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
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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3. As shown in Figure 2, in $\triangle A B C$, $\angle B$ is a right angle, the angle bisector of $\angle A$ is $A D$, and the median on side $B C$ is $A E$. Points $D$ and $E$ divide $B C$ into three segments in the ratio $1: 2: 3$. Then $\sin \angle B A C=(\quad)$.
(A) $\frac{12}{13}$
(B) $\frac{4 \sqrt{3}}{9}$
(C) $... | 3.C.
Let $B D=x, A B=y$, then $D E=2 x, E C=3 x$.
By $\frac{B D}{D C}=\frac{A B}{A C}$, we get $A C=5 y$.
Also, $A B^{2}+B C^{2}=A C^{2}$, which means $y^{2}+(6 x)^{2}=(5 y)^{2}$.
Therefore, $x^{2}=\frac{2 y^{2}}{3}, \sin \angle B A C=\frac{6 x}{5 y}=\frac{2 \sqrt{6}}{5}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,306 |
2. Let $a$, $b$, $c$ be integers, and for all real numbers $x$,
$$
(x-a)(x-8)+1=(x-b)(x-c)
$$
always holds. Then the value of $a+b+c$ is $\qquad$. | 2.20 or 28 .
$$
\begin{array}{l}
\text { Given } x^{2}-(8+a) x+8 a+1 \\
=x^{2}-(b+c) x+b c
\end{array}
$$
always holds, so,
$$
\left\{\begin{array}{l}
8+a=b+c, \\
8 a+1=b c .
\end{array}\right.
$$
Eliminating $a$ yields $b c-8(b+c)=-63$, which simplifies to $(b-8)(c-8)=1$.
Since $b$ and $c$ are integers, we have $b-8... | 20 \text{ or } 28 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,307 |
3. As shown in Figure 2, in two concentric circles with center $O$, $M N$ is the diameter of the larger circle, intersecting the smaller circle at points $P$ and $Q$. The chord $M C$ of the larger circle intersects the smaller circle at points $A$ and $B$. If $O M=2$, $O P=1$, and $M A=A B=B C$, then the area of $\tria... | 3. $\frac{3 \sqrt{15}}{8}$.
Let $M A=x$.
From $M A \cdot M B=M P \cdot M Q$, we get $x \cdot 2 x=1 \times 3$.
Solving for $x$ yields $x=\sqrt{\frac{3}{2}}$.
Connect $C N$. In the right triangle $\triangle M C N$,
$$
M C=3 x=3 \sqrt{\frac{3}{2}}, M N=4 \text {. }
$$
Therefore, $N C=\sqrt{N M^{2}-M C^{2}}=\sqrt{\frac{5... | \frac{3 \sqrt{15}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,308 |
4. From $1,2, \cdots, 2006$, at least
$\qquad$ odd numbers must be taken to ensure that there are definitely two numbers whose sum is 2008. | 4.503.
From $1,2, \cdots, 2006$, selecting two odd numbers whose sum is 2008, there are a total of 501 pairs as follows:
$$
3+2005,5+2003, \cdots, 1003+1005 \text {. }
$$
Since 1 added to any of these odd numbers will not equal 2008, therefore, at least 503 odd numbers must be selected to ensure that there are defini... | 503 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,309 |
One, (20 points) Real numbers $x, y, z, w$ satisfy $x \geqslant y \geqslant z \geqslant w \geqslant 0$, and $5 x+4 y+3 z+6 w=100$. Find the maximum and minimum values of $x+y+z+w$. | Let $z=w+a, y=w+a+b, x=w+$ $a+b+c$. Then $a, b, c \geqslant 0$, and
$$
x+y+z+w=4 w+3 a+2 b+c .
$$
Therefore, $100=5(w+a+b+c)+4(w+a+$
$$
\begin{array}{l}
\quad b)+3(w+a)+6 w \\
=18 w+12 a+9 b+5 c \\
=4(4 w+3 a+2 b+c)+(2 w+b+c) \\
\geqslant 4(x+y+z+w) .
\end{array}
$$
Thus, $x+y+z+w \leqslant 25$.
When $x=y=z=\frac{25}... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,310 |
II. (25 points) As shown in Figure 3, in the right triangle $\triangle ABC$, $\angle B = 90^{\circ}$, its incircle is tangent to sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Connect $AD$ and let it intersect the incircle at another point $P$. Connect $PC$, $PE$, and $PF$. Given that $PC \perp P... | (1) As shown in Figure 8, connect $D F$. Then $\triangle B D F$ is an isosceles right triangle. Therefore,
$$
\begin{array}{l}
\angle F P D=\angle F D B \\
=45^{\circ} .
\end{array}
$$
Thus, $\angle D P C=45^{\circ}$.
Since $\angle P D C=\angle P F D$, we have
$\triangle P F D \backsim \triangle P D C$.
Therefore, $\f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,311 |
$$
\begin{array}{r}
\text { Three. (25 points) In }\left[\frac{1^{2}}{2008}\right],\left[\frac{2^{2}}{2008}\right], \cdots, \\
{\left[\frac{2008^{2}}{2008}\right] \text {, how many different integers are there (where }[x]}
\end{array}
$$
indicates the greatest integer not greater than $x$)? | Three, let $f(n)=\frac{n^{2}}{2008}$.
When $n=2,3, \cdots, 1004$, we have
$$
\begin{array}{l}
f(n)-f(n-1) \\
=\frac{n^{2}}{2008}-\frac{(n-1)^{2}}{2008}=\frac{2 n-1}{2008}1 .
\end{array}
$$
And $f(1005)=\frac{1005^{2}}{2008}=\frac{(1004+1)^{2}}{2008}$
$$
=502+1+\frac{1}{2008}>503,
$$
Therefore, $\left[\frac{1005^{2}}{... | 1507 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,312 |
1. Let $a_{n}=7^{n}+9^{n}\left(n \in \mathbf{N}_{+}\right)$. Then the remainder when $a_{2008}$ is divided by 64 is ( ).
(A) 0
(B) 2
(C) 16
(D) 22 | $\begin{array}{l}\text { i.1.B. } \\ a_{2008}=7^{2008}+9^{2008} \\ =(8-1)^{2008}+(8+1)^{2008} \\ =64 k+2 .\end{array}$ | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,313 |
3. Polynomial
$$
\begin{array}{l}
|x+1|+|x+2|+\cdots+|x+2008|+ \\
|x-1|+|x-2|+\cdots+|x-2008|
\end{array}
$$
The minimum value of the above expression is ( ).
(A) 4034072
(B) 4030056
(C) 2008
(D) 0 | 3. A.
From the problem, we know that within $-1 \leqslant x \leqslant 1$, the polynomial has the
$$
\begin{array}{l}
\text { minimum value } \\
2(1+2+\cdots+2008) \\
=2008 \times 2009=4034072 .
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,315 |
4. Four students participate in a competition. The competition rules stipulate: each student must choose one question to answer from question A or B. Choosing question A, a correct answer earns 100 points, and a wrong answer earns -100 points; choosing question B, a correct answer earns 90 points, and a wrong answer ea... | 4.C.
(1) If all four people choose A (or B), among them, two are correct and two are wrong, there are $2 \mathrm{C}_{4}^{2}=12$ ways;
(2) If two people choose A and two choose B, for each question, one person is correct and one is wrong, there are $2^{2} \mathrm{C}_{4}^{2}=24$ ways.
In summary, there are 36 different ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,316 |
4. Express 2008 as the sum of $k\left(k \in \mathbf{N}_{+}\right)$ distinct square numbers. Then the minimum value of $k$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 4.B.
Since the square of an odd number is congruent to 1 modulo 8, and the square of an even number is congruent to 0 modulo 4, if 2008 is the sum of two squares, i.e., $a^{2}+b^{2}=2008$, then $a$ and $b$ must both be even. Let $a=2a_{1}, b=2b_{1}$, transforming it into $a_{1}^{2}+b_{1}^{2}=502$, then $a_{1}$ and $b_... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,317 |
5. Indeterminate Equation
$$
\begin{array}{l}
\frac{1}{2}(x+y)(y+z)(z+x)+(x+y+z)^{3} \\
=1-x y z
\end{array}
$$
The number of integer solutions to the equation is ( ).
(A) 2
(B) 4
(C) 5
(D) 6 | 5.D.
Let $x+y=u, y+z=v, z+x=w$.
Then the equation transforms to
$$
\begin{array}{l}
4 \text { uvw }+(u+v+w)^{3} \\
=8-(u+v-w)(u-v+w) . \\
\quad(-u+v+w) .
\end{array}
$$
Simplifying, we get
$$
4\left(u^{2} v+v^{2} w+w^{2} u+u v^{2}+v w^{2}+u u^{2}\right)+8 u w=8 \text {, }
$$
which is
$$
u^{2} v+v^{2} w+w^{2} u+w w^{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,318 |
1. Add the same integer $a(a>0)$ to the numerator and denominator of $\frac{2008}{3}$, making the fraction an integer. Then the integer $a$ added has $\qquad$ solutions. | $=1.3$.
From the problem, we know that $\frac{2008+a}{3+a}$ should be an integer, which means $\frac{2008+a}{3+a}=\frac{2005}{3+a}+1$ should be an integer. Therefore, $(3+a) \mid 2005=5 \times 401$.
Since 2005 has 4 divisors, at this point, $a$ can take the values $2002$, $398$, $2$, and one divisor corresponds to a va... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,320 |
3. The equation about $x$
$$
\frac{x-a}{b c}+\frac{x-b}{a c}+\frac{x-c}{a b}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \text {, }
$$
where $a+b+c \neq 0$, the solution of the equation is $\qquad$ . | 3. $x=a+b+c$. Transform the original equation into
$$
\begin{array}{l}
\left(\frac{x-a}{b c}-\frac{1}{b}-\frac{1}{c}\right)+\left(\frac{x-b}{a c}-\frac{1}{a}-\frac{1}{c}\right)+ \\
\left(\frac{x-c}{a b}-\frac{1}{a}-\frac{1}{b}\right)=0 .
