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3. (1) Write down four consecutive positive integers, each of which is a multiple of a perfect square other than 1, and specify which perfect square each is a multiple of;
(2) Write down six consecutive positive integers, each of which is a multiple of a perfect square other than 1, and specify which perfect square eac... | 3. (1) $242, 243, 244, 245$ are four consecutive positive integers, 242 is a multiple of $11^{2}$, 243 is a multiple of $3^{2}$, 244 is a multiple of $2^{2}$, and 245 is a multiple of $7^{2}$.
(2) $2348124, 2348125, 2348126,$
$2348127, 2348128, 2348129$
are six consecutive positive integers, where 2348124 is a multipl... | 2348124, 2348125, 2348126, 2348127, 2348128, 2348129 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,645 |
$$
\begin{array}{r}
\text { 1. If } 2008=a_{n}(-3)^{n}+a_{n-1}(-3)^{n-1}+ \\
\cdots+a_{1}(-3)+a_{0}\left(a_{i}=0, \pm 1, \pm 2, i=0,1\right. \text {, }
\end{array}
$$
$\cdots, n)$, then $a_{n}+a_{n-1}+\cdots+a_{1}+a_{0}=$ $\qquad$ | 1.0 or $\pm 4$ or $\pm 8$.
$$
\begin{aligned}
2008= & 2(-3)^{6}-2(-3)^{5}-2(-3)^{3}+ \\
& (-3)^{2}+1,
\end{aligned}
$$
At this point, $a_{n}+a_{n-1}+\cdots+a_{0}=0$;
$$
\begin{aligned}
2008= & 2(-3)^{6}-2(-3)^{5}-2(-3)^{3}+ \\
& (-3)^{2}-(-3)-2,
\end{aligned}
$$
At this point, $a_{n}+a_{n-1}+\cdots+a_{0}=-4$;
$$
\beg... | 0 \text{ or } \pm 4 \text{ or } \pm 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,646 |
2. The number of values of $n$ $\left(n \in \mathbf{N}_{+}\right)$ for which the equation $x^{2}-6 x-2^{n}=0$ has integer solutions is $\qquad$ | 2.1.
$x=3 \pm \sqrt{3^{2}+2^{n}}$, where $3^{2}+2^{n}$ is a perfect square. Clearly, $n \geqslant 2$.
When $n \geqslant 2$, we can assume
$$
2^{n}+3^{2}=(2 k+1)^{2}\left(k \in \mathbf{N}_{+}, k \geqslant 2\right),
$$
i.e., $2^{n-2}=(k+2)(k-1)$.
It is evident that $k-1=1, k=2, n=4$.
The number of values of $n$ that mak... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,647 |
4. Given that $a$ is an integer, the equation concerning $x$
$$
\frac{x^{2}}{x^{2}+1}-\frac{4|x|}{\sqrt{x^{2}+1}}+2-a=0
$$
has real roots. Then the possible values of $a$ are | $$
4.0,1,2 \text {. }
$$
Let $y=\frac{|x|}{\sqrt{x^{2}+1}}$. Then $0 \leqslant y<1$. From
$$
\begin{array}{l}
y^{2}-4 y+2-a=0 \Rightarrow(y-2)^{2}=2+a \\
\Rightarrow 1<2+a \leqslant 4 \Rightarrow-1<a \leqslant 2 .
\end{array}
$$
Therefore, the possible values of $a$ are $0, 1, 2$. | 0, 1, 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,649 |
II. (25 points) As shown in Figure 2, the incircle of $\triangle ABC$ touches $AB$ and $AC$ at $E$ and $F$ respectively, $D$ is the midpoint of $BC$, and the angle bisectors of $\angle B$ and $\angle C$ intersect the line $EF$ at $N$ and $M$ respectively. Prove that $DM = DN$.
保留了原文的换行和格式。 | As shown in Figure 4, let the incenter be $I$, and connect $IA$, $IE$, $IF$, $BM$, and $CN$. Then $A$, $E$, $I$, and $F$ are concyclic, so
$$
\begin{array}{l}
\angle IEF = \angle IAF = \frac{\angle A}{2}, \\
\angle BEM = \frac{\pi}{2} + \frac{\angle A}{2} \\
= \angle BIC.
\end{array}
$$
Therefore, $B$, $E$, $M$, and $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,650 |
5. If a number can be expressed as the sum of the digits of some multiple of 91, then this number is called a "harmonious number". Then, among $1,2, \cdots, 2008$, the number of harmonious numbers is $\qquad$ | 5.2007 .
Notice that $91=7 \times 13$.
Numbers with a digit sum of 1 are not multiples of 91.
$1001, 10101, 10011001, 101011001$,
$100110011001, 1010110011001, \cdots$
are all multiples of 91, and their digit sums are 2, 3, $4, 5, 6, 7, \cdots$ respectively. Therefore, among $1, 2, \cdots, 2008$, the number of numbers... | 2007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,651 |
6. It is known that the selling price of a certain model of car is 230,000 yuan per unit. A factory's total cost for producing this model of car in a year consists of fixed costs and production costs. The fixed cost for one year is 70,000,000 yuan. When producing $x$ units of this car in a year, the production cost for... | 6.318.
If the sales revenue of this type of car produced by the factory in a year is not less than the total cost, then
$$
\begin{array}{l}
23 x-\left[7000+\frac{70-\sqrt{x}}{3 \sqrt{x}} \cdot x\right] \geqslant 0 \\
\Rightarrow x-\sqrt{x}-300 \geqslant 0 \Rightarrow \sqrt{x} \geqslant \frac{1+\sqrt{1201}}{2} \\
\Righ... | 318 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,652 |
7. If 2008 numbers $a_{1}, a_{2}, \cdots, a_{2008}$ satisfy
$$
a_{1}=2, a_{n}^{2}-\left(\frac{a_{n-1}}{2008}+\frac{1}{a_{n-1}}\right) a_{n}+\frac{1}{2008}=0,
$$
where, $n=2,3, \cdots, 2008$, then the maximum value that $a_{2008}$ can reach is | 7. $\frac{2008^{2006}}{2}$.
From the given information, we have
$$
a_{n}=\frac{1}{2} \frac{a_{n \cdot 1}}{2008} \text { or } a_{n} \stackrel{1}{2}=\frac{1}{a_{n \cdot 1}} \text {, }
$$
This means that $a_{n-1}$ can only be transformed into $a_{n}(n=2,3, \cdots, 2008)$ through either the (1)st type of transformation o... | \frac{2008^{2006}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,653 |
8. Given that $\odot O$ is tangent to line $l$ at point $M$, and a fixed point $A$ outside $\odot O$ and $\odot O$ are on the same side of line $l$. The distance from point $A$ to line $l$ is greater than the diameter of $\odot O$, and point $B$ is on $\odot O$. Draw a perpendicular line $A N$ from point $A$ to line $l... | 8. Segment $A M$ inside.
Let the other intersection point of line $A B$ with $\odot O$ be $D$, and assume point $B$ is between points $A$ and $D$. Draw a perpendicular line $D E$ from point $D$ to line $A C$, with the foot of the perpendicular being $E$. Then,
$$
\begin{array}{l}
A B \cdot A D=k \ (k \text{ is a const... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,654 |
9. On the two legs $AB, AC$ of isosceles $\triangle ABC$ with base angles equal to $80^{\circ}$, points $D, E$ are taken such that $\angle BDC=$ $50^{\circ}, \angle BEC=40^{\circ}$. Then $\angle ADE=$ $\qquad$ | $9.50^{\circ}$.
Given $\angle B A C=20^{\circ}, \angle B C D=50^{\circ}$, hence
$$
\begin{array}{l}
B C=B D, \\
\angle C B E=60^{\circ}, \angle A B E=20^{\circ} .
\end{array}
$$
Take a point $F$ on $C E$ such that $\angle C B F=20^{\circ}$, then
$$
\begin{array}{l}
\angle E B F=40^{\circ}, \\
B F=F E, \\
\angle D B F=... | 50^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,655 |
10. Select 1004 numbers from $1,2, \cdots, 2008$ such that their total sum is 1009000, and the sum of any two of these 1004 numbers is not equal to 2009. Then the sum of the squares of these 1004 numbers is $\qquad$
Reference formula:
$$
1^{2}+2^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1) \text {. }
$$ | 10.1351373940.
Divide $1,2, \cdots, 2008$ into 1004 groups:
$$
\{1,2008\},\{2,2007\}, \cdots,\{1004,1005\} \text {. }
$$
By the problem's setup, exactly one number is taken from each group. Replace 2, 4, $\cdots, 2008$ with 1004, 1006, 1008, 1010 by their counterparts in the same group, 1005, 1003, 1001, 999, respect... | 1351373940 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,656 |
1. If $a, b$ are real numbers, and satisfy $(a-1)^{2}+$ $\sqrt{b-2}=0$, then
$$
\begin{array}{l}
\frac{1}{a b}+\frac{1}{(a+1)(b+1)}+\frac{1}{(a+2)(b+2)}+ \\
\cdots+\frac{1}{(a+2008)(b+2008)} \\
=(\quad) .
