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Let sets $A$ and $B$ be sets composed of positive integers, $|A|=10, |B|=9$, and set $A$ satisfies the following condition:
If $x, y, u, v \in A, x+y=u+v$, then $\{x, y\}=\{u, v\}$.
Let $A+B=\{a+b \mid a \in A, b \in B\}$.
Prove: $|A+B| \geqslant 50$ ( $|X|$ represents the number of elements in set $X$).
(2005, Nation... | Prove: Let $A=\left\{a_{1}, a_{2}, \cdots, a_{10}\right\}$,
$$
\begin{array}{l}
B=\left\{b_{1}, b_{2}, \cdots, b_{9}\right\}, \\
C_{i}=\left\{a_{1}+b_{i}, a_{2}+b_{i}, \cdots, a_{10}+b_{i}\right\}
\end{array}
$$
where $a_{i}(i=1,2, \cdots, 10)$ are distinct.
Prove by contradiction.
$$
\left|C_{i} \cap C_{j}\right| \le... | 54 | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,664 |
1. Given $\triangle A B C$. If for any $t \in \mathbf{R}, \mid B A- t B C|\geqslant| A C \mid$, then $\triangle A B C(\quad)$.
(A) must be an acute triangle
(B) must be an obtuse triangle
(C) must be a right triangle
(D) the answer is uncertain | -1.C.
Let $\angle ABC = \alpha$, and draw $AD \perp BC$ at point $D$.
From $|BA - tBC| \geq |AC|$, we get
$|BA|^2 - 2tBA \cdot BC + t^2|BC|^2 \geq |AC|^2$.
Let $t = \frac{BA \cdot BC}{|BC|^2}$, substituting into the above equation gives
$|BA|^2 - 2|BA|^2 \cos^2 \alpha + \cos^2 \alpha |BA|^2 \geq |AC|^2$,
which simplifi... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,665 |
2. Let $\log _{x}\left(2 x^{2}+x-1\right)>\log _{x} 2-1$. Then the range of $x$ is $(\quad)$.
(A) $\frac{1}{2}\frac{1}{2}, x \neq 1$
(C) $x>1$
(D) $0<x<1$ | 2. B.
Since $\left\{\begin{array}{l}x>0, x \neq 1, \\ 2 x^{2}+x-1>0,\end{array}\right.$ therefore, $x>\frac{1}{2}, x \neq 1$.
From $\log _{x}\left(2 x^{2}+x-1\right)>\log _{x} 2-1$, we deduce
$$
\log _{x}\left(2 x^{3}+x^{2}-x\right)>\log _{x} 2 \text {, }
$$
further leading to
$$
\left\{\begin{array} { l }
{ 0 < x <... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,666 |
3. Given the set $A=\{x \mid 5 x-a \leqslant 0\}, B=$ $\{x \mid 6 x-b>0\}, a, b \in \mathbf{N}$, and $A \cap B \cap \mathbf{N}=$ $\{2,3,4\}$. Then the number of integer pairs $(a, b)$ is ( ).
(A) 20
(B) 25
(C) 30
(D) 42 | 3.C.
From $5 x-a \leqslant 0$, we get $x \leqslant \frac{a}{5}$;
From $6 x-b>0$, we get $x>\frac{b}{6}$.
To make $A \cap B \cap \mathbf{N}=\{2,3,4\}$, it should be
$$
\left\{\begin{array}{l}
1 \leqslant \frac{b}{6} < 2, \\
4 \leqslant \frac{a}{5} < 5,
\end{array} \text { i.e., } \left\{\begin{array}{l}
6 \leqslant b<1... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,667 |
Example 4 As shown in Figure $4, \odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$. A tangent line to $\odot O_{2}$ through point $A$ intersects $\odot O_{1}$ at point $C$. Line $CB$ intersects $\odot O_{2}$ at point $D$, and line $DA$ intersects $\odot O_{1}$ at point $E$. Connect $CE$.
Find this: $DA \c... | Proof: As shown in Figure 4, connect $A B$. Since $A C$ is the tangent of $\odot O_{2}$, we have $\angle E A C = \angle A B D = \angle E$. Therefore, $C E = C A$.
Obviously, $C A^{2} = C B \cdot C D$, so $C E^{2} = C B \cdot C D$.
It is easy to see that $D A \cdot D E = D B \cdot D C = D C(D C - B C) = D C^{2} - D C \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,668 |
4. In the right triangular prism $A_{1} B_{1} C_{1}-A B C$, $\angle B A C = \frac{\pi}{2}, A B = A C = A A_{1} = 1$. Given that $G$ and $E$ are the midpoints of $A_{1} B_{1}$ and $C C_{1}$ respectively, and $D$ and $F$ are moving points on segments $A C$ and $A B$ (excluding the endpoints). If $G D \perp E F$, then the... | 4. A.
Taking $A$ as the origin, $AB$ as the $x$-axis, $AC$ as the $y$-axis, and $AA_1$ as the $z$-axis to establish a spatial rectangular coordinate system, we have
$$
\begin{array}{l}
F\left(t_{1}, 0,0\right)\left(0<t_{1}<1\right), E\left(0,1, \frac{1}{2}\right), \\
G\left(\frac{1}{2}, 0,1\right), D\left(0, t_{2}, 0\... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,669 |
5. Let $f(x)=x^{3}+\log _{2}\left(x+\sqrt{x^{2}+1}\right)$. Then for any real numbers $a, b, a+b \geqslant 0$ is a ( ) for $f(a)+f(b) \geqslant 0$.
(A) sufficient and necessary condition
(B) sufficient but not necessary condition
(C) necessary but not sufficient condition
(D) neither sufficient nor necessary condition | 5.A.
Obviously, $f(x)=x^{3}+\log _{2}\left(x+\sqrt{x^{2}+1}\right)$ is an odd function and monotonically increasing. Therefore, if $a+b \geqslant 0$, then $a \geqslant-b$, so $f(a) \geqslant f(-b)$, which means $f(a) \geqslant-f(b)$. Thus, $f(a)+f(b) \geqslant 0$.
Conversely, if $f(a)+f(b) \geqslant 0$, then $f(a) \g... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,670 |
7. Let $f(x)=\sin ^{4} x-\sin x \cdot \cos x+\cos ^{4} x$. Then the range of $f(x)$ is $\qquad$ . | $$
\begin{array}{l}
\text { II.7.[0, } \left.\frac{9}{8}\right] . \\
f(x)=\sin ^{4} x-\sin x \cdot \cos x+\cos ^{4} x \\
=1-\frac{1}{2} \sin 2 x-\frac{1}{2} \sin ^{2} 2 x
\end{array}
$$
Let $t=\sin 2 x$, then
$$
\begin{array}{l}
f(x)=g(t) \\
=1-\frac{1}{2} t-\frac{1}{2} t^{2}=\frac{9}{8}-\frac{1}{2}\left(t+\frac{1}{2}... | 0 \leqslant f(x) \leqslant \frac{9}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,672 |
8. If for all $\theta \in \mathbf{R}$, the complex number
$$
z=(a+\cos \theta)+(2 a-\sin \theta) \mathrm{i}
$$
has a modulus not exceeding 2, then the range of the real number $a$ is
$\qquad$. | 8. $\left[-\frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right]$.
According to the problem, we have
$$
\begin{array}{l}
|z| \leqslant 2 \Leftrightarrow(a+\cos \theta)^{2}+(2 a-\sin \theta)^{2} \leqslant 4 \\
\Leftrightarrow 2 a(\cos \theta-2 \sin \theta) \leqslant 3-5 a^{2} \\
\Leftrightarrow-2 \sqrt{5} a \sin (\theta-\varph... | \left[-\frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,673 |
9. Given the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ with its left and right foci being $F_{1}$ and $F_{2}$, respectively, point $P$ lies on the line $l: x-\sqrt{3} y+8+2 \sqrt{3}=$ 0. When $\angle F_{1} P F_{2}$ takes its maximum value, the ratio $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}$ is $\qquad$. | 9. $\sqrt{3}-1$.
From plane geometry, to maximize $\angle F_{1} P F_{2}$, the circle passing through points $F_{1}$, $F_{2}$, and $P$ must be tangent to line $l$ at point $P$. Suppose line $l$ intersects the $x$-axis at point $A(-8-2 \sqrt{3}, 0)$, then $\angle A P F_{1} = \angle A F_{2} P$, i.e., $\triangle A P F_{1}... | \sqrt{3} - 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,674 |
10. A cylindrical container with a base radius of $1 \mathrm{~cm}$ contains four solid iron balls with a radius of $\frac{1}{2} \mathrm{~cm}$. The four balls are pairwise tangent, and the two balls at the bottom are tangent to the bottom of the container. Now water is poured into the container so that the water level j... | 10. $\left(\frac{1}{3}+\frac{\sqrt{2}}{2}\right) \pi$.
