also } \frac{\pi}{2}<2<\pi, \text { so } \cos 2<0 .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,974 |
4. Given $\sin \theta+\cos \theta=\frac{1}{3}, \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the value of $\theta$ is ( ).
(A) $-\arccos \frac{\sqrt{21}+1}{9}$
(B) $-\arccos \frac{\sqrt{21}-1}{9}$
(C) $-\arcsin \frac{\sqrt{17}+1}{6}$
(D) $-\arcsin \frac{\sqrt{17}-1}{6}$ | 4.D.
Given $\sin \theta+\cos \theta=\frac{1}{3}$, squaring both sides and simplifying yields $\sin \theta \cdot \cos \theta=-\frac{4}{9}$.
Considering $\sin \theta, \cos \theta$ as the roots of the quadratic equation $x^{2}-\frac{1}{3} x- \frac{4}{9}=0$, we solve to get $x_{1,2}=\frac{1 \pm \sqrt{17}}{6}$.
Since $\th... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,975 |
5. Suppose that $a \in \mathbf{R}$, line $(1-a) x+$ $(a+1) y-4(a+1)=0$, always passes through a fixed point $P$, and point $Q$ is on the curve $x^{2}-$ $x y+1=0$. Then the range of slope of a line passing through $P$ and $Q$ is ( ).
(A) $[-2,+\infty)$
(B) $[-3,+\infty)$
(C) $(1,+\infty)$
(D) $(3,+\infty)$
Note: fixed p... | 5.B.
The problem states: For any real number $a$, the line $(1-a) x + (a+1) y - 4(a+1) = 0$ always passes through a fixed point $P$. Point $Q$ lies on the curve $x^2 - xy + 1 = 0$. The range of the slope of the line passing through points $P$ and $Q$ is ( ).
(A) $[-2, +\infty)$
(B) $[-3, +\infty)$
(C) $(1, +\infty)$
(... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,976 |
6. The domain of the function $y=\sqrt{5-4 x-x^{2}}+\log _{\frac{1}{2}}(\cos 2 x+$ $\sin x-1)$ is ( ).
(A) $\left(0, \frac{1}{2}\right)$
(B) $\left[-5,-\frac{7 \pi}{6}\right) \cup\left(0, \frac{\pi}{6}\right)$
(C) $\left(0, \frac{\pi}{6}\right)$
(D) $\left(-\frac{7 \pi}{6},-\pi\right) \cup\left(0, \frac{\pi}{6}\right)$ | 6.D.
From the problem, we have $\left\{\begin{array}{l}5-4 x-x^{2} \geqslant 0, \\ \cos 2 x+\sin x-1>0 .\end{array}\right.$
Therefore, $\left\{\begin{array}{l}-5 \leqslant x \leqslant 1, \\ 0<\sin x<\frac{1}{2} .\end{array}\right.$
Thus, $\left\{\begin{array}{l}-5 \leqslant x \leqslant 1, \\ 2 k \pi<x<2 k \pi+\frac{\p... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,977 |
Five. (20 points) Suppose a line intersects the graph of the function $y=x^{4}-x^{2}+x$ at exactly two distinct points. Find all such line equations.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
---
Five. (20 points) Suppo... | Obviously, the line $x=a$ intersects the graph of the function $y=x^{4}-x^{2}+x$ at only one point. Therefore, let the equation of the line be $y=p x+q$. Substituting it into $y=x^{4}-x^{2}+x$, we get
$$
x^{4}-x^{2}+(1-p) x-q=0 \text {. }
$$
Equation (1) has exactly two distinct real roots, which can occur in the foll... | y=x-\frac{1}{4}, y=\left(1 \pm \frac{2 \sqrt{6}}{9}\right) x+\frac{1}{12} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,978 |
7. For the equation $\frac{x^{2}}{\sin \alpha}+\frac{y^{2}}{\cos \alpha}=\tan \alpha(\alpha$ is a constant and $\left.\alpha \neq \frac{k \pi}{2}, k \in \mathbf{Z}\right)$, which of the following conclusions is incorrect? ( ).
(A) It can represent a hyperbola
(B) It can represent an ellipse
(C) It can represent a circl... | 7.D.