\end{array}
$$
(A) $\frac{2006}{2007}$
(B) $\frac{2007}{2008}$
(C) $\frac{2008}{2009}$
(D) $\frac{2009}{2010}$ | $\begin{array}{l}\text { I.1.D. } \\ \text { Given }(a-1)^{2} \geqslant 0, \sqrt{b-2} \geqslant 0, \\ (a-1)^{2}+\sqrt{b-2}=0 \\ \Rightarrow a=1, b=2 . \\ \text { Therefore, the original expression }=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{2009 \times 2010} \\ =\left(1-\frac{1}{2}\right)+\left(\frac{1}... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,657 |
2. In an acute $\triangle A B C$, the measures of the three interior angles are prime numbers, and the length of the shortest side is 1. The number of non-congruent triangles that satisfy these conditions is ( ).
(A) 1
(B) 2
(C) 3
(D) More than 3 | 2. A.
Let the measures of the three interior angles be $x$, $y$, and $z$ degrees. Then $x$, $y$, and $z$ are prime numbers, and $x + y + z = 180$. Therefore, at least one of $x$, $y$, and $z$ must be an even prime number. Without loss of generality, let $x$ be the even prime number. Then $x = 2$. Thus, $y + z = 178$.
... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,658 |
3. Given point $A(a, \sqrt{2})$ is the intersection point of the graphs of the two functions $y=k x-2$ and $y=(\sqrt{2}-1) x$. Then the real number $k$ equals ( ).
(A) $-\sqrt{2}$
(B) $1-\sqrt{2}$
(C) $\sqrt{2}-1$
(D) 1 | 3.D.
From the condition, we know $\sqrt{2}=(\sqrt{2}-1) a$, then
$$
\begin{array}{l}
a=\frac{\sqrt{2}}{\sqrt{2}-1}=2+\sqrt{2} . \\
\text { Also } \sqrt{2}=k a-2 \text {, i.e., } \sqrt{2}=k(\sqrt{2}+2)-2 .
\end{array}
$$
Solving for $k$ gives $k=1$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,659 |
4. In rhombus $A B C D$, $\angle A B C=60^{\circ}, A B=$ $1, E$ is the midpoint of side $B C$. Then the minimum value of the sum of the distances from a moving point $P$ on diagonal $B D$ to points $E$ and $C$ is ( ).
(A) $\frac{\sqrt{3}}{4}$
(B) $\frac{\sqrt{3}}{3}$
(C) $\frac{\sqrt{3}}{2}$
(D) $\sqrt{3}$ | 4.C.
As shown in Figure 3, construct the symmetric point $E_{1}$ of $E$ with respect to $BD$. Since $BD$ is the axis of symmetry of the rhombus $ABCD$, $E_{1}$ is the midpoint of side $AB$. Connect $CE_{1}$ and let it intersect $BD$ at point $P_{1}$.
Since $PE + PC = PE_{1} + PC \geqslant CE_{1}$, the equality holds ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,660 |
Three, (25 points) Divide $1,2, \cdots, 9$ into three groups, each containing three numbers, such that the sum of the numbers in each group is a prime number.
(1) Prove that there must be two groups with equal sums;
(2) Find the number of all different ways to divide them.
| Three, (1) Since the sum of three different numbers in $1,2, \cdots, 9$ is between 6 and 24, the prime numbers among them are only 7, 11, 13, 17, 19, 23, these six. Now, these six numbers are divided into two categories based on their remainders when divided by 3:
$A=\{7,13,19\}$, where each number leaves a remainder o... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,661 |
6. As shown in Figure 1, in the convex quadrilateral $A B C D$, $E$ and $F$ are the midpoints of sides $A B$ and $C D$ respectively. $A F$ and $D E$ intersect at point $G$, and $B F$ and $C E$ intersect at point $H$. The area of quadrilateral $E G F H$ is 10. Then the sum of the areas of $\triangle A D G$ and $\triangl... | 6. B.
Solution 1: Specialization.
From $A B / / B C$, we get
$$
S_{\triangle A X C}=S_{\triangle E C F}, S_{\triangle B H C}=S_{\triangle E H F}.
$$
Therefore, $S_{\triangle U D G}+S_{\triangle B M C}=S_{\triangle N C F}+S_{\triangle E H F}$
$$
=S_{\text {quadrilateral } E C F H}=10.
$$
Solution 2:
As shown in Figur... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,663 |
1. Given real numbers $a$, $b$, $c$ satisfy $(a+b)(b+c)(c+a)=0$ and $abc<0$. Then the value of the algebraic expression $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}$ is | $=.1 .1$.
From $(a+b)(b+c)(c+a)=0$, we know that at least two of $a, b, c$ are opposite numbers. Also, since $abc<0$, it follows that among $a, b, c$, there must be two positive and one negative.
Thus, the value of $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}$ is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,664 |
2. In $\triangle A B C$, $A B=5, A C=12, C B=$ $13, D, E$ are points on side $B C$, satisfying $B D=1, C E=$ 8. Then the degree of $\angle D A E$ is | $$
\begin{array}{l}
\mathbf{2 . 4 5} . \\
\text { As shown in Figure } 5, \text { let } \angle A D C \\
=\alpha, \angle A E B=\beta .
\end{array}
$$
As shown in Figure 5, let $\angle A D C$
Since $A B^{2}+A C^{2}$
$$
=5^{2}+12^{2}=13^{2}=C B^{2} \text {. }
$$
Therefore, $\angle B A C=90^{\circ}$.
From the given cond... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,665 |
3. In Rt $\triangle A B C$, $F$ is the midpoint of the hypotenuse $A B$, and $D, E$ are points on sides $C A, C B$ respectively, such that $\angle D F E=$ $90^{\circ}$. If $A D=3, B E=4$, then the length of segment $D E$ is $\qquad$ | 3.5 .
As shown in Figure 6, extend $D F$ to point $G$ such that $D F = F G$, and connect $G B$ and $G E$.
Given $A F = F B$, we have
$\triangle A D F \cong \triangle B G F$
$\Rightarrow B G = A D = 3$
$\Rightarrow \angle A D F = \angle B G F$
$\Rightarrow A D \parallel G B$
$\Rightarrow \angle G B E + \angle A C B = 1... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,666 |
4. Divide the positive integers $1,2, \cdots, 10$ into two groups, $A$ and $B$, where group $A$: $a_{1}, a_{2}, \cdots, a_{m}$; group $B$: $b_{1}, b_{2}, \cdots$, $b_{n}$. Now, take one number from each of the groups $A$ and $B$, and multiply the two numbers taken. Then the maximum value of the sum of all different pro... | 4.756 .
From the condition, the sum of all different products of two numbers is $S=\left(a_{1}+\cdots+a_{m}\right)\left(b_{1}+\cdots+b_{n}\right)$.
$$
\begin{array}{l}
\text { Let } x=a_{1}+\cdots+a_{m}, y=b_{1}+\cdots+b_{n} . \text { Then } \\
x+y=1+2+\cdots+10=55, \\
S=x y=\frac{1}{4}\left[(x+y)^{2}-(x-y)^{2}\right]... | 756 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,667 |
Three. (20 points) Given that $a$, $b$, and $c$ are real numbers. Prove: Among the values of the algebraic expressions $(a+b+c)^{2}$, $(a+b-c)^{2}$, $(b+c-a)^{2}$, and $(c+a-b)^{2}$, at least one is not less than $a^{2}+b^{2}+c^{2}$, and at least one is not greater than $a^{2}+b^{2}+c^{2}$. | $$
\begin{array}{l}
\text { Three, let } A=(a+b+c)^{2}, B=(a+b-c)^{2}, \\
C=(b+c-a)^{2}, D=(c+a-b)^{2} .
\end{array}
$$
Then $A=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$,
$$
\begin{array}{l}
B=a^{2}+b^{2}+c^{2}+2 a b-2 b c-2 c a, \\
C=a^{2}+b^{2}+c^{2}-2 a b+2 b c-2 c a, \\
D=a^{2}+b^{2}+c^{2}-2 a b-2 b c+2 c a .
\end{arra... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,668 |
Four. (25 points) As shown in Figure 2, in the right trapezoid $ABCD$, $\angle ABC = \angle BAD = 90^{\circ}$, $AB = 16$. The diagonals $AC$ and $BD$ intersect at point $E$. A line $EF \perp AB$ is drawn through $E$ at point $F$, and $O$ is the midpoint of side $AB$, with $FE + EO = 8$. Find the value of $AD + BC$. | Let $O F=x$.
Then $F B=8-x, F A=8+x$.
Given that $D A / / E F / / C B$, we have $\frac{F E}{A D}=\frac{F B}{A B}$, which means
$$
A D=\frac{16}{8-x} E F \text {. }
$$
Similarly, $B C=\frac{16}{8+x} E F$.
