problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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Example 10 Proof: In a non-obtuse $\triangle ABC$, $\sin A+\sin B+\sin C>2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Example 10 Proof: In a non-obtuse $\triangle ABC$, $\sin A+\sin B+\sin C>2$. | Proof: Let $x=\tan \frac{A}{2}, y=\tan \frac{B}{2}, z=\tan \frac{C}{2}$, we need to prove $\frac{2 x}{1+x^{2}}+\frac{2 y}{1+y^{2}}+\frac{2 z}{1+z^{2}}>2$
$$
\begin{array}{l}
\Leftrightarrow \frac{4}{(x+y)(y+z)(z+x)}>2 \\
\Leftrightarrow(x+y)(y+z)(z+x)<2 \\
\Leftrightarrow x+y+z-x y z<2 .
\end{array}
$$
Since $\frac{\a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,317 |
Example 11 Given that the side lengths $a, b, c$ of $\triangle ABC$ satisfy:
$$
\begin{array}{l}
a^{2}<b^{2}+c^{2}+bc \\
b^{2}<c^{2}+a^{2}+ca \\
c^{2}<a^{2}+b^{2}+ab
\end{array}
$$
Prove: $\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}<\frac{4 \sqrt{3}}{3}$. | Proof: From the given conditions, each interior angle of $\triangle ABC$ is less than $120^{\circ}$. Let $x=\tan \frac{A}{2}, y=\tan \frac{B}{2}, z=\tan \frac{C}{2}$.
Since $\frac{\angle A}{2}, \frac{\angle B}{2}, \frac{\angle C}{2} \in \left(0, \frac{\pi}{3}\right)$, it follows that $x, y, z \in (0, \sqrt{3})$.
Thus, ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,318 |
Example 12 Let $\angle A, \angle B, \angle C$ be the interior angles of $\triangle ABC$. Prove: for any real numbers $u, v, w$, we have
$$
\begin{array}{l}
u^{2}+v^{2}+w^{2} \\
\geqslant 2 u v \sin \frac{C}{2}+2 w u \sin \frac{B}{2}+2 v w \sin \frac{A}{2} .
\end{array}
$$ | Prove: Let $x=\tan \frac{A}{2}, y=\tan \frac{B}{2}, z=\tan \frac{C}{2}$, i.e., to prove
$$
\begin{array}{l}
u^{2}+v^{2}+w^{2} \\
\geqslant 2 w \frac{z}{\sqrt{1+z^{2}}}+2 u v \frac{y}{\sqrt{1+y^{2}}}+2 v w \frac{x}{\sqrt{1+x^{2}}}
\end{array}
$$
Notice that
$$
\begin{array}{l}
2 u v \frac{z}{\sqrt{1+z^{2}}}=2 z \cdot \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,319 |
Example 13 Let $\alpha, \beta, \gamma$ be acute angles, and $\cos ^{2} \alpha+$ $\cos ^{2} \beta+\cos ^{2} \gamma=1$. Prove:
$\cot \alpha \cdot \cot \beta+\cot \beta \cdot \cot \gamma+\cot \gamma \cdot \cot \alpha \leqslant \frac{3}{2}$. | Prove: Let $\cos ^{2} \alpha=y z, \cos ^{2} \beta=z x, \cos ^{2} \gamma=x y$. Then $x y+y z+z x=1$, and
$$
\begin{array}{l}
\sin \alpha=\sqrt{1-y z}=\sqrt{x y+z x}, \\
\sin \beta=\sqrt{y z+x y}, \sin \gamma=\sqrt{z x+y z} .
\end{array}
$$
Notice that
$$
\begin{array}{l}
\cot \alpha \cdot \cot \beta=\frac{\cos \alpha \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,320 |
Example 14 Try to prove the cosine rule using the system method, that is, in $\triangle ABC$, we have $a^{2}=b^{2}+c^{2}-2bc\cos A$. | Proof: Since $a=2 R \sin A, b=2 R \sin B$, $c=2 R \sin C$, we need to prove
$\sin ^{2} A=\sin ^{2} B+\sin ^{2} C-2 \sin B \cdot \sin C \cdot \cos A$
Let $x=\cot A, y=\cot B, z=\cot C$, then we have $\sin ^{2} A+\sin ^{2} B+\sin ^{2} C-2 \sin B \cdot \sin C \cdot \cos A$ $=\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,321 |
Example 15 Let $x, y, z$ be non-negative real numbers, and $x+y+z=1$. Prove that: $0 \leqslant xy+yz+zx-2xyz \leqslant \frac{7}{27}$. | Proof: Let
$$
f(x, y, z)=x y+y z+z x-2 x y z \text {. }
$$
When one of $x, y, z$ is 0, the conclusion is obvious.
Now consider the case where $x, y, z$ are all positive.
$$
\begin{array}{l}
f(x, y, z) \\
=(x y+y z+z x)(x+y+z)-2 x y z \\
=x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)+x y z \\
\geqslant 0 .
\end{array}
$$
Next, we ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,322 |
Example 3 In a non-obtuse $\triangle A B C$, the circumradius is $R$, and $m_{a}, m_{b}, m_{c}$ are the lengths of the three medians of the triangle. Then $4 R \leqslant m_{a}+m_{b}+m_{c} \leqslant \frac{9}{2} R$. | Proof: Since $\triangle ABC$ is a non-obtuse triangle, the circumcenter $O$ is inside or on the boundary of $\triangle ABC$. Connecting $AG$, $BG$, $CG$, then point $O$ must be inside or on the boundary of a certain triangle (let's assume it is $\triangle GBC$).
Thus, we have $GB + GC \geqslant OB + OC$, which means
$$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,323 |
Proposition: Divide the major axis of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ into $n(n \in \mathbf{N}$, and $n>1)$ equal parts. Draw perpendiculars to the $x$-axis through each division point, intersecting the upper half (or lower half) of the ellipse at points $P_{1}, P_{2}, \cdots, P_{n-1}$. L... | Proof: As shown in Figure 1, by the symmetry of the ellipse,
$$
\begin{array}{r}
P_{1} \text { and } P_{n-1}, \\
P_{2} \text { and } P_{n-2}, \cdots . .
\end{array}
$$
are symmetric with respect to the $y$-axis.
Thus, the distances from $P_{1}, P_{2}, \cdots, P_{n-1}$ to the right focus (or left focus) are equal to th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,324 |
Proposition (1) $\sum \frac{h_{a}+h_{b}}{r_{a}+r_{b}}=2+\frac{2 r}{R}$;
(2) $\prod \frac{h_{b}+h_{c}}{r_{a}+h_{a}}=\frac{2 r}{R}$.
Where, the three sides of $\triangle A B C$ are $a, b, c$, the semi-perimeter is $p$, the area is $S$, the circumradius and inradius are $R, r$ respectively, the exradii are $r_{a}, r_{b},... | Prove: (1) Since $S=\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=r p$, then $h_{a}=\frac{2 r p}{a}, h_{b}=\frac{2 r p}{b}$.
Also, $S=(p-a) r_{a}=(p-b) r_{b}=r p$, then $r_{a}=\frac{p p}{p-a}, r_{b}=\frac{p p}{p-b}$.
And $a b c=4 R r p$,
$(p-a)(p-b)(p-c)=r^{2} p$,
thus $\frac{h_{a}+h_{b}}{r_{a}+r_{b}}=\frac{\frac{2 r p}{a}+\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,325 |
1. Calculate the value of $(\sqrt{30}+\sqrt{21}-3)(\sqrt{3}+\sqrt{10}-\sqrt{7})$ equals ( ).
(A) $6 \sqrt{7}$
(B) $-6 \sqrt{7}$
(C) $20 \sqrt{3}+6 \sqrt{7}$
(D) $20 \sqrt{3}-6 \sqrt{7}$ | $\begin{array}{l}\text { I. A. } \\ \text { Original expression }=\sqrt{3}(\sqrt{10}+\sqrt{7}-\sqrt{3})(\sqrt{3}+\sqrt{10}-\sqrt{7}) \\ =\sqrt{3}[\sqrt{10}+(\sqrt{7}-\sqrt{3})][\sqrt{10}-(\sqrt{7}-\sqrt{3})] \\ =\sqrt{3}\left[(\sqrt{10})^{2}-(\sqrt{7}-\sqrt{3})^{2}\right] \\ =\sqrt{3}(10-7+2 \sqrt{21}-3)=6 \sqrt{7} .\e... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,326 |
2. If real numbers $x, y$ make three of the four numbers $x+y, x-y, xy, \frac{x}{y}$ equal, then the value of $|y|-|x|$ is ( ).
(A) $-\frac{1}{2}$
(B) 0
(C) $\frac{1}{2}$
(D) $\frac{3}{2}$ | 2.C.
Obviously, $x+y \neq x-y$, otherwise, $y=0$, at this time, $\frac{x}{y}$ is undefined. Therefore, it must be that $x y=\frac{x}{y}$. Thus, we have
$$
x\left(y^{2}-1\right)=0 \text {. }
$$
Solving this, we get $x=0$ or $y= \pm 1$.
(1) When $x=0$, if $x y=x+y$ or $x y=x-y$, both lead to $y=0$, making $\frac{x}{y}$... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,327 |
3. If real numbers $a, b, c$ satisfy the condition $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ $=\frac{1}{a+b+c}$, then among $a, b, c$ ( ).
(A) there must be two numbers that are equal
(B) there must be two numbers that are opposites
(C) there must be two numbers that are reciprocals
(D) no two numbers are equal | 3. B.
From the given, we have $(a+b+c)(bc+ca+ab)=abc$.