When $\alpha$ is in the first quadrant and $\alpha \neq 2 k \pi+\frac{\pi}{4}(k \in Z)$, the equation can represent an ellipse;
When $\alpha=2 k \pi+\frac{\pi}{4}(k \in Z)$, the equation represents a circle;
When $\alpha$ is in the second or fourth quadrant, the equation can represent a hyperbola;
When $\alpha$ i... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,979 |
8. Given $F_{1}$ and $F_{2}$ are the foci of an ellipse, $P$ is a point on the ellipse, $\angle F_{1} P F_{2}=90^{\circ}$, and $\left|P F_{2}\right|<\left|P F_{1}\right|$, the eccentricity of the ellipse is $\frac{\sqrt{6}}{3}$. Then $\angle P F_{1} F_{2}: \angle P F_{2} F_{1}$ $=(\quad)$.
(A) $1: 5$
(B) $1: 3$
(C) $1:... | 8. A.
Let $\left|P F_{2}\right|=x$, then $\left|P F_{1}\right|=2 a-x$.
Given $\angle F_{1} P F_{2}=90^{\circ}$, we know $\sin \angle P F_{1} F_{2}=\frac{x}{2 c}, \cos \angle P F_{1} F_{2}=\frac{2 a-x}{2 c}$. Therefore, $\left(\frac{x}{2 c}\right)^{2}+\left(\frac{2 a-x}{2 c}\right)^{2}=1$. Given $e=\frac{\sqrt{6}}{3}$,... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,980 |
9. The equation about $x$ is $\left|\mathrm{e}^{\ln x \mid}-2\right|=t(0<t<$ 1), where $t$ is a constant. Then the number of roots of the equation is ( ).
(A) 2
(B) 3
(C) 4
(D) Cannot be determined
Translate the above text into English, please keep the original text's line breaks and format, and output the translation... | 9. C.
When $x \geqslant 2$, $\left|\mathrm{e}^{\ln x \mid}-2\right|=\left|\mathrm{e}^{\ln x}-2\right|=|x-2|=$ $x-2=t$, then $x=2+t$ is a root of the equation;
When $1 \leqslant x<2$, $\left|\mathrm{e}^{\ln x \mid}-2\right|=\left|\mathrm{e}^{\ln x}-2\right|=|x-2|$ $=2-x=t$, then $x=2-t$ is a root of the equation;
Whe... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,981 |
10. If the curve symmetric to the hyperbola $x^{2}-y^{2}=a^{2}(a>0)$ with respect to the line $y=x-2$ is tangent to the line $2 x+3 y-6=0$, then the value of $a$ is ( ).
(A) $\frac{4 \sqrt{5}}{5}$
(B) $\frac{8 \sqrt{5}}{5}$
(C) $\frac{12 \sqrt{5}}{5}$
(D) $\frac{16 \sqrt{5}}{5}$ | 10.B.
It is easy to find the equation of the symmetric curve as
$$
(y+2)^{2}-(x-2)^{2}=a^{2} \text { . }
$$
Solving the system of equations $\left\{\begin{array}{l}(y+2)^{2}-(x-2)^{2}=a^{2}, \\ 2 x+3 y-6=0 .\end{array}\right.$
Eliminating $x$ yields $-\frac{5}{4} y^{2}+7 y+3-a^{2}=0$.
From $\Delta=64-5 a^{2}=0$, solv... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,982 |
11. The point $P$ on the line $3 x+2 y=1$ is equidistant from points $A(2,1)$ and $B(1,-2)$. Then the coordinates of point $P$ are $\qquad$. | $$
=11 \cdot\left(\frac{3}{7},-\frac{1}{7}\right) \text {. }
$$
Let $P(x, y)$. From $|P A|=|P B|$, we know $x+3 y=0$.
Also, $3 x+2 y=1$, solving the system of equations yields
$$
x=\frac{3}{7}, y=-\frac{1}{7} \text {. }
$$ | \left(\frac{3}{7},-\frac{1}{7}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,983 |
12. Given vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ satisfy $|\boldsymbol{a}|=2,|\boldsymbol{b}|=1$, and the angle between them is $60^{\circ}$. Then the range of real number $\lambda$ that makes the angle between vectors $\boldsymbol{a}+\lambda \boldsymbol{b}$ and $\lambda \boldsymbol{a}-2 \boldsymbol{b}$ obtuse i... | 12. $(-1-\sqrt{3},-1+\sqrt{3})$.
From the problem, we get $(a+\lambda b) \cdot(\lambda a-2 b)=\lambda^{2}+2 \lambda-2<0$.