Thus, $A D+B C=\left(\frac{16}{8-x}+\frac{16}{8+x}\right) E F$
$$
=\frac{16 \times 16}{8^{2}-x^{2}} E F \text {. }
... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,669 |
Five. (25 points) In the expression “ $\square 1 \square 2 \square 3 \square 4 \square 5 \square 6 \square 7$ $\square 8 \square 9$ ”, fill in the small squares with “+” or “-” signs. If the algebraic sum can be $n$, then the number $n$ is called a "representable number"; otherwise, it is called an "unrepresentable num... | (1) Since $+1-2-3+4+5-6+7-8$ $+9=7$, therefore, 7 is a number that can be represented.
$$
\text { Also, }+1+2+3+4+5+6+7+8+9=45
$$
is an odd number, and for any two integers $a$ and $b$, $a+b$ and $a-b$ have the same parity. Therefore, no matter how the “ $+\cdots \times$ - ” signs are filled, the algebraic sum must be ... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,670 |
2. If $x^{2}-(m+1) x+1$ is a perfect square, then the value of $m$ is ( ).
(A) -1
(B) 1
(C) 1 or -1
(D) 1 or -3 | 2.D.
From $m+1= \pm 2$, we get $m=1$ or $m=-3$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,671 |
1. Given four operations with non-zero real numbers $x$ and $y$:
(1) $x y=1$,
(2) $|x| y=1$,
(3) $x|y|=1$,
(4) $|x||y|=1$.
Then $y$ is a function of $x$ in ( ).
$(\mathrm{A})(1)(2) \quad(\mathrm{B})(2)(3) \quad(\mathrm{C})(3)(4) \quad(\mathrm{D})(1)(4)$ | - 1.A.
According to the definition of a function, if for "every" non-zero $x$, there is a "unique" $y$ corresponding to it, then $y$ is a function of $x$. (1) and (2) meet this requirement; however, (3) and (4) do not meet this requirement, because "every" non-zero $x$ has two $y$ values corresponding to it. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,672 |
3. As shown in Figure 1, points $A$, $B$, and $C$ are sequentially on line $l$, $M$ is the midpoint of segment $AC$, and $N$ is the midpoint of segment $BC$. To find the length of $MN$, which of the following conditions is sufficient?
Figure 1
(A) $AB=12$
(B) $BC=4$
(C) $AM=5$
(D) $CN=2$ | 3.A.
Since $M N=M C-N C=\frac{1}{2} A C-\frac{1}{2} B C$
$$
=\frac{1}{2}(A C-B C)=\frac{1}{2} A B,
$$
Therefore, we only need to know $A B$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,673 |
4. In the Cartesian coordinate system $x 0 y$, it is known that point $A(3,-3)$, and $P$ is a point on the $y$-axis. Then the number of points $P$ that make $\triangle A O P$ an isosceles triangle is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 4.C.
Considering two cases with points $A$, $O$, and $P$ as the apex of an isosceles triangle respectively | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,674 |
5. Given the equation $(2 a+b) x-1=0$ has no solution for $x$. Then the value of $a b$ is ( ).
(A) negative
(B) positive
(C) non-negative
(D) non-positive | 5.D.
From the problem, we know that $2a + b = 0$.
Therefore, we have $a = b = 0$ or $a$ and $b$ have opposite signs. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,675 |
6. The graph of the linear function $y=k(x-1)$ passes through the point $M(-1,-2)$. Then its intersection with the $y$-axis is ( ).
(A) $(0,-1)$
(B) $(1,0)$
(C) $(0,0)$
(D) $(0,1)$ | 6. A.
From the problem, we know that $k(-1-1)=-2$, which means $k=1$. Therefore, the function expression is $y=x-1$. Substituting $x=0$ into it, we get $y=-1$. Hence, the intersection point of its graph with the $y$-axis is $(0,-1)$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,676 |
7. As shown in Figure 2, construct two equilateral triangles $\triangle A B C$ and $\triangle C D E$ on the same side of line segment $A E$ $(\angle A C E<120^{\circ})$, and let $P$ and $M$ be the midpoints of line segments $B E$ and $A D$, respectively. Then $\triangle C P M$ is ( ).
(A) obtuse triangle
(B) right tria... | 7.C.
It is easy to see that $\triangle A C D \cong \triangle B C E$.
Therefore, $\triangle B C E$ can be considered as $\triangle A C D$ rotated $60^{\circ}$ clockwise around point $C$. Since $M$ is the midpoint of segment $A D$ and $P$ is the midpoint of segment $B E$, $C P$ is obtained by rotating $C M$ $60^{\circ}$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,677 |
8. A junior high school sports team needs to purchase a batch of sports shoes to prepare for the school sports meet. It is known that the sports team has 20 students, and the statistical table is as shown in Table 1. Due to an accident, the two data points in the shaded area of Table 1 are no longer visible.
Table 1
\b... | 8.C.
(1) From the meaning of median and mode, as well as Table 1, when the median of this set of data is 40, the mode must be 40. Therefore, option (A) is incorrect.
(2) When the number of people wearing size 39 and size 40 shoes is both 5, the median and mode are not equal, so option (B) is incorrect.
(3) Assuming the... | C | Other | MCQ | Yes | Yes | cn_contest | false | 720,678 |
9. As shown in Figure $3, A$ and $B$ are two points on the graph of the function $y=\frac{k}{x}$. Points $C, D, E,$ and $F$ are on the coordinate axes, forming a square and a rectangle with points $A, B,$ and $O$. If the area of square $O C A D$ is 6, then the area of rectangle $O E B F$ is ( ).
(A) 3
(B) 6
(C) 9
(D) 1... | \begin{array}{l}=\frac{1}{2}|k|=6 . \\ \text { Then } S_{\text {K } / \text { shape }}=\frac{1}{2} O E \cdot O F=\frac{1}{2}\left|x_{B} y_{B}\right| \\ =\frac{1}{2}|k|=6 . \\\end{array} | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,679 |
10. A store has 5 bags of flour, each weighing between 25 to $30 \mathrm{~kg}$. The store has a scale, but it only has weights that can measure $50 \sim$ $70 \mathrm{~kg}$. Now, to determine the weight of each bag of flour, at least ( ) times of weighing are needed.
(A) 4
(B) 5
(C) 6
(D) 7 | 10. B. $x \mathrm{~kg}, y \mathrm{~kg}, z \mathrm{~kg}$, two items are weighed together, the recorded weights are $a \mathrm{~kg}, b \mathrm{~kg}, c \mathrm{~kg}$, then the system of equations is
$$
\left\{\begin{array}{l}
x+y=a, \\
y+z=b, \\
z+x=c
\end{array}\right.
$$
To find $x, y, z$, you can weigh two items toget... | B | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 720,680 |
11. If the system of inequalities $\left\{\begin{array}{l}x-1>0, \\ x-a<0\end{array}\right.$ has no solution, then the range of values for $a$ is $\qquad$ | $$
\text { II, 11. } a \leqslant 1 \text {. }
$$
Solve the system of inequalities $\left\{\begin{array}{l}x-1>0, \\ x-a1, \\ x<a .\end{array}\right.$
Since the original system of inequalities has no solution, it must be that $a \leqslant 1$ | a \leqslant 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,681 |
12. Given $a-b=1, a^{2}-b^{2}=-1$. Then $a^{2008}-b^{2008}=$ | 12. -1 .
Given $a^{2}-b^{2}=(a+b)(a-b)=-1$, and $a-b=1$, then $a+b=-1$.
Therefore, $\left\{\begin{array}{l}a+b=-1 \\ a-b=1 .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}a=0, \\ b=-1 .\end{array}\right.$ Hence, $a^{2 \alpha R}-b^{20 R}=0^{2008}-(-1)^{2008}=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,682 |
2. For any real numbers $m$, $n$, $p$, it is required:
(1) to write the corresponding quadratic equation;
(2) to make the equation have $n^{2}-4 m p$ as the discriminant.
Among the following answers:
$$
\begin{array}{l}
m x^{2}+n x+p=0, \\
p x^{2}+n x+m=0, \\
x^{2}+n x+m p=0, \\
\frac{1}{a} x^{2}+n x+a m p=0(a \neq 0)... | 2.B.
When $m=0$, equation (1) is not a quadratic equation; when $p=0$, equation (2) is not a quadratic equation. Equations (3) and (4) are quadratic equations, and both have $n^{2}-4 m p$ as their discriminant.
Explanation: (1) For a quadratic equation in one variable $a x^{2}+b x+c=0$ to be a quadratic equation, the... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,683 |
13. As shown in Figure 4, in rhombus $A B C D$, $A E \perp B C$ at point $E$. If $\cos B$
$$
=\frac{4}{5}, E C=2 \text {, }
$$
$P$ is a moving point on side $A B$,
then the minimum length of line segment $P E$ is $\qquad$ | 13.4.8.
Let the side length of square $ABCD$ be $x$. Then $AB=BC=x$.
Given $EC=2$, we know $BE=x-2$.
Since $AE \perp BC$ at point $E$, in the right triangle $\triangle ABE$, $\cos B=\frac{x-2}{x}$.
Given $\cos B=\frac{4}{5}$, we have $\frac{x-2}{x}=\frac{4}{5}$.
Solving for $x$ gives $x=10$, so $AB=10$.