\end{array}
$$
From this, we can solve for $x=a+b+c$. | x=a+b+c | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,322 |
One. (20 points) Arrange natural numbers in a triangular number table as shown in Figure 4, following the principle of increasing from top to bottom and from left to right (each row has one more number than the previous row). Let $a_{i j}\left(i, j \in \mathbf{N}_{+}\right)$ be the number located in the $j$-th position... | (1) In the triangular number table, the first $n$ rows contain
$$
1+2+\cdots+n=\frac{n(n+1)}{2}
$$
numbers, i.e., the last number in the $i$-th row is $\frac{i(i+1)}{2}$.
Therefore, the $i$ that makes $a_{i j}=2.008$ is the smallest positive integer solution to the inequality $\frac{i(i+1)}{2} \geqslant 2008$.
Since ... | \frac{5}{2}-\frac{1}{n}-\frac{1}{n+1} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,324 |
II. (25 points) Let $l$ be a line in the plane of acute $\triangle ABC$, and let the lines symmetric to $l$ with respect to the three sides of $\triangle ABC$ intersect each other in pairs at points $A^{\prime} 、 B^{\prime} 、 C^{\prime}$. Prove: The incenter of $\triangle A^{\prime} B^{\prime} C^{\prime}$ lies on the c... | As shown in Figure 7, let line $l$ intersect sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. The reflections of line $l$ across $BC$, $CA$, and $AB$ are lines $k$, $m$, and $n$, respectively. Let the intersections of lines $m$ and $n$, $n$ and $k$, and $k$ and $m$ be points $A'$, $B'$, and $C'$, r... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,325 |
Three. (25 points) If placing the natural number $N$ to the right of any natural number always results in a new number that is divisible by $N$, then $N$ is called a "magic number." Find all magic numbers.
---
Translation:
Three. (25 points) If placing the natural number $N$ to the right of any natural number always... | Let $N$ be a magic number, with the number of digits being $m$. Then, the new number obtained by placing $N$ to the right of 1 is $10^{m}+N$.
Since $N \perp\left(10^{m}+N\right)$, it follows that $N \perp 10^{m}$. Therefore, $N=2^{a} \cdot 5^{b}$ (where $a$ and $b$ are non-negative integers not exceeding $m$), which m... | 2 \cdot 10^{k}, 5^{r} \cdot 10^{k} \text{ (where } r=0,1,2,3 \text{ and } k \text{ is a non-negative integer)} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,326 |
1. A certain road consists of three equal-length sections: uphill, flat, and downhill. It is known that a car travels at average speeds of $v_{1} 、 v_{2} 、 v_{3}$ on each section, respectively. Then the average speed of this car over the entire road is ( ).
(A) $\frac{v_{1}+v_{2}+v_{3}}{3}$
B) $\frac{\frac{1}{v_{1}}+\f... | - 1.D.
Solution 1: Let the total length of the road be $3s$, then the lengths of the three different sections are all $s$. The time taken by the car to travel each section is $t_{i}=\frac{s}{v_{i}}(i=1,2,3)$, and the average speed of the car over the entire road is
$$
\bar{v}=\frac{3 s}{t_{1}+t_{2}+t_{3}}=\frac{3 s}{\... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,327 |
5. In rectangle $A B C D$, $A B=56, A D=35$. Using lines parallel to $A B$ and $A D$, it is divided into $35 \times$ 56 unit squares. The number of unit squares that are internally crossed by diagonal $A C$ is ( ) .
(A) 84
(B) 85
(C) 91
(D) 92 | 5.A.
Take $A$ as the origin, the lines $AB$ and $AD$ as the $x$-axis and $y$-axis, respectively. Apart from the coordinate axes, there are 35 horizontal lines and 56 vertical lines. The diagonal $AC$ intersects with each of these lines, resulting in $35+56=91$ intersection points (including coincident points).
Next, ... | 84 | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,328 |
2. It is known that the distance between Xiao Wang's and Xiao Li's homes is $16 \mathrm{~km}$. Xiao Wang rides a bicycle from home at a uniform linear speed of $12 \mathrm{~km} / \mathrm{h}$, and Xiao Li rides an electric bike from home at the same time at a uniform linear speed of $24 \mathrm{~km} / \mathrm{h}$. After... | 2.D.
After moving for $20 \mathrm{~min}$, Xiao Wang walked $4 \mathrm{~km}$, and Xiao Li walked $8 \mathrm{~km}$. Location $M$ is on the circle "with Xiao Wang's home as the center and a radius of $4 \mathrm{~km}$", and location $N$ is on the circle "with Xiao Li's home as the center and a radius of $8 \mathrm{~km}$",... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,329 |
3. There are 20 students standing in a row. If they count off every other person from left to right, Xiao Li reports number 8; if they count off every third person from right to left, Xiao Chen reports number 6. Then, starting from Xiao Chen and counting to Xiao Li, the number Xiao Li reports is ( ).
(A) 11
(B) 12
(C) ... | 3.A.
Solution 1: According to the problem, Xiao Li is in the 2 × 7 + 1 = 15th position from left to right in the queue, and Xiao Chen is in the 3 × 5 + 1 = 16th position from right to left, which is the 20 - 16 + 1 = 5th position from left to right. As shown in Figure 3.
Figure 3
Therefore, starting from Xiao Chen and... | A | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 719,330 |
4. If $M=(|a+b|-|a|+|b|)(|a+b|$ $-|a|-|b|)$, then the range of values for $M$ is ( ).
(A) all positive numbers
(B) all non-positive numbers
(C) all negative numbers
(D) all non-negative numbers | 4.B.
$$
\begin{aligned}
\text { Solution } 1: M= & (|a+b|-|a|+|b|) . \\
& (|a+b|-|a|-|b|) \\
= & (|a+b|-|a|)^{2}-|b|^{2} \\
= & (a+b)^{2}-2|a(a+b)|+a^{2}-b^{2} \\
= & a^{2}+2 a b+b^{2}-2|a(a+b)|+a^{2}-b^{2} \\
= & 2 a(a+b)-2|a(a+b)| \leqslant 0 .
\end{aligned}
$$
Solution 2: Take \(a=b=0\), then \(M=0\), which can eli... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,331 |
6. Given that the three sides of $\triangle A B C$ are $8, 12, 18$, and that $\triangle A_{1} B_{1} C_{1}$ also has one side of length 12, and is similar but not congruent to $\triangle A B C$. Then the number of such $\triangle A_{1} B_{1} C_{1}$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 6.C.
Let $A B=8, B C=12, A C=18$. By the similarity of two triangles that are not congruent, we can set
$$
\frac{A B}{A_{1} B_{1}}=\frac{B C}{B_{1} C_{1}}=\frac{A C}{A_{1} C_{1}}=q \neq 1,
$$
i.e., $\frac{8}{A_{1} B_{1}}=\frac{12}{B_{1} C_{1}}=\frac{18}{A_{1} C_{1}}=q \neq 1$.
Since $B_{1} C_{1} \neq 12$, it can only... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,333 |
1. A bicycle tire, if installed on the front wheel, will wear out after traveling $5000 \mathrm{~km}$; if installed on the back wheel, it will wear out after traveling $3000 \mathrm{~km}$. If the front and back tires are swapped after traveling a certain distance, so that a pair of new tires wear out simultaneously, th... | II. 1.3750.
Assume the total wear of each new tire when it is scrapped is $k$, then the wear per kilometer for a tire installed on the front wheel is $\frac{k}{5000}$, and the wear per kilometer for a tire installed on the rear wheel is $\frac{k}{3000}$. Let a pair of new tires travel $x \mathrm{~km}$ before swapping p... | 3750 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,334 |
2. Connect the diagonals $AC, BD$ of rectangle $ABCD$ intersecting at point $E$, draw a perpendicular from $E$ to $AB$ intersecting $AB$ at $F$; connect $DF$ intersecting $AC$ at $E_{1}$, draw a perpendicular from $E_{1}$ to $AB$ intersecting $AB$ at $F_{1}$; connect $DF_{1}$ intersecting $AC$ at $E_{2}$, draw a perpen... | 2. $\frac{1}{2008}$.
As shown in Figure 6. By the Midline Theorem, we have
$$
\begin{array}{l}
A F=\frac{1}{2} A B, \\
E F=\frac{1}{2} A D .
\end{array}
$$
$$
\text { Also, } \triangle A E_{1} F_{1} \backsim
$$
$\triangle A E F, \triangle A E_{1} D \backsim \triangle E E_{1} F$, thus
$$
\frac{A F_{1}}{F_{1} F}=\frac{A... | \frac{1}{2008} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,335 |
$\begin{array}{l}\text { 3. } \frac{1}{3}+\frac{2}{2^{4}+2^{2}+1}+\frac{3}{3^{4}+3^{2}+1}+\cdots+ \\ \frac{100}{100^{4}+100^{2}+1}=\end{array}$ | 3. $\frac{5050}{10101}$.
Notice that
$$
\frac{k}{k^{4}+k^{2}+1}=\frac{1}{2}\left(\frac{1}{k^{2}-k+1}-\frac{1}{k^{2}+k+1}\right) \text {. }
$$
Thus, the original expression is
$$
\begin{aligned}
= & \frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\cdots+\right. \\
& \left.\left(\frac{1... | \frac{5050}{10101} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,336 |
4. As shown in Figure 1, given that the two medians $A D$ and $B E$ of $\triangle A B C$ intersect at point $G$, we can obtain 8 figures:
$\triangle A B D$, $\triangle A C D$,
$\triangle B A E$, $\triangle B C E$,
$\triangle G A B$, $\triangle G A E$,
$\triangle G B D$, quadrilateral $C E G D$.
