problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
5. Given a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, define points $B_{1}$, $B_{2}, \cdots, B_{n-1}$ as follows:
(1) If $i=1$ or $i=n-1$, then $B_{i}$ is the midpoint of side $A_{i} A_{i+1}$;
(2) If $i \neq 1, i \neq n-1, S_{i}$ is the intersection of $A_{1} A_{i+1}$ and $A_{n} A_{i}$, then $B_{i}$ is the i... | 5. First, prove the following lemma.
Lemma: Let quadrilateral $ABCD$ be an isosceles trapezoid with bases $AB$ and $CD$, and diagonals $AC$ and $BD$ intersecting at point $S$. Let $M$ be the midpoint of side $BC$, and the angle bisector of $\angle BSC$ intersects $BC$ at $N$ (as shown in Figure 3). Then $\angle AMD = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,936 |
6. Let $P$ be a convex polygon. Prove: There exists a convex hexagon inside $P$ whose area is at least $\frac{3}{4}$ of the area of $P$. | 6. Take any two parallel supporting lines $s, t$ of a convex polygon $P$ (i.e., $P$ is on the same side of the lines $s, t$, and $P$ has common points with the lines $s, t$), and let $S, T$ be the intersection points of the convex polygon $P$ with the lines $s, t$, respectively. Thus, the line segment $ST$ divides $P$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,937 |
Example 6 If $m$ takes any real number not equal to 0 or 1, the parabola
$$
y=\frac{m-1}{m} x^{2}+\frac{2}{m} x-\frac{m-3}{m}
$$
in the Cartesian coordinate system always passes through two fixed points, then, the distance between these two fixed points is $\qquad$ . | $$
y=\frac{m-1}{m} x^{2}+\frac{2}{m} x-\frac{m-3}{m}
$$
Reversing the roles, rearranging as a linear equation in $m$, we get
$$
\left(x^{2}-y-1\right) m+\left(-x^{2}+2 x+3\right)=0 \text {. }
$$
Given that when $m$ is any real number not equal to 0 or 1, the parabola passes through two fixed points, we have
$$
\left\... | 4 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,938 |
7. Given $\triangle A B C$, point $X$ is a moving point on line $B C$, and point $C$ is between points $B$ and $X$. Also, the incircles of $\triangle A B X$ and $\triangle A C X$ intersect at two distinct points $P$ and $Q$. Prove: $P Q$ passes through a fixed point independent of point $X$. | 7. First, prove a lemma.
Lemma Given that $L$ and $K$ are the midpoints of sides $AB$ and $AC$ of $\triangle ABC$, and the incircle of $\triangle ABC$ touches sides $BC$ and $CA$ at points $D$ and $E$ respectively. Then the intersection point of $KL$ and $DE$ lies on the angle bisector of $\angle ABC$.
Proof of the le... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,939 |
8. Given a cyclic quadrilateral $A B C D$, the lines $A D$ and $B C$ intersect at point $E$, and point $C$ is between points $B$ and $E$. The diagonals $A C$ and $B D$ intersect at point $F$. Let point $M$ be the midpoint of side $C D$, and point $N$ be a point on the circumcircle of $\triangle A B M$ different from $M... | 8. First prove that on the circumcircle of $\triangle A B M$, there is a unique point $N \neq M$ such that $\frac{A N}{B N}=\frac{A M}{B M}$.
If $\lambda=\frac{A M}{B M} \neq 1$, then $M 、 N$ are the intersection points of the circumcircle of $\triangle A B M$ and the locus of points $X$ satisfying $\frac{A X}{B X}=\l... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,940 |
1. If $x=\frac{a-b}{a+b}$, and $a \neq 0$, then $\frac{b}{a}$ equals ( ).
(A) $\frac{1-x}{1+x}$
(B) $\frac{1+x}{1-x}$
(C) $\frac{x-1}{x+1}$
(D) $\frac{x+1}{x-1}$ | - 1.A.
Since $a \neq 0$, we have $x=\frac{a-b}{a+b}=\frac{1-\frac{b}{a}}{1+\frac{b}{a}}$. Let $\frac{b}{a}=y$, then $x=\frac{1-y}{1+y}$.
Also, since $a \neq 0$, then $x \neq -1$. Therefore, $y=\frac{1-x}{1+x}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,941 |
2. As shown in Figure 1, each face of the cube has a positive integer. It is known that the sum of the numbers on opposite faces is equal. If the numbers opposite to $13, 9, 3$ are $a, b, c$ respectively, then the value of $a^{2}+b^{2}+c^{2}-a b-b c-c a$ is ( ).
(A) 48
(B) 76
(C) 96
(D) 152 | 2. B.
According to the problem, we have
$$
13+a=9+b=3+c \text {. }
$$
Thus, $a-b=-4, b-c=-6, c-a=10$.
Then $a^{2}+b^{2}+c^{2}-a b-b c-c a$
$$
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=76 .
$$ | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 715,942 |
3. Given a triangle with three sides of positive integer lengths. If only one side length is 5, and it is not the shortest side, then the number of triangles that satisfy the condition is ( ).
(A) 4
(B) 6
(C) 8
(D) 10 | 3.D.
If 5 is the longest side, the side lengths of the triangles that satisfy the conditions are
$$
2,4,5 ; 3,4,5 ; 3,3,5 ; 4,4,5 \text {. }
$$
If 5 is the second longest side, the side lengths of the triangles that satisfy the conditions are
$$
2,5,6 ; 3,5,6 ; 3,5,7 ; 4,5,6 ; 4,5,7 ; 4,5,8 \text {. }
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,943 |
4. As shown in Figure 2, in $\square A B C D$, if points $E$ and $F$ on side $A B$ satisfy $A E=E F=F B, C E$ intersects $D F$ and $D B$ at points $M$ and $N$ respectively, then $E M: M N: N C$ equals ( ).
(A) $2: 1: 4$
(B) $4: 3: 5$
(C) $5: 3: 12$
(D) $5: 4: 12$ | 4.C.
Let $A E=E F=F B=1$, set $E M=x, M N=y, N C=$ $z$. From $\triangle C D M \subset \triangle E F M$, we get $\frac{x}{y+z}=\frac{1}{3}$, i.e., $3 x-y-z=0$.
From $\triangle C D N \backsim \triangle E B N$, we get $\frac{x+y}{z}=\frac{2}{3}$, i.e., $3 x+3 y-2 z=0$.
Then we have $x=\frac{5}{12} z, y=\frac{1}{4} z$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,944 |
5. As shown in Figure 3. Given that three circles with centers at $A$, $B$, and $C$ are tangent to each other and all are tangent to line $l$. If the radii of $\odot A$, $\odot B$, and $\odot C$ are $a$, $b$, and $c$ ($0<c<a<b$), then $a$, $b$, and $c$ must satisfy the relationship ( ).
(A) $2 b=a+c$
(B) $2 \sqrt{b}=\s... | 5.D.
As shown in Figure 6, draw perpendiculars from points $A$, $B$, and $C$ to line $l$, with the feet of the perpendiculars being $A_{1}$, $B_{1}$, and $C_{1}$, respectively. Draw $A B_{2} \perp B B_{1}$, with the foot of the perpendicular being $B_{2}$, and connect $A B$. In
Rt $\triangle A B B_{2}$, we have
$$
\be... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,945 |
6. Given that $a$ and $b$ are positive integers, and the parabola $y = a x^{2} + b x + 1$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, then the minimum value of $a + b$ is ( ).
(A) 16
(B) 10
(C) 4
(D) 1 | 6. B.
Since $a, b$ are positive integers, we have
$$
-\frac{b}{2 a} < 0 \text{. }
$$
The parabola $y=a x^{2}+b x+1$ intersects the $x$-axis at two distinct points $A$ and $B$, and the distances from points $A$ and $B$ to the origin are both less than 1. Therefore, points $A$ and $B$ lie between 0 and -1. Thus, $a$ an... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,946 |
8. There are two people, A and B. A, on a car, notices that B is walking in the opposite direction. $1 \mathrm{~min}$ later, A gets off the car to chase B. If A's speed is twice that of B, but slower than the car's speed by $\frac{4}{5}$, then the time it takes for A to catch up with B after getting off the car is $\ma... | 8.11.
Let the speed of A be $x \mathrm{~m} / \mathrm{min}$, then according to the problem, the speed of B is $\frac{x}{2} \mathrm{~m} / \mathrm{min}$, and the speed of the car is $5 x \mathrm{~m} / \mathrm{min}$.
Let the time it takes for A to catch up with B be $t \mathrm{~min}$. According to the problem, we have $\f... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,948 |
Example 7 As shown in Figure 5, given points $A(0,3)$, $B(-2, -1)$, $C(2, -1)$, and $P\left(t, t^{2}\right)$ as a moving point on the parabola $y=x^{2}$ located within $\triangle A B C$ (including the boundary), the line $B P$ intersects $A C$ at point $E$, and the line $C P$ intersects $A B$ at point $F$. Express $\fr... | Given that the line $AB$ is the graph of the linear function $y = ax + b$. Substituting $A(0,3)$ and $B(-2,-1)$, we solve to get
$$
a=2, b=3.
$$
Therefore, the line $AB$ is the graph of the linear function $y = 2x + 3$. The x-coordinates of the intersection points of this line with the parabola $y = x^2$ satisfy the e... | \frac{t^2 + 2t + 5}{t^2 - 2t + 5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,949 |
9. As shown in Figure 4, in quadrilateral $ABCD$, $AC$ bisects $\angle BAD$, $CE \perp AB$ at point $E$, and $AE = \frac{1}{2}(AB + AD)$. If $\angle D = 120^{\circ}$, then $\angle B$ equals | $9.60^{\circ}$.
As shown in Figure 7, on $AB$, take $AF = AD$, and connect $CF$.
