problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
3. The value of $x$ when the algebraic expression $4-x^{2}-\sqrt{1-x^{2}}$ reaches its minimum is $\qquad$ ; the value of $x$ when the algebraic expression $4-x^{2}-\sqrt{1-x^{2}}$ reaches its maximum is $\qquad$ . | 3. When the minimum value is reached, $x= \pm \frac{\sqrt{3}}{2} ;$ when the maximum value is reached, $x= \pm 1,0$.
Notice that
$$
\begin{array}{l}
4-x^{2}-\sqrt{1-x^{2}}=3+\left(1-x^{2}\right)-\sqrt{1-x^{2}} \\
=\left(\sqrt{1-x^{2}}-\frac{1}{2}\right)^{2}+\frac{11}{4} .
\end{array}
$$
When $\sqrt{1-x^{2}}=\frac{1}{2... | x= \pm \frac{\sqrt{3}}{2}; x= \pm 1,0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,785 |
4. Given that $\odot O_{1}$ and $\odot O_{2}$ are externally tangent, their radii are $112$ and $63$, respectively. The segment $A B$ is intercepted by their two external common tangents on their internal common tangent. Then, the length of $A B$ is $\qquad$ . | 4.168.
From the tangent theorem, we easily know that
$$
\begin{array}{l}
A B=\text { length of the external common tangent }=\sqrt{(112+63)^{2}-(112-63)^{2}} \\
=2 \sqrt{112 \times 63}=168 .
\end{array}
$$ | 168 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,786 |
5. Given in the Cartesian coordinate system $x O y$, the vertices of square $A B C D$ are $A(-1,1)$, vertex $C(1,1+2 \sqrt{3})$. Then, the coordinates of vertices $B$ and $D$ are $\qquad$
$\qquad$ | 5. $(\sqrt{3}, \sqrt{3})$ and $(-\sqrt{3}, 2+\sqrt{3})$.
It is easy to know that the midpoint $T(0,1+\sqrt{3})$ of $A C$.
Translate the square $A B C D$ so that its center $T$ moves to the origin $O$, then we have $A^{\prime}(-1,-\sqrt{3}), C^{\prime}(1, \sqrt{3})$.
Rotate $A^{\prime} \leqslant C^{\prime}$ by $90^{\ci... | (\sqrt{3}, \sqrt{3}) \text{ and } (-\sqrt{3}, 2+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,787 |
6. In an $m$ row, $n$ column grid, there are $m n$ small squares with a side length of 1. Each small square is colored with one of three colors: red, yellow, or blue. It is known that each row of the grid has 6 red small squares, each column has 8 yellow small squares, and the entire grid has 15 blue small squares. If ... | $$
\text { 6. } m=17, n=13 \text {. }
$$
From the problem, we have $6 m+8 n+15=m n$. Therefore,
$$
\begin{array}{l}
(m-8)(n-6)=48+15=63 \\
=1 \times 63=3 \times 21=7 \times 9 .
\end{array}
$$
Since $n$ is a two-digit prime number, we have $n=13, m=17$. | m=17, n=13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,788 |
7. The solution to the equation $\sqrt{\frac{2 x-6}{x-11}}=\frac{3 x-7}{x+6}$ is | 7. $x_{1}=19, x_{2}=\frac{13+5 \sqrt{2}}{7}$.
Squaring both sides of the equation and simplifying, we get
$$
\begin{array}{l}
7 x^{3}-159 x^{2}+511 x-323=0, \\
(x-19)\left(7 x^{2}-26 x+17\right)=0 .
\end{array}
$$
Solving, we find $x_{1}=19, x_{2,3}=\frac{13 \pm 5 \sqrt{2}}{7}$.
By the definition of the equation, $x>... | x_{1}=19, x_{2}=\frac{13+5 \sqrt{2}}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,789 |
8. Let $a$ and $b$ be constants, and $b<0$. The graph of the parabola $y=a x^{2}+b x+a^{2}+\sqrt{2} a-4$ is one of the four graphs in Figure 1. Then $a=$ $\qquad$ . | 8. $a=\sqrt{2}$.
From $b0$, we get $a$ $>0$. Therefore, $a=\sqrt{2}$. | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,790 |
2. As shown in Figure 1, in the obtuse triangle $\triangle ABC$, $BC=1, \angle A = 30^{\circ}, D$ is the midpoint of side $BC$, and $G$ is the centroid of $\triangle ABC$. If $B$ and $C$ are fixed points, when point $A$ moves, the range of the length of segment $GD$ is ( ).
(A) $0 < GD \leqslant \frac{\sqrt{13}}{6}$
(B... | 2.B.
Since $\angle A=30^{\circ}$, and $\triangle A B C$ is an obtuse triangle, point $A$ is on the arc $\overparen{A_{1} B}$ or $\overparen{A_{2} C}$ (excluding the endpoints) as shown in Figure 4 (where $A_{1} B \perp B C, A_{2} C \perp B C$).
Let $G_{1}$ be the centroid of $\mathrm{Rt} \triangle A_{1} B C$, then
$$... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,791 |
9. As shown in Figure 2, in $\triangle A B C$, $A B=B C=5, A C$ $=7, \triangle A B C$ has an inscribed circle $\odot O$ that is tangent to side $A C$ at point $M$. A line $M N$ parallel to side $B C$ is drawn through point $M$ and intersects $\odot O$ at point $N$. A tangent to $\odot O$ is drawn through point $N$ and ... | 9.0.6.
As shown in Figure 4, connect $P O$ intersecting $B C$ at point $D$. By the tangent property, we know
$$
P O \perp M N \text {. }
$$
Since $M N \parallel B C$, it follows that $P O \perp B C$.
Therefore, point $D$ is the point of tangency between $\odot O$ and $B C$.
Also, $C D=C M=3.5, P C=\frac{C D}{\cos C}=... | 0.6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,792 |
10. It is known that someone bought a car for 121,000 yuan. If the annual insurance, maintenance, and gasoline costs total 10,000 yuan, and the car repair costs are 0 yuan in the first year, and from the second year onwards, increase by 0.2 ten thousand yuan each year. Then, after using the car for $\qquad$ years, it s... | $10.11,3.1$.
After using the car for $x$ years, the annual average cost is
$$
\begin{array}{l}
y=\frac{1}{x}\left\{12.1+x \times 1+\frac{x}{2}[0+0.2(x-1)]\right\} \\
=0.9+0.1 x+\frac{12.1}{x} \\
\geq 0.9+2 \sqrt{0.1 \times 12.1}=3.1 .
\end{array}
$$
The minimum annual average cost of the car is 3.1 ten thousand yuan, ... | 11, 3.1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,793 |
1. In a right-angled triangle, the square of the hypotenuse is exactly equal to twice the product of the two legs. Then, the ratio of the three sides of this triangle is ( ).
(A) $3: 4: 5$
(B) $1: 1: 1$
(C) $2: 3: 4$
(D) $1: 1: \sqrt{2}$ | $-1 . \mathrm{D}$
Let the two legs be $a$ and $b$. Then $a^{2}+b^{2}=2 a b \Rightarrow a=b$. Therefore, the ratio of the three sides of this right triangle is $1: 1: \sqrt{2}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,794 |
2. The number of integers $x$ that satisfy the inequality $|x-2006|+|x| \leqslant 9999$ is ( ) .
(A) 9998
(B) 9999
(C) 10000
(D) 10001 | 2.B.
When $x \geqslant 2006$, we have $x-2006+x \leqslant 9999$, which means $x \leqslant \frac{2006+9999}{2}$, or $x \leqslant 6002$;
When $0 \leqslant x<2006$, we have $2006-x+x \leqslant 9999$, which is obviously true;
When $x<0$, we have $2006-x-x \leqslant 9999$, which means $x \geqslant \frac{2006-9999}{2}$, o... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,795 |
3. From the 14 natural numbers $1,2, \cdots, 14$, take out $k$ numbers to ensure that there are two numbers, one of which is twice the other. Then the minimum value of $k$ is ( ).
(A) 8
(B) 9
(C) 10
(D) 11 | 3. C.
Divide $\{1,2, \cdots, 14\}$ into
$$
\{1,2\},\{3,6\},\{4,8\},\{5,10\},\{7,14\},\{9,11,12,13\} \text {. }
$$
If 10 numbers are taken from $\{1,2, \cdots, 14\}$, then at least 6 numbers must be taken from the first 5 groups, meaning that two numbers from the same group are taken, satisfying the problem's requirem... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,796 |
4. A natural number $q$, if any two digits are taken, and the left digit is greater than the right digit, then this number is said to have one inversion. Use $N X(q)$ to denote the number of inversions in $q$ (for example, $N X(3214)=3, N X(12344)=0$). Then the remainder when $N X$ (324 167 895) is divided by 4 is ( ).... | 4. A.
$$
\begin{array}{l}
N X(324167895)=N X(32416785)+1 \\
=N X(32415)+4=N X(3241)+4=4+4=8 .
