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
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Example 5 In a convex quadrilateral $A B C D$, let $O$ be the intersection of $A C$ and $B D$. If $B O$ is the symmedian of $\triangle A B C$, and $D O$ is the symmedian of $\triangle A D C$, prove: $A O$ is the symmedian of $\triangle A B D$.
(2006, Romanian National Team Training Test Question) | Proof As shown in Figure 5, let $T_{1}$ be the intersection of the tangent line through point $D$ to the circumcircle of $\triangle A D C$ with $A C$, and $T_{2}$ be the intersection of the tangent line through point $B$ to the circumcircle of $\triangle A B C$ with $A C$.
Since $B O$ is the symmedian of $\triangle A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,622 |
Example 6 In isosceles $\triangle ABC$, $AB=AC$, and $M$ is the midpoint of $BC$. Find the locus of point $P$ inside the triangle that satisfies
$$
\angle BPM + \angle CPA = \pi
$$
(2006, Romanian National Team Selection Test) | Proof As shown in Figure 6, let $A P$ intersect $B C$ at point $S$.
Since $\angle S P C$
$$
\begin{array}{l}
=\pi-\angle C P A \\
=\angle B P M,
\end{array}
$$
Therefore, $P S$ is the symmedian of $\triangle B P C$.
Next, we prove that the locus of point $P$ is the arc of the circle inside $\triangle A B C$ that make... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,623 |
1. If $\frac{a}{b}=20, \frac{b}{c}=10$, then the value of $\frac{a+b}{b+c}$ is ().
(A) $\frac{11}{21}$
(B) $\frac{21}{11}$
(C) $\frac{110}{21}$
(D) $\frac{210}{11}$ | - 1. D.
From the given, we have
$$
\frac{a+b}{b+c}=\frac{\frac{a}{b}+1}{1+\frac{c}{b}}=\frac{20+1}{1+\frac{1}{10}}=\frac{210}{11} .
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,624 |
2. If real numbers $a, b$ satisfy $\frac{1}{2} a - a b + b^{2} + 2 = 0$, then the range of values for $a$ is ( ).
(A) $a \leqslant -2$
(B) $a \leqslant -2$ or $a \geqslant 4$
(C) $a \geqslant 4$
(D) $-2 \leqslant a \leqslant 4$ | 2. B.
Since $b$ is a real number, the discriminant of the quadratic equation in $b$, $b^{2}-a b+\frac{1}{2} a+2=0$, is
$$
\Delta=(-a)^{2}-4 \times 1 \times\left(\frac{1}{2} a+2\right) \geqslant 0 .
$$
Solving this, we get $a \leqslant-2$ or $a \geqslant 4$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,625 |
3. As shown in Figure 1, in quadrilateral $A B C D$, $\angle B=$
$$
\begin{array}{l}
135^{\circ}, \angle C=120^{\circ}, A B \\
=2 \sqrt{3}, B C=4-2 \sqrt{2}, \\
C D=4 \sqrt{2} \text {. Then the length of side } A D
\end{array}
$$
is ( ).
(A) $2 \sqrt{6}$
(B) $4 \sqrt{6}$
(C) $4+\sqrt{6}$
(D) $2+2 \sqrt{6}$ | 3. D.
As shown in Figure 8, draw $A E$ and $D F$ perpendicular to line $B C$, with the feet of the perpendiculars being $E$ and $F$ respectively.
From the given information, we have
$$
\begin{array}{c}
B E=A E=\sqrt{6}, \\
C F=2 \sqrt{2}, D F=2 \sqrt{6} .
\end{array}
$$
Thus, $E F=4+\sqrt{6}$.
Draw $A G \perp D F$, w... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,626 |
4. In a sequence of numbers $x_{1}, x_{2}, \cdots$, it is known that $x_{1}=1$, and for $k \geqslant 2$,
$$
x_{k}=x_{k-1}+1-4\left(\left[\frac{k-1}{4}\right]-\left[\frac{k-2}{4}\right]\right),
$$
where $[a]$ denotes the greatest integer not exceeding the real number $a$ (for example, $[2.6]=2,[0.2]=0$). Then $x_{2010}... | 4. B.
From the given conditions, we have
$$
\begin{array}{l}
x_{1}=1, x_{2}=2, x_{3}=3, x_{4}=4, \\
x_{5}=1, x_{6}=2, x_{7}=3, x_{8}=4, \\
\cdots \cdots .
\end{array}
$$
Thus, in this sequence,
$$
x_{4 k+i}=x_{i}(k \in \mathbf{Z}, i=1,2,3,4) \text {. }
$$
Since $2010=4 \times 502+2$, we have $x_{2010}=2$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,627 |
5. As shown in Figure 2, in the Cartesian coordinate system $x O y$, the vertices of the isosceles trapezoid $A B C D$ have coordinates $A(1,1)$, $B(2,-1)$, $C(-2,-1)$, and $D(-1,1)$. A point $P(0,2)$ on the $y$-axis is rotated $180^{\circ}$ about point $A$ to get point $P_{1}$, $P_{1}$ is rotated $180^{\circ}$ about p... | 5. B.
From the given, we have points $P_{1}(2,0)$ and $P_{2}(2,-2)$. Let $P_{2}\left(a_{2}, b_{2}\right)$, where $a_{2}=2, b_{2}=-2$. According to the symmetry relationship, we sequentially obtain
$$
\begin{array}{l}
P_{3}\left(-4-a_{2},-2-b_{2}\right), \\
P_{4}\left(2+a_{2}, 4+b_{2}\right), \\
P_{5}\left(-a_{2},-2-b_... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,628 |
6. Given $a=\sqrt{5}-1$. Then the value of $2 a^{3}+7 a^{2}-2 a-12$ is $\qquad$ .
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | $=, 6.0$.
From the given, we have $(a+1)^{2}=5$.
Rearranging gives $a^{2}+2 a=4$.
$$
\begin{array}{l}
\text { Therefore, } 2 a^{3}+7 a^{2}-2 a-12 \\
=2 a^{3}+4 a^{2}+3 a^{2}-2 a-12 \\
=3 a^{2}+6 a-12=0 \text {. }
\end{array}
$$ | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,629 |
7. There is a bus, a truck, and a car each traveling in the same direction at a constant speed on a straight road. At a certain moment, the bus is in front, the car is at the back, and the truck is exactly in the middle between the bus and the car. After $10 \mathrm{~min}$, the car catches up with the truck; after anot... | 7. 15 .
Suppose at a certain moment, the distances between the truck and the bus, and the truck and the car are both $S \mathrm{~km}$. The speeds of the car, truck, and bus are $a, b, c(\mathrm{~km} / \mathrm{min})$, respectively, and it takes the truck $x \mathrm{~min}$ to catch up with the bus.
From the problem, we ... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,630 |
8. As shown in Figure 3, in the Cartesian coordinate system $x O y$,
the vertices of the polygon $O A B C D E$ are
$O(0,0), A(0,6)$,
$B(4,6), C(4,4)$,
$D(6,4), E(6,0)$. If
the line $l$ passes through the point $M(2,3)$, and divides the polygon $O A B C D E$
into two parts of equal area, then the function expressi... | 8. $y=-\frac{1}{3} x+\frac{11}{3}$
As shown in Figure 9, extend $BC$ to intersect the $x$-axis at point $F$, connect $OB$ and $AF$, and connect $CE$ and $DF$ to intersect at point $N$.
From the given information, $M(2,3)$ is the midpoint of $OB$ and $AF$, meaning $M$ is the center of rectangle $ABFO$. Therefore, line... | y=-\frac{1}{3} x+\frac{11}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,631 |
9. As shown in Figure 4, rays $A M$ and $B N$ are both perpendicular to line segment $A B, E$ is a point on $A M$, and a perpendicular line $A C$ is drawn from point $A$ to $B E$, intersecting $B E$ and $B N$ at points $F$ and $C$, respectively. A perpendicular line $C D$ is drawn from point $C$ to $A M$, with the foot... | 9. $\frac{\sqrt{5}-1}{2}$.
Let $F C=m, A F=n$.
Since Rt $\triangle A F B \backsim$ Rt $\triangle A B C$, we have $A B^{2}=A F \cdot A C$.
Also, since $F C=D C=A B$, we have $m^{2}=n(n+m)$,
which simplifies to $\left(\frac{n}{m}\right)^{2}+\frac{n}{m}-1=0$.
Solving this, we get $\frac{n}{m}=\frac{\sqrt{5}-1}{2}$ or $\... | \frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,632 |
10. For $i=2,3, \cdots, k$, the remainder when the positive integer $n$ is divided by $i$ is $i-1$. If the smallest value of $n$, $n_{0}$, satisfies $2000<n_{0}<3000$, then the smallest value of the positive integer $k$ is | 10.9.
Since $n+1$ is a multiple of $2,3, \cdots, k$, the smallest value of $n$, $n_{0}$, satisfies
$$
n_{0}+1=[2,3, \cdots, k],
$$
where $[2,3, \cdots, k]$ represents the least common multiple of $2,3, \cdots, k$.
$$
\begin{array}{l}
\text { Since }[2,3, \cdots, 8]=840, \\
{[2,3, \cdots, 9]=2520,} \\
{[2,3, \cdots, 1... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,633 |
11. A. As shown in Figure 5, the parabola $y=a x^{2}+b x(a>0)$ intersects the hyperbola $y=\frac{k}{x}$ at points $A$ and $B$. It is known that point $A(1,4)$, point $B$ is in the third quadrant, and the area of $\triangle A O B$ is 3 (where $O$ is the origin).