It is easy to ... | 4.8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,684 |
14. Xiaoding, Xiaoming, and Xiaoqian are playing a game together and need to determine the order of play. They agree to use "scissors, paper, rock" to decide. What is the probability that all three of them will choose "paper" in one round? $\qquad$ . | 14. $\frac{1}{27}$.
Use the tree diagram shown in Figure 8 to list all possible outcomes of a round: the gestures made by the person.
Figure 8
Figure 8 only shows part of the tree diagram (lists 9 possible outcomes, please list 9 outcomes; finally, change “hammer” to list 9 outcomes as well. Therefore, there are a tot... | \frac{1}{27} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,685 |
15. Given that $a$ and $b$ are real numbers, and $ab=1, a \neq 1$, let $M=\frac{a}{a+1}+\frac{b}{b+1}, N=\frac{1}{a+1}+\frac{1}{b+1}$. Then the value of $M-N$ is $\qquad$. | 15.0.
Given $a b=1, a \neq 1$, so,
$$
\begin{array}{l}
M=\frac{a}{a+1}+\frac{b}{b+1}=\frac{a}{a+a b}+\frac{b}{b+a b} \\
=\frac{a}{a(1+b)}+\frac{b}{b(1+a)}=\frac{1}{a+1}+\frac{1}{b+1}=N .
\end{array}
$$
Thus, $M-N=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,686 |
16. As shown in Figure 5, in $\triangle A B C$, $A B=A C=\sqrt{5}$, $B C=2$, a circle $\odot O$ with $A B$ as its diameter intersects $A C$ and $B C$ at points $D$ and $E$ respectively. Then the area of $\triangle C D E$ is $\qquad$. | 16. $\frac{2}{5}$.
Connect $A E$ and $B D$, and draw $D F \perp E C$ at $F$. Since $A B$ is the diameter of $\odot O$, we have $\angle A D B = \angle A E B = 90^{\circ}$. Also, since $A B = A C$, then $C E = \frac{1}{2} B C = 1$. Therefore, $A E = \sqrt{A C^{2} - C E^{2}} = 2$. From $\frac{1}{2} B C \cdot A E = \frac{... | \frac{2}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,687 |
18. Given the line $y=b$ (where $b$ is a real number) intersects the graph of the function $y=\left|x^{2}-4 x+3\right|$ at least at three points. Then the range of the real number $b$ is | $18.0<b \leqslant 1$.
Week 9
Notice that $y=x^{2}-4 x+3=(x-2)^{2}-1$, the rough graph of this function is shown in Figure 10(a).
Therefore, the graph of the function $y=\left|x^{2}-4 x+3\right|$ is as shown in Figure 10(b).
And when $b$ takes all real numbers, $y=b$ represents all lines perpendicular to the $y$-axis... | 0<b \leqslant 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,689 |
19. A large supermarket holds a promotional event during the New Year's Day holiday. It is stipulated that no discount is given for a single purchase not exceeding 100 yuan, a 10% discount is applied for a single purchase exceeding 100 yuan but not exceeding 300 yuan, and for purchases exceeding 300 yuan, the first 300... | $$
\begin{array}{l}
\text { Three, 19. Note that } \\
100 \times 0.9=90<94.5<100, \\
300 \times 0.9=270<282.8 .
\end{array}
$$
Let the original price of Xiao Mei's second purchase be $x$ yuan. Then
$$
(x-300) \times 0.8+300 \times 0.9=282.8 \text {. }
$$
Solving for $x$ gives $x=316$.
Below, we discuss two scenarios:... | 366.8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,690 |
20. As shown in Figure 7, the side length of square $ABCD$ is 1, point $F$ moves on segment $CD$, and $AE$ bisects $\angle BAF$ intersecting side $BC$ at point $E$.
(1) Prove:
$$
AF = DF + BE \text{.}
$$
(2) Let $DF = x (0 \leqslant x \leqslant 1)$, and let $S$ be the sum of the areas of $\triangle ADF$ and $\triangle ... | 20. (1) As shown in Figure 11, extend $C B$ to point $G$ such that $B G = D F$, and connect $A G$.
Since quadrilateral $A B C D$ is a square, in the right triangles $\triangle A D F$ and $\triangle A B G$,
$$
\begin{array}{l}
A D = A B, D F = B G, \\
\angle A D F = \angle A B G = 90^{\circ}.
\end{array}
$$
Therefore,... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,691 |
Find the lengths of the three sides of a triangle that satisfy the following conditions.
(1) The side lengths are integers;
(2) The perimeter is an integer multiple of the area. | Solution: Let the three sides of the triangle be $a, b, c$, and the perimeter is $k$ (where $k$ is an integer) times the area. According to Heron's formula, we have
\[a+b+c=k \sqrt{\frac{a+b+c}{2} \cdot \frac{b+c-a}{2} \cdot \frac{c+a-b}{2} \cdot \frac{a+b-c}{2}}\]
Then, \(4 \cdot \frac{a+b+c}{2}=k^{2} \cdot \frac{b+c-... | (6,25,29,1),(7,15,20,1),(9,10,17,1),(5,12,13,1),(6,8,10,1),(3,4,5,2) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,692 |
In the square $A B C D$, as shown in Figure 2, with point $A$ as the center and $A B$ as the radius, draw $\overparen{B D}$. With point $B$ as the center and $\frac{1}{2} A B$ as the radius, draw arcs intersecting $A B$, $B C$, and $\overparen{B D}$ at points $E$, $F$, and $G$, respectively. Line $C G$ intersects $A B$... | Proof: As shown in Figure 2, draw perpendiculars from point $G$ to $AB$ and $BC$, with the feet of the perpendiculars being $R$ and $Q$, respectively, and connect $AG$ and $BG$.
In $\triangle ABG$, by the cosine rule, we have
$\cos \angle ABG = \frac{4a^2 + a^2 - 4a^2}{2 \times 2a \times a} = \frac{1}{4}$,
which means... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,693 |
3. Four Olympic medalists came to Xiao Wang's school to meet the teachers and students, two of whom were once Xiao Wang's classmates (the other two were not classmates). When the four medalists walked onto the stage one by one, the probability that the second person is Xiao Wang's classmate is ( ).
(A) $\frac{1}{6}$
(B... | 3. D.
Solution 1: Use "○" to represent classmate Xiao Wang, then all possible outcomes of the 4 people walking on stage in sequence can be enumerated using the tree diagram shown in Figure 3. There are 6 possible outcomes in total, and 3 of these outcomes have "○" as the second person, so the probability that "the sec... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,694 |
239 Given that $a, b, c, m$ are positive numbers. Prove:
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant \frac{a+m}{b+m}+\frac{b+m}{c+m}+\frac{c+m}{a+m} \text {. }
$$ | Proof: Without loss of generality, let $a \geqslant c, b \geqslant c$. Then
$$
\begin{array}{l}
\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3 \\
=\frac{a}{b}+\frac{b}{a}-2+\left(\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1\right) \\
=\frac{(a-b)^{2}}{a b}+\frac{(a-c)(b-c)}{a c} \\
\geqslant \frac{(a-b)^{2}}{(a+m)(b+m)}+\frac{(a-c)(b-... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,695 |
In the acute triangle $\triangle ABC$, $O$ is the circumcenter, $H$ is the orthocenter, $AD$ is the median, and $OH \perp AD$. Prove that $\frac{1}{S_{\triangle HAB}}$, $\frac{1}{S_{\triangle HBC}}$, $\frac{1}{S_{\triangle HCA}}$ form an arithmetic sequence. | Proof: As shown in Figure 3, construct the three altitudes $A A^{\prime}, B B^{\prime}, C C^{\prime}$ of $\triangle A B C$, and let $H$ be their intersection, the orthocenter. Let $O H$ intersect $A D$ at point $G$. According to the definition of the Euler line of a triangle, $G$ is the centroid of $\triangle A B C$. A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,696 |
4. From the numbers $1,2,3,4,5$, select several (at least 2) without repetition and find their sum, the number of different sums that can be obtained is ( ) .
(A) 26
(B) 17
(C) 13
(D) 12 | 4.C.
The sum formed by $1,2,3,4,5$ will not be less than $1+2$ $=3$, and will not be greater than $1+2+3+4+5=15$. Therefore, the number of different sums will not exceed $15-3+1=13$. Furthermore, since these five numbers are consecutive integers starting from 1, each of the 13 different sums can be achieved by substit... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,697 |
For any real numbers $a, b$, construct the algebraic expression
$$
M=a^{2}+a b+b^{2}-a-b+\frac{1}{2} \text {. }
$$
Then the minimum value of $M$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{6}$
(D) 0 | 5.C.
With $a$ as the main element, we complete the square to get
$$
\begin{array}{l}
M=a^{2}+(b-1) a+b^{2}-b+\frac{1}{2} \\
=a^{2}+2 \cdot \frac{b-1}{2} \cdot a+\frac{(b-1)^{2}}{4}+\frac{3 b^{2}-2 b+1}{4} \\
=\left(a+\frac{b-1}{2}\right)^{2}+\frac{3}{4}\left(b-\frac{1}{3}\right)^{2}+\frac{1}{6} \geqslant \frac{1}{6} .... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,698 |
6. As shown in Figure 1, point $C$ is any point on segment $A B$, $\triangle D A C$ and $\triangle E C B$ are both equilateral triangles, and they are on the same side of $A B$. Connect $A E$ to intersect $C D$ at point $M$, and connect $B D$ to intersect $C E$ at point $N$.