\end{array}
$$ | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,797 |
5. As shown in Figure $1, P$ is a point on the graph of the function $y=\frac{1}{2 x}(x>0)$, the line $y=-x+1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Draw $P M \perp x$-axis at point $M$, intersecting $A B$ at point $E$, and draw $P N \perp y$-axis at point $N$, intersecting $A B$ at ... | 5.C.
Let $P(x, y)$.
From $F N / / O A$, we get $\frac{A F}{A B}=\frac{O N}{O B}$, which means $A F=\sqrt{2} y$.
Similarly, $B E=\sqrt{2} x$.
Therefore, $A F \cdot B E=2 x y=1$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,798 |
6. If 5 consecutive natural numbers are all composite, then this group of numbers is called a "twin 5 composite". So, among the natural numbers not exceeding 100, there are $\qquad$ groups of twin 5 composite. | Ni, 6.10.
It is easy to know that the prime numbers not exceeding 100 are
$$
\begin{array}{l}
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47, \\
53,59,61,67,71,73,79,83,89,97 .
\end{array}
$$
There are 10 groups of twin 5 composites, namely
$$
\begin{array}{l}
24,25,26,27,28 ; 32,33,34,35,36 ; \\
48,49,50,51,52 ; 54,55,56,5... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,799 |
7. In $\triangle A B C$, $A C=B C, \angle A C B=90^{\circ}$, $D 、 E$ are points on side $A B$, $A D=3, B E=4$, $\angle D C E=45^{\circ}$. Then the area of $\triangle A B C$ is $\qquad$ | 7:36.
As shown in Figure 5, rotate $\triangle C E B$ $90^{\circ}$ clockwise to get $\triangle C E^{\prime} A$. Connect $E^{\prime} D$.
It is easy to see that $A E^{\prime}=B E=4$, $\angle E^{\prime} A D=90^{\circ}$, so
$$
\begin{array}{l}
E^{\prime} D=\sqrt{A E^{\prime 2}+A D^{2}} \\
=\sqrt{4^{2}+3^{2}}=5 .
\end{array... | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,800 |
3. Let $a, b$ be positive integers, and $a+b, a+5, b-2$ are the lengths of the three sides of a right-angled triangle. Then the number of positive integer pairs $(a, b)$ is $(\quad)$.
(A) 0
(B) 1
(C) 2
(D) 3 | 3.A.
If $a+b$ is the length of the hypotenuse, then
$$
(a+b)^{2}=(a+5)^{2}+(b-2)^{2} \text {, }
$$
which simplifies to $2(ab-5a+2b)=29$.
The left side of the equation is even, while the right side is odd, leading to a contradiction.
If $a+5$ is the length of the hypotenuse, then
$$
(a+5)^{2}=(a+b)^{2}+(b-2)^{2} \text... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,802 |
9. As shown in Figure 2, in rectangle $A B C D$, $A B=7, A D$ $=24, P$ is a moving point on side $B C$, and
$P E \perp A C$ at point $E$,
$P F \perp B D$ at point $F$.
Then $P E+P F=$ $\qquad$ | 9.6.72.
Let the intersection point of diagonals $AC$ and $BD$ be $O$. Connect $OP$.
From $AB=7, AD=24$, we get $OB=OC=\frac{25}{2}$.
From $S_{\triangle OBC}=S_{\triangle OPB}+S_{\triangle OPC}$, we get $\frac{1}{2} \times \frac{1}{2} AB \cdot BC=\frac{1}{2} OB \cdot PF + \frac{1}{2} OC \cdot PE$,
which means $PE + PF ... | 6.72 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,803 |
10. There are two sets of square paper pieces of the same size and the same number, one set black and one set white. Xiao Zhang first uses the white paper pieces to form a rectangle without any gaps in the middle, then uses the black paper pieces to surround the already formed white rectangle to create a larger rectang... | 10.350.
Let the first white rectangle be $a \times b$, then the first black rectangle is $(a+2)(b+2)$, the second white rectangle is $(a+4) \times (b+4), \cdots \cdots$, and the fifth black rectangle is $(a+18)(b+18)$.
Obviously, the difference in perimeter between each black rectangle and its inner white rectangle (... | 350 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,804 |
Three, (15 points) In the pentagram $A B C D E$, the intersection points of the intersecting line segments are labeled as shown in Figure 3. It is given that
$$
\begin{array}{l}
A Q=Q C, \\
B R=R D, \\
C R=R E, \\
D S=S A .
\end{array}
$$
Prove: $B T=T P=P E$. | Three, as shown in Figure 7,
Connect $A E$, $A B$,
$B C$, $C D$, $D E$,
$T R$, $A R$, and let the intersection of $A R$
and $B E$ be
$O$.
Since $B R=$
$$
R D, C R=R E,
$$
Therefore, quadrilateral $B C D E$ is a parallelogram.
Thus, $D E \mathbb{\|} B C$.
Since $B R=R D, D S=S A$, according to the midline theorem of tr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,805 |
Four, (15 points) Three distinct positive integers, if the sum of the product of any two and 1 is exactly divisible by the third number, are called a "Linglong Triplet".
(1) Prove that the three positive integers in a Linglong Triplet are pairwise coprime;
(2) Find all Linglong Triplets. | Let three distinct positive integers be $a, b, c$, satisfying $c \mid (ab + 1)$, $b \mid (ca + 1)$, and $a \mid (bc + 1)$.
Next, we prove: $a, b, c$ must be pairwise coprime.
If not, assume $(a, b) > 1 \Rightarrow (ca, b) = d > 1$, at this time, $ca + 1$ cannot be divided by $d$, i.e., $d \nmid (ca + 1)$, but $d \mid b... | (1, 2, 3) \text{ and } (2, 3, 7) | Number Theory | proof | Yes | Yes | cn_contest | false | 717,806 |
Five. (10 points) As shown in Figure 4, there are $m$ points inside $\triangle ABC$. Some line segments are drawn between these points and between these points and the points $A$, $B$, and $C$. These line segments have no common points inside the triangle other than these $m$ points, and they exactly divide $\triangle ... | (1) Let the total number of small triangles be $n$. Among the sides of these $n$ small triangles, 3 sides are the original triangle's sides $AB$, $BC$, and $CA$. Therefore, the number of sides of the small triangles located inside is $3n-3$. Moreover, each of these sides belongs to two small triangles, meaning each sid... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,807 |
2. Which of the following calculations is correct? ( ).
(A) $\left(a b^{4}\right)^{4}=a b^{8}$
(B) $(-3 p q)^{2}=-6 p^{2} q^{2}$
(C) $x^{2}-\frac{1}{2} x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^{2}$
(D) $3\left(a^{2}\right)^{3}-6 a^{6}=-3 a^{6}$ | 2.D.
$$
\begin{array}{l}
\left(a b^{4}\right)^{4}=a^{4} b^{16},(-3 p q)^{2}=9 p^{2} q^{2}, \\
\left(x-\frac{1}{2}\right)^{2}=x^{2}-x+\frac{1}{4} .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,809 |
3. As shown in Figure 1, let the areas of the semicircles with diameters as the three sides of the right triangle $\triangle A B C$ be $S_{1}$, $S_{2}$, and $S_{3}$, and the area of the right triangle $\triangle A B C$ be $S$. Then the relationship between them is ( ).
(A) $S=S_{1}$
(B) $S_{1}=S_{2}+S_{3}$
(C) $S=S_{1}... | 3. B.
$$
\begin{array}{l}
S_{1}=\frac{\pi c^{2}}{2}, S_{2}=\frac{\pi a^{2}}{2}, S_{3}=\frac{\pi b^{2}}{2}, S=\frac{a b}{2}, \text { and } \\
c^{2}=a^{2}+b^{2} .
\end{array}
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,810 |
5. As shown in Figure 3, in a square with side length $a$, a smaller square with side length $b$ is removed $(a>b)$, and the remaining part is cut and rearranged into a rectangle. By calculating the area of the figure (shaded part), the equation that can be verified is ( ).
(A) $a^{2}-b^{2}=(a+b)(a-b)$
(B) $(a+b)^{2}=a... | 5.A.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,812 |
4. Given the parabola $y=a x^{2}+b x+c(a>0)$ and the line $y=k(x-1)-\frac{k^{2}}{4}$. Regardless of the value of $k$, the parabola and the line have only one common point. Then, the equation of the parabola is ( ).
(A) $y=x^{2}$
(B) $y=x^{2}-2 x$
(C) $y=x^{2}-2 x+1$
(D) $y=2 x^{2}-4 x+2$ | 4.C.