(1) Find the values of the real numbers $a$, $b$, and $k$;... | 11. A. (1) Since point $A(1,4)$ lies on the hyperbola $y=\frac{k}{x}$, we have $k=4$.
Therefore, the function expression of the hyperbola is $y=\frac{4}{x}$.
Let point $B\left(t, \frac{4}{t}\right)(t>0)$, so,
$$
\left\{\begin{array}{l}
a + b = 4, \\
4a - 2b = -2
\end{array} \Rightarrow \left\{\begin{array}{l}
a=1, \\
b... | E(8,-2) \text{ or } (2,-8) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,634 |
11. B. Let real numbers $a, b$ satisfy
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {. }
$$
Find the minimum value of $u=9 a^{2}+72 b+2$. | 11. B. From
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {, }
$$
we get $(a-2 b)(3 a-4 b+5)=0$.
Therefore, $a-2 b=0$ or $3 a-4 b+5=0$.
(1) When $a-2 b=0$,
$$
\begin{array}{l}
u=9 a^{2}+72 b+2=36 b^{2}+72 b+2 \\
=36(b+1)^{2}-34 .
\end{array}
$$
Thus, when $b=-1$, the minimum value of $u$ is -34.
(2) When $3 a-4 b+5=0$,
... | -34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,635 |
12. A. As shown in Figure 6, $\triangle ABC$ is an acute triangle, and $P, Q$ are two points on side $BC$. The circumcenters of $\triangle ABP$ and $\triangle ACQ$ are $O_{1}$ and $O_{2}$, respectively. Determine whether the intersection point $D$ of the extension of $BO_{1}$ and the extension of $CO_{2}$ lies on the c... | 12. A. The answer is negative, that is, the intersection point $D$ of the extension of $B O_{1}$ and the extension of $\mathrm{CO}_{2}$ cannot lie on the circumcircle of $\triangle A B C$.
As shown in Figure 11, let the intersection point of line $\mathrm{BO}_{1}$ and $\mathrm{CO}_{2}$ be $D$.
Then $\angle O_{1} B P$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,636 |
12. B. As shown in Figure 7, $\triangle ABC$ is an isosceles triangle, $AP$ is the altitude on the base $BC$, point $D$ is a point on the line segment $PC$, $BE$ and $CF$ are the diameters of the circumcircles of $\triangle ABD$ and $\triangle ACD$ respectively, and $EF$ is connected. Prove:
$$
\tan \angle PAD = \frac{... | 12. B. As shown in the figure,
12, connect $E D$ and $F D$.
Since $B E$ and $C F$ are both diameters, we have,
$$
\begin{array}{l}
E D \perp B C, \\
F D \perp B C .
\end{array}
$$
Therefore, points $D$, $E$, and $F$ are collinear.
Draw $A H \perp E F$, with the foot of the perpendicular at $H$. Then $A H = P D$.
Conn... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,637 |
Example 6 Divide the plane into congruent hexagonal rooms, and open doors according to the following method: if three walls meet at a point, then open one door on each of two walls, and leave the third wall without a door. Prove: no matter how winding the route taken to return to the original room, the number of doors ... | Prove that two rooms with a common door are called "adjacent". Color these rooms on the plane with two different colors so that adjacent rooms have different colors.
Notice that, to move from a room of one color to a room of the same color, one must pass through a room of the other color. Clearly, to move from any roo... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,638 |
13. A. Real numbers $a, b$ make the system of equations about $x, y$
$$
\left\{\begin{array}{l}
x y - x^{2} = 1, \\
x y^{2} + a x^{2} + b x + a = 0
\end{array}\right.
$$
have real solutions $(x, y)$.
(1) Prove: $|y| \geqslant 2$;
(2) Find the minimum value of $a^{2} + b^{2}$. | 13. A. (1) From equation (1), we know $x \neq 0$, and $y=x+\frac{1}{x}$.
Therefore, $|y|=\left|x+\frac{1}{x}\right|=|x|+\frac{1}{|x|} \geqslant 2$.
(2) Substituting $x^{2}=x y-1$ into equation (2) gives $x\left(y^{2}+a y+b\right)=0$.
Thus, $y^{2}+a y+b=0$.
Since the system of equations has real solutions, the equation
... | \frac{16}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,639 |
13. B. Find all prime numbers $p$ and positive integers $m$ that satisfy $2 p^{2}+p+8=m^{2}-2 m$. | 13. B. From the given, we have
$$
p(2 p+1)=(m-4)(m+2) \text {. }
$$
Thus, $p \mid(m-4)(m+2)$.
Since $p$ is a prime number, we have
$$
p \mid(m-4) \text { or } p \mid(m+2) \text {. }
$$
(1) If $p \mid(m-4)$, let $m-4=k p$ (where $k$ is a positive integer). Then, $m+2>k p$.
Thus, $3 p^{2}>p(2 p+1)$
$$
=(m-4)(m+2)>k^{2} ... | p=5, m=9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,640 |
14. A. From the 2010 positive integers $1,2, \cdots, 2010$, what is the maximum number of integers that can be selected such that the sum of any three selected numbers is divisible by 33? | 14. A. First, the following 61 numbers: $11, 11+33, 11+2 \times 33, \cdots, 11+60 \times 33$ (i.e., 1991) satisfy the conditions of the problem.
On the other hand, let $a_{1}<a_{2}<\cdots<a_{n}$ be the numbers selected from 1, 2, $\cdots, 2010$ that satisfy the conditions of the problem. For any four numbers $a_{i}, a... | 61 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,641 |
14. B. Color the five sides and five diagonals of the convex pentagon $A B C D E$, such that any two segments sharing a common vertex are of different colors. Find the minimum number of colors needed. | 14. B. Since vertex $A$ is the common point of four segments $A B$, $A C$, $A D$, and $A E$, at least four colors are needed.
If only four colors are used, let's assume they are red, yellow, blue, and green. Then, the four segments extending from each vertex must include one each of red, yellow, blue, and green. There... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,642 |
1. Simplify $\frac{\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}$ to get ( ).
(A) $\sqrt{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) 2
(D) $\frac{1}{2}$ | - 1. D.
Notice
$$
\begin{array}{l}
13+\sqrt{48}=(2 \sqrt{3})^{2}+1+4 \sqrt{3} \\
=(1+2 \sqrt{3})^{2}, \\
5-(1+2 \sqrt{3})=4-2 \sqrt{3}=(\sqrt{3}-1)^{2}, \\
3+(\sqrt{3}-1)=2+\sqrt{3}=\frac{4+2 \sqrt{3}}{2} \\
=\frac{(\sqrt{3}+1)^{2}}{2}=\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)^{2} .
\end{array}
$$
Therefore, the origi... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,643 |
2. As shown in Figure 1, $\triangle A B C$ is an isosceles right triangle, and quadrilateral $D E F G$ is an inscribed square. $H$ is the intersection of the diagonals of the square. Then, among the triangles formed by the line segments in the figure, the number of pairs of congruent triangles is $(\quad)$.
(A) 12
(B) ... | 2. C.
Let $AB=3$. In Figure 1, all triangles are isosceles right triangles, where there are 5 triangles with a hypotenuse of length 1, forming 10 pairs of congruent triangles; there are 6 triangles with a hypotenuse of length $\sqrt{2}$, forming 15 pairs of congruent triangles; there are 2 triangles with a hypotenuse ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,644 |
$$
\begin{array}{ll}
& \text { 3. Let } a b \neq 0 \text {, and the function } \\
& f_{1}(x)=x^{2}+2 a x+4 b \\
\text { and } & f_{2}(x)=x^{2}+4 a x+2 b
\end{array}
$$
have the same minimum value $u$, and the functions
$$
\begin{aligned}
f_{3}(x) & =-x^{2}+2 b x+4 a \\
\text { and } f_{4}(x) & =-x^{2}+4 b x+2 a
\end{... | 3. C.
$$
\begin{array}{c}
\text { Given } f_{1}(x)=(x+a)^{2}+4 b-a^{2} \geqslant 4 b-a^{2}, \\
f_{2}(x)=(x+2 a)^{2}+2 b-4 a^{2} \geqslant 2 b-4 a^{2},
\end{array}
$$
Also, $4 b-a^{2}=u=2 b-4 a^{2}$, then
$$
-2 b=3 a^{2} \text {. }
$$
Given $f_{3}(x)=-(x-b)^{2}+4 a+b^{2} \leqslant 4 a+b^{2}$,
$$
f_{4}(x)=-(x-2 b)^{2}+... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,645 |
4. If the equation with respect to $x$
$$
x^{2}+2 a x+7 a-10=0
$$
has no real roots, then the equation that must have real roots is ( ).
(A) $x^{2}+2 a x+3 a-2=0$
(B) $x^{2}+2 a x+5 a-6=0$
(C) $x^{2}+2 a x+10 a-21=0$
(D) $x^{2}+2 a x+2 a+3=0$ | 4. A.
From the equation $x^{2}+2 a x+7 a-10=0$ having no real roots, we get its discriminant $\Delta<0$, but this does not guarantee that $\Delta_{B} 、 \Delta_{C} 、 \Delta_{D}$ are non-negative.