The following equations are obtained:
(1) $A... | 6. A.
Since $\triangle A C E$ can be obtained by rotating $\triangle D C B$ clockwise around point $C$ by $60^{\circ}$, we have
$$
A E=B D, \angle C A M=\angle C D N .
$$
From $A C=D C, \angle A C M=\angle D C N=60^{\circ}$, we get $\triangle A C M \cong \triangle D C N$.
Thus, all four equations hold.
However, when ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,699 |
$\begin{array}{l}\text { 1. } \frac{1}{2}+\frac{5}{2^{2}+2}+\frac{11}{3^{2}+3}+\cdots+ \\ \frac{100^{2}+100-1}{100^{2}+100}=\end{array}$ | $$
=1 . \frac{100^{2}}{101} \text {. }
$$
Solution 1: Notice that
$$
\begin{array}{l}
\frac{k^{2}+k-1}{k^{2}+k}=1-\frac{1}{k(k+1)} \\
=1-\left(\frac{1}{k}-\frac{1}{k+1}\right) .
\end{array}
$$
Therefore, the original expression $=100-\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\right.$
$$
\l... | \frac{100^{2}}{101} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,700 |
3. A student obtained a perfect square when substituting a natural number into one of the quadratic functions $f_{k}(x)(k=1,2,3,4)$. The function he used is $(\quad)$.
(A) $f_{1}(x)=x^{2}+5 x+7$
(B) $f_{2}(x)=x^{2}+7 x+10$
(C) $f_{3}(x)=x^{2}+9 x+18$
(D) $f_{4}(x)=x^{2}+11 x+2$ | 3.D.
When $x$ is a natural number, we have
$$
\begin{array}{l}
(x+2)^{2}<x^{2}+5 x+7<(x+3)^{2}, \\
(x+3)^{2}<x^{2}+7 x+10<(x+4)^{2}, \\
(x+4)^{2}<x^{2}+9 x+18<(x+5)^{2} .
\end{array}
$$
That is, when $x$ is a natural number, the functions $f_{1}(x)$, $f_{2}(x)$, and $f_{3}(x)$ all lie between two consecutive square n... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,701 |
2. In $\triangle A B C$, $\angle B=2 \angle C, A D$ is the angle bisector of $\angle A$, $m A B=n B D(n>m>0)$. Then $\cos C=$ $\qquad$ . | 2. $\frac{n+m}{2 n}$.
As shown in Figure 4, extend $C B$ to $E$ such that $B E=A B$. In the isosceles $\triangle A B E$, we have
$$
\begin{array}{l}
\angle B E A=\angle B A E \\
=\frac{1}{2} \angle A B C=\angle C,
\end{array}
$$
which means $\triangle A E C$ is an isosceles triangle, so $A C=A E$.
Furthermore, in $\t... | \frac{n+m}{2 n} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,702 |
3. A frog starts from point $A(-6,3)$ and jumps to point $B(-2,5)$, then from point $B$ to point $C$ on the $y$-axis, continues from point $C$ to point $D$ on the $x$-axis, and finally returns from point $D$ to point $A$. When the total distance of the frog's four jumps is minimized, the positions of points $C$ and $D$... | 3. As shown in Figure 5, the steps are as follows:
(1) Draw a Cartesian coordinate system, and points $A(-6,3)$, $B(-2,5)$;
(2) Draw the point $A'$, which is the reflection of point $A$ over the $x$-axis;
(3) Draw the point $B'$, which is the reflection of point $B$ over the $y$-axis;
(4) Connect $A' B'$ to intersect t... | 2 \sqrt{5} + 8 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,703 |
4. There are 7 students, and it is known that each of them has at least 3 classmates among the remaining 6. Then, the number of pairs of classmates among these 7 students is $\qquad$ .
| 4.7.
Proof: 7 people are all classmates.
Take any student $A$, according to the problem, among the remaining 6 people, there are at least 3 classmates, denoted as $A_{1} 、 A_{2} 、 A_{3}$.
Next, we prove: the remaining 3 people $B_{1} 、 B_{2} 、 B_{3}$ are also classmates of $A$.
If not, there must exist $B_{i}(1 \leq... | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,704 |
One. (20 points) Given the parabola $y=x^{2}+p x+q$ passes through the points $\left(x_{1}, 0\right), \left(x_{2}, 0\right), (1, a), (2, b), 1<x_{1}$ $<x_{2}<2$. Prove:
(1) If $x_{1}=1.4, x_{2}=1.6$, then $a, b$ are both positive numbers less than 0.25;
(2) If $a \geqslant 0.25$, then $b<0.25$. | (1) According to the problem, we have
$$
x^{2}+p x+q=\left(x-x_{1}\right)\left(x-x_{2}\right) \text {. }
$$
Substitute $x=1,2$. By $10, \\
b=4+2 p+q=\left(2-x_{1}\right)\left(2-x_{2}\right)>0 .
\end{array}
$$
Take $x_{1}=1.4, x_{2}=1.6$ and substitute into equations (1) and (2) respectively, we get
$$
\begin{array}{l... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,705 |
$$
\begin{aligned}
& \text { II. (25 points) As shown in Figure 2, in } \triangle A B C, \angle A C B \\
= & \angle C A B + 30^{\circ} \\
= & \angle A B C + 60^{\circ}, \text{ a point } D \text{ is taken on side } A B, \text{ and a point } E \text{ is taken on the extension of } C A, \text{ such that } A C \cdot C E + ... | (1) Suppose in $\triangle A B C$, $A C=a$. From the given, $\angle A B C=30^{\circ}, \angle C A B=60^{\circ}, \angle A C B=90^{\circ}$. Therefore, $A B=2 a, B C=\sqrt{3} a$.
As shown in Figure 6, construct $\angle C B F=60^{\circ}$, intersecting the extension of $C A$ at $F$. Then, Rt $\triangle B F C \backsim \mathrm... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,706 |
Three. (25 points) At the table tennis tournament opening on October 4th, everyone has designated 104 as the "lucky number." The organizers plan to select doubles "lucky players" from 100 participating athletes according to the following procedure for a performance at the opening ceremony:
(1) The 100 athletes form a c... | Three, obviously, programs (1) and (2) have no obstacles.
Below is the proof: at least two pairs of doubles lucky players can be selected by program (3).
Notice that the numbers on the cards the organizer has are: $1, 4, 7, \cdots$, 97, 100 (all numbers that leave a remainder of 1 when divided by 3). These numbers can... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,707 |
1. Given $x\left(y-\frac{1}{x}\right)=2008$, where $x$ and $y$ are positive integers. Then the sum of the maximum and minimum values of $x+y$ is ( ).
(A) 2010
(B) 2100
(C) 2008
(D) 2000 | -.1.B.
From the given, we have
$$
\begin{array}{l}
2008=x\left(y-\frac{1}{x}\right) \\
\Rightarrow x y=2009=1 \times 2009=7 \times 7 \times 41 .
\end{array}
$$
Thus, $(x+y)_{\text {max }}=1+2009=2010$,
$$
(x+y)_{\text {min }}=49+41=90 \text {. }
$$
Therefore, the sum of the maximum and minimum values of $x+y$ is 2100... | 2100 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,708 |
2. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then
$$
\begin{array}{l}
{\left[\frac{2009 \times 83}{2009}\right]+\left[\frac{2010 \times 83}{2009}\right]+\cdots+\left[\frac{4017 \times 83}{2009}\right]} \\
=(\quad) .
\end{array}
$$
(A) 249075
(B) 250958
(C) 174696
(D) 25... | 2.A.
Original expression
$$
\begin{aligned}
= & {\left[\frac{(2009+0) \times 83}{2009}\right]+\left[\frac{(2009+1) \times 83}{2009}\right]+} \\
& \cdots+\left[\frac{(2009+2008) \times 83}{2009}\right] \\
= & 83+\left[\frac{0 \times 83}{2009}\right]+83+\left[\frac{1 \times 83}{2009}\right]+\cdots+ \\
& 83+\left[\frac{2... | 249075 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,709 |
3. On a $2 \times 3$ rectangular grid paper, the vertices of each small square are called lattice points. Then the number of isosceles right triangles with lattice points as vertices is ( ).
(A) 24
(B) 38
(C) 46
(D) 50 | 3. D.
The lengths of line segments with grid points as vertices can take 8 values: $1, \sqrt{2}, 2, \sqrt{5}, 2 \sqrt{2}, 3, \sqrt{10}, \sqrt{13}$. The isosceles right triangles formed by these line segments can be classified into 4 cases based on their side lengths:
$$
1,1, \sqrt{2} ; \sqrt{2}, \sqrt{2}, 2 ; 2,2,2 \s... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,710 |
4. As shown in Figure 1, in equilateral $\triangle A B C$, $E$ is the midpoint of $A B$, and $D$ is on $A C$ such that $\frac{A D}{A C}=\frac{1}{3}$. Then ( ).