From $\left\{\begin{array}{l}y=a x^{2}+b x+c, \\ y=k(x-1)-\frac{k^{2}}{4}\end{array}\right.$ we get
$$
a x^{2}+(b-k) x+c+k+\frac{k^{2}}{4}=0 \text {. }
$$
By the problem statement, equation (1) has two equal real roots, so
$$
\Delta=(b-k)^{2}-4 a\left(c+k+\frac{k^{2}}{4}\right)=0 \text {, }
$$
which simplifies ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,813 |
7. In Rt $\triangle A B C$, $\angle C=90^{\circ}$. Which of the following equations is not necessarily true? ( ).
(A) $\sin A=\sin B$
(B) $\cos A=\sin B$
(C) $\sin A=\cos B$
(D) $\sin (A+B)=\sin C$ | 7.A. When $\angle A \neq \angle B$, $\sin A \neq \sin B$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,815 |
8. In the physical education class, all the girls in the class took a 100-meter test, with the standard time being $18 \mathrm{~s}$. Table 1 records the performance of 8 girls in the first group, where the plus sign indicates a time greater than $18 \mathrm{~s}$, and the minus sign indicates a time less than $18 \mathr... | 8.C.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Other | MCQ | Yes | Yes | cn_contest | false | 717,816 |
13. A piece of clothing is marked at 132 yuan. If it is sold at a 90% of the marked price, a profit of $10 \%$ can still be made. Then the purchase price of this piece of clothing is ( ) yuan.
(A) 106
(B) 105
(C) 118
(D) 108 | 13.D.
$$
132 \times 90\% \div (1+10\%) = 108 \text{ (yuan). }
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,821 |
14. If the value of the fraction $\frac{3 x^{2}-12}{x^{2}+4 x+4}$ is 0, then the value of $x$ is ( ).
(A) 2
(B) $\pm 2$
(C) -2
(D) $\pm 4$ | 14.A. From $\left\{\begin{array}{l}3 x^{2}-12=0, \\ x^{2}+4 x+4 \neq 0,\end{array}\right.$ we get $x=2$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,822 |
15. If $x^{2}-2(k+1) x+4$ is a perfect square, then the value of $k$ is ( ).
(A) $\pm 1$
(B) $\pm 3$
(C) -1 or 3
(D) 1 or -3 | 15. D.
Let $x=k+1$, then $4-(k+1)^{2}$ is a perfect square, so $4-(k+1)^{2}=0,1,4$.
Solving gives $k=-3,1,-1$.
Upon verification, $k=-3,1$ meet the requirements. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,823 |
5. If $x$ is a real number, denote $\{x\}=x-[x]$ (where $[x]$ represents the greatest integer not exceeding $x$), then the number of real roots of the equation
$$
2006 x+\{x\}=\frac{1}{2007}
$$
is ( ).
(A) 0
(B) 1
(C) 2
(D) an integer greater than 2 | 5.C.
Since $x=[x]+\{x\}$, the original equation can be transformed into $2006[x]+2007\{x\}=\frac{1}{2007}$.
Also, $0 \leqslant 2007\{x\}<2007$, so, $[x]=-1$ or $[x]=0$.
If $[x]=-1$, then
$$
\begin{array}{l}
\{x\}=\frac{2006 \times 2007+1}{2007^{2}}=\frac{2007^{2}-2007+1}{2007^{2}} \\
=1-\frac{2006}{2007^{2}}<1 .
\end... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,824 |
16. As shown in Figure 9, in $\triangle A B C$, $P$ is a point on $A B$. The following four conditions are given:
(1) $\angle A C P=\angle B$;
(2) $\angle A P C=\angle A C B$;
(3) $A C^{2}=A P \cdot A B$;
(4) $A B \cdot C P=A P \cdot C B$
Among these, the conditions that satisfy the similarity of $\triangle A P C$ and ... | 16. D.
From the similarity of $\triangle A P C$ and $\triangle A C B$, and
$$
\angle A=\angle A, \angle A C P \neq \angle A C B \text {, }
$$
it follows that $\angle A C P=\angle B, \angle A P C=\angle A C B$.
Therefore, $\frac{A P}{A C}=\frac{A C}{A B}=\frac{C P}{C B}$.
Hence, $A C^{2}=A P \cdot A B ; A B \cdot C P=... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,825 |
17. Given in a circle $\odot O$ with radius 2, the side $AB=2\sqrt{3}$ of the inscribed $\triangle ABC$. Then the degree of $\angle C$ is ).
(A) $60^{\circ}$
(B) $30^{\circ}$
(C) $60^{\circ}$ or $120^{\circ}$
(D) $30^{\circ}$ or $150^{\circ}$ | 17. C.
From $\sin C=\frac{AB}{2R}=\frac{\sqrt{3}}{2}$, we get $\angle C=60^{\circ}$ or $120^{\circ}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,826 |
18. If the three sides of a right-angled triangle are $a, b, c, \angle B=90^{\circ}$, then, the nature of the roots of the equation
$$
a\left(x^{2}-1\right)-2 c x+b\left(x^{2}+1\right)=0
$$
with respect to $x$ is ( ).
(A) It has two equal real roots
(B) It has two unequal real roots
(C) It has no real roots
(D) Cannot... | 18. A.
$$
\begin{array}{l}
\Delta=(2 c)^{2}-4(a+b)(b-a) \\
=4\left(c^{2}+a^{2}-b^{2}\right)=0 .
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,827 |
19. Given point $P(9+\sqrt{-a},-3+a)$. Then the quadrant in which point $P$ lies is ( ).
(A) First quadrant
(B) Second quadrant
(C) Third quadrant
(D) Fourth quadrant | 19.D.
From $9+\sqrt{-a}>0$, and $a \leqslant 0$, we know $-3+a<0$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,828 |
20. If the graph of the function $y=k x^{2 k^{2}+k-2}$ is a hyperbola, and it is in the second and fourth quadrants, then $k=(\quad)$.
(A) $\frac{1}{2}$
(B) -1
(C) $-\frac{3}{2}$
(D) 1 | 20. B.
From $\left\{\begin{array}{l}k<0, \\ 2 k^{2}+k-2=-1,\end{array}\right.$ we solve to get $k=-1$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,829 |
21. If the length of the upper base of a trapezoid is $l$, and the length of the line segment connecting the midpoints of the two non-parallel sides is $m$, then the length of the line segment connecting the midpoints of the two diagonals is ( ).
(A) $m-2 l$
(B) $\frac{m}{2}-l$
(C) $2 m-l$
(D) $m-l$ | 21. D.
Let the lower base of the trapezoid be $a$, and the length of the required line segment be $b$. Then $\frac{a+l}{2}=m, \frac{a-l}{2}=b$.
Solving for $b$ gives $b=m-l$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,830 |
22. One side of a rhombus is equal to the leg of an isosceles right triangle. If one angle of the rhombus is $60^{\circ}$, then the ratio of the area of the rhombus to the area of the isosceles right triangle is ( ).
(A) $\sqrt{3}: 2$
(B) $\sqrt{3}: 1$
(C) $1: \sqrt{3}$
(D) $\sqrt{3}: 4$ | 22. B.
Let the side length of the rhombus be $a$, and the areas of the rhombus and the isosceles right triangle be $S_{1}$ and $S_{2}$, respectively. Then
$$
S_{1}=a^{2} \sin 60^{\circ}=\frac{\sqrt{3} a^{2}}{2}, S_{2}=\frac{a^{2}}{2} .
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,831 |
23. If the two real roots of the equation $8 x^{2}+2 k x+k-1=0$ are $x_{1} 、 x_{2}$, and they satisfy $x_{1}^{2}+x_{2}^{2}=1$, then the value of $k$ is ( ).
(A) -2 or 6
(B) -2
(C) 6
(D) 4 | 23. B.
Given $x_{1}+x_{2}=-\frac{k}{4}, x_{1} x_{2}=\frac{k-1}{8}$, then
$$
x_{1}^{2}+x_{2}^{2}=\left(-\frac{k}{4}\right)^{2}-2 \times \frac{k-1}{8}=\frac{k^{2}-4 k+4}{16}=1 \text {. }
$$
Solving for $k$ gives $k=-2$ or 6.
$$
\text { Also, } \Delta=\left(-\frac{k}{4}\right)^{2}-4 \times \frac{k-1}{8}=\frac{k^{2}-8 k+... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,832 |
24. Given that the radius of $\odot O$ is $10 \mathrm{~cm}, A$ is a point on $\odot O$, $B$ is the midpoint of $O A$, and the distance between point $B$ and point $C$ is $5 \mathrm{~cm}$. Then the positional relationship between point $C$ and $\odot O$ is ( ).
(A) Point $C$ is inside $\odot O$
(B) Point $C$ is on $\odo... | 24. D.
$$
O C \leqslant O B + B C = 10 .
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,833 |
25. Given that $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, the common chord intersects the line segment $O_{1} O_{2}$ at point $G$. If $AB=48$, the radii of $\odot O_{1}$ and $\odot O_{2}$ are $30$ and $40$ respectively, then the area of $\triangle A O_{1} O_{2}$ is ( ).
(A) 600
(B) 300 or 168
(C) ... | 25.D.
Since $O_{1} G=\sqrt{A O_{1}^{2}-A G^{2}}=\sqrt{30^{2}-24^{2}}=18$,
$$
O_{2} G=\sqrt{A O_{2}^{2}-A G^{2}}=\sqrt{40^{2}-24^{2}}=32 \text {, }
$$
then $S_{\triangle 1 O_{1} O_{2}}=\frac{1}{2} A G\left(O_{2} G \pm O_{1} G\right)=600$ or 168. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,834 |
6. As shown in Figure 2, square $ABCD$ is inscribed in $\odot O$, $P$ is a point on the minor arc $\overparen{CD}$, $PA$ intersects $BD$ at point $M$, $PB$ intersects $AC$ at point $N$, and let $\angle PAC = \theta$. If $MN \perp PA$, then the value of $2 \cos^2 \theta - \tan \theta$ is ( ).