Therefore, option $(A)$ must have real roots, while the other equations may not necessarily have real roots. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,646 |
5. As shown in Figure 2, in square $ABCD$, $E$ and $F$ are points on sides $AB$ and $BC$ respectively. $DE$ intersects $AC$ at point $M$, and $AF$ intersects $BD$ at point $N$. If $AF$ bisects $\angle BAC$, $DE \perp AF$, and let $x=\frac{BE}{OM}$, $y=\frac{BN}{ON}$, $z=\frac{CF}{BF}$, then ( ).
(A) $x>y>z$
(B) $x=y=z$... | 5. D.
By the Angle Bisector Theorem, we have
$$
\frac{B N}{O N}=\frac{A B}{A O}=\sqrt{2}=\frac{A C}{A B}=\frac{C F}{B F},
$$
which means $y=z=\sqrt{2}$.
Since the angle bisector of $\triangle A M E$ coincides with its altitude, $\triangle A M E$ is an isosceles triangle, i.e., $A M=A E$.
As shown in Figure 4, draw $O... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,647 |
6. Place the numbers $1,2, \cdots, 8$ at the eight equally spaced points on a circle such that the sum of any two adjacent numbers is a prime number. If rotations of the circle that result in the same arrangement are considered identical, then the number of different arrangements is ( ).
(A) 4
(B) 8
(C) 12
(D) 16 | 6. A.
If the sum of two adjacent numbers is an odd prime, then the numbers on the circle alternate between odd and even.
Thus, the numbers adjacent to 8 must be 3, 5, 7, and the numbers adjacent to 3, 5, 7 must be 4, 6. The remaining two numbers, 1 and 2, must be adjacent to each other, and 1 must be adjacent to eith... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 722,648 |
1. If the sum of $k$ consecutive positive integers is 2010, then the maximum value of $k$ is | Ni. 1.60 .
Let $2010=(n+1)+(n+2)+\cdots+(n+k)$.
Then $k(2 n+k+1)=4020$.
Notice that $k<2 n+k+1$, and
$$
4020=2^{2} \times 3 \times 5 \times 67 \text {, }
$$
To maximize the value of $k$, 4020 should be expressed as the product of the closest pair of factors, which is $4020=60 \times 67$.
Thus, $k_{\max }=60$. | 60 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,649 |
3. For the cyclic quadrilateral $ABCD$, the lengths of the four sides in sequence are $AB=2, BC=7, CD=6, DA=9$. Then the area of the quadrilateral is $\qquad$ . | 3. 30 .
Since $7^{2}+6^{2}=85=9^{2}+2^{2}$, that is,
$$
B C^{2}+C D^{2}=D A^{2}+A B^{2} \text {, }
$$
thus, $\triangle B C D$ and $\triangle D A B$ are both right triangles.
Therefore, the area of the quadrilateral is
$$
S_{\triangle B C D}+S_{\triangle D A B}=\frac{1}{2}(7 \times 6+9 \times 2)=30 \text {. }
$$ | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,651 |
4. In $\pm 1 \pm 2 \pm 3 \pm 5 \pm 20$, by appropriately choosing + or -, different algebraic sums can be obtained $\qquad$.
| 4.24.
Among $1,2,3,5,20$, there are three odd numbers, so their algebraic sum must be odd.
Observing, we see that from $1,2,3,5$ we can obtain all odd numbers with absolute values not exceeding 11.
According to the problem, the expression must include 1, 2, 3, 5, and 20. Therefore, the integers that can be obtained ... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,652 |
One, (20 points) For a right-angled triangle with integer side lengths, if the lengths of its two legs are the roots of the equation
$$
x^{2}-(k+2) x+4 k=0
$$
find the value of $k$ and determine the lengths of the three sides of the right-angled triangle. | Given the right-angled sides are $a$ and $b$ (both have the same parity), so,
$$
\left\{\begin{array} { l }
{ k - 6 + n = 1 6 , } \\
{ k - 6 - n = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
k-6+n=8, \\
k-6-n=4 .
\end{array}\right.\right.
$$
Solving gives $k=15$ or 12.
When $k=15$,
$$
\begin{array}{l}
a+b=17... | k=15 \text{ or } 12, \text{ sides: } (5, 12, 13) \text{ or } (6, 8, 10) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,653 |
II. (25 points) As shown in Figure 3, from any point $P$ inside $\triangle ABC$, draw $PD \perp BC, PE \perp CA, PF \perp AB$. If $BD = BF, CD = CE$, prove: $AE = AF$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | II. Note the following fact: If the two diagonals of a quadrilateral are perpendicular to each other, then the sum of the squares of the two pairs of opposite sides are equal.
As shown in Figure 5, connect
$P A, P B, P C$. Then
$$
\begin{array}{l}
P A^{2}+B F^{2} \\
=P B^{2}+A F^{2}, \\
P B^{2}+C D^{2} \\
=P C^{2}+B D^... | null | Geometry | proof | Yes | Yes | cn_contest | false | 722,654 |
Three. (25 points) Given that $a$, $b$, and $c$ are positive integers, and $\frac{\sqrt{3} a+b}{\sqrt{3} b+c}$ is a rational number. Prove: $\frac{a^{2}+b^{2}+c^{2}}{a+b+c}$ is an integer. | Three, since $\sqrt{3}$ is an irrational number, therefore, $\sqrt{3} b-c \neq 0$.
$$
\begin{array}{l}
\text { and } \frac{\sqrt{3} a+b}{\sqrt{3} b+c}=\frac{(\sqrt{3} a+b)(\sqrt{3} b-c)}{3 b^{2}-c^{2}} \\
=\frac{3 a b-b c+\sqrt{3}\left(b^{2}-a c\right)}{3 b^{2}-c^{2}}
\end{array}
$$
If it is a rational number, then $b... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,655 |
1. Let $a_{1}, a_{2}, \cdots, a_{10} \in(1,+\infty)$. Then
$$
\frac{\log _{a_{1}} 2009+\log _{a_{2}} 2009+\cdots+\log _{a_{10}} 2009}{\log _{a_{1,2} \cdots a_{10}} 2009}
$$
the minimum value is | -1.100 .
$$
\begin{array}{l}
\text { Original expression }=\left(\sum_{i=1}^{10} \frac{\lg 2009}{\lg a_{i}}\right) \frac{\lg \left(\prod_{i=1}^{10} a_{i}\right)}{\lg 2009} \\
=\left(\sum_{i=1}^{10} \frac{1}{\lg a_{i}}\right)\left(\sum_{i=1}^{10} \lg a_{i}\right) \\
\geqslant 10^{2},
\end{array}
$$
Equality holds if an... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,656 |
2. Given $x, y \in \mathbf{N}_{+}$, and
$$
1+2+\cdots+y=1+9+9^{2}+\cdots 9^{x-1} \text {. }
$$
Then express $y$ as a function of $x$, the analytical expression is
$$
y=
$$
$\qquad$ | 2. $\frac{3^{x}-1}{2}$.
From the given equation, we have
$$
\frac{y(y+1)}{2}=\frac{9^{x}-1}{9-1} \text {. }
$$
Thus, $y(y+1)=\frac{3^{x}+1}{2} \cdot \frac{3^{x}-1}{2}$
$$
\Rightarrow y=\frac{3^{x}-1}{2} .
$$ | \frac{3^{x}-1}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,657 |
3. Given the function $f(x)=\left|x^{2}-2\right|$. If $f(a)=$ $f(b)$, and $0<a<b$, then the range of values for $a b$ is | 3. $(0,2)$.
From Figure 2, based on the graph of $y=f(x)$, the only values that satisfy $f(a)=f(b)$ and $0<a<b$ are
$$
\begin{array}{c}
0<a<\sqrt{2} \\
<b<2,
\end{array}
$$
and $f(a)=2-a^{2}, f(b)=b^{2}-2$.
Thus, $2-a^{2}=b^{2}-2$, which simplifies to $a^{2}+b^{2}=4$.
Therefore, $a$ and $b$ can be expressed as
$$
b=2... | (0,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,658 |
4. All real pairs $(x, y)=$ $\qquad$ that satisfy the equation,
$$
\log _{2}\left[2 \cos ^{2}(x y)+\frac{1}{2 \cos ^{2}(x y)}\right]=-y^{2}+y+\frac{3}{4}
$$ | 4. $\left(k \pi+\frac{\pi}{2}, \frac{1}{2}\right)(k \in \mathbf{Z})$.
From $2 \cos ^{2}(x y)>0$, we get
the left side $\geqslant \log _{2} 2=1$.
And the right side $=-\left(y-\frac{1}{2}\right)^{2}+1 \leqslant 1$.
Therefore, the left side must equal the right side, both equal to 1.
$$
\begin{array}{l}
\text { Hence } ... | \left(k \pi+\frac{\pi}{2}, \frac{1}{2}\right)(k \in \mathbf{Z}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,660 |
5. If $[a]$ denotes the greatest integer not exceeding the real number $a$, then the solution to the equation $[\tan x]=2 \sin ^{2} x$ is $\qquad$ . | 5. $x=k \pi$ or $l \pi+\frac{\pi}{4}(k, l \in \mathbf{Z})$.