Expanding and rearranging, we get
$$
\begin{array}{l}
a^{2} c+a^{2} b+b^{2} c+b^{2} a+c^{2} b+c^{2} a+2 a b c=0 \\
\Rightarrow a c(a+b)+a b(a+b)+b c(a+b)+c^{2}(a+b)=0 \\
\Rightarrow a(a+b)(b+c)+c(a+b)(b+c)=0 \\
\Rightarrow(a+b)(b+c)(c+a)=0 .
\end{array}
$$
Thus, ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,328 |
4. As shown in Figure 1, in trapezoid $A B C D$, $A D / / B C$, $A D \perp C D, B C=C D=2 A D, E$ is a point on $C D$, $\angle A B E=45^{\circ}$. Then the value of $\tan \angle A E B$ is ( ).
(A) $\frac{3}{2}$
(B) 2
(C) $\frac{5}{2}$
(D) 3 | 4.D.
As shown in Figure 6, extend $DA$ to point $F$ such that $AF = AD$, and connect $BF$. From the given information, quadrilateral $FBCD$ is a square, hence $BF = BC$.
Extend $AF$ to point $G$ such that $FG = CE$, and connect $BG$. Then, Rt $\triangle BFG \cong$ Rt $\triangle BCE$. Therefore, $BG = BE$, and $\angle... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,329 |
5. Use several small cubes of the same size, with surfaces all white or all red, to form a large cube $A B C D-E F G H$ as shown in Figure 2. If the small cubes used on the diagonals $A G$, $B H$, $C E$, and $D F$ of the large cube are all red, and a total of 41 are used, and the rest of the large cube is made up of sm... | 5.C.
For a large cube, the number of small cubes used on each of the four space diagonals is the same, so the total number of small cubes used on the four space diagonals should be a multiple of 4. However, according to the given condition, the four space diagonals actually use 41 small cubes, which is an odd number. ... | 1290 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,330 |
6. The students of Class 2, Grade 8 participated in a community public welfare activity - collecting used batteries. Among them, the students in Group A collected an average of 17 each, the students in Group B collected an average of 20 each, and the students in Group C collected an average of 21 each. If these three g... | 6.A.
Let the number of students in groups A, B, and C be $x$, $y$, and $z$ respectively. From the problem, we have
$$
17 x+20 y+21 z=233 \text {. }
$$
Since $233=17 x+20 y+21 z>17 x+17 y+17 z$, then $x+y+z<\frac{233}{17}=13 \frac{12}{17}$, and $x+y+z>\frac{233}{21}=11 \frac{2}{21}$.
Thus, $11 \frac{2}{21}<x+y+z<13 \f... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,331 |
7. If the graph of the inverse proportion function $y=\frac{k}{x}$ intersects the graph of the linear function $y=a x+b$ at points $A(-2, m)$ and $B(5, n)$, then the value of $3 a+b$ is
保留了源文本的换行和格式。 | Ni.7.0.
Since points $A(-2, m)$ and $B(5, n)$ lie on the graph of an inverse proportion function, we have
$$
\left\{\begin{array}{l}
m=-\frac{k}{2}, \\
n=\frac{k}{5} .
\end{array}\right.
$$
Since points $A(-2, m)$ and $B(5, n)$ also lie on the line $y=a x+b$, we get
$$
\left\{\begin{array}{l}
m=-2 a+b \\
n=5 a+b
\end... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,332 |
8. Given real numbers $a, b, c$ satisfy
$$
a-b+c=7, a b+b c+b+c^{2}+16=0 \text {. }
$$
Then the value of $\frac{b}{a}$ is $\qquad$ . | 8. $-\frac{4}{3}$.
From $a-b+c=7$, we get $a=b-c+7$. Substituting into $a b+b c+b+c^{2}+16=0$, we get $(b-c+7) b+b c+b+c^{2}+16=0$. Simplifying, we get $(b+4)^{2}+c^{2}=0$. Therefore, $b=-4, c=0, a=3$. Thus, $\frac{b}{a}=-\frac{4}{3}$. | -\frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,333 |
4 Stewart's Theorem
Theorem Let $D$ be any point on the side $BC$ of $\triangle ABC$. Then
$$
A D^{2}=\frac{b^{2} B D+c^{2} D C}{a}-B D \cdot D C \text {. }
$$ | Proof: As shown in Figure 4, draw \( AE \perp BC \) with \( E \) as the foot of the perpendicular. Let \( DE = x \), \( BD = u \), \( DC = v \), and \( AD = t \). Then,
\[
\begin{array}{l}
AE^{2} = b^{2} - (v - x)^{2} \\
= c^{2} - (u + x)^{2} = t^{2} - x^{2}.
\end{array}
\]
Thus, \( t^{2} = b^{2} - v^{2} + 2ux \), \( ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,334 |
9. As shown in Figure 3, in $\triangle A B C$, $A D$ intersects side $B C$ at point $D, \angle B=45^{\circ}, \angle A D C=60^{\circ}, D C=2 B D$. Then $\angle C$ equals $\qquad$ | $9.75^{\circ}$.
As shown in Figure 7, draw $C E \perp A D$ at point $E$, and connect $B E$.
Since $\angle A D C=60^{\circ}$, therefore, $\angle E C D=30^{\circ}$, and
$$
D C=2 D E .
$$
Also, $D C=2 B D$, so
$$
\begin{aligned}
D E & =B D, \\
\angle D B E & =\angle D E B=30^{\circ} .
\end{aligned}
$$
Thus, $\angle D B... | 75^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,335 |
10. As shown in Figure 4, given that the radii of $\odot O_{1}$ and $\odot O_{2}$ are $r_{1}$ and $r_{2}$ respectively, $\odot O_{2}$ passes through point $O_{1}$, and the two circles intersect at points $A$ and $B$. Point $C$ is on $\odot O_{2}$, and line segment $A C$ intersects $\odot O_{1}$ at point $D$. Then, line... | 10. (1) (2) (4).
As shown in Figure 8, extend $C B$ to intersect $\odot O_{1}$ at point $E$, and connect $A E$.
Since $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, we have
$$
\angle C = \angle A O_{2} O_{1}.
$$
Similarly, $\angle E = \angle A O_{1} O_{2}$.
Since $\angle B D C$ is the exterior angl... | (1) (2) (4) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,336 |
11. Let $n$ be a positive integer, and $n^{2}+1085$ is a positive integer power of 3. Then the value of $n$ is $\qquad$ | 11.74.
Generally, the unit digit of $n^{2}$ (where $n$ is a positive integer) can only be 0, 1, 4, 5, 6, 9, so the unit digit of $n^{2}+1085$ can only be 5, 6, 9, 0, 1, 4; while the unit digit of $3^{m}$ (where $m$ is a positive integer) can only be 1, 3, 7, 9.
From the given, let $n^{2}+1085=3^{m}$ (where $n, m$ are... | 74 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,337 |
12. There are four distinct positive integers. Any two of them are taken to form a pair, and in the same pair, the larger number is subtracted by the smaller number, then the differences of each pair are added together, and the sum is exactly 18. If the product of these four numbers is 23100, find these four numbers. | Three, 12. Let's assume four different positive integers are $a_{1}, a_{2}, a_{3},$ $a_{4}\left(a_{1}>a_{2}>a_{3}>a_{4}\right)$.
According to the problem, any two numbers chosen from the four form a group, making a total of 6 groups. Thus,
$$
\begin{array}{l}
\left(a_{1}-a_{2}\right)+\left(a_{1}-a_{3}\right)+\left(a_{... | 10,11,14,15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,338 |
13. In the figure, $BC$ is the diameter of the semicircle $\odot O$, $D$ is the midpoint of $\overparen{AC}$, and the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $E$.
(1) Prove: $AC \cdot BC = 2 BD \cdot CD$;
(2) If $AE = 3$, $CD = 2\sqrt{5}$, find the lengths of chord $AB$ and diameter $BC$. | 13. (1) Proof 1: As shown in Figure 9, extend $BA$ and $CD$ to intersect at point $G$. Since $BC$ is the diameter of the semicircle $\odot O$, we have
$$
\angle BAC = \angle BDC = 90^{\circ}.
$$
Thus, $\triangle CAG$ and $\triangle BDC$ are both right triangles.
Since $D$ is the midpoint of $\overparen{AC}$, then $\o... | AB = 6, BC = 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,339 |
14. Given that $a$, $b$, and $c$ are positive integers, and the graph of the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from points $A$ and $B$ to the origin $O$ are both less than 1, find the minimum value of $a+b+c$. | 14. Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, where $x_{1}$ and $x_{2}$ are the roots of the equation $a x^{2}+b x+c=0$.
Given that $a$, $b$, and $c$ are positive integers, we have
$x_{1}+x_{2}=-\frac{b}{a} < 0$.
Thus, the equation $a x^{2}+b x+c=0$ has two negative real roots, i.e., $x_{1} < 0$.
Solvi... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,340 |
1. (i) (Grade 10) Let the sets be
$$
A=\left\{a^{2}+8 \mid a \in \mathbf{N}\right\}, B=\left\{b^{2}+29 \mid b \in \mathbf{N}\right\} \text {. }
$$
If $A \cap B=P$, then $P$ contains $(\quad)$ elements.
(A) 0
(B) 1
(C) 2
(D) at least 3
(ii) (Grade 11) Three non-coincident planes can divide space into $n$ parts. Then al... | -1. (i) C.
From $a^{2}+8=b^{2}+29$, we get $a^{2}-b^{2}=21$, which is $(a-b)(a+b)=21$.
By $\left\{\begin{array}{l}a-b=1, \\ a+b=21,\end{array}\right.$ we solve to get $\left\{\begin{array}{l}a=11, \\ b=10 .\end{array}\right.$
By $\left\{\begin{array}{l}a-b=3, \\ a+b=7,\end{array}\right.$ we solve to get $\left\{\begin{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,341 |
2.(i) (Grade 11) If the lengths of the three altitudes of a triangle are $12, 15, 20$, then the shape of the triangle is ( ).