Solving this, we get $-1-\sqrt{3}<\lambda<-1+\sqrt{3}$. | (-1-\sqrt{3},-1+\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,984 |
13. Given the inequality $|a x-3| \leqslant b$ has the solution set $\left[-\frac{1}{2}, \frac{7}{2}\right]$. Then $a+b=$ $\qquad$ . | 13.6.
From $|a x-3| \leqslant b$, we get $3-b \leqslant a x \leqslant 3+b$. It is easy to see that $a \neq 0$, so $\frac{3-b}{a}+\frac{3+b}{a}=-\frac{1}{2}+\frac{7}{2}$. Solving this, we get $a=2$.
Therefore, $\frac{3-b}{a}=-\frac{1}{2}$. Solving this, we get $b=4$. Hence, $a+b=6$. | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,985 |
14. The solution set of the inequality $(2+\sqrt{3})^{x}+(2-\sqrt{3})^{x}>8$ is $\qquad$ . | $$
\begin{array}{l}
\text { 14. }\left(-\infty, \log _{(2+\sqrt{3})}(4-\sqrt{15})\right) \cup \\
\left(\log _{(2+\sqrt{3})}(4+\sqrt{15}),+\infty\right) .
\end{array}
$$
Let $t=(2+\sqrt{3})^{x}>0$. Then the original inequality can be transformed into $t+\frac{1}{t}>8$.
Solving this, we get $t>4+\sqrt{15}$ or $0<t<4-\sq... | \left(-\infty, \log _{(2+\sqrt{3})}(4-\sqrt{15})\right) \cup \left(\log _{(2+\sqrt{3})}(4+\sqrt{15}),+\infty\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,986 |
15. The equation $(\arccos x)^{2}+(2-t) \arccos x+4$ $=0$ has real solutions. Then the range of values for $t$ is $\qquad$. | 15. $[6,+\infty)$.
Let $a=\arccos x$. Then the original equation can be transformed into $a^{2}+(2-t) a+4=0$, which has real solutions in $[0, \pi]$.
By $\Delta=t^{2}-4 t-12 \geqslant 0$, solving yields $t \geqslant 6$ or $t \leqslant-2$.
Since the product of the two roots is 4, it is known that the two roots have the... | [6,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,987 |
16. In $\triangle A B C$, the three interior angles are $\angle A$, $\angle B$, and $\angle C$, and $2 \angle C - \angle B = 180^{\circ}$. Additionally, the ratio of the perimeter of $\triangle A B C$ to its longest side is $m$. Therefore, the maximum value of $m$ is | 16. $\frac{9}{4}$.
From $2 \angle C-\angle B=180^{\circ}$, we know $120^{\circ}>\angle C>90^{\circ}$, so $\angle C$ is the largest. Its longest side is $c$, and we have $-\frac{1}{2}<\cos C<0$.
$$
\begin{array}{l}
\text { Also, } \angle A=180^{\circ}-\angle B-\angle C=360^{\circ}-3 \angle C, \text { then } \\
m=\frac{... | \frac{9}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,988 |
One, (50 points) In a convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $P$. If there exists a line $t$ passing through point $P$ such that $t$ simultaneously bisects a pair of opposite sides of quadrilateral $ABCD$, determine the shape of quadrilateral $ABCD$.
| I. The quadrilateral is a parallelogram or a trapezoid.
Let the line through point $P$ intersect $A B, C D$ at $M, N$, with $A M=$ $M B, C N=N D$.
Draw $E F / / C D$ through point $M$, intersecting lines $P A$ and $P B$ at $E$ and $F$. Then points $E$ and $F$ cannot be on the same side of $A B$.
If points $E$ and $F$... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,989 |
17. The equation of the directrix of the hyperbola $x(y+1)=1$ is
$\qquad$ | 17. $y=-x+\sqrt{2}-1$ or $y=-x-\sqrt{2}-1$.
The hyperbola $x(y+1)=1$ can be viewed as the hyperbola $xy=1$ translated one unit downward along the y-axis. The equations of the two directrices of $xy=1$ are $y=-x+\sqrt{2}$ or $y=-x-\sqrt{2}$. Translating these two lines downward by 1 unit gives the equations of the dire... | y=-x+\sqrt{2}-1 \text{ or } y=-x-\sqrt{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,990 |
18. The inequality $x+2 \sqrt{2 x y} \leqslant a(x+y)$ holds for all positive numbers $x, y$. Then the minimum value of the real number $a$ is $\qquad$ | 18.2.
