problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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2. As shown in Figure 1, in
$\triangle A B C$, $\angle A B D$
$=\angle D B E=\angle E B C$,
$\angle A C D=\angle D C E=$
$\angle E C B$. If $\angle B E C$
$=145^{\circ}$, then $\angle B D C$
is equal to ( ).
$\begin{array}{ll}\text { (A) } 100^{\circ} & \text { (B) } 105^{\circ}\end{array}$
(C) $110^{\circ}$
(D) $115^{... | 2.C.
From $\frac{1}{3}(\angle B+\angle C)=180^{\circ}-145^{\circ}=35^{\circ}$, we have $\frac{2}{3}(\angle B+\angle C)=70^{\circ}$.
Then $\angle B D C=180^{\circ}-70^{\circ}=110^{\circ}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,492 |
3. If $3 x^{3}-k x^{2}+4$ is divided by $3 x-1$ and the remainder is 3, then the value of $k$ is ( ).
(A) 2
(B) 4
(C) 9
(D) 10 | 3.D.
Given that $3 x^{3}-k x^{2}+1$ can be divided by $3 x-1$, then $3 \times\left(\frac{1}{3}\right)^{3}-k\left(\frac{1}{3}\right)^{2}+1=0$.
Thus, $k=10$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,493 |
4. The number of roots of the equation $|1-| x||+\sqrt{|x|-2}$ $=x$ with respect to $x$ is ( ).
(A) 0
(B) 1
(C) 3
(D) 4 | 4.B.
It is clear that $x \geqslant 0$, so $|1-x|+\sqrt{x-2}=x$. And $x \geqslant 2$. Thus, $x-1+\sqrt{x-2}=x$. Therefore, $\sqrt{x-2}=1$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,494 |
5. As shown in Figure 2, in equilateral $\triangle A B C$, $B D=2 D C$, $D E \perp B E, C E$ and $A D$ intersect at point $P$. Then ( ).
(A) $A P>A E>E P$
(B) $A E>A P>E P$
(C) $A P>E P>A E$
(D) $E P>A E>A P$ | 5.A.
From $B E=\frac{1}{2} B D=D C$, we have $\triangle B E C \cong \triangle C D A$.
Then $\angle A P E=\angle P A C+\angle A C P=\angle P C D+\angle A C P=60^{\circ}$.
Also, $\angle E A P A E>E P$.
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,495 |
6. If $a$ and $b$ are the roots of the equation
$$
(x+c)(x+d)=1
$$
then, $(a+c)(b+c)$ equals ( ).
(A) 1
(B) -1
(C) 0
(D) $c^{2}$ | 6. B.'
From the given, we have $\left\{\begin{array}{l}a+b=-(c+d), \\ a b=c d-1 .\end{array}\right.$
$$
\begin{array}{l}
\text { Then }(a+c)(b+c)=a b+(a+b) c+c^{2} \\
=a d-1-(c+d) c+c^{2}=-1 .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,496 |
3. The cost price of each product is 120 yuan. During the trial sales phase, the relationship between the selling price $x$ (yuan) of each product and the daily sales volume $y$ (units) is shown in Table 2. If the daily sales volume $y$ is a linear function of the selling price $x$, to maximize profit, what should the ... | (Answer: When the selling price is set to 160 yuan, the maximum daily profit is 1600 yuan.) | 1600 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,498 |
8. If $a+b+c=0, \frac{1}{a+1}+\frac{1}{b+2}+$ $\frac{1}{c+3}=0$, then, $(a+1)^{2}+(b+2)^{2}+(c+3)^{2}$ is equal to (.
(A) 36
(B) 16
(C) 14
(D) 3 | 8. A.
From the problem, we have
$$
\begin{array}{l}
(a+1)(b+2)+(b+2)(c+3)+(c+3)(a+1)=0 \text {. } \\
\text { Then }(a+1)^{2}+(b+2)^{2}+(c+3)^{2} \\
=(a+b+c+6)^{2}-2[(a+1)(b+2)+ \\
(b+2)(c+3)+(c+3)(a+1)]=36 .
\end{array}
$$ | 36 | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,499 |
10. As shown in Figure 3, in quadrilateral $ABCD$, $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively. $AN$, $BN$, $DM$, and $CM$ divide the quadrilateral into 7 regions with areas $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}$, $S_{5}$, $S_{6}$, and $S_{7}$. Therefore, the relationship that always holds is ( ).
(A) $S_{2... | 10. B.
Let the distances from $A$, $M$, and $B$ to $DC$ be $h_{a}$, $h_{m}$, and $h_{b}$, respectively. It is easy to see that $h_{a} + h_{b} = 2 h_{m}$. Then
$$
\begin{array}{l}
S_{\triangle D N} + S_{\triangle B N C} \\
=\frac{1}{2}\left(h_{a} + h_{b}\right) \cdot \frac{1}{2} D C = \frac{1}{2} h_{m} \cdot D C \\
= S... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,501 |
$13.2 x^{2}+4 x y+5 y^{2}-4 x+2 y-5$ can achieve the minimum value of $\qquad$ . | 13. -10 .
Original expression $=(x+2 y)^{2}+(x-2)^{2}+(y+1)^{2}-10$. When $x=2, y=-1$, it has the minimum value -10. | -10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,504 |
15.1 A can of coffee is shared by person A and person B, and they finish it together in 10 days. If person A drinks it alone, it takes 12 days. 1 pound of tea is shared by person A and person B, and they finish it together in 12 days. If person B drinks it alone, it takes 20 days. Assuming that person A will never drin... | Three, 15. It is known that A drinks $\frac{1}{30}$ of 1 catty of tea every day, and B drinks $\frac{1}{60}$ of 1 can of coffee every day.
After 30 days, A finishes the tea while B only drinks half a can of coffee, and the remaining half can of coffee is drunk by A and B together in 5 days. Therefore, it takes a total ... | 35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,506 |
1. The quadratic function that satisfies the condition $f\left(x^{2}\right)=[f(x)]^{2}$ is ( ).
(A) $f(x)=x^{2}$
(B) $f(x)=a x^{2}+5$
(C) $f(x)=x^{2}+x$
(D) $f(x)=-x^{2}+2004$ | -1.A.
Let $f(x)=a x^{2}+b x+c$. Substitute, expand and determine $a=1$, $b=0$, $c=0$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,508 |
Example 1 Try to find the volume of the edge-tangent sphere (a sphere that is tangent to all edges) of a regular tetrahedron with edge length $a$.
| Solution 1: As shown in Figure 1, let $O$ be the center of the regular tetrahedron $ABCD$. In the right triangle $\triangle OBF$, by the Pythagorean theorem, the circumradius of this regular tetrahedron $AO=BO=\frac{\sqrt{6}}{4} a$. Therefore, the radius of the edge-tangent sphere $r=OP=\frac{\sqrt{2}}{4} a$ (the inrad... | \frac{\sqrt{2}}{24} \pi a^{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,509 |
3. The equation ||$x|-1|=a$ has exactly three real solutions. Then $a$ equals ( ).
(A) 0
(B) 0.5
(C) 1
(D) $\sqrt{2}$ | 3.C.
Graphical method: Draw the graph of the function $y=|| x | -11$ (as shown in Figure 5). When a line parallel to the $O x$ axis, $y=a$, intersects the graph of the function $y=|| x|-1|$ at exactly three points, $a=1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,511 |
4. Real numbers $a, b, c$ satisfy $a+b>0, b+c>0$, $c+a>0, f(x)$ is an odd function on $\mathbf{R}$, and is a strictly decreasing function, i.e., if $x_{1}<x_{2}$ then $f(x_{1})>f(x_{2})$
(D) $f(a)+2 f(b)+f(c)=2004$ | 4.B.
$x \in \mathbf{R}, f(x)$ is an odd function, so $f(0)=0$. Since $f(x)$ is a strictly decreasing function, for $x>0$, we have $f(x)<0$, then
$$
\begin{array}{l}
a>-b \Rightarrow f(a)<f(-b)=-f(b) \\
\Rightarrow f(a)+f(b)<0 .
\end{array}
$$
Similarly, $f(b)+f(c)<0, f(c)+f(a)<0$.
Adding them up, we get $2[f(a)+f(b)+f... | B | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,512 |
5. Given that among the four positive integers $a, b, c, d$, $a$ leaves a remainder of $1$ when divided by $9$, $b$ leaves a remainder of $3$ when divided by $9$, $c$ leaves a remainder of $5$ when divided by $9$, and $d$ leaves a remainder of $7$ when divided by $9$. Then the two numbers that cannot be perfect squares... | 5.B.
Let the integer $x$ be divided by 9 with a remainder of $0, \pm 1, \pm 2, \pm 3, \pm 4$, then $x^{2}$ divided by 9 has a remainder of $0,1,4,7$. Therefore, positive integers that leave a remainder of $2,3,5,6,8$ when divided by 9 cannot be perfect squares, so $b$ and $c$ cannot be perfect squares. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 715,513 |
6. In the sequence of positive real numbers $a_{1} 、 a_{2} 、 a_{3} 、 a_{4} 、 a_{5}$, $a_{1} 、$ $a_{2} 、 a_{3}$ form an arithmetic sequence, $a_{2} 、 a_{3} 、 a_{4}$ form a geometric sequence, and the common ratio is not equal to 1. Also, the reciprocals of $a_{3} 、 a_{4} 、 a_{5}$ form an arithmetic sequence. Then ( ).