Let the centers of the four solid iron balls be $\mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}, \mathrm{O}_{4}$, where $O_{1}, O_{2}$ are the centers of the two balls on the lower layer, and $A, B, C, D$ are the projections of the four centers on the bottom plane... | \left(\frac{1}{3}+\frac{\sqrt{2}}{2}\right) \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,675 |
11. The number of real solutions to the equation $\left(x^{2006}+1\right)\left(1+x^{2}+x^{4}+\cdots+\right.$ $\left.x^{2004}\right)=2006 x^{2005}$ is $\qquad$ | 11.1.
$$
\begin{array}{l}
\left(x^{2006}+1\right)\left(1+x^{2}+x^{4}+\cdots+x^{2004}\right)=2006 x^{2005} \\
\Leftrightarrow\left(x+\frac{1}{x^{2005}}\right)\left(1+x^{2}+x^{4}+\cdots+x^{2004}\right)=2006 \\
\Leftrightarrow x+x^{3}+x^{5}+\cdots+x^{2005}+\frac{1}{x^{2008}}+\frac{1}{x^{2003}} \\
\quad+\cdots+\frac{1}{x}=... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,676 |
12. A bag contains 8 white balls and 2 red balls. Each time, one ball is randomly taken out, and then one white ball is put back. What is the probability that all the red balls are exactly taken out by the 4th draw? $\qquad$ . | 12.0 .0434$.
The probability of exactly finishing all the red balls on the 4th draw is
$$
\begin{array}{l}
\frac{2}{10} \times\left(\frac{9}{10}\right)^{2} \times \frac{1}{10}+\frac{8}{10} \times \frac{2}{10} \times \frac{9}{10} \times \frac{1}{10}+ \\
\left(\frac{8}{10}\right)^{2} \times \frac{2}{10} \times \frac{1}{... | 0.0434 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,677 |
13. Given an integer $n(n \geqslant 2)$, let $M_{0}\left(x_{0}, y_{0}\right)$ be an intersection point of the parabola $y^{2}=n x-1$ and the line $y=x$. Prove: For any positive integer $m$, there must exist an integer $k$ $(k \geqslant 2)$, such that $\left(x_{0}^{m}, y_{0}^{m}\right)$ is an intersection point of the p... | Three, 13. The intersection point of $y^{2}=n x-1$ and $y=x$ is
$$
x_{0}=y_{0}=\frac{n \pm \sqrt{n^{2}-4}}{2} \text {, }
$$
Clearly, we have $x_{0}+\frac{1}{x_{0}}=n$.
If $\left(x_{0}^{m}, y_{0}^{m}\right)$ is an intersection point of the parabola $y^{2}=k x-1$ and the line $y=x$, then $k=x_{0}^{m}+\frac{1}{x_{0}^{m}}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,678 |
Example 5 Let $P$ be a point on the bisector of $\angle A O B$, and let a circle be drawn through points $O$ and $P$, intersecting $O A$ and $O B$ at points $M$ and $N$ respectively. Prove:
$$
O P^{2}=P M^{2}+O M \cdot O N .
$$ | Proof: As shown in Figure 5, take a point $S$ on $OB$ such that $OS = OM$. Construct the circumcircle of $\triangle PNS$ intersecting $OP$ at point $Q$, and connect $SP$, $SQ$, and $PN$.
Obviously, point $S$ is symmetric to point $M$ with respect to $OP$, so we have $PS = PM = PN$.
Thus, $\angle PSN = \angle PNS = \a... | OP^2 = PM^2 + OM \cdot ON | Geometry | proof | Yes | Yes | cn_contest | false | 717,679 |
14. Express 2006 as the sum of 5 positive integers $x_{1}, x_{2}$, $x_{3}, x_{4}, x_{5}$. Let $S=\sum_{1 \leqslant i<j \leqslant 5} x_{i} x_{j}$. Find:
(1) When $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ take what values, $S$ reaches its maximum value;
(2) Suppose for any $1 \leqslant i, j \leqslant 5$, there is $\left|x_{i}-... | 14. (1) First, the set of such $S$ values is bounded, so there must exist a maximum and a minimum value.
If $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=2006$, and let $S=$ $\sum_{1 \leqslant i x_{1} x_{2}$.
\end{aligned}
$$
Rewrite $S$ as
$$
\begin{aligned}
S= & \sum_{1 \in i0$, which contradicts the assumption that $S$ reaches i... | 402,402,402,400,400 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,680 |
15. Let $f(x)=x^{2}+a$. Denote
$$
\begin{array}{l}
f^{1}(x)=f(x), \\
f^{n}(x)=f\left(f^{n-1}(x)\right)(n=2,3, \cdots),
\end{array}
$$
$M=\left\{a \in \mathbf{R} \mid\right.$ for all positive integers $\left.n,\left|f^{n}(0)\right| \leqslant 2\right\}$. Prove: $M=\left[-2, \frac{1}{4}\right]$. | 15. (1) If $a2, a \notin M$.
(2) If $-2 \leqslant a \leqslant \frac{1}{4}$, by the problem, we have
$$
f^{1}(0)=a, f^{n}(0)=\left(f^{n-1}(0)\right)^{2}+a(n=2,3, \cdots) \text {. }
$$
(i) When $0 \leqslant a \leqslant \frac{1}{4}$, $\left|f^{n}(0)\right| \leqslant \frac{1}{2}(\forall n \geqslant 1)$.
In fact, when $n=1... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,681 |
One, (50 points) As shown in Figure 1, an ellipse with foci at $B_{0}$ and $B_{1}$ intersects the sides $A B_{i}$ of $\triangle A B_{0} B_{1}$ at points $C_{i} (i=0,1)$. On the extension of $A B_{0}$, take any point $P_{0}$, and draw a circular arc $\overparen{P_{0} Q_{0}}$ with $B_{0}$ as the center and $B_{0} P_{0}$ ... | (1) Obviously, $B_{0} P_{0}=B_{0} Q_{0}$. The circles are internally tangent at points $Q_{0}, P_{1}, Q_{1}$, thus we have
$$
\begin{array}{l}
C_{1} B_{0}+B_{0} Q_{0}=C_{1} P_{1}, \\
B_{1} C_{1}+C_{1} P_{1}=B_{1} C_{0}+C_{0} Q_{1}, \\
C_{0} Q_{1}=C_{0} B_{0}+B_{0} P_{0}^{\prime} .
\end{array}
$$
Adding the above three... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,682 |
II. (50 points) Given the infinite sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{0}=x, a_{1}=y, \\
a_{n+1}=\frac{a_{n} a_{n-1}+1}{a_{n}+a_{n-1}}(n=1,2, \cdots) .
\end{array}
$$
(1) For which real numbers $x, y$, does there always exist a positive integer $n_{0}$, such that when $n \geqslant n_{0}$, ... | II. (1) Notice that
$$
a_{n}-a_{n+1}=a_{n}-\frac{a_{n} a_{n-1}+1}{a_{n}+a_{n-1}}=\frac{a_{n}^{2}-1}{a_{n}+a_{n-1}}(n=1 \text {, }
$$
$2, \cdots)$.
If for some positive integer $n$, we have $a_{n+1}=a_{n}$, then it must be that $a_{n}^{2}=1$, and $a_{n}+a_{n-1} \neq 0$.
If $n=1$, we get
$|y|=1$ and $x \neq -y$.
If $n>1$... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,683 |
Three, (50 points) Solve the system of equations:
$$
\left\{\begin{array}{l}
x-y+z-w=2 \\
x^{2}-y^{2}+z^{2}-w^{2}=6 \\
x^{3}-y^{3}+z^{3}-w^{3}=20 \\
x^{4}-y^{4}+z^{4}-w^{4}=66
\end{array}\right.
$$ | Let $p=x+z, q=x z$, then we have
$$
\begin{array}{l}
p^{2}=x^{2}+z^{2}+2 q, \\
p^{3}=x^{3}+z^{3}+3 p q, \\
p^{4}=x^{4}+z^{4}+4 p^{2} q-2 q^{2} .
\end{array}
$$
Similarly, let $s=y+w, t=y w$, then we have
$$
\begin{array}{l}
s^{2}=y^{2}+w^{2}+2 t, \\
s^{3}=y^{3}+w^{3}+3 s t \\
s^{4}=y^{4}+w^{4}+4 s^{2} t-2 t^{2} .
\end... | (3,2,1,0),(3,0,1,2),(1,2,3,0),(1,0,3,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,684 |
1. The real number $m=2005^{3}-2005$. Among the following numbers, which one cannot divide $m$? ( ).
(A) 2006
(B) 2005
(C) 2004
(D) 2003 | \(\begin{array}{l}\text {-, 1.D. } \\ m=2005^{3}-2005=2005\left(2005^{2}-1\right) \\ =2005 \times 2004 \times 2006\end{array}\) | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,685 |
2. Given that $a$, $b$, $c$, and $d$ are distinct positive integers, and $a b c d=441$. Then the value of $a+b+c+d$ is ( ).
(A) 30
(B) 32
(C) 34
(D) 36 | 2.B.
Since $441=1 \times 3 \times 7 \times 21$, therefore,
$$
a+b+c+d=1+3+7+21=32
$$ | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,686 |
4. Given that $m, n$ are real numbers, and satisfy $m^{2}+2 n^{2}+m-\frac{4}{3} n+\frac{17}{36}=0$. Then the square root of $-m n^{2}$ is ( ).
(A) $\frac{\sqrt{2}}{6}$
(B) $\pm \frac{\sqrt{2}}{6}$
(C) $\frac{1}{6}$
(D) $\pm \frac{1}{6}$ | 4.B.