If two figures are rando... | 4. $\frac{2}{7}$.
From 8 figures, choosing any 2 figures has $\frac{8 \times 7}{2}=28$ ways, among which there are three cases where the areas are equal:
(1) Triangles with an area of $\frac{1}{2} S_{\triangle A B C}$ are 4 $(\triangle A B D, \triangle A C D, \triangle B A E, \triangle B C E)$, resulting in $\frac{4 \... | \frac{2}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,337 |
One, (20 points) Given the parabola $y=x^{2}$ and the moving line $y=(2 t-1) x-c$ have common points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, and $x_{1}^{2}+x_{2}^{2}=t^{2}+2 t-3$.
(1) Find the range of values for $t$;
(2) Find the minimum value of $c$, and determine the value of $t$ when $c$ is at ... | I. By combining $y=x^{2}$ and $y=(2 t-1) x-c$, eliminating $y$ yields the quadratic equation
$$
x^{2}-(2 t-1) x+c=0
$$
with real roots $x_{1}$ and $x_{2}$. By Vieta's formulas, we have
$$
x_{1}+x_{2}=2 t-1, \quad x_{1} x_{2}=c.
$$
Thus, $c=x_{1} x_{2}=\frac{1}{2}\left[\left(x_{1}+x_{2}\right)^{2}-\left(x_{1}^{2}+x_{2... | \frac{11-6 \sqrt{2}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,338 |
6. If positive integers $a, b, c, x, y, z$ satisfy
$$
a x=b+c, b y=a+c, c z=a+b,
$$
then the number of possible values for the product $x y z$ is ( ).
(A) 2
(B) 3
(C) 4
(D) infinitely many | 6. B.
If the three numbers $a, b, c$ are all equal, then $x=y=z=2$, at this time, $xyz=8$.
If the three numbers $a, b, c$ are not all equal, let's assume $c$ is the largest number, then $a+b<2c$. According to $z=\frac{a+b}{c}$ being a positive integer, we get $z=1, c=a+b$.
$$
\begin{array}{l}
\text { Hence } x=\frac{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,339 |
II. (25 points) In an equilateral $\triangle ABC$, take any point $P$ inside it, and connect $PA$, $PB$, and $PC$. Prove that there must exist two segments among $PA$, $PB$, and $PC$ with lengths $m$ and $n$ such that
$$
\frac{\sqrt{5}-1}{2}<\frac{m}{n} \leqslant 1 \text {. }
$$ | As shown in Figure 7, $\triangle P A B$ is rotated $60^{\circ}$ clockwise around point $B$, resulting in $\triangle P_{1} C B \cong \triangle P A B$. Connect $P P_{1}$. Then $P_{1} C = P A$, and the isosceles $\triangle B P_{1} P$ is an equilateral triangle, with $P_{1} P = B P_{1} = P B$. Therefore, a triangle $\trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,340 |
Three. (25 points) Assign 18 table tennis players numbered $1, 2, \cdots, 18$ to 9 tables for singles matches, with the rule that the sum of the numbers of the two players on each table must be a perfect square greater than 4. Can this rule be implemented? If the rule cannot be implemented, provide a proof; if the rule... | Three, Solution 1: Since the sum of the two largest numbers is $18+17=35<36$, the sum of the numbers of two players on the same table can only be one of the three square numbers: $25, 16, 9$. Let the number of pairs with sums of $25, 16, 9$ be $x$, $y$, and $z$ respectively $(x, y, z$ are non-negative integers). Accord... | (18,7),(17,8),(16,9),(15,1),(14,2),(13,3),(12,4),(11,5),(10,6) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,341 |
1. Among the positive integers $1,2, \cdots, 2008$, the number of integers that can be expressed in the form $\frac{m n-1}{m-n}\left(m, n \in \mathbf{N}_{+}\right)$ is ( ).
(A) 2008
(B) 2006
(C) 2007
(D) 2004 | - 1.C.
If $\frac{m n-1}{m-n}=1$, rearranging gives $(m+1)(n-1)=0$.
Solving, we get $m=-1$ or $n=1$.
Therefore, when $n=1, m>1$, $\frac{m n-1}{m-n}=\frac{m-1}{m-1}=1$.
If $\frac{m n-1}{m-n}=2$, rearranging gives $(m+2)(n-2)=-3=3 \times(-1)$.
Solving, we get $m=1, n=1$.
But $m \neq n$, thus, $\frac{m n-1}{m-n} \neq 2$.
... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,342 |
2. Given that $a, b, c$ satisfy
$$
\begin{array}{l}
|2 a-4|+|b+2|+\sqrt{(a-3) b^{2}}+a^{2}+c^{2} \\
=2+2 a c .
\end{array}
$$
Then the value of $a-b+c$ is ( ).
(A) 4
(B) 6
(C) 8
(D) 4 or 8 | 2.D.
When $b=0$, the original equation becomes
$$
|2 a-4|+(a-c)^{2}=0 \text {. }
$$
Solving this, we get $a=c=2$.
Therefore, $a-b+c=4$.
When $b \neq 0$, $b^{2}>0, a \geqslant 3$.
Thus, $2 a-4 \geqslant 2$.
The original equation becomes
$$
2 a-4+|b+2|+\sqrt{(a-3) b^{2}}+(a-c)^{2}=2 \text {, }
$$
which simplifies to
$... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,343 |
3. As shown in Figure 1, in Rt $\qquad$ $A B C$, $\angle B A C=90^{\circ}, A B=4$, $B C=12, A D \perp B C$ at point $D$, the angle bisector of $\angle A B C$ intersects $A C$ and $A D$ at points $E$ and $F$ respectively. A line through $F$ parallel to $B C$ intersects $A C$ at point $G$. Then the length of $F G$ is ( )... | 3. C.
In the right triangle $\triangle ABC$, by the projection theorem, we have $AB^2 = BD \cdot BC$.
Therefore, $BD = \frac{AB^2}{BC} = \frac{4^2}{12} = \frac{4}{3}$. Hence, $DC = BC - BD = 12 - \frac{4}{3} = \frac{32}{3}$.
Since $BF$ is the angle bisector of $\angle ABC$, we have $\frac{AF}{FD} = \frac{AB}{BD} = 3$.... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,344 |
5. As shown in Figure 2, in pentagon $A B C D E$, $\angle A=$ $120^{\circ}, \angle B=\angle E=$ $90^{\circ}, A B=B C=1, A E$ $=D E=2$. Find points $M$ and $N$ on $B C$ and $D E$ respectively, such that the perimeter of $\triangle A M N$ is minimized. Then the minimum perimeter of $\triangle A M N$ is ( ).
(A) $2 \sqrt{... | 5.B.
As shown in Figure 5, extend $AB$ to $P$ such that $PB = AB$; extend $AE$ to $Q$ such that $EQ = AE$. Connect $PQ$ to intersect $BC$ and $DE$ at points $M$ and $N$, respectively. Then the perimeter of $\triangle AMN$ is minimized, and the minimum perimeter is the length of segment $PQ$. Draw $PF \perp AE$ at $F$.... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,346 |
6. Given that $2^{n}\left(n \in \mathbf{N}_{+}\right)$ can divide $2007^{2018}-1$. Then the maximum value of $n$ is ( ).
(A) 12
(B) 13
(C) 14
(D) 15 | 6. C.
Let $2007=a$. Then
$$
\begin{array}{l}
2007^{2048}-1=a^{2048}-1^{2} \\
=\left(a^{2^{10}}+1\right)\left(a^{2^{9}}+1\right) \cdots\left(a^{2^{1}}+1\right) . \\
(a+1)(a-1) .
\end{array}
$$
For any positive integer $k$, we have
$$
a^{2^{k}}+1 \equiv(-1)^{2^{k}}+1 \equiv 2(\bmod 4) \text {, }
$$
which means $a^{2^... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,347 |
1. Given $x$ is a real number,
$$
\sqrt{x^{3}+2020}-\sqrt{2030-x^{3}}=54 \text {. }
$$
Then $28 \sqrt{x^{3}+2020}+27 \sqrt{2030-x^{3}}=$ | II. 1.2007.
Let $x^{3}+2020=a, 2030-x^{3}=b$. Then $a+b=4050$.
From the problem, we have
$$
\begin{array}{l}
\sqrt{a}-\sqrt{b}=54 \\
\Rightarrow a+b-2 \sqrt{a b}=54^{2}=2916 \\
\Rightarrow 2 \sqrt{a b}=(a+b)-2916 \\
\quad=4050-2916=1134 \\
\Rightarrow(\sqrt{a}+\sqrt{b})^{2}=(a+b)+2 \sqrt{a b} \\
\quad=4050+1134=5184 \\... | 2007 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,348 |
2. As shown in Figure 3, the diagonals of quadrilateral $ABCD$ intersect at point $O, \angle BAD=\angle BCD=60^{\circ}$, $\angle CBD=55^{\circ}, \angle ADB=$ $50^{\circ}$. Then the degree measure of $\angle AOB$ is | 2. $80^{\circ}$.
As shown in Figure 6, draw perpendiculars from point $C$ to $A B$, $B D$, and $A D$, with the feet of the perpendiculars being $E$, $F$, and $G$ respectively.
It is easy to see that $\angle A B D=70^{\circ}$, $\angle C B D=\angle C B E=55^{\circ}$, $\angle B D C=\angle C D G=65^{\circ}$.