Since $AC$ bisects $\angle BAD$, then $\angle DAC = \angle FAC$.
Also, $AC$ is a common side, so,
$\triangle DAC \cong \triangle FAC$.
Thus, $\angle AFC = \angle D = 120^{\circ}$.
And $\angle AFC + \angle CFE = 180^{\circ}$, hence
$\ang... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,950 |
10. If a convex $n$-sided polygon has exactly 4 obtuse interior angles, then the maximum number of sides $n$ of this polygon is $\qquad$ . | 10.7.
Since a convex $n$-sided polygon has exactly 4 obtuse interior angles, the sum of these 4 angles is greater than $360^{\circ}$ and less than $720^{\circ}$. The other $n-4$ angles are right angles or acute angles, so the sum of these $n-4$ angles is no more than $(n-4) \times 90^{\circ}$ and greater than $0^{\cir... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,951 |
11. If in a $4 \times 4$ grid of 16 cells, each cell is filled with a number such that the sum of the numbers in all adjacent cells of each cell is 1, then the sum of the 16 numbers in the grid is $\qquad$
(Note: Adjacent cells refer to cells that share exactly one edge) | 11.6.
First, fill the 16 squares with the English letters $A, B, C$, $\cdots, O, P$ (as shown in Figure 8). Given that for each square, the sum of the numbers in all adjacent squares is 1, so,
$$
\begin{aligned}
16= & 2(A+D+M+P)+3(B+C+E+H+ \\
& I+L+N+O)+4(F+G+J+K) \\
= & 2(A+B+C+\cdots+O+P)+(B+C+E+ \\
& H+I+L+N+O)+2(F... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,952 |
12. Given the quadratic function
$$
y=a(a+1) x^{2}-(2 a+1) x+1,
$$
where $a$ is a positive integer.
(1) If the graph of the function $y$ intersects the $x$-axis at points $A$ and $B$, find the length of segment $AB$;
(2) If $a$ takes the values $1, 2, \cdots, 2005$ successively, the 2005 line segments intercepted by t... | Three, 12. (1) Let the graph of the function $y$ intersect the $x$-axis at two points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$, then $x_{1}, x_{2}$ are the two real roots of the equation
$$
a(a+1) x^{2}-(2 a+1) x+1=0
$$
Given that $y$ is a quadratic function, we have $a \neq 0$ and $a \neq -1$.
Also, $\Delta=(... | \frac{2005}{2006} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,953 |
13. Given that $p$, $q$, $\frac{2 q-1}{p}$, and $\frac{2 p-1}{q}$ are all integers, and $p>1$, $q>1$. Find the value of $p+q$. | 13. If $\frac{2 q-1}{p} \geqslant 2, \frac{2 p-1}{q} \geqslant 2$, then
$$
2 q-1 \geqslant 2 p, 2 p-1 \geqslant 2 q \text {. }
$$
Adding the two inequalities gives $2 p+2 q-2 \geqslant 2 p+2 q$. This is clearly a contradiction.
Therefore, at least one of $\frac{2 q-1}{p}$ and $\frac{2 p-1}{q}$ must be less than 2. Sup... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,954 |
14. As shown in Figure $5, \triangle A B C$ is inscribed in $\odot O, A C > B C$, and point $D$ is the midpoint of $\overparen{A C B}$. Prove: $A D^{2}=A C \cdot B C+C D^{2}$. | 14. As shown in Figure 9, draw $D E \perp A C$, with the foot of the perpendicular at point $E$. In the right triangle $\triangle A E D$,
$$
A D^{2}=A E^{2}+E D^{2} ;
$$
In the right triangle $\triangle C E D$,
$$
C D^{2}=C E^{2}+E D^{2} \text {. }
$$
Thus, we have
$$
\begin{aligned}
& A D^{2}-C D^{2} \\
= & A E^{2}-... | A D^{2}=A C \cdot B C+C D^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 715,955 |
1. Given real numbers $a$, $b$, $c$ satisfy $a^{2}+2 b=7$, $b^{2}-2 c=-1$, $c^{2}-6 a=-17$. Then the value of $a+b+c$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | -、1.B.
Adding the three equations and completing the square yields
$$
(a-3)^{2}+(b+1)^{2}+(c-1)^{2}=0 .
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,956 |
2. As shown in Figure 1, in $\triangle A B C$, $A C=$ $B C, \angle A C B=90^{\circ}, A D$ bisects $\angle B A C, B E \perp$ $A D$ intersects the extension of $A C$ at point $F$, and the foot of the perpendicular is $E$. Then the conclusions:
(1) $A D=B F$,
(2) $C F=C D$,
(3) $A C+C D=A B$,
(4) $B E=C F$,
(5) $B F=2 B E... | 2.D.
From $\triangle A C D \cong \triangle B C F$, $A D$ bisects $\angle B A C$, and $A D \perp B E$, we get (1)(2)(3)(5) correct. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,957 |
3. The average of five unequal natural numbers is 15, and the median is 18. Then the maximum value of the largest number among these five numbers is ( ).
(A) 35
(B) 36
(C) 37
(D) 38 | 3. C.
Let the other four numbers among these five numbers be $a, b, c, d$, and $a<b<18<c<d$. Then, $a+b+c+d=57$.
To maximize $d$, $a, b, c$ need to take their minimum values, so let $a=0, b=1, c=19$. In this case, $d=37$. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 715,958 |
4. As shown in Figure 2, given the slant height of the cone $O A=6$, and the radius of the base circle is 2. A small insect starts at point $A$ on the base of the cone, travels around the lateral surface of the cone, and returns to point $A$. The shortest distance the insect travels is ( ).
(A) 12
(B) $4 \pi$
(C) $6 \s... | 4.D.
Draw the lateral development (as shown in Figure 7). Connect $A A^{\prime}$, then $A A^{\prime}$ is the shortest distance. Let $\angle A O A^{\prime}=n^{\circ}$. Since $\overparen{A A^{\prime}}=4 \pi$, then $4 \pi=\frac{6 n \pi}{180}$, so $n=120$. Therefore, $A A^{\prime}=6 \sqrt{3}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,959 |
Example 10 Find $\sum_{k=0}^{n} \frac{1}{2^{k}} C_{n+k}^{k}$.
| Solution: Let the original expression be $f(n)$, then $f(0)=1$. By the identity (II), we have
$$
\begin{array}{l}
f(n+1)=\sum_{k=0}^{n+1} \mathrm{C}_{n+1+k}^{k} \cdot \frac{1}{2^{k}} \\
=\sum_{k=0}^{n+1} \mathrm{C}_{n+k}^{k} \cdot \frac{1}{2^{k}}+\sum_{k=1}^{n+1} \mathrm{C}_{n+k}^{k-1} \cdot \frac{1}{2^{k}} .
\end{arra... | 2^n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,960 |
5. Among the positive integers less than 1000, the number of integers that are divisible by 5 or 7, but not by 35, is ( ).
(A) 285
(B) 313
(C) 341
(D) 369 | 5.A.
If $[x]$ represents the integer part of the positive number $x$, then among the 999 positive integers less than 1000, there are $\left[\frac{999}{5}\right]$ numbers divisible by 5, and $\left[\frac{999}{7}\right]$ numbers divisible by 7. The numbers that are divisible by both 5 and 7 are $\left[\frac{999}{5 \time... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 715,961 |
6. Given that the price of a certain commodity increases by $x$ percent (1 percent being $10 \%$), the sales volume will decrease by $\frac{5}{6} x$ percent. To achieve the maximum turnover, the price should be increased by ( ) percent.
(A) 1
(B) 2
(C) 3
(D) 4 | 6.A.
Let the original selling price be $a$, and the original sales volume be $b$. Then the revenue is
$$
\begin{array}{l}
y=a(1+10\% x) \cdot b\left(1-\frac{5}{6} x \times 10\%\right) \\
=a b\left(1+\frac{x}{60}-\frac{x^{2}}{120}\right) .
\end{array}
$$
Since $a>0, b>0$, the revenue is maximized when $x=-\frac{\frac{... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,962 |
7. One day, a group of monkeys went to pick peaches. When dividing the peaches they picked, they found that if each monkey got 4, there would be 52 left; if each monkey got 6, then one of the monkeys would get less than 6 peaches. Then the total number of peaches picked by this group of monkeys is peaches. | II. 7.160 or 164.
Let there be $x$ monkeys, then the peaches they pick are $4x + 52$.
From the problem, we have $0 < 4x + 52 - 6(x - 1) < 6$.
Solving this, we get $x = 27$ or 28.
Therefore, $4x + 52 = 160$ or 164. | 160 \text{ or } 164 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,963 |
9. As shown in Figure 3, eight circles with a radius of 1 are placed inside a rectangle. Adjacent circles are tangent to each other, and the circles at the top-left and bottom-right corners are tangent to two sides of the rectangle, while the other circles are tangent to one side of the rectangle. The area of the recta... | $9.18+9 \sqrt{3}$.
As shown in Figure 9, connecting the centers of any two adjacent circles, by the properties of circles, $\triangle O_{1} O_{2} O_{3}$ is an equilateral triangle with a side length of 2, and the height of the triangle can be calculated as $\sqrt{3}$. Using the tangency of circles and lines, the length... | 18+9 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,965 |
11. A box contains balls of red, yellow, and white. If the number of white balls is at most $\frac{1}{2}$ of the number of yellow balls, and at least $\frac{1}{3}$ of the number of red balls, and the total number of yellow and white balls is no more than 55, then the box can contain at most $\qquad$ red balls. | 11.54.
Suppose there are $x$ white balls, $y$ yellow balls, and $z$ red balls. According to the problem, we can set up the following system of equations:
$$
\left\{\begin{array}{l}
x \leqslant \frac{1}{2} y, \\
x \geqslant \frac{1}{3} z, \\
y + x \leqslant 55 .