(... | 6.A.
From the given, we have $\left\{\begin{array}{l}2 a_{2}=a_{1}+a_{3}, \\ a_{3}^{2}=a_{2} a_{4}, \\ a_{3}^{-1}+a_{5}^{-1}=2 a_{4}^{-1} \text {. }\end{array}\right.$
From equation (3), we get $\frac{a_{3}+a_{5}}{a_{3} a_{5}}=\frac{2}{a_{4}}$.
From equation (2), we get $a_{4}=\frac{a_{3}^{2}}{a_{2}}$, substituting th... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,514 |
8. Given $a=1+2+\cdots+2004$. Then the remainder when $a$ is divided by 17 is | 8.1.
$$
\begin{array}{l}
a=1+2+\cdots+2004 \\
=\frac{2004 \times 2005}{2}=2009010 .
\end{array}
$$
Then 2009010 divided by 17 gives a quotient of 118177, with a remainder of 1. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,516 |
9. Given $f(x)=x^{2}+x-1$. If $a b^{2} \neq 1$, and $f\left(a^{-1}\right)=f\left(b^{2}\right)=0$, then $\frac{a}{1+a b^{2}}=$ $\qquad$ . | 9. -1 .
Given $f(x)=x^{2}+x-1, f\left(a^{-1}\right)=f\left(b^{2}\right)=0, a b^{2} \neq 1$, we know that $a^{-1}$ and $b^{2}$ are the two real roots of $f(x)=x^{2}+x-1$.
By Vieta's formulas, we get $\frac{1}{a}+b^{2}=-1, \frac{b^{2}}{a}=-1$. Thus, $\frac{1}{a}+b^{2}=\frac{b^{2}}{a}=-1$.
Therefore, we have $1+a b^{2}=-... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,517 |
11. If $a, b \in \mathbf{R}$ and $a^{2}+b^{2}=10$, then the range of values for $a-b$ is $\qquad$ | 11. $[-2 \sqrt{5}, 2 \sqrt{5}]$.
Given $a, b \in \mathbf{R}$ and $a^{2}+b^{2}=10$, we have
$$
\begin{array}{l}
(a-b)^{2}=2\left(a^{2}+b^{2}\right)-(a+b)^{2} \\
\leqslant 2\left(a^{2}+b^{2}\right)=20 .
\end{array}
$$
Thus, $|a-b| \leqslant 2 \sqrt{5}$. | [-2 \sqrt{5}, 2 \sqrt{5}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,519 |
Example 2 Rotate the closed figure formed by the parabola $y=x^{2}$ and the line $y=t$ around the $y$-axis once, and find the volume of the resulting solid of revolution. | Solution: Place the solid of revolution on a plane, and cut this solid with a plane parallel to the plane and at a distance $h$ from it (as shown in Figure 2). It is easy to see that the area of the circular cross-section is
$$
S_{\text {cut }}=\pi(\sqrt{h})^{2}=\pi h .
$$
This indicates that the cross-section of the ... | \frac{1}{2} \pi t^{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 715,520 |
12. Given that $a$ and $b$ are two real roots of the equation $x^{4} + m = 9 x^{2}$, and they satisfy $a + b = 4$. Then the value of $m$ is $\qquad$. | 12.12.25.
The equation can be transformed into $x^{4}-9 x^{2}+m=0$. Then $a^{2}, b^{2}$ are the roots of $y^{2}-$ $9 y+m=0$, where $y=x^{2}$. By Vieta's formulas, we get
$$
a^{2}+b^{2}=9, a^{2} b^{2}=m .
$$
Given $a+b=4$, then $(a+b)^{2}=16$, we can obtain $a b=\frac{7}{2}$. | 12.25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,521 |
13. $\cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 60^{\circ} \cdot \cos 80^{\circ}=$ | $\begin{array}{l}\text { 13. } \frac{1}{16} \cdot \\ \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 60^{\circ} \cdot \cos 80^{\circ} \\ =\frac{2 \sin 20^{\circ} \cdot \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 60^{\circ} \cdot \cos 80^{\circ}}{2 \sin 20^{\circ}} \\ =\frac{2 \sin 40^{\circ} \cdot \cos 40^{\circ}... | \frac{1}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,522 |
1. Given the function $f(x)=\frac{x^{2}}{1+x^{2}}$. Then, $f\left(\frac{1}{2004}\right)+f(1)+f(2004)=$ $\qquad$ . | $-1 . \frac{3}{2}$.
If $a b=1$, then
$$
\begin{array}{l}
f(a)+f(b)=\frac{a^{2}}{1+a^{2}}+\frac{b^{2}}{1+b^{2}} \\
=\frac{1+a^{2}+1+b^{2}}{1+a^{2}+b^{2}+a^{2} b^{2}}=1 .
\end{array}
$$
Therefore, $f(1)=\frac{1}{2}, f(2004)+f\left(\frac{1}{2004}\right)=1$.
Thus, $f\left(\frac{1}{2004}\right)+f(1)+f(2004)=\frac{3}{2}$. | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,524 |
2. Simplify $\left(\log _{3} 4+\log _{2} 9\right)^{2}-\left(\log _{3} 4-\log _{2} 9\right)^{2}$ $=$ $\qquad$ | 2.16.
$$
\begin{array}{l}
(\log 4+\log 9)^{2}-(\log 4-\log 9)^{2} \\
=4 \log 4 \cdot \log 9=4 \times \frac{\log 4}{\log 3} \times \frac{\log 9}{\log 2} \\
=4 \times \frac{2 \log 2}{\log 3} \times \frac{2 \log 3}{\log 2}=16 .
\end{array}
$$ | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,525 |
3. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{2}=2$, $a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+a_{n+2}$, and $a_{n+1} a_{n+2} \neq$ 1. Then $a_{1}+a_{2}+\cdots+a_{2004}=$ $\qquad$ | 3.4008.
Substituting $a_{1}=1, a_{2}=2$ into $a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+$ $a_{n+2}$, we get $a_{3}=3$.
$$
\begin{array}{l}
\text { From } a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+a_{n+2}, \\
a_{n+1} a_{n+2} a_{n+3}=a_{n+1}+a_{n+2}+a_{n+3},
\end{array}
$$
Subtracting the two equations, we get $\left(a_{n+3}-a_{n}... | 4008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,526 |
4. As shown in Figure 2, the radii of $\odot O_{1}$ and $\odot O_{2}$ are 2 and 4 respectively, and $O_{1} O_{2}=10$. Then the area of $\triangle M N P$ formed by the two inner common tangents and one outer common tangent of the two circles is $\qquad$ . | 4. $\frac{32}{3}$.
As shown in Figure 6, the external common tangent $AB = 4 \sqrt{6}$, the internal common tangents $CD = EF = 8$, $O_{1}P = \frac{10}{3}$, $O_{2}P = \frac{20}{3}$. By the Pythagorean theorem, we get
$$
PC = PF = \frac{8}{3}, PD = PE = \frac{16}{3}.
$$
Let $MA = MF = x$, $ND = NB = y$.
By the tangent... | \frac{32}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,527 |
```
(15 points) Given $a b c \neq 0$. Prove:
$$
\begin{array}{l}
\frac{a^{4}}{4 a^{4}+b^{4}+c^{4}}+\frac{b^{4}}{a^{4}+4 b^{4}+c^{4}}+ \\
\frac{c^{4}}{a^{4}+b^{4}+4 c^{4}} \leqslant \frac{1}{2} .
\end{array}
$$
``` | $$
\geqslant 2 a^{4}+2 a^{2} b^{2}+2 a^{2} c^{2},
$$
Therefore, $\frac{a^{4}}{4 a^{4}+b^{4}+c^{4}} \leqslant \frac{a^{4}}{2 a^{4}+2 a^{2} b^{2}+2 a^{2} c^{2}}$
$$
=\frac{a^{2}}{2\left(a^{2}+b^{2}+c^{2}\right)} \text {. }
$$
Similarly, we get $\frac{b^{4}}{a^{4}+4 b^{4}+c^{4}} \leqslant \frac{b^{2}}{2\left(a^{2}+b^{2}... | \frac{1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 715,529 |
Three. (15 points) Given $\alpha \neq \frac{\pi}{2}+2 k \pi(k \in \mathbf{Z})$. Prove:
that at least one of the following quadratic equations in $x$ has two distinct real roots:
$$
\begin{array}{l}
x^{2}-\left(1-\cos ^{3} \alpha\right) x+\cos \alpha=0, \\
x^{2}-\left(1-\sin ^{3} \alpha\right) x+\sin \alpha=0, \\
x^{2}-... | Three discriminants of the equations are denoted as $\Delta_{1}, \Delta_{2}, \Delta_{3}$, respectively, then
$$
\begin{array}{l}
\Delta_{1}=\left(1-\cos ^{3} \alpha\right)^{2}-4 \cos \alpha, \\
\Delta_{2}=\left(1-\sin ^{3} \alpha\right)^{2}-4 \sin \alpha, \\
\Delta_{3}=\left(\sqrt{\frac{1+\cos \alpha}{1-\sin \alpha}}\r... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,530 |
Example 1 In a square $ABCD$ with side length 1, take point $P$ on side $AB$, point $Q$ on side $BC$, point $M$ on side $CD$, and point $N$ on side $AD$. If $AP + AN + CQ + CM = 2$, prove that $PM \perp QN$. | Proof: As shown in Figure 2, if the square $ABCD$ is rotated $90^{\circ}$ clockwise around point $A$, then the square $ABCD$ will be transformed to the position of square $AD C_{1} D_{1}$. Here, $A \rightarrow A, B \rightarrow D, Q \rightarrow Q_{1}, C \rightarrow C_{1}, D \rightarrow D_{1}, N \rightarrow N_{1}$, and t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,531 |
Example: $2 P$ is a point inside a regular $\triangle A B C$, $\angle A P B=113^{\circ}, \angle A P C=123^{\circ}$. Prove that a triangle can be formed with $A P, B P, C P$ as sides, and determine the degree measures of the interior angles of the formed triangle. | Proof: As shown in Figure 3, with point $C$ as the center, rotate $\triangle A P C$ counterclockwise by $60^{\circ}$. Since $A C = B C$, $\angle A C P = \angle B C P_{1} = 60^{\circ} - \angle P C B$, $C P = C P_{1}$, and $\angle P C P_{1} = 60^{\circ}$, we have:
$\triangle A P C \cong \triangle B P_{1} C$.