(A) Acute triangle
(B) Right triangle
(C) Obtuse triangle
(D) Shape uncertain
(ii) (Grade 12) The vertex of the parabola is at the origin, the axis of symmetry is the $x$-axis, and the focus is ... | 2. (i) B.
The ratio of the three sides of the triangle is $\frac{1}{12}: \frac{1}{15}: \frac{1}{20}=5: 4: 3$. Therefore, it is a right triangle.
(ii)D.
The focus of the parabola is the intersection point of the line $3 x-4 y=12$ with the $x$-axis, which is $(4,0)$. Suppose the equation of the parabola is $y^{2}=2 p x$... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,342 |
3. (i) (Grade 11) Let $f(x)=\frac{1+x}{1-x}$, and denote $f_{1}(x)=f(x)$. If $f_{n+1}(x)=f\left(f_{n}(x)\right)$, then $f_{2006}(x)=$ ( ).
(A) $x$
(B) $-\frac{1}{x}$
(C) $\frac{1+x}{1-x}$
(D) $\frac{x-1}{x+1}$
(ii) (Grade 12) The base $ABCD$ of the pyramid $P-ABCD$ is a unit square $(A, B, C, D$ are arranged counterclo... | 3. (i) B.
It is easy to find that $f_{4 n+1}(x)=\frac{1+x}{1-x}, f_{4 n+2}(x)=-\frac{1}{x}$,
$$
f_{4 n+3}(x)=\frac{x-1}{x+1}, f_{4 n+4}(x)=x \text {. }
$$
Therefore, $f_{2000}(x)=-\frac{1}{x}$.
(ii) C.
$$
\begin{array}{l}
P A=\sqrt{1^{2}+(\sqrt{3})^{2}}=2, \\
P D=\sqrt{B D^{2}+P B^{2}}=\sqrt{2+3}=\sqrt{5} .
\end{arra... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,343 |
4. If $a=\sin \theta+\tan \theta, b=\cos \theta+\cot \theta$, then which of the following expressions is incorrect? ( ).
(A) $\sin \theta=\frac{a b-1}{b+1}$
(B) $\cos \theta=\frac{1-a b}{a+1}$
(C) $\tan \theta+\cot \theta=\frac{(a+b+1)^{2}+1-2 a b}{(a+1)(b+1)}$
(D) $\tan \theta-\cot \theta=\frac{(a-b)(a+b+2)}{(a+1)(b+1... | 4.B.
From the given, we have
$$
\begin{array}{l}
\sin \theta \cdot \cos \theta = a \cos \theta - \sin \theta, \\
\sin \theta \cdot \cos \theta = b \sin \theta - \cos \theta,
\end{array}
$$
Thus, \( a \cos \theta - \sin \theta = b \sin \theta - \cos \theta \).
Therefore, \((b+1) \sin \theta = (a+1) \cos \theta\), whic... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,344 |
5 Steiner Theorem
Theorem If $P$ is any point inside $\triangle A B C$, draw $P D \perp B C$, intersecting $B C$ at point $D$, draw $P E \perp C A$ at point $E$, and draw $P F \perp A B$ at point $F$. Then
$$
A F^{2}+B D^{2}+C E^{2}=A E^{2}+C D^{2}+B F^{2} .
$$ | Proof: $A F^{2}+B D^{2}+C E^{2}$
$$
\begin{array}{l}
=P A^{2}-P F^{2}+P B^{2}-P D^{2}+P C^{2}-P E^{2} \\
=P A^{2}-P E^{2}+P C^{2}-P D^{2}+P B^{2}-P F^{2} \\
=A E^{2}+C D^{2}+B F^{2}
\end{array}
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,345 |
$5.1003^{2000}$ 's last digit is ( ).
(A) 1
(B) 3
(C) 7
(D) 9 | 5.D.
It is easy to know that the last digit of $1003^{n}$ is the same as that of $3^{n}$. The last digits of $3^{4 k}, 3^{4 k+1}$, $3^{4 k+2}, 3^{4 k+3}$ are $1,3,9,7$, and since 2006 is a number of the form $4 n+2$, the solution follows. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,346 |
6. Let $a, b, c \in \mathbf{R}_{+}$, and $a b + b c + c a = 108$. Then the minimum value of $\frac{a b}{c} + \frac{b c}{a} + \frac{c a}{b}$ is ( ).
(A) 6
(B) 12
(C) 18
(D) 36 | 6.C.
$$
\begin{array}{l}
S=\frac{(a b)^{2}+(b c)^{2}+(c a)^{2}}{a b c} \\
\geqslant \frac{a b \cdot b c+b c \cdot c a+c a \cdot a b}{a b c} \\
=a+b+c=\sqrt{(a+b+c)^{2}} \\
\geqslant \sqrt{3(a b+b c+c a)}=\sqrt{324}=18 .
\end{array}
$$ | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 718,347 |
7. (i) (Grade 11) Let $M=\{1,2, \cdots, 100\}, A$ be a subset of $M$, and $A$ contains at least one cube number. Then the number of such subsets $A$ is $\qquad$
(ii) (Grade 12) Two players, A and B, are playing a table tennis singles final, using a best-of-five format (i.e., the first to win three games wins the champi... | (ii) $2^{100}-2^{96}$ (i.e., $\left.15 \times 2^{96}\right)$.
$M$ contains 4 cube numbers, which are $1^{3}, 2^{3}, 3^{3}, 4^{3}$. The total number of subsets of $M$ is $2^{100}$, among which, the number of subsets that do not contain cube numbers is $2^{96}$. Therefore, the number of subsets that contain cube numbers ... | 2^{100}-2^{96} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,348 |
8. (i) (Grade 11) In an isosceles right triangle, the right-angle vertex $A(1,0)$, and the centroid $G(2,0)$. Then the coordinates of the other two vertices $B$ and $C$ are $\qquad$
(ii) (Grade 12) The area of the orthogonal projection of a regular tetrahedron with edge length 1 on a horizontal plane is $S$. Then the m... | 8. (i) $\left(\frac{5}{2}, \frac{3}{2}\right)$ and $\left(\frac{5}{2},-\frac{3}{2}\right)$.
Let the midpoint of the hypotenuse be $D$.
From $|A G|=2|G D|$, we get $D\left(\frac{5}{2}, 0\right)$.
Let the other two vertices be $B$ and $C$, then $B D$ is perpendicular and equal to $A D$.
Thus, we get $B\left(\frac{5}{2}, ... | \left(\frac{5}{2}, \frac{3}{2}\right) \text{ and } \left(\frac{5}{2}, -\frac{3}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,349 |
9. (i) (Grade 11) Given $\sin \theta+\cos \theta=\frac{\sqrt{2}}{5}$ $\left(\frac{\pi}{2}<\theta<\pi\right)$. Then $\tan \theta-\cot \theta=$ $\qquad$ .
(ii) (Grade 12) The function $f(x)=\frac{1}{\sin ^{2} x}+\frac{2}{\cos ^{2} x}$ $\left(0<x<\frac{\pi}{2}\right)$ has a minimum value of $\qquad$ . | 9. (i) $-\frac{8 \sqrt{6}}{23}$.
Squaring the condition $\sin \theta + \cos \theta = \frac{\sqrt{2}}{5}$, we get
$$
1 + 2 \sin \theta \cdot \cos \theta = \frac{2}{25}.
$$
Therefore, $\sin \theta \cdot \cos \theta = -\frac{23}{50}$,
$$
(\sin \theta - \cos \theta)^2 = 1 - 2 \sin \theta \cdot \cos \theta = \frac{48}{25}... | 3 + 2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,350 |
10. If the curve $y=\left|x^{2}-3\right|$ intersects the line $y=2 x+$ $k$ at exactly three points, then the value of $k$ is $\qquad$ | $10.4,2 \sqrt{3}$.
(1) If the line $y=2 x+k$ is tangent to the curve $y=-\left(x^{2}-3\right)$ at point $P$ (figure omitted), then $2 x+k=3-x^{2}$, which means $x^{2}+2 x+k-3=0$ has a discriminant of 0. Thus,
$$
2^{2}=4(k-3) \text {. }
$$
Solving for $k$ gives $k=4$.
(2) If the line $y=2 x+k$ passes through $Q(-\sqrt{... | 4, 2\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,351 |
11. The terms of the sequence $\left\{a_{n}\right\}$ are positive, and the sum of its first $n$ terms $S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right)$. Then $a_{n}=$ $\qquad$ | 11. $\sqrt{n}-\sqrt{n-1}$.
From $a_{1}=S_{1}=\frac{1}{2}\left(a_{1}+\frac{1}{a_{1}}\right)$, we get $a_{1}=\frac{1}{a_{1}}$.
Given $a_{1}>0$, we have $a_{1}=1, S_{1}=1$.
When $n>1$, from
$$
S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right) \text {, }
$$
we know $S_{n-1}+a_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\... | \sqrt{n}-\sqrt{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,352 |
12. At a party, 9 celebrities performed $n$ "trio dance" programs. If in these programs, any two people have collaborated exactly once, then $n=$ $\qquad$ | 12.12.
Consider 9 people as 9 points. If two people have performed in the same group, then a line segment is drawn between the corresponding two points. Thus, every pair of points is connected, resulting in a total of 36 line segments. Each trio dance corresponds to a triangle, which has three sides. When all sides ap... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,353 |
13. (i) (Grade 11) In the arithmetic sequence $\left\{a_{n}\right\}: a_{n}=4 n -1\left(n \in \mathbf{N}_{+}\right)$, after deleting all numbers that can be divided by 3 or 5, the remaining numbers are arranged in ascending order to form a sequence $\left\{b_{n}\right\}$. Find the value of $b_{2006}$.