\end{array}\right.
$$
From equation (1), we get $y \geqs... | 54 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,967 |
13. In a football tournament with 10 teams, a double round-robin system is used, meaning any two teams play each other twice, once at home and once away. A win earns 3 points, a draw earns 1 point, and a loss earns 0 points.
(1) How many matches were played in total?
(2) If every match results in the maximum points, wh... | (1) Using a list or formula calculation, it can be known that a total of 90 matches were played in this competition.
(2) The highest score for a single match is 3 points. If the maximum score is achieved in every match, the total points of all teams in this competition would be $3 \times 90 = 270$ (points): The lowest ... | 38 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,969 |
14. The quality of a product can be divided into 6 different levels. If the working hours remain unchanged, 40 pieces of the lowest level product can be produced daily; if the level is increased by one, the profit per piece can increase by 1 yuan, but 2 fewer pieces of the product will be produced daily.
(1) If the pro... | 14. (1) Let the profit be maximized when producing the $x$-th grade product. Clearly, $1 \leqslant x \leqslant 6$, and $x$ is a natural number. Let the daily profit be $y$ yuan. Then
$$
\begin{array}{l}
y=[40-2(x-1)][16+(x-1)] \\
=-2 x^{2}+12 x+630=-2(x-3)^{2}+648 .
\end{array}
$$
Thus, when $x=3$, $y$ is maximized.
T... | x=\left\{\begin{array}{ll}
6, & 8 \leqslant a \leqslant 10 ; \\
\frac{22-a}{2} \text { (rounded to the nearest integer) }, & 10<a<20 ; \\
1, & 20 \leqslant a \leqslant 24 .
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,970 |
15. As shown in Figure 6, in the right triangle $\triangle ABC$, $\angle C=90^{\circ}$, $AC=4$, $BC=3$, where $\odot O_{1}, \odot O_{2}, \cdots, \odot O_{n}$ are $n(n \geqslant 2)$ equal circles, $\odot O_{1}$ is externally tangent to $\odot O_{2}$, $\odot O_{2}$ is externally tangent to $\odot O_{3}$, $\cdots \cdots$,... | 15. As shown in Figure 12, connect $A O_{1}$ and extend it to intersect $B C$ at point $D$. Connect $B O_{n}$ and extend it to intersect $A D$ and $A C$ at points $O$ and $E$, respectively. Draw perpendiculars from points $O_{1}$ and $O_{n}$ to $A B$, and let $F$ and $G$ be the feet of these perpendiculars.
Since $\od... | r = \frac{5}{2 n + 3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,972 |
1. The positive integer pairs $(x, y)$ that satisfy $y=\sqrt{x+1}+\sqrt{x+2005}$ ( ).
(A) There is only one pair
(B) There are exactly two pairs
(C) There are at least three pairs
(D) There are none | -.1.B.
Let $a^{2}=x+1, b^{2}=x+2005$, where $a, b$ are both natural numbers. Then
$$
\begin{array}{l}
y=a+b, \\
b^{2}-a^{2}=(b+a)(b-a)=2004=2^{2} \times 3 \times 167 .
\end{array}
$$
Since $b+a$ and $b-a$ have the same parity, and $b+a$
$$
\begin{array}{l}
>b-a, \text { so, } \\
\left\{\begin{array} { l }
{ b + a = 1... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,973 |
2. In Rt $\triangle A B C$, $a$, $b$, and $c$ represent the lengths of the three sides. If $\frac{a+b+c}{a+c}=\sqrt{2}$, then the ratio of the three sides of Rt $\triangle A B C$ is $(\quad)$.
(A) $1: 2: \sqrt{5}$
(B) $1: 1: \sqrt{2}$
(C) $1: \sqrt{2}: \sqrt{3}$
(D) $1: \sqrt{3}: 2$ | 2.B.
From $\frac{a+b+c}{a+c}=\sqrt{2}$, we get $\sqrt{2}-1=\frac{b}{a+c}$, which means $a+c=(\sqrt{2}+1) b$.
If $b$ is the hypotenuse, then $a+c<2 b$. This contradicts the given condition, so $b$ is not the hypotenuse.
Among $a$ and $c$, let's assume $c$ is the hypotenuse, thus,
$$
a^{2}+b^{2}=c^{2}=[(\sqrt{2}+1) b-a]... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,974 |
3. Given real numbers $a, b, c$ satisfy the inequalities
$$
|a| \geqslant|b+c|,|b| \geqslant|c+a|,|c| \geqslant |a+b|
$$
then the value of $a+b+c$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 3. A.
From the given conditions, we have
$$
\begin{array}{l}
a^{2} \geqslant(b+c)^{2}=b^{2}+2 b c+c^{2}, \\
b^{2} \geqslant(c+a)^{2}=c^{2}+2 c a+a^{2}, \\
c^{2} \geqslant(a+b)^{2}=a^{2}+2 a b+b^{2} .
\end{array}
$$
Adding the three inequalities, we get
$$
a^{2}+b^{2}+c^{2} \geqslant 2\left(a^{2}+b^{2}+c^{2}\right)+2 ... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 715,975 |
4. Given that $a$ is a natural number, the equation with respect to $x$
$$
2 x-a \sqrt{1-x}-a+4=0
$$
has at least one integer root. Then the number of possible values of $a$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 4.C.
From $\sqrt{1-x} \geqslant 0$ we know $x \leqslant 1$.
Also, $2 x+4=a(\sqrt{1-x}+1) \geqslant 0$, so $x \geqslant-2$. When $x=1,0,-2$, $a$ takes the values $6,2,0$ respectively. When $x=-1$, $a$ is not an integer.
Therefore, there are 3 possible values for $a$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,976 |
5. As shown in Figure 1, the convex quadrilateral $ABCD$ is inscribed in a circle, with $CB=CD$. Two moving points $E$ and $F$ are on $AC$ and $AD$ respectively, and satisfy $EF \parallel BD$. Let $BE$ intersect $CF$ at point $P$, then the geometric position of point $P$ is ( ).
(A) outside the circle
(B) inside the ci... | 5.C.
As shown in Figure 6, let $AC$ intersect $BD$ at $K$. Given $CB = CD$, we know $\angle BAC = \angle CAD$.
Thus, $\triangle ABC \backsim \triangle AKD$.
Therefore, $\frac{AB}{AC} = \frac{AK}{AD} = \frac{AE}{AF}$.
Hence, $\triangle ABE \backsim \triangle ACF$, $\angle 1 = \angle 2$.
At this point, $\angle BPC = 18... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,977 |
6. For any integer $n(n \geqslant 2)$, satisfying $a^{n}=a+1, b^{2 n}=b+3 a$
the size relationship of the positive numbers $a$ and $b$ is ( ).
(A) $a>b>1$
(B) $b>a>1$
(C) $a>1,01$ | 6.A.
From $a^{n}=a+1>1$, we know $a>1$.
Also, $b^{2 n}=b+3 a>3 a>3$, thus, $b>1$.
Therefore, $a$ and $b$ are both greater than 1.
Since $a^{2 n}-b^{2 n}$
$$
\begin{array}{l}
=(a+1)^{2}-(b+3 a)=a^{2}-a-b+1, \\
\text { and } a^{2 n}-b^{2 n} \\
=(a-b)\left(a^{2 n-1}+a^{2 n-2} b+\cdots+b^{2 n-1}\right),
\end{array}
$$
th... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,978 |
1. In a convex quadrilateral $ABCD$,
$$
\angle A-\angle B=\angle B-\angle C=\angle C-\angle D>0 \text {. }
$$
If one of the interior angles is $84^{\circ}$, then the other three interior angles are $\qquad$ | Let $\angle A-\angle B=\angle B-\angle C=\angle C-\angle D=x>0$, then $\angle C=\angle D+x, \angle B=\angle D+2 x$, $\angle A=\angle D+3 x$.
Also, $\angle A+\angle B+\angle C+\angle D=360^{\circ}$, so $4 \angle D+6 x=360^{\circ}$.
Therefore, $\angle D+\frac{3}{2} x=90^{\circ}>84^{\circ}$.
Clearly, $\angle D+\frac{3}{2... | \angle C=88^{\circ}, \angle B=92^{\circ}, \angle A=96^{\circ} \text{ or } \angle D=72^{\circ}, \angle B=96^{\circ}, \angle A=108^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,979 |
2. Given that $a$ and $b$ are non-zero constants, and
$$
\frac{a^{2}}{\sin ^{2} \theta}+\frac{b^{2}}{\cos ^{2} \theta}=(a+b)^{2} \text {. }
$$
Then $\frac{a^{3}}{\sin ^{4} \theta}+\frac{b^{3}}{\cos ^{4} \theta}=$ (express in terms of $a$ and $b$). | 2. $(a+b)^{3}$.
Given the equation, after removing the denominator, we get
$$
a^{2} \cos ^{2} \theta+b^{2}\left(1-\cos ^{2} \theta\right)=(a+b)^{2} \cos ^{2} \theta\left(1-\cos ^{2} \theta\right) \text {, }
$$
which simplifies to $\left[(a+b) \cos ^{2} \theta-b\right]^{2}=0$.
However, $a+b \neq 0$ (otherwise, from th... | (a+b)^{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,980 |
3. The graph of the quadratic function $y=a x^{2}+b x+c$ is shown in Figure 2. Let
$$
k=\frac{b(2 a+b)(a+b+c)}{c(2 a-b)(a-b+c)} .
$$
Then the sign of $k$ is | 3. $k>0$.