If $[\tan x]=0$, then
$$
\sin x=0 \Rightarrow x=k \pi(k \in \mathbf{Z}) ;
$$
If $[\tan x]=1$, then
$$
\sin x= \pm \frac{\sqrt{2}}{2} \Rightarrow x=l \pi+\frac{\pi}{4}(l \in \mathbf{Z}) ;
$$
If $[\tan x]=2$, then
$$
\sin x= \pm 1,
$$
the original equation h... | x=k \pi \text { or } l \pi+\frac{\pi}{4}(k, l \in \mathbf{Z}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,661 |
6. The solution set of the inequality $2^{2 x} \leqslant 3 \times 2^{x+\sqrt{x}}+4 \times 2^{2 \sqrt{x}}$ is $\qquad$ . | $\begin{array}{l}\text { 6. }[0,4] \\ 4 \times 2^{2 \sqrt{x}}+3 \times 2^{x+\sqrt{x}} \geqslant 2^{2 x} \\ \Leftrightarrow 4+3 \times 2^{x-\sqrt{x}} \geqslant 2^{2 x-2 \sqrt{x}} \\ \Leftrightarrow\left(2^{x-\sqrt{x}}\right)^{2}-3 \times 2^{x-\sqrt{x}}-4 \leqslant 0 \\ \Leftrightarrow-1 \leqslant 2^{x-\sqrt{x}} \leqslan... | [0,4] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,662 |
7. Let $A$ be the set of all positive integers not exceeding 2009, i.e., $A=\{1,2, \cdots, 2009\}$, and let $L \subseteq A$, where the difference between any two distinct elements of $L$ is not equal to 4. Then the maximum possible number of elements in the set $L$ is | 7. 1005.
Divide the set $A$ into the following 1005 subsets:
$$
\begin{array}{l}
A_{4 k+i}=\{8 k+i, 8 k+i+4\}(i=1,2,3,4 ; \\
k=0,1, \cdots, 250), \\
A_{1005}=\{2009\} .
\end{array}
$$
If the number of elements in $L$ is greater than 1005, then at least one of the first 1004 subsets is a subset of $L$, meaning there ... | 1005 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,663 |
8. In a plane, given a convex decagon and all its diagonals, in such a graph, the number of triangles that have at least two vertices as vertices of the convex decagon is $\qquad$ (answer with a number). | 8. 960 .
The number of triangles with all three vertices being vertices of a convex decagon is $\mathrm{C}_{10}^{3}$.
The number of triangles with only two vertices being vertices of the convex decagon, and the other vertex being the intersection of two diagonals, can be determined as follows: Two diagonals define fo... | 960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,664 |
9. (14 points) Let the function $f(x)$ be defined on the closed interval
$[0,1]$, satisfying $f(0)=0, f(1)=1$, and for any $x, y \in[0,1](x \leqslant y)$, we have
$$
f\left(\frac{x+y}{2}\right)=\left(1-a^{2}\right) f(x)+a^{2} f(y),
$$
where the constant $a$ satisfies $0<a<1$. Find the value of $a$. | $$
\begin{array}{l}
f\left(\frac{1}{2}\right)=f\left(\frac{0+1}{2}\right) \\
=\left(1-a^{2}\right) f(0)+a^{2} f(1)=a^{2}, \\
f\left(\frac{1}{4}\right)=f\left(\frac{0+\frac{1}{2}}{2}\right) \\
=\left(1-a^{2}\right) f(0)+a^{2} f\left(\frac{1}{2}\right)=a^{4}, \\
f\left(\frac{3}{4}\right)=f\left(\frac{\frac{1}{2}+1}{2}\ri... | a=\frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,665 |
10. (14 points) As shown in Figure 1, $A$ is the right vertex of the hyperbola $\frac{x^{2}}{4}-y^{2}=1$. Two perpendicular lines passing through $A$ intersect the right branch of the hyperbola at points $M$ and $N$, respectively. Is the line $MN$ guaranteed to pass through a fixed point on the $x$-axis? If such a fixe... | 10. Clearly, the right vertex $A(2,0)$.
Translate the $y$-axis to the right by 2 units, making $A$ the origin of the new rectangular coordinate system. In the new coordinate system, the equation of the hyperbola is
$$
\frac{\left(x^{\prime}+2\right)^{2}}{4}-y^{2}=1 \text {, }
$$
which simplifies to $4 y^{2}-x^{\prime... | \left(\frac{10}{3}, 0\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,666 |
11. (16 points) Let $A$ and $B$ be two different subsets of the set $\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$, such that $A$ is not a subset of $B$, and $B$ is not a subset of $A$. Find the number of different ordered pairs $(A, B)$. | 11. The set $\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$ has $2^{5}$ subsets, and the number of different ordered pairs $(A, B)$ is $2^{5}\left(2^{5}-1\right)$.
If $A \subset B$, and suppose $B$ contains $k(1 \leqslant k \leqslant 5)$ elements. Then the number of ordered pairs $(A, B)$ satisfying $A \subset B$ i... | 570 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,667 |
12. (16 points) Let the sequence of positive integers $\left\{a_{n}\right\}$ satisfy
$$
a_{10 k-9}+a_{10 k-8}+\cdots+a_{10 k} \leqslant 19
$$
Let the sum $\sum_{p=i+1}^{j} a_{p}$ be $S(i, j)$, where $i, j \in \mathbf{N}_{+}$, and $i<j$. Prove: The set of all $S(i, j)$ is equal to $\mathbf{N}_{+}$. | 12. Clearly, $S(i, j) \in \mathbf{N}_{+}$.
Next, we prove: For any $n_{0} \in \mathbf{N}_{+}$, there exists
$$
S(i, j)=n_{0} \text {. }
$$
Let $S_{n}$ denote the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Consider the first $10 n_{0}+10$ partial sums
$$
S_{1}i$, and
$$
n_{0}=S_{j}-S_{i}=S(i, j... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,668 |
1. Let the set
$$
A=\left\{5, \log _{2}(a+3)\right\}, B=\{a, b\}(a, b \in \mathbf{R}) \text {. }
$$
If $A \cap B=\{1\}$, then $A \cup B=$ $\qquad$ | $-、 1 .\{5,1,-1\}$.
Since $A \cap B=\{1\}$, therefore, $\log _{2}(a+3)=1$
Solving for $a$ gives $a=-1$. Thus, $b=1$.
Hence, $A \cup B=\{5,1,-1\}$. | \{5,1,-1\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,669 |
2. As shown in Figure 1, in the first quadrant, the three vertices $A, B, C$ of rectangle $A B C D$ are on the graphs of the functions $y=$ $\log _{\frac{\sqrt{2}}{2}} x$, $y=x^{\frac{1}{2}}$, and $y=\left(\frac{\sqrt{3}}{2}\right)^{x}$, respectively, and the sides of the rectangle are parallel to the coordinate axes. ... | 2. $\left(\frac{1}{2}, \frac{9}{16}\right)$.
Substitute $y_{A}=2$ into $y=\log _{\frac{\sqrt{2}}{2}} x$, we get $x_{A}=\frac{1}{2}$.
Thus, $x_{D}=\frac{1}{2}$.
Since $A B / / x$ axis, therefore, $y_{B}=y_{A}=2$.
Substitute into $y=x^{\frac{1}{2}}$, we get $x_{B}=4$. Thus, $x_{C}=4$.
Substitute $x_{c}=4$ into $y=\left(... | \left(\frac{1}{2}, \frac{9}{16}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,670 |
3. Four congruent right triangles can be assembled to form the square string figure shown in Figure 2(a), or the rhombus shown in Figure 2(b). If the area of the large square in Figure 2(a) is 100, and the area of the small square is 4, then the cosine value of one acute angle in the rhombus in Figure 2(b) is | 3. $\frac{7}{25}$.
Let the three sides of a right-angled triangle be $a, b, c$ $(a<b<c)$, and the smallest acute angle be $\theta$. Then the acute angle of the rhombus is $2 \theta$.
According to the problem, we have
$$
\left\{\begin{array} { l }
{ c ^ { 2 } = 1 0 0 , } \\
{ ( b - a ) ^ { 2 } = 4 , } \\
{ a ^ { 2 } +... | \frac{7}{25} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,671 |
4. The body diagonal of a rectangular prism is 10, and the projection of this diagonal on one of the surfaces of the prism is 8. Then the maximum volume of this rectangular prism is $\qquad$ . | 4. 192 .
According to the problem, the height of the cuboid is $\sqrt{10^{2}-8^{2}}=6$. Let the side lengths of the base of the cuboid be $a$ and $b$. Then $a^{2}+b^{2}=64$.
Thus, the volume of the cuboid is $V=6 a b \leqslant 3\left(a^{2}+b^{2}\right)=192$.
The equality holds if and only if $a=b=4 \sqrt{2}$. Therefor... | 192 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,672 |
5. Choose three vertices from the eight vertices of a cube. Then the probability that these three points form a right-angled triangle is | 5. $\frac{6}{7}$.
Solution 1: From the eight vertices of a cube, three can be chosen, with a total of $\mathrm{C}_{8}^{3}=56$ ways to choose.
A cube has six faces and six diagonal planes, all of which are rectangles (including squares). In each rectangle, there are four right-angled triangles, so there are a total of... | \frac{6}{7} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,673 |
6. Let $a$, $b$, $c$ be three prime numbers, and satisfy $a b c=5(a+b+c)$.
Then these three numbers in ascending order are $\qquad$ | 6. $2,5,7$.
From $a b c=5(a+b+c)$ and $a, b, c$ are all prime numbers, we know that one of these three numbers must be equal to 5.
Without loss of generality, let $a=5$, and $b>c$. Then $b c=5+b+c$,
which simplifies to $(b-1)(c-1)=6$.