Since $x+2 \sqrt{2 x y} \leqslant x+(x+2 y)=2(x+y)$, then the minimum value of $a$ is 2. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,991 |
19. As shown in Figure 1, a
small boat is traveling northward at a speed of $10 \mathrm{~m} / \mathrm{s}$ over the lake, and on a bridge $20 \mathrm{~m}$ above the lake, a car is moving eastward at a speed of $20 \mathrm{~m} / \mathrm{s}$. Currently, the boat is $40 \mathrm{~m}$ south of point $P$ on the water, and t... | 19.30.
Assuming after $t \mathrm{~s}$, the distance between the boat and the car is $y \mathrm{~m}$, then
$$
\begin{array}{l}
y=\sqrt{(40-10 t)^{2}+(-30+20 t)^{2}+400} \\
=\sqrt{500(t-2)^{2}+900} \text {. } \\
\end{array}
$$
When $t=2 \mathrm{~s}$, $y$ has a minimum value of $30 \mathrm{~m}$. | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,992 |
20. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, the set of points on the surface of the cube that are a distance of $\frac{2 \sqrt{3}}{3}$ from point $A$ forms a curve (this curve may not lie in a single plane). Then the length of this curve is $\qquad$ | 20. $\frac{5 \sqrt{3} \pi}{6}$.
This curve is distributed across six surfaces. The parts within the plane $A C$, plane $A B_{1}$, and plane $A D_{1}$ are all circular arcs with $A$ as the center, a radius of $\frac{2 \sqrt{3}}{3}$, and a central angle of $\frac{\pi}{6}$. The total length of these three segments is $\f... | \frac{5 \sqrt{3} \pi}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,993 |
21. Let $S_{n}$ be the sum of the first $n$ terms of an arithmetic sequence. Assume that $S_{3}=9, S_{20}$ $>0$, and $S_{21}<0$. Then the range of the common difference $d$ is $\qquad$ , the maximum term of the sequence $S_{1}, S_{2}, S_{3}, \cdots$ is $\qquad$
Note: term - one item, arithmetic sequence - arithmetic s... | $\equiv 21 \cdot\left(-\frac{6}{17},-\frac{1}{3}\right), S_{10}$.
The problem states: Let $S_{n}$ represent the sum of the first $n$ terms of an arithmetic sequence, $S_{3}=9, S_{20}>0, S_{21}0$,
$S_{21}=21 a_{1}+210 d0, a_{11}<0$.
Therefore, starting from the 11th term, it is less than 0. Hence, $S_{10}$ is the maximu... | S_{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,994 |
22. If $x, y \in \mathbf{R}$, and satisfy $\sqrt{x+2}+\sqrt{y-5}$ $=6$, then the minimum value of $x+2 y$ is $\qquad$ , and the maximum value is $\qquad$ _. | 22.32,80.
Let $\sqrt{x+2}=a, \sqrt{y-5}=b$, then $a+b=6$.
Thus, $x+2 y=a^{2}-2+2\left(b^{2}+5\right)$
$$
=3(b-2)^{2}+32(0 \leqslant b \leqslant 6) \text {. }
$$
When $b=2$, $x+2 y$ has a minimum value of 32;
When $b=6$, $x+2 y$ has a maximum value of 80. | 32,80 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,995 |
23. The line $l$ passing through point $E\left(-\frac{p}{2}, 0\right)$ intersects the parabola $C: y^{2}=2 p x(p>0)$ at points $A$ and $B$, and the inclination angle of line $l$ is $\alpha$. Then the range of $\alpha$ is $\qquad$; $F$ is the focus of the parabola, and the area of $\triangle A B F$ is $\qquad$ (expresse... | 23. $\left(0, \frac{\pi}{4}\right) \cup\left(\frac{3 \pi}{4}, \pi\right), \frac{p^{2} \sqrt{\cos 2 \alpha}}{\sin \alpha}$.
Let the equation of line $l$ be $y=k\left(x+\frac{p}{2}\right)$.
Solving the system of equations $\left\{\begin{array}{l}y=k\left(x+\frac{p}{2}\right) \\ y^{2}=2 p x\end{array}\right.$, we get
$$
... | \left(0, \frac{\pi}{4}\right) \cup\left(\frac{3 \pi}{4}, \pi\right), \frac{p^{2} \sqrt{\cos 2 \alpha}}{\sin \alpha} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,996 |
24. There are 10 circles on a sphere, these 10 circles can divide the sphere into at least $\qquad$ regions, and at most $\qquad$ regions. | 24. 11, 92.