From Figure 2, we can see that the intersection point $(0, c)$ of the function $y=a x^{2}+b x+c$ with the $y$-axis is below the $x$-axis. Therefore, $c<0$;
When $x=-1$, $y=a-b+c<0$. | k>0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,981 |
Example 12 Proof: $\sum_{k=s}^{n} \mathrm{C}_{k}^{s} \mathrm{C}_{n-k}^{t}=\mathrm{C}_{n+1}^{s+t+1}$. | Proof: From equation (1), considering the generating functions $(1-x)^{-s-1}$ and $(1-x)^{-t-1}$ corresponding to $\left\{C_{k}^{s}\right\}$ and $\left\{C_{n-k}^{t}\right\}$ respectively $(s \leqslant k \leqslant n, 0 \leqslant t \leqslant n-k)$. Also,
$$
\begin{aligned}
& (1-x)^{-s-1}(1-x)^{-t-1} \\
& =(1-x)^{-(s+t+1)... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,982 |
4. As shown in Figure $3, A B$ is the diameter of a semicircle, points $P$ and $Q$ are on the semicircle, $P E \perp A B$ at $E, Q F \perp A B$ at $F$. If $A P \cdot A Q=E F \cdot A B$, then $\frac{A E}{E F}=$ $\qquad$ . | 4. $\frac{\sqrt{5}-1}{2}$.
It is known that $A P^{2}=A E \cdot A B, A Q^{2}=A F \cdot A B$.
Therefore, $(A P \cdot A Q)^{2} A A E \cdot A F \cdot A B^{2}$.
And it is given that $A P \cdot A Q=E F \cdot A B$, thus,
$$
E F^{2}=A E \cdot A F \text {. }
$$
Hence, point $E$ is the golden section of $A F$ $(E F>A E)$.
So, ... | \frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,983 |
One, (20 points) Class A and Class B set out from School $A$ simultaneously to a military camp $B$ for military training, which is $75 \mathrm{~km}$ away from the school. The walking speed of Class A students is $4 \mathrm{~km} / \mathrm{h}$, and the walking speed of Class B students is $5 \mathrm{~km} / \mathrm{h}$. T... | As shown in Figure 7, let the students of Class A travel by car from point $A$ to point $E$, then walk, covering a distance of $a \text{ km}$ by car, and the car returns empty to point $C$; the students of Class B board the car at point $C$, by which time they have already walked $b \text{ km}$.
The time taken by the ... | 6 \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,984 |
II. (25 points) As shown in Figure 4, in $\triangle ABC$, $AB > AC$, $AM$ is the median, and point $N$ is on side $BC$. It is known that $\frac{BN}{CN} = \frac{AB^2}{AC^2}$. Prove that $\angle BAM = \angle CAN$. | As shown in Figure 8, construct the circumcircle of $\triangle A M N$ intersecting $A B$ and $A C$ at $P$ and $Q$, respectively, and connect $P Q$. Since
$$
\begin{array}{l}
B M \cdot B N=B P \cdot B A, \\
C N \cdot C M=C Q \cdot C A,
\end{array}
$$
we have, $\frac{B M \cdot B N}{C N \cdot C M}=\frac{B P \cdot B A}{C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,985 |
Three. (25 points) As shown in Figure 5, a particle moves within the region $\{(x, y) \mid x \geqslant 0, y \geqslant 0\}$. In the first second, it moves from the origin to point $B_{1}(0,1)$, then from point $B_{1} \rightarrow C_{1}$ $\rightarrow A_{1}$, and subsequently moves along the $x$-axis, $y$-axis, and their p... | Three, suppose the time taken for a particle to reach points $A_{n}, B_{n}, C_{n}$ from the origin are $a_{n}, b_{n}, c_{n}$, respectively. Then it is clear that
$$
\begin{array}{l}
a_{1}=3, a_{2}=a_{1}+1, \\
a_{3}=a_{1}+12=a_{1}+3 \times 4, a_{4}=a_{3}+1, \\
a_{5}=a_{3}+20=a_{3}+5 \times 4, a_{6}=a_{5}+1, \\
\cdots \c... | 2008 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,986 |
1. Given that the hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ is a regular hexagon inscribed in a circle $\odot O$ with radius $R$. Establish a mapping $f: P \rightarrow f(P)$ from a point $P$ on the plane $\alpha$ of $\odot O$ to the set of real numbers $\mathbf{R}$, where $f(P)=\left|\sum_{i=1}^{6} P A_{i}\right|$.... | 1.1.B.
Since the hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ is a regular hexagon with center $O$, we have $\sum_{i=1}^{6} O A_{i}=0$.
$$
\begin{array}{l}
\text { Hence } f(P)=\left|\sum_{i=1}^{6} P A_{i}\right|=\left|\sum_{i=1}^{6}\left(P O+\boldsymbol{O A}_{i}\right)\right| \\
=\left|\sum_{i=1}^{6} \boldsymbol{P O}... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,987 |
$$
\begin{array}{l}
\text { 2. Given } a>b>c>1, M=a-\sqrt{c}, \\
N=a-\sqrt{b}, P=2\left(\frac{a+b}{2}-\sqrt{a b}\right), \\
Q=3\left(\frac{a+b+c}{3}-\sqrt[3]{a b c}\right) .
\end{array}
$$
Then the smallest among $M, N, P, Q$ is $(\quad)$. | 2.C.
Obviously, $M>N$.
Since $Q-P=c+2 \sqrt{a b}-3 \sqrt[3]{a b c}$
$$
>3 \sqrt[3]{c \cdot \sqrt{a b} \cdot \sqrt{a b}}-3 \sqrt[3]{a b c}=0 \text {, }
$$
Therefore, $Q>P$.
$$
\begin{array}{l}
\text { Also, } P-N=b-2 \sqrt{a b}+\sqrt{b} \\
=\sqrt{b}[(\sqrt{b}-\sqrt{a})+(1-\sqrt{a})]<0,
\end{array}
$$
Hence, $P<N$.
Th... | C | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,988 |
3. Given the sequence $0,1,1,2,2,3,3,4,4, \cdots$ with the sum of the first $n$ terms as $f(n)$. For any positive integers $p, q(p>q)$, $f(p+q)-f(p-q)=(\quad)$.
(A) $p^{2}+q^{2}$
(B) $2\left(p^{2}+q^{2}\right)$
(C) $p q$
(D) $2 p q$ | 3.C.
It is known that, $f(n)=\left\{\begin{array}{ll}\frac{n^{2}-1}{4}, & n \text { is odd; } \\ \frac{n^{2}}{4}, & n \text { is even. }\end{array}\right.$
Since $p+q$ and $p-q$ have the same parity, we consider two cases.
(1) If $p+q$ and $p-q$ are both odd, then
$$
\begin{array}{l}
f(p+q)-f(p-q) \\
=\frac{(p+q)^{2}-... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,989 |
4. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0), F_{1}$ and $F_{2}$ are its left and right foci, and $P$ is any point on the ellipse. If $\triangle P F_{1} F_{2}$ is a right triangle, then the number of possible points $P$ is ( ) .
(A) 4
(B) 6
(C) 4 or 6
(D) 4, 6, or 8 | 4.D.
(1) If $\angle P F_{1} F_{2}=90^{\circ}$, obviously, there are 2 points $P$.
Similarly, if $\angle P F_{2} F_{1}=90^{\circ}$, there are 2 points $P$.
(2) If $\angle F_{1} P F_{2}=90^{\circ}$, then we have
$\left|F_{1} F_{2}\right|^{2}=\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2} \geqslant 2\left(\frac{\left|... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,990 |
For any $x \in \mathbf{R}$, the inequality
$$
\sin ^{2} x+m \sin x+\frac{m^{2}-3}{m} \leqslant 0
$$
always holds. Then the range of $m$ is $(\quad)$.
(A) $m \leqslant-\frac{3}{2}$
(B) $0<m \leqslant 1$
(C) $0<m \leqslant 3$
(D) $m \leqslant-\frac{3}{2}$ or $0<m \leqslant 3$ | 5. B.
Let $t=\sin x$, and denote $g(t)=t^{2}+m t+\frac{m^{2}-3}{m}$.
Then the given condition is transformed into, when $t \in[-1,1]$, $g(t) \leqslant 0$ always holds. This is equivalent to
$$
\left\{\begin{array}{l}
g(1)=1+m+\frac{m^{2}-3}{m}=\frac{2 m^{2}+m-3}{m} \leqslant 0, \\
g(-1)=1-m+\frac{m^{2}-3}{m}=\frac{m-3... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 715,991 |
6. Given that $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a cube with edge length $a$, $E$ and $F$ are moving points on edges $A B$ and $B C$ respectively, and $A E=B F$. If the angle between $A_{1} E$ and $C_{1} F$ is minimized, then $(\quad)$.
(A) $A E=B F=\frac{1}{4} a$
(B) $A E=B F=\frac{1}{3} a$
(C) $A E=B F=\frac{2}{5} ... | 6.D.
As shown in Figure 1, extend $C_{1} F$, $A_{1} E$ to intersect the extension of $B_{1} B$ at points $P, Q$ respectively. Draw $P M / / A_{1} E$ intersecting $A_{1} B_{1}$ at point $M$, and connect $C_{1} M$. Then $\angle M P C_{1}$ is the angle formed by $A_{1} E$ and $C_{1} F$, denoted as $\angle M P C_{1} = \th... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,992 |
Example 8 The profits from operating and selling two types of goods, A and B, are $p$ (ten thousand yuan) and $q$ (ten thousand yuan) respectively, and the relationship between sales profit and invested capital $x$ (ten thousand yuan) is given by the empirical formulas: $p=\frac{1}{5} x, q=\frac{3}{5} \sqrt{x}$. Given ... | Explanation: Let the amount of money invested in product B be $x$ million yuan, then the amount of money invested in product A is $(3-x)$ million yuan. Let the total profit from products A and B be $y$, then
$$
\begin{array}{l}
y=\frac{1}{5}(3-x)+\frac{3}{5} \sqrt{x} \\
=-\frac{1}{5} x+\frac{3}{5} \sqrt{x}+\frac{3}{5}(... | 1.05 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,993 |
Example 9 As shown in Figure 6, it is known that a corner of a square steel plate with a side length of 4 is rusted, where $A F=2, B F=1$. To make reasonable use of this steel plate, a rectangular block MPND will be cut from the pentagon $E A B C D$, with point $P$ located on $A B$, and it is required that the area be ... | As shown in Figure 6, point $P$ is on line segment $AB$. Draw $PM \perp DE$ and $PN \perp CD$, with the feet of the perpendiculars being $M$ and $N$ respectively. Extend $NP$ to intersect $EF$ at $Q$. Let $PQ = x$, and the area of quadrilateral $MPND$ be $y$. It is easy to see that
then $\triangle APQ \sim \triangle A... | 80\% | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,994 |
5. Given that the graph of a quadratic function opens upwards and does not pass through the origin $O$, the vertex coordinates are $(1,-2)$, it intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$, and satisfies the relationship $|O C|^{2}=|O A| \cdot|O B|$.
