problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Example 1 Let $A M$ be the median on side $B C$ of $\triangle A B C$, and let any line intersect $A B$, $A C$, and $A M$ at $P$, $Q$, and $N$ respectively. Prove that $\frac{A B}{A P}$, $\frac{A M}{A N}$, and $\frac{A C}{A Q}$ form an arithmetic sequence ${ }^{[1]}$.
(1978, Liaoning Province High School Mathematics Com... | Analysis: As shown in Figure 1, each pair of the line segments $AB, AC, AM$ are two sides of $\triangle ABC$, $\triangle ACM$, and $\triangle ABM$, respectively. Each pair of the line segments $AP, AN, AQ$ are two sides of $\triangle APN$, $\triangle AQN$, and $\triangle APQ$, respectively, and
$$
\begin{array}{l}
S_{\... | \frac{AB}{AP} + \frac{AC}{AQ} = \frac{2AM}{AN} | Geometry | proof | Yes | Yes | cn_contest | false | 719,749 |
Example 2 As shown in Figure 2, given that $\triangle ABC$ is an acute triangle inscribed in $\odot O$, the distances from point $O$ to the three sides $a, b, c$ of $\triangle ABC$ are $h_{a}, h_{b}, h_{c}$, with the feet of the perpendiculars being $H_{a}, H_{b}, H_{c}$. If $R$ is the radius of $\odot O$, prove that $... | Proof: Let $A O, B O, C O$ intersect $\odot O$ at points $M, N, P$ respectively, and connect $A P, B P, B M, C M, C N, A N$.
Since $\angle A B M=90^{\circ}$, we have $M B \perp A B$.
Also, $O H_{c} \perp A B$, thus, $O H_{c} / / M B$.
Noting that $O$ is the midpoint of $A M$, by the median theorem of a triangle, we kno... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,750 |
Example 3 In $\triangle A B C$, $A^{\prime} 、 B^{\prime} 、 C^{\prime}$ are on $B C 、 C A 、 A B$ respectively. Given that $A A^{\prime} 、 B B^{\prime} 、 C C^{\prime}$ concur at $O$, and $\frac{A O}{O A^{\prime}}+\frac{B O}{O B^{\prime}}+\frac{C O}{O C^{\prime}}=92$. Find the value of $\frac{A O}{O A^{\prime}} \cdot \fra... | Let $\angle A O B^{\prime}=\theta_{1}, \angle A O C^{\prime}=\theta_{2}$.
As shown in Figure 3, from $S_{\triangle A O B}+$
$S_{\triangle C O B^{\prime}}=S_{\triangle M O C}$, we get
$$
\begin{array}{l}
\frac{1}{2} \rho_{1} \rho_{2} \sin \theta_{1}+ \\
\frac{1}{2} \rho_{1} \rho_{6} \sin \left[180^{\circ}-\right. \\
\le... | 94 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,751 |
Example 4 Given that the acute triangle $\triangle ABC$ is inscribed in the unit circle $\odot O$, and the distances from the center $O$ to the three sides of $\triangle ABC$ are $h_{a}, h_{b}, h_{c}$. Prove: $h_{a}^{2}+h_{b}^{2}+h_{c}^{2}+2 h_{a} h_{b} h_{c}=1$.