(A) 1
(B) $\frac{\sqrt{2}}{... | 6. A.
Let the radius of $\odot O$ be 1, then $A C=2$.
As shown in Figure 2, connect $P C$. Then $\angle A P C=90^{\circ}$. Therefore, $P A=A C \cos \theta=2 \cos \theta$.
In the right triangle $\triangle A O M$, $A M=\frac{O A}{\cos \theta}=\frac{1}{\cos \theta}$;
In the right triangle $\triangle A M N$, $M N=A M \tan... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,835 |
26. Among the numbers $2004, 2005, 2006, 2007$, the number that cannot be expressed as the difference of squares of two integers is ( ).
(A) 2004
(B) 2005
(C) 2006
(D) 2007 | 26. C.
From $x^{2}-y^{2}=(x+y)(x-y)$, where $x+y$ and $x-y$ have the same parity, we know that $x^{2}-y^{2}$ does not leave a remainder of 2 when divided by 4. Since 2006 leaves a remainder of 2 when divided by 4, 2006 cannot be expressed as the difference of squares of two integers. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,836 |
27. As shown in Figure $10, B C$ is the diameter of the semicircle $\odot O$, $E F \perp$ $B C$ at point $F, \frac{B F}{F C}=5$. Given $A B=8, A E=2$. Then the length of $A D$ is $(\quad$.
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{1+\sqrt{3}}{2}$
(C) $1+\sqrt{3}$
(D) $1+\sqrt{2}$ | 27. B.
Connect $B E$. Since $B C$ is the diameter, we know $\angle B E C=90^{\circ}$. Therefore,
$$
B E=\sqrt{A B^{2}-A E^{2}}=\sqrt{8^{2}-2^{2}}=2 \sqrt{15} \text {. }
$$
Also, from Rt $\triangle B F E \backsim \mathrm{Rt} \triangle E F C$, we have
$$
\frac{B E}{E C}=\frac{B F}{E F}=\frac{E F}{F C} \Rightarrow \frac... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,837 |
28. As shown in Figure 11, $\triangle A B C$ is translated along side $A B$ to the position of $\triangle A^{\prime} B^{\prime} C^{\prime}$. The overlapping part (the shaded area in the figure) is half the area of $\triangle A B C$. If $A B=\sqrt{2}$, then the distance the triangle has moved, $A A^{\prime}$, is ( ).
(A... | 28. A.
Let $A^{\prime} C^{\prime}$ intersect $B C$ at point $O$. Then $\triangle A B C \backsim \triangle A^{\prime} B O$. Also, $\frac{S_{\triangle A^{\prime} B O}}{S_{\triangle A B C}}=\frac{1}{2}$, so $\frac{A^{\prime} B}{A B}=\frac{1}{\sqrt{2}} \Rightarrow A^{\prime} B=1$.
Thus, $A A^{\prime}=\sqrt{2}-1$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,838 |
29. As shown in Figure 12, the areas of the two triangles formed by the diagonals and the two bases of trapezoid $ABCD$ are $p^{2}$ and $q^{2}$, respectively. Then the area of the trapezoid is ( ).
(A) $2\left(p^{2}+q^{2}\right)$
(B) $(p+q)^{2}$
(C) $p^{2}+q^{2}+p q$
(D) $p^{2}+q^{2}+\frac{p^{2} q^{2}}{p^{2}+q^{2}}$ | 29. B.
Let $A C$ and $B D$ intersect at point $O$. Then $\triangle O C D \backsim \triangle O A B$
$$
\begin{array}{l}
\text { Also, } \frac{S_{\triangle O C D}}{S_{\triangle O C B}}=\frac{q^{2}}{p^{2}} \Rightarrow \frac{O C}{O A}=\frac{O D}{O B}=\frac{q}{p}, \\
\text { hence } \frac{S_{\triangle O W D}}{S_{\triangle ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,839 |
30. The sum of the two diagonals of a rhombus is $l$, and the area is $S$. Then its side length is ( ).
(A) $\frac{1}{2} \sqrt{4 S-l^{2}}$
(B) $\frac{1}{2} \sqrt{4 S+l^{2}}$
(C) $\frac{1}{2} \sqrt{l^{2}-4 S}$
(D) $\frac{1}{2} \sqrt{l^{2}+4 S}$ | 30.C.
Let the lengths of the diagonals of the rhombus be $a$ and $b$, then $a+b=l$, $\frac{1}{2} a b=S$.
Therefore, the side length of the rhombus is $\frac{1}{2} \sqrt{a^{2}+b^{2}}=\frac{1}{2} \sqrt{l^{2}-4 S}$. Note: The reference answer is provided by Song Qiang, the editor of our journal.
（Provided by Nü Dao） | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,840 |
As shown in Figure 3, $\triangle ABC$ is inscribed in $\odot O$, with $AC > BC$. Points $D_{1}$ and $D_{2}$ are on $\overparen{AC}$, and $\overparen{AD_{1}} = \overparen{BCD_{2}}$. Connect $AD_{1}$, $AD_{2}$, $CD_{1}$, and $CD_{2}$. Prove that: $AD_{1} \cdot AD_{2} = AC \cdot BC + CD_{1} \cdot CD_{2}$. | Proof: As shown in Figure 2, connect $B D_{2}$, connect $D_{1} D_{2}$ and extend it to intersect the extension of $B C$ at point $E$.
Since $\overparen{A D_{1}}=\overparen{B C D_{2}}$, we have
$$
A D_{1}=B D_{2} \text {. }
$$
In $\triangle A C D_{2}$ and $\triangle D_{1} C E$, we have
$$
\angle C A D_{2}=\angle C D_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,841 |
192 In the square $A B C D$, points $E$ and $F$ are on sides $A B$ and $C D$ respectively, and $A E=\frac{1}{m} A B, D F=\frac{1}{n} D C, D E$ intersects $A F$ at point $G, G H \perp A B$, with the foot of the perpendicular being $H$. Try to express $\frac{A H}{A B}$ in terms of $m$ and $n$. | Solution: As shown in Figure 4, let the line $DE$ intersect the line $BC$ at point $L$, and the line $AF$ intersect the line $BC$ at point $K$. Draw a line through point $D$ parallel to $AK$ intersecting the line $BC$ at point $P$.
Given $AE=\frac{1}{m} AB, DF=\frac{1}{n} DC$, we know
$EB=\frac{m-1}{m} AB, FC=\frac{n-1... | \frac{1}{m+n} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,842 |
In $\triangle A B C$, prove that:
$$
\frac{\cos ^{4} A}{1+\cos ^{2} A}+\frac{\cos ^{4} B}{1+\cos ^{2} B}+\frac{\cos ^{4} C}{1+\cos ^{2} C} \geqslant \frac{3}{20} .
$$ | Proof: Notice that
$$
\begin{array}{l}
\frac{\cos ^{4} A}{1+\cos ^{2} A} \geqslant \frac{1}{25}\left(9 \cos ^{2} A-1\right) \\
\Leftrightarrow 25 \cos ^{4} A \geqslant\left(9 \cos ^{2} A-1\right)\left(1+\cos ^{2} A\right) \\
\Leftrightarrow 16 \cos ^{4} A-8 \cos ^{2} A+1 \geqslant 0 \\
\Leftrightarrow\left(4 \cos ^{2} ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,843 |
1. Given that $x$ and $y$ are real numbers, and satisfy
$$
\left(x+\sqrt{x^{2}+2008}\right)\left(y+\sqrt{y^{2}+2008}\right)=2008 \text {. }
$$
Then the value of $x^{2}-3 x y-4 y^{2}-6 x-6 y+2008$
is $\qquad$ | II, 1.2008.
From the given, we have
$$
x+\sqrt{x^{2}+2008}=\frac{2008}{y+\sqrt{y^{2}+2008}} .
$$
Rationalizing the denominator, we get
$$
x+\sqrt{x^{2}+2008}=\sqrt{y^{2}+2008}-y .
$$
Similarly, $y+\sqrt{y^{2}+2008}=\sqrt{x^{2}+2008}-x$.
$$
\begin{array}{l}
\text { (1) }+ \text { (2) gives } x+y=0 \text {. } \\
\text ... | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,845 |
3. The number of integer solutions to the equation $|x y|+|x+y|=1$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | 3.C.
Since $x, y$ are integers, then $|xy|, |x+y|$ are non-negative integers. Therefore, one of $|xy|, |x+y|$ is 0, and the other is 1. Considering the cases, we get 6 solutions. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,846 |
2. For real numbers $a, b, c$ satisfying
$$
a+b+c=0, \, abc=2 \text{.}
$$
then $u=|a|^{3}+|b|^{3}+|c|^{3}$ has the minimum value of
$\qquad$. | 2.10.
From the problem, we know that $a$, $b$, and $c$ must be one positive and two negative.