Therefore, $... | 63^{\circ}, 64^{\circ}, 53^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,532 |
3. In pentagon $A B C D E$, $A B=A E, B C+D E=$ $C D, \angle A B C+\angle A E D=180^{\circ}$, connect $A D$. Prove: $A D$ bisects $\angle C D E$. | (Rotation hint: Rotate $\triangle A D E$ clockwise so that $A E$ coincides with $A B$.) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,533 |
5. Given that point $O$ is inside a convex quadrilateral $A B C D$, $\angle A O B$ $=\angle C O D=120^{\circ}, A O=O B$, and $C O=O D, K$ is the midpoint of $A B$, $L$ is the midpoint of $B C$, and $M$ is the midpoint of $C D$. Prove that $\triangle K L M$ is an equilateral triangle. | (Tip: $\triangle A O C$ rotates counterclockwise $120^{\circ}$ around point $O$, thus coinciding with $\triangle B O D$, knowing that $A C = B D$, and the angle between $A C$ and $B D$ is $60^{\circ}$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,534 |
Example 1 In an exam, there are 5 multiple-choice questions, each with 4 different options, and each person selects exactly 1 option for each question. In 2000 answer sheets, it is found that there exists an $n$, such that in any $n$ answer sheets, there are 4 sheets where any 2 sheets have at most 3 answers the same. ... | Solution: Let the 4 possible answers for each question be denoted as $1, 2, 3, 4$, and the answers on each test paper be denoted as $(g, h, i, j, k)$, where $g, h, i, j, k \in \{1,2,3,4\}$. Let $\{(1, h, i, j, k), (2, h, i, j, k), (3, h, i, j, k), (4, h, i, j, k)\}, h, i, j, k = 1, 2, 3, 4$, yielding a total of 256 qua... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,535 |
Example 2 In a carriage, any $m(m \geqslant 3)$ passengers have a unique common friend (if A is a friend of B, then B is also a friend of A, and no one is a friend of themselves), how many friends does the person with the most friends have in this carriage?
| Solution: Let the person $A$ with the most friends have $k$ friends, denoted as $B_{1}, B_{2}, \cdots, B_{k}$, and let $S=\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$.
Obviously, $k \geqslant m$.
If $k>m$, let $\left\{B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}\right\}$ be any $(m-1)$-element subset of $S$, then $A, B_{i_... | k=m | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,536 |
Example 3 Let $n \in \mathbf{N}_{+}, n \geqslant 2, S$ be an $n$-element set. Find the smallest positive integer $k$, such that there exist subsets $A_{1}, A_{2}, \cdots, A_{k}$ of $S$ with the following property: for any two distinct elements $a, b$ in $S$, there exists $j \in\{1,2, \cdots, k\}$, such that $A_{j} \cap... | Solution: Let $S=\{1, 2, \cdots, k\}$. Construct the table
$1:$
If $i \in A_{j}$, then mark the cell at the intersection of the row for $A_{j}$ and the column for $i$ with 1, and mark the rest of the cells with 0.
Consider the sequence of columns in Table 1, $P_{1}, P_{2}, \cdots, P_{n}$.
We will prove: The subsets $A_... | k=\left\{\begin{array}{ll}
\log _{2} n, & \text { when } \log _{2} n \in \mathbf{N}_{+} ; \\
{\left[\log _{2} n\right]+1,} & \text { when } \log _{2} n \notin \mathbf{N}_{+}
\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,537 |
Example 4 Given $n$ points in the plane with no three points collinear, $m$ line segments are drawn with these $n$ points as endpoints. It is known that for any two points $A$ and $B$ among these $n$ points, there is a point $C$ such that $C$ is connected to both $A$ and $B$ by line segments. Find the minimum value of ... | Let the $n$ points be denoted as $A_{1}, A_{2}, \cdots, A_{n}$. First, consider an example.
If $n$ is odd, connect the line segments $A_{1} A_{2}, A_{1} A_{3}, \cdots$,
$$
A_{1} A_{n} ; A_{2} A_{3}, A_{4} A_{5}, \cdots, A_{n-1} A_{n} \text {. }
$$
If $n$ is even, connect the line segments $A_{1} A_{2}, A_{1} A_{3}, \c... | \left\lfloor\frac{3 n-3}{2}\right\rfloor | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,538 |
Example 5 Given a set of points $P=\left\{P_{1}, P_{2}, \cdots, P_{1924}\right\}$ on a plane, and no three points in $P$ are collinear. Divide all the points in $P$ into 83 groups arbitrarily, such that each group has at least three points, and each point belongs to exactly one group. Then, connect any two points in th... | Solution: (1) Let $m(G)=m_{0}, G$ is obtained from the groups $X_{1}, X_{2}, \cdots, X_{88}$, where $X_{i}$ is the set of points in the $i$-th group, $i=1,2, \cdots, 83$.
Let $\left|X_{i}\right|=x_{i}, i=1,2, \cdots, 83$, then we have
$$
x_{1}+x_{2}+\cdots+x_{83}=1994 \text {, }
$$
and $m_{0}=C_{x_{1}}^{3}+C_{x_{2}}^{... | 168544 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,539 |
Example 6 Let $n$ be a fixed positive even number. Consider an $n \times n$ square board, which is divided into $n^{2}$ unit square cells. Two different cells on the board are called adjacent if they share a common edge. Mark $N$ unit square cells on the board so that every cell on the board (marked or unmarked) is adj... | Solution: Let $n=2k$.
First, color the square board in a checkerboard pattern like a chessboard. Let $f(n)$ be the minimum value of $N$ sought, $f_{a}(n)$ be the minimum number of white squares that must be marked so that every black square has a marked white square adjacent to it. Similarly, define $f_{b}(n)$ as the m... | k(k+1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,540 |
1. After 20 figure skaters have performed, 9 judges respectively assign them rankings from 1 to 20. It is known that the difference between any rankings given to each athlete does not exceed 3. If the sum of the rankings received by each athlete is arranged in an increasing sequence, $c_{1} \leqslant c_{2} \leqslant$ $... | (Tip: Let the set of athletes who have won first place be $A$. If $|A|=1$, then $c_{1}=9$; If $|A|=2$, then one of them wins no less than 5 first places, by the problem's condition we know $c_{1} \leqslant 5 \times 1+4 \times 4=21$; If $|A|=3$, similarly, we have $c_{1}+c_{2}+c_{3} \leqslant 1 \times 9+3 \times 9+4 \ti... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,541 |
2. In space, there are 10 points, all connected pairwise, and these line segments are colored with red and blue. Among them, the line segments connected from point $A$ are all red. Among the triangles formed with these 10 points as vertices, how many triangles with three sides of the same color are there at least?