When these 10 circles do not intersect each other, the number of regions on the sphere is the least, which is 11.
When these 10 circles intersect each other pairwise and no three circles intersect at the same point, the number of regions on the sphere is the most. Each circle is divided into 18 segments, ... | 11, 92 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,997 |
25. The coordinates of point $P(x, y)$ satisfy the following relations:
$$
\left\{\begin{array}{l}
2 x+y \geqslant 15, \\
x+3 y \geqslant 27, \\
x \geqslant 2, \\
y \geqslant 3,
\end{array}\right.
$$
and $x, y$ are both integers. Then the minimum value of $x+y$ is
$\qquad$, and at this time, the coordinates of point $... | $25.12,(3,9)$ or $(4,8)$.
Adding equations (1), (2), and (3), we get $x+y \geqslant 11$.
If $x+y=11$, then the equalities in equations (1), (2), and (3) hold, and the system of equations has no integer solutions.
If $x+y=12$, from $2 x+y \geqslant 15$, we get $x \geqslant 3$;
from $x+3 y \geqslant 27$, we get $2 y \geq... | 12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,998 |
In the semicircle $\odot O$ with diameter $AB$, take a point $C$. Draw $CD \perp AB$ at $D$. $CD$ divides the semicircle $\odot O$ into two figures, the incircles of which touch $AB$ at $E$ and $F$ respectively. Prove that: $\frac{AE \cdot EB}{AF \cdot FB}=\frac{DE}{DF}$. | Proof: Let $\odot O_{1}$ be tangent to $\overparen{A C}$ at point $G$, and tangent to $C D$ at point $H$. Auxiliary lines are shown in Figure 2.
It is easy to see that $O, O_{1}, G$ are collinear.
From $O B = O G$, we get $\angle O B G = \angle O G B$;
From $O_{1} H = O_{1} G$, we get $\angle O_{1} H G = \angle O_{1}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,999 |
II. (50 points) Given a positive real number $a$ and a positive integer $n$. Find:
$S_{n}(a)=\sum_{j=0}^{n-1}\left[\sqrt{a+\frac{j}{n}}\right]$ (expressed in terms of $a$ and $n$). | Let $k^{2} \leqslant [a]+1j_{0}$, then $k^{2} \leqslant a+\frac{j}{n}<k^{2}+\frac{j}{n}<k^{2}+1$.
Therefore, when $j=0,1, \cdots, j_{0}$, $\left[\sqrt{a+\frac{j}{n}}\right]=k-1$:
When $j=j_{0}+1, j_{0}+2, \cdots, n-1$, $\left[\sqrt{a+\frac{j}{n}}\right]=k$.
Then $S_{n}(a)=\sum_{j=0}^{n-1}\left[\sqrt{a+\frac{j}{n}}\righ... | S_{n}(a)=\left\{\begin{array}{ll}
n[\sqrt{a}], & \text{if } [a]+1 \text{ is not a perfect square; } \\
n[\sqrt{[a]+1}]+[n|a|]-n, & \text{if } [a]+1 \text{ is a perfect square. }
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,000 |
Initially, $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at $D$, and an isosceles right $\triangle A C B$ is inscribed in $\odot O_{1}$, with the tangency point $D$ on the semicircle $\overparen{A B}$. Through points $A$, $B$, and $C$, the tangents $A M$, $B N$, and $C P$ to $\odot O_{2}$ are drawn, with $M$, ... | Prove: As shown in Figure 3, draw line $O_{1} O_{2}$ intersecting $\odot O_{1}$ and $\odot O_{2}$ at points $E$ and $F$, respectively, then the point of tangency $D$ lies on the line segment $O_{1} O_{2}$. Let the radii of $\odot O_{1}$ and $\odot O_{2}$ be $r$ and $R$, respectively, then $E D=2 r$, $D F=2 R$.