From the problem, we know that $\left(m+\frac{1}{2}\right)^{2}+2\left(n-\frac{1}{3}\right)^{2}=0$.
Solving, we get $m=-\frac{1}{2}, n=\frac{1}{3}$. Then $-m n^{2}=\frac{1}{18}$.
Therefore, the square root of $-m \pi^{2}$ is $\pm \sqrt{\frac{1}{18}}= \pm \frac{\sqrt{2}}{6}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,687 |
5. The number of students in the first and second grades of a school is the same, and the number of students in the third grade is $\frac{4}{5}$ of the number of students in the second grade. It is known that the number of boys in the first grade is the same as the number of girls in the second grade, and the number of... | 5.G.
According to the problem, the data in the question can be represented by Table 1.
From the problem, we know that $x = \frac{1}{4}(b + a - b + x)$.
Solving this, we get $x = \frac{a}{3}$.
\begin{tabular}{|c|c|c|c|}
\hline & Grade 7 & Grade 8 & Grade 9 \\
\hline Boys & $b$ & $a-b$ & $x$ \\
\hline Girls & $a-b$ & $... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,688 |
6. As shown in Figure 1, points $E$, $F$, $G$, $H$, $M$, $N$ are on the sides $BC$, $AC$, $AB$ of $\triangle ABC$, respectively, and $NH \parallel MG \parallel BC$, $ME \parallel NF \parallel AC$, $GF \parallel EH \parallel AB$. There are two ants, one black and one white, starting from point $F$ at the same time and s... | 6.C.
In Figure 1, quadrilaterals $C F N H$, $A H E M$, and $B M G F$ are all parallelograms. Therefore, the distance crawled by the black ant is
$$
\begin{array}{l}
F N+N H+H E+E M+M G+G F \\
=C H+C F+A M+A H+B F+B M \\
=(C H+A H)+(C F+B F)+(A M+B M) . \\
=A C+B C+A B=\text { the distance crawled by the white ant. }
\... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,689 |
Example 6 As shown in Figure 6, let O be the circumcenter of acute $\triangle ABC$, and the extensions of $BO$ and $CO$ intersect $AC$ and $AB$ at points $D$ and $E$ respectively. If $\angle A = 60^{\circ}$, prove:
$$
\begin{array}{l}
A B \cdot B E+A C \cdot C D \\
=B C^{2} .
\end{array}
$$ | Proof: As shown in Figure 6, let the angle bisector of $\angle B A C$ intersect $B C$ at point $F$. Given that $\angle B O C=2 \angle B A C=120^{\circ}$, we know that $\angle D B C=\angle E C B=30^{\circ}$.
Since $\angle F A C=30^{\circ}=\angle D B C$, we have $\triangle F A C \backsim \triangle D B C$.
Therefore, $C D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,690 |
7. After cutting off a corner of a convex polygon, the sum of the interior angles of the resulting polygon is $2520^{\circ}$. Then the number of sides of the original polygon is ( ).
(A) 14
(B) 15
(C) 15 or 16
(D) 15 or 16 or 17 | 7.D.
$$
2520^{\circ}=(16-2) \times 180^{\circ} \text {. }
$$
After cutting off one corner, the number of sides of the polygon may remain unchanged, may increase, or may decrease. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,691 |
8. Let $a$ be the integral part of $\sqrt{2}$ and $b$ be its decimal part. Let $c$ be the integral part of $\pi$ and $d$ be the decimal part. If $a d-b c=m$, then ( ).
(A) $-2<m<-1$
(B) $-1<m<0$
(C) $0<m<1$
(D) $1<m<2$
Note: integral part - integer part, decimal part - decimal part | 8.A.
The problem states: Given that $a$ and $b$ are the integer and fractional parts of $\sqrt{2}$, respectively, and $c$ and $d$ are the integer and fractional parts of $\pi$, respectively. If $a d - b c = m$, then the range of $m$ is $(\quad$.
(A) $-2 < m < -1$
(B) $-1 < m < 0$
(C) $0 < m < 1$
(D) $1 < m < 2$
Since ... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,692 |
9. For real numbers $a, b$, define the operation * as follows: $a * b=\left\{\begin{array}{ll}a^{2} b, & \text { when } a \geqslant b \text {; } \\ a b^{2}, & \text { when } a<b \text { . }\end{array}\right.$ Given that $3 * m=36$. Then the real number $m$ equals ( )
(A) $2 \sqrt{3}$
(B) 4
(C) $\pm 2 \sqrt{3}$
(D) 4 or... | 9. A.
When $3 \geqslant m$, $3 * m=9 m=36$, then $m=4$, which contradicts $3 \geqslant m$, so we discard it.
When $3 < m$, $3 * m=9 m=36$, hence $m=2 \sqrt{3}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,693 |
10 . Rearrange the sequence of consecutive natural numbers $1,2, \cdots, n(n \geqslant 3)$ into a new sequence $a_{1}, a_{2}, \cdots, a_{n}$. If $\left(a_{1}-1\right)\left(a_{2}-2\right) \cdots\left(a_{n}-n\right)$ is exactly an odd number, then $n$ ( ).
(A) must be even
(B) must be odd
(C) can be either odd or even
(C... | 10. A.
From $\left(a_{1}-1\right)\left(a_{2}-2\right) \cdots\left(a_{n}-n\right)$ being odd, we know that $a_{1}-1, a_{2}-2, \cdots, a_{n}-n$ are all odd. If $n$ is odd, then
$$
\left(a_{1}-1\right)+\left(a_{2}-2\right)+\cdots+\left(a_{n}-n\right)
$$
is odd. However,
$$
\begin{array}{l}
\left(a_{1}-1\right)+\left(a_{... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,694 |
12. Calculate:
\[
\begin{array}{l}
\sqrt{2005 \times 2006 \times 2007 \times 2008+1}-2006^{2} \\
=
\end{array}
\] | 12.2005 .
Let $n=2005$. Then
$$
\begin{array}{l}
\text { Original expression }=\sqrt{n(n+1)(n+2)(n+3)+1}-(n+1)^{2} \\
=\sqrt{\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1}-(n+1)^{2} \\
=\sqrt{\left(n^{2}+3 n+1\right)^{2}}-(n+1)^{2}=n=2005 .
\end{array}
$$ | 2005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,696 |
13. Given $x=2 \sqrt{2}+1$. Then the fraction
$$
\frac{x^{2}-2 x-9}{x^{3}-11 x-15}=
$$
$\qquad$ | 13.2
Given $x=2 \sqrt{2}+1$, we know $x-1=2 \sqrt{2}$, then $x^{2}-2 x-7=0$. Therefore, $\frac{x^{2}-2 x-9}{x^{3}-11 x-15}=\frac{\left(x^{2}-2 x-7\right)-2}{(x+2)\left(x^{2}-2 x-7\right)-1}=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,697 |
14. The lengths of the sides of a rectangle are positive integers, and the numerical value of its area is equal to twice the numerical value of its perimeter. There are $\qquad$ such rectangles.
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。
Note: The blank space represented by $\qquad$ is kept as is in the translation. | 14.3.
Let the sides of the rectangle be $a$ and $b$ ($a \geqslant b$). Then $ab = 2(2a + 2b)$, which simplifies to $(a-4)(b-4) = 16$.
Since $a$ and $b$ are positive integers, we have $\left\{\begin{array}{l}a-4=16,8,4, \\ b-4=1,2,4 .\end{array}\right.$ Solving this, we get $\left\{\begin{array}{l}a=20,12,8, \\ b=5,6,8... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,698 |
15. Suppose that in Fig. 2, the length of side of square $A B C D$ is $1, E$ and $F$ are mid-points of $C D$ and $A D$ respectively, $B E$ and $C F$ intersect at a point $P$. Then the length of line segment $C P$ is $\qquad$ | 15. $\frac{\sqrt{5}}{5}$.
The problem states: As shown in Figure 2, the side length of square $ABCD$ is $1$, and $E$, $F$ are the midpoints of sides $CD$, $AD$ respectively. $BE$ and $CF$ intersect at point $P$. Then the length of segment $CP$ is $\qquad$
From the problem, we know $\triangle CPE \sim \triangle CDF$, s... | \frac{\sqrt{5}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,699 |
For example, $7 A C$ is the longer diagonal of $\square A B C D$, $O$ is a point inside $\square A B C D$, $O E \perp A B$ at point $E$, $O F \perp A D$ at point $F$, $O G \perp A C$ at point $G$. Prove:
$$
A E \cdot A B+A F \cdot A D=A G \cdot A C \text {. }
$$ | Proof: As shown in Figure 7, draw perpendiculars from points $B$, $C$, and $D$ to line $AO$, with $M$, $Q$, and $N$ being the feet of the perpendiculars, respectively. Draw perpendiculars from points $B$ and $D$ to $AC$, with $L$ and $K$ being the feet of the perpendiculars, respectively.
Clearly, points $A$, $E$, $O$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,701 |
17. As shown in Figure 3, in trapezoid $A B C D$, $F$ is the midpoint of $C D$, $A F \perp A B$, $E$ is a point on side $B C$ such that $A E=B E$. If $A B=m$ (where $m$ is a constant), then the length of $E F$ is $\qquad$ | 17. $\frac{m}{2}$.