Without loss of generality, let $a>0$, $b<0$, and $c<0$.
Since $b+c=-a$ and $bc=\frac{2}{a}$, it follows that $b$ and $c$ are the two negative roots of the equation $x^{2}+a x+\frac{2}{a}=0$. Therefore, we have
$$
\Delta=a^{2... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,847 |
3. In the rectangular coordinate plane, given $A(-\sqrt{3}, 0)$, $B(\sqrt{3}, 0)$, point $P$ moves on the line $y=\frac{\sqrt{3}}{3}(x+4)+1$. When $\angle A P B$ is maximized, the value of $\frac{P A}{P B}$ is $\qquad$. | 3. $\sqrt{3}-1$.
As shown in Figure 5, let the intersection point of the line with the $x$-axis be $M$. By plane geometry knowledge, to maximize $\angle A P B$, the circle passing through points $A$, $B$, and $P$ must be tangent to the line at point $P$.
Since $\angle M P A = \angle M B P$, we have $\triangle M P A \s... | \sqrt{3} - 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,848 |
4. Let the three sides of the right triangle $\triangle ABC$ be $a$, $b$, and $c$, with $a < b < c$. If $\frac{b}{c+a} + \frac{a}{c+b} = \frac{17}{20}$. Then $a: b: c$ $=$ $~$. $\qquad$ | 4.8:15:17.
Since $c^{2}-a^{2}=b^{2}, c^{2}-b^{2}=a^{2}$, we have
$$
\begin{array}{l}
\frac{17}{20}=\frac{b}{c+a}+\frac{a}{c+b}=\frac{b(c-a)}{c^{2}-a^{2}}+\frac{a(c-b)}{c^{2}-b^{2}} \\
=\frac{c-a}{b}+\frac{c-b}{a}=\frac{c(a+b)-\left(a^{2}+b^{2}\right)}{a b} \\
=\frac{c(a+b-c)}{a b} .
\end{array}
$$
Also, $a b=\frac{(a... | 8:15:17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,849 |
One, (20 points) Given that $m$ and $n$ are positive integers, and $m > n, 2006 m^2 + m = 2007 n^2 + n$. Is $m - n$ a perfect square? Prove your conclusion.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $-、 m-n$ is a perfect square.
Proof as follows:
Let $m=n+k$ ($k$ is a positive integer).
Substitute into $2006 m^{2}+m=2007 n^{2}+n$, we get
$$
n^{2}-2 \times 2006 k n-\left(2006 k^{2}+k\right)=0 \text {. }
$$
Since $n$ is a positive integer, therefore,
$$
\Delta=4(2006 k)^{2}+4\left(2006 k^{2}+k\right)
$$
is a perfe... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,850 |
II. (25 points) As shown in Figure 3, in trapezoid $ABCD$, $AB \parallel CD$, $AD=12$, and $E$ is a point on side $CD$ such that $\frac{CE}{ED} = \frac{5}{4}$. Let the radius of the circle $\odot O_{1}$ passing through points $A, B, C, E$ be $R_{1}$, and the radius of the circle $\odot O_{2}$ passing through points $A,... | (1) Since $BC$ is the tangent of $\odot O_{2}$, therefore, $\angle ACB = \angle CDA$.
Also, $AB // CD$, so $\angle BAC = \angle ACD$.
Thus, $\triangle ABC \backsim \triangle CAD$.
Hence, $\angle ABC = \angle CAD$.
Therefore, $AD$ is the tangent of $\odot O_{1}$.
By the secant-tangent theorem, we have $AD^2 = DE \cdot D... | \frac{1}{3} < \frac{R_{1}}{R_{2}} < \frac{5}{3}, \frac{R_{1}}{R_{2}} \neq 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,851 |
Three. (25 points) Find the value of the real number $a$ such that the graph of the function
$$
f(x)=(x+a)(|x-a+1|+|x-3|)-2 x+4 a
$$
is centrally symmetric. | For the convenience of narration, use
$$
\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\} 、 \min \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}
$$
to represent the maximum and minimum numbers among $a_{1}, a_{2}, \cdots, a_{n}$, respectively.
When $x \leqslant \min \{a-1,3\}$, we have
$$
\begin{array}{l}
f(x)=(x+a)(-2 x+a+... | -\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,852 |
1. Given the following 4 propositions:
(1) If $m, n$ are known numbers, the sum of the monomials $2 x^{5} y^{n-2}$ and $(m+5) x^{1 m-n+4 \mid} y$ is a monomial, then the value of $m+n$ is -3 or 7.
(2) If $M, N$ are polynomials containing only one variable $x$, with degrees 6 and 3 respectively, then $M - N^{2}$ is a po... | -、1.D.
(1) When $m=-5, n>2$, the sum of the monomial $2 x^{5} y^{n-2}$ and 0 is still $2 x^{5} y^{n-2}$. At this time, $m+n$ can be all integers greater than -3, so statement (1) is incorrect.
(2) When $M=x^{6}+1, N=x^{3}+1$, $M-N^{2}=-2 x^{3}$. And $-2 x^{3}$ is a cubic monomial, so statement (2) is incorrect.
(3) Whe... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,853 |
2. As shown in Figure 1, $ABCD$ is a square with a side length of 1. On the line containing the diagonal $AC$, there are two points $M$ and $N$ such that $\angle MBN=135^{\circ}$. Then the minimum value of $MN$ is ( ).
(A) $1+\sqrt{2}$
(B) $2+\sqrt{2}$
Figure 1
(C) $3+\sqrt{2}$
(D) $2 \sqrt{2}$ | 2.B.
Let $A M=x$. It is easy to prove that $\triangle A B M \backsim \triangle C N B$.
Therefore, $\frac{A B}{C N}=\frac{A M}{C B}$, which means $\frac{1}{C N}=\frac{x}{1}$, or $C N=\frac{1}{x}$.
Thus, $M N=A M+A C+C N=x+\sqrt{2}+\frac{1}{x}$ $=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+2+\sqrt{2} \geqslant 2+\sqrt{... | 2+\sqrt{2} | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,854 |
3. Given real numbers $a, b, c$ satisfy
$$
\left(-\frac{1}{a}+b\right)\left(\frac{1}{a}+c\right)+\frac{1}{4}(b-c)^{2}=0 \text {. }
$$
Then the value of the algebraic expression $a b+a c$ is ( ).
(A) -2
(B) -1
(C) 1
(D) 2 | 3. A.
The given equation can be rewritten as
$$
4(a b+1)(a c+1)+(a b-a c)^{2}=0,
$$
which simplifies to $(a b+a c)^{2}+4(a b+a c)+4=0$, or equivalently $[(a b+a c)+2]^{2}=0$.
Thus, $a b+a c=-2$. | -2 | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,855 |
4. As shown in Figure 2, in $\triangle A B C$, $\angle B A C=$ $60^{\circ}, B C=18, D$ is a point on $A B$, $A C=B D, E$ is the midpoint of $C D$. Then the length of $A E$ is ( ).
(A) 12
(B) 9
(C) $9 \sqrt{3}$
(D) None of the above | 4.B.
As shown in Figure 12, extend $AC$ to point $F$ such that $CF = AD$. Connect $BF$, and draw $CG \parallel AB$ intersecting $BF$ at point $G$. Connect $DG$ and $AG$.
Since $AC = DB$,
$$
CF = AD \text{, }
$$
Therefore, $AC + CF = DB + AD$, which means $AF = AB$.
Also, $\angle BAC = 60^{\circ}$, so $\triangle ABF$ ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,856 |
4. $a, b$ are the roots of the equation $x^{2}+(m-5) x+7=0$. Then $\left(a^{2}+m a+7\right)\left(b^{2}+m b+7\right)=$ ( ).
(A) 365
(B) 245
(C) 210
(D) 175 | $\begin{array}{l}\text { 4.D. } \\ \text { Given } a b=7, a^{2}+m a+7=5 a, b^{2}+m b+7=5 b, \text { then } \\ \text { therefore, }\left(a^{2}+m a+7\right)\left(b^{2}+m b+7\right)=25 a b=175 \text {. }\end{array}$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,857 |
5. Given real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
2005 a^{3}=2006 b^{3}=2007 c^{3}=2008 d^{3} \\
\sqrt[3]{2005 a^{2}+2006 b^{2}+2007 c^{2}+2008 d^{2}} \\
=\sqrt[3]{2005}+\sqrt[3]{2006}+\sqrt[3]{2007}+\sqrt[3]{2008} .
\end{array}
$$
Then the value of $a^{-1}+b^{-1}+c^{-1}+d^{-1}$ is ( ).
(A) 1
(B) 0
(C) -1... | 5.D.
Let $x=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$,
$2005 a^{3}=2006 b^{3}=2007 c^{3}=2008 d^{3}=k^{3}$.