Connect... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,001 |
Given a semicircle $\odot O_{1}$ with diameter $AB$, point $D$ lies on $O_{1}B$ and does not coincide with $O_{1}$ or $B$. A line through point $D$ perpendicular to $AB$ intersects the semicircle $\odot O_{1}$ at point $C$. Circle $\odot O_{2}$ is internally tangent to the semicircle $\odot O_{1}$ at $F$, tangent to $C... | Proof: As shown in Figure 4, connect $\mathrm{O}_{2} \mathrm{M}, \mathrm{O}_{2} \mathrm{N}, \mathrm{CO}_{1}$. Let the radius of the semicircle $\odot O_{1}$ be $R$, and the radius of $\odot O_{2}$ be $r$, and $\angle C A B = \theta \left(\theta \in \left(0, \frac{\pi}{2}\right)\right)$.
Since $\angle C A B = \theta$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,002 |
178 Find all functions \( f: \mathbf{R} \rightarrow \mathbf{R} \) such that for all \( x, y \in \mathbf{R} \),
\[
f\left((x+y)^{2}\right)=(x+y)(f(x)+f(y)).
\] | Solution: Taking $x=y=0$ in equation (1), we get
$$
f(0)=0 \text {. }
$$
Taking $y=0$ in equation (1) and using equation (2), we get
$$
f\left(x^{2}\right)=x f(x) \text {. }
$$
From equations (1) and (3), we have
$$
(x+y)(f(x)+f(y))=(x+y) f(x+y) \text {. }
$$
Thus, $f(x+y)=f(x)+f(y)(\forall(x+y) \neq 0)$.
From equat... | f(x)=kx | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,003 |
Three, (50 points) Given a positive integer $s(s \geqslant 5)$, decompose $s$ into the sum of several distinct positive integers, and let the product of these positive integers be $F$. For all different decompositions, find the maximum value of $F$.
| (3) If there exists some $i$ such that $a_{i+1}-a_{i}>2$, then let
$$
a_{i}^{\prime}=a_{i}+1, a_{i+1}^{\prime}=a_{i+1}-1 \text {. }
$$
By $a_{i+1}^{\prime} a_{i}^{\prime}-a_{i+1} a_{i}=a_{i+1}-a_{i}-1 \geqslant 3-1=2>0$, we know $F^{\prime}>F$, a contradiction.
(4) If $a_{1}=4$, then by $s \geqslant 5$ we know that $a... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,004 |
1. Let the function $f(x)=\sqrt{\frac{1}{2}-x}+\sqrt{x-\frac{1}{3}}$ have a maximum value of $a$ and a minimum value of $b$. Then the value of $a+b$ is ( ).
(A) $\frac{\sqrt{3}}{6}(3+\sqrt{3})$
(B) $\frac{\sqrt{3}}{6}(2+\sqrt{2})$
(C) $\frac{\sqrt{3}}{6}(3+\sqrt{2})$
(D) $\frac{\sqrt{3}}{6}(2+\sqrt{3})$ | 1.B.
Let $u=\sqrt{-\frac{1}{2}-x}, v=\sqrt{x-\frac{1}{3}}$, then $u^{2}+v^{2}=\frac{1}{6}$. Since $u^{2}+v^{2} \leqslant(u+v)^{2} \leqslant 2\left(u^{2}+v^{2}\right)$, we have $a=\sqrt{\frac{1}{3}}, b=\sqrt{\frac{1}{6}}$.
Thus, $a+b=\frac{\sqrt{3}}{6}(2+\sqrt{2})$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,005 |
2. The cross-section obtained by cutting a cylinder with a plane at an angle of $30^{\circ}$ to the cylinder's axis is an ellipse. The eccentricity of this ellipse is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{3}}{2}$
(D) $\frac{\sqrt{3}}{4}$ | 2.C.
Let the radius of the cylinder's base be $r$, and the semi-major and semi-minor axes of the ellipse be $a, b$. Then $a=2r, b=r, c=\sqrt{3}r$. Therefore, $e=\frac{c}{a}=\frac{\sqrt{3}}{2}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,006 |
3. In the tetrahedron $P-ABC$, $BC=3$, $CA=4$, $AB=5$. If the dihedral angles between the three lateral faces and the base are $A-BC-P$ at $45^{\circ}$, $B-CA-P$ at $45^{\circ}$, and $C-AB-P$ at $45^{\circ}$, then the volume of the tetrahedron $P-ABC$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 3. B.
Draw $P O \perp$ plane $A B C$ at $O, O D \perp B C$ at $D, O E \perp C A$ at $E, O F \perp A B$ at $F$, let $O P=h$. Then
$\angle P D O=45^{\circ}, \angle P E O=45^{\circ}, \angle P F O=45^{\circ}$.