Thus, $(b, c)=(4,3)$ (discard), $(7,2)$.
Therefore, these three numbers in ascendin... | 2,5,7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,674 |
7. If the equation about $x$
$$
x^{3}+a x^{2}+b x-4=0\left(a 、 b \in \mathbf{N}_{+}\right)
$$
has a positive integer solution, then $|a-b|=$ | 7.1.
Solution 1 Let $m$ be a positive integer solution of the equation.
If $m \geqslant 2$, then $a m^{2}+b m=4-m^{3}<0$, which contradicts that $a$ and $b$ are both positive integers.
Therefore, only $m=1$. Substituting it in, we get $a+b=3$.
Since $a, b \in \mathbf{N}_{+}$, $\{a, b\}=\{1,2\}$.
Thus, $|a-b|=1$.
Solut... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,675 |
8. For any positive integer $n$, we have
$$
\begin{array}{l}
\frac{1}{n}\left[\sqrt{1-\left(\frac{1}{n}\right)^{2}}+\sqrt{1-\left(\frac{2}{n}\right)^{2}}+\cdots+\sqrt{1-\left(\frac{n-1}{n}\right)^{2}}\right] \\
<a<\frac{1}{n}\left[1+\sqrt{1-\left(\frac{1}{n}\right)^{2}}+\cdots+\sqrt{1-\left(\frac{n-1}{n}\right)^{2}}\ri... | 8. $\frac{\pi}{4}$.
Construct the function $y=\sqrt{1-x^{2}}(x \geqslant 0)$, its graph is the part of the unit circle in the first quadrant, as shown in Figure 4.
Let $A(1,0)$ and $B(0,1)$. Divide the line segment $O A$ into $n$ equal parts, and draw perpendiculars from each of these points and point $A$ to the $x$-... | \frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 722,676 |
9. (16 points) In $\triangle A B C$, $\overrightarrow{A B} \cdot \overrightarrow{A C}=8$, let $\angle B A C=\theta$, the area of $\triangle A B C$ is $S$, and it satisfies
$$
4(2-\sqrt{3}) \leqslant S \leqslant 4 \sqrt{3} \text { : }
$$
(1) Find the range of $\theta$;
(2) Find the maximum and minimum values of the func... | (1) From $\overrightarrow{A B} \cdot \overrightarrow{A C}=8$, we get $|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \cos \theta=8$.
Since $4(2-\sqrt{3}) \leqslant \frac{1}{2}|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \sin \theta \leqslant 4 \sqrt{3}$,
it follows that $2-\sqrt{3} \leqslant \tan \theta \leqs... | f(\theta)_{\max }=3, \quad f(\theta)_{\min }=2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,677 |
10. (20 points) Let the center of a moving circle be on the parabola $y=\frac{1}{4} x^{2}$, and the radius of the circle be equal to the y-coordinate of the center of the circle. Find the set of all points on such circles.
保留源文本的换行和格式，翻译结果如下：
10. (20 points) Let the center of a moving circle be on the parabola $y=\fr... | 10. Let the center of a moving circle be \( C\left(a, \frac{1}{4} a^{2}\right) \), and \( P(x, y) \) be any point on \( \odot C \). Then
\[
(x-a)^{2}+\left(y-\frac{1}{4} a^{2}\right)^{2}=\left(\frac{1}{4} a^{2}\right)^{2},
\]
which simplifies to \((2-y) a^{2}-4 x a+2\left(x^{2}+y^{2}\right)=0\).
When \( y=2 \), \( a=\f... | \left\{(x, y) \mid x^{2}+(y-1)^{2} \geqslant 1, y \geqslant 0\right\} \backslash\{(0,2)\} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,678 |
One, (30 points) As shown in Figure 3, in the convex quadrilateral $A B C D$, $\angle A B C=\angle A D C$, $E, F, G, H$ are the midpoints of $A C, B D, A D, C D$ respectively. Prove:
$$
\text { (1) } E, F, G, H
$$
are concyclic;
$$
\begin{array}{l}
\text { (2) } \angle A E F \\
=\angle A C B-\angle A C D .
\end{array}... | (1) As shown in Figure 5, connect $E G$, $E H$, $F G$, $F H$, and $G H$. Then,
$F G \parallel B A$,
$F H \parallel B C$.
Therefore, $\angle G F H = \angle A B C$.
Similarly, $\angle G H F = \angle A C B$.
Since quadrilateral $D G E H$ is a parallelogram, therefore,
$\angle G E H = \angle A D C = \angle A B C = \angle G... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,679 |
2. Prove: In a plane that is colored with two colors, there must exist a rectangle whose four vertices are the same color. | Take three horizontal lines and nine vertical lines. Consider only the 27 intersection points of these lines. Since each column of three points is colored in two colors, there are $2^{3}=8$ different coloring methods, so among the nine vertical lines, there exist two lines whose intersection points have the same colori... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,680 |
II. (30 points) Given the function
$$
f(x)=\log _{a} \frac{x-3}{x+3}(a>0 \text {, and } a \neq 1) \text {. }
$$
If there exist real numbers $m, n (m<n)$ and $a$, such that the domain of $f(x)$ is $(m, n)$, and the range is
$$
\left(1+\log _{a}(n-1), 1+\log _{a}(m-1)\right),
$$
find the ranges of $m$ and $a$ respectiv... | From $\frac{x-3}{x+3}>0$, we get the range of $x$ as
$(-\infty,-3) \cup(3,+\infty)$.
Since the domain of $f(x)$ is $(m, n)$, and $m>1$, $n>1$, it follows that $m>3$.
Also, $m-13$.
The range of $f(x)$ is
$$
\left(1+\log _{a}(n-1), 1+\log _{a}(m-1)\right) \text {, }
$$
Then $\log _{a} a(n-1)0, \\
g(3)=12 a>0, \\
-\frac{... | m \in (3,+\infty), a \in \left(0, \frac{2-\sqrt{3}}{4}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,681 |
Three. (40 points) Remove all terms containing the digit 9 from the sequence of positive integers $1, 2, \cdots$, and let the remaining terms, in their original order, form a new sequence denoted as $a_{1}, a_{2}, \cdots, a_{n}, \cdots$. Prove:
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<80 .
$$ | Three, it is known that the number of natural numbers from 1 to $10^{n}$ that do not contain the digit 9 is $9^{n}$. Therefore, the number of natural numbers from $10^{n-1}$ to $10^{n}-1$ that do not contain the digit 9 is $8 \times 9^{n-1}$.
Let the sum of the reciprocals of the natural numbers from $10^{n-1}$ to $10... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,682 |
1. If $x^{2}-x-2=0$, then
$$
\frac{x^{2}-x+2 \sqrt{3}}{x^{4}-2 x^{3}+x^{2}-1+\sqrt{3}}
$$
is equal to ( ).
(A) $\frac{2 \sqrt{3}}{3}$
(B) $\frac{\sqrt{3}}{3}$
(C) $\sqrt{3}$
(D) $\sqrt{3}$ or $\frac{\sqrt{3}}{3}$ | - 1. A.
From $x^{2}-x=2$, we get
the original expression $=\frac{x^{2}-x+2 \sqrt{3}}{\left(x^{2}-x\right)^{2}-1+\sqrt{3}}=\frac{2 \sqrt{3}}{3}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,683 |
2. Let $a$, $b$, $c$ be integers, and for all real numbers $x$, we have
$$
(x-a)(x-2009)-2=(x-b)(x-c)
$$
holds. Then the number of all such ordered triples $(a, b, c)$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 2. B.
From the problem, we have
$$
\begin{array}{l}
x^{2}-(a+2009) x+2009 a-2 \\
=(x-b)(x-c)=0 . \\
\text { Hence } \Delta=(a+2009)^{2}-4(2009 a-2) \\
=(a-2009)^{2}+8
\end{array}
$$
is a perfect square, let it be $n^{2}\left(n \in \mathbf{N}_{+}\right)$. Then
$$
(n-a+2009)(n+a-2009)=8 \text {. }
$$
Since $n-a+2009$ ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,684 |
3. In an acute triangle $\triangle ABC$, $BE$ and $CF$ are two altitudes. Let the area of $\triangle AEF$ be $S_{1}$, and the area of quadrilateral $BCEF$ be $S_{2}$. If the diameter of the circumcircle of $\triangle AEF$ is exactly equal to the length of $BC$, then the relationship between $S_{1}$ and $S_{2}$ is ( ).
... | 3. B.
As shown in Figure 2, draw the altitude $A D$, and let $H$ be the orthocenter. It is easy to see that points $A, F, H, E$ are concyclic, with $A H$ as the diameter of the circle. Then
$$
\begin{array}{l}
A H=B C . \\
\text { Hence } \triangle A H E \cong \triangle B C E \\
\Rightarrow A E=B E \Rightarrow \angle ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,685 |
4. The real number $b$ satisfies $|b|<3$, and there exists a real number $a$, such that $a<b$ always holds. Then the range of values for $a$ is ( ).
(A) $(-\infty, 3]$
(B) $(-\infty, 3)$
(C) $(-\infty,-3]$
(D) $(-\infty,-3)$ | 4. C.
From $|b|<3 \Rightarrow -3<b<3$.
Since $a<b$ always holds, hence $a \leqslant -3$. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 722,686 |
5. The minimum value of the algebraic expression $\sqrt{x^{2}+4}+\sqrt{x^{2}-24 x+153}$ is ( ).
(A) 12
(B) 13
(C) 14
(D) 15 | 5. B.