Obviously, $a, b, c, d, k$ have the same sign and are not zero, then
$$
\begin{array}{l}
2005 a^{2}=\frac{k^{3}}{a}, 2006 b^{2}=\frac{k^{3}}{b}, \\
2007 c^{2}=\frac{k^{3}}{c}, 2008 d^{2}=\frac{k^{3}}{d} ;... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,858 |
6. Given in the Cartesian coordinate system $x O y$, the line $y=2 k x+3-4 k$ intersects the positive semi-axis of $x$ and the positive semi-axis of $y$ at points $A$ and $B$ respectively, and $P$ is a point on the line segment $A B$. $P M \perp x$ axis at point $M$, and $P N \perp y$ axis at point $N$. Then the maximu... | 6.C.
Let the coordinates of point $P$ be $\left(x_{0}, y_{0}\right)$, and the area of rectangle OMPN be $S$. Then $x_{0}>0, y_{0}>0, S=x_{0} y_{0}$.
Since point $P\left(x_{0}, y_{0}\right)$ lies on $y=2 k x+3-4 k$, we have $y_{0}=2 k x_{0}+3-4 k$.
Thus, $S=x_{0}\left(2 k x_{0}+3-4 k\right)=2 k x_{0}^{2}+(3-4 k) x_{0}$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,859 |
1. As shown in Figure $3, A D$ is the angle bisector of $\triangle A B C$,
$$
\begin{array}{l}
\angle C=2 \angle B, A B=a^{2}- \\
4 b+4, A C=8 c-27- \\
2 b^{2}, C D=9+4 a-c^{2}
\end{array}
$$
Then $B C=$ . $\qquad$ | II. $1 \cdot \frac{7}{3}$.
Extend $AC$ to point $E$ such that $CE = CD$, and connect $DE$. Then we have $\angle E = \angle CDE = \frac{1}{2} \angle ACD = \angle B$.
Since $AD$ is the angle bisector of $\angle BAC$, then
$$
\angle BAD = \angle EAD, \frac{AB}{BD} = \frac{AC}{CD}.
$$
Therefore, $\triangle ABD \cong \tri... | \frac{7}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,860 |
2. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a-b+c=7, \\
a b+b c+b+c^{2}+16=0 .
\end{array}
$$
Then the value of $\left(a^{-1}-b^{-1}\right)^{a k c}(a+b+c)^{a+b+c}$ is $\qquad$ . | 2. -1 .
From equation (1), we get
$$
(-b)+(a+c+1)=8 \text {. }
$$
From equation (2), we get
$$
(a+c+1)(-b)=c^{2}+16 \text {. }
$$
Therefore, $a+c+1$ and $-b$ are the two roots of the equation
$$
x^{2}-8 x+c^{2}+16=0
$$
Thus, we have $\Delta=(-8)^{2}-4\left(c^{2}+16\right) \geqslant 0$.
Hence, $c^{2} \leqslant 0$.
I... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,861 |
3. As shown in Figure 4, the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $O$, and $E$, $F$, and $G$ are the midpoints of $AB$, $OC$, and $OD$ respectively. Given that $OA = AD$, $OB = BC$, and $CD = \sqrt{3} AB$. Then the degree measure of $\angle FEG$ is $\qquad$ | 3. $120^{\circ}$.
As shown in Figure 13, connect $A G$, $B F$, and $F G$, and draw $E P \perp F G$ at point $P$.
Let $A B=2 a$, then $C D=\sqrt{3} A B=2 \sqrt{3} a$.
Since $O A=A D$ and $G$ is the midpoint of $O D$, we have
$A G \perp O D$.
Therefore, $\angle A G B=90^{\circ}$.
Similarly, $\angle A F B=90^{\circ}$.
Th... | 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,862 |
4. As shown in Figure 5, the quadrilateral $ABCD$ is a schematic diagram of a desert area, with points $A$ and $B$ on the $x$-axis, $E(2,6)$, $F(3,4)$. The broken line $OFE$ is a water channel flowing through this desert. The desert to the east of the water channel is contracted for greening by person A, and the desert... | 4. $\left(\frac{5}{3}, 0\right)$.
As shown in Figure 14, connect $O E$, draw $F P / / O E$ intersecting $A B$ at point $P$, connect $E P$ intersecting $O F$ at point $G$.
Since $O E / / P F$, then
$S_{\triangle O E F}=S_{\triangle O B P}$.
Thus, $S_{\triangle O E P}-S_{\triangle O E C}$
$$
=S_{\triangle O B P}-S_{\tri... | \left(\frac{5}{3}, 0\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,863 |
One. (20 points) There are three cylindrical containers $M$, $N$, and $P$, whose axial cross-sections are shown in figures 6(a), (b), and (c), respectively. The internal base areas are $S_{1} \mathrm{~cm}^{2}$, $S_{2} \mathrm{~cm}^{2}$, and $S_{3} \mathrm{~cm}^{2}$, and the internal heights are $h_{1} \mathrm{~cm}$, $h... | (1) From Figure 7, we know that filling containers $M$, $N$, and $P$ takes $60 \mathrm{~s}$. From Figure 8, we know that filling two of the three containers $M$, $N$, and $P$ takes $54 \mathrm{~s}$. Therefore, filling the top container in Figure 8 takes $60-54=6(\mathrm{~s})$.
Similarly, filling the top container in F... | 720 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,864 |
II. (25 points) As shown in Figure 10, the distance between two parallel lines $l_{1}$ and $l_{2}$ is 6. There is a fixed circle $\odot O$ with a radius of 1 between $l_{1}$ and $l_{2}$, tangent to line $l_{2}$ at point $A$. $P$ is a moving point on line $l_{1}$. Two tangents $PB$ and $PC$ are drawn from $P$ to $\odot ... | As shown in Figure 15, draw $PD \perp l_{2}$ at point $D$, and connect $OA, OB, OC, OM, ON,$ and $OP$. Then
$$
\begin{array}{l}
PD=6, \\
OA=OB=OC=1 .
\end{array}
$$
Let $AM=m, AN=n, PC=p, DN=x$, then
$$
DM=m+n-x .
$$
From the problem, $\odot O$ is the incircle of $\triangle PMN$, so
$$
\begin{array}{l}
BM=AM=m, CN=AN... | 1.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,865 |
Three. (25 points) Given that $k$ is a constant, the quadratic equation in $x$
$$
\left(k^{2}-2 k\right) x^{2}+(4-6 k) x+8=0
$$
has solutions that are all integers. Find the value of $k$. | When $k=0$, the original equation becomes $4 x+8=0$, solving for $x$ gives $x=$ -2. Therefore, when $k=0$, the solutions to the original equation are all integers.
When $k=2$, the original equation becomes $-8 x+8=0$, solving for $x$ gives $x=$ 1. Therefore, when $k=2$, the solutions to the original equation are all i... | -2,0,1,2,\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,866 |
1. Given that there are 6 parallel network lines between locations $A$ and $B$, and their maximum information capacities are $1,1,2,2,3,4$. Now, if we randomly select three network lines such that the sum of their maximum information capacities is greater than or equal to 6, the number of ways to do this is ( ) ways.
(... | -,1.C.
As shown in Figure 6, according to the problem, at least one of the fifth and sixth network cables must be used.
If the sixth network cable is chosen, any two of the first five can be selected, which gives a total of 10 ways;
If the fifth network cable is chosen, then selecting the third and fourth (3, 2, 2) h... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,867 |
5. As shown in Figure 1, in the right triangle $\triangle ABC$, the hypotenuse $BC=4$, $\angle ABC=30^{\circ}$, and circles are drawn with $AB$ and $AC$ as diameters, respectively. Then the area of the common part of these two circles is ( )
(A) $\frac{2 \pi}{3}+\frac{\sqrt{3}}{2}$
(B) $\frac{5 \pi}{6}-\frac{2 \sqrt{3}... | 5.C.
Let the area of the common part of the two circles be $S$. As shown in Figure 4, it is easy to see that
$$
\begin{array}{l}
=\frac{\pi}{3} \times 1^{2}+\frac{\pi}{6} \times(\sqrt{3})^{2}-\frac{1}{2} \times 2 \times \sqrt{3}=\frac{5 \pi}{6}-\sqrt{3} . \\
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,868 |
2. Let $a, b, c$ be any distinct positive numbers,
$$
x=\frac{b^{2}+1}{a}, y=\frac{c^{2}+1}{b}, z=\frac{a^{2}+1}{c} \text {. }
$$
Then the three numbers $x, y, z$ ( ).
(A) are all not greater than 2
(B) at least one is greater than 2
(C) are all not less than 2
(D) at least one is less than 2 | 2. B.
Notice that
$$
\begin{array}{l}
x y z=\frac{b^{2}+1}{a} \cdot \frac{c^{2}+1}{b} \cdot \frac{a^{2}+1}{c} \\
=\frac{b^{2}+1}{b} \cdot \frac{c^{2}+1}{c} \cdot \frac{a^{2}+1}{a} \\
=\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\left(c+\frac{1}{c}\right) .
\end{array}
$$
Since \(a, b, c\) are distinct positiv... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,869 |
3. In the convex quadrilateral $A B C D$, $\angle B=\angle D = 90^{\circ}$, $A D=D C$. If $S_{\text {quadrilateral } A B C D}=12$, then $A B+B C$ $=(\quad)$.
(A)6
(B) $4 \sqrt{3}$
(C) $5 \sqrt{2}$
(D) $3 \sqrt{6}$ | 3. B.