As shown in Figure 3, extend $A F$ to intersect the extension of $B C$ at point $G$. It is easy to prove that $\triangle A D F \cong \triangle G C F$. Therefore, $A F = G F$.
Since $\angle B A G = 90^{\circ}$ and $A E = B E$, $A E$ is the median of the hypotenuse $B G$ in the right triangle $\trian... | \frac{m}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,702 |
18. $A 、 n$ are natural numbers, and
$$
A=n^{2}+15 n+26
$$
is a perfect square. Then $n$ equals $\qquad$ | 18.23.
Let $A=n^{2}+15 n+26=(n+2)(n+13)$ be a perfect square, we can set $n+2=a^{2}, n+13=b^{2}(a, b$ are positive integers and $b$ $>a)$. Therefore, $11=b^{2}-a^{2}=(b+a)(b-a)$. Then we have $\left\{\begin{array}{l}b+a=11, \\ b-a=1 .\end{array}\right.$ Solving this, we get $\left\{\begin{array}{l}a=5, \\ b=6 .\end{ar... | 23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,703 |
19. A rectangular prism has integer length, width, and height, and its volume is exactly $2006 \mathrm{~cm}^{3}$. After painting its surface red, it is then cut into small cubes with edge lengths of $1 \mathrm{~cm}$. If there are 178 small cubes with three red faces, then the number of small cubes with exactly two red ... | 19.1824 .
From the problem, we know that one of the edges of the cuboid must be $1 \mathrm{~cm}$, otherwise, only 8 small cubes would have three faces painted red. Let's assume the length, width, and height of the cuboid are $x \mathrm{~cm}, y \mathrm{~cm}, 1 \mathrm{~cm}$, respectively. Then we have
$$
\left\{\begin{... | 1824 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,704 |
21. A school has six elementary school grades, with 8 classes in each grade; three junior high school grades, with 8 classes in each grade; and three senior high school grades, with 12 classes in each grade. Now, 27 classes are to be selected for a survey, ensuring that the proportion of classes selected from each type... | Three, 21.2, 2.
Using stratified sampling.
The ratio of the number of classes in primary, junior high, and senior high school in the school is
$$
48: 24: 36=4: 2: 3 \text {, }
$$
Then, 12 classes are sampled from primary school, $27 \times \frac{4}{9}=12$ classes, distributed to six grades, with 2 classes sampled from... | 2, 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,706 |
23. Given that $m$, $n$, and $l$ are all two-digit positive integers, and they are not all equal, their least common multiple is 385. Then the maximum value of $m+n+l$ is $\qquad$, and the minimum value is $\qquad$ | 23. 209,57.
Given $385=5 \times 11 \times 7$, to maximize $m+n+l$, we can take $m=n=77, l=55$; to minimize $m+n+l$, we can take $m=n=11, l=35$. | 209,57 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,707 |
24. The construction cost of a project must not exceed 1.9 million yuan. If the project is undertaken by Company A, it will take 20 days, with a daily fee of 100,000 yuan; if undertaken by Company B, it will take 30 days, with a daily fee of 60,000 yuan. To shorten the construction period, it is decided that Company A ... | 24.10, 25.
Let the remaining work be completed by company B in $x$ days. Then we have
$$
\left\{\begin{array}{l}
\frac{m}{20}+\frac{x}{30}=1, \\
10 m+6 x \leqslant 190 .
\end{array}\right.
$$
From equation (1), we get $x=\frac{60-3 m}{2}$.
Substituting equation (3) into equation (2) yields $m \leqslant 10$.
The total... | 10, 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,708 |
25. Express 2006 as the sum of $n(n \geqslant 3)$ consecutive natural numbers. Please write two expressions. | $$
\begin{array}{l}
25.500+501+502+503,110+111+112+\cdots+ \\
126,5+6+7+\cdots+63 .
\end{array}
$$
(Provided by Hong Wangbao) | 500+501+502+503, 110+111+112+\cdots+126, 5+6+7+\cdots+63 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,709 |
2. Given that the decimal part of $\lg x$ is $a$. Then the decimal part of $\lg \frac{1}{x^{2}}$ is ( ).
(A) $-2 a$
(B) $1-2 a$
(C) $2-2 a$
(D) None of the above | 2.D.
Let $\lg x=N+a(N$ be an integer, $0 \leqslant a<1)$, then $\lg \frac{1}{x^{2}}$ $=-2 N-2 a$ and the fractional part of $\lg \frac{1}{x^{2}}$ is
$$
\left\{\begin{array}{ll}
-2 a, & a=0 ; \\
1-2 a, & 0<a \leqslant \frac{1}{2} \\
2-2 a, & \frac{1}{2}<a<1 .
\end{array}\right.
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,711 |
Example 8 Let $D$ be a point on the base $BC$ of isosceles $\triangle ABC$, and $F$ be a point on the arc inside $\triangle ABC$ of the circle passing through $A, D, C$. The circle passing through $B, D, F$ intersects side $AB$ at point $E$. Prove that $CD \cdot EF + DF \cdot AE = BD \cdot AF$.
(1st China Southeast Mat... | Proof: As shown in Figure 8, let the circle passing through points $B$, $D$, and $F$ intersect line $AF$ at point $K$. Connect $KB$, $KC$, and $KD$.
From $\angle KBD = \angle KFD = \angle ACB$, we know that
$BK \parallel AC$.
Thus, $S_{\triangle X C} = S_{\triangle B C}$.
Therefore, $S_{\triangle D D} + S_{\triangle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,712 |
3. Draw the tangent lines to the parabola $y=x^{2}+a x+4 a^{2}$ from the origin $O$. When $a$ varies, the two points of tangency lie on the parabolas ( ).
(A) $y=\frac{1}{2} x^{2}, y=\frac{3}{2} x^{2}$
(B) $y=\frac{3}{2} x^{2}, y=\frac{5}{2} x^{2}$
(C) $y=x^{2}, y=3 x^{2}$
(D) $y=3 x^{2}, y=5 x^{2}$ | 3.B.
Let the required tangent line equation be $y=k x$. Substituting it into the parabola $y=x^{2}+a x+4 a^{2}$, we get
$$
x^{2}+(a-k) x+4 a^{2}=0 .
$$
By $\Delta=(a-k)^{2}-16 a^{2}=0$, which means
$$
(k+3 a)(k-5 a)=0,
$$
we solve to get $k=-3 a, k=5 a$.
If $k=-3 a$, then the x-coordinate of the tangent point is $x=... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,713 |
4. Given $\triangle A B C$ is an isosceles right triangle, $\angle C$ $=90^{\circ}, D 、 E$ are two points on side $A B$, and point $D$ is between $A E$, $\angle D C E=45^{\circ}$. Then the triangle formed by $A D 、 D E 、 E B$ as side lengths is ( ).
(A) acute triangle
(B) obtuse triangle
(C) right triangle
(D) cannot b... | 4.C.
Rotate $\triangle C A D$ counterclockwise by $90^{\circ}$ around point $C$, to get $\triangle C B F$, then
$$
B F=A D, \angle E B F=90^{\circ} \text {. }
$$
Since $\triangle C D E \cong \triangle C F E$, we have $E F=D E$.
Thus, $A D^{2}+E B^{2}=D E^{2}$.
Therefore, the triangle formed by the side lengths $A D, ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,714 |
5. Write positive integers starting from 1 in a continuous line, the 2006th digit is ( ).
(A) 0
(B) 5
(C) 7
(D) None of the above
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 5.A.
Since there are 9 one-digit numbers, 90 two-digit numbers, and 900 three-digit numbers, the 2006th digit is a digit of a three-digit number. Therefore,
$$
(2006-9-2 \times 90) \div 3=1817 \div 3=605 \frac{2}{3} \text {. }
$$
Thus, the required digit is the tens digit of the 606th three-digit number, which is the... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,715 |
6. Given that the line connecting the vertex $V$ and the center $O$ of the base circle of a cone is perpendicular to the base, a plane passing through the midpoint $M$ of $VO$ is tangent to $\odot O$ and intersects the cone in an ellipse. If the radius of $\odot O$ is 1, then the length of the minor axis of the ellipse... | 6.B.
Let the center of the ellipse be $S$, and the plane through the midpoint $M$ of $VO$ intersects $\odot O$ at point $A$. $MA$ intersects the ellipse at point $B$, then $AB$ is the major axis of the ellipse. Let $AC$ be the diameter of $\odot O$, then $O S \Perp \frac{1}{2} B C$. Therefore,
$$
S M=B M=\frac{1}{2} S... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,716 |
7. Let the first term and common difference of an arithmetic sequence be positive integers, the number of terms be a prime number no less than 3, and the sum of all terms be 2006. Then the number of such sequences is $\qquad$.
| Ni, 7.15.
Let the first term and common difference of an arithmetic sequence be $a$ and $d$, and the number of terms be $n$, then
$$
n a+\frac{1}{2} n(n-1) d=2006,
$$
i.e., $n[2 a+(n-1) d]=2 \times 2006=2 \times 2 \times 17 \times 59$.
Since $2 a+(n-1) d \geqslant 2+n-1=n+1>n$, hence $n=17,59$.
(1) If $n=17$, then $a+... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,717 |
8. Given real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
(x-11)^{5}+15(x-11)=5, \\
(y-4)^{5}+15(y-4)=-5 .