(2000, Chizhou City, Anhui Province High School Mathemat... | Prove: As shown in Figure 4,
extend the radii $A O$, $B O$,
$C O$ to intersect $\odot O$ at
$A_{1}$, $B_{1}$, $C_{1}$. Then
$$
\begin{array}{c}
B A_{1}=2 \sin \alpha_{1}, \\
C A_{1}=2 \sin \alpha_{2}, \\
C B_{1}=2 \sin \beta_{1}, \\
A B_{1}=2 \sin \beta_{2}, \\
A C_{1}=2 \sin \gamma_{1}, B C_{1}=2 \sin \gamma_{2}, \\
h... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,752 |
Question 3: Find all pairs of coprime positive integers $(x, y)$ such that
$$
x\left|\left(y^{2}+210\right), y\right|\left(x^{2}+210\right)^{[4]} \text {. }
$$
(2005, China National Team Training Problem)
The solution given in [4] states that $(x, y)$ holds for $(1,1),(1,211)$, and also discusses the case $x \leqslant ... | Prove: For pairs of positive integers $(x, y)$, if
$$
x\left|\left(y^{2}+210\right), y\right|\left(x^{2}+210\right),(x, y)=1 \text {, }
$$
then it is equivalent to
$$
\begin{array}{l}
x y \mid\left(x^{2}+y^{2}+210\right)((x, y)=1) \\
\Leftrightarrow x^{2}+y^{2}+210=k x y\left((x, y)=1, k \in \mathbf{N}_{+}\right) . \\... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,753 |
For each square number $k^{2}(k \geqslant 2)$, construct all pairs of non-zero natural numbers $(a, b)$ such that $\frac{a^{2}+b^{2}}{a b+1}=k^{2}$.
This problem is an adaptation of the 6th question from the 29th IMO, and it is also the 5th question from the 1998 Canadian National Mathematical Competition. | Solution: Transform the equation $\frac{a^{2}+b^{2}}{a b+1}=k^{2}$ into an indeterminate equation similar to the proposition:
$$
a^{2}+b^{2}-k^{2}=k^{2} a b \text {. }
$$
If the greatest common divisor of $a$ and $b$ is $t\left(t \in \mathbf{N}_{+}\right)$, then from equation (1) we know $t \mid k$. Therefore,
$$
\beg... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,754 |
1. Given that $a$, $b$, and $c$ are all positive numbers not equal to 1, and $a^{-2}=b^{3}=c^{6}$. Then the value of $a b c$ is ( ).
(A) 3
(B) 2
(C) 1
(D) $\frac{1}{2}$ | -1.C.
From $a^{-2}=b^{3}=c^{6}$, we get $a=c^{-3}, b=c^{2}$.
Then $a b c=c^{-3} c^{2} c=1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,755 |
Example 4 As shown in Figure 4, let $P$ be a point inside $\triangle ABC$, and $AP$, $BP$, $CP$ intersect the opposite sides at $D$, $E$, $F$ respectively. Prove:
$$
\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}=1 .
$$ | Proof: Note that
$$
\begin{array}{l}
\frac{A F}{F B}=\frac{S_{\triangle A C F}}{S_{\triangle B C F}}=\frac{S_{\triangle A P F}}{S_{\triangle B P F}} \\
=\frac{S_{\triangle A C F}-S_{\triangle A P F}}{S_{\triangle B C F}-S_{\triangle B P F}}=\frac{S_{\triangle A P C}}{S_{\triangle B P C}} .
\end{array}
$$
$$
\frac{C E}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,756 |
3. A set of mutually unequal data has a median of 80, the average of the numbers less than the median is 70, and the average of the numbers greater than the median is 96. Let the average of this set of data be $\bar{x}$, then $($.
(A) $\bar{x}=82$
(B) $\bar{x}=83$
(C) $80 \leqslant \bar{x} \leqslant 82$
(D) $82 \leqsla... | 3.D.
If this set of data has $2 n$ numbers, then
$$
\bar{x}=\frac{70 n+96 n}{2 n}=83 ;
$$
If this set of data has $2 n+1$ numbers, then
$$
\bar{x}=\frac{70 n+80+96 n}{2 n+1}=83-\frac{3}{2 n+1} \text {. }
$$
Since $n \geqslant 1$, therefore, $82 \leqslant \bar{x} \leqslant 83$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,758 |
4. In Rt $\triangle A B C$, $\angle C=90^{\circ}, B C<A C$, $B C \cdot A C=\frac{1}{4} A B^{2}$. Then $\angle A$ equals $($ ).