As shown in Figure 7, rotate $\triangle D A B$ to the position of $\triangle D C K$.
Since $A D=D C$, point $A$ must coincide with point $C$.
Since $\angle B=\angle D=90^{\circ}$,
it follows that $\angle A+\angle C=180^{\circ}$.
Therefore, points $B$, $C$, and $K$ are collinear.
At this point, $\triangle D B K... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,870 |
4. Given the quadratic function $f(x)=a x^{2}+b x+c$ whose graph is shown in Figure 1. Let
$$
\begin{aligned}
p= & |a-b+c|+ \\
& |2 a+b|, \\
q= & |a+b+c|+ \\
& |2 a-b| .
\end{aligned}
$$
Then ( ).
(A) $p>q$
(B) $p=q$
(C) $p<q$
(D) The relationship between $p$ and $q$ cannot be determined | 4.C.
From Fig. 1, we know that $f(0)=c=0, f(1)=a+b+c>0$, then
$$
a+b>0, f(-1)=a-b+c=a-b1$, we get
$$
2 a+b>0 \text {. }
$$
And $2 a-b=a+(a-b)<0$, so,
$$
\begin{array}{l}
p=-(a-b)+(2 a+b)=a+2 b, \\
q=(a+b)-(2 a-b)=-a+2 b .
\end{array}
$$
Since $p-q=(a+2 b)-(-a+2 b)=2 a<0$, hence $p<q$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,871 |
5. If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are distinct positive odd numbers, and satisfy
$$
\begin{array}{l}
\left(2005-x_{1}\right)\left(2005-x_{2}\right)\left(2005-x_{3}\right) . \\
\left(2005-x_{4}\right)\left(2005-x_{5}\right)=24^{2},
\end{array}
$$
then the last digit of $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x... | 5.A.
Since $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are distinct positive odd numbers, the 5 factors on the left side of the given equation are distinct even numbers. And $24^{2}$ can be decomposed into the product of 5 distinct even numbers in only one unique form:
$$
24^{2}=2 \times(-2) \times 4 \times 6 \times(-6),
$$
... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,872 |
6. As shown in Figure 2, in trapezoid $A B C D$, $D C / / A B$, a line $M N$ $/ / A B$ is drawn through the intersection of $A C$ and $B D$, with points $M$ and $N$ on $A D$ and $B C$ respectively. The relationship satisfied by $A B$, $D C$, and $M N$ is ( ).
(A) $A B+D C=\sqrt{5} M N$
(B) $A B \cdot D C=M N^{2}$
(C) $... | 6.D.
Let the intersection point of $A C$ and $B D$ be $O$.
By the properties of similar triangles, we have
$$
\frac{M O}{A B}=\frac{D O}{D B}=\frac{C O}{C A}=\frac{O N}{A B} \text {. }
$$
Therefore, $M O=O N, M N=2 M O$.
Also, $\frac{M O}{D C}+\frac{M O}{A B}=\frac{A M}{A D}+\frac{D M}{A D}=1$, thus
$$
\frac{1}{A B}+... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,873 |
1. The second number in the $n$-th row of the number array shown in Figure 3
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | $$
\text { II. 1. } n^{2}-2 n+3
$$
The first number in the $n$-th row is
$$
\begin{array}{l}
3+3+5+7+\cdots+(2 n-3) \\
=2+(1+3+5+7+\cdots+2 n-3) \\
=2+\frac{(1+2 n-3)(n-1)}{2}=n^{2}-2 n+3 .
\end{array}
$$ | n^{2}-2 n+3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,874 |
2. For the quadratic equation in $x$
$$
m^{2} x^{2}+(2 m+3) x+1=0
$$
there are two real roots whose product is 1; for the quadratic equation in $x$
$$
x^{2}+(2 a+m) x+2 a+1-m^{2}=0
$$
there is a real root that is greater than 0 and less than 4. Then the integer value of $a$ is . $\qquad$ | 2, -1.
According to the problem, we have $m^{2}=1, m= \pm 1$, and
$$
\Delta=(2 m+3)^{2}-4 m^{2}=12 m+9>0 \text {. }
$$
Thus, $m=1$.
Let $f(x)=x^{2}+(2 a+1) x+2 a$, then we should have
$$
f(0) f(4)=2 a(20+10 a)<0,
$$
which means $a(a+2)<0$.
Therefore, $-2<a<0$. Hence, $a=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,875 |
3. In a certain basketball tournament, Xiao Ming played 10 games. In the 6th, 7th, 8th, and 9th games, he scored 23 points, 14 points, 11 points, and 20 points, respectively. His average score in the first 9 games was higher than his average score in the first 5 games. If the average score of the 10 games he played exc... | 3.29.
Let the average score of the first 5 games be $x$, then the average score of the first 9 games is
$$
\frac{5 x+23+14+11+20}{9}=\frac{5 x+68}{9} .
$$
According to the problem, $\frac{5 x+68}{9}>x$.
Solving this, we get $x<17$.
Therefore, the total score of the first 5 games is at most $5 \times 17-1=84$ points.
... | 29 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,876 |
4. As shown in Figure $4, \odot O$ is the circumcircle of $\triangle A B C$, $B C=$ $a, C A=b$, and $\angle A-\angle B$ $=90^{\circ}$. Then the radius of $\odot O$ is $\qquad$ | 4. $\frac{1}{2} \sqrt{a^{2}+b^{2}}$.
Draw the diameter $C D$ of $\odot O$, and connect $D B$.
Since $\angle A=\angle B+90^{\circ}=\angle B+\angle C B D=\angle A B D$,
we have $\overparen{C D B}=\overparen{A C D}$.
Thus, $\overparen{A C}=\overparen{B D}, B D=A C=b$.
At this point, $C D=\sqrt{a^{2}+b^{2}}$.
Therefore, t... | \frac{1}{2} \sqrt{a^{2}+b^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,877 |
One, (20 points) Prove: There exists a positive constant $c$, such that for all real numbers $x, y, z$, we have
$$
\begin{array}{l}
1+|x+y+z|+|x y+y z+z x|+|x y z| \\
>c(|x|+|y|+|z|) .
\end{array}
$$ | Let the left side of the inequality be denoted as $M$. Clearly, for any $x, y, z$, we have
$$
\begin{array}{l}
M^{2}>(x+y+z)^{2}+2|x y+y z+z x| \\
=x^{2}+y^{2}+z^{2}+2(x y+y z+z x+|x y+y z+z x|) \\
\geqslant x^{2}+y^{2}+z^{2} .
\end{array}
$$
Since $x^{2}+y^{2} \geqslant 2|x y|, y^{2}+z^{2} \geqslant 2|y z|, x^{2}+z^{... | c=\frac{\sqrt{3}}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 717,878 |
6. Choose $k$ different numbers from $1,2, \cdots, 13$, such that the difference between any two of these $k$ numbers is neither 5 nor 8. Then the maximum value of $k$ is $(\quad$.
(A) 5
(B)6
(C) 7
(D) 8 | 6. B.
Arrange these 13 numbers in a circle such that the difference between any two adjacent numbers is 5 or 8 (as shown in Figure 5). If more than 6 numbers are taken, there must be 2 numbers that are adjacent on the circle. On the other hand, 6 numbers that meet the condition can be taken (by selecting any 6 numbers... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,879 |
II. (25 points) As shown in Figure 5, with the side $AB$ of the acute triangle $\triangle ABC$ as the diameter, a semicircle $\odot O$ is drawn intersecting sides $BC$ and $CA$ at points $E$ and $F$, respectively. Tangents to $\odot O$ are drawn through points $E$ and $F$, intersecting at point $P$. Prove that:
$$
CP \... | II. As shown in Figure 8, connect $A E$ and $B F$ to get the intersection point $Q$. Clearly, point $Q$ is the orthocenter of $\triangle A B C$, and thus
$$
C Q \perp A B \text {. }
$$
Extend $F P$ to point $K$ such that $P K = P F$, and connect $E F$ and $K E$.
It is easy to see that $\angle P E F = \angle P F E$
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,880 |
Three. (25 points) Given the quadratic function
$$
f(x)=a x^{2}+b x+c
$$
the graph passes through points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, and satisfies
$$
a^{2}+\left(y_{1}+y_{2}\right) a+y_{1} y_{2}=0 \text {. }
$$
(1) Prove: $y_{1}=-a$ or $y_{2}=-a$;
(2) Prove: The graph of the function... | Three, (1) From $a^{2}+\left(y_{1}+y_{2}\right) a+y_{1} y_{2}=0$, we get $\left(a+y_{1}\right)\left(a+y_{2}\right)=0$.
Solving, we get $y_{1}=-a$ or $y_{2}=-a$.
(2) When $a>0$, the graph of the quadratic function $f(x)$ opens upwards, and the y-coordinates of points $A$ and $B$ on the graph are at least one of them is ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,881 |
1. If the two roots of the equation $x^{2}-3 x+1=0$ are also roots of the equation $x^{4}-p x^{2}+q=0$, then the units digit of $(p+q)^{2008}$ is ( ).