\end{array}\right.
$$
Then $x+y$ equals $\qquad$ . | 8. 15 .
Since the function $f(t)=t^{5}+15 t$ is strictly monotonically increasing in the real number domain, and $f(x-11)=f(4-y)$, then $x-11=4-y$. Therefore, $x+y=15$. | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,718 |
9. The number of intersection points of all diagonals inside a regular octagon is $\qquad$ .
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 9.49.
Since any four points determine one intersection, there are $C_{B}^{A}=70$ intersections in total. Also, because the 6 intersections of the 4 longest diagonals coincide into 1 point; the 3 intersections of one longest and two medium-length diagonals coincide into 1 point, with a total of 8 such cases.
Therefore,... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,719 |
10. If $x, y$ are real numbers, and $x^{2}+x y+y^{2}=3$, then the maximum and minimum values of $x^{2}-x y+y^{2}$ are $\qquad$. | $10.9,1$.
Let $x^{2}-x y+y^{2}=m$, then
$$
x+y= \pm \sqrt{\frac{9-m}{2}}, x y=\frac{3-m}{2} \text {. }
$$
Therefore, $x, y$ are the two real roots of the equation $t^{2} \pm \sqrt{\frac{9-m}{2}} t+\frac{3-m}{2}=0$. By $\Delta \geqslant 0$, we get $m \geqslant 1$; by $\frac{9-m}{2} \geqslant 0$, we get $m \leqslant 9$. | 9,1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,720 |
11. The 8 vertices of a cube can form
$\qquad$ non-equilateral triangles. | 11.48.
Each vertex leads to three edges whose endpoints can form an equilateral triangle, making a total of 8 equilateral triangles. Therefore, the answer is $\mathrm{C}_{8}^{3}-8=48$. | 48 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,721 |
1. $E D$ is tangent to the circle with diameter $A B$ at point $D, C$ is a point on $A B$, $C E \perp E D, E$ is the foot of the perpendicular.
Prove:
$$
C E \cdot A B=A C \cdot C B+C D^{2}
$$ | (提示: 辅助线如图 9,则 $\triangle D F G \sim \triangle C D E$. 故
$$
\begin{aligned}
& A C \cdot C B+C D^{2} \\
= & C D \cdot C G+C D^{2}=C D(C G+C D) \\
= & C D \cdot C D=C E \cdot D F=C E \cdot A B .)
\end{aligned}
$$
(提示: As shown in Figure 9, $\triangle D F G \sim \triangle C D E$. Therefore,
$$
\begin{aligned}
& A C \cdot... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,723 |
13. There are 2006 distinct complex numbers, such that the product of any two of them (including self-multiplication) is one of these 2006 numbers. Find the sum of these 2006 numbers. | Three, 13. Let there be $n$ distinct complex numbers $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$.
If one of them is 0, without loss of generality, let $\alpha_{n}=0$, then the set
$$
\left\{\alpha_{1}^{2}, \alpha_{1} \alpha_{2}, \cdots, \alpha_{1} \alpha_{n-1}\right\}=\left\{\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n-1... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,724 |
14. Find
$$
2 \sum_{k=1}^{n} k^{3} \mathrm{C}_{n}^{k}-3 n \sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}+n^{2} \sum_{k=1}^{n} k \mathrm{C}_{n}^{k}
$$
the value. | 14. Solution 1: Since $k \mathrm{C}_{n}^{k}=n \mathrm{C}_{n-1}^{k-1}$, we have:
$$
\begin{array}{l}
\sum_{k=1}^{n} k \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} n \mathrm{C}_{n-1}^{k-1}=n 2^{n-1} \\
\sum_{k=1}^{n} k^{2} \mathrm{C}_{n}^{k}=\sum_{k=1}^{n} k n \mathrm{C}_{n-1}^{k-1} \\
=n \sum_{k=1}^{n}(k-1) \mathrm{C}_{n-1}^{k-1}+... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,725 |
15. Given the sequence $\left\{a_{n}\right\}(n \geqslant 0)$ satisfies $a_{0}=0$, for all non-negative integers $n$, we have
$$
a_{n+1}=2 \sqrt{30 a_{n}\left(a_{n}+1\right)}+11 a_{n}+5 .
$$
Find the general term formula for $a_{n}$. | 15. $a_{n+1}=2 \sqrt{30 a_{n}\left(a_{n}+1\right)}+11 a_{n}+5>a_{n}$.
Subtract $11 a_{n} + 5$ from both sides of $a_{n+1}-11 a_{n}-5=2 \sqrt{30 a_{n}\left(a_{n}+1\right)}$ and square both sides, then rearrange to get
$$
\begin{array}{l}
a_{n}^{2}+a_{n+1}^{2}-22 a_{n} a_{n+1}-10 a_{n}-10 a_{n+1}+25=0 . \\
\text { Also,... | a_{n}=-\frac{1}{2}+\frac{1}{4}(11+2 \sqrt{30})^{n}+\frac{1}{4}(11-2 \sqrt{30})^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,726 |
1. Given that $a$, $b$, and $c$ are pairwise distinct real numbers. Then the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are ( ).
(A) must have two distinct real roots
(B) have no real roots
(C) must have two equal real roots
(D) the roots of the equation may take the values $a$, $b$, $c$ | 1.A.
The original equation can be transformed into
$$
3 x^{2}-2(a+b+c) x+a b+b c+c a=0 .
$$
Its discriminant is
$$
\begin{array}{l}
\Delta=4(a+b+c)^{2}-4 \times 3(a b+b c+c a) \\
=2\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] .
\end{array}
$$
Since $a, b, c$ are pairwise unequal, then $\Delta>0$, so the equation must h... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,727 |
2. In a circle with radius 1, the area of the figure formed by all chords starting from point $A$ and having a length not less than the side length $a$ of the inscribed equilateral $\triangle ABC$ is ( ).
(A) $\frac{\pi}{2}+\frac{\sqrt{2}}{3}$
(B) $\frac{\pi}{3}+\frac{\sqrt{2}}{2}$
(C) $\frac{\pi}{2}+\frac{\sqrt{3}}{3}... | 2.D.
From the given conditions, it is easy to know that the figure formed by these chords is precisely the equilateral $\triangle A B C$ and its corresponding segments. Let the center of $\triangle A B C$ be $O$, then the area of the small sector $\overparen{B O C}$ is $\frac{\pi}{3}$. And
$$
S_{\triangle A O B}=S_{\t... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,728 |
3. Given that $a$ and $b$ are real numbers, and $b-a=2$006. If the roots of the quadratic equation $x^{2}+a x+b=0$ are all integers, then the number of sets of roots for this equation is ( ).
(A) 4
(B) 6
(C) 8
(D) 10 | 3.B.
By Vieta's formulas, we have $x_{1}+x_{2}=-a, x_{1} x_{2}=b$. Then $x_{1}+x_{2}+x_{1} x_{2}=2006$.
Therefore, $\left(x_{1}+1\right)\left(x_{2}+1\right)=2007=9 \times 223$
$$
\begin{array}{l}
=-9 \times(-223)=3 \times 669=-3 \times(-669) \\
=1 \times 2007=(-1) \times(-2007) .
\end{array}
$$
It is easy to see that... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,729 |
4. Figure 1 is a triangular number table, from top to bottom called the first row, second row, …. It is known that each “ ” in the triangular number table is the reciprocal of a positive integer, and is equal to the sum of the two numbers directly below it. If the number in the first row is $\frac{1}{1}$, then the numb... | 4.C.
Let the two numbers in the second row be $m, n$. Then
$$
\frac{1}{m}+\frac{1}{n}=1\left(m, n \in \mathbf{N}_{+}\right) \text {. }
$$
Thus, $m=\frac{n}{n-1}=1+\frac{1}{n-1}$. Solving this, we get $n-1=1$. Therefore, $m=2$, and $n=2$, meaning the numbers in the second row can only be $\frac{1}{2}, \frac{1}{2}$.
Le... | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 717,730 |
5. In $\triangle A B C$, $A B<B C<C A$, and $A C-$ $A B=2, A D$ is the angle bisector of $\angle B A C$, $E$ is a point on side $A C$, and line $B E$ intersects $A D$ at point $G$, and $\frac{A C}{C D}=\frac{A E}{B D}=$ $2, \frac{A G}{G D}=2007$. Then the length of side $B C$ is $(\quad$.
(A) 2008
(B) 2007
(C.) 2006
(D... | 5.B.
As shown in Figure 4, draw $EF \parallel AD$ intersecting $CD$ at point $F$, and let $AB = x$. Then,
$$
\frac{AC}{AE} = \frac{CD}{DF} = \frac{CD}{BD}.
$$
Since $BD = DF$, $DG$ is the midline of $\triangle BEF$. Therefore, $BG = GE$. Also, $\angle BAG = \angle EAG$, so $AB = AE = x$. Thus, $CE = AC - AE = AC - AB... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,731 |
6. In a mathematics competition, there are three types of questions: multiple-choice questions (6 questions), fill-in-the-blank questions (4 questions), and problem-solving questions (3 questions). Among them, each multiple-choice question and each fill-in-the-blank question is worth 7 points, and the problem-solving q... | G.D.