In ... | (Tip: Let the number of same-colored triangles be $x$, then the number of differently colored triangles is $\mathrm{C}_{10}^{3}-x$. Let the total number of same-colored angles be $y$, then $y=3 x+\left(\mathrm{C}_{10}^{3}-\right.$ $x)=120+2 x$. When $y$ reaches its minimum value, $x$ also reaches its minimum value. The... | 30 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,542 |
Example: 31 red right-angled triangular paper pieces with a hypotenuse of 29, 1 blue right-angled triangular paper piece with a hypotenuse of 49, and 1 yellow square paper piece, are used to form a right-angled triangle as shown in Figure 4. Question: What is the sum of the areas of the red and blue triangular paper pi... | Solution: As shown in Figure 5, rotate the right triangle $\triangle BDE$ counterclockwise around point $D$ by $90^{\circ}$. Since $DE = DF$, point $E$ clearly coincides with point $F$. Given $\angle DEB = 90^{\circ} = \angle DFC$, $EB$ falls along $FC$, and point $B$ lands on point $G$ on $FC$. At this moment, the rig... | 710.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,543 |
3. In a $13 \times 13$ square grid, select the centers of $k$ small squares such that no four of these points form the vertices of a rectangle (with sides parallel to those of the original square). Find the maximum value of $k$ that satisfies this condition. | (Let the $i$-th column have $x_{i}$ points $(i=1,2, \cdots, 13)$, then $\sum_{i=1}^{13} x_{i}=k$, the $x_{i}$ points in the $i$-th column form $\mathrm{C}_{x_{i}}^{2}$ different point pairs (if $x_{1}<2$, then $\mathrm{C}_{x_{1}}^{2}=0$). Add a column to the side of the $13 \times 13$ square, and each point pair projec... | 52 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,544 |
4. A test paper has 4 multiple-choice questions, each with three options (A), (B), (C). Several students take the exam, and after grading, it is found that: any 3 students have 1 question where their answers are all different. How many students can take the exam at most? | (提示: If 10 people take the exam, then for the 1st question, at least 7 people choose two options: for the 2nd question, at least 5 people among these 7 choose two options; for the 3rd question, at least 4 people among these 5 choose two options; for the 4th question, at least 3 people among these 4 choose two options. ... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,545 |
5. Given $n$ points $A_{1}, A_{2}, \cdots, A_{n}(n \geqslant 3)$ in the plane, no three of which are collinear. By selecting $k$ pairs of points, determine $k$ lines (i.e., draw a line through each pair of the $k$ pairs of points), such that these $k$ lines do not form a triangle with all three vertices being given poi... | (提示: If a line $l$ connects two points (let's denote them as $\left.A_{1} 、 A_{2}\right)$, by the problem's condition, these two points cannot simultaneously connect to point $A_{1}(i \geqslant 3)$, meaning the number of lines passing through at least one of $A_{1} 、 A_{2}$ is at most $n-1$ (including line $l$). Simila... | \frac{n^2}{4} \text{ if } n \text{ is even; } \frac{n^2-1}{4} \text{ if } n \text{ is odd.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,546 |
Question 2 As shown in Figure 3, given a regular $\triangle ABC$, $D$ is any point on side $BC$, the circumcenter and incenter of $\triangle ABD$ are $O_{1}$ and $I_{1}$, the circumcenter and incenter of $\triangle ACD$ are $O_{2}$ and $I_{2}$, and the center of $\triangle ABC$ is $O$. Prove:
(1) $\mathrm{OO}_{1}+\math... | Proof: Without loss of generality, let $\angle A D B \geqslant 90^{\circ}$. Because
$$
\begin{array}{l}
\angle A O_{2} D=2 \angle C=120^{\circ}, \\
\angle A I_{2} D=90^{\circ}+\frac{1}{2} \angle C=120^{\circ}, \\
\angle B=60^{\circ},
\end{array}
$$
Therefore, $A 、 O_{2} 、 I_{2} 、 D 、 B$ are concyclic with the center a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,547 |
Question 3 As shown in Figure 4, given an equilateral $\triangle ABC$, $D$ is any point on side $BC$, and the incenter of $\triangle ABD$ and $\triangle ACD$ are $I_{1}$ and $I_{2}$, respectively. A regular $\triangle I_{1} I_{2} E$ is constructed with $I_{1} I_{2}$ as a side (points $D$ and $E$ are on opposite sides o... | Proof: Take the circumcenters $O_{1}, O_{2}$ of $\triangle A B D$ and $\triangle A C D$, respectively, and draw auxiliary lines as shown in Figure 4.
From Problem 2, we know that $A, O_{1}, I_{1}, D$ and $A, O_{2}, I_{2}, D$ are concyclic, and the radii of the two circles are equal. Therefore,
$$
\begin{array}{l}
\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,548 |
Question 4 As shown in Figure 5, given an equilateral $\triangle ABC$, $D$ is any point on side $BC$, and the incenters of $\triangle ABD$ and $\triangle ACD$ are $I_{1}$ and $I_{2}$, respectively. Take two points $E$ and $F$ on $AD$ such that $AE=CD$ and $AF=BD$. Prove:
(1) $I_{1} I_{2}^{2}=I_{1} E^{2}+I_{2} F^{2}$;
(... | Proof: (1) Let the incircle of $\triangle A B D$ touch $A D$ at point $H$, and connect $D I_{1}$, $D I_{2}$, and $I_{1} H$.
Since $D E=A D-A E=A D-C D$
$$
=A D-(A B-B D)=2 D H \text {, }
$$
Therefore, $E H=D H$.
Combining with $I_{1} H \perp D E$, we know $I_{1} E=I_{1} D$.
Similarly, $I_{2} F=D I_{2}$.
Thus, $I_{1} I... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,549 |
Question 5 As shown in Figure 6, given an equilateral $\triangle ABC$, $D$ is any point on side $BC$, the incircles of $\triangle ABD$ and $\triangle ACD$ are $\odot I_{1}$ and $\odot I_{2}$, respectively. The external common tangent of these two circles, other than $BC$, intersects $AD$, $AB$, and $AC$ at $F$, $M$, an... | Proof: Obviously, the conclusion holds when $M N / / B C$.
Assume $N M$ intersects $C B$ at point $Q$, clearly $I_{2} I_{1}$ passes through point $Q$. Let $M I_{1}$ and $N I_{2}$ intersect at point $P$, it is evident that $P$ is the excenter of $\triangle A M N$, hence $A P$ bisects $\angle B A C$. Let $A P$ intersect ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,550 |
A circle passes through the vertices $A$ and $B$ of $\triangle ABC$, intersecting the segments $AC$ and $BC$ at points $D$ and $E$, respectively. The line $BA$ and $ED$ intersect at point $F$, and the line $BD$ and $CF$ intersect at point $M$. Prove that $MF = MC$ if and only if
$$
MB \cdot MD = MC^2.
$$
(2003, United ... | Proof: As shown in Figure 1, connect $A E$. In $\triangle C B F$, since $A C$, $B M$, and $E F$ intersect at point $D$, by Ceva's Theorem, we have
$$
\begin{array}{l}
\frac{C E}{B E} \cdot \frac{A B}{A F} \cdot \frac{M F}{M C} \\
=1 .
\end{array}
$$
Sufficiency.
From $M B \cdot M D=M C^{2} \Rightarrow \frac{M C}{M D}=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,551 |
Given $a, b, c$ are positive real numbers. Prove:
$$
\begin{array}{l}
\left(a^{5}-a^{2}+3\right)\left(b^{5}-b^{2}+3\right)\left(c^{5}-c^{2}+3\right) \\
\geqslant(a+b+c)^{3} .
\end{array}
$$
(2004, USA Mathematical Olympiad)
Studying this problem, the author found that it can be generalized.
Proposition If $a_{i} \in \m... | Proof: Since $a_{i} \in \mathbf{R}_{+}, i=1,2, \cdots, n$, we have:
$$
\begin{array}{l}
\left(a_{i}^{n}-1\right)\left(a_{i}^{n-1}-1\right) \geqslant 0\left(n \in \mathbf{N}_{+}\right) \\
\Leftrightarrow a_{i}^{2 n-1}-a_{i}^{n}-a_{i}^{n-1}+1 \geqslant 0 \\
\Leftrightarrow a_{i}^{2 n-1}-a_{i}^{n-1}+n \geqslant a_{i}^{n}+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,552 |
Let $\angle X O Y=90^{\circ}, P$ be a point inside $\angle X O Y$, and $O P=1, \angle X O P=30^{\circ}$. Draw an arbitrary line through point $P$ that intersects rays $O X$ and $O Y$ at points $M$ and $N$, respectively. Find the maximum value of $O M+O N-M N$.
(2004, IMO China National Training Team Selection Exam) | Solution: As shown in Figure 1, let
$$
\angle P M O=\theta\left(0^{\circ}<\theta\right.
$$
$<90^{\circ}$ ), then
$$
\begin{array}{l}
O M+O N-M N \\
=\frac{\sqrt{3}}{2}+\frac{1}{2} \cot \theta+ \\
\frac{1}{2}+\frac{\sqrt{3}}{2} \tan \theta-\frac{1}{2 \sin \theta}-\frac{\sqrt{3}}{2 \cos \theta} \text {. } \\
\text { Let ... | \sqrt{3}+1-\sqrt[4]{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,553 |
Example 4 As shown in Figure 6, construct squares $ABDE$ and $CAFG$ outward on the sides $AB$ and $AC$ of $\triangle ABC$. Connect $EF$, and draw a perpendicular from point $A$ to $BC$, intersecting $EF$ at point $M$. Prove: $EM=FM$. | Proof: As shown in Figure 7, rotate $\triangle A B C$ 90 degrees clockwise around point $A$ to the position of $\triangle A E C_{1}$. It is easy to see that points $C_{1}$, $A$, and $F$ are collinear, $A C_{1} = A F$, and $A$ is the midpoint of side $C_{1} F$ of $\triangle F E C_{1}$.
It is evident that $\angle E A M$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,554 |
Let $\odot O$ be a circle with an inscribed convex quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ intersect at point $P$. The circle $\odot O_{1}$ passing through points $P$ and $B$ intersects the circle $\odot O_{2}$ passing through points $P$ and $A$ at points $P$ and $Q$. Furthermore, $\odot O_{1}$ and $\odot O_... | As shown in Figure 1.