(1) Find the analytical ex... | (Answer: (1) $y=x^{2}-2 x-1$ or $y=(1+\sqrt{2}) x^{2}-$
$$
\begin{array}{l}
(2+2 \sqrt{2}) x+\sqrt{2}-1 \text {; (2) } S_{\triangle A B C}=\sqrt{2} \text { or }(\sqrt{2}-1) \cdot \\
\sqrt{2(\sqrt{2}-1)} .)
\end{array}
$$ | S_{\triangle A B C}=\sqrt{2} \text { or } (\sqrt{2}-1) \cdot \sqrt{2(\sqrt{2}-1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,995 |
4. Given that the complex number $z$ satisfies $z^{2005}+\lambda z^{2004} \mathrm{i}-\lambda z-\mathrm{i}$ $=0$, where $\lambda$ is a real constant, and $\lambda^{2} \leqslant 1$. Then $|z|=$ ( ).
(A) $\frac{\sqrt{2}}{2}$
(B) $\left|\frac{\lambda+1}{\lambda-1}\right|$
(C) 1
(D) $\sqrt{1+\lambda^{2}}$ | 4.C.
From the given, we have $z^{2 \omega 0}=\frac{\lambda z+\mathrm{i}}{z+\lambda \mathrm{i}}$, let $z=a+b \mathrm{i}$. When $a^{2}+b^{2}>1$, we have $\lambda^{2}\left(a^{2}+b^{2}\right)+2 \lambda b+1>a^{2}+b^{2}+2 b \lambda+\lambda^{2}$
which simplifies to $\left(a^{2}+b^{2}-1\right)\left(\lambda^{2}-1\right)>0$.
Ho... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,996 |
5. Given the equation $a \sin x + b \cot x + c \csc x = 0$ has two distinct roots $\alpha, \beta$ satisfying $|\alpha \pm \beta| \neq k \pi$, where $k \in \mathbf{Z}$. Then $\cos ^{2} \alpha + \cos ^{2} \beta = (\quad)$.
(A) $2 + \frac{b^{2} + 2 a c}{a^{2}}$
(B) $\frac{b+c}{a}$
(C) $\frac{b^{2} + 4 a}{c^{2}}$
(D) $\fra... | 5. A.
It is known that $\sin x \neq 0$, so we have $a \sin ^{2} x + b \cos x + c = 0$, which can be rewritten as $a \cos ^{2} x - b \cos x - (a + c) = 0$.
Given that $\alpha, \beta$ are the two roots, we know that $\cos \alpha, \cos \beta$ are the two distinct roots of the equation
$$
a x^{2} - b x - (a + c) = 0
$$
(... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,997 |
6. Given $x, y, t \in \mathbf{R}_{+}$, let
$$
M=\frac{x}{x^{2}+t}+\frac{y}{y^{2}+t}, N=\frac{4(x+y)}{(x+y)^{2}+4 t} \text {. }
$$
Then among the following 4 propositions:
(1) If $x y=3 t$, then $M=N$;
(2) If $x y>3 t$, then $M \leqslant N$;
(3) If $x y<3 t$, then $M \geqslant N$;
(4) If $x y=3 t, M=N$ does not necessa... | 6. A.
$$
\begin{array}{l}
M-N=\frac{x}{x^{2}+t}+\frac{y}{y^{2}+t}-\frac{4(x+y)}{(x+y)^{2}+4 t} \\
=\frac{(x+y)(x y+t)}{x^{2} y^{2}+\left(x^{2}+y^{2}\right) t+t^{2}}-\frac{4(x+y)}{(x+y)^{2}+4 t} .
\end{array}
$$
Let $P=\frac{x y+t}{x^{2} y^{2}+\left(x^{2}+y^{2}\right) t+t^{2}}$.
Substituting $x y=3 t$ into the above eq... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 715,998 |
1. Given $t \in \mathbf{R}_{+}$. Then the minimum value of $\frac{1}{\sqrt{1+3 t^{4}}}+\frac{t^{3}}{\sqrt[4]{12}}$ is $\qquad$ . | $$
=1 \cdot \frac{2 \sqrt{2}}{3} \text {. }
$$
From $1+3 t^{4} \geqslant 2 \sqrt{3 t^{4}}=\sqrt{12 t^{4}}$, we know
$$
\begin{array}{l}
\sqrt{1+3 t^{4}} \geqslant \sqrt{12 t} . \\
\text { Therefore, } \frac{1}{\sqrt{1+3 t^{4}}}+\frac{t^{3}}{\sqrt{12}} \geqslant \frac{1+t^{4}}{\sqrt{1+3 t^{4}}} . \\
=\frac{\sqrt{1+3 t^... | \frac{2 \sqrt{2}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,999 |
2. Given the equation of $\odot O_{1}$ is $(x-8)^{2}+y^{2}=4$, and the equation of $\odot O_{2}$ is $(x-12)^{2}+y^{2}=9$, points $A$ and $B$ are on $\odot O_{1}$ and $\odot O_{2}$ respectively, and $A B$ is the external common tangent of the two circles. A moving line $l \parallel A B$ intersects $\odot O_{1}$ at point... | 2. $x^{2}+y^{2}=90$ H $x \neq \frac{75}{8}$.
From 1 $O O_{1}=8$,
$O O_{2}=12$ we know $O, A$.
$B$. Points are collinear. Let $A B$
$\operatorname{l} O_{1} O_{2}$ intersect at point $O$,
with $O$ as the origin.
$O A$ as the $x$-axis, establish
a rectangular coordinate system as shown in Figure 6.
Then the equation of $\... | x^{2}+y^{2}=90 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,000 |
4. Let $\alpha, \beta, \gamma \in \mathbf{R}$, and satisfy
$$
\begin{array}{l}
\sin \alpha \cdot \cos \beta+|\cos \alpha \cdot \sin \beta| \\
\quad=\sin \alpha \cdot|\cos \alpha|+|\sin \alpha| \cdot \cos \beta .
\end{array}
$$
Then the minimum value of $(\tan \gamma-\sin \alpha)^{2}+(\cot \gamma-\cos \beta)^{2}$ is | $4.3-2 \sqrt{2}$.
Given the equation factorization
$(\sin \alpha-|\sin \beta|)(\cos \beta-|\cos \alpha|)=0$.
Let $x=\sin \alpha, y=\cos \beta$. Then
$\left(x-\sqrt{1-y^{2}}\right)\left(y-\sqrt{1-x^{2}}\right)=0$.
Thus, $x=\sqrt{1-y^{2}}$ or $y=\sqrt{1-x^{2}}$.
The point $(\tan \gamma, \cot \gamma)$ is on the hyperbola ... | 3-2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,002 |
5. Define the sequence $\left\{a_{n}\right\}: a_{n}=4+n^{3}, n \in \mathbf{N}_{+}$. Let $d_{n}=\left(a_{n}, a_{n+1}\right)$, i.e., $d_{n}$ is the greatest common divisor of $a_{n}$ and $a_{n+1}$. Then the maximum value of $d_{n}$ is $\qquad$ | 5.433.
Given) $d_{n} 1\left(n^{3}+4,(n+1)^{3}+4\right)$, we know $d_{n} \mid\left(4+n^{3}, 3 n^{2}+3 n+1\right)$.
Thus, we have
$$
d_{n} \mid\left[-3\left(4+n^{3}\right)+n\left(3 n^{2}+3 n+1\right), 3 n^{2}+3 n+1\right] \text {, }
$$
which means
$$
d_{n} \mid\left(3 n^{2}+n-12,3 n^{2}+3 n+1\right) \text {. }
$$
Ther... | 433 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,003 |
6. Let $f(x)=a x^{2}+b x+c$ and three different points $P_{i}\left(x_{i}, y_{i}\right)(i=1,2,3)$ form the circumradius $R$ of $\triangle P_{1} P_{2} P_{3}$, where $a, b, c$ are given real numbers. Then $R \quad \frac{1}{2|a|}($ fill in “ $>$ ”, “ $=$ ”, or “ $<$ ” $)$. | 6. $>$.
Let the three sides of $\triangle P_{1} P_{2} P_{3}$ be $a^{\prime}, b^{\prime}, c^{\prime}$.
From $a^{\prime} b^{\prime} c^{\prime}=4 R S_{2} P_{1} P_{2} r_{3}$, we only need to prove
$$
a^{\prime} b^{\prime} c^{\prime}>\frac{2 S_{\triangle P_{1} P_{2} \mu_{3}}}{|a|} \text {. }
$$
By $y_{i}=a x_{i}^{2}+b x_{... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,004 |
Three. (20 points) Given that $a$ and $b$ are integers, and satisfy $\sqrt{9-8 \sin 50^{\circ}}=a+b \csc 50^{\circ}$. Find all possible values of $a$ and $b$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, from the triple angle formula, we have
$$
\sin 3 \times 50^{\circ}=3 \sin 50^{\circ}-4 \sin ^{3} 50^{\circ}=\frac{1}{2} \text {. }
$$
Thus, $-8 \sin ^{3} 50^{\circ}+6 \sin 50^{\circ}-1=0$, which means
$\left(9-8 \sin 50^{\circ}\right) \sin ^{2} 50^{\circ}=\left(3 \sin 50^{\circ}-1\right)^{2}$.