(A)2
(B) 4
(C) 6
(D) 8 | -1 1.C.
Let the two roots of the equation $x^{2}-3 x+1=0$ be $\alpha, \beta$, then $\alpha^{4}-p \alpha^{2}+q=0, \beta^{4}-p \beta^{2}+q=0$. Also, $\Delta=(-3)^{2}-4=5>0$, so $\alpha \neq \beta$, which means $\alpha^{2} \neq \beta^{2}$. From equation (1), we get
$$
p=\frac{\alpha^{4}-\beta^{4}}{\alpha^{2}-\beta^{2}}=\a... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,882 |
2. The function $y=\frac{1}{x^{2}+a x+b}$ (where $a, b$ are non-zero constants) achieves its maximum value under the condition ( ).
(A) $a^{2}-4 b \geqslant 0$
(B) $a^{2}-4 b \neq 0$
(C) $a^{2}-4 b<0$
(D) It depends on the values of $a, b$, cannot be determined | 2.C.
Notice that
$$
y=\frac{1}{x^{2}+a x+b}=\frac{4}{4\left(x+\frac{a}{2}\right)^{2}+4 b-a^{2}} \text {. }
$$
(1) If $a^{2}-4 b \geqslant 0$, then $x^{2}+a x+b=0$ has real roots, in which case, $y$ has no maximum value;
(2) If $a^{2}-4 b<0$, then $0<y \leqslant \frac{4}{4 b-a^{2}}$.
In summary, when $a^{2}-4 b<0$, $y_... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,883 |
3. Let the inradius of $\triangle A B C$ be $r, B C=a$, $A C=b, A B=c$, and the altitudes be $h_{a} 、 h_{b}$ 、 $h_{c}$, satisfying $h_{a}+h_{b}+h_{c}=9 r$. Then the shape of $\triangle A B C$ is ( ).
(A) must be an obtuse triangle
(B) must be an equilateral triangle
(C) must not be an acute triangle
(D) may not be a ri... | 3. B.
It is known that $S_{\triangle A B C}=\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=\frac{1}{2} c h_{c}$, so we have $h_{a}+h_{b}+h_{c}=2 S_{\triangle I B C}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.
Since $S_{\triangle A B C}=\frac{1}{2} r(a+b+c)$,
$$
h_{a}+h_{b}+h_{c}=9 r \text {, }
$$
then we have $(a+b+c... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,884 |
4. If three distinct non-zero real numbers $x$, $y$, $z$ satisfy the relation
$$
x(y-z)=\frac{y(z-x)}{q}=\frac{z(y-x)}{q^{2}},
$$
then the value of $q$ is ( ).
(A) $\frac{-1 \pm \sqrt{5}}{2}$
(B) $\frac{1 \pm \sqrt{5}}{2}$
(C) $\pm \frac{-1+\sqrt{5}}{2}$
(D) $\pm \frac{1-\sqrt{5}}{2}$ | 4.B.
From the given conditions, we know $q \neq \pm 1$.
By the properties of geometric sequences, we have
$$
\frac{x(y-z)+y(z-x)}{1+q}=\frac{z(y-x)}{q^{2}}.
$$
Also, $x(y-z)+y(z-x)=z(y-x)$, so $1+q=q^{2}$, which simplifies to $q^{2}-q-1=0$. Solving this, we get $q=\frac{1 \pm \sqrt{5}}{2}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,885 |
5. If one of the legs of a right triangle is 12, and the other two sides are integers, then the number of right triangles that meet this condition is ( ).
(A) 1
(B) 6
(C) 4
(D) infinitely many | 5.C.
Let $a=12, c$ be the hypotenuse, then we have
$$
c^{2}-b^{2}=a^{2}=144 \text {. }
$$
Since $144=2^{4} \times 3^{2}$, we have
$$
\begin{array}{l}
(c+b)(c-b)=72 \times 2 ; \\
(c+b)(c-b)=36 \times 4 ; \\
(c+b)(c-b)=18 \times 8 ; \\
(c+b)(c-b)=16 \times 9 ; \\
(c+b)(c-b)=48 \times 3 ; \\
(c+b)(c-b)=24 \times 6 .
\en... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,886 |
6. As shown in Figure 1, in $\triangle A B C$, $\angle A>\angle B>$ $\angle C, I$ is the incenter. Three routes are given:
(1) $I \rightarrow A \rightarrow C \rightarrow B \rightarrow I$;
(2) $I \rightarrow C \rightarrow B \rightarrow A \rightarrow I$;
(3) $I \rightarrow B \rightarrow A \rightarrow C \rightarrow I$.
If... | 6.C.
Since $\angle A > \angle B > \angle C$, then $a > b > c$.
Let $A I = x, B I = y, C I = z$ (it is easy to know $xI D$,
$$
i.e., $x + (a - b) > y$.
Therefore, $b + y < a + x$.
Similarly, $c + z < b + y$.
Thus, $c + z < b + y < a + x$.
From this, comparing $l_{1}, l_{2}, l_{3}$, we can see that $l_{3}$ is the short... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,887 |
$$
\left\{\begin{array}{l}
2006(x-y)+2000(y-z)+2008(z-x)=0, \\
2000^{2}(x-y)+2000^{2}(y-z)+2000^{2}(z-x)=2008 .
\end{array}\right.
$$
Then the value of $z-y$ is $\qquad$ . | $=、 1.2008$
Let $z-y=t$.
From the first equation, we get $z-2 x+y=0$.
Thus, $z-x=\frac{t}{2}$, and consequently, $x-y=\frac{t}{2}$.
Therefore, from the second equation, we can find that
$$
t=z-y=2008 .
$$ | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,888 |
2. As shown in Figure 2, in quadrilateral $ABCD$, $AB=10$, $BC=17$, $CD=13$, $DA$ $=20$, $AC=21$. Then $BD=$ | $2.10 \sqrt{5}$.
As shown in Figure 6, draw $B E \perp A C$ at point $E$, $D F \perp A C$ at point $F$, and draw $B G / / A C$ intersecting the extension of $D F$ at point $G$. Then, quadrilateral $B E F G$ is a rectangle, and $\triangle B D G$, $\triangle A B E$, and $\triangle C D F$ are all right triangles.
In $\tri... | 10 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,889 |
1. If the quadratic equation with integer coefficients
$$
x^{2}+(a+3) x+2 a+3=0
$$
has one positive root $x_{1}$ and one negative root $x_{2}$, and $\left|x_{1}\right|<\left|x_{2}\right|$, then
$$
a=
$$
$\qquad$ | $=、 1 .-2$.
Since the two roots of the equation are not equal, we have $\Delta>0$, that is
$$
(a+3)^{2}>4(2 a+3) \text {. }
$$
Solving this, we get $a>3$ or $a-3, a<-\frac{3}{2}$, which means $-3<a<-\frac{3}{2}$.
Since $a$ is an integer, then $a=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,890 |
3. A three-digit number $x y z$ (where $x$, $y$, and $z$ are distinct), rearranging its digits to form the largest and smallest possible three-digit numbers. If the difference between the largest and smallest three-digit numbers is equal to the original three-digit number, then this three-digit number is $\qquad$ | 3.495.
Let the three-digit number $\overline{y z}$, after rearrangement, form the largest three-digit number $\overline{a b c}(a>b>c)$, then the smallest three-digit number is $\overline{c b a}$.
Since $1 \leqslant a \leqslant 9,1 \leqslant b \leqslant 9,1 \leqslant c \leqslant 9$, and
$$
\begin{array}{l}
\overline{a ... | 495 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,891 |
4. As shown in Figure 3, in the isosceles right triangle $\triangle ABC\left(\angle C=90^{\circ}\right)$, take a point $P$ inside, and $AP=AC=$ $a, BP=CP=b(a>b)$. Then $\frac{a^{2}+b^{2}}{a^{2}-b^{2}}=$ $\qquad$ | 4. $\sqrt{3}$.
From the problem, we know that $\angle B P C$ and $\angle C A P$ are supplementary.
As shown in Figure 7, extend $B P$ to intersect $A C$
at point $K$, then
$$
P C=P K=b, B K=2 b .
$$
Since $\triangle P C K \sim \triangle A C P$, we have
$$
C K=\frac{b^{2}}{a} \text {. }
$$
Also, $B K^{2}=B C^{2}+C K^... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,892 |
One. (20 points) If the lengths of the two legs $a, b (a \neq b)$ of a right triangle are both integers, and satisfy
$$
\left\{\begin{array}{l}
a+b=m+2, \\
a b=4 m .
\end{array}\right.
$$
Find the lengths of the three sides of this right triangle. | Since $a$ and $b$ are positive integers, $m$ is also a positive integer. Therefore, $a$ and $b$ are two distinct integer solutions of the quadratic equation $x^{2}-(m+2)x+4m=0$.
Thus, $\Delta=(m+2)^{2}-16m=m^{2}-12m+4$ must be a perfect square.
Let $m^{2}-12m+4=k^{2}$ (where $k$ is a positive integer), i.e.,
$$
m^{2}-... | 5, 12, 13 \text{ or } 6, 8, 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,893 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.