Since $\angle P J F=\angle P A F=\angle C A F=\angle C D F$,
thus, $P J \parallel C D$.
Similarly, $I P \parallel C D$.
Therefore, points $I, P, J$ are collinear.
Also, $\angle E F D=180^{\circ}-\angle E C D=180^{\circ}-\angle E I J$,
thus, points $E, F, J, I$ are concyclic.
Hence, by the Radical ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,555 |
Given a positive integer $n(n \geqslant 2)$, find the largest $\lambda$ such that: if there are $n$ bags, each containing some balls of integer powers of 2 grams, and the total weight of the balls in each bag is the same, then there must be a certain weight of balls whose total number is at least $\lambda$. (The same b... | Let's assume the heaviest ball weighs 1. First, we prove:
$$
\lambda_{\max } \geqslant\left[\frac{n}{2}\right]+1 \text {. }
$$
Let the total weight of the balls in each bag be $G$, then $G \geqslant 1$.
Assume the total number of balls of any weight is less than or equal to $\left[\frac{n}{2}\right]$. Considering the ... | \left[\frac{n}{2}\right]+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,556 |
Three, $n$ is a positive integer, $a_{j}(j=1,2, \cdots, n)$ are complex numbers, and for any non-empty subset $I$ of the set $\{1,2, \cdots, n\}$, we have
$$
\left|\prod_{j \in I}\left(1+a_{j}\right)-1\right| \leqslant \frac{1}{2} .
$$
Prove: $\sum_{j=1}^{n}\left|a_{j}\right| \leqslant 3$.
(Supplied by Hua-Wei Zhu) | Three, let $1+a_{j}=r_{j} \mathrm{e}^{\mathrm{i} \theta_{j}},\left|\theta_{j}\right| \leqslant \pi, j=1,2, \cdots, n$. Then the given condition becomes
$$
\left|\prod_{j \in I} r_{j} \cdot \mathrm{e}^{i \sum_{j \in I_{j}}^{j}}-1\right| \leqslant \frac{1}{2} .
$$
First, we prove the following lemma.
Lemma: Let $r, \the... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,557 |
Let $a_{1}, a_{2}, \cdots, a_{6} ; b_{1}, b_{2}, \cdots, b_{6} ; c_{1}$, $c_{2}, \cdots, c_{6}$ all be permutations of $1,2, \cdots, 6$. Find the minimum value of $\sum_{i=1}^{6} a_{i} b_{i} c_{i}$.
(Xiong Bin) | Let $S=\sum_{i=1}^{6} a_{i} b_{i} c_{i}$. By the AM-GM inequality, we have
$$
\begin{array}{l}
S \geqslant 6 \sqrt[6]{\prod_{i=1}^{6} a_{i} b_{i} c_{i}}=6 \sqrt[6]{(6!)^{3}} \\
=6 \sqrt{6!}=72 \sqrt{5}>160 .
\end{array}
$$
Next, we prove that $S>161$.
Since the geometric mean of the 6 numbers $a_{1} b_{1} c_{1}, a_{2}... | 162 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,558 |
Five, let $n$ be any given positive integer, and $x$ be a positive real number. Prove:
$$
\sum_{k=1}^{n}\left(x\left[\frac{k}{x}\right]-(x+1)\left[\frac{k}{x+1}\right]\right) \leqslant n,
$$
where $[a]$ denotes the greatest integer not exceeding the real number $a$.
(Yu Hongbing) | First, we prove a lemma.
Lemma For any real numbers $\alpha, \beta > 0$, there exist an integer $u$ and a real number $\gamma$ such that
$$
a = \beta u + \gamma,
$$
where $0 \leq \gamma < \beta$.
Let $f(k) = k - c_{k} - 1$, since when $k \in I$, we have
$$
k = c_{k}(x + 1) + d_{k} > c_{k} + 1,
$$
thus $0 < f(k) < n$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,559 |
Six, let $\alpha$ be a given positive real number. Find all functions $f: \mathbf{N}_{+} \rightarrow \mathbf{R}$, such that for any positive integers $k, m$ satisfying the condition
$$
\alpha m \leqslant k < (\alpha+1) m
$$
we have
$$
f(k+m)=f(k)+f(m).
$$
(Proposed by Yonggao Chen) | Six $f(n)=b n, b$ is any given real number.
First, prove: when $n \geqslant 2 \alpha+3$, we have
$$
f(n+1)-f(n)=f(n)-f(n-1) \text {. }
$$
It is sufficient to prove that there exists a positive integer $u$, such that
$$
\begin{array}{l}
f(n+1)-f(n)=f(u+1)-f(u), \\
f(n)-f(n-1)=f(u+1)-f(u) .
\end{array}
$$
Equation (1) ... | f(n)=b n, b \text{ is any given real number} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,560 |
Let $D$ be a point on side $B C$ of $\triangle A B C$, and point $P$ lies on line segment $A D$. A line through point $D$ intersects line segments $A B$ and $P B$ at points $M$ and $E$, respectively, and intersects the extensions of line segments $A C$ and $P C$ at points $F$ and $N$, respectively. If $D E = D F$, prov... | For $\triangle A M D$ and line $B E P$, $\triangle A F D$ and line $N C P$, $\triangle A M F$ and line $B D C$, by Menelaus' theorem, we respectively get
$$
\begin{array}{l}
\frac{A P}{P D} \cdot \frac{D E}{E M} \cdot \frac{M B}{B A}=1 . \\
\frac{A C}{C F} \cdot \frac{F N}{N D} \cdot \frac{D P}{P A}=1 . \\
\frac{A B}{B... | D M = D N | Geometry | proof | Yes | Yes | cn_contest | false | 715,561 |
Three, (1) Does there exist an infinite sequence of positive integers $\left\{a_{n}\right\}$, such that for any positive integer $n$, we have
$$
a_{n+1}^{2} \geqslant 2 a_{n} a_{n+2} .
$$
(2) Does there exist an infinite sequence of positive irrational numbers $\left\{a_{n}\right\}$, such that for any positive integer ... | (1) Suppose there exists a sequence of positive integers $\left\{a_{n}\right\}$ satisfying the condition.
Since $a_{n+1}^{2} \geqslant 2 a_{n} a_{n+2}, a_{n}>0$, we have
$$
\begin{array}{l}
\quad \frac{a_{n}}{a_{n-1}} \leqslant \frac{1}{2} \cdot \frac{a_{n-1}}{a_{n-2}} \leqslant \frac{1}{2^{2}} \cdot \frac{a_{n-2}}{a_{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,562 |
Given a positive integer $n$ greater than 2004, fill the numbers $1, 2, \cdots, n^2$ into the squares of an $n \times n$ chessboard (consisting of $n$ rows and $n$ columns) such that each square contains exactly one number. If a number in a square is greater than the numbers in at least 2004 squares in its row and at l... | For the convenience of narration, if a number in a cell is greater than the numbers in at least 2004 other cells in its row, then this cell is called “row superior”. Since in each row, the 2004 cells with the smallest numbers are not “row superior”, there are $n-2004$ “row superior” cells in each row. A cell that is “s... | n(n-2004) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,563 |
Example 5 As shown in Figure 8, in quadrilateral $A B C D$, $\angle A B C=30^{\circ}$, $\angle A D C=60^{\circ}, A D$ $=D C$. Prove: $B D^{2}=A B^{2}+B C^{2}$. | Proof: As shown in Figure 9, connect $AC$. Since $AD = DC$ and $\angle ADC = 60^{\circ}$, $\triangle ADC$ is an equilateral triangle. Therefore, we have
$$
DC = CA = AD.
$$
Rotate $\triangle DCB$ $60^{\circ}$ clockwise around point $C$ to the position of $\triangle ACE$, and connect $EB$. At this time,
$$
\begin{array... | BD^2 = AB^2 + BC^2 | Geometry | proof | Yes | Yes | cn_contest | false | 715,564 |
$$
\begin{array}{l}
\sqrt{2}(2 a+3) \cos \left(\theta-\frac{\pi}{4}\right)+\frac{6}{\sin \theta+\cos \theta}-2 \sin 2 \theta \\
<3 a+6
\end{array}
$$
For $\theta \in\left[0, \frac{\pi}{2}\right]$, the inequality always holds. Find the range of values for $a$.
(Zhang Hongcheng, provided) | Let $\sin \theta + \cos \theta = x$. Then
$$
\begin{array}{l}
\cos \left(\theta - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} x \\
\sin 2 \theta = x^{2} - 1, \quad x \in [1, \sqrt{2}].
\end{array}
$$
Thus, the original inequality can be transformed into
$$
(2a + 3)x + \frac{6}{x} - 2(x^{2} - 1) > 0.
$$
Rearranging, we g... | a > 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,565 |
1. As shown in Figure 1, there is a rectangular piece of paper $A B C D, A B=8$, $A D=6$. The paper is folded so that the edge $A D$ lies on the edge $A B$, with the fold line being $A E$. Then, $\triangle A E D$ is folded along $D E$ to the right, and the intersection of $A E$ and $B C$ is point $F$. The area of $\tri... | - .1.A.