Since $9>8 \sin 5... | a=3, b=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,005 |
Example 1 Let $X_{n}=\{1,2, \cdots, n\}$, for any non-empty subset $A$ of $X_{n}$, let $T(A)$ be the product of all numbers in $A$. Find $\sum_{A \subseteq X_{n}} T(A)$. | Explanation: Constructive Function
$$
\begin{array}{l}
f(x)= \sum_{1 \leq X_{n}} T(A) x^{n-|A|} \\
=(1+2+\cdots+n) x^{n-1}+x^{n-2} \sum_{1 \leqslant i<j \leqslant n} \ddot{j}+ \\
\quad x^{n-3} \sum_{1 \leqslant i<j<k \leqslant n} \ddot{j} k+\cdots+n!x^{0} \\
=(x+1)(x+2) \cdots(x+n)-x^{n},
\end{array}
$$
where $|A|$ d... | (n+1)!-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,006 |
Four. (20 points) Let $S$ be the set of real numbers $x$ that satisfy the inequality $x^{2}+$ $\sqrt{2-x} \leqslant 2$. Find $S$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | The original inequality is
$$
\sqrt{2-x} \leqslant 2-x^{2} \text {. }
$$
By definition, we know
$-\sqrt{2} \leqslant x \leqslant \sqrt{2}$,
and $y=2-x^{2}$
is the inverse function of $y=\sqrt{2-x}$ within $[-\sqrt{2}, \sqrt{2}]$. Draw the function graphs, as shown in Figure 9.
Let $A\left(x_{1}, y_{1}\right)$ and $B\... | S=\left\{x \left\lvert\, \frac{1-\sqrt{5}}{2} \leqslant x \leqslant 1\right.\right\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,007 |
$$
\begin{array}{c}
\text { Five. (20 points) Given } \theta_{1}, \theta_{2}, \cdots, \theta_{n} \in \mathbf{R} \text {, and } \\
f=\sin \left(\theta_{1}-\theta_{2}\right)+\sin \left(\theta_{2}-\theta_{3}\right)+\cdots+\sin \left(\theta_{n}-\theta_{1}\right),
\end{array}
$$
Here $n \geqslant 3$.
(1) When $\left\{\thet... | (1) Suppose $|\theta_{k}|$ is decreasing, then $\theta_{i}-\theta_{i+1} \in [0, 2\pi)$.
For $z_{k} = \mathrm{e}^{\mathrm{i} \theta_{k}}$, we have
$\sin \left(\theta_{k}-\theta_{k+1}\right) = \operatorname{Im}\left(z_{k} \overline{z_{k+1}}\right)$.
Thus, $f = \operatorname{Im}\left(\sum_{k=1}^{n} z_{k} \overline{z_{k+1}... | f_{\text{max}} = n \sin \frac{2\left[\frac{n+1}{4}\right] \pi}{n}, f_{\text{min}} = -n \sin \frac{2\left[\frac{n+1}{4}\right] \pi}{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,008 |
Given $a, b, x, y$ are non-negative real numbers, and $a+b=27$. Try to find the maximum value of $\lambda$, such that the inequality
$$
\left(a x^{2}+b y^{2}+4 x y\right)^{3} \geqslant \lambda\left(a x^{2} y+b x y^{2}\right)^{2}
$$
always holds, and find the conditions for equality. | Let $a=0, b=27, x=27, y=2$, then the original inequality is $\lambda \leqslant 4$.
We will now prove that $\lambda=4$ makes the given inequality always true.
It is only necessary to prove under the original conditions that
$$
\left(a x^{2}+b y^{2}+4 x y\right)^{3} \geqslant 4(a x+b y)^{2} x^{2} y^{2}.
$$
When $x$ or $... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,009 |
II. (50 points) As shown in Figure 3, quadrilateral $ABCD$ is a cyclic quadrilateral inscribed in $\odot M$, and the opposite sides intersect at points $E$ and $F$, respectively. $O$ is the midpoint of $EF$, and $OH$ is tangent to $\odot M$ at point $H$. Prove:
$$
H E \perp H F .
$$ | As shown in Figure 10, take point $I$ on $EF$ such that $\angle C I F = \angle C B E$. Then
$$
F C \cdot F B = E F \cdot F.
$$
$$
\text{Also, } \angle C I E = \angle C B A
$$
$= \angle C D F$, then
$$
E C \cdot E D = E F \cdot I E.
$$
$$
\begin{array}{l}
\text{(1) } + \text{ (2) gives} \\
E H^{2} = E C \cdot E D + F C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,010 |
Three, (50 points) A school has $n$ classes ($n$ is a given positive integer), and the number of boys and girls in each class differs by at most 1. Now the school is holding a table tennis tournament, with the following rules: players from the same class do not compete, while every two players from different classes co... | Three, suppose there are $m$ classes of Zhi Duowei with an odd number of students, $n=k^{2}+r \cdot k$, where $k$ is a natural number, and $r$ is a non-negative integer. $0 \leqslant r \leqslant 2 k$.
When $r=0,2,2 k-1$, $m=n$;
When $r=1$, $m=n-1$;
When $r \neq 2 k-1$ and $3 \leqslant r \leqslant 2 k$, $m=n+2-r$.
Let t... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,011 |
Initially 161 points $P$, $Q$, $R$ are on the sides $BC$, $CA$, $AB$ of $\triangle ABC$ respectively, and they trisect the perimeter. Let the lengths of the three sides of $\triangle ABC$ be $a$, $b$, $c$. Find the maximum value of $S=AQ \cdot BP + BR \cdot CQ + CP \cdot AR$. | Solution: As shown in Figure 2, let $A Q=$
$$
\begin{array}{l}
x, B R=y, C P=z \text {. Let } t \\
=\frac{1}{3}(a+b+c) \text {, then } \\
S=x(t-y)+ \\
y(t-z)+ \\
z(t-x) \\
=(x+y+z) t-(x y+y z+z x) \text {. } \\
\end{array}
$$
Since $b=x+(t-z), c=y+(t-x)$, then
$$
y=x+c-t, z=x+t-b \text {. }
$$
Thus, $S=-3 x^{2}+(a+3 ... | \frac{1}{6} \sum a^{2}+\frac{1}{36}\left(\sum a\right)^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,012 |
Does there exist a permutation $a_{1}, a_{2}, \ldots, a_{2005}$ of $1,2, \ldots, 2005$ such that
$$
f(n)=n+a_{n} \quad (n=1,2, \ldots, 2005)
$$
are all perfect squares? | Solution: The conclusion is affirmative.
Below is the construction of a permutation of $1,2, \ldots, 2005$ that satisfies the given conditions.
Since $45^{2}=2025, 2025-2005=20$, for $20 \leq n \leq 2005$, we take $a_{n}=2025-n$, and we have
$$
f(n)=n+a_{n}=2025=45^{2}
$$
which satisfies the given conditions.
Next, w... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,013 |
There are $161 n$ singers participating in the performance of an art festival, and the organizers have arranged $m$ performances for them. It is required that each performance is presented by 4 of the singers, and the number of times any two of the $n$ singers perform together is the same. Please design a scheme so tha... | Solution: Let the number of times each pair of singers performs together be $r$.
(1) $n=6$. At this time, 6 singers can form $\mathrm{C}_{6}^{2}$ pairs of singers, each pair performing $r$ times, for a total of $r C_{6}^{2}=15 r$ performances. On the other hand, $m$ performances have $m \mathrm{C}_{4}^{2}=6 m$ pairs of... | 15, 7, 18 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,014 |
Given an acute triangle $\triangle ABC$ with a fixed shape, two moving points $D, E$ are on sides $AB, AC$ respectively, and $DF \perp BC, EG \perp BC$, with $F, G$ being the feet of the perpendiculars. Let $B E \cap C D=O, F E \cap G D=P$. Prove: the line $OP$ always passes through a certain fixed point.
Translate th... | Proof: As shown in Figure 3, draw the altitude $AQ$, then the foot of the perpendicular $Q$ is a fixed point. Below, we will prove that the line $OP$ passes through point $Q$, i.e., points $O$, $P$, and $Q$ are collinear. Considering $\triangle ABE$ being intersected by the line $DOC$, using Menelaus' theorem, we get
$... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,015 |
Example 2 Let $X_{n}=\{1,2, \cdots, n\}$. Find $\sum_{A \subseteq X_{n}} S(A)$, where $S(A)$ denotes the sum of all elements in $A$, and $S(\varnothing)=0$. | Let $A=\left\{a_{1}, a_{2}, \cdots, a_{t}\right\}$ be any subset of $X_{n}$, then $S(A)=a_{1}+a_{2}+\cdots+a_{t}$. Hence,
$$
\sum_{A \subseteq X_{n}} S(A)=\sum_{\left\{a_{1}, a_{2}, \cdots, a_{t} \subseteq X_{n}\right\}}\left(a_{1}+a_{2}+\cdots+a_{t}\right) \text {. }
$$
Consider the number of times $i(i=1,2, \cdots, ... | n(n+1) 2^{n-2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,016 |
Example 3 Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of $1,2, \cdots, n$. For a number $a_{i}$ in the permutation, if it satisfies: either $i=n$, or for all $j(ia_{j}$, then $a_{i}$ is called a "large number" in this permutation. Let the number of large numbers in the permutation $\sigma=$ $\left(a_{1}, a_{2}, ... | Let $g(i)$ denote the number of permutations where $i$ is the largest number, then
$$
\sum_{\sigma} r(\sigma)=\sum_{i=1}^{n} g(i) \text {. }
$$
Consider any permutation $\sigma$ where $i$ is the largest number, and we determine the order of $1,2, \cdots, n$ in the permutation $\sigma$.