(ii) (Grade 12) Gi... | (i) Since $a_{n+15}-a_{n}=60$, $a_{n}$ is a multiple of 3 or 5 if and only if $a_{n+15}$ is a multiple of 3 or 5.
Now, divide the positive direction of the number line into a series of intervals of length 60:
$$
(0,+\infty)=(0,60] \cup(60,120] \cup(120,180] \cup \cdots \text {. }
$$
Note that the first interval contai... | 15043 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,354 |
14. As shown in Figure 1, quadrilateral $ABCD$ is inscribed in a circle, $P$ is the midpoint of $AB$, $PE \perp AD$, $PF \perp BC$, $PG \perp CD$, and $M$ is the intersection of line segments $PG$ and $EF$. Prove that:
$$
ME = MF.
$$ | 14. As shown in Figure 2, draw $A F_{1}$
$$
\perp B C, B E_{1} \perp A D\left(E_{1}\right. \text {, }
$$
$F_{1}$ are the feet of the perpendiculars). Then
$$
\begin{array}{l}
P E_{1}=\frac{1}{2} A B \\
=P F_{1} .
\end{array}
$$
Let $P G \cap E_{1} F_{1}=K$.
Since $A, B, F_{1}, E_{1}$
are concyclic, then
$$
\angle C F_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,355 |
Example 4 Find a point $P$ inside an acute $\triangle ABC$ such that $A F^{2}$ $+B D^{2}+C E^{2}$ is minimized, where points $D$, $E$, and $F$ are the projections of point $P$ onto $BC$, $CA$, and $AB$, respectively.
(28th IMO Preliminary Problem) | Solution: Let the three sides of $\triangle A B C$ be $a, b, c$, $B L=x, C M=y, A N=z$. By Steiner's theorem, we have
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2} \\
=\frac{1}{2}\left[x^{2}+(a-x)^{2}+y^{2}+(b-y)^{2}+z^{2}+(c-z)^{2}\right] \\
\geqslant \frac{1}{4}\left(a^{2}+b^{2}+c^{2}\right) .
\end{array}
$$
Equality holds ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,356 |
15. If positive integers $m, n, k$ satisfy $m n=k^{2}+1$, prove: there exist $a, b, c, d \in \mathbf{N}$, such that
$$
m=a^{2}+b^{2}, n=c^{2}+d^{2}, k=a c+b d
$$
hold simultaneously. | 15. Let $m \leqslant n$. We use mathematical induction on $k$.
When $k=1$, since $m n=2$, then $m=1, n=2$. At this time, we have $m=0^{2}+1^{2}, n=1^{2}+1^{2}, k=0 \times 1+1 \times 1$, so the conclusion holds.
Assume that when $k=0$, hence $s-t>0, t<r$.
By the induction hypothesis, there exist $a_{1}, b_{1}, c_{1}, d... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,357 |
1. Let $n(n \geqslant 2)$ be a given positive integer, and $a_{1}, a_{2}$, $\cdots, a_{n} \in(0,1)$. Find the maximum value of $\sum_{i=1}^{n} \sqrt[6]{a_{i}\left(1-a_{i+1}\right)}$, where $a_{n+1}=a_{1}$.
(Provided by Hua-Wei Zhu) | 1. By the arithmetic-geometric mean inequality, we have
$$
\begin{array}{l}
\sqrt[6]{a_{i}\left(1-a_{i+1}\right)} \\
=2^{\frac{4}{6}} \sqrt[6]{a_{i}\left(1-a_{i+1}\right) \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}} \\
\leqslant 2^{\frac{2}{3}} \times \frac{1}{6}\left(a_{i}+1-a_{i+1}+2\r... | \frac{\sqrt[3]{4} n}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 718,358 |
2. Find the smallest positive real number $k$ such that for any 4 distinct real numbers $a, b, c, d$ not less than $k$, there exists a permutation $p, q, r, s$ of $a, b, c, d$ such that the equation $\left(x^{2}+p x+q\right)\left(x^{2}+r x+s\right)=0$ has 4 distinct real roots. (Feng Zhigang) | 2. On one hand, if $k4(d-a)>0, c^{2}-4 b>4(c-b)$ $>0$, therefore, equations (1) and (2) both have two distinct real roots.
Secondly, if equations (1) and (2) have a common real root $\beta$, then
$$
\left\{\begin{array}{l}
\beta^{2}+d \beta+a=0, \\
\beta^{2}+\phi \beta+b=0 .
\end{array}\right.
$$
Subtracting the two e... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,359 |
3. As shown in Figure 1, in $\triangle P B C$, $\angle P B C=60^{\circ}$. A tangent line to the circumcircle $\odot O$ of $\triangle P B C$ is drawn through point $P$, intersecting the extension of $C B$ at point $A$. Points $D$ and $E$ lie on line segment $P A$ and $\odot O$, respectively, such that $\angle D B E=90^{... | 3. (1) When $B F$ bisects $\angle P B C$, since $\angle D B E=90^{\circ}$, therefore, $B D$ bisects $\angle P B A$. Thus,
$$
\frac{P F}{F C} \cdot \frac{C B}{B A} \cdot \frac{A D}{D P}=\frac{P B}{B C} \cdot \frac{B C}{B A} \cdot \frac{A B}{P B}=1 .
$$
By the converse of Ceva's Theorem, $A F$, $B P$, and $C D$ are conc... | \tan \alpha=\frac{6+\sqrt{3}}{11} | Geometry | proof | Yes | Yes | cn_contest | false | 718,360 |
4. Let $a$ be a positive integer that is not a perfect square. Prove that for every positive integer $n$,
$$
S_{n}=\{\sqrt{a}\}+\{\sqrt{a}\}^{2}+\cdots+\{\sqrt{a}\}^{n}
$$
is always an irrational number. Here $\{x\}=x-[x]$, where $[x]$ denotes the greatest integer not exceeding $x$.
(Supplied by Tao Pingsheng) | 4. Let $c^{2}0, y_{2 k}0$.
Also, since $y_{2}^{2}-y_{1}^{2}>0$, we have $\left|y_{2 k-1}\right|0, \\
y_{2 k+1}+y_{2 k}=-(2 c-1) y_{2 k}+\left(a-c^{2}\right) y_{2 k-1}>0,
\end{array}
$$
Multiplying these, we get $y_{2 k+1}^{2}-y_{2 k}^{2}>0$, i.e., $\left|y_{2 k}\right|0 \text {. }
$$
Therefore,
$$
\begin{array}{l}
T... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,361 |
5. Let $S=\{n \mid n-1, n, n+1$ can all be expressed as the sum of two positive integers squared $\}$. Prove: If $n \in S$, then $n^{2} \in S$
(Wang Jianwei, problem contributor) | 5. Note that, if $x, y$ are integers, then by parity analysis we know
$$
x^{2}+y^{2} \equiv 0,1,2(\bmod 4) \text {. }
$$
If $n \in S$, then from the above, $n=1(\bmod 4)$. Therefore, we can set
$$
\begin{array}{l}
n-1=a^{2}+b^{2}(a \geqslant b), \\
n=c^{2}+d^{2}(c>d), \\
n+1=e^{2}+f^{2}(e \geqslant f),
\end{array}
$$
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,362 |
6. As shown in Figure $2, A B$ is the diameter of $\odot O$, $C$ is a point on the extension of $A B$, a secant line is drawn through point $C$, intersecting $\odot O$ at points $D$ and $E$. $O F$ is the diameter of the circumcircle $\odot O_{1}$ of $\triangle B O D$, and $C F$ is extended to intersect $\odot O_{1}$ at... | 6. As shown in Figure 3, connect $A D$, $D G$, $G A$, $G O$, $E A$, and $E O$.
Since $O F$ is the diameter of the circumcircle of isosceles $\triangle D O B$, $O F$ bisects $\angle D O B$, i.e., $\angle D O B=2 \angle D O F$.
Also, $\angle D A B=\frac{1}{2} \angle D O B$, so,
$\angle D A B=\angle D O F$.
Since $\angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,363 |
7. Let $k$ be a positive integer no less than 3, and $\theta$ be a real number. Prove: If $\cos (k-1) \theta$ and $\cos k \theta$ are both rational numbers, then there exists a positive integer $n (n > k)$, such that $\cos (n-1) \theta$ and $\cos n \theta$ are both rational numbers.
(Li Weiguo) | 7. First, prove the following conclusion:
Let $\alpha$ be a real number. If $\cos \alpha$ is a rational number, then for any positive integer $m, \cos m \alpha$ is a rational number.
Use mathematical induction on $m$.
Since $\cos \alpha$ is a rational number, we have $\cos 2 \alpha=2 \cos ^{2} \alpha-1$ is also a rati... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,364 |
8. Given a positive integer $n(n \geqslant 2)$. Find the minimum value of $|X|$, such that for any $n$ two-element subsets $B_{1}$, $B_{2}, \cdots, B_{n}$ of the set $X$, there exists a subset $Y$ of $X$ satisfying:
(1) $|Y|=n$;
(2) for $i=1,2, \cdots, n$, we have $\left|Y \cap B_{i}\right| \leqslant 1$.
Here, $|A|$ de... | 8. $|X|_{\min }=2 n-1$.
(1) When $|X|=2 n-2$, it is not necessarily true that there exists a $Y$ that satisfies the conditions.
In fact, let $X=\{1,2, \cdots, 2 n-2\}$, and consider a partition of $X$:
$$
\begin{array}{l}
B_{1}=\{1,2\}, B_{2}=\{3,4\}, \cdots \cdots \\
B_{n-1}=\{2 n-3,2 n-2\} .
\end{array}
$$
Since $|... | 2n-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,365 |
1. Given that $a$, $b$, $c$, $p$, and $q$ are positive integers, $p$ and $q$ are the remainders when $a$ and $b$ are divided by $c$, respectively. If $a > b$, and $a + b = 2(p + q)$, then $a + b + c$ must have a factor of ( ).