From the folding process, we know $D E=A D=6, \angle D A E=\angle C E F=$ $45^{\circ}$. Therefore, $\triangle C E F$ is an isosceles right triangle, and $E C=8-6$ $=2$. Hence, $S_{\triangle C E F}=2$. | 2 | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,566 |
2. If $M=3 x^{2}-8 x y+9 y^{2}-4 x+6 y+13$ (where $x, y$ are real numbers), then the value of $M$ must be ( ).
(A) positive
(B) negative
(C) zero
(D) integer | 2. A.
Notice that
$$
\begin{aligned}
M & =3 x^{2}-8 x y+9 y^{2}-4 x+6 y+13 \\
& =2(x-2 y)^{2}+(x-2)^{2}+(y+3)^{2} \geqslant 0,
\end{aligned}
$$
and $x-2 y, x-2, y+3$ these 3 numbers cannot be 0 at the same time, so, $M>0$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,567 |
3. Given that point $I$ is the incenter of acute $\triangle A B C$, $A_{1} 、 B_{1}$ 、 $C_{1}$ are the reflections of point $I$ over sides $B C 、 C A 、 A B$ respectively. If point $B$ lies on the circumcircle of $\triangle A_{1} B_{1} C_{1}$, then $\angle A B C$ equals ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\ci... | 3.C.
Since $I A_{1}=I B_{1}=I C_{1}=2 r$ (where $r$ is the inradius of $\triangle A B C$), point $I$ is also the circumcenter of $\triangle A_{1} B_{1} C_{1}$. As shown in Figure 5, let the intersection of $I A_{1}$ and $B C$ be point $D$. Then $I B=I A_{1}=2 I D$, so $\angle I B D=30^{\circ}$. Similarly, $\angle I B ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,568 |
4. Given
$$
A=48 \times\left(\frac{1}{3^{2}-4}+\frac{1}{4^{2}-4}+\cdots+\frac{1}{100^{2}-4}\right) \text {. }
$$
The positive integer closest to $A$ is ( ).
(A) 18
(B) 20
(C) 24
(D) 25 | 4.D.
For positive integers $n(n \geqslant 3)$, we have $\frac{1}{n^{2}-4}=\frac{1}{4}\left(\frac{1}{n-2}-\frac{1}{n+2}\right)$.
Then $A=48 \times \frac{1}{4}\left[\left(1+\frac{1}{2}+\cdots+\frac{1}{8}\right)-\left(\frac{1}{5}+\frac{1}{6}+\cdots+\frac{1}{R C}\right)\right]$
$$
\begin{array}{l}
=12\left(1+\frac{1}{2}+\... | 25 | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,569 |
5. Let $a, b$ be positive integers, and satisfy $56 \leqslant a+b \leqslant 59, 0.9<\frac{a}{b}<0.91$. Then $b^{2}-a^{2}$ equals ( ).
(A) 171
(B) 177
(C) 180
(D) 182 | 5.B.
From the given, we have
$$
0.9 b+b56 \text {, }
$$
then $29<b<32$. Therefore, $b=30,31$.
When $b=30$, from $0.9 b<a<0.91 b$, we get $27<a<28$, there is no such positive integer $a$;
When $b=31$, from $0.9 b<a<0.91 b$, we get $27<a<29$, so $a=28$.
Thus, $b^{2}-a^{2}=177$. | 177 | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,570 |
6. On the face of a circular clock, $O A$ represents the second hand, $O B$ represents the minute hand ($O$ is the center of rotation for both hands). If the current time is exactly 12 o'clock, then after $\qquad$ seconds, the area of $\triangle O A B$ will reach its maximum for the first time. | Ni.6. $15 \frac{15}{59}$.
Let the height on side $O A$ be $h$, then $h \leqslant O B$, so,
$S_{\triangle O A B}=\frac{1}{2} O A \cdot h \leqslant \frac{1}{2} O A \cdot O B$.
The equality holds when $O A \perp O B$.
At this moment, the area of $\triangle O A B$ is maximized.
Suppose after $t \mathrm{~s}$, $O A$ and $O B... | 15 \frac{15}{59} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,571 |
Example 6 As shown in Figure 10, in $\triangle ABC$, $AB=AC$,
$\angle BAC=120^{\circ}$,
$\triangle ADE$ is an equilateral
triangle, point $D$ is on
side $BC$. It is known that
$BD: DC=2: 3$. When the area of $\triangle ABC$ is $50 \mathrm{~cm}^{2}$, find the area of $\triangle ADE$.
(7th Japan Arithmetic Olympiad (Fina... | Analysis: Directly solving the problem is difficult. By rotating $\triangle A B C$ counterclockwise around point $A$ by $120^{\circ}$ and $240^{\circ}$ to form a regular $\triangle M B C$ (as shown in Figure 11), the regular $\triangle A D E$ becomes regular $\triangle A D_{1} E_{1}$ and regular $\triangle A D_{2} E_{2... | 14 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,572 |
7. In the Cartesian coordinate system, the parabola
$$
y=x^{2}+m x-\frac{3}{4} m^{2}(m>0)
$$
intersects the $x$-axis at points $A$ and $B$. If the distances from points $A$ and $B$ to the origin are $O A$ and $O B$, respectively, and satisfy $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, then the value of $m$ is $\qquad$. | 7. 2 .
Let the roots of the equation $x^{2}+m x-\frac{3}{4} m^{2}=0$ be $x_{1}$ and $x_{2}$, and $x_{1}<0<x_{2}$.
From $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, we know $O A>O B$.
Also, since $m>0$, the axis of symmetry of the parabola is to the left of the $y$-axis, thus,
$$
O A=\left|x_{1}\right|=-x_{1}, O B=x_{2} ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,573 |
8. There are two decks of playing cards, each arranged in the following order: the 1st card is the Big Joker, the 2nd card is the Small Joker, followed by the four suits of Spades, Hearts, Diamonds, and Clubs, with each suit arranged in the order of $\mathrm{A}, 2,3, \cdots, \mathrm{J}, \mathrm{Q}, \mathrm{K}$. Someone... | 8. The 6 of Diamonds in the second deck.
According to the problem, if the number of cards is $2, 2^{2}, 2^{3}, \cdots$, $2^{n}$, then, according to the specified operation method, the last remaining card will be the last card of these cards. For example, if you have only 64 cards, according to the specified operation ... | 6 of Diamonds in the second deck | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,574 |
9. Given that $D$ and $E$ are points on the sides $BC$ and $CA$ of $\triangle ABC$, respectively, and $BD=4$, $DC=1$, $AE=5$, $EC=2$. Connecting $AD$ and $BE$, they intersect at point $P$. Through point $P$, draw $PQ \parallel CA$ and $PR \parallel CB$, which intersect side $AB$ at points $Q$ and $R$, respectively. The... | 9. $\frac{400}{1089}$.
As shown in Figure 6, draw $E F / / A D$, intersecting side $B C$ at point $F$. Then,
$$
\frac{C F}{F D}=\frac{C E}{E A}=\frac{2}{5} .
$$
Thus, $F D=\frac{5}{7}$.
Since $P Q / / C A$, we have,
$$
\frac{P Q}{E A}=\frac{B P}{B E}=\frac{B D}{B F}=\frac{28}{33} \text {. }
$$
Therefore, $P Q=\frac{... | \frac{400}{1089} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,575 |
10. Given $x_{1}, x_{2}, \cdots, x_{40}$ are all positive integers, and $x_{1}+$ $x_{2}+\cdots+x_{40}=58$. If the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ is $A$, and the minimum value is $B$, then the value of $A+B$ is $\qquad$ | 10.494.
Since the number of ways to write 58 as the sum of 40 positive integers is finite, the minimum and maximum values of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ exist.
Assume without loss of generality that $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{40}$.
If $x_{1}>1$, then $x_{1}+x_{2}=\left(x_{1}-1\rig... | 494 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,576 |
11. 8 people are traveling to the railway station in two cars with the same speed, with 4 people in each car (excluding the driver). One of the cars breaks down 15 km away from the railway station, and there are still 42 minutes left before the ticket check stops. At this point, the only available means of transportati... | Three, 11. [Plan One] When the car breaks down, the 4 people in this car get out and walk, while the other car takes the 4 people inside to the train station, immediately returns to pick up the 4 people walking, and takes them to the train station.
Let the distance walked by the 4 people in the broken-down car be $x \... | 37(\mathrm{~min})<42(\mathrm{~min}) | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,577 |
12. As shown in Figure 2, two circles with unequal radii intersect at points $A$ and $B$. Line segment $CD$ passes through point $A$ and intersects the two circles at points $C$ and $D$ respectively. Connect $BC$ and $BD$, and let $P$, $Q$, and $K$ be the midpoints of $BC$, $BD$, and $CD$ respectively. Let $M$ and $N$ ... | 12. (1) As shown in Figure 8, since $M$ is the midpoint of $\overparen{B C}$ and $P$ is the midpoint of $B C$, we have
$$
\begin{array}{l}
M P \perp B C, \\
\angle B P M=90^{\circ} .