First, arrange $i+1, i+2, \cdot... | n!\sum_{i=1}^{n} \frac{1}{i} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,017 |
Example 4 Let $X_{n}=\{1,2, \cdots, n\}$, for any subset $A$ of $X_{n}$, denote $t(A)$ as the smallest element in $A$. Find $t=$ $\sum_{A \subseteq X_{n}} t(A)$. | Explanation: Classification of the values of $t(A)$:
Let $t(A)=i(i=1,2, \cdots, n)$, then
$$
\sum_{1 \leq X_{n}} t(A)=\sum_{i=1}^{n} \sum_{t(A)=i, A \subseteq X_{n}} i \text {. }
$$
When $t(A)=i$, $A$ does not contain any number smaller than $i$, and the $n-i$ numbers larger than $i$ can either belong to $A$ or not, e... | 2^{n+1}-2-n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,018 |
Example 5 Let $A$ be any permutation of $X_{n}=\{1,2, \cdots, n\}$, define $f(A)=\sum_{i=1}^{n}|i-A(i)|$, where $A(i)$ is the number at the $i$-th position of permutation $A$. Prove:
$$
\sum_{A} f(A)=\frac{1}{3} n!\left(n^{2}-1\right)
$$ | Explanation: Notice that
$$
S=\sum_{A} f(A)=\sum_{A} \sum_{i=1}^{n}|i-A(i)| .
$$
We classify the values of $A(i)$:
When $A(i)=j(j=1,2, \cdots, n)$,
$$
|i-A(i)|=|i-j| \text {. }
$$
Next, we discuss the number of times $|i-j|$ appears in $S$.
Since $A(i)=j$, i.e., $j$ is in the $i$-th position of $A$, and the other $n-... | \frac{1}{3} n!\left(n^{2}-1\right) | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,019 |
Example 6 For any subset $A$ of $X_{n}=\{1,2, \cdots, n\}$, denote the sum of all elements in $A$ as $S(A), S(\varnothing)=0$, and the number of elements in $A$ as $|A|$. Find $\sum_{A \subseteq X_{n}} \frac{S(A)}{|A|}$. | Explanation: First classify $|A|$:
Let $|A|=k(k=1,2, \cdots, n)$, then
$$
\begin{array}{l}
\sum_{A \leq X_{n}} \frac{S(A)}{|A|}=\sum_{k=1}^{n} \sum_{|A|=k} \frac{S(A)}{|A|} \\
=\sum_{k=1}^{n} \frac{1}{k} \sum_{|A|=k} S(A) .
\end{array}
$$
Next, for each element $i$ in $X_{n}$, consider the number of times $i$ appears ... | \frac{1}{2}(n+1)\left(2^{n}-1\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,020 |
Example 8 For any non-empty subset $A$ of $X_{n}$, define $f(A)$ as the sum of the smallest and largest elements in $A$. Find the average value of all $f(A)$ for $X_{n}$. | For a non-empty subset $A$ of $X_{n}$, let $A=\left\{a_{1}<a_{2}<\cdots<a_{i}\right\}$.
(1) If $a_{1}+a_{t}=n+1$, then $f(A)=n+1$.
(2) If $a_{1}+a_{t} \neq n+1$, then let
$$
A^{\prime}=\left\{n+1-a_{1}, n+1-a_{2}, \cdots, n+1-a_{t}\right\} \text {. }
$$
We now prove that $A \neq A^{\prime}$.
In fact, if $A=A^{\prime}$... | n+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,021 |
1. Let $n(n>2), Y_{n}=\{2,3, \cdots, n\}$. For each non-empty subset $A$ of $Y_{n}$, let $T(A)$ be the product of all elements in $A$. Find $\sum_{A \subseteq Y_{n}} T(A)$. | (提示: Construct the function $f(x)=\sum_{1 \leq Y_{n}} T(A) x^{n-|A|-1}$. Then $\left.\sum_{A \leq Y_{n}} T(A)=f(1)=\frac{1}{2}(n+1)!-1.\right)$ | \frac{1}{2}(n+1)!-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,022 |
2. For any subset $A(|A|>1)$ of $X_{n}=\{1,2, \cdots, n\}$, let the sum of the products of every two elements in $A$ be $f(A)$, the sum of the sums of every two elements in $A$ be $g(A)$, and the sum of the absolute values of the differences of every two elements in $A$ be $h(A)$. Find $\sum_{A \leq X_{n}} f(A), \sum_{... | $\begin{array}{l}\text { (Hint: } \sum_{1 \leq x_{n}} f(A)=2^{n-2} \sum_{1 \leqslant i<j \leqslant n} i j \\ =2^{n-5} \cdot \frac{n(n+1)(3 n+2)(n-1)}{3}, \\ \sum_{1 \leq x_{n}} g(A)=2^{n-2} \sum_{1<i<j<n}(i+j)=2^{n-3} n\left(n^{2}-1\right), \\ \left.\sum_{1 \leq x_{n}} h(A)=2^{n-2} \sum_{1<i<j<n}|i-j|=2^{n-2} C_{n+1}^{... | \sum_{1 \leq x_{n}} f(A)=2^{n-5} \cdot \frac{n(n+1)(3 n+2)(n-1)}{3}, \sum_{1 \leq x_{n}} g(A)=2^{n-3} n\left(n^{2}-1\right), \sum_{1 \leq x_{n}} h(A)=2^{n-2} C_{n+1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,023 |
Example 10 As shown in Figure 7, in $\triangle ABC$, $BC=6$, $AC=4\sqrt{2}$, $\angle C=45^{\circ}$, $P$ is a moving point on side $BC$. A line through point $P$ parallel to $AB$ intersects $AC$ at point $D$. Connect $AP$, the areas of $\triangle ABP$, $\triangle APD$, and $\triangle CDP$ are denoted as $S_{1}$, $S_{2}$... | Explanation: (1) Draw $P M \perp A C, A N \perp B C$. From the problem, we know that $B P=x(0<x<6)$ and $\frac{A D}{B P}=\frac{A C}{B C}$. Therefore,
$$
\begin{array}{l}
A D=\frac{A C}{B C} \cdot B P=\frac{4 \sqrt{2}}{6} x=\frac{2 \sqrt{2}}{3} x, \\
S_{1}=S_{\triangle A B P}=\frac{1}{2} B P \cdot A N=2 x, \\
S_{2}=S_{\... | S_{1}=S_{2}+S_{3}, S_{2}=S_{3}, S_{1}=2 S_{2}, S_{1}=2 S_{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,024 |
3. For any $r$-element subset $A$ of $X_{n}=\{1,2, \cdots, n\}$, let the smallest number in $A$ be $t(A)$. Find $\sum t(A)$. | (Let $t(A)=i$. Consider the number of $r$-element subsets with $i$ as the smallest element, obtaining $\sum_{t=1} t(A)=\mathrm{C}_{n+1}^{r+1}$. ) | \mathrm{C}_{n+1}^{r+1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,025 |
4. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers, and $S_{k}$ be the sum of the products of any $k$ numbers chosen from $a_{1}, a_{2}, \cdots, a_{n}$. Prove:
$$
S_{k} S_{n-k} \geqslant\left(\mathrm{C}_{n}^{k}\right)^{2} a_{1} a_{2} \cdots a_{n}
$$ | $\begin{array}{l}\text { (Hint: } S_{k}=\sum_{1<i_{1}<i_{2}<\cdots<i_{k}<n} a_{i_{1}} a_{i_{2}} \cdots a_{i_{k}} \geqslant \mathrm{C}_{n}^{k} \\ \left.\sqrt[n]{\prod_{1<i_{1}<i_{2}<\cdots<i_{k} \leqslant n} a_{i_{1}} a_{i_{2}} \cdots a_{i_{k}}}=\mathrm{C}_{n}^{k} \sqrt[c]{\left(a_{1} a_{2} \cdots a_{n}\right)_{n-1}^{\m... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,026 |
5. For any subsets $A, B$ of $X_{n}$ ( $A, B$ can be the same), find $\sum_{A, B \subseteq X}|A \cap B|$, where $|A|$ denotes the number of elements in $A$.
| (The number of set pairs $(A, B)$ such that $|A \cap B|=k$ is $\mathrm{C}_{n}^{k} 3^{n-k}$, from which we get $\left.\sum_{A, B \subseteq X_{n}}|A \cap B|=n \times 4^{n-1}.\right)$ | n \times 4^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,027 |
Example 1 As shown in Figure 1, let $A D$ be the altitude from vertex $A$ to the base $B C$ of acute $\triangle A B C$, and let $H$ be any point on $A D$ different from $A$ and $D$. The extensions of $\mathrm{BH}$ and $\mathrm{CH}$ intersect $A C$ and $A B$ at points $E$ and $F$, respectively. Prove:
$$
\angle E D H=\a... | Proof 1: Line $F H C$ intersects $\triangle A B E$. By Menelaus' theorem and the area formula of a triangle, we have
$$
\begin{aligned}
1= & \frac{A F}{F B} \cdot \frac{B H}{H E} \cdot \frac{E C}{C A}=\frac{S_{\triangle A F D}}{S_{\triangle B D}} \cdot \frac{S_{\triangle H B D}}{S_{\triangle A D E}} \cdot \frac{S_{\tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,028 |
Example 3 As shown in Figure 3, take points $B$ and $D$ on the sides $AE$ and $AF$ of $\triangle AEF$, respectively, such that $BF \cap ED = C$, and $AC \cap BD = P$. Draw $PO \perp EF$ at point $O$. Prove:
$$
\angle BOP = \angle DOP.