(A) 2
(B) 3
(C) 5
(D) Cannot be determined | -、1.B.
Let $a=m c+p, b=n c+q$.
Since $a+b=2(p+q)$, we have
$$
\begin{array}{l}
(m c+p)+(n c+q)=2(p+q), \\
(m+n) c=p+q<2 c,
\end{array}
$$
Therefore, $m+n<2$.
Since $a, b, c, p, q$ are positive integers, $m, n \in \mathbf{N}$. When $m+n=0$, $m=n=0$, then $a+b=p+q$. Also, $a+b=2(p+q)$, so $a=b=0$, which is a contradicti... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,366 |
Example 5 As shown in Figure 5, in $\triangle A B C$, $O$ is the circumcenter, the three altitudes $A D$, $B E$, and $C F$ intersect at point $H$, line $E D$ intersects $A B$ at point $M$, and $F D$ intersects $A C$ at point $N$. Prove:
(1) $O B \perp F D$, $O C \perp E D ;$
(2) $O H \perp M N$.
保留源文本的换行和格式如下：
Exampl... | Proof: (1) Since points $A, C, D, F$ are concyclic, then $\angle B D F=\angle B A C$.
Also, $\angle O B C=\frac{1}{2}\left(180^{\circ}-\angle B O C\right)$
$$
=90^{\circ}-\angle B A C \text {, }
$$
Therefore, $O B \perp D F$.
Similarly, $O C \perp D E$.
(2) From $C F \perp M A, B E \perp N A, D A \perp B C, O B$ $\per... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,367 |
2. Given that $\alpha$ is a solution to the equation $x^{3}+3 x-1=0$. Then the line $y=\alpha x+1-\alpha$ definitely does not pass through ( ).
(A) the first quadrant
(B) the second quadrant
(C) the third quadrant
(D) the fourth quadrant | 2. D.
When $x \leqslant 0$, $x^{3}+3 x-10$. From $x^{3}+3 x-1=0$, we get $\alpha^{2}=\frac{1}{\alpha}-3$.
Therefore, $\frac{1}{\alpha}-3>0$, which means $\alpha<\frac{1}{3}$. Thus, $0<\alpha<\frac{1}{3}$.
Hence, the line $y=\alpha x+1-\alpha$ definitely does not pass through the fourth quadrant. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,368 |
3. Given points $E$ and $F$ are on the sides $BC$ and $CD$ of square $ABCD$ respectively, $\angle EAF=45^{\circ}$, and $S_{\text {square } ABCD}$ : $S_{\triangle AEF}=5: 2$. Then $AB: EF=(\quad$.
(A) $5: 2$
(B) $25: 4$
(C) $\sqrt{5}: 2$
(D) $5: 4$ | 3. D.
As shown in Figure 4, if $\triangle A D F$ is rotated $90^{\circ}$ clockwise around point $A$ to get $\triangle A B G$, then points $G$, $B$, and $C$ are collinear, and
$$
\begin{array}{l}
A G=A F, \\
\angle G A B=\angle F A D, \\
\angle G A F=90^{\circ} .
\end{array}
$$
Therefore, $\angle G A E=\angle G A F-\a... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,369 |
4. From the 20 positive integers $1, 2, \cdots, 20$, each time take 3 numbers to form an ordered triplet $(a, b, c)$, such that $b$ is the mean proportional between $a$ and $c$. Then the number of different ordered triplets $(a, b, c)$ is $(\quad)$.
(A) 22
(B) 20
(C) 16
(D) 11 | 4.A.
Since $2^{2}=1 \times 4,3^{2}=1 \times 9,4^{2}=1 \times 16=2 \times 8$,
$$
\begin{array}{l}
6^{2}=2 \times 18=3 \times 12=4 \times 9,8^{2}=4 \times 16, \\
10^{2}=5 \times 20,12^{2}=9 \times 16=8 \times 18,
\end{array}
$$
and each equation generates two sets of ordered real number pairs $(a, b, c)$, so there are ... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,370 |
5. Given that $A$ and $B$ are two fixed points on a circle $\odot O$ with radius 5, and $P$ is a moving point on $\odot O$. If $AB=6$, let the maximum value of $PA+PB$ be $s$, and the minimum value be $t$, then the value of $s+t$ is $(\quad)$.
(A) $6 \sqrt{10}$
(B) $8 \sqrt{10}$
(C) $6 \sqrt{10}+6$
(D) $6 \sqrt{10}+3$ | 5.C.
As shown in Figure 5, let the diameter $CD$ of $\odot O$ be perpendicular to the chord $AB$ at point $M$, and let $P$ be any point on $\odot O$ different from point $C$. Draw $CE \perp PB$ at point $E$ and $CF \perp PA$ at point $F$. Then,
$$
\begin{array}{l}
OM=4, CM=9, \\
CA=CB=\sqrt{3^{2}+9^{2}} \\
=3 \sqrt{10... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,371 |
6. As shown in Figure 1, in the Cartesian coordinate system $x O y$, $l_{A B}: y=k x+1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Through $B$, $B C \perp A B$ is drawn, intersecting the $x$-axis at point $C$. Through $C$, $C D \perp B C$ is drawn, intersecting the $y$-axis at point $D$. ... | 6. A.
From the problem, we know $\triangle A B C \backsim \triangle C D E$. Therefore,
$$
\frac{C D}{A B}=\frac{C E}{A C}=2 \text {, }
$$
which means $C D=2 A B$.
Also, $\triangle A O B \backsim \triangle C O D$, so,
$$
\frac{A O}{C O}=\frac{A B}{C D}=\frac{1}{2}, C O=2 A O \text {. }
$$
Given $O B=1$, and $O B^{2}=... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,372 |
1. Arrange the numbers $1,2, \cdots, 13$ in a row $a_{1}, a_{2}$, $\cdots, a_{13}$, where $a_{1}=13, a_{2}=1$, and ensure that $a_{1}+a_{2}+$ $\cdots+a_{k}$ is divisible by $a_{k+1}(k=1,2, \cdots, 12)$. Then the value of $a_{4}$ $+a_{5}+\cdots+a_{12}$ is $\qquad$ . | II. 1.68.
Since $a_{1}+a_{2}+\cdots+a_{12}$ is divisible by $a_{13}$, then $a_{1}+a_{2}+\cdots+a_{12}+a_{13}$ is also divisible by $a_{13}$, meaning $a_{13}$ is a factor of $a_{1}+a_{2}+\cdots+a_{12}+a_{13}=13 \times 7$. Given that $a_{1}=13$, $a_{2}=1$, we have $a_{13}=7$.
Also, since $a_{1}+a_{2}$ is divisible by $a... | 68 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,373 |
2. As shown in Figure 2, in rhombus $A B C D$, $\angle A B C = 120^{\circ}, A B = \sqrt{3}$, $E$ is a point on the extension of $B C$, $A E$ intersects $C D$ at point $F$, and $B F$ is extended to intersect $D E$ at point $G$. Then the range of the length of segment $B G$ is $\qquad$ | 2. $\sqrt{3}<BG \leqslant 2$.
As shown in Figure 6, extend $BG$ and $AD$ to intersect at point $M$, and connect $BD$. Thus, $\triangle BCD$ is an equilateral triangle, and
$$
\frac{BC}{DM}=\frac{CF}{DF}=\frac{CE}{AD}.
$$
Since $AD=BC=CD$, we have $\frac{BD}{DM}=\frac{CE}{CD}$.
Since $\angle BDM=\angle DCE=120^{\circ}... | \sqrt{3}<BG \leqslant 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,374 |
3. In rectangle $A B C D$, $A B=4, A D=3$, a ray of light is emitted from the midpoint $P_{0}$ of $A B$ to point $P_{1}$ on side $B C$, reflects to point $P_{2}$ on side $C D$, then reflects to point $P_{3}$ on side $D A$, and finally reflects to point $P_{4}$ on side $A B$. If $P_{0} P_{4}=1$, then $\tan \angle B P_{0... | 3. $\frac{6}{7}$ or $\frac{2}{3}$.
When point $P_{4}$ is to the left of point $P_{0}$, by symmetry we get Figure 7.
Since $P_{0} P_{4}=1, P_{0}$ is the midpoint of $A B$, and $A B=4, A D=3$, then $A P_{4}=1, E A=1, E F=6, F G=10, P_{0} H=7$. Therefore, $\tan \angle B P_{0} P_{1}=\frac{6}{7}$.
Similarly, when point $P_... | \frac{6}{7} \text{ or } \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,375 |
4. Let $a$, $b$, $c$ be positive integers, and satisfy
$$
a^{2}+b^{2}+c^{2}-a b-b c-c a=19 \text {. }
$$
Then the minimum value of $a+b+c$ is $\qquad$ | 4. 10 .
By the cyclic symmetry of $a, b, c$, without loss of generality, assume $a \geqslant b \geqslant c$.
Let $a-b=m, b-c=n$, then $a-c=m+n$, and
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[m^{2}+n^{2}+(m+n)^{2}\right],
\end{array}
$$
which is $m^{2}+m n+n^{2}-19=0$.
For the equation in ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,376 |
One, (20 points) Given the equation $a x^{2}+4 x+b=0$ $(a<0)$ has two real roots $x_{1} 、 x_{2}$, and the equation $a x^{2}+3 x+b=0$ has two real roots $\alpha 、 \beta$.
(1) If $a 、 b$ are both negative integers, and $|\alpha-\beta|=1$, find the values of $a 、 b$;
(2) If $\alpha<1<\beta<2, x_{1}<x_{2}$, prove:
$-2<x_{1... | (1) From the problem, we have $\alpha+\beta=-\frac{3}{a}, \alpha \beta=\frac{b}{a}$.