(A) $\triangle A E D \backsim \triangle B E D$ (B) $\triangle A E D \backsim \triangle C B D$ (C) $\triangle A E D \backsim \triangle A B D$ (D) $\triangle B A ... | 4.B.
Consider option (A): In $\triangle A E D$, $\angle A=60^{\circ}$, while $\triangle B E D$ does not have a $60^{\circ}$ angle, so we exclude it.
Consider option (C): Since $\angle E A D=\angle B A D$, and $\angle A E D>\angle E B D$, if option $(\mathrm{C})$ holds, then it must be that $\angle A E D=\angle A D B$... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,711 |
4. If a regular hexagon $T_{1}$ is inscribed in $\odot O$, and a regular hexagon $T_{2}$ is circumscribed about $\odot O$, then the ratio of the area of $T_{1}$ to the area of $T_{2}$ is ( ).
(A) $2: 3$
(B) $3: 4$
(C) $4: 5$
(D) $5: 6$ | 4.B.
As shown in Figure 3, $T_{1}$ and $T_{2}$ are respectively divided into isosceles triangles with a vertex angle of $120^{\circ}$. Then, $T_{1}$ contains 18 isosceles triangles, and $T_{2}$ contains 24 isosceles triangles, with the ratio being $18: 24=3: 4$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,712 |
5. Given four points in the Cartesian coordinate system $A(-2,4)$, $B(-2,0)$, $C(2,-3)$, and $D(2,0)$. Let $P$ be a point on the $x$-axis, and after connecting $P A$ and $P C$, the two triangles formed with $A B$, $C D$, and the $x$-axis (i.e., $\triangle P A B$ and $\triangle P C D$) are similar. Then the number of al... | 5.C.
(1) $A B$ and $C D$ are corresponding sides.
(i) $P$ is the intersection point of $A C$ and the $x$-axis. The equation of $A C$ is $y=-\frac{7}{4} x+\frac{1}{2}$, giving $x=\frac{2}{7}, y=0$.
(ii) Take the point $C$'s reflection about the $x$-axis, $C^{\prime}(2,3)$, $P$ is the intersection point of $A C^{\prime}$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,713 |
6. The positive integer solutions of the equation $-m^{4}+4 m^{2}+2^{n} m^{2}+2^{n}+5=0$ are ( ) groups.
(A) 1
(B) 2
(C) 4
(D) Infinite | 6. A.
From the factorization of the left side of the equation, we get
$$
\left(m^{2}+1\right)\left(-m^{2}+2^{n}+5\right)=0 \text {. }
$$
Therefore, $-m^{2}+2^{n}+5=0$.
So, $2^{n}+4=(m+1)(m-1)$.
Since $m+1$ and $m-1$ have the same parity, they are both even, which means $m$ is odd.
Let $m=2k+1$. Then $2^{n-2}+1=k(k+1)... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,714 |
1. As shown in Figure 2, quadrilateral $ABCD$ is a rectangle, and $AB = 2BC$. Points $M$ and $N$ are the midpoints of sides $BC$ and $CD$, respectively. $AM$ intersects $BN$ at point $E$. If the area of the shaded region is $a$, then the area of rectangle $ABCD$ is $\qquad$ | 2.1. $5 a$.
As shown in Figure 3, draw $E F \perp B C$ at point $F$.
Let $A B=4 x, E F =h$. Then
$$
C N=2 x, B M=x \text {. }
$$
From $B C=C N$, we know $\angle N B C=45^{\circ}$.
Thus, $B F=E F=h$.
So $M F=x-h$.
By $\triangle M F E \backsim \triangle M B A$ we get
$$
\begin{array}{l}
\frac{E F}{A B}=\frac{M F}{M B} \... | 5a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,715 |
2. Let $a$ and $b$ be integers, and the equation
$$
a x^{2}+b x+1=0
$$
has two distinct positive roots both less than 1. Then the minimum value of $a$ is | 2. 5 .
Let the two roots of the equation be $x_{1}, x_{2}$.
From $x_{1} x_{2}=\frac{1}{a}>0$, we know $a>0$.
Also, $f(0)=1>0$, so according to the problem, we have
$$
\left\{\begin{array}{l}
\Delta=b^{2}-4 a>0, \\
00 .
\end{array}\right.
$$
Since $a$ is a positive integer, from equations (2) and (3), we get
$-(a+1)<b... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,716 |
3. The equation about $x$
$$
x^{2}+a|x|+a^{2}-3=0\left(a \in \mathbf{R}_{+}\right)
$$
has a unique real solution. Then $a=$ $\qquad$ | 3. $\sqrt{3}$.
For the equation $x^{2}+a|x|+a^{2}-3=0$, if $x$ is a solution, then $-x$ is also a solution. According to the problem, it has a unique solution, so $x=-x$, which means the only real solution must be 0. Therefore, $a^{2}-3=0$ and $a>0$, hence $a=\sqrt{3}$. | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,717 |
4. Divide the 20 positive integers from $1 \sim 20$ into two groups $A$ and $B$, such that the sum of all numbers in group $A$ equals $N$, and the product of all numbers in group $B$ also equals $N$. Then all possible values of $N$ are $\qquad$ | 4.180, 182, 192.
Let the numbers in group $B$ be $a, b, c, d, \cdots$, and the numbers are in ascending order according to their letter sequence. Then,
$$
N=a b c d \cdots=210-(a+b+c+d+\cdots) \text {. }
$$
Therefore, if $N5$, then
$$
N>1 \times 2 \times 3 \times 4 \times 5 \times 6=720 \text {, }
$$
which contradic... | 180, 182, 192 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,718 |
One. (20 points) Find all ordered integer pairs $(a, b)$ such that the roots of the equation
$$
x^{4}+\left(2 b-a^{2}\right) x^{2}-2 a x+b^{2}-1=0
$$
are all integers. | $$
\begin{array}{l}
\Rightarrow\left\{\begin{array}{l}
x_{1}=\frac{-a-1}{2}, \\
x_{2}=\frac{-a+1}{2}, \\
x_{3}=\frac{a+3}{2}, \\
x_{4}=\frac{a-3}{2} ;
\end{array}\right. \\
\left\{\begin{array}{l}
x_{1}=\frac{-a+1}{2}, \\
x_{2}=\frac{-a-1}{2}, \\
x_{3}=\frac{a-3}{2}, \\
x_{4}=\frac{a+3}{2}
\end{array}\right. \\
\left\{... | a=2n-1, b=n^2-n-1 (n \in \mathbf{Z}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,719 |
II. (25 points) Let the reflections of the circumcenter of $\triangle ABC$ over the three sides be $A^{\prime}, B^{\prime}, C^{\prime}$. Prove:
(1) $A A^{\prime}, B B^{\prime}, C C^{\prime}$ intersect at a point $P$;
(2) Let the midpoints of the sides of $\triangle ABC$ be $A_{1}, B_{1}, C_{1}$, then $P$ is the circumc... | (1) As shown in Figure 4, let $O$ be the circumcenter of $\triangle ABC$. Connect $OA$, $OB$, $OC$, and the hexagon $A C' B A' C B'$.
Since $OA = OB = OC$ and $O, C'$ are symmetric with respect to $AB$, quadrilateral $OAC'B$ is a rhombus.
Similarly, quadrilaterals $OBA'C$ and $OCB'A$ are also rhombuses.
Thus, the side... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,720 |
Three. (25 points) There are 2009 athletes, numbered $1, 2, \cdots, 2009$. Some of them are selected to participate in the honor guard, but it must be ensured that among the remaining athletes, no one's number is equal to the product of the numbers of any two other people. How many athletes must be selected for the hon... | Three, since $45^{2}=2025>2009$, if both people's numbers are greater than or equal to 45, then their product is greater than 2009 and cannot be another person's number. If one of the two numbers is 1, then the product of 1 and the other number equals that number, which does not meet the requirement. Therefore, if the ... | 43 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,721 |
1. If $m^{2}=n+2, n^{2}=m+2(m \neq n)$, then the value of $m^{3}-2 m n+n^{3}$ is ( ).
(A) 1
(B) 0
(C) -1
(D) -2 | -.1.D.
Since $m^{2}-n^{2}=n-m \neq 0$, therefore,
$$
\begin{array}{l}
m+n=-1 . \\
\text { Also } m^{3}=m^{2} \cdot m=(n+2) m=m n+2 m, \\
n^{3}=n^{2} \cdot n=(m+2) n=m n+2 n .
\end{array}
$$
Then $m^{3}-2 m n+n^{3}=2(m+n)=-2$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,722 |
5. Let $a, b, c, d, e$ be distinct positive odd numbers. If the equation
$$
(x-a)(x-b)(x-c)(x-d)(x-e)=2009
$$
has an integer root $x$, then the last digit of $a+b+c+d+e$ is ( ).
(A) 1
(B) 3
(C) 7
(D) 9 | 5.I).
For any integer $x, x-a, x-b, x-c, x-$ $d, x-e$ are $I$ distinct integers. And expressing 2009 as the product of 7 $I$ distinct integers has only one unique form:
$$
2009=1 \times(-1) \times 7 \times(-7) \times 41.
$$
From this, $x$ is even.