(1) The possible scores for multiple-choice and fill-in-the-blank questions are $0,7,14, \cdots, 70$, a total of 11 possibilities. The possible scores for problem-solving questions are $0,5,10, \cdots, 70$, a total of 15 possibilities. Therefore, there can be $11 \times 15=165$ results.
(2) The following 23 scores... | 117 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,732 |
1. Given that $a$, $b$, and $c$ are non-zero real numbers, satisfying
$$
\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c}{c} \text {. }
$$
Then the value of $\frac{(a+b)(b+c)(c+a)}{a b c}$ is $\qquad$ | 2. -1 or 8.
Let $\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c}{c}=k$, then
$$
\begin{array}{l}
b+c=(k+1) a, c+a=(k+1) b, \\
a+b=(k+1) c .
\end{array}
$$
Thus, $2(a+b+c)=(k+1)(a+b+c)$.
Hence $a+b+c=0$ or $b+c=2 a, c+a=2 b, a+b$ $=2 c$.
Therefore, $\frac{(a+b)(b+c)(c+a)}{a b c}=-1$ or 8. | -1 \text{ or } 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,733 |
3. In $\triangle A B C$, $\angle B=45^{\circ}, A C<B C$, draw $A D \perp B C$ at point $D$, take a point $E$ on $A C$ such that $D E=A D$. Prove:
$$
B C \cdot C D=A C \cdot C E+B C \cdot A D \text {. }
$$ | $$
\begin{array}{l}
\quad DF=AD \\
AC \cdot CE=CF \cdot CB \\
=BC(CD-DF) \\
=BC \cdot CD-BC \cdot DF \\
=BC \cdot CD-BC \cdot AD .)
\end{array}
$$
As shown in Figure 11, let the circle passing through points $A$, $B$, and $E$ intersect $DC$ at point $F$, and connect $EF$. $D$ is the center of the circle. Thus, | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,734 |
3. Arrange $n$ squares of different sizes without overlapping, so that the total area of the resulting figure is exactly 2,006. The minimum value of $n$ is $\qquad$ $\therefore$. | 3.3.
Let the side lengths of $n$ squares be $x_{1}, x_{2}, \cdots, x_{n}$, then $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=2006$.
Since $x_{i}^{2} \equiv 0$ or $1(\bmod 4)$, and $2006 \equiv 2(\bmod 4)$, there must be at least two odd numbers among $x_{i}$.
If $n=2$, then $x_{1}$ and $x_{2}$ are both odd, let them be $2 p+... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,736 |
4. As shown in Figure 2, two congruent equilateral triangles $\triangle A_{1} B_{1} C_{1}$ and $\triangle A_{2} B_{2} C_{2}$ with side lengths as positive integers have their centers coinciding, and satisfy $A_{1} B_{1} \perp A_{2} C_{2}$. If the area of the hexagon $A B C D E F$ is $S=\frac{1}{m}-\frac{\sqrt{3}}{n}$, ... | 4. $\frac{1}{3}$.
By symmetry, we know that $A_{1} B_{1} \perp A_{2} C_{2}, B_{1} C_{1} \perp A_{2} B_{2}, C_{1} A_{1} \perp B_{2} C_{2}$, which means the right triangles $\triangle A_{1} A F, \operatorname{Rt} \triangle A_{2} A B, \operatorname{Rt} \triangle B_{1} C B, \operatorname{Rt} \triangle B_{2} C D, \operator... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,737 |
One, (20 points) Prove: Two right-angled triangles are congruent if their areas and perimeters are respectively equal.
The above text has been translated into English, retaining the original text's line breaks and format. | In right triangles $\triangle ABC$ and $\operatorname{Rt} \triangle A_{1} B_{1} C_{1}$, the legs are $AC=b, BC=a$ and $A_{1} C_{1}=b_{1}, B_{1} C_{1}=a_{1}$, and the hypotenuses are $AB=c$ and $A_{1} B_{1}=c_{1}$. According to the problem, we have
$$
\left\{\begin{array}{l}
c^{2}=a^{2}+b^{2}, \\
c_{1}^{2}=a_{1}^{2}+b_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,738 |
Given that $k$ and $a$ are positive integers, and $2004k + a$, $2004(k+1) + a$ are both perfect squares.
(1) How many ordered pairs of positive integers $(k, a)$ are there?
(2) Identify the minimum value of $a$ and explain your reasoning. | II. (1) Let $2004k + a = m^2$,
(1)
$$
2004(k+1) + a = n^2 \text{, }
$$
where $m, n$ are positive integers, then
$$
n^2 - m^2 = 2004 \text{. }
$$
Thus, $(n+m)(n-m) = 2004 = 2 \times 2 \times 3 \times 167$.
Noting that $m+n$ and $n-m$ have the same parity, we have
$$
\left\{\begin{array}{l}
n + m = 1002, \\
n - m = 2
\... | 137 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,739 |
Three. (25 points) As shown in Figure 3, given a cyclic quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $N$, point $M$ is on diagonal $BD$ and satisfies $\angle BAM = \angle DAN$, $\angle BCM = \angle DCN$. Prove:
(1) $M$ is the midpoint of $BD$;
(2) $\frac{AN}{CN} = \frac{AM}{CM}$. | Three, according to the theorem that angles subtended by the same arc are equal, we get
$$
\angle D A N=\angle D B C, \angle D C N=\angle D B A \text {. }
$$
Since $\angle D A N=\angle B A M, \angle B C M=\angle D C N$,
thus, $\angle B A M=\angle M B C, \angle A B M=\angle B C M$.
Therefore, $\triangle B A M \backsim ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,740 |
1. Among the 8 vertices and the center of a cube, a total of 9 points, the number of sets of four coplanar points is ( ).
(A) 28
(B) 32
(C) 36
(D) 40 | 1.C.
A cube has 6 sides forming 6 different sets of coplanar four points. Additionally, a cube has 6 diagonal sections, each of which, together with the center 0 of the cube, can produce 5 different sets of coplanar four points. Thus, this results in $6 \times 5+6=36$ different sets of coplanar four points. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,741 |
2. Regarding the set
$$
S=\left\{a \mid a \in \mathbf{Z} \text {, and } \frac{a^{2}-a+88}{a^{3}+1} \in \mathbf{Z}\right\}
$$
the correct statement about the number of elements $|S|$ is ( ).
(A) $|S|=2$
(B) $|S|=3$
(C) $|S|=4$
(D) $|S| \geqslant 5$ | 2.B.
From the given condition, we have
$(a+1)\left(a^{2}-a+1\right) \mid\left(a^{2}-a+88\right)$.
Then $\left(a^{2}-a+1\right) \mid\left(a^{2}-a+88\right)=\left(a^{2}-a+1\right)+87$.
Thus, $\left(a^{2}-a+1\right) \mid 87=3 \times 29$.
Since $a^{2}-a+1>0$, from equation (1) we get
$a^{2}-a+1=1,3,29,87$.
When $a^{2}-a+1... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,742 |
$$
\text { 3. } \arccos \frac{1}{3}+\frac{1}{2} \arccos \frac{7}{9}=(\quad) \text {. }
$$
(A) $\frac{3 \pi}{8}$
(B) $\frac{2 \pi}{3}$
(C) $\frac{\pi}{2}$
(D) $\arcsin \frac{8}{9}$ | 3. C.
Let $\theta=\arccos \frac{1}{3}+\frac{1}{2} \arccos \frac{7}{9}$, then $\frac{\pi}{6}<\theta<\frac{\pi}{2}+\frac{1}{2} \times \frac{\pi}{6}=\frac{\pi}{2}+\frac{\pi}{12}$, i.e., $\frac{\pi}{3}<2 \theta<\pi+\frac{\pi}{6}$.
Since $\cos \left(2 \arccos \frac{1}{3}\right)=2\left(\frac{1}{3}\right)^{2}-1=-\frac{7}{9}... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,743 |
4. Let $a_{1}=1, S_{n+1}=2 S_{n}-\frac{n(n+1)}{2}+1$, where $S_{n}=a_{1}+a_{2}+\cdots+a_{n}(n=1,2, \cdots)$. If we define $\Delta a_{n}=a_{n+1}-a_{n}$, then the set
$$
S=\left\{n \mid n \in \mathbf{N}_{+}, \Delta\left(\Delta a_{n}\right) \geqslant-2006\right\}
$$
has the number of elements $|S|$ as ( ).
(A)9
(B) 10
(C... | 4.C.
From the problem, we have
$$
\begin{array}{l}
S_{n+1}=2 S_{n}-\frac{n(n+1)}{2}+1, \\
S_{n}=2 S_{n-1}-\frac{(n-1) n}{2}+1(n \geqslant 2) .
\end{array}
$$
Subtracting (2) from (1) gives
$$
a_{n+1}=2 a_{n}-n(n \geqslant 2) \text {. }
$$
Taking $n=1$ in the given equation, we get
$$
a_{1}+a_{2}=2 a_{1}-1+1 \text {, ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,744 |
1. Let $\sqrt{3+\sqrt{2}+\sqrt{3}+\sqrt{6}}=\sqrt{x}+\sqrt{y}+\sqrt{z}$, and $x, y, z$ are rational numbers. Then $x y z=(\quad)$.
(A) $\frac{3}{4}$
(B) $\frac{5}{6}$
(C) $\frac{7}{12}$
(D) $\frac{13}{18}$ | 1. A.