(A) $15^{\circ}$
(B) $18^{\circ}$
(C) $20^{\circ}$
(D) $25^{\circ}$ | 4.A.
Draw $C D \perp A B$ at point $D$.
From $A B \cdot C D=A C \cdot B C=\frac{1}{4} A B^{2}$, we get
$$
C D=\frac{1}{4} A B \text {. }
$$
Take the midpoint $M$ of $A B$, and connect $C M$. Then $C M=\frac{1}{2} A B$. Therefore, $C D=\frac{1}{2} C M$.
Thus, $\angle C M D=30^{\circ}, \angle A=\frac{1}{2} \angle C M D... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,759 |
5. Given a sequence of numbers $a_{1}, a_{2}, \cdots, a_{100}$, where $a_{3} = 9, a_{7} = -7, a_{98} = -1$, and the sum of any three consecutive numbers is a constant. Then the value of $a_{1} + a_{2} + \cdots + a_{100}$ is ( ).
(A) 0
(B) 40
(C) 32
(D) 26 | 5.D.
According to the problem, we have
$$
\begin{array}{l}
a_{1}=a_{4}=a_{7}=\cdots=a_{100}=-7, \\
a_{2}=a_{5}=a_{8}=\cdots=a_{98}=-1, \\
a_{3}=a_{6}=a_{9}=\cdots=a_{99}=9 .
\end{array}
$$
Then $\bar{a}_{1}+a_{2}+\cdots+a_{100}$
$$
=33(9-7-1)-7=26 \text {. }
$$ | 26 | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,760 |
6. As shown in Figure 2, fold the hexagon $A B C D E F$ along the line $G H$ so that points $A$ and $B$ fall inside the hexagon $C D E F G H$. Let $\angle C+\angle D+\angle E+\angle F = \alpha$. Then which of the following conclusions must be true? ( ).
(A) $\angle 1+\angle 2=900^{\circ}-2 \alpha$
(B) $\angle 1+\angle ... | 6.B.
As shown in Figure 9, let the extension of FA intersect the extension of CB at point P, and the extension of GA' intersect the extension of HB' at point P'.
By symmetry, we know
$$
\begin{array}{c}
\angle 1=2 \angle A P P^{\prime}, \\
\angle 2=2 \angle B P P^{\prime}. \\
\text { Then } \angle 1+\angle 2=2 \angle ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,761 |
8. As shown in Figure 5, in rhombus $A B C D$, $\angle A B C=120^{\circ}$, $F$ is the midpoint of $D C$, and the extension of $A F$ intersects the extension of $B C$ at point $E$. Then the measure of the acute angle formed by lines $B F$ and $D E$ is ( ).
(A) $30^{\circ}$
(B) $40^{\circ}$
(C) $50^{\circ}$
(D) $60^{\cir... | 8.D.
Extend $B F$ to intersect $D E$ at point $M$, and connect $B D$. Then $\triangle B C D$ is an equilateral triangle, $F$ is the midpoint of $C D$, and $B F$ bisects $\angle C B D$. Therefore, $\angle M B C=30^{\circ}$.
Since $F$ is the midpoint of $C D$ and $A D / / C E$, $\triangle A D F$ and $\triangle E C F$ a... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,763 |
10. Given $y=\sqrt{x-1}+\sqrt{4-x}$ (where $x$ and $y$ are real numbers). Then the difference between the maximum and minimum values of $y$ is ( ).
(A) $\sqrt{6}-3$
(B) 3
(C) $\sqrt{5}-\sqrt{3}$
(D) $\sqrt{6}-\sqrt{3}$ | 10.D.
$$
\begin{array}{l}
y \geqslant 0, 1 \leqslant x \leqslant 4 . \\
y^{2}=3+2 \sqrt{-x^{2}+5 x-4} \\
=3+2 \sqrt{-\left(x-\frac{5}{2}\right)^{2}+\frac{9}{4}},
\end{array}
$$
Then $3 \leqslant y^{2} \leqslant 6$.