\end{array}
$$
Similarly,
$$
\angle N Q B=90^{\circ} .
$$
Connecting $A B$, we get
$$
\begin{array}{l}
\angle P B M=\frac{1}{2} \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,578 |
13. Given that $p$ and $q$ are both prime numbers, and that the quadratic equation in $x$, $x^{2}-(8 p-10 q) x+5 p q=0$, has at least one positive integer root. Find all prime number pairs $(p, q)$. | 13. From the sum of the two roots of the equation being $8 p-10 q$, we know that if one root is an integer, the other root is also an integer. From the product of the two roots being $5 p q$, we know that the other root is also a positive integer.
Let the two positive integer roots of the equation be $x_{1}$ and $x_{2... | (7,3) \text{ or } (11,3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,579 |
14. From the 200 positive integers $1, 2, \cdots, 205$, what is the maximum number of integers that can be selected such that for any three selected numbers $a, b, c (a < b < c)$, we have $a b \neq c$? | 14. First, the 193 numbers $1, 14, 15, \cdots, 205$ satisfy the conditions of the problem.
In fact, if $a, b, c (a < b < c)$, then $a b \geqslant 14 \times 15 = 210 > c$.
On the other hand, consider the following 12 sets of numbers:
$(2, 25, 2 \times 25), (3, 24, 3 \times 24), \cdots$,
$(13, 14, 13 \times 14)$.
These ... | 193 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,580 |
11. In a school's spring sports meeting, several students form a rectangular formation with 8 columns. If 120 more students are added to the original formation, a square formation can be formed; if 120 students are removed from the original formation, a square formation can also be formed. How many students are there i... | 11. Let the original rectangular formation have $8x$ students. From the given conditions, we know that $8x+120$ and $8x-120$ are both perfect squares. Therefore, we can set
$$
\left\{\begin{array}{l}
8x+120=m^{2}, \\
8x-120=n^{2}.
\end{array}\right.
$$
where $m$ and $n$ are positive integers, and $m > n$.
Subtracting ... | 136 \text{ or } 904 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,581 |
12. As shown in Figure 3, $\triangle ABC$ is an acute triangle. A circle $\odot O$ is drawn with $BC$ as its diameter, and $AD$ is a tangent to $\odot O$. A perpendicular line is drawn from a point $E$ on $AB$ to $AB$, intersecting the extension of $AC$ at point $F$.
If $\frac{AB}{AF}=\frac{AE}{AC}$, prove that $AD=AE$... | 12. As shown in Figure 9. Let $\odot O$ intersect $A B$ at point $G$, and connect $C G$. Since $B C$ is the diameter of $\odot O$, we have $C G \perp A B$.
Also, $E F \perp A B$, so $C G / / E F$. Therefore,
$$
\frac{A G}{A E}=\frac{A C}{A F},
$$
which implies $A C \cdot A E=A G \cdot A F$.
Since $A D$ is a tangent to... | A D=A E | Geometry | proof | Yes | Yes | cn_contest | false | 715,582 |
Example 7 As shown in Figure 12, in the convex hexagon $A B C D E F$, $B C = C D, E F = F A, \angle B C D = \angle E F A = 60^{\circ}$. Let $G, H$ be two points inside this hexagon such that $\angle B G D = \angle A H E = 120^{\circ}$. Prove:
$$
\begin{array}{l}
B G + G D + G H + \\
H A + H E \geqslant C F .
\end{arra... | Prove: Connect $B D, E A$. From $B C=C D, \angle B C D=60^{\circ}$, we know that $\triangle B C D$ is an equilateral triangle.
Since $\angle B G D=120^{\circ}$, points $B, C, D, G$ are concyclic. Therefore,
$$
\angle B C G=\angle B D G.
$$
With point $B$ as the center of rotation, rotate $\triangle B C G$ clockwise b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,583 |
13. Given a quadratic equation in $x$
$$
(6-k)(9-k) x^{2}-(117-15 k) x+54=0
$$
both roots of which are integers. Find all real values of $k$ that satisfy the condition. | 13. The original equation can be transformed into
$$
[(6-k) x-9][(9-k) x-6]=0 \text {. }
$$
Since this equation is a quadratic equation in $x$, we have $-k \neq 6, k \neq 9$. Therefore, we have
$$
x_{1}=\frac{9}{6-k}, x_{2}=\frac{6}{9-k} \text {. }
$$
Eliminating $k$, we get $x_{1} x_{2}-2 x_{1}+3 x_{2}=0$, which sim... | k=7, \frac{15}{2}, \frac{39}{5}, \frac{33}{4}, \frac{21}{2}, 15,3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,584 |
14. As shown in Figure 4, with the sides $AB$, $BC$, and $CA$ of the acute $\triangle ABC$ as the hypotenuses, construct isosceles right triangles $\triangle DAB$, $\triangle EBC$, and $\triangle FAC$ outward. Prove:
(1) $AE = DF$;
(2) $AE \perp DF$. | 14. (1) As shown in Figure 10, extend $BD$ to point $P$ such that $DP = BD$, and connect $AP$ and $CP$.
Since $\triangle DAB$ is an isosceles right triangle, we have:
$$
\begin{array}{l}
\angle ADB = 90^{\circ}, \\
AD = BD, \\
\frac{AB}{BD} = \sqrt{2}.
\end{array}
$$
Since $DP = BD$, we have $\frac{AB}{BP} = \frac{\s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,585 |
1. Given real numbers $a, b, c$ satisfy:
(1) $\sqrt{\frac{a}{b}}(\sqrt{a b}+2 b)=2 \sqrt{a b}+3 b$;
(2) $a=b c$.
Then the number of values for $c$ is ( ).
(A) 1
(B) 2
(C) 3
(D) infinitely many | - 1.C.
Obviously, $ab>0$.
When $a>0, b>0$, from (1) we get $a=3b \Rightarrow c=\frac{a}{b}=3$. When $a<0, b<0$, from (1) we get
$$
a+4 \sqrt{ab}+3b=0,
$$
which simplifies to $\left(\sqrt{\frac{a}{b}}\right)^{2}-4 \sqrt{\frac{a}{b}}+3=0$. Solving this, we get $\sqrt{\frac{a}{b}}=1$ or 3. Therefore, $c=1$ or 9. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,586 |
2. If the remainders when $n^{2}+4$ and $7 n-2$ are divided by 3 are the same for two unequal positive integers $n$ not exceeding 100, then the number of such integers $n$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 2. A.
From $\left\{\begin{array}{l}1 \leqslant n^{2}+4 \leqslant 100, \\ 1 \leqslant 7 n-2 \leqslant 100,\end{array}\right.$ we get $\frac{3}{7} \leqslant n \leqslant \sqrt{96}$.
Since $n$ is an integer, the possible values of $n$ are
$1,2,3,4,5,6,7,8,9$.
Because $\left(n^{2}+4\right)-(7 n-2)=(n-1)(n-6)$ is a multiple... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 715,587 |
3. Given real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=4$. Then the maximum value of $(2 x-y)^{2}+(2 y-z)^{2}+(2 z-x)^{2}$ is ( ).
(A) 12
(B) 20
(C) 28
(D) 36 | 3.C.
From the problem, we have
$$
\begin{array}{l}
(2 x-y)^{2}+(2 y-z)^{2}+(2 z-x)^{2} \\
\approx 20-4(x y+y z+z x) \\
=28-2(x+y+z)^{2} \leqslant 28 .
\end{array}
$$
When $x+y+z=0$, the equality holds.
Thus, the maximum value sought is 28. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,588 |
4. In an acute triangle $\triangle A B C$, $A B=A C>B C$, point $D$ is on side $A B$, and $A D=B C$. If $\angle B D C=30^{\circ}$, then $\angle A=(\quad)$.
(A) $10^{\circ}$
(B) $20^{\circ}$
(C) $15^{\circ}$
(D) $22.5^{\circ}$ | 4.B.
As shown in Figure 5, construct an equilateral $\triangle A D E$ outside $\triangle A B C$ with $A D$ as a side, and connect $E C$. It is easy to see that
$$
\begin{array}{l}
\triangle A D C \cong \triangle E D C \\
\Rightarrow \angle A C D=\angle E C D, A C=E C \\
\Rightarrow \triangle A B C \cong \triangle C A... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,589 |
5. If $\alpha, \beta$ are the two real roots of the quadratic equation in $x$
$$
x^{2}-2 m x+1=m^{2}
$$
where $m$ is a real number, then the minimum value of $\alpha^{2}+\beta^{2}$ is ( ).
(A)3
(B) 4
(C) 1
(D) 2 | 5.C.
Since $\alpha+\beta=2 m, \alpha \beta=1-m^{2}$, therefore, $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=6 m^{2}-2$.
Also, from $\Delta=4 m^{2}-4\left(1-m^{2}\right) \geqslant 0$, solving gives $m^{2} \geqslant \frac{1}{2}$. Thus, $\alpha^{2}+\beta^{2} \geqslant 6 \times \frac{1}{2}-2=1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,590 |
6. As shown in Figure 1, point $C$ is a trisection point of the semicircular arc $\overparen{A B}$ with a radius of 1. Two semicircles are constructed outward on the chords $A C$ and $B C$, and points $D$ and $E$ are also trisection points of these two semicircular arcs. Four more semicircles are then constructed outwa... | 6. B.