Therefore, $B... | 80^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,349 |
1. Let $|a|>1$, simplify
$$
\left(a+\sqrt{a^{2}-1}\right)^{4}+2\left(1-2 a^{2}\right)\left(a+\sqrt{a^{2}-1}\right)^{2}+3
$$
the result is $\qquad$ . | Let $x_{0}=a+\sqrt{a^{2}-1}$. Clearly, $x_{0}$ is a root of the equation $x^{2}-2 a x+1=0$, and thus it is also a root of the equation
$$
\left(x^{2}-2 a x+1\right)\left(x^{2}+2 a x+1\right)=0
$$
Then $\left(x_{0}^{2}+1\right)^{2}-4 a^{2} x_{0}^{2}=0$, which means
$$
x_{0}^{4}+2\left(1-2 a^{2}\right) x_{0}^{2}+1=0 \te... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,350 |
3. Given that $n$ is a natural number, $9 n^{2}-10 n+2009$ can be expressed as the product of two consecutive natural numbers. Then the maximum value of $n$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.2007 .
Let $9 n^{2}-10 n+2009=m(m+1)$, where $m$ is a natural number. Then
$$
9 n^{2}-10 n+\left(2009-m^{2}-m\right)=0 \text {. }
$$
Considering equation (1) as a quadratic equation in the natural number $n$, its discriminant should be the square of a natural number. Let's assume $\Delta=t^{2}(t \in \mathbf{N})$, t... | 2007 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,351 |
4. Figure 4 is a schematic diagram of a clock hanging on a wall. $O$ is the center of rotation of its second hand, $M$ is the other end of the second hand, $O M$ $=10 \mathrm{~cm}, l$ is a vertical line passing through point $O$. There is an ant $P$ crawling on the second hand $O M$, the distance from ant $P$ to point ... | $4.20 \pi$.
As shown in Figure 7, with point $O$ as the center and $10 \, \text{cm}$ as the radius, draw $\odot O$. Draw $MN \perp l$ at point $N$, and draw the perpendicular from $O$ to $l$ intersecting $\odot O$ at points $Q_{1}$ and $Q_{2}$. Connect $PQ_{1}$. Then,
$$
\begin{array}{l}
MN \parallel OQ_{1}, \\
\angle ... | 20 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,352 |
One. (20 points) Let $k$ be a constant. The equation with respect to $x$
$$
x^{2}-2 x+\frac{3 k^{2}-9 k}{x^{2}-2 x-2 k}=3-2 k
$$
has four distinct real roots. Find the range of values for $k$. | Let $y=x^{2}-2 x-2 k$. Then the original equation can be transformed into
$$
\begin{array}{l}
y+2 k+\frac{3 k^{2}-9 k}{y}=3-2 k \\
\Rightarrow y^{2}+(4 k-3) y+3 k^{2}-9 k=0 \\
\Rightarrow(y+3 k)(y+k-3)=0 \\
\Rightarrow y=-3 k \text { or } y=-k+3 \\
\Rightarrow x^{2}-2 x+k=0
\end{array}
$$
$$
\text { or } x^{2}-2 x-k-3=... | -4 < k < 1 \text{ and } k \neq 0 \text{ and } k \neq -\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,353 |
II. (25 points) Given that $O$ is the circumcenter of $\triangle A B C$, $\angle B A C=45^{\circ}$, extend $B C$ to $D$ such that $C D=\frac{1}{2} B C$, and $A D / / O C$. Find the measure of $\angle A B C$. | $$
\begin{array}{l}
\angle B A C=45^{\circ} \\
\Rightarrow \angle B O C=90^{\circ} \\
\Rightarrow \angle O B C=\angle O C B=45^{\circ} .
\end{array}
$$
Given $A D / / O C$
$$
\Rightarrow O E \perp A D, \frac{B O}{O E}=\frac{B C}{C D}=2
$$
$$
\begin{array}{l}
\Rightarrow \angle A B E=\frac{1}{2} \angle A O E=30^{\circ}... | 75^{\circ} \text{ or } 15^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,354 |
Three, (25 points) During the New Year, the grandmother gave three grandsons a total of 400 yuan as New Year's money. There are 50 yuan, 20 yuan, and 10 yuan banknotes available in various quantities for the three grandsons to choose from, but each can only take banknotes of the same denomination. One of them takes a n... | Three people take money in the amounts of $x y$, $x$, and $y$, with denominations of $a$ yuan, $b$ yuan, and $c$ yuan respectively, where $x \geqslant 1, y \geqslant 1$. Then $a x y + b x + c y = 400$.
(1) When $a = b = c$, i.e., the denominations chosen by the three people are the same, then
$$
x y + x + y = \frac{400... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,355 |
1. The sum of all natural numbers $x$ that make the algebraic expression $y=\frac{x^{2}+11}{x+1}$ an integer is ( ).
(A) 5
(B) 6
(C) 12
(D) 22 | - 1.D.
Since $y=\frac{x^{2}+11}{x+1}=x-1+\frac{12}{x+1}$, to make $\frac{12}{x+1}$ an integer, the natural number $x$ can take $0,1,2,3,5$, 11, the sum of which is 22. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,356 |
2. The number of real roots of the equation $|x|-\frac{4}{x}=\frac{3|x|}{x}$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 2.A.
When $x>0$, $x-\frac{4}{x}=3, x^{2}-3 x-4=0$, $x_{1}=4, x_{2}=-1$ (discard);
When $x<0$, $-x-\frac{4}{x}=-3, x^{2}-3 x+4=0, \Delta<0$, no real roots.
In summary, the original equation has only one real root $x=4$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,357 |
$$
\begin{array}{l}
\text { 3. As shown in figure } 1, \angle X O Y \\
=90^{\circ}, \quad O W \text { bisects } \\
\angle X O Y, P A \perp O X, P B \\
\perp O Y, P C \perp O W \text {. If } O A \\
+O B+O C=1 \text {, then } O C \\
=(\quad) .
\end{array}
$$
3. As shown in figure $1, \angle X O Y$
(A) $2-\sqrt{2}$
(B) $\... | 3. B.
Extend $C P$ to intersect $O Y$ at point $D$, it is easy to know that $B D=P B=O A$.
Then $O A+O B=O B+B D=O D=\sqrt{2} O C$. Therefore, $1=O A+O B+O C=(\sqrt{2}+1) O C$, which means $O C=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,358 |
4. For each $x$, the function $y$ is
$$
y_{1}=2 x, y_{2}=x+2, y_{3}=-\frac{3}{2} x+12
$$
the minimum of these three functions. Then the maximum value of the function $y$ is ( ).
(A) 4
(B) 6
(C) 8
(D) $\frac{48}{7}$ | 4.B.
Solving the systems of equations $y_{1} 、 y_{2}, y_{1} 、 y_{3}, y_{2} 、 y_{3}$ yields the intersection points $A(2,4), B\left(\frac{24}{7}, \frac{48}{7}\right), C(4,6)$.
When $x \leqslant 2$, $\min \left\{y_{1}, y_{2}, y_{3}\right\}=y_{1}=2 x \leqslant$ 4, the maximum value is 4;
When $24$ (it seems there is a m... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,359 |
5. Within the range of the independent variable $x$, $59 \leqslant x \leqslant 60$, the number of integer values of the quadratic function $y=x^{2}+x+\frac{1}{2}$ is ( ).
(A) 59
(B) 120
(C) 118
(D) 60 | 5.B.
Notice that $y=x^{2}+x+\frac{1}{2}=\left(x+\frac{1}{2}\right)^{2}+\frac{1}{4}$. When $x>-\frac{1}{2}$, $y$ increases as $x$ increases.
Since $59 \leqslant x \leqslant 60$, then
$$
59^{2}+59+\frac{1}{2} \leqslant y \leqslant 60^{2}+60+\frac{1}{2},
$$
which means $3540 \frac{1}{2} \leqslant y \leqslant 3660 \frac{... | 120 | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,360 |
2. $a_{1}, a_{2}, \cdots, a_{10}$ represent the ten digits $1,2,3,4,5,6$, $7,8,9,0$, respectively, to form two five-digit numbers
$$
m=\overline{a_{1} a_{2} a_{3} a_{4} a_{5}}, n=\overline{a_{6} a_{7} a_{8} a_{9} a_{10}}(m>n) .
$$
Then the minimum value of $m-n$ is | 2.247.
Since $m>n$, we have $a_{1}>a_{6}$.
To make $m-n$ as small as possible, we should take $a_{1}-a_{6}=1$, the last four digits of $m$ should be the smallest, and the last four digits of $n$ should be the largest. The largest four-digit number that can be formed from the different digits 1,2,3,4,5,6,7,8,9,0 is 987... | 247 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,361 |
6. As shown in Figure $2, \odot O_{1}$ and $\odot O_{2}$ intersect at points $E$ and $F$, $AB$ and $CD$ are two common tangents, line $EF$ intersects $AB$ and $CD$ at points $P$ and $Q$ respectively. The relationship between $AB$, $PQ$, and $EF$ is ( ).
(A) $2 AB=PQ+EF$
(B) $AB^2=PQ \cdot EF$
(C) $AB^2+EF^2=PQ^2$
(D) $... | 6.C.
Let $P E=Q F=a, E F=b$. Then $P Q^{2}=(2 a+b)^{2}=4 a^{2}+4 a b+b^{2}$.