Let $x=2 m$. Then
$$
\begin{array}{l}
\{2 m-a, 2 m-b, 2 m-c, 2 m-d, 2 ... | 9 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,723 |
2. The integer solutions of the inequality $0 \leqslant a x+5 \leqslant 4$ are 1, 2, 3, 4. Then the range of values for $a$ is ( ).
(A) $a \leqslant-\frac{5}{4}$
(B) $a<-1$
(C) $-\frac{5}{4} \leqslant a<-1$
(D) $a \geqslant-\frac{5}{4}$ | 2.C.
It is known that $-\frac{1}{a} \leqslant x \leqslant-\frac{5}{a}$.
According to the problem, we should have
$$
\begin{array}{l}
0-\frac{5}{a} \geqslant 4 . \\
\text { Therefore, }-\frac{5}{4} \leqslant a<-1 .
\end{array}
$$ | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 720,724 |
3. As shown in Figure 1, in the convex quadrilateral $A B C D$, $A D=D C$ $=C B, A C \neq B D$, and $\angle D A B+\angle C B A=$ $120^{\circ}$. A circle $\omega$ is drawn through points $A, B, D$. Then the point $C'$, which is the reflection of $C$ over $A B$, is ( ).
(A) on the circle $\omega$
(B) outside the circle $... | 3. A.
As shown in Figure 4, let
$$
\begin{array}{c}
\angle D A C \\
=\angle D C A=x, \\
\angle C D B \\
=\angle C B D=y . \\
\text { By } \angle D A B+ \\
\angle C B A=120^{\circ} \text {, we get } \\
\angle A D C+\angle D C B=240^{\circ} . \\
\text { Therefore }, 2(x+y)=120^{\circ}, x+y=60^{\circ} .
\end{array}
$$
T... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,725 |
4. If a sequence of numbers $a_{1}, a_{2}, \cdots$ satisfies that for any positive integer $n$ there is $a_{1}+a_{2}+\cdots+a_{n}=n^{3}$, then the value of $\frac{1}{a_{2}-1}+\frac{1}{a_{3}-1}+\cdots+\frac{1}{a_{100}-1}$ is ( ).
(A) $\frac{33}{100}$
(B) $\frac{11}{100}$
(C) $\frac{11}{99}$
(D) $\frac{33}{101}$ | 4. A.
When $n \geqslant 2$, we have
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=n^{3}, \\
a_{1}+a_{2}+\cdots+a_{n-1}=(n-1)^{3} .
\end{array}
$$
Subtracting the two equations gives $a_{n}=3 n^{2}-3 n+1$.
$$
\begin{array}{l}
\text { Then } \frac{1}{a_{n}-1}=\frac{1}{3 n(n-1)} \\
=\frac{1}{3}\left(\frac{1}{n-1}-\frac{1... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,726 |
5. Given that $x$ and $y$ are unequal natural numbers, satisfying $x^{3}+19 y=y^{3}+19 x$.
Then the integer part of $\sqrt{x^{2}+y^{2}}$ is ( ).
(A) 4
(B) 3
(C) 2
(D) 1 | 5. B.
Assume $x>y$. From the given, we have $x^{3}-y^{3}=19(x-y)$.
Since $x$ and $y$ are both natural numbers, we have $x^{2}+x<x^{2}+xy+y^{2}=19<3x^{2}$. This leads to $x=3, y=2$.
Therefore, $\sqrt{x^{2}+y^{2}}=\sqrt{13}$, and its integer part is 3. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,727 |
6. Given a convex quadrilateral $A B C D$ with $D C / / A B, A C$ $\perp B D$. Then ( ).
(A) $A B+D C < A D+B C$ | 6. B.
(1) When $D C \neq A B$, as shown in Figure 5(a), quadrilateral $A B C D$ is a trapezoid. Draw the midline $M N$. It is easy to see that
$$
A B+D C=2 M N \text {. }
$$
And $A D=2 P M, B C=2 P N(P$ is the intersection of $A C$ and $B D$), at this time,
$$
\begin{array}{l}
A B+D C=2 M N \\
<2(P M+P N)=A D+B C .
\e... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,728 |
1. Given $S=1^{2}-2^{2}+3^{2}-4^{2}+\cdots-100^{2}+$ $101^{2}$. Then the remainder when $S$ is divided by 103 is | $=.1 .1$.
Notice that
$$
\begin{array}{l}
S=1+\left(3^{2}-2^{2}\right)+\left(5^{2}-4^{2}\right)+\cdots+\left(101^{2}-100^{2}\right) \\
=1+2+3+\cdots+100+101 \\
=\frac{101 \times 102}{2}=5151=50 \times 103+1 .
\end{array}
$$
Therefore, the required remainder is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,729 |
2. The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c=$ $\qquad$ | 2. -1 .
Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$. Since $\triangle ABC$ is a right triangle, it is known that $x_{1}$ and $x_{2}$ must have opposite signs, so $x_{1} x_{2}=\frac{c}{a}<0$. By the projection theorem, we know that $|O C|^{2}=|A O| \cdot|B O|$, which means $c^{2}=\left|x_{1}\right| \cdot\... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,730 |
4. As shown in Figure 2, the diameter $A B=2 R$ of the semicircle $\odot O$, a moving point $C$ is on the semicircle. A circle is drawn with $C$ as the center and is tangent to $A B$. Then the maximum value of the shaded area $S$ enclosed by the circle, $A B$, $A C$, and $B C$ is $\qquad$ | 4. $\frac{R^{2}}{\pi}$.
As shown in Figure 6, let $A C = b, B C = a$, and construct $C D \perp A B$ at $D$. Then
$$
C D = \frac{a b}{2 R}.
$$
Thus, $S = \frac{1}{2} a b - \frac{1}{4} \pi \left( \frac{a b}{2 R} \right)^{2}$
$$
= -\frac{\pi}{16 R^{2}} (a b)^{2} + \frac{1}{2} a b.
$$
When $a b = -\frac{\frac{1}{2}}{2 \... | \frac{R^{2}}{\pi} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,732 |
One, (20 points) Given the sides of a triangle are equal to $a, b, c$. Prove:
$$
\frac{3}{2} \leqslant \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2 .
$$ | $$
\begin{array}{l}
\text { From } a<b+c, b<c+a, c<a+b \text {, we get } \\
\frac{a}{b+c}<1, \frac{b}{c+a}<1, \frac{c}{a+b}<1 . \\
\text { Then } \frac{a^{2}}{b+c}<a, \frac{b^{2}}{c+a}<b, \frac{c^{2}}{a+b}<c . \\
\text { Also, }\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)(a+b+c) \\
=\frac{a(a+b+c)}{b+c}+\frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,733 |
6. If the three sides of a right-angled triangle are positive integers, and the numerical value of its perimeter is equal to the numerical value of its area, then it is called a "standard right-angled triangle". Then, the number of standard right-angled triangles is ( ).
(A) 0
(B) 1
(C) 2
(D) infinitely many | 6. C.
Let the three sides of the right triangle $\triangle ABC$ be $x, y, z$ $(0 < x \leqslant y < z)$. Then,
$$
x + y + z = \frac{xy}{2}, \quad x^2 + y^2 = z^2.
$$
Therefore, $x^2 + y^2 = \left(\frac{xy}{2} - x - y\right)^2$.
This simplifies to $\frac{xy}{4} + 2 = x + y$, or $(x-4)(y-4) = 8$.
Thus, $x-4=1, y-4=8$
or... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,734 |
II. (25 points) As shown in Figure 3, in $\triangle ABC$, $AB=AC$, $AD$ is the altitude, points $B$ and $C$ lie on line $l_{1}$, and point $A$ lies on line $l_{2}$. It is known that $\odot O_{1}$ is tangent to $AB$, $l_{1}$, and $l_{2}$, and $\odot O_{2}$ is tangent to $AC$, $l_{1}$, and $l_{2}$. Prove that the sum of ... | $$
\begin{array}{l}
\text{Let } O_{1} E=r_{1}, O_{2} F=r_{2}. \\
\text{Let } B C=a, A B=A C=b, B E=x, C F=
\end{array}
$$
$y$. Take the incenter $I$ of $\triangle A B C$, $I$ must be on $A D$.
Let the inradius of $\triangle A B C$ be $r$. Then $I D=r$.
Also, $B D=D C=\frac{1}{2} a$.
Let $A B$ touch $\odot O_{1}$ at $K$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,735 |
Three. (25 points) Given that $p$ is a prime number greater than 3. Prove: The square of $p$ leaves a remainder of 1 when divided by 24. | Proof 1: It suffices to prove that $p^{2}-1=(p-1)(p+1)$ is divisible by 24. Since $p$ is a prime number greater than 3, $p$ is odd. Therefore, $p-1$ and $p+1$ are two consecutive even numbers, and one of them is a multiple of 4. Hence, $(p-1)(p+1)$ is divisible by 8. Furthermore, among the three consecutive integers $p... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,736 |
1. Given real numbers $x, y, z$ satisfy
$$
\left(2 x^{2}+8 x+11\right)\left(y^{2}-10 y+29\right)\left(3 z^{2}-18 z+32\right) \leqslant 60 \text {. }
$$
Then the value of $x+y-z$ is ( ).