From $|\alpha-\beta|=1 \Rightarrow(\alpha+\beta)^{2}-4 \alpha \beta=1$
$$
\Rightarrow \frac{9}{a^{2}}-\frac{4 b}{a}=1 \Rightarrow a(a+4 b)=9 \text {. }
$$
Since $a$ and $b$ are both negative integers, we have $a=-1, a+4 b=-9$. Therefo... | a=-1, b=-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,377 |
II. (25 points) As shown in Figure 3, in $\triangle ABC$, isosceles $\triangle DAB$ is constructed outward with $AB$ as the base, and isosceles $\triangle FAC$ is constructed inward with $AC$ as the base, such that $\angle DAB = \angle CAF = \alpha$ (a constant). Prove:
(1) The angle formed between line $DF$ and $BC$ i... | As shown in Figure 8, construct an isosceles $\triangle B C E$ outside the shape with $B C$ as the base, such that $\angle C B E=\alpha$, and connect $E F$. Clearly,
$\triangle A D B \backsim \triangle A F C$
$\sim \triangle B E C$.
Therefore, $\frac{A F}{A C}=\frac{A D}{A B}$,
$\angle D A F=\angle D A B+\angle B A F=\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,379 |
Three. (25 points) In a population of 1000 individuals numbered $0,1,2, \cdots, 999$, they are sequentially divided into 10 groups: Group 0 includes numbers $0,1, \cdots, 99$, Group 1 includes numbers $100,101, \cdots, 199$, and so on. Now, one number is to be selected from each group to form a sample of size 10, with ... | (1) When $x=24, a=33$, the values of $x+a k$ corresponding to $k$ taking the values $0,1, \cdots, 9$ are
$$
24,57,90,123,156,189,222,255,288,321 \text {. }
$$
Therefore, the 10 numbers of the sample drawn are
$$
24,157,290,323,456,589,622,755,888,921 \text {. }
$$
(2) Since a number with the last two digits 87 can app... | 585 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,380 |
1. Let $0<x<y<\frac{\pi}{2}$, and satisfy $\tan y=\tan x+\frac{1}{\cos x}$. Then, the value of $y-\frac{x}{2}$ is ( ).
(A) $\frac{\pi}{12}$
(B) $\frac{\pi}{6}$
(C) $\frac{\pi}{4}$
(D) $\frac{\pi}{3}$ | $-1 . C$.
$$
\begin{array}{l}
\tan x + \frac{1}{\cos x} = \frac{1 + \sin x}{\cos x} = \frac{1 + \cos \left(\frac{\pi}{2} - x\right)}{\sin \left(\frac{\pi}{2} - x\right)} \\
= \frac{2 \cos^2 \left(\frac{\pi}{4} - \frac{x}{2}\right)}{2 \sin \left(\frac{\pi}{4} - \frac{x}{2}\right) \cdot \cos \left(\frac{\pi}{4} - \frac{x... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,381 |
2. For the function $f(x)=\lg \frac{1+x}{1-x}$, there are three numbers satisfying $|a|<1,|b|<1,|c|<1$, and
$$
f\left(\frac{a+b}{1+a b}\right)=1, f\left(\frac{b-c}{1-b c}\right)=2 .
$$
Then, the value of $f\left(\frac{a+c}{1+a c}\right)$ is $(\quad)$.
(A) -1
(B) $\lg 2$
(C) $\sqrt{10}$
(D) 3 | 2.A.
Given $|a|<1,|b|<1,|c|<1$, we know
$$
\begin{array}{l}
\left|\frac{a+b}{1+a b}\right|<1,\left|\frac{b-c}{1-b c}\right|<1,\left|\frac{a+c}{1+a c}\right|<1 . \\
\text { and } f\left(\frac{a+b}{1+a b}\right)=\lg \frac{1+\frac{a+b}{1+a b}}{1-\frac{a+b}{1+a b}} \\
=\lg \frac{a b+1+a+b}{a b+1-a-b}=\lg \frac{(a+1)(b+1)}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,382 |
3. The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}=\frac{5 a_{n-1}-2}{a_{n-1}-5}\left(n \geqslant 2, n \in \mathbf{N}_{+}\right),
$$
and the sum of the first 2008 terms is 401. Then the value of $\sum_{i=1}^{2008} a_{i} a_{i+1}$ is ( ).
(A) -4
(B) -6
(C) 6
(D) 4 | 3. B.
Let $a_{1}=m$, then we have
$$
a_{2}=\frac{5 m-2}{m-5}, a_{3}=\frac{5 a_{2}-2}{a_{2}-5}=m=a_{1} .
$$
It is easy to see that the period of $a_{n}$ is 2.
Thus, $a_{2009}=a_{2007}=\cdots=a_{1}$.
Notice that $a_{n+1}=\frac{5 a_{n}-2}{a_{n}-5}$, so,
$$
\begin{array}{l}
a_{n} a_{n+1}=5\left(a_{n}+a_{n+1}\right)-2 . \... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,383 |
4. For the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, the left focus is $F$, and the intersection point of the line $y=\sqrt{3} x$ with the ellipse in the first quadrant is $M$. A tangent line to the ellipse at $M$ intersects the $x$-axis at point $N$. Then the area of $\triangle M N F$ is ( ).
(A) $\sqrt{3}+\frac{\s... | 4.C.
As shown in Figure 4, from the ellipse equation $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, we get $a^{2}=4, b^{2}=3, c^{2}=1$.
Therefore, $F(-1,0)$.
Let $M\left(x_{0}, y_{0}\right)$, it is easy to know that the tangent line equation through point $M$ is
$$
\frac{x_{0} x}{4}+\frac{y_{0} y}{3}=1 .
$$
Let $y=0$, solving ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,384 |
5. Let the function $f(n)=k$, where $n \in \mathbf{N}_{+}, k$ is the $n$-th digit after the decimal point of 0.14951728901, for example, $f(3)=9$. Let $f_{n+1}(x)=f\left(f_{n}(x)\right)$, then the value of $f_{2007}(6)$ is $(\quad)$.
(A) 1
(B) 2
(C) 4
(D) 7 | 5.A.
Notice
$$
\begin{array}{l}
f(6)=7, f_{2}(6)=f(7)=2, f_{3}(6)=f(2)=4, \\
f_{4}(6)=f(4)=5, f_{5}(6)=f(5)=1, \\
f_{6}(6)=f(1)=1, \cdots \cdots f_{2007}(6)=1 .
\end{array}
$$ | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,385 |
6. Given two unequal acute angles $\alpha, \beta$, satisfying
$$
\begin{array}{l}
x \sin \beta + y \cos \alpha = \sin \alpha, \\
x \sin \alpha + y \cos \beta = \sin \beta,
\end{array}
$$
where $\alpha + \beta \neq \frac{\pi}{2}$, and $x, y \in \mathbf{R}$. Then the value of $x^{2} - y^{2}$ is ( ).
(A) -1
(B) 0
(C) 1
(... | 6.C.
From the problem, we get
$$
\begin{array}{l}
x=\frac{\sin \beta \cdot \cos \alpha-\sin \alpha \cdot \cos \beta}{\sin \alpha \cdot \cos \alpha-\sin \beta \cdot \cos \beta} \\
=\frac{2 \sin (\beta-\alpha)}{\sin 2 \alpha-\sin 2 \beta}=\frac{2 \sin (\beta-\alpha)}{2 \cos (\alpha+\beta) \cdot \sin (\alpha-\beta)} \\
=... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,386 |
1. As shown in Figure 1, label the six vertices of a regular hexagon with the numbers $0,1,2,3,4,5$ in sequence. An ant starts from vertex 0 and crawls counterclockwise, moving 1 edge on the first move, 2 edges on the second move, $\cdots \cdots$ and $2^{n-1}$ edges on the $n$-th move. After 2005 moves, the value at th... | Ni.1.1.
According to the problem, after 2005 movements, the ant has crawled over
$$
1+2+\cdots+2^{2004}=2^{2005}-1
$$
edges.
Let $x=2^{2005}-1$. Consider the remainder of $x$ modulo 6.
Since $x \equiv 1(\bmod 2), x \equiv 1(\bmod 3)$, we have
$$
3 x \equiv 3(\bmod 6), 2 x \equiv 2(\bmod 6), 5 x \equiv 5(\bmod 6) \text... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,387 |
2. Given $x \in\left(0, \frac{\pi}{2}\right), \sin ^{2} x, \sin x \cdot \cos x$, and $\cos ^{2} x$ cannot form a triangle, and the length of the interval of $x$ that satisfies the condition is $\arctan k$. Then $k$ equals $\qquad$ (Note: If $x \in(a, b), b>a$, then the length of such an interval is $b-a$). | 2.2.
First, find the range of $x$ that can form a triangle.
(1) If $x=\frac{\pi}{4}$, then $\sin ^{2} \frac{\pi}{4} 、 \cos ^{2} \frac{\pi}{4} 、 \sin \frac{\pi}{4} \cdot \cos \frac{\pi}{4}$ can form a triangle.
(2) If $x \in\left(0, \frac{\pi}{4}\right)$, at this time,
$$
\sin ^{2} x\cos ^{2} x$, that is,
$$
\tan ^{2} ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,388 |
1. If the irreducible proper fraction $\frac{a}{b}$ satisfies $a+b=2007$ $\left(a, b \in \mathbf{N}_{+}\right)$, then it is called an "ideal fraction". Then, the number of ideal fractions is ( ).
(A) 337
(B) 666
(C) 674
(D) 1332 | - 1.B.
Given $a+b=2007$ and $a1$, then $d \mid 2007$.
Since $2007=3^{2} \times 223$, we have $3 \mid d$ or $223 \mid d$.