It is easy to know that points $D$, $C$, and $E$ are collinear,
\[
\begin{array}{l}
A C=\frac{1}{2} A B=1, B C=A B \cos 30^{\circ}=\sqrt{3}, \\
B E=B C \cos 30^{\circ}=\frac{3}{2}, C E=D C=\frac{\sqrt{3}}{2}, A D=\frac{1}{2},
\end{array}
\]
and quadrilateral $A B E D$ is a right trapezoid, with the outer 4 semi... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,591 |
1. Define an operation “ * ”: For any pair of real numbers $(x, y)$, it always holds that
$$
(x, y) *(x, y)=\left(x+y+1, x^{2}-y-1\right) \text {. }
$$
If real numbers $a$ and $b$ satisfy $(a, b) *(a, b)=(b, a)$, then $a=$ $\qquad$ ,$b=$ $\qquad$ . | $$
\text { II. 1. }-1,1 \text {. }
$$
According to the problem, we get
$$
\left\{\begin{array}{l}
a+b+1=b, \\
a^{2}-b-1=a .
\end{array} \text { Solving, we get } a=-1, b=1\right. \text {. }
$$ | a=-1, b=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,592 |
2. Given line segment $A B=6$, there is a moving point $P$ on the plane that always satisfies $P A-P B=4$. Draw the perpendicular line from point $A$ to the angle bisector of $\angle A P B$, with the foot of the perpendicular being $M$. Then the maximum value of the area of $\triangle A M B$ is $\qquad$
Translate the ... | 2.6.
As shown in Figure 6, extend $AM$ and $PB$, and let the extensions intersect at point $C$. From the given conditions, it is easy to see that $\triangle AMP \cong \triangle MPP$. Then
$$
\begin{array}{l}
PC = PA, \\
AM = MC. \\
\text{Also, } PA - PB = 4,
\end{array}
$$
which means $PC - PB = BC = 4$.
Let the heig... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 715,593 |
Example 8 As shown in Figure 13, in the convex hexagon $A B C D E F$, $A B=B C=C D, D E=E F=F A, \angle B C D=$ $\angle E F A=60^{\circ}$. Let $G$ and $H$ be two points inside this hexagon such that $\angle A G B=\angle D H E=120^{\circ}$. Prove:
$$
A G+G B+G H+D H+H E \geqslant C F .
$$
(38th IMO) | In fact, by constructing the axisymmetric hexagon $D B C_{1} A E F_{1}$ of hexagon $A B C D E F$ with respect to the line $l$ where $B E$ lies, according to the result of Example 7, we have
$$
A G+G B+G H+D H+H E \geqslant C_{1} F_{1} .
$$
Noting that $C_{1} F_{1}=C F$, thus,
$$
A G+G B+G H+D H+H E \geqslant C F \text... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,594 |
3. A city's No. 10 Middle School has built a new six-story enclosed student dormitory, with 15 rooms on each floor, and this dormitory has only one entrance and exit safety door. The safety inspection department has tested the normal passage of this safety door: in an emergency, the safety door can function normally fo... | 3.168.
Assuming that under normal circumstances, this safety door can allow an average of $x$ people to pass through per minute. According to the problem, we have
$$
8 \times 15 \times 6 \leqslant x(1+90\%+80\% \times 3) .
$$
Solving this, we get $167.4 \approx \frac{720}{4.3} \leqslant x$.
Therefore, the minimum int... | 168 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,595 |
One. (20 points) As shown in Figure 3, given that point $C$ is the midpoint of the minor arc $\overparen{A B}$ of $\odot O$, point $D$ is on $\overparen{A C}$, and $A C=2, A D + B D = \sqrt{6} + \sqrt{2}$. Find the degree measure of $\angle D A C$. | Obviously, $B C=A C$.
As shown in Figure 7, with $C$ as the center and $C A$ as the radius, draw $\odot C$, which must pass through point $B$. Extend $B C$ and $B D$ to intersect $\odot C$ at points $E$ and $F$, respectively, and connect $A E$, $A F$, $E F$, and $C F$. Then
$$
\begin{array}{l}
\angle D A C=\angle D B C... | 15^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,597 |
For every positive integer $x$, obtain the number $z$ according to the following process:
(i) Move the last digit of $x$ to the first position to get $y$;
(ii) Then take the square root of $y$ to get its arithmetic square root $z$.
For example, if $x=9, y=9, z=\sqrt{9}=3$;
and if $x=5002, y=2500$,
$$
z=\sqrt{2500}=50 ... | (1) If $x$ is a one-digit number, then $y=x, z=\sqrt{y}=\sqrt{x}$.
Also, $z=\frac{1}{2} x$, so $\sqrt{x}=\frac{1}{2} x$.
Solving this, we get $x_{1}=4, x_{2}=0$ (discard).
Thus, $x=4$.
(2) If $x$ is a two-digit number, let $x=10 a+b, a, b$ be integers, and $0<a \leqslant 9,0 \leqslant b \leqslant 9$. Then
$$
y=10 b+a,... | 4 \text{ or } 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,598 |
Three. (25 points) As shown in Figure 4, given the line $l: y = kx + 2 (k < 0)$ intersects the $y$-axis at point $A$ and the $x$-axis at point $B$. The circle $\odot P$ with $OA$ as its diameter intersects $l$ at another point $D$. When the minor arc $\overparen{A D}$ is flipped along the line $l$, it intersects $OA$ a... | Three, (1) As shown in Figure 8, connect
$O D, D E$. Then
$$
\begin{array}{l}
\angle D E O \\
=\angle E A D+\angle A D E \\
=\frac{\overparen{A E D^{\circ}}}{2}=\frac{\overparen{A m D}^{\circ}}{2} \\
=\angle A O D .
\end{array}
$$
Therefore, $O D=D E$.
When $k=-2$, it is easy to get $A(0,2), B(1,0), O A=2, O B$ $=1$. ... | \frac{4}{5}, k=-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,599 |
1. The function $f(x)$ defined on $\mathbf{R}$ satisfies
$$
f(x+3)+f(x)=0,
$$
and the function $f\left(x-\frac{3}{2}\right)$ is an odd function. Given the following 3 propositions:
(1) The period of the function $f(x)$ is 6;
(2) The graph of the function $f(x)$ is symmetric about the point $\left(-\frac{3}{2}, 0\right... | $-1 . \mathrm{A}$.
From $f(x+3)=-f(x)$, we know $f(x+6)=f(x)$. Therefore, (1) is correct; Shifting the graph of $f\left(x-\frac{3}{2}\right)$ to the left by $\frac{3}{2}$ units results in the graph of $f(x)$, and the graph of $f\left(x-\frac{3}{2}\right)$ is symmetric about the origin, so, (2) is correct; From (2), we ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,600 |
2. If $\alpha \in\left(\frac{\pi}{2}, \pi\right)$, then the range of real numbers $x$ that satisfy the equation
$$
\log _{2}\left(x^{2}-x+2\right)=\sin \alpha-\sqrt{3} \cos \alpha
$$
is ( ).
(A) $-1<x<2$
(B) $-1<x<0$ or $1<x<2$
(C) $0 \leqslant x \leqslant 1$
(D) $-1 \leqslant x<0$ or $1<x \leqslant 2$ | 2.D.
Since $\frac{\pi}{2}<\alpha<\pi$, therefore, $\frac{\pi}{6}<\alpha-\frac{\pi}{3}<\frac{2 \pi}{3}$.
Thus, $\sin \alpha-\sqrt{3} \cos \alpha=2 \sin \left(\alpha-\frac{\pi}{3}\right) \in(1,2]$.
Hence, $1<\log _{2}\left(x^{2}-x+2\right) \leqslant 2$. This gives $2<x^{2}-x+2 \leqslant 4$.
Solving, we get $-1 \leqslan... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,601 |
3. As shown in Figure 1, in Yang
Hui's triangle, above the diagonal line $l$,
starting from 1, the numbers indicated by
the arrow form a zigzag sequence:
$1,3,3,4,6,5,10$, $\cdots$, let the sum of the first $n$ terms be $S_{n}$. Then the value of $S_{17}$ is ( ).
(A) 172
(B) 217
(C) 228
(D) 283 | 3. B.
By the properties of combinations, the sequence $1,3,3,4,6,5,10, \cdots$ is actually composed of $\mathrm{C}_{2}^{2}, \mathrm{C}_{3}^{1}, \mathrm{C}_{3}^{2}, \mathrm{C}_{4}^{1}, \mathrm{C}_{4}^{2}, \mathrm{C}_{5}^{1}, \mathrm{C}_{5}^{2}, \cdots$. Therefore,
$$
\begin{array}{l}
S_{17}=C_{2}^{2}+C_{3}^{1}+C_{3}^{2... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,602 |
4. Given point $P$ is inside or on the circumference of a fixed circle $\odot O$, and a moving circle $\odot C$ passes through point $P$ and is tangent to $\odot O$. Then the locus of the center of $\odot C$ could be ( ).
(A) circle, ellipse, or hyperbola
(B) two rays, circle, or parabola
(C) two rays, circle, or ellip... | 4.C.
If point $P$ is on $\odot O$, then the locus of the center of $\odot C$ is two rays, as shown in Fig. 3(a); if point $P$ coincides with the center of $\odot O$, then the locus of the center of $\odot C$ is a circle with $O$ as the center and $\frac{r}{2}$ as the radius, as shown in Fig. 3(b); if point $P$ is insi... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,603 |
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