(A) 3
(B) 2
(C) 1
(D) 0 | -.1.D.
From the given equation, we have
$$
\left[2(x+2)^{2}+3\right]\left[(y-5)^{2}+4\right]\left[3(z-3)^{2}+5\right] \leqslant 60 \text {. }
$$
Therefore, $x=-2, y=5, z=3$.
Thus, $x+y-z=-2+5-3=0$. | 0 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,737 |
2. Given that $m, n$ can each take one of $1,2, \cdots, 2009$. Then the number of pairs $(m, n)$ that make the equation $x^{2}-m x+n=0$ have real roots is ( ).
(A)676 211570
(B)676 211571
(C)676 211572
(D)676211578 | 2.A.
From the problem, we have
$$
\Delta=(-m)^{2}-4 n \geqslant 0 \text {, }
$$
which means $n \leqslant \frac{1}{4} m^{2}$.
(1) If $m$ is odd, let $m=2 k-1(k=1,2$, $\cdots, 1005$ ). Then
$$
n \leqslant \frac{1}{4}(2 k-1)^{2}=k^{2}-k+\frac{1}{4} \text {. }
$$
Thus, $n$ can take $1,2, \cdots, k^{2}-k$.
So $n$ can tak... | 676211570 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,738 |
3. As shown in Figure 1, given that $E$ is the midpoint of side $AD$ of rectangle $ABCD$, $F$ is a point on $AB$, and $EF \perp CE$. If $AB$ : $BC = \sqrt{3} : 2$, then the number of pairs of similar triangles in the figure (there are 4 triangles) is ( ).
(A) 2
(B) 3
(C) 4
(D) 6 | 3.D.
Let $BC = 2a$. Then $AE = ED = a$.
From the problem, we know $AB = CD = \sqrt{3} a$,
$$
CE = 2a, \angle ECD = 30^{\circ} \text{.}
$$
Since $CE = CB$, then $\angle BCF = \angle ECF = 30^{\circ}$.
Thus, $\angle AEF = 30^{\circ}$.
Therefore, the four triangles in the figure are pairwise similar.
Hence, there are 6 ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,739 |
4. Given integers $a, b, c, d$ satisfy
$$
27\left(3^{a}+3^{b}+3^{c}+3^{d}\right)=6888 \text {. }
$$
Then the value of $a+b+c+d$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 4.C.
Assume $a \leqslant b \leqslant c \leqslant d$. From the given equation, we have
$$
3^{a+3}\left(1+3^{b-a}+3^{c-a}+3^{d-a}\right)=3 \times 2296 \text {. }
$$
(1) If $3^{a+3}=1$, i.e., $a=-3$, then
$$
31\left(1+3^{b+3}+3^{c+3}+3^{d+3}\right)=3 \times 2296 \text {. }
$$
Since $3^{b+3} \leqslant 3^{c+3} \leqslant 3... | 6 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,740 |
6. Given positive integers $x_{1}, x_{2}, \cdots, x_{10}$ satisfying $x_{1}<x_{2}<\cdots<x_{10}$,
and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2} \leqslant 2009$.
Then the maximum value of $x_{8}-x_{5}$ is ( ).
(A) 17
(B) 18
(C) 19
(D) 20 | 6. B.
$$
\begin{array}{l}
\text { Clearly } x_{i} \geqslant i(i=1,2, \cdots, 7), \\
x_{8}+1 \leqslant x_{9}, x_{8}+2 \leqslant x_{10} .
\end{array}
$$
From the problem, we have
$$
\begin{array}{l}
1^{2}+2^{2}+\cdots+7^{2}+x_{8}^{2}+\left(x_{8}+1\right)^{2}+\left(x_{8}+2\right)^{2} \\
\leqslant 2009 .
\end{array}
$$
S... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 720,742 |
2. Given that point $E$ is the midpoint of side $B C$ of rhombus $A B C D$, $\angle A B C=30^{\circ}$, and $P$ is a point on diagonal $B D$ such that $P C + P E = \sqrt{13}$. Then the maximum area of rhombus $A B C D$ is $\qquad$ | $2.10+4 \sqrt{3}$.
As shown in Figure 3, connect $A C$, $A E$, and $A P$, and draw $A F \perp B C$ at point $F$.
From the problem, we have $A B=B C$, and $B D$ is the perpendicular bisector of $A C$. Therefore, $P A=P C$.
Also, $A E \leqslant P A+P E=P C+P E=\sqrt{13}$, so the maximum value of $A E$ is $\sqrt{13}$.
Let... | 10+4 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,744 |
1. When $x=\frac{\sqrt{21}-5}{2}$, the value of the algebraic expression
$$
x(x+1)(x+2)(x+3)(x+4)(x+5)
$$
is $\qquad$ . | $$
\text { II. 1. }-15 \text {. }
$$
From $2 x+5=\sqrt{2 \mathrm{i}}$, after rearrangement, we get
$$
\begin{array}{l}
x^{2}+5 x+1=0 . \\
\text { Therefore, } x(x+5)=-1 . \\
\text { Also, } (x+1)(x+4)=x^{2}+5 x+4=3, \\
(x+2)(x+3)=x^{2}+5 x+6=5 .
\end{array}
$$
Thus, multiplying the equations, the value sought is -15. | -15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,745 |
3. The real number $x$ satisfies
$$
\sqrt{x^{2}+3 \sqrt{2} x+2}-\sqrt{x^{2}-\sqrt{2} x+2}=2 \sqrt{2} x \text {. }
$$
Then the value of the algebraic expression $x^{4}+x^{-4}$ is $\qquad$ | 3.14.
Obviously $x \neq 0$.
Let $\sqrt{x^{2}+3 \sqrt{2} x+2}=a$,
$$
\sqrt{x^{2}-\sqrt{2} x+2}=b \text {. }
$$
Then $a^{2}-b^{2}=(a+b)(a-b)=4 \sqrt{2} x$.
Also, $a-b=2 \sqrt{2} x$, so, $a+b=2$.
Thus, $a=\sqrt{2} x+1$.
Then $(\sqrt{2} x+1)^{2}=x^{2}+3 \sqrt{2} x+2$.
Rearranging gives $x^{2}-\sqrt{2} x-1=0$, which is $x... | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,746 |
4. Given Rt $\triangle A B C$ has two medians of lengths 3 and 4. Then the length of the third median is $\qquad$ . | 4. $\sqrt{5}$ or $\sqrt{29}$.
Let the two legs of the right triangle $ABC$ be $a$ and $b$, and the hypotenuse be $c$. Then the lengths of the medians to the two legs are $\sqrt{\left(\frac{a}{2}\right)^{2}+b^{2}}, \sqrt{a^{2}+\left(\frac{b}{2}\right)^{2}}$, and the length of the median to the hypotenuse is $\frac{c}{2... | \sqrt{5} \text{ or } \sqrt{29} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,747 |
One, (20 points) If the inequality about $x$
$$
a x^{2}-4|x+1|+2 a<0
$$
has no real solutions, find the range of values for $a$.
| Given the problem, for all real numbers $x$, we have $a x^{2}-4|x+1|+2 a \geqslant 0$.
Clearly, $a \neq 0$, so $a>0$.
Let $y=a x^{2}-4|x+1|+2 a$. Then
$$
y=\left\{\begin{array}{ll}
a x^{2}-4 x+2 a-4, & x \geqslant-1 ; \\
a x^{2}+4 x+2 a+4, & x<-1.
\end{array}\right.
$$
For the function
$$
y=a x^{2}-4 x+2 a-4 \quad (x \... | a \geqslant \sqrt{3}+1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,748 |
II. (25 points) Given that $E$ is a point on side $AB$ of quadrilateral $ABCD$, $AC$ bisects $\angle ECD$, $\angle BAC = \angle BDC$, and $BD$ intersects $CE$ at point $F$. Prove:
$$
CD \cdot EF + CF \cdot EF = BF \cdot DF \text{. }
$$ | As shown in Figure 4, let $AC$ intersect $BD$ at point $M$.
$$
\text{Given } \angle BAC = \angle BDC \text{,}
$$
we know that points $A, B, C, D$ are concyclic. Therefore,
$$
\begin{array}{l}
\angle ABD = \angle ACD. \\
\text{Also, } \angle ACE = \angle ACD,
\end{array}
$$
Thus, $\angle ABD = \angle ACE$, which means
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,749 |
Three. (25 points) A positive integer $M$ has 8 positive divisors. Xiao Feng, while calculating the sum of the 8 positive divisors of $M$, forgot to add one of the divisors, resulting in a sum of 2776. Find all possible values of the positive integer $M$.
| Three, if $M$ has at least four different prime factors, then the number of positive divisors of $M$ is at least $(1+1)^{4}=16$, which contradicts the problem statement.
Therefore, $M$ has at most three different prime factors.
If all prime factors of $M$ are odd, then all positive divisors of $M$ are odd. In this case... | 2008 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,750 |
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