In the set $M=\{1,2, \cdots, 1003\}$, the numbers divisible by 3 are $\left[\frac{1003}{3}\right]=334$, the numbers divisible by 223 are $\left[\frac{1003}{223}\right]=4$, and the numbers divisible b... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,389 |
2. The two roots of the equation
$$
\frac{x^{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}\right) x+\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=0
$$
are the eccentricities of two conic sections, which are ( ).
(A) ellipse and hyperbola
(B) hyperbola and parabola
(C) parabola and ellipse
(D) elli... | 2.C.
Let $e_{1}$ and $e_{2}$ represent these two eccentricities. Then, from
$$
e_{1} e_{2}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}<1
$$
we know that one of them must be an ellipse. Furthermore, from
$$
\begin{array}{l}
\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}+\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}} \\
... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,390 |
5. For a given positive integer $n$, the number of integer points in the closed region (including the boundary) formed by the line $y=$ $n^{2}$ and the parabola $y=x^{2}$ is $\qquad$ | 5. $\frac{1}{3}(2 n+1)\left(2 n^{2}-n+3\right)$.
As shown in Figure 5, the line $y=n^{2}$ intersects the parabola $y=x^{2}$ at points $A\left(n, n^{2}\right)$ and $B\left(-n, n^{2}\right)$. Let the segment on the line $x=k$ within the region be $C D$, with endpoints $C\left(k, n^{2}\right)$ and $D\left(k, k^{2}\right)... | \frac{1}{3}(2 n+1)\left(2 n^{2}-n+3\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,391 |
11. (16 points) For a convex $n$-sided polygon $(n \geqslant 3)$ with unequal side lengths, color each side using one of three colors: red, yellow, or blue, but no two adjacent sides can have the same color. How many different coloring methods are there? | 11. Let the number of different coloring methods be $p_{n}$. It is easy to know that $p_{3}=6$.
When $n \geqslant 4$, first for edge $a_{1}$, there are 3 different coloring methods. Since the color of edge $a_{2}$ is different from that of edge $a_{1}$, there are 2 different coloring methods for edge $a_{2}$. Similarl... | p_{n}=2^{n}+(-1)^{n} \times 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,392 |
12. (16 points) Let $a, b \in [0,1]$. Find
$$
S=\frac{a}{1+b}+\frac{b}{1+a}+(1-a)(1-b)
$$
the maximum and minimum values. | 12. Since $S=\frac{a}{1+b}+\frac{b}{1+a}+(1-a)(1-b)$ $=\frac{1+a+b+a^{2} b^{2}}{(1+a)(1+b)}=1-\frac{a b(1-a b)}{(1+a)(1+b)} \leqslant 1$,
thus, when $a b=0$ or $a b=1$, the equality holds.
Therefore, the maximum value of $S$ is 1.
Let $T=\frac{a b(1-a b)}{(1+a)(1+b)}, x=\sqrt{a b}$. Then
$$
\begin{array}{l}
T=\frac{a b... | \frac{13-5 \sqrt{5}}{2} \text{ and } 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,393 |
1. Let $[x]$ denote the greatest integer not greater than $x$. Let the set $A=\left\{x \mid x^{2}-[x]=2\right\}, B=\{x|| x \mid<$ $2\}$. Then $A \cap B=(\quad)$.
(A) $(-2,2)$
(B) $[-2,2]$
(C) $\{\sqrt{3},-1\}$
(D) $\{-\sqrt{3}, 1\}$ | - 1.C.
Since $x=0 \notin A$, options $(\mathrm{A})$ and $(\mathrm{B})$ are eliminated.
Also, $x=-1, \sqrt{3}$ satisfy the condition, hence the answer is $(\mathrm{C})$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,394 |
2. If $f(x)=\left(2 x^{5}+2 x^{4}-53 x^{3}-57 x+54\right)^{2006}$, then $f\left(\frac{\sqrt{111}-1}{2}\right)=(\quad)$.
(A) -1
(B) 1
(C) 2005
(D) 2007 | 2. B.
Let $t=\frac{\sqrt{111}-1}{2}$, then $2 t^{2}+2 t-55=0$. Therefore, $f(t)=\left[\left(2 t^{2}+2 t-55\right)\left(t^{3}+t-1\right)-1\right]^{2006}$ $=(-1)^{2006}=1$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,395 |
3. The vertices of a quadrilateral lie on the sides of a square with side length 1. If the sum of the squares of the four sides is $t$, then the range of values for $t$ is ( ).
(A) $[1,2]$
(B) $[2,4]$
(C) $[1,3]$
(D) $[3,6]$ | 3. B.
When the vertices of the quadrilateral coincide with the vertices of the square, we know $t=4$, thus eliminating options (A) and (C); when taking the midpoints of each side, we get $t=2$, thus eliminating option (D). | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,396 |
4. As shown in Figure 1, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $P$ is a point on edge $A B$. A line $l$ is drawn through point $P$ in space such that $l$ makes a $30^{\circ}$ angle with both plane $A B C D$ and plane $A B C_{1} D_{1}$. The number of such lines is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 4. B.
Since the dihedral angle $C_{1}-A B-D$ has a plane angle of $45^{\circ}$, therefore, within this dihedral angle and its "opposite" dihedral angle, there does not exist a line passing through point $P$ that forms a $30^{\circ}$ angle with both plane $A B C D$ and plane $A B C_{1} D_{1}$. Instead, consider its sup... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,397 |
5. In an isosceles right triangle $\triangle ABC$, the hypotenuse $BC=4\sqrt{2}$. An ellipse has $C$ as one of its foci, the other focus lies on the line segment $AB$, and the ellipse passes through points $A$ and $B$. The standard equation of the ellipse (with foci on the $x$-axis) is ( ).
(A) $\frac{x^{2}}{6+4 \sqrt{... | 5.A.
Since $B C=4 \sqrt{2}$, let the other focus of the ellipse be $D$. Establish a rectangular coordinate system with $D C$ as the $x$-axis and the midpoint of $D C$ as the origin. Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. Therefore,
$$
|A D|+|B D|+|A C|+|B C|=4 a,
$$
whi... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,398 |
6. Divide each side of a square into 8 equal parts, and take the division points as vertices (excluding the vertices of the square), the number of different triangles that can be formed is ( ).
(A) 1372
(B) 2024
(C) 3136
(D) 4495 | 6. C.
Solution 1: First, note that the three vertices of the triangle are not on the same side of the square. Choose any three sides of the square so that each of the three vertices lies on one of them, which can be done in 4 ways.
Next, select a point on each of the chosen three sides, which can be done in $7^{3}$ w... | 3136 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,399 |
7. The sum of the first $m$ terms of the arithmetic sequence $\left\{a_{n}\right\}$ is 90, and the sum of the first $2 m$ terms is 360. Then the sum of the first $4 m$ terms is $\qquad$ . | $=\mathbf{7 . 1 4 4 0}$.
Let $S_{k}=a_{1}+a_{2}+\cdots+a_{k}$, it is easy to know that $S_{m} 、 S_{2 m}-S_{m}$ 、 $S_{3 m}-S_{2 m}$ form an arithmetic sequence. Therefore, $S_{3 m}=810$.
It is also easy to know that $S_{2 m}-S_{m} 、 S_{3 m}-S_{2 m} 、 S_{4 m}-S_{3 m}$ form an arithmetic sequence, then $S_{4 m}=3 S_{3 m}... | 1440 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,400 |
8. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
4 y^{3}+\frac{1}{2} \sin 2 y+a=0 .
\end{array}\right.
$$
then the value of $\cos (x+2 y)$ is | 8.1.
Let $f(t)=t^{3}+\sin t$. Then $f(t)$ is monotonically increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
From the original system of equations, we get $f(x)=f(-2 y)=2 a$, and since $x$ and $-2 y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, it follows that $x=-2 y$, hence $x+2 y=0$. Therefore, $\cos (... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,401 |
6. If the six edges of a tetrahedron are $2,3,4,5$, 6,7, then there are $\qquad$ different shapes (if two tetrahedra can be made to coincide by appropriate placement, they are considered the same shape). | 6.10.
Let the edge of length $k$ be denoted as $l_{k}(k \in\{2,3,4,5,6,7\})$. Consider $l_{2}$ and $l_{3}$.
(1) If $l_{2}$ and $l_{3}$ are coplanar, then the other side of the plane must be $l_{4}$.
(i) If $l_{2}$, $l_{3}$, and $l_{4}$ form a triangle in a clockwise direction (all referring to the direction of the thr... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,402 |
9.100 Arrange chairs in a circle, with $n$ people sitting on the chairs, such that when one more person sits down, they will always be sitting next to one of the original $n$ people. The minimum value of $n$ is $\qquad$ . | 9.34.
From the problem, we know that after $n$ people sit down, there are at most two empty chairs between any two people. If we can arrange it so that there are exactly two empty chairs between every two people, then $n$ is minimized. In this case, if we number the chairs where people are sitting, we get an arithmeti... | 34 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,403 |
10. In $\triangle A B C$, $A B=\sqrt{30}, A C=\sqrt{6}, B C$ $=\sqrt{15}$, there is a point $D$ such that $A D$ bisects $B C$ and $\angle A D B$ is a right angle, the ratio $\frac{S_{\triangle A D B}}{S_{\triangle A B C}}$ can be written as $\frac{m}{n}$. ($m, n$ are coprime positive integers). Then $m+n=$ | 10.65.
Let the midpoint of $BC$ be $E$, and $AD=\frac{x}{2}$.
By the median formula, we get $AE=\frac{\sqrt{57}}{2}$.
By the Pythagorean theorem, we get $120-15+57=2 \sqrt{57} x$, solving for $x$ gives $x=\frac{81}{\sqrt{57}}$.
Thus, $\frac{m}{n}=\frac{AD}{2AE}=\frac{x}{2\sqrt{57}}=\frac{27}{38}$.