And $A B^{2}=4 P A^{2}=4 P E \cdot P F$
$$
=4 a(a+b)=4 a^{2}+4 a b \text {, }
$$
Thus $A B^{2}+E F^{2}=4 a^{2}+4 a b+b^{2}=P Q^{2}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,362 |
1. Given
$$
\frac{y+z-x}{x+y+z}=\frac{z+x-y}{y+z-x}=\frac{x+y-z}{z+x-y}=p \text {. }
$$
Then $p^{3}+p^{2}+p=$ $\qquad$ . | Notice
$$
\begin{array}{l}
p^{2}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x}=\frac{z+x-y}{x+y+z}, \\
p^{3}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x} \cdot \frac{x+y-z}{z+x-y} \\
=\frac{x+y-z}{x+y+z}, \\
p^{3}+p^{2}+p \\
=\frac{x+y-z}{x+y+z}+\frac{z+x-y}{x+y+z}+\frac{y+z-x}{x+y+z} \\
=\frac{x+y+z}{x+y+z}=1 .
\end{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,363 |
2. Let the equation concerning $x$ be
$$
a x^{2}+(a+2) x+9 a=0
$$
has two distinct real roots $x_{1}, x_{2}$, and $x_{1}<1<x_{2}$. Then, the range of values for $a$ is $\qquad$. | 2. $-\frac{2}{11}<a<0, \\ \Delta=(a+2)^{2}-4 \times a \times 9 a>0, \\ f(1)=a+a+2+9 a0, \\ f(1)=a+a+2+9 a>0 .\end{array}\right.$
Solving the inequality set (1), we know that the range of $a$ in (1) and (2) has no common part, therefore, (1) has no solution;
Solving the inequality set (2) yields $\left\{\begin{array}{l}... | -\frac{2}{11}<a<0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,364 |
One, (20 points) A company spent 4.8 million yuan to purchase the production technology of a certain product, and then invested another 15.2 million yuan to buy production equipment to process and produce the product. It is known that the cost of producing each piece of this product is 40 yuan. Market research has foun... | $-、(1) y=\left\{\begin{array}{ll}-\frac{2}{25} x+28, & 100 \leqslant x \leqslant 200 \\ -\frac{1}{10} x+32, & 200<x \leqslant 300\end{array}\right.$
(2) When $100 \leqslant x \leqslant 200$, $w=x y-40 y-(1520+480)$.
Substituting $y=-\frac{2}{25} x+28$ into equation (1) gives $w=x\left(-\frac{2}{25} x+28\right)-40\left(... | 190 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,367 |
II. (25 points) As shown in Figure 4, $\triangle A^{\prime} B C \backsim \triangle A B^{\prime} C \backsim \triangle A B C^{\prime}$. Prove: $A A^{\prime} 、 B B^{\prime} 、 C C^{\prime}$ are concurrent. | As shown in Figure 7, let $A A^{\prime}$ and $B B^{\prime}$ intersect at point $O$, and connect $O C$ and $O C^{\prime}$.
From $\triangle A^{\prime} B C \sim \triangle A B^{\prime} C$, it is easy to deduce that
$\triangle A A^{\prime} C \sim \triangle B^{\prime} B C$.
Thus, $\angle A A^{\prime} C = \angle B^{\prime} B ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,368 |
Three. (25 points) Semicircle
The diameter $AB$ of semicircle $\odot O$ is the hypotenuse of isosceles right $\triangle ABC$, point $P$ is on ray $BA$, $PQ$ is tangent to semicircle $\odot O$ at point $Q$, the angle bisector of $\angle BPQ$ intersects $AC$ and $BC$ at points $E$ and $F$. Prove: $AE^2 + BF^2 = EF^2$.
F... | Three, as shown in Figure 8, connect $Q A$ and $Q B$ intersecting $P F$ at points $X$ and $Y$, and connect $Q E$ and $Q F$. It is easy to see that
$$
\begin{array}{l}
\angle P Q A=\angle P B Q, \\
\angle X Q Y=\angle A Q B=90^{\circ} .
\end{array}
$$
$$
\text { Figure } 8
$$
Since $\angle Q X Y=\frac{1}{2} \angle B P ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,369 |
1. Given that $m$ and $n$ are two consecutive positive integers, $m<n$, and $a=m n$, let $x=\sqrt{a+n}+\sqrt{a-m}$, $y=\sqrt{a+n}-\sqrt{a-m}$. Which of the following statements is correct? ( ).
(A) $x$ is odd, $y$ is even
(B) $x$ is even, $y$ is odd
(C) Both $x$ and $y$ are odd
(D) Both $x$ and $y$ are even | -、1.G.
From the problem, we know that $a=m(m+1)=n(n-1)$. Then,
$$
\begin{array}{l}
\sqrt{a+n}=\sqrt{n(n-1)+n}=n, \\
\sqrt{a-m}=\sqrt{m(m+1)-m}=m .
\end{array}
$$
Therefore, $x=n+m=m+m+1=2 m+1$, $y=n-m=1$.
Thus, both $x$ and $y$ are odd numbers. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,370 |
2. Let $a$, $b$, $c$, and $S$ be the lengths of the three sides and the area of a triangle, respectively. For the equation in $x$
$$
b^{2} x^{2}+\left(b^{2}+c^{2}-a^{2}\right) x+c^{2}=0
$$
the discriminant is $\Delta$. Then the relationship between $\Delta$ and $S$ is ( ).
(A) $\Delta=16 S^{2}$
(B) $\Delta=-16 S^{2}$
... | 2.B.
Since $\Delta=\left(b^{2}+c^{2}-a^{2}\right)^{2}-4 b^{2} c^{2}$
$$
\begin{array}{l}
=\left(b^{2}+c^{2}-a^{2}+2 b c\right)\left(b^{2}+c^{2}-a^{2}-2 b c\right) \\
=\left[(b+c)^{2}-a^{2}\right]\left[(b-c)^{2}-a^{2}\right] \\
=(b+c+a)(b+c-a)(b-c+a)(b-c-a) .
\end{array}
$$
Let $p=\frac{1}{2}(a+b+c)$, so,
$$
\begin{ar... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,371 |
3. As shown in Figure 3, in the right triangle $\triangle ABC$, $BC=3, AC=$ $4, AB=5$, and its incircle is $\odot O$. Through $OA, OB, OC$ and the intersection points $M, N, K$ with $\odot O$, draw the tangents to $\odot O$, which intersect the three sides of $\triangle ABC$ at $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}... | $3.2 \sqrt{2}+\sqrt{5}+\frac{2 \sqrt{10}}{3}-\frac{11}{3}$.
As shown in Figure 7, let the radius of $\odot O$ be $r$, and $\odot O$ is tangent to $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Connect $OD$, $OE$, and $OF$. Then
$$
\begin{array}{c}
S_{\triangle ABC}=6 . \\
\text { By } \frac{1}{2} r(AB+... | 2 \sqrt{2}+\sqrt{5}+\frac{2 \sqrt{10}}{3}-\frac{11}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,372 |
3. Let $a$ be the fractional part of $\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}$, and $b$ be the fractional part of $\sqrt{6+3 \sqrt{3}}-\sqrt{6-3 \sqrt{3}}$. Then the value of $\frac{2}{b}-\frac{1}{a}$ is ( ).
(A) $\sqrt{6}+\sqrt{2}-1$
(B) $\sqrt{6}-\sqrt{2}+1$
(C) $\sqrt{6}-\sqrt{2}-1$
(D) $\sqrt{6}+\sqrt{2}+1$ | 3. B.
Since $\sqrt{3 \pm \sqrt{5}}=\sqrt{\frac{6 \pm 2 \sqrt{5}}{2}}=\frac{\sqrt{5} \pm 1}{\sqrt{2}}$, we have
$$
\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\frac{(\sqrt{5}+1)-(\sqrt{5}-1)}{\sqrt{2}}=\sqrt{2} \text {. }
$$
Therefore, $a=\sqrt{2}-1$.
Also, $\sqrt{6 \pm 3 \sqrt{3}}=\sqrt{\frac{3}{2}}(\sqrt{3} \pm 1)$, so
$$
\... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,373 |
4. As shown in Figure $1, D$ and $E$ are points on the sides $AB$ and $AC$ of $\triangle ABC$, respectively. The perimeters of $\triangle ACD$ and $\triangle BCD$ are equal, and the perimeters of $\triangle ABE$ and $\triangle CBE$ are equal. Let the area of $\triangle ABC$ be $S$. If $\angle ACB=90^{\circ}$, then the ... | 4. A.
Let $B C=a, C A=b, A B=c$. From the problem, we know $A D+A C=B C+C E=\frac{1}{2}(a+b+c)$. Therefore, $A D=\frac{1}{2}(a+c-b), C E=\frac{1}{2}(b+c-a)$. Thus, $A D \cdot C E=\frac{1}{4}(a+c-b)(b+c-a)$ $=\frac{1}{4}\left[c^{2}-(a-b)^{2}\right]$ $=\frac{1}{4}\left(c^{2}-a^{2}-b^{2}\right)+\frac{1}{2} a b$.
Given $... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,374 |
5. As shown in Figure 2, in $\triangle A B C$, $A B=8, B C=7$, $A C=6$, extend side $B C$ to point $P$, such that $\triangle P A B$ is similar to $\triangle P C A$. Then the length of $P C$ is ( ).
(A) 7
(B) 8
(C) 9
(D) 10 | 5. C.
From the problem, we know it can only be $\triangle P A B \backsim \triangle P C A$. Therefore, we have $\frac{P A}{P C}=\frac{P B}{P A}=\frac{A B}{A C}=\frac{8}{6}=\frac{4}{3}$.
Thus, $P B=\frac{4}{3} P A, P B=P C+B C=P C+7$, $P A=\frac{4}{3} P C$.
Also, $P A^{2}=P B \cdot P C \Rightarrow\left(\frac{4}{3} P C\r... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,375 |
6. As shown in Figure 3, a circle with diameter $P Q=2 r(r \in \mathbf{Q})$ is tangent to a circle with radius $R(R \in \mathbf{Q})$ at point $P$. The vertices $A$ and $B$ of square $A B C D$ lie on the larger circle, and the smaller circle is outside the square and tangent to side $C D$ at point $Q$. If the side lengt... | 6.D.