Since $y \geqslant 0$, therefore, $\sqrt{3} \leqslant y \leqslant \sqrt{6}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,765 |
11. Given positive numbers $a, b, c, d, e, f$ satisfy
$$
\begin{array}{l}
\frac{b c d e f}{a}=4, \frac{a c d e f}{b}=9, \frac{a b d e f}{c}=16, \\
\frac{a b c e f}{d}=\frac{1}{4}, \frac{a b c d f}{e}=\frac{1}{9}, \frac{a b c d e}{f}=\frac{1}{16} .
\end{array}
$$
Then the value of $(a+c+e)-(b+d+f)$ is
$\qquad$ | 11. $-2 \frac{7}{12}$.
Since $(a b c d e f)^{4}=1$, and $a, b, c, d, e, f$ are all positive, we have $a b c d e f=1$.
Thus, $\frac{1}{a^{2}}=4 \Rightarrow a=\frac{1}{2}$.
Similarly, $b=\frac{1}{3}, c=\frac{1}{4}, d=2, e=3, f=4$.
Therefore, $(a+c+e)-(b+d+f)=-2 \frac{7}{12}$. | -2 \frac{7}{12} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,766 |
3.1 Problems related to area
Example $5 \triangle A B C$ has three sides $a, b, c$ satisfying $1 \leqslant a \leqslant 3 \leqslant b \leqslant 5 \leqslant c \leqslant 7$.
Find the perimeter of $\triangle A B C$ when its area is maximized. | Solution: Note that
$$
S_{\triangle A B C}=\frac{1}{2} a b \sin C \leqslant \frac{1}{2} a b \leqslant \frac{1}{2} \times 3 \times 5,
$$
equality holds if and only if $a=3, b=5, \angle C=90^{\circ}$. At this time, $c=\sqrt{a^{2}+b^{2}}=\sqrt{34}$ satisfies $5 \leqslant c \leqslant 7$.
Therefore, when $a=3, b=5, c=\sqr... | 8+\sqrt{34} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,767 |
12. As shown in Figure 7, $\triangle A B D$ and $\triangle C E D$ are both equilateral triangles, $A C=B C, A C \perp B C$. If $B E$ $=\sqrt{2}$, then $C D=$ $\qquad$ | 12. $\sqrt{3}-1$.
Extend $DC$ to intersect $AB$ at point $M$.
Since $AD=BD, AC=BC$, line $DC$ is the perpendicular bisector of $AB$. It is easy to prove that $\triangle ADC \cong \triangle BDE$.
Thus, $AC=BE=\sqrt{2}, CM=\frac{\sqrt{2}}{2} AC=1$, $AB=\sqrt{2} AC=2$.
Therefore, $DM=\sqrt{3}, CD=\sqrt{3}-1$. | \sqrt{3}-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,768 |
13. Ordered pairs of positive integers $(a, b)(a<b)$ satisfy $a+b=2008$, and $a, b$ are coprime. Then the number of pairs $(a, b)$ that satisfy the condition is $\qquad$. | 13.500.
Since $2008=2^{3} \times 251$, therefore, $a$ cannot be an even number, and it cannot be a multiple of 251. And $a \leqslant 1003$, so the number of $a$ is $1003-501-2=500$. | 500 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,769 |
14. As shown in Figure 8, in the right trapezoid $A B C D$, $\angle A$ is a right angle, $A B / / C D, A B=7$, $C D=5, A D=2$. A moving line $l$ intersects $A B$ at point $P$, and intersects $C D$ at point $Q$, and divides the trapezoid $A B C D$ into two parts of equal area. Then the maximum value of the distance from... | 14. $\sqrt{10}$.
According to the problem, we have $A P+D Q=6$.
Let the midpoints of $A D$ and $P Q$ be $M$ and $N$ respectively, then $M N$