Thus, $O D=h \cot 45^{\circ}=h, O E=h \cot 45^{\circ}=h$, $O F=h \cot 45^{\circ}=h$.
In $\triangle A B C$, $3 O D... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,007 |
3. A person rolls a die twice, obtaining the numbers $m, n$ in succession, which are used as the coefficients of the quadratic equation $x^{2}+m x+n=0$. The probability that the equation has real roots is ( ).
(A) $\frac{1}{2}$
(B) $\frac{5}{9}$
(C) $\frac{17}{36}$
(D) $\frac{19}{36}$ | 3. D.
From the problem, we know that $m, n \in \{1,2, \cdots, 6\}$.
Thus, the total number of events is 36.
The equation has real roots if and only if $m^{2} \geqslant 4 n$, which is equivalent to $n \leqslant \frac{m^{2}}{4}$.
According to this, we can list:
Values of $n$: $1,2,3,4,5,6$;
Number of $m$: $5,4,3,3,2,2$.... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,008 |
4. Let $A=\{a, b, c, d\}, B=\{1,2,3\}$. If the mapping $f: A \rightarrow B$, such that
$$
f(a)+f(b)+f(c)+f(d)=8,
$$
then the number of such mappings is ( ).
(A) 16
(B) 17
(C) 19
(D) 24 | 4.C.
Consider the number of integer solutions to $x_{1}+x_{2}+x_{3}+x_{4}=8\left(1 \leqslant x_{i} \leqslant 3, i=1,2,3\right.$, $4)$, which is equivalent to the number of integer solutions to $y_{1}+y_{2}+y_{3}+y_{4}=4\left(0 \leqslant y_{i} \leqslant 2\right.$, $i=1,2,3,4)$. It equals $\mathrm{C}_{7}^{3}-4 \mathrm{C... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,009 |
5. Given $\sin 2 \alpha=m, \cos 2 \alpha=n, m, n \in$ $(0,1)$. Among the following four values:
(1) $\frac{1+m}{n}$,
(2) $\frac{n}{1-m}$,
(3) $\frac{n+m+1}{n-m+1}$,
(4) $\frac{m-n+1}{m+n-1}$,
which can be the value of $\tan \left(\alpha+\frac{\pi}{4}\right)$? ( )
(A) 1
(B) 2
(C) 3
(D) 4 | 5.D.
$$
\begin{array}{l}
\tan \left(\alpha+\frac{\pi}{4}\right)=\frac{1+\tan \alpha}{1-\tan \alpha}=\frac{\cos \alpha+\sin \alpha}{\cos \alpha-\sin \alpha} \\
=\frac{(\cos \alpha+\sin \alpha)^{2}}{\cos ^{2} \alpha-\sin ^{2} \alpha}=\frac{1+\sin 2 \alpha}{\cos 2 \alpha}=\frac{1+m}{n}, \\
\tan \left(\alpha+\frac{\pi}{4}\... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,010 |
6. Given positive integers $a$ and $b$ satisfy the equation $a+2a^2=$ $b+3b^2$. Then the numbers $\sqrt{1+2a+2b}$ and $\sqrt{1+3a+3b}$ are ( ).
(A) both rational numbers
(B) both irrational numbers
(C) one is a rational number, the other is an irrational number
(D) any of the above three situations may occur | 6.A.
From the given equation, we have $\frac{a}{b}=\frac{3 b+1}{2 a+1}$.
By the properties of proportions, we have
$$
\frac{a}{b}=\frac{3 a+3 b+1}{2 a+3 b+1}, \frac{a}{b}=\frac{2 a+3 b+1}{2 a+2 b+1} \text {. }
$$
Therefore, $\frac{3 a+3 b+1}{2 a+3 b+1}=\frac{2 a+3 b+1}{2 a+2 b+1}$, which means
$$
(2 a+2 b+1)(3 a+3 b+... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,011 |
1. Use $1,2,3,4,5$ to form a five-digit number, such that the difference between any two adjacent digits is at least 2. Then the number of such five-digit numbers is $\qquad$ . | 2.1.14.
Consider $\square$ classified by the middle number $a$:
(1) If $a=1$, then 2 and 3 are on opposite sides of $a$, and 4 and 5 are on opposite sides of $a$, resulting in 4 permutations:
$$
24135,24153,35142,53142 \text {; }
$$
(2) If $a=2$, then 4 and 5 are on opposite sides of $a$, and 1 and 3 are not adjacent t... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,012 |
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