Auxiliary lines as shown in Figure 4.
From the problem, we know
$$
O A^{2}=O E^{2}+A E^{2} \text {. }
$$
Let $A B=2 x$, then
$$
A E=x \text {. }
$$
Thus, $R^{2}=[2 x-(R-2 r)]^{2}+x^{2}$.
Simplifying, we get
$$
5 x^{2}-4(R-2 r) x+4\left(r^{2}-R r\right)=0 \text {. }
$$
For $A B$ to be a rational number, it is s... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,376 |
1. Given that the two roots of the equation $x^{2}+x-1=0$ are $\alpha, \beta$. Then the value of $\frac{\alpha^{3}}{\beta}+\frac{\beta^{3}}{\alpha}$ is $\qquad$ | Let $A=\frac{\alpha^{3}}{\beta}+\frac{\beta^{3}}{\alpha}, B=\frac{\alpha^{3}}{\alpha}+\frac{\beta^{3}}{\beta}=\alpha^{2}+\beta^{2}$.
From the given information,
$$
\begin{array}{l}
\alpha+\beta=-1, \alpha \beta=-1 . \\
\text { Hence } B=(\alpha+\beta)^{2}-2 \alpha \beta=1+2=3 . \\
\text { Also, } \alpha^{3}+\beta^{3}=(... | -7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,377 |
2. Divide the positive integers $1,2, \cdots, 2008$ into 1004 groups: $a_{1}, b_{1} ; a_{2}, b_{2} ; \cdots ; a_{1004}, b_{1004}$, and satisfy
$$
a_{1}+b_{1}=a_{2}+b_{2}=\cdots=a_{1004}+b_{1004} \text {. }
$$
For all $i(i=1,2, \cdots, 1004), a_{i} b_{i}$ the maximum value is | 2.1009020 .
Notice that
$$
\begin{array}{l}
a_{i} b_{i}=\frac{1}{4}\left[\left(a_{i}+b_{i}\right)^{2}-\left(a_{i}-b_{i}\right)^{2}\right], \\
a_{i}+b_{i}=\frac{(1+2008) \times 1004}{1004}=2009 .
\end{array}
$$
To maximize the value of $a_{i} b_{i}$, the value of $a_{i}-b_{i}$ must be minimized, and the minimum value ... | 1009020 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,378 |
3. $A D 、 B E 、 C F$ are the angle bisectors of $\triangle A B C$. If $B D+B F=C D+C E=A E+A F$, then the degree measure of $\angle B A C$ is $\qquad$ . | 3. $60^{\circ}$.
Let $B C=a, C A=b, A B=c$.
By the Angle Bisector Theorem, we have
$$
\begin{aligned}
B D & =\frac{a c}{b+c}, C D=\frac{a b}{b+c}, \\
B F & =\frac{a c}{a+b}, C E=\frac{a b}{a+c} .
\end{aligned}
$$
From $B D+B F=C D+C E$, we know
$$
\frac{c}{b+c}+\frac{c}{a+b}=\frac{b}{b+c}+\frac{b}{a+c} \text {. }
$$
... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,379 |
4. Among the following four propositions:
(1) A quadrilateral with one pair of opposite sides equal and one pair of opposite angles equal is a parallelogram;
(2) A quadrilateral with one pair of opposite sides equal and one diagonal bisecting the other diagonal is a parallelogram;
(3) A quadrilateral with one pair of o... | 4. (4).
Propositions (1), (2), and (3) can be refuted with the following counterexamples:
Proposition (1): The quadrilateral $ABCD$ in Figure 5(a), where $\triangle ABD \cong \triangle CDE$.
Proposition (2): As shown in Figure 5(b), construct an isosceles $\triangle ADE$, extend the base $ED$ to any point $O$, and us... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,380 |
One, (20 points) In $\triangle A B C$, it is known that $\angle A>\angle B>$ $\angle C$, and $\angle A=2 \angle B$. If the three sides of the triangle are integers, and the area is also an integer, find the minimum value of the area of $\triangle A B C$. | Let $B C=a, C A=b, A B=c$.
As shown in Figure 6, construct the angle bisector $A D$ of $\angle B A C$, then
$$
\begin{array}{l}
\angle B A D=\angle D A C \\
= \angle B, \\
\angle A D C=\angle B+\angle B A D \\
= 2 \angle B .
\end{array}
$$
Figure 6
Therefore, $\triangle A C D \backsim \triangle B C A$. Thus,
$$
\frac... | 132 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,381 |
II. (25 points) Given that $G$ is any point inside $\triangle ABC$, and $BG$, $CG$ intersect $AC$, $AB$ at points $E$, $F$ respectively. Find the minimum value of $k$ such that the inequality $S_{\triangle BGF} \cdot S_{\triangle OCE} \leqslant k S_{\triangle ABC}^{2}$ always holds. | As shown in Figure 7, let
$$
\frac{A F}{A B}=x, \frac{A E}{A C}=y \text {. }
$$
If $0$ has real roots, then
$$
\Delta=(t-2)^{2}-8 t \geqslant 0
$$
$\Rightarrow t \geqslant 6+4 \sqrt{2}$ or $t \leqslant 6-4 \sqrt{2}$.
Thus, $t \leqslant 6-4 \sqrt{2}$.
Therefore, $t_{\max }=6-4 \sqrt{2}$, at this time $u=2 \sqrt{2}-2$.
... | 17-12 \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,382 |
4. If 6 pieces of $1 \times 2$ paper are used to cover a $3 \times 4$ grid, the number of different ways to cover it is. | 4.11.
As shown in Figure 8, the cells of the grid are numbered. Let $M(a, b)$ denote the number of ways to cover the entire grid when the cells numbered $a$ and $b$ (which are adjacent) are covered by the same piece of paper. We focus on the covering of cell 8. It is given that $M(8,5) = 2, M(8,11) = 3, M(8,7) = M(8,9... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,383 |
$$
\begin{array}{l}
\text { Three. (25 points) Given } \\
\left(x+\sqrt{y^{2}+1}\right)\left(y+\sqrt{x^{2}+1}\right)=1 .
\end{array}
$$
Prove: $x+y=0$. | (1) First, prove that when $x=0$, $y=0$, or when $y=0$, $x=1$. If not, assume that when $x=0$, $y>0$. Then
$$
\begin{array}{l}
\left(x+\sqrt{y^{2}+1}\right)\left(y+\sqrt{x^{2}+1}\right) \\
=\sqrt{y^{2}+1}(y+1)>1,
\end{array}
$$
which contradicts the given condition.
When $x=0, y>0$, or $x>0, y>0$,
$$
\left(x+\sqrt{y^{... | x+y=0 | Algebra | proof | Yes | Yes | cn_contest | false | 719,384 |
Example 1 Convert the rational number $\frac{67}{29}$ to a continued fraction. | Explanation: Since $\frac{67}{29}=2+\frac{9}{29}, \frac{29}{9}=3+\frac{2}{9}$, $\frac{9}{2}=4+\frac{1}{2}$, therefore,
$$
\begin{array}{l}
\frac{67}{29}=2+\frac{1}{3+\frac{1}{4+\frac{1}{2}}}=2+\frac{1}{3}+\frac{1}{4}+\frac{1}{2} \\
=[2,3,4,2] .
\end{array}
$$ | [2,3,4,2] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,394 |
Example 2 Find the convergent fraction of $3+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=[3,2,3, 4,5]$. | Explanation: First, list $p_{-1}=0, p_{0}=1, q_{-1}=1, q_{0}=$ $0, a_{1}=3, a_{2}=2, a_{3}=3, a_{4}=4, a_{5}=5$, in Table 1:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline$n$ & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline$a_{n}$ & & & 3 & 2 & 3 & 4 & 5 \\
\hline$p_{n}$ & 0 & 1 & & & & & \\
\hline$q_{n}$ & 1 & 0 & & & & & \\... | \frac{539}{157} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,395 |
Three. (50 points) There are 2008 students participating in a large public welfare activity. If two students know each other, then these two students are considered as a cooperative group.
(1) Find the minimum number of cooperative groups $m$, such that no matter how the students know each other, there exist three stud... | (1) Let $n=1004$.
The following proof: $m=n^{2}+1$.
Divide the students into two large groups, each with $n$ students, and no students in the same large group know each other, while each student knows every student in the other large group, thus forming $n^{2}$ cooperative groups, but there do not exist three students ... | 1008017 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,397 |
Initially 217 As shown in Figure $3, \odot O$ has a radius equal to $R$, the vertices $B$ and $C$ of $\triangle A B C$ are both on $\odot O$, point $A$ is outside $\odot O$, $A C$ and $A B$ intersect $\odot O$ at points $E$ and $F$ respectively, $B E$ intersects $C F$ at point $P$, and a point $Q$ is taken on ray $O P$... | Proof: As shown in Figure 3, connect $O B, O C, O E, O F, Q B, Q C, Q E, Q F$.
Since $O B^{2}=R^{2}=O P \cdot O Q$, that is,
$$
\frac{O Q}{O B}=\frac{O B}{O P}, \angle Q O B=\angle B O P,
$$
Therefore, $\triangle Q O B \backsim \triangle B O P$.
Thus, $\angle O Q B=\angle O B P=\angle O B E=\angle O E B$.
Hence, $B, O... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,398 |
Initial 218 In a rhombus flower bed with a perimeter of 16 and an included angle of $60^{\circ}$, 10 flowers are planted. Prove: the distance between at least two of them is no more than $\sqrt{3}$.