As shown in Figure 3, let $AB = 12$, $DA \perp AB$, $EB \perp AB$, and $DA = 2$, $EB = 3$. Then,
$$
\begin{aligned}
DC & =\sqrt{x^{2}+2^{2}}, \\
CE & =\sqrt{(x-12)^{2}+3^{2}} .
\end{aligned}
$$
Construct the point $D'$, which is the reflection of point $D$ over $AB$. Then,
$$
D' C = D C.
$$
Clearly, when point... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,687 |
6. $\odot O$ has a radius of $r$, and its circumscribed right trapezoid $A B C D$ has two bases $A B=a, D C=b$. Then the relationship between $r, a, b$ is ( ).
(A) $r=|a-b|$
(B) $r=\frac{1}{2} \sqrt{\left|a^{2}-b^{2}\right|}$
(C) $r=\frac{1}{2} \sqrt{a b}$
(D) $\frac{1}{r}=\frac{1}{a}+\frac{1}{b}$ | 6. D.
As shown in Figure 4, draw the altitude $C E$. Then
$$
\begin{array}{l}
C E=2 r, \\
E B=|a-b|, \\
B C=a+b-2 r . \\
\text { By }(a+b-2 r)^{2} \\
=B C^{2}=C E^{2}+E B^{2}=(2 r)^{2}+(a-b)^{2},
\end{array}
$$
we get $r(a+b)=a b$.
Thus, $\frac{1}{r}=\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,688 |
2. Given $|x| \leqslant 1,|y| \leqslant 1$, and
$$
k=|x+y|+|y+1|+|2 y-x-4| \text {. }
$$
Then the sum of the maximum and minimum values of $k$ is $\qquad$ | 2. 10 .
From the given conditions, we know
$$
-1 \leqslant x \leqslant 1, -1 \leqslant y \leqslant 1 \text{. }
$$
Thus, $y+1 \geqslant 0, 2y-x-40$ when,
$$
k=x+y+y+1-(2y-x-4)=2x+5 \text{. }
$$
Therefore, $3 \leqslant k \leqslant 7$.
Hence, the maximum value of $k$ is 7, and the minimum value is 3. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,690 |
Example 1 Given real numbers $x, y, z$ satisfy $x+y=5$ and $z^{2}=x y+y-9$.
Then $x+2 y+3 z=$ $\qquad$ . | Solve: From $x+y=5$, we can set $x=\frac{5}{2}+t, y=\frac{5}{2}-t$.
Substitute into $z^{2}=x y+y-9$ and simplify to get $z^{2}+\left(t+\frac{1}{2}\right)^{2}=0$.
Thus, $z=0, t=-\frac{1}{2}$.
Therefore, $x=2, y=3$.
Hence, $x+2 y+3 z=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,691 |
Example 2 Given that $x, y$ are positive integers, and $x y+x+y=23, x^{2} y+x y^{2}=120$. Then $x^{2}+y^{2}=$ $\qquad$ | Solve: From $x y+x+y=23$, we can set $x y=\frac{23}{2}+t, x+y=\frac{23}{2}-t$.
Multiplying the two equations gives
$$
x^{2} y+x y^{2}=\left(\frac{23}{2}\right)^{2}-t^{2}=120 \text{. }
$$
Solving for $t$ yields $t= \pm \frac{7}{2}$.
Since $x$ and $y$ are both positive integers, when $t=\frac{7}{2}$, we have $x+y=8, x y... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,692 |
Example 11 Let real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x+y=z-1, \\
x y=z^{2}-7 z+14 .
\end{array}\right.
$$
Question: What is the maximum value of $x^{2}+y^{2}$? For what value of $z$ does $x^{2}+y^{2}$ achieve its maximum value? | Solve: From $x+y=z-1$, we can set
$$
x=\frac{z-1}{2}+t, y=\frac{z-1}{2}-t \text {. }
$$
Substituting into $x y=z^{2}-7 z+14$ and simplifying, we get
$$
3 z^{2}-26 z-55=-4 t^{2} \text {. }
$$
Therefore, $3 z^{2}-26 z-55 \leqslant 0$.
Solving this, we get $\frac{11}{3} \leqslant z \leqslant 5$.
$$
\begin{array}{l}
\tex... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,693 |
4. Given real numbers $a$ and $b$ satisfy $ab = 2a + 2b - 3$. Then the minimum value of $a^{2} + b^{2}$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 4. B.
Let $x=a+b, y=a b$. Then $y=2 x-3$. Therefore,
$$
\begin{array}{l}
a^{2}+b^{2}=(a+b)^{2}-2 a b \\
=x^{2}-2 y=x^{2}-2(2 x-3) \\
=x^{2}-4 x+6=(x-2)^{2}+2 \geqslant 2 .
\end{array}
$$
When $x=2$, i.e., $y=1, a=b=1$, the equality holds. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,694 |
5. In the right triangle $ABC (AC > BC)$, $\angle C = 90^{\circ}$, point $D$ is on $AC$, and $CD = CB = 2$, $DE \parallel BC$ intersects $AB$ at point $E$, $F$ is the midpoint of $CD$, and $\angle BEF = \angle B$. Then the length of $DE$ is ( ).
(A) $\frac{2}{3}$
(B) 1
(C) $\frac{4}{3}$
(D) $\frac{5}{3}$ | 5. C.
As shown in Figure 2, take the midpoint $G$ of $BE$, and connect $GF$. Then
$GF \parallel BC \parallel DE$,
and $GF=\frac{1}{2}(DE + BC)$.
Therefore, $\angle FGE$
$$
=\angle B=\angle BEF \text{. }
$$
Thus, $GF=EF$.
Let $EF=x$. Then $DE=2GF-BC=2x-2$.
In the right triangle $\triangle DEF$, $DF=\frac{1}{2}CD=1$.
... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,695 |
6. Given that $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at point $P, AB$ is a common external tangent of the two circles, point $A$ is on $\odot O_{1}$, point $B$ is on $\odot O_{2}$, $AC$ is a diameter of $\odot O_{1}$, and $CD$ is a tangent of $\odot O_{2}$, with $D$ being the point of tangency. Then the... | 6. B.
As shown in Figure 3, connect $A P$, $B P$, and $C P$. Then
$$
\begin{array}{l}
\angle A P C=90^{\circ}, \\
\angle A P B=90^{\circ}. \\
\text { Therefore, } \angle C P B \\
=\angle A P C+ \\
\angle A P B \\
=180^{\circ}.
\end{array}
$$
Thus, points $C$, $B$, and $P$ are collinear.
By the tangent property, $A C ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,696 |
1. Given that $a$, $b$, $c$, and $d$ are prime numbers, and $a b c d$ is the sum of 77 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$ | Given 77 consecutive positive integers, with the middle one being $x$. Then $a b c d=77 x=7 \times 11 x$, and $x \geqslant 39$.
Since $a, b, c, d$ are prime numbers, thus, $x$ can be decomposed into the product of two prime numbers, and the sum of these two prime numbers is minimized.
It is easy to verify that when $x... | 32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,697 |
2. Given an equilateral $\triangle A B C$ with side length $2, P$ is a point inside $\triangle A B C$, and the distances from point $P$ to the three sides $B C, A C, A B$ are $x, y, z$ respectively, and their product is $\frac{\sqrt{3}}{9}$. Then the sum of the squares of $x, y, z$ is | 2. 1 .
As shown in Figure 4, connect
$$
P A, P B, P C \text{. }
$$
It is easy to see that, in the equilateral $\triangle A B C$, we have
$$
\begin{array}{l}
x+y+z \\
=\frac{\sqrt{3}}{2} A B=\sqrt{3} .
\end{array}
$$
Thus, $\sqrt{3}=x+y+z$
$$
\geqslant 3 \sqrt[3]{x y z}=3 \sqrt[3]{\frac{\sqrt{3}}{9}}=\sqrt{3} \text{.... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,698 |
3. Given the quadratic function
$$
y=3 a x^{2}+2 b x-(a+b) \text {, }
$$
when $x=0$ and $x=1$, the value of $y$ is positive. Then, when $0<x<1$, the parabola intersects the $x$-axis at $\qquad$ points. | 3. 2 .
From the given, we have $\left\{\begin{array}{l}-(a+b)>0, \\ 3 a+2 b-(a+b)>0 .\end{array}\right.$
Solving, we get $a>0$.
Let $x=\frac{1}{2}$, then
$$
y=\frac{3 a}{4}+b-(a+b)=-\frac{a}{4}<0 .
$$
By the Intermediate Value Theorem, $3 a x^{2}+2 b x-(a+b)=0$ has one root when $0<x<\frac{1}{2}$, and one root when $... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,699 |
4. In concave quadrilateral $A B C D$, $\angle A=\angle C=$ $40^{\circ}, \angle B=50^{\circ}, E, F, G, H$ are the midpoints of sides $A B, B C$, $C D, D A$ respectively. Then the shape of quadrilateral $E F G H$ is $\qquad$ | 4. Rectangle.
As shown in Figure 5, connect $AC$ and $BD$. By the Midline Theorem of a triangle, we have
$$
\begin{array}{l}
E H \perp \frac{1}{2} B D, \\
F G \perp \frac{1}{2} B D .
\end{array}
$$
Thus, $E H \perp F G$.