Squaring both sides, we get
$$
3+\sqrt{2}+\sqrt{3}+\sqrt{6}=x+y+z+2 \sqrt{x y}+2 \sqrt{y z}+2 \sqrt{x z} \text{. }
$$
Based on the symmetry of the rational numbers $x, y, z$, we can consider the system of equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
2 \sqrt{x y}=\sqrt{2}, \\
2 \sqrt{y z}=\sqrt{3}, \\
2 \sqr... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,745 |
2. Let the quadratic function $f(x)=a x^{2}+a x+1$ have a graph that opens downwards, and satisfy $f(f(1))=f(3)$. Then the value of $2 a$ is ( ).
(A) -3
(B) -5
(C) -7
(D) -9 | 2.B.
Notice
$$
\begin{array}{l}
f(1)=2 a+1, f(3)=12 a+1, \\
f(f(1))=a(2 a+1)^{2}+a(2 a+1)+1 .
\end{array}
$$
From $f(f(1))=f(3)$, we get
$$
(2 a+1)^{2}+(2 a+1)=12 \text{. }
$$
Therefore, $2 a+1=3$ or -4.
Since $a<0$, hence $2 a=-5$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,746 |
One, (20 points) Solve the equation:
$$
(12 x+5)^{2}(6 x-1)(x+1)=\frac{55}{2} \text {. }
$$ | One, transforming the original equation, we get
$$
(12 x+5)^{2}(12 x-2)(12 x+12)=660 \text {. }
$$
Let $12 x+5=t$, then $t^{2}(t-7)(t+7)=660$, which is $t^{4}-49 t^{2}=660$.
Solving this, we get $t^{2}=60$ or $t^{2}=-11$ (discard).
Thus, we have $t= \pm 2 \sqrt{15}$, which means $12 x+5= \pm 2 \sqrt{15}$.
Therefore, t... | x_{1,2}=\frac{-5 \pm 2 \sqrt{15}}{12} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,747 |
4.5 soccer teams are conducting a round-robin tournament (each pair of teams plays one match). It is known that Team A has played 3 matches, Team B has played more matches than Team A, Team C has played fewer matches than Team A, and Team D and Team E have played the same number of matches, but Team D and Team E have n... | 4.6.
Team B has played 4 matches. If Team C has only played 1 match, then Team C hasn't played against Team A, and Team A must have played against Teams D and E, so Teams D and E have each played 2 matches. Therefore, the total number of matches is
$$
(3+4+1+2+2) \div 2=6 \text {. }
$$
If Team C has played 2 matches,... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,748 |
One, (20 points) Given that $t$ is a root of the quadratic equation $x^{2}-$ $x-1=0$. For any rational number $a$, there exist rational numbers $b$ and $c$ such that $(a t+1)(b t+c)=1$.
(1) Find $b$ and $c$ (expressed in terms of $a$);
(2) Does there exist a rational number $a$ such that at least one of $b$ or $c$ equa... | (1) Since $t$ is a root of $x^{2}-x-1=0$, then $t^{2}=t+1$. From $(a t+1)(b t+c)=1$, we get
$$
a b t^{2}+a c t+b t+c-1=0 \text {, }
$$
which means
$$
a b(t+1)+a c t+b t+c-1=0 \text {. }
$$
Thus, $(a b+a c+b) t+a b+c-1=0$.
Since $t$ is an irrational number and $a, b, c$ are rational numbers, we have
$$
\left\{\begin{a... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,749 |
II. (25 points) A group purchases tickets to a park, with ticket prices as shown in Table 1:
Table 1
\begin{tabular}{|c|c|c|c|}
\hline Number of ticket buyers & $1 \sim 50$ & $51 \sim 100$ & Over 100 \\
\hline Price per ticket & 13 yuan & 11 yuan & 9 yuan \\
\hline
\end{tabular}
There are two tour groups, Group A and ... | If the number of people does not exceed 100, the cost is at most 1300 yuan, so the total number of people in the two tour groups exceeds 100.
Also, $1008 \div 9=112$, so the total number of people in the two tour groups is 112.
Let the number of people in the two tour groups be $x$ and $y$ respectively. From $x+y=112$,... | 41, 71 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,750 |
Three. (25 points) As shown in Figure 3, $\triangle A B C$ is divided into six smaller triangles by three concurrent lines $A D$, $B E$, and $C F$. If the areas of $\triangle B P F$, $\triangle C P D$, and $\triangle A P E$ are all 1, find the areas of $\triangle A P F$, $\triangle D P B$, and $\triangle E P C$. | Three, let the areas of $\triangle A P F$, $\triangle D P B$, and $\triangle E P C$ be $x$, $y$, and $z$ respectively.
Since $\frac{S_{\triangle D P B}}{S_{\triangle D P C}}=\frac{B D}{D C}, \frac{S_{\triangle D A B}}{S_{\triangle A K C}}=\frac{B D}{D C}$,
then $\frac{S_{\triangle D P B}}{S_{\triangle D P C}}=\frac{S_{... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,751 |
II. (25 points) As shown in Figure 2, in quadrilateral $ABCD$, $\angle ACB = \angle ADB = 90^{\circ}$. From the intersection point $N$ of diagonals $AC$ and $BD$, draw $NM \perp AB$ at point $M$. Line segments $AC$ and $MD$ intersect at point $E$, and $BD$ and $MC$ intersect at point $F$. $P$ is any point on line segme... | As shown in Figure 7, it is evident that points $A, B, C, D$ are concyclic, and points $B, C, N, M$ are also concyclic. Therefore, $\angle ACD = \angle ABD = \angle MCN$.
Hence, $AC$ bisects $\angle DCM$.
Similarly, $BD$ bisects $\angle CDM$.
As shown in Figure 7, let $PH \perp MC$ at point $H$, $PG \perp MD$ at point ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,758 |
9. Given real numbers $x, y$ satisfy
$$
(2 x+1)^{2}+y^{2}+(y-2 x)^{2}=\frac{1}{3} \text {. }
$$
Then the value of $x+y$ is . $\qquad$ | 9. $-\frac{2}{3}$ | -\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,761 |
1. A building has 4 elevators, each of which can stop at three floors (not necessarily consecutive floors, and not necessarily the lowest floor). For any two floors in the building, there is at least one elevator that can stop at both. How many floors can this building have at most? | 2. Let the building have $n$ floors, then the number of floor pairs is $\frac{n(n-1)}{2}$. Each elevator stops at 3 floors, which gives $\frac{3 \times 2}{2}=3$ floor pairs, so, $4 \times 3 \geqslant$ $\frac{n(n-1)}{2}$. Therefore, $n \leqslant 5$.
When $n=5$, the floors served by the four elevators are $(1,4,5),(2,4,... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,765 |
3. Divide the numbers $1,2, \cdots, 30$ into $k$ groups (each number can only appear in one group) such that the sum of any two different numbers in each group is not a perfect square. Find the minimum value of $k$.
Put the above text into English, please keep the original text's line breaks and format, and output the... | 3. First consider the numbers $6,19,30$.
Since $6+19=5^{2}, 6+30=6^{2}, 19+30=7^{2}$, these 3 numbers must be in 3 different groups. Therefore, $k \geqslant 3$.
Next, divide the 30 numbers $1,2, \cdots, 30$ into the following 3 groups:
$$
\begin{array}{l}
A_{1}=\{3,7,11,15,19,23,27,4,8,16,24\}, \\
A_{2}=\{1,5,9,13,17,... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,767 |
1. The teacher said: "To place the largest possible circle inside a triangle with side lengths of $2$, $2$, and $2x$, what should the value of the positive real number $x$ be?"
Student A said: "I think $x=1$."
Student B said: "I believe $x=\sqrt{2}$."
Student C said: "Your answers are both incorrect."
Who among the thr... | 1. On one hand, the area of the triangle is equal to $r(2+x)$; on the other hand, the height of the triangle on its base is $\sqrt{4-x^{2}}$, so the area of the triangle is equal to $x \sqrt{4-x^{2}}$. Therefore, $r=\frac{x \sqrt{4-x^{2}}}{2+x}$.
When $x=1$, $r=\frac{1}{\sqrt{3}}\frac{16}{27}>\frac{1}{1.7}$.
Thus, $x=\... | x=\sqrt{5}-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,768 |
Three. (25 points) A rectangular glass tabletop has shattered into several small glass fragments, each of which is a convex polygon. After reassembling the fragments into the original rectangle, there are 30 intersection points, 20 of which are on the boundary of the original rectangle (including the four vertices of t... | Three, suppose among all the fragments, there are $a_{3}$ triangles, $a_{4}$ quadrilaterals, $\cdots \cdots, k$-sided polygons $a_{k}$ $\left(a_{3}, a_{4}, \cdots, a_{k}\right.$ are non-negative integers
$$
S_{\text {angle }}=a_{3} \times \pi+a_{4} \times 2 \pi+\cdots+a_{k}(k-2) \pi \text {. }
$$
On the other hand, th... | 36 \text{ fragments, either 34 triangles and 2 quadrilaterals; or 35 triangles and 1 pentagon.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,769 |
2. A triangle can be divided into two isosceles triangles. One of the interior angles of the original triangle is $36^{\circ}$, find all possible values of the largest interior angle of the original triangle. | 2. Let's assume $\angle B=36^{\circ}$.