Therefore, $m+n=27+3... | 65 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,404 |
11. Let $A B C D-A_{1} B_{1} C_{1} D_{1}$ be a cube with edge length 1. Then the minimum distance between a point $P$ on the incircle of the top face $A B C D$ and a point $Q$ on the circle passing through the vertices $A, B, C_{1}, D_{1}$ is $\qquad$. | 11. $\frac{\sqrt{3}-\sqrt{2}}{2}$.
Let point $O$ be the center of the cube. It is easy to see that
$$
O Q=\frac{\sqrt{3}}{2}, O P=\frac{\sqrt{2}}{2} \text {. }
$$
By the triangle inequality, we have $P Q \geqslant O Q-O P=\frac{\sqrt{3}-\sqrt{2}}{2}$, with equality if and only if points $O$, $P$, and $Q$ are collinea... | \frac{\sqrt{3}-\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,405 |
12. The rules of a "level-passing game" stipulate: on the $n$th level, a die must be rolled $n$ times, and if the sum of the points that appear in these $n$ rolls is greater than $2^{n}$, it counts as passing the level. Therefore, the probability of consecutively passing the first 3 levels is $\qquad$ | 12. $\frac{100}{243}$.
Since the dice are uniform cubes, the probability of each number appearing after a throw is equal.
Let event $A_{n}$ be “failure at the $n$-th level”, then the complementary event $B_{n}$ is “success at the $n$-th level”. In the $n$-th game, the total number of basic events is $6^{n}$.
First L... | \frac{100}{243} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,406 |
13. (16 points) Does there exist a smallest positive integer $t$, such that the inequality
$$
(n+t)^{n+t}>(1+n)^{3} n^{n} t^{t}
$$
holds for any positive integer $n$? Prove your conclusion. | $$
\text { Three, 13. Take }(t, n)=(1,1),(2,2),(3,3) \text {. }
$$
It is easy to verify that when $t=1,2,3$, none of them meet the requirement.
When $t=4$, if $n=1$, equation (1) obviously holds.
If $n \geqslant 2$, we have
$$
\begin{array}{l}
4^{4} n^{n}(n+1)^{3}=n^{n-2}(2 n)^{2}(2 n+2)^{3} \times 2^{3} \\
\leqslant\... | 4 | Inequalities | proof | Yes | Yes | cn_contest | false | 718,407 |
14. (18 points) Let $x, y, z$ be positive real numbers. Find the minimum value of the function
$$
f(x, y, z)=\frac{(1+2 x)(3 y+4 x)(4 y+3 z)(2 z+1)}{x y z}
$$ | 14. Given a fixed $y$, we have
$$
\begin{array}{l}
\frac{(1+2 x)(3 y+4 x)}{x}=\frac{8 x^{2}+(6 y+4) x+3 y}{x} \\
=8 x+\frac{3 y}{x}+6 y+4 \\
\geqslant 2 \sqrt{24 y}+6 y+4=(\sqrt{6 y}+2)^{2},
\end{array}
$$
with equality holding if and only if $x=\sqrt{\frac{3 y}{8}}$.
Similarly, $\frac{(4 y+3 z)(2 z+1)}{z}=6 z+\frac{4... | 194+112 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,408 |
15. (22 points) Let $A$ and $B$ be the common left and right vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Let $P$ and $Q$ be moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
... | 15. (1) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}}.
$$
Similarly,
$$
k_{3}+k_{4}=-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}}.
$$
Let $O$ be the... | 8 | Geometry | proof | Yes | Yes | cn_contest | false | 718,409 |
16. (22 points) Arrange $m$ guests of the same gender to stay in $A_{1}, A_{2}, \cdots, A_{n}$, these $n$ rooms as follows: First, arrange 1 guest and $\frac{1}{7}$ of the remaining guests in room $A_{1}$; then, from the remaining guests, arrange 2 guests and $\frac{1}{7}$ of the remaining guests in room $A_{2}$; and s... | 16. Let the number of guests remaining after arranging the $k$-th room $A_{k}$ be $a_{k}$, then $a_{0}=m, a_{n-1}=n$.
Since the number of guests staying in the $k$-th room $A_{k}$ is $k + \frac{a_{k-1}-k}{7}$, we have $a_{k-1}-a_{k}=k+\frac{a_{k-1}-k}{7}$, which simplifies to
$a_{k}=\frac{6}{7}\left(a_{k-1}-k\right)$.... | 36 \text{ guests, 6 rooms, each room accommodates 6 guests} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,410 |
1. The functions $y=f(x)$ and $y=g(x)$ have domains and ranges of $\mathbf{R}$, and both have inverse functions. Then the inverse function of
$$
y=f^{-1}\left(g^{-1}(f(x))\right)
$$
is $(\quad)$.
(A) $y=f\left(g\left(f^{-1}(x)\right)\right)$
(B) $y=f\left(g^{-1}\left(f^{-1}(x)\right)\right)$
(C) $y=f^{-1}(g(f(x)))$
(D... | $-1 . \mathrm{C}$.
From $y=f^{-1}\left(g^{-1}(f(x))\right)$, we sequentially get
$$
\begin{array}{l}
f(y)=g^{-1}(f(x)), g(f(y))=f(x), \\
f^{-1}(g(f(y)))=x .
\end{array}
$$
Interchanging $x$ and $y$ yields $y=f^{-1}(g(f(x)))$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,411 |
2. The set $M$ consists of functions $f(x)$ that satisfy the following condition: when $x_{1}, x_{2} \in[-1,1]$, we have
$$
\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \leqslant 4\left|x_{1}-x_{2}\right| \text {. }
$$
For the two functions
$$
f_{1}(x)=x^{2}-2 x+5, f_{2}(x)=\sqrt{|x|} \text {, }
$$
the follow... | 2.D.
Notice
$$
\begin{array}{l}
\left|f_{1}\left(x_{1}\right)-f_{1}\left(x_{2}\right)\right|=\left|x_{1}^{2}-x_{2}^{2}-2\left(x_{1}-x_{2}\right)\right| \\
=\left|x_{1}-x_{2}\right|\left|x_{1}+x_{2}-2\right| \leqslant 4\left|x_{1}-x_{2}\right|, \\
\left|f_{2}\left(x_{1}\right)-f_{2}\left(x_{2}\right)\right|=\left|\sqrt... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,412 |
Three. (20 points) Determine whether there exist 2007 real numbers $a_{1}, a_{2}, \cdots, a_{2007}$, satisfying:
(1) $\left|a_{i}\right|<1, i=1,2, \cdots, 2007$;
(2) $\sum_{i=1}^{2007}\left|a_{i}\right|-\left|\sum_{i=1}^{2007} a_{i}\right|=2006$. | Three, if there exist 2007 real numbers $a_{1}$, $a_{2}, \cdots, a_{2007}$ that satisfy the given conditions, let $\left|\sum_{i=1}^{2007} a_{i}\right|=t \geqslant 0$.
Removing the absolute value symbol and separating the positive and negative parts, we can write
$$
\left|\sum_{i=1}^{2007} a_{i}\right|=\sum_{i=1}^{k} x... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,413 |
3. Points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the parabola $y=2 x^{2}$ are symmetric with respect to the line $y=x+m$. If $2 x_{1} x_{2}=-1$, then the value of $2 m$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 3. A.
$$
\begin{array}{l}
\text { Given } \frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-1, \\
\frac{y_{1}+y_{2}}{2}=\frac{x_{1}+x_{2}}{2}+m, \\
2 x_{1} x_{2}=-1 \text { and } y_{1}=2 x_{1}^{2}, y_{2}=2 x_{2}^{2},
\end{array}
$$
we get
$$
\begin{array}{l}
x_{2}-x_{1}=y_{1}-y_{2}=2\left(x_{1}^{2}-x_{2}^{2}\right) \Rightarrow x_{1}+x... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,414 |
4. In $\triangle A B C$, $B C=a, A C=b, A B=c$. Then $(b+c-a):(a+c-b):(a+b-c)$ equals ( ).
(A) $\sin \frac{A}{2}: \sin \frac{B}{2}: \sin \frac{C}{2}$
(B) $\cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2}$
(C) $\tan \frac{A}{2}: \tan \frac{B}{2}: \tan \frac{C}{2}$
(D) $\cot \frac{A}{2}: \cot \frac{B}{2}: \cot \frac{... | 4.D.
As shown in Figure 3, it is easy to know
$$
\begin{array}{l}
b+c-a=2 A D \\
=2 r \cot \frac{A}{2}, \\
a+c-b=2 r \cot \frac{B}{2}, \\
a+b-c=2 r \cot \frac{C}{2} .
\end{array}
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,415 |
5. For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the center, right focus, right vertex, and the intersection of the right directrix with the $x$-axis are denoted as $O, F, G, H$ respectively. Then the maximum value of $\frac{|F G|}{|O H|}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{... | 5.C.
$$
\frac{F G}{O H}=\frac{a-c}{\frac{a^{2}}{c}}=\frac{a c-c^{2}}{a^{2}}=e-e^{2}=e(1-e) \leqslant \frac{1}{4},
$$
The equality holds if and only if $e=\frac{1}{2}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,416 |
6. The range of the function $f(x)=\sqrt{x-3}+\sqrt{12-3 x}$ is ( ).
(A) $[1, \sqrt{2}]$
(B) $[1, \sqrt{3}]$
(C) $\left[1, \frac{3}{2}\right]$
(D) $[1,2]$ | 6.D.
The domain of $f(x)$ is $3 \leqslant x \leqslant 4$, so $0 \leqslant x-3 \leqslant 1$.
Let $x-3=\sin ^{2} \theta\left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)$, then
$$
\begin{array}{l}
f(x)=\sqrt{x-3}+\sqrt{3(4-x)} \\
=\sin \theta+\sqrt{3\left(1-\sin ^{2} \theta\right)}=\sin \theta+\sqrt{3} \cos \theta \... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,417 |
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