$$
Note, when $AC \perp EF$, this problem becomes Example 1. Therefore, when solving... | Prove: For $\triangle A B D$ and point $C$, applying Ceva's Theorem, we have
$$
\begin{aligned}
1= & \frac{A E}{E B} \cdot \frac{B P}{P D} \cdot \frac{D F}{F A}=\frac{S_{\triangle A E O}}{S_{\triangle B E O}} \cdot \frac{S_{\triangle P B O}}{S_{\triangle A O D}} \cdot \frac{S_{\triangle D O F}}{S_{\triangle A O F}} \\
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,029 |
Example 4 In $\triangle A B C$, take a point $F$ on side $A B$ and a point $E$ on side $A C$, such that $B E \cap C F=M$. Draw $M D \perp B C$ at point $D$. Given that $\angle F D M=\angle M D E$. Prove: $A 、 M$ 、 $D$ are collinear. | Proof: As shown in Figure 5, extend $DM$ to intersect the broken line $BAC$ at a point $A'$. Without loss of generality, assume that point $A'$ lies on side $AB$. (Including point $A$, but if it passes through point $A$, the conclusion is already established, so we can assume it does not pass through point $A$). Connec... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,030 |
Example 5: On side $BC$ of $\triangle ABC$, take a point $D$. With $D$ as the center, construct a semicircle inside the triangle that is tangent to sides $AB$ and $AC$ at points $F$ and $E$ respectively. Let $CF \cap BE = H$, and draw $HG \perp BC$ at point $G$. Prove that $HG$ bisects $\angle FGE$.
This is a selected... | Prove: Connecting $D A, D E, D F$, and drawing $A G^{\prime} \perp B C$ at point $G^{\prime}$, we have
$\triangle B D F \backsim \triangle B A G^{\prime}, \triangle C D E \backsim \triangle C A G^{\prime}$. Since $D E=D F$, it follows that,
$$
\frac{B G^{\prime}}{B F}=\frac{A G^{\prime}}{D F}=\frac{A G^{\prime}}{D E}=\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,031 |
Example 6 As shown in Figure 7, in the acute $\triangle ABC$, $AD$ is the internal angle bisector of $\angle A$, point $D$ is on side $BC$, and through point $D$, $DE \perp AC$ and $DF \perp AB$ are drawn, with the feet of the perpendiculars being $E$ and $F$ respectively. Connecting $BE$ and $CF$, they intersect at po... | Right triangle.
A discerning eye can see that the figure in this problem is almost identical to the figure in Example 5. Therefore, we can immediately prove that $BE$, $CF$, and the altitude from $A$ to side $BC$ are concurrent at point $H$. The proof method is exactly the same. Once this is proven, the remaining part ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,032 |
Example 8 As shown in Figure 9, the acute $\triangle A B C$ is inscribed in $\odot O$. Two tangents to $\odot O$ are drawn through points $B$ and $C$, respectively, and intersect the tangent to $\odot O$ through point $A$ at points $M$ and $N$. $A D$ is the altitude from $A$ to side $B C$. Prove that $A D$ bisects $\an... | Proof: As shown in Figure 9, let $\angle A D M=\alpha$, $\angle A D N=\beta$. Denote $A B \cap M D=F, A C \cap N D=E$, then we have
$$
\begin{array}{l}
\frac{A F \cdot C E}{F B \cdot E A}=\frac{S_{\triangle A F D}}{S_{\triangle B F D}} \cdot \frac{S_{\triangle D C E}}{S_{\triangle D A E}} \\
=\frac{A D \sin \alpha}{B D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,033 |
Example 1: On a leftover piece of an acute triangle, process it into a square part so that the four vertices of the square are all on the sides of the triangle. If the three sides of the triangle are $a$, $b$, and $c$, and $a > b > c$. Which side should two vertices of the square be placed on to maximize the area of th... | Solution: As shown in Figure 2, from equation (1) we get $\frac{A E}{A D}=\frac{P S}{B C}$, which means
$$
P S=\frac{A D \cdot B C}{A D+B C}
$$
As shown in Figure 3, from equation (1) we get $M N=\frac{A B \cdot C G}{A B+C G}$.
Let the area of $\triangle A B C$ be $S$. Note that
$$
A B \cdot C G=B C \cdot A D=2 S \tex... | M N > P S | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,034 |
Example 11 As shown in Figure 8, sector $A O B$ is one quarter of a unit circle. The center of the semicircle $\odot O_{1}$, $O_{1}$, is on $O A$ and is internally tangent to $\overparen{A B}$ at point $A$. The center of the semicircle $\odot O_{2}$, $O_{2}$, is on $O B$ and is internally tangent to $\overparen{A B}$ a... | (1) As shown in Figure 8, let the radii of $\odot O_{1}$ and $\odot O_{2}$ be $R$ and $r$ respectively. Then,
$$
\begin{array}{l}
y=\frac{1}{2} \pi\left(R^{2}+r^{2}\right) \\
=\frac{1}{2} \pi\left[(R+r)^{2}-2 R r\right] .
\end{array}
$$
Connecting $O_{1} O_{2}$, the line connecting the centers must pass through the po... | (3-2 \sqrt{2}) \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,035 |
Example 2 Let $\triangle ABC$ have three inscribed squares (one side on one side of the triangle, the other two vertices on the other two sides) of equal area. Prove: $\triangle ABC$ is an equilateral triangle.
(19th Jiangsu Province Junior High School Mathematics Competition) | Proof: From Example 1, we know that $S_{\text {square PORS }}=S_{\text {square MNKL }}$ if and only if $a=c$. Since the areas of the three inscribed squares on the sides of $\triangle A B C$ are equal, $\triangle A B C$ is an equilateral triangle.
untranslated text remains the same in terms of line breaks and formatt... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,036 |
Example 4 As shown in Figure $5, \triangle A B C$'s area is three times the area of its inscribed rectangle $P Q R S$, and the values of side $B C$ and height $A D$ are rational numbers. Under what conditions is the perimeter of rectangle $P Q R S$ a rational number? Under what conditions is it an irrational number? | Solution: Let $BC=a$, $AD=h$, $PQ=m$, $PS=n$.
From equation (1), we get $\frac{h-n}{h}=\frac{m}{a}$, which means $mh + an = ah$.
Also, $mn = \frac{1}{3} \times \frac{1}{2} ah = \frac{1}{6} ah$, so,
$mhan = \frac{1}{6} a^2 h^2$.
From equations (1) and (2), we know that $hm$ and $an$ are the two roots of the equation $x^... | 2(m+n) = a + h \pm \frac{\sqrt{3}}{3}(a - h) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,038 |
Example 2 Let point $A(a, b)$ be an external point of the circle $x^{2}+y^{2}=R^{2}$, and let $A P$ and $A Q$ be the tangents from $A$ to the circle. Then the equation of the line $P Q$ is
$$
a x+b y=R^{2} \text {. }
$$
The proof of this conclusion is quite interesting, and it left a deep impression on me from my days... | Proof 1: As shown in Figure 1, let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, then the tangent line equations through points $P$ and $Q$ are
$$
\begin{array}{l}
A P: x_{1} x+y_{1} y=R^{2}, \\
A Q: x_{2} x+y_{2} y=R^{2}.
\end{array}
$$
Since point $A$ lies on both lines, we have
$$
\begin{array}{l}
x_{1}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,039 |
Example 3 Let point $A(a, b)$ be an external point of the circle $x^{2}+y^{2}=R^{2}$, and let $B$ and $C$ be the points where the circle intersects the $x$-axis. Connecting $A B$ and $A C$ sequentially intersects the circle at points $F$ and $E$. Draw the tangents $A P$ and $A Q$ from $A$ to the circle, then $P Q$, $B ... | ```
Proof: As shown in Figure 2, from Example 2, the equation of the line }PQ\mathrm{ is
ax+by = R R
The equations of BE and CF are respectively
(a-R)(x+R)+by=0,
(a+R)(x-R)+by=0,
i.e., }(a-R)x+by=\mp@subsup{R}{}{2}-aR\mathrm{,
Adding these equations gives the equation of the line pas... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,040 |
Example 4 Given the equation $|x|=a x+1$ has one negative root and no positive root. Then, the range of values for $a$ is ( ).
(A) $a>-1$
(B) $a=1$
(C) $a \geqslant 1$
(D) None of the above
(1987, National Junior High School Mathematics League)
To solve correctly, a case-by-case discussion is required. First, the absol... | Solution: It is clear that the roots of the equation are not equal to 0.
The following is completed in three steps.
(1) Find the range of values for $a$ when the equation has negative roots.
Let $x-1$.
For $a>-1$, the equation has negative roots, but it may also have positive roots, so we need to remove those $a$ that... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,041 |
Example 7 As shown in Figure 7(b), in the "windmill triangle," $A A^{\prime}=B B^{\prime}=C C^{\prime}=2, \angle A O B^{\prime}=\angle B O C^{\prime}=$ $\angle C O A^{\prime}=60^{\circ}$. Prove:
$$
S_{\triangle A B^{\prime}}+S_{\triangle B A C^{\prime}}+S_{\triangle O A^{\prime}}<\sqrt{3} .
$$ | Proof 1: As shown in Figure 7(a), translate the figure into an equilateral $\triangle O P Q$ with side length 2, then
$$
\begin{array}{l}
S_{\triangle A O B}+S_{\triangle B A C^{\prime}}+S_{\triangle O O A^{\prime}} \\
<S_{\triangle O P P}=\frac{\sqrt{3}}{4} O P^{2}=\sqrt{3} .
\end{array}
$$
Proof 2: Let $O A=a, O B=b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,042 |
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