Therefore, quadrilateral $E F G H$ is a parallelogram.
$$
\begin{array}{l}
\text { By } \angle B ... | Rectangle | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,700 |
One. (20 points) Consider the equation in $x$:
$$
x^{2}-8 x+a|x-4|+12-2 a=0
$$
which has exactly two real roots. Find the range of values for $a$. | $$
\begin{array}{l}
|x-4|^{2}+a|x-4|-2(a+2)=0 . \\
\text { Let } y=|x-4|(y \geqslant 0) . \\
\text { Then } y^{2}+a y-2(a+2)=0 \\
\Rightarrow[y+(a+2)](y-2)=0 \\
\Rightarrow y_{1}=-(a+2), y_{2}=2,
\end{array}
$$
That is, $|x-4|=-(a+2)$ or $|x-4|=2$.
Therefore, the original equation has exactly two real roots
$$
\Leftri... | -(a+2)=2 \text{ or } -(a+2)<0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,701 |
II. (25 points) In the right triangle $\triangle ABC$, $\angle C = 90^{\circ}$, $AC = BC$, $D$ is the midpoint of $AB$, and $E$, $F$ are points on $AC$, $BC$ respectively, such that $\angle EDF = 45^{\circ}$. If
$$
AB = 4 \sqrt{2}, \quad EF = \frac{5}{3},
$$
find the length of $AE$. | In the right triangle $ABC$, it is easy to know that
$$
AC = BC = 4.
$$
In $\triangle ADE$ and $\triangle BFD$,
$$
\begin{aligned}
& \angle ADE \\
= & 180^{\circ} - \angle EDF - \angle BDF \\
= & 180^{\circ} - 45^{\circ} - \angle BDF \\
= & 180^{\circ} - \angle B - \angle BDF \\
= & \angle BFD.
\end{aligned}
$$
Since... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,702 |
Three. (25 points) From the natural numbers $1, 2, \cdots, 2010$, take $n$ numbers such that the sum of any three of the taken numbers is divisible by 21. Find the maximum value of $n$.
| $$
\begin{array}{l}
\Rightarrow\left(x^{2}+16+\frac{64}{x^{2}}\right)-\left(8 x+\frac{64}{x}\right)+\frac{119}{9}=0 \\
\Rightarrow\left(x+\frac{8}{x}\right)^{2}-8\left(x+\frac{8}{x}\right)+\frac{119}{9}=0 \\
\Rightarrow\left(x+\frac{8}{x}-\frac{17}{3}\right)\left(x+\frac{8}{x}-\frac{7}{3}\right)=0 \\
\Rightarrow x=\fra... | 96 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,703 |
1. Let $A \cup B=\{1,2, \cdots, 10\},|A|=|B|$. Then the number of all possible ordered pairs of sets $(A, B)$ is | - 1. 9953.
If $|A|=k(k=5,6, \cdots, 10)$, then $A$ has $\mathrm{C}_{10}^{k}$ possibilities.
For each possibility of $A$, since $A \cup B=\{1,2, \cdots, 10\}$, we know that $C_{A \cup B} A \subseteq B$.
Thus, $B$ already has $10-k$ elements.
To ensure $|A|=|B|$, it is necessary to select $2 k-10$ elements from $A$ to ... | 8953 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,705 |
3. Let $x, y, z \in \mathbf{R}_{+}$. Then the minimum value of $\frac{\left(x^{2}+y^{2}\right)^{3}+z^{6}}{2 x^{3} y^{3}+y^{3} z^{3}+z^{3} x^{3}}$ is $\qquad$ . | 3.2.
Let $x=y=1, z=\sqrt[3]{2}$, we get
$$
\frac{\left(x^{2}+y^{2}\right)^{3}+z^{6}}{2 x^{3} y^{3}+y^{3} z^{3}+z^{3} x^{3}}=2 \text {. }
$$
Notice that
$$
\begin{array}{l}
\left(x^{2}+y^{2}\right)^{3}+z^{6} \\
=\left(x^{6}+y^{6}+3 x^{4} y^{2}+3 x^{2} y^{4}\right)+z^{6} \\
=\left(2 x^{4} y^{2}+2 x^{2} y^{4}\right)+\le... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,707 |
4. Equation
$$
\begin{array}{l}
\frac{x_{1} x_{2} \cdots x_{2010}}{x_{2011}}+\frac{x_{1} x_{2} \cdots x_{2009} x_{2011}}{x_{2010}}+\cdots+ \\
\frac{x_{2} x_{3} \cdots x_{2011}}{x_{1}}=2011
\end{array}
$$
The number of different ordered integer solutions $\left(x_{1}, x_{2}, \cdots, x_{2011}\right)$ is | 4. $2^{2010}$.
The equation can be transformed into
$$
x_{1} x_{2} \cdots x_{2011}\left(\frac{1}{x_{2011}^{2}}+\frac{1}{x_{2010}^{2}}+\cdots+\frac{1}{x_{1}^{2}}\right)=2011 \text {. }
$$
From this, we know that $x_{1} x_{2} \cdots x_{2011}>0$.
$$
\begin{array}{l}
\text { Also, } \frac{2011}{x_{1} x_{2} \cdots x_{2011... | 2^{2010} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,708 |
5. Fill $1,2, \cdots, n^{2}$ into an $n \times n$ chessboard, with each cell containing one number, and each row forming an arithmetic sequence with a common difference of 1. If any two of the $n$ numbers on the chessboard are neither in the same row nor in the same column, then the sum of these $n$ numbers is called a... | 5.1.
The numbers in the $k$-th row of the number table are
$$
(k-1) n+1,(k-1) n+2, \cdots,(k-1) n+n \text {. }
$$
Let the number taken from the $k$-th row be
$$
(k-1) n+a_{k} \text {. }
$$
Since these numbers are from different columns, $a_{1}, a_{2}, \cdots, a_{n}$ is a permutation of $1,2, \cdots, n$.
Therefore, t... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,709 |
6. In the spatial quadrilateral $ABCD$, $AB \perp BC, BC$ $\perp CD, CD \perp DA, DA \perp AB$. If $BD=1$, then the maximum value of the area of $\triangle ABC$ is $\qquad$ | 6. $\frac{1}{4}$.
First, prove that quadrilateral $ABCD$ is a rectangle, which only requires proving that points $A, B, C, D$ are coplanar.
Use proof by contradiction.
As shown in Figure 1, assume point $D$ is outside the plane $ABC$, and draw $DD_{1} \perp$ plane $ABC$ at point $D_{1}$, and connect $AD_{1}, CD_{1}$. ... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,710 |
7. Let $a_{1}=1, a_{n+1}=\frac{1-n}{n+1} a_{n}+\frac{(-1)^{n-1}}{n(n+1)}$. Then for any positive integer $n, a_{n}=$ | 7. $\left\{\begin{array}{ll}1, & n=1 ; \\ \frac{(-1)^{n}}{n}, & n>1\end{array}\right.$
Notice that
$$
\frac{(-1)^{n-1}}{n(n+1)}=(-1)^{n-1}\left(\frac{1}{n}-\frac{1}{n+1}\right) \text {. }
$$
Thus, rearranging terms we get
$$
\begin{array}{l}
a_{n+1}+\frac{(-1)^{n}}{n}=\frac{1-n}{n+1} a_{n}-\frac{(-1)^{n-1}}{n+1} \\
=... | a_{n}=\left\{\begin{array}{ll}1, & n=1 ; \\ \frac{(-1)^{n}}{n}, & n>1\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,711 |
8. There are 9 students participating in a math competition in the same classroom, with seats arranged in 3 rows and 3 columns, represented by a $3 \times 3$ grid, where each cell represents a seat. To prevent cheating, three types of exams, $A$, $B$, and $C$, are used, and it is required that any two adjacent seats (c... | 8. 246.
Let $a_{ij}$ denote the square at the $i$-th row and $j$-th column. First, consider the number of ways to distribute type $A$ papers at $a_{22}$.
$$
\begin{array}{l}
\text { Let } M=\left\{a_{12}, a_{21}, a_{32}, a_{23}\right\}, \\
N=\left\{a_{11}, a_{31}, a_{33}, a_{13}\right\} .
\end{array}
$$
Consider the ... | 246 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,712 |
9. (16 points) Let the range of the function $f(x)$ be $[1,2]$, and let $g(x)=f(x)+\frac{a}{f(x)}$. Find the minimum value of $p(a)=\max g(x)-\min g(x)$. | 9. Let $q(t)=t+\frac{a}{t}(t>0)$. Then $q^{\prime}(t)=1-\frac{a}{t^{2}}$.
When $a<0$, $q^{\prime}(t)>0$, thus, $q(t)$ is increasing on $(0,+\infty)$.
When $a \geqslant 0$, let $1-\frac{a}{t^{2}}>0$, we get $t>\sqrt{a}$.
Thus, $q(t)$ is decreasing on $(0, \sqrt{a}]$ and increasing on $[\sqrt{a},+\infty)$.