Plant 10 flowers in a rhombus flower bed with a perimeter of 16 and an included angle of $60^{\circ}$. Prove: the distan... | Proof 1: As shown in Figure 4, construct the drawers. Quadrilateral $C E D F$ is a rhombus, $\angle E=60^{\circ}, C E=4$. With center $I$ and diameter $\sqrt{3}$, draw $\odot I$. Draw $I L \perp C E$ at point $L$, intersecting $\odot I$ at points $W$ and $T$, and intersecting $D F$ at point $S$.
Draw $I R \perp C F$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,399 |
Let $x_{i}>1, i=1,2, \cdots, n, n \in \mathbf{N}$, $n \geqslant 3$. Prove:
$$
\sum_{i=1}^{n} \frac{x_{i}^{n}}{\left(x_{i+1}-1\right)^{2}} \geqslant \frac{n^{n+1}}{4(n-2)^{n-2}},
$$
where $x_{n+1}=x_{1}$. | $$
\begin{array}{l}
\frac{x_{i}^{n}}{\left(x_{i+1}-1\right)^{2}}+\frac{n^{n}}{8(n-2)^{n-3}}\left(x_{i+1}-1\right)+ \\
\frac{n^{n}}{8(n-2)^{n-3}}\left(x_{i+1}-1\right)+ \\
{[\underbrace{\frac{n^{n}}{4(n-2)^{n-2}}+\cdots+\frac{n^{n}}{4(n-2)^{n-2}}}_{n-3 \uparrow}]} \\
\geqslant n \sqrt[n]{x_{i}^{n} \cdot \frac{n^{n(n-1)}... | \sum_{i=1}^{n} \frac{x_{i}^{n}}{\left(x_{i+1}-1\right)^{2}} \geqslant \frac{n^{n+1}}{4(n-2)^{n-2}} | Inequalities | proof | Yes | Yes | cn_contest | false | 719,400 |
As high as 218 As shown in Figure 6, three circular arcs are externally tangent to each other at points $A$, $B$, and $C$. Points $D$, $E$, and $F$ are the midpoints of $\overparen{B C}$, $\overparen{C A}$, and $\overparen{A B}$, respectively. Prove that the lines $A D$, $B E$, and $C F$ are concurrent if and only if a... | Proof: First, we prove a lemma.
Lemma: Let $\alpha, \beta, \gamma$ be the three interior angles of $\triangle ABC$. Then, $\tan \frac{2 \alpha+\beta}{4} \cdot \tan \frac{2 \beta+\gamma}{4} \cdot \tan \frac{2 \gamma+\alpha}{4}=1$ holds if and only if at least two of $\alpha, \beta, \gamma$ are equal.
Proof of the lemma:... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,401 |
Example 5 Prove: A triangle with an area of 1 cannot be covered by a parallelogram with an area less than 2.
| Proof: (1) Assuming that $\triangle ABC$ with an area of 1 has two vertices coinciding with the vertices of the parallelogram DEFG covering it, it is evident that, regardless of whether these two vertices are endpoints of a side or endpoints of a diagonal of the parallelogram, we have
$$
S_{\square D E F G} \geqslant 2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,402 |
Example 6 There is a loop made of silk thread on the table, with a perimeter of $2 a$. Prove:
(1) When the loop is formed into a parallelogram, it can be completely covered by a circular paper with a diameter of $a$;
(2) Regardless of the shape of the loop, it can be completely covered by this circular paper. [2]
(1964... | Proof: (1) As shown in Figure 4, let the diagonals of the parallelogram $ABCD$ intersect at point $O$. Clearly,
$$
O B=O D=\frac{1}{2} B D<\frac{1}{2}(B C+C D)=\frac{a}{2} \text {. }
$$
Similarly, $O A=O C<\frac{a}{2}$.
Thus, by placing the center of the circle at point $O$, the circle can cover the parallelogram $ABC... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,403 |
Example 7 In $\square A B C D$, $\triangle A B D$ is an acute triangle, $A B=a, A D=1, \angle B A D=\alpha$. Prove that the necessary and sufficient condition for four circles $\odot A$, $\odot B$, $\odot C$, $\odot D$ with centers at $A$, $B$, $C$, $D$ and radius 1 to cover $\square A B C D$ is $a \leqslant \cos \alph... | Proof: As shown in Figure 6, let the circumradius of $\triangle ABD$ be $r$, and the circumcenter be $O$. By the Law of Sines and the Law of Cosines, we have
$$
\begin{array}{l}
2 r=\frac{BD}{\sin \alpha}, \quad BD=\sqrt{a^{2}+1-2 a \cos \alpha} . \\
\text { Therefore, } r=\frac{\sqrt{a^{2}+1-2 a \cos \alpha}}{2 \sin \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,404 |
Question 2 If it is only required that no 5 stones appear consecutively in horizontal or vertical lines; then, on an $m \times n$ $(m \geqslant 5, n \geqslant 5)$ board, what is the minimum number of stones that need to be removed to possibly meet the requirement? | Reviewing the previous scenarios (1) and (2), we find that their discussion process is still valid for question 2. For scenario (3), on a $7 \times 8$ chessboard, we can only prove: at least 10 chess pieces need to be removed. Therefore, on an $m \times n$ chessboard, the minimum number of chess pieces that need to be ... | \left\{\begin{array}{ll}
{\left[\frac{m n}{5}\right]-1,} & \text { if } m \text { and } n \text {, when divided by } 5, \\
{\left[\frac{m n}{5}\right],} & \text { otherwise. }
\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,407 |
In the acute triangle $\triangle ABC$, $AB < AC$, $AD$ is the altitude on side $BC$, $P$ is a point on segment $AD$, $PE \perp AC$, with the foot of the perpendicular being $E$, and $PF \perp AB$, with the foot of the perpendicular being $F$. $O_{1}$ and $O_{2}$ are the circumcenters of $\triangle BDF$ and $\triangle C... | I- 4 -
| proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,408 |
Second question As shown in Figure 8, in a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center point of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected". Now, some of the 56 chess pieces are removed ... | Solution: The minimum number of chess pieces to be removed is 11.
Define the square at the $x$-th column and $y$-th row as $(x, y)$ $(1 \leqslant x \leqslant 8,1 \leqslant y \leqslant 7, x, y \in \mathbf{Z})$.
A group of 5 consecutive chess pieces is called a "good group". The problem requires that no chess pieces for... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,409 |
Third question Let the set $P=\{1,2,3,4,5\}$. For any $k \in P$ and positive integer $m$, denote
$$
f(m, k)=\sum_{i=1}^{5}\left[m \sqrt{\frac{k+1}{i+1}}\right],
$$
where $[a]$ represents the greatest integer not exceeding $a$.
Prove that for any positive integer $n$, there exist $k \in P$ and a positive integer $m$ su... | Proof: Consider another function
$$
g(t)=\sum_{i=1}^{5}\left[\sqrt{\frac{t}{i+1}}\right](\iota=2,3, \cdots) .
$$
Next, we prove:
$g(t+1)-g(t) \leqslant 1$
In fact, we have
$g(2)=1, g(3)=2, g(4)=3$,
$g(5)=4, g(6)=5$.
Thus, when $t=2,3,4,5$, we have
$g(t+1)-g(t)=1$.
Equation (1) holds.
When $t \geqslant 6$,
$g(t+1)-g(t)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,410 |
Example 3 Find the integer solutions of the equation $205 x+93 y=7$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Explanation: First, write $\frac{205}{93}$ in the form of a continued fraction:
$$
\begin{aligned}
\frac{205}{93} & =2+\frac{19}{93}, \frac{93}{19}=4+\frac{17}{19}, \\
\frac{19}{17} & =1+\frac{2}{17}, \frac{17}{2}=8+\frac{1}{2}, \\
\text { we get } \quad \frac{205}{93} & =2+\frac{1}{4}+\frac{1}{1}+\frac{1}{8}+\frac{1}{... | x = -308 - 93 t, y = 679 + 205 t (t \in \mathbf{Z}) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,411 |
1. The equation
$$
\sqrt[3]{(x+1)(x-4)}+\sqrt[3]{(x+2)(5-x)}=6
$$
has ( ) real solutions.
(A) 0
(B) 1
(C) 2
(D) More than 2 | - 1.A.
Let $a=\sqrt[3]{(x+1)(x-4)}$,
$$
b=\sqrt[3]{(x+2)(5-x)} \text {. }
$$
Then $a+b=6, a^{3}+b^{3}=6$.
Thus, $a^{2}+b^{2}-a b=1$. Therefore, $a b=\frac{35}{3}$.
Hence, $a, b$ are the two real roots of the equation $t^{2}-6 t+\frac{35}{3}=0$. However, $\Delta=-\frac{32}{3}<0$, no real solutions. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,412 |
2. A regular 2007-gon $P$ is divided by some of its non-intersecting diagonals into several regions, each of which is a triangle. Then the number of acute triangles is ( ).
(A) 0
(B) 1
(C) greater than 1
(D) depends on the method of division
直接输出了翻译结果，保留了源文本的换行和格式。 | 2.B.
Only the triangle containing the center $O$ of the regular 2007-gon is an acute triangle, so there is only one. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,413 |
3. Given the quadratic function about the parameter $a(a>0)$: $y=a x^{2}+\sqrt{1-a^{2}} x+a^{2}-3 a-\frac{1}{4}+\frac{1}{4 a}(x \in \mathbf{R})$, the minimum value of which is a function $f(a)$ of $a$. Then the minimum value of $f(a)$ is ( ).
(A) -2
(B) $-\frac{137}{64}$
(C) $-\frac{1}{4}$
(D) None of the above results... | 3.A.
When $x=-\frac{\sqrt{1-a^{2}}}{2 a}$, the minimum value of $y$ is $f(a)=a^{2}-\frac{11}{4} a-\frac{1}{4}(0<a \leqslant 1)$.