(1) If the dividing line does not pass through point $B$. Let's assume the dividing line is $A D$, at this time, $\triangle B A D$ is a $\left(36^{\circ}, 36^{\circ}, 108^{\circ}\right)$ or $\left(36^{\circ}, 72^{\circ}, 72^{\circ}\right)$ triangle.
If $\triangle B A D$ is a $\le... | 72^{\circ}, 90^{\circ}, 108^{\circ}, 126^{\circ}, 132^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,770 |
5. "Lucky numbers" refer to positive integers that are equal to 19 times the sum of their digits (in decimal). Find all the lucky numbers.
| 5. Let $10a + b$ be a number with at most two digits. The equation $10a + b = 19(a + b)$ holds if and only if $a = b = 0$. Therefore, all lucky numbers must have at least three digits.
Assume a lucky number is an $m (m \geqslant 4)$-digit number, then the number is at least $10^{m-1}$, and the sum of its digits is at ... | 114, 133, 152, 171, 190, 209, 228, 247, 266, 285, 399 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,773 |
6. Two players, A and B, are playing a number-filling game on an $n \times n$ grid. Each turn, a player is allowed to fill a cell with the number 0 or 1 (each cell can only contain one number). Player A starts, and they take turns filling in numbers until every cell in the grid is filled. If the sum of the numbers in e... | 6. (1) When $n=2006$, the second player (乙) has a winning strategy.
Use $1 \times 2$ dominoes to partition the table such that each row consists of 1003 dominoes. When player 甲 fills a number in one of the cells of a domino, player 乙 also fills a number in the same domino, ensuring that the sum of the two numbers in t... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,774 |
7. Let $n$ be any odd positive integer. Prove:
$$
1596^{n}+1000^{n}-270^{n}-320^{n}
$$
is divisible by 2006. | 7. Since $2006=2 \times 17 \times 59$, to prove the conclusion, it is only necessary to prove that for odd positive integers $n$,
$$
1596^{n}+1000^{n}-270^{n}-320^{n}
$$
is divisible by $2$, $17$, and $59$.
It is clear that the expression is divisible by 2.
Using the formulas, for odd $n$,
$$
\begin{array}{l}
a^{n}+b^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 717,775 |
8. Remove all positive integers that are divisible by 4 and those that leave a remainder of 1 when divided by 4, and arrange the remaining numbers in ascending order to form a sequence $\left\{a_{n}\right\}$:
$$
2,3,6,7,10,11, \cdots \text {. }
$$
The sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ i... | 8. It is known that $a_{2 n-1}=4 n-2, a_{2 n}=4 n-1(n=1,2$, $\cdots$ ). Therefore,
$$
\begin{aligned}
& S_{2 n}=\left(a_{1}+a_{2}\right)+\left(a_{3}+a_{4}\right)+\cdots+\left(a_{2 n-1}+a_{2 n}\right) \\
= & 5+13+21+\cdots+(8 n-3) \\
= & \frac{(5+8 n-3) n}{2}=(2 n)^{2}+n, \\
& S_{2 n-1}=S_{2 n}-a_{2 n}=4 n^{2}+n-(4 n-1)... | 2013021 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,776 |
9. As shown in Figure 7, the areas of equilateral triangles $\triangle ABC$ and $\triangle PQR$ are both equal to 1, and the center $M$ of $\triangle PQR$ lies on the boundary of $\triangle ABC$. If the area of the overlapping part of these two triangles is $S$, find the minimum value of $S$. | 9. At the three vertices of the equilateral $\triangle PQR$, three congruent equilateral triangles are cut off, resulting in a regular hexagon with an area of $\frac{2}{3}$. $M$ is the center of this regular hexagon.
If point $M$ coincides with one of the vertices of $\triangle ABC$ (as shown in Figure 13(a)), it is e... | \frac{1}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,777 |
10. Find all four-digit numbers $m$, such that $m<2006$, and there exists a positive integer $n$, such that $m-n$ is a prime number, and $m n$ is a perfect square. | 10. From the given conditions, we know that $m-n=p$ (where $p$ is a prime number), so $m=n+p$.
Let $mn=n(n+p)=x^2$, where $x$ is a positive integer. Then, $4n^2+4pn=4x^2$, which means
$$
(2n+p)^2 - p^2 = (2x)^2.
$$
Thus, $(2n-2x+p)(2n+2x+p) = p^2$.
Since $p$ is a prime number, we have
$$
\left\{\begin{array}{l}
2n-2x+... | 1156, 1296, 1369, 1600, 1764 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,778 |
1. (1) If
$$
4 x^{3}-x^{2}-4 x+2=(2 x+\alpha)^{2}(x+\beta)+\gamma
$$
is an identity, find the constants $\alpha$, $\beta$, and $\gamma$;
(2) Given $|x| \leqslant 1$. Find the maximum and minimum values of the expression $4 x^{3}-x^{2}-4 x+2$. | 1. (1) Notice that
$$
4 x^{3}-x^{2}-4 x+2=(2 x+\alpha)^{2}(x+\beta)+\gamma \text{, }
$$
Therefore, $4 x^{3}-x^{2}-4 x+2$
$$
=4 x^{3}+4(\alpha+\beta) x^{2}+\left(\alpha^{2}+4 \alpha \beta\right) x+\alpha^{2} \beta+\gamma \text{. }
$$
By comparison, we get $\left\{\begin{array}{l}4(\alpha+\beta)=-1, \\ \alpha^{2}+4 \al... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,779 |
1. If $x, y, z$ are all real numbers, and satisfy
$$
\begin{array}{l}
\frac{x y}{(y+z)(z+x)}+\frac{y z}{(z+x)(x+y)}+ \\
\frac{z x}{(x+y)(y+z)}=1,
\end{array}
$$
then the values of $x, y, z$ are ( ).
(A) all positive
(B) all non-negative
(C) all negative
(D) exactly one is zero | $-1 . \mathrm{D}$.
Obviously, $(x+y)(y+z)(z+x) \neq 0$.
Eliminating the denominator, we get
$$
\begin{array}{l}
x y(x+y)+y z(y+z)+z x(z+x) \\
=(x+y)(y+z)(z+x) .
\end{array}
$$
Simplifying and organizing, we get $x y z=0$.
Therefore, at least one of $x, y, z$ is zero.
If two or three of $x, y, z$ are zero, then at leas... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,780 |
2. Let the incircle $\odot O$ of $\triangle ABC$ intersect the median $BM$ on side $CA$ at points $G$ and $H$, with point $G$ between points $B$ and $H$. Given that $BG=HM$ and $AB=2$. What are the values of $BC$ and $CA$ when the length of segment $GH$ is maximized? Also, find the maximum value of $GH$.
---
The tran... | 2. Suppose $BC > AB$. As shown in Figure 3, let $\odot O$ be tangent to sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. Then $AE = AF$.
By the tangent-secant theorem, we have
$$
\begin{array}{l}
EM = \sqrt{HM(HM + GH)} \\
= \sqrt{BG(BG + GH)} = BF.
\end{array}
$$
Therefore, $AM = AB = 2$, $AC = 2A... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,781 |
3. Given a sequence of positive integers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$, where $a_{1}=2^{2006}$, and for each positive integer $i, a_{i+1}$ is equal to the square of the sum of the digits of $a_{i}$. Find the value of $a_{2006}$. | 3. Since the remainders when $2, 2^{2}, 2^{3}, 2^{4}, 2^{5}, 2^{6}, \cdots$ are divided by 9 are $2, 4, 8, 7, 5, 1, \cdots$, therefore, the remainders when $2^{m+6}$ and $2^{m}$ are divided by 9 are equal.
But $2006 = 334 \times 6 + 2$, so the remainder when $a_{1}$ is divided by 9 is 4. Thus, the remainder when the s... | 169 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,782 |
1. Given
$$
\begin{array}{l}
12 a^{2}+7 b^{2}+5 c^{2} \\
\leqslant 12 a|b|-4 b|c|-16 c-16 .
\end{array}
$$
Then $a=$ $\qquad$ ,$b=$ $\qquad$ ,$c=$ $\qquad$ . | 1. $a=\frac{1}{2}, b=-1, c=-2$.
From the problem, we have
$$
\begin{array}{l}
3\left(4 a^{2}-4 a|b|+b^{2}\right)+\left(4 b^{2}+4 b|c|+c^{2}\right)+ \\
4\left(c^{2}+4 c+4\right) \leqslant 0,
\end{array}
$$
which simplifies to $3(2 a-|b|)^{2}+(2 b+|c|)^{2}+4(c+2)^{2} \leqslant 0$.
Thus, $\left\{\begin{array}{l}2 a-|b|=... | a=\frac{1}{2}, b=-1, c=-2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,783 |
2. Given $a, b, c \in \mathbf{R}$, and satisfy $a>b>c$, $a+b+c=0$. Then, the range of $\frac{c}{a}$ is $\qquad$ | 2. $\left(-2,-\frac{1}{2}\right)$.
From $a>b>c$ and $a+b+c=0$, we know $a>0, c-2 ; \\
0=a+b+c>a+2 c \Rightarrow-\frac{a}{2}>c \Rightarrow \frac{c}{a}<-\frac{1}{2} .
\end{array}
$ | \left(-2,-\frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,784 |
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