Let $t=f(x)$.... | 3-2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,713 |
10. (20 points) Let \( a_{1}=\frac{11}{7}, a_{n+1}=1+\frac{2}{a_{n}} \). Prove:
$$
(-1) a_{1}+(-1)^{2} a_{2}+\cdots+(-1)^{n} a_{n}<1
$$ | 10. Since $a_{n+1}=1+\frac{2}{a_{n}}$, we have:
$$
\begin{array}{l}
\frac{a_{n}+1}{a_{n}-2}=\frac{1+\frac{2}{a_{n-1}}+1}{1+\frac{2}{a_{n-1}}-2} \\
=\frac{2 a_{n-1}+2}{2-a_{n-1}}=(-2) \frac{a_{n-1}+1}{a_{n-1}-2} \\
\Rightarrow \frac{a_{n}+1}{a_{n}-2}=\frac{a_{1}+1}{a_{1}-2} \cdot(-2)^{n-1} \\
=\frac{11+7}{11-14}(-2)^{n-... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,714 |
11. (20 points) Let $AB$ be a chord of an ellipse parallel to the major axis. Through the midpoint $M$ of $AB$, draw two chords $CD$ and $EF$ of the ellipse. Connect $DE$ and $CF$, intersecting $AB$ at points $P$ and $Q$ respectively. Prove: $PM - QM$ is a constant. | 11. Taking $M$ as the origin and the line $AB$ as the $x$-axis, we establish a rectangular coordinate system.
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{(y-p)^{2}}{b^{2}}=1$, that is,
$$
b^{2} x^{2}+a^{2}(y-p)^{2}-a^{2} b^{2}=0 \text {. }
$$
The equations of the two lines $CD$ and $EF$ are
$$
\left(... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,716 |
一、(40 points) In quadrilateral $ABCD$, $BC = CD$, $AB \perp BC$, $AD \perp BD$. Let $P$ be the midpoint of $BD$. Prove: $AB \cdot AP = AC \cdot AD$.
| As shown in Figure 4, let $C P$ intersect $A B$ at point $O$.
Since $B C = C D$, we have $C P \perp B D$.
Also, $A D \perp B D$, so $O P \parallel A D$.
Since $P$ is the midpoint of $B D$, $O$ is the midpoint of $A B$.
In the right triangle $\triangle O B C$, because $B P \perp O C$, we have
$$
B C^{2} = C P \cdot C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,717 |
$$
\begin{array}{l}
\text { II. (40 points) Let } A=\left\{a_{1}, a_{2}, \cdots, a_{2010}\right\}, \text { where, } \\
a_{n}=n+\left[\sqrt{n}+\frac{1}{2}\right](1 \leqslant n \leqslant 2010) .
\end{array}
$$
A positive integer $k$ has the following property: there exists a positive integer $m$, such that $m+1, m+2, \c... | For $2 \leqslant n \leqslant 2010$, we have
$$
\begin{array}{l}
0<a_{n}-a_{n-1} \\
=1+\left[\sqrt{n}+\frac{1}{2}\right]-\left[\sqrt{n-1}+\frac{1}{2}\right] \\
<\left(1+\sqrt{n}+\frac{1}{2}\right)-\left(\sqrt{n-1}+\frac{1}{2}-1\right) \\
=2+\sqrt{n}-\sqrt{n-1}=2+\frac{1}{\sqrt{n}+\sqrt{n-1}}<3 .
\end{array}
$$
Therefor... | \{2 i \mid i=1,2, \cdots, 44\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,718 |
Three, (50 points) Find the smallest positive integer $k$, such that $625^{k} \equiv 1(\bmod 343)$ | $$
\left(5^{4}\right)^{k} \equiv 1\left(\bmod 7^{3}\right),
$$
which can be transformed into $(89 \times 7+2)^{k}-1 \equiv 0\left(\bmod 7^{3}\right)$.
Expanding it, we get
$$
\begin{array}{l}
\mathrm{C}_{k}^{2}(89 \times 7)^{2} \times 2^{k-2}+\mathrm{C}_{k}^{1}(89 \times 7) \times 2^{k-1}+2^{k}-1 \\
\equiv 0\left(\bmo... | 147 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,719 |
Four. (50 points) $A$, $B$, and $C$ are playing table tennis. When two of them play, the other one acts as the referee. The loser of this match will be the referee in the next match, and the other two will continue playing. After several matches, it is known that $A$ has played a total of $a$ matches, and $B$ has playe... | Let $C$ have competed in $c$ matches. Then the total number of player appearances is $a+b+c$. However, each match produces 2 player appearances, so the total number of matches is $\frac{a+b+c}{2}$.
Therefore, the number of matches where $C$ was the referee is
$$
\frac{a+b+c}{2}-c=\frac{a+b-c}{2} \text{. }
$$
Since if ... | \left[\frac{a+b}{3}\right]+r | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 722,720 |
The 277th National Junior High School Mathematics Competition consists of 14 questions (5 multiple-choice questions, 5 fill-in-the-blank questions, 4 problem-solving questions), with a full score of 150 points. Among them, each correct answer for multiple-choice and fill-in-the-blank questions earns 7 points, and a wro... | The score can be selected from 16 5s and 10 7s, let the total number of 5s and 7s selected be $k$.
Since selection and non-selection are relative, i.e., the score obtained by selecting $m$ 5s and $n$ 7s is the same as the score obtained by selecting $16-m$ 5s and $10-n$ 7s, which sums up to 150, we only need to consid... | 127 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,721 |
Initially 278 Use three types of regular polygon tiles with equal side lengths to pave the ground, their vertices fit together, just enough to completely cover the ground. Find the number of sides of these three polygons. | Let the number of sides of these three polygons be $y$, $z$, and assume that at a vertex there are $k_{1}$ $x$-sided polygons, $k_{2}$ $y$-sided polygons, and $k_{3}$ $z$-sided polygons. Thus,
$$
\begin{array}{l}
k_{1} \cdot \frac{(x-2) 180^{\circ}}{x}+k_{2} \cdot \frac{(y-2) 180^{\circ}}{y}+ \\
k_{3} \cdot \frac{(z-2)... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,722 |
277 As shown in Figure $3, D$ is any point inside $\triangle A B C$. A line through $D$ intersects the circumcircles $\odot O_{1}, \odot O_{2}, \odot O_{3}$ of $\triangle A B D, \triangle B C D$, and $\triangle C A D$ at points $E, F$, and $G$ respectively. $L$ is any point on the circumcircle $\odot O_{4}$ of $\triang... | $$
\begin{array}{l}
\text { Hence } \frac{A M}{M B} \cdot \frac{B N}{N C} \cdot \frac{C O}{O A} \\
=\frac{S_{\triangle E A L}}{S_{\triangle E B L}} \cdot \frac{S_{\triangle B F L}}{S_{\triangle C L F}} \cdot \frac{S_{\triangle G G L}}{S_{\triangle G A L}} \\
=\frac{S_{\triangle E A L}}{S_{\triangle C L F}} \cdot \frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,723 |
Let $n$ be a positive integer. Prove:
$$
\frac{\mathrm{C}_{n}^{0}}{1}+\frac{\mathrm{C}_{n}^{1}}{3}+\frac{\mathrm{C}_{n}^{2}}{5}+\cdots+\frac{\mathrm{C}_{n}^{n}}{2 n+1} \leqslant 2^{n} \cdot \frac{n+1}{2 n+1} .
$$ | Proof: Let $i \in \mathbf{N}_{+}$ and $i \leqslant n+1$. Then
$$
\begin{aligned}
(2 i & -1)[2(n+2-i)-1] \\
& =[1+2(i-1)][1+2(n+1-i)] \\
& =1+2 n+4(i-1)(n+1-i) \\
& \geqslant 2 n+1, \\
& \frac{1}{2 i-1}+\frac{1}{2(n+2-i)-1} \\
& =\frac{2(n+1)}{(2 i-1)[2(n+2-i)-1]} \\
& \leqslant \frac{2(n+1)}{2 n+1} .
\end{aligned}
$$
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,724 |
1. Let $\left\{a_{n}\right\}$ be a convex sequence, $S_{n}=\sum_{i=1}^{n} a_{i}$. Prove that for $k<m<n$, we have
$$
\frac{n-m}{k} S_{k}+\frac{m-k}{n} S_{n}+\frac{k-n}{m} S_{m} \geqslant 0 .
$$ | Hint: Use property 6 and property 3.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | proof | Yes | Yes | cn_contest | false | 722,725 |
3. Given real numbers $a, b, c$ satisfy
$$
a^{2}-b c-6 a+1=0, b^{2}+c^{2}+b c-2 a-1=0 \text {. }
$$
Find the range of values for $a$. | Answer: $\frac{10-\sqrt{94}}{3} \leqslant a \leqslant \frac{10+\sqrt{94}}{3}$. | \frac{10-\sqrt{94}}{3} \leqslant a \leqslant \frac{10+\sqrt{94}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,726 |
3. Let the sequence of positive numbers $\left\{a_{n}\right\}$ satisfy
$$
a_{k-1} a_{k+1} \leqslant a_{k}^{2}(k=1,2, \cdots) \text {. }
$$
Prove: When $n \geqslant 2$, we have
$$
\frac{\sum_{i=0}^{n} a_{i}}{n+1} \cdot \frac{\sum_{i=1}^{n-1} a_{i}}{n-1} \geqslant \frac{\sum_{i=0}^{n-1} a_{i}}{n} \cdot \frac{\sum_{i=1}^... | Prompt: It is only necessary to prove
$$
\left(\sum_{i=0}^{n} a_{i}\right)\left(\sum_{i=1}^{n-1} a_{i}\right) \geqslant\left(n^{2}-1\right) a_{0} a_{n} \text {. }
$$
From the problem, we know that $\left.\mid-\lg a_{n}\right\}$ is a convex sequence.
Then by property 4, we get
$$
\begin{array}{l}
\frac{\sum_{i=0}^{n}\l... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 722,728 |
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