Since the axis of symmetry is $a=\frac{11}{8}$, when $a=1$, the minimum value of $f(a)$ is -2. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,414 |
4. Given $a, b (a \leqslant b)$ are positive integers, and real numbers $x, y$ satisfy $x+y=4(\sqrt{x+a}+\sqrt{y+b})$. If the maximum value of $x+y$ is 40, then the number of pairs $(a, b)$ that satisfy the condition is $(\quad)$.
(A) 1
(B) 3
(C) 5
(D) 7 | 4.C.
Since $(m+n)^{2} \leqslant 2\left(m^{2}+n^{2}\right)$, we have
$$
\begin{array}{l}
x+y=4(\sqrt{x+a}+\sqrt{y+b}) \\
\leqslant 4 \sqrt{2(x+a+y+b)} . \\
\text { Therefore, }(x+y)^{2}-32(x+y)-32(a+b) \leqslant 0 .
\end{array}
$$
Thus, $x+y \leqslant 16+4 \sqrt{16+2(a+b)}$.
From $16+4 \sqrt{16+2(a+b)}=40$, we get
$$
... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,415 |
5. Define the length of intervals $(c, d)$, $[c, d)$, $(c, d]$, and $[c, d]$ to be $d-c$ ($d>c$). Given that the real number $a>b$. Then the length sum of the intervals formed by $x$ satisfying $\frac{1}{x-a}+\frac{1}{x-b} \geqslant 1$ is $(\quad$.
(A) 1
(B) $a-b$
(C) $a+b$
(D) 2 | 5.D.
The original inequality is equivalent to $\frac{2 x-(a+b)}{(x-a)(x-b)} \geqslant 1$.
When $x>a$ or $x>0$.
Let the two roots of $f(x)=0$ be $x_{1}$ and $x_{2}$ $\left(x_{1}<x_{2}\right)$, then the interval of $x$ that satisfies $f(x) \leqslant 0$ is $\left(a, x_{2}\right]$, and the length of the interval is $x_{2}... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,416 |
6. Through the vertex $D$ of the tetrahedron $ABCD$, a sphere with radius 1 is drawn. This sphere is tangent to the circumscribed sphere of the tetrahedron $ABCD$ at point $D$, and is also tangent to the plane $ABC$. If $AD=2\sqrt{3}$, $\angle BAD=\angle CAD=45^{\circ}$, $\angle BAC=60^{\circ}$, then the radius $r$ of ... | 6. C.
Draw a perpendicular from point $D$ to plane $ABC$, with the foot of the perpendicular being $H$. Draw $DE \perp AB$, with the foot of the perpendicular being $E$, and draw $DF \perp AC$, with the foot of the perpendicular being $F$. Then $HE \perp AB, HF \perp AC$, and
$$
AE = AF = AD \cos 45^{\circ} = \sqrt{6}... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,417 |
7. If the system of equations concerning $x$ and $y$ $\left\{\begin{array}{l}a x+b y=1, \\ x^{2}+y^{2}=10\end{array}\right.$ has solutions, and all solutions are integers, then the number of ordered pairs $(a, b)$ is $\qquad$ | $=7.32$.
Since the integer solutions of $x^{2}+y^{2}=10$ are
$$
\begin{array}{l}
(1,3),(3,1),(1,-3),(-3,1), \\
(-1,3),(3,-1),(-1,-3),(-3,-1),
\end{array}
$$
Therefore, the number of lines connecting these eight points that do not pass through the origin is 24, and the number of tangents passing through these eight poi... | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,418 |
8. All positive integer solutions of the equation $x^{2}+3 y^{2}=2007$ are $\qquad$ . | 8. $x=42, y=9$.
Since $a^{2} \equiv 0,1(\bmod 3), 2007 \equiv 0(\bmod 3)$, therefore, $x \equiv 0(\bmod 3)$.
Let $x=3 x_{1}$, similarly we get $y \equiv 0(\bmod 3)$.
Let $y=3 y_{1}$, then the original equation becomes $x_{1}^{2}+3 y_{1}^{2}=$
223. From $1 \leqslant x_{1}<\sqrt{223}$, we get $1 \leqslant x_{1} \leqslan... | x=42, y=9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,419 |
9. Given that $D$ is a point on the side $BC$ of the equilateral $\triangle ABC$ with side length 1, the inradii of $\triangle ABD$ and $\triangle ACD$ are $r_{1}$ and $r_{2}$, respectively. If $r_{1}+r_{2}=\frac{\sqrt{3}}{5}$, then there are two points $D$ that satisfy this condition, denoted as $D_{1}$ and $D_{2}$. T... | 9. $\frac{\sqrt{6}}{5}$.
Let $B D=x$. By the cosine theorem, we get $A D=\sqrt{x^{2}-x+1}$.
On one hand, $S_{\triangle B D D}=\frac{1}{2} x \cdot \frac{\sqrt{3}}{2}$;
On the other hand,
$S_{\triangle A B D}=\frac{1}{2}\left(1+x+\sqrt{x^{2}-x+1}\right) r_{1}$.
Solving this, we get $r_{1}=\frac{\sqrt{3}}{6}\left(1+x-\sq... | \frac{\sqrt{6}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,420 |
10. Equation
$$
\begin{array}{l}
\frac{(x-1)(x-4)(x-9)}{(x+1)(x+4)(x+9)}+ \\
\frac{2}{3}\left[\frac{x^{3}+1}{(x+1)^{3}}+\frac{x^{3}+4^{3}}{(x+4)^{3}}+\frac{x^{3}+9^{3}}{(x+9)^{3}}\right]=1
\end{array}
$$
The number of distinct non-zero integer solutions is | 10.4 .
Using $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$, the original equation is
$$
\begin{array}{l}
\frac{(x-1)(x-4)(x-9)}{(x+1)(x+4)(x+9)}+1+\frac{2}{3}\left[\frac{x^{3}+1}{(x+1)^{3}}-\right. \\
\left.1+\frac{x^{3}+4^{3}}{(x+4)^{3}}-1+\frac{x^{3}+9^{3}}{(x+9)^{3}}-1\right]=0 \\
\Leftrightarrow \frac{x^{3}+49 x... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,421 |
Example 4 Convert the pure periodic continued fraction
$$
2+\frac{1}{3}+\ldots=2+\frac{1}{3}+\frac{1}{2}+\frac{1}{3}+\cdots=[\overline{2,3}]
$$
into a quadratic irrational number. | Explanation: Let $x=[\overline{2,3}]$, then $x=2+\frac{1}{3}+\frac{1}{x}$. Therefore, we have Table 4.
Table 4
\begin{tabular}{|c|c|c|c|c|c|}
\hline$a_{n}$ & & & 2 & 3 & $x$ \\
\hline$p_{n}$ & 0 & 1 & 2 & 7 & $7 x+2$ \\
\hline$q_{n}$ & 1 & 0 & 1 & 3 & $3 x+1$ \\
\hline
\end{tabular}
Thus, $x=\frac{7 x+2}{3 x+1}$, whic... | x=\frac{3+\sqrt{15}}{3} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,422 |
11. Let the set $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$,
$$
B=\left\{a_{1}^{2}, a_{2}^{2}, a_{3}^{2}, a_{4}^{2}, a_{5}^{2}\right\},
$$
where $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are 5 different positive integers, and
$$
\begin{array}{l}
a_{1}<a_{2}<a_{3}<a_{4}<a_{5}, \\
A \cap B=\left\{a_{1}, a_{4}\right\}... | 11.2.
Since $a_{1}^{2}=a_{1}$, therefore, $a_{1}=1, a_{4}=9$.
Since $B$ contains 9, $A$ contains 3.
If $a_{3}=3$, then $a_{2}=2$.
Thus, $a_{5}+a_{5}^{2}=146$. No positive integer solution.
If $a_{2}=3$, since $10 \leqslant a_{5} \leqslant 11$, then $a_{2}^{2} \neq a_{5}$.
Thus, $a_{3}+a_{3}^{2}+a_{5}+a_{5}^{2}=152$.
Al... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,423 |
12. In the Cartesian coordinate system, the traffic distance between points $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$ is defined as
$$
d(P, Q)=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right| .
$$
If the traffic distance from $C(x, y)$ to points $A(1,3)$ and $B(6,9)$ is equal, where the real numbe... | $12.5(\sqrt{2}+1)$.
From the condition, we have
$$
|x-1|+|y-3|=|x-6|+|y-9| \text {. }
$$
When $y \geqslant 9$,
$$
|x-1|+y-3=|x-6|+y-9 \text {, }
$$
which simplifies to $|x-1|+6=|x-6|$, with no solution;
When $3 \leqslant y \leqslant 9$,
$$
|x-1|+y-3=|x-6|+9-y \text {, }
$$
which simplifies to $2 y-12=|x-6|-|x-1|$.
W... | 5(\sqrt{2}+1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,424 |
13. Given that the circumcenter of $\triangle A B C$ is $O, \angle A<90^{\circ}$, $P$ is any point on the circumcircle of $\triangle O B C$ and inside $\triangle A B C$, the circle with $O A$ as diameter intersects $A B$ and $A C$ at points $D$ and $E$, $O D$ and $O E$ intersect $P B$ and $P C$ or their extensions at p... | Three, 13. Connect $A G$, intersecting $B P$ at point $F^{\prime}$. From $O E \perp A E$, we know $\triangle G A C$ is an isosceles triangle.
Thus, $\angle G A C = \angle G C A$.
$$
\begin{array}{l}
\text { Also, } 2 \angle B A C = \angle B O C = \angle B P C \\
= \angle B A C + \angle G C A + \angle P B A,
\end{array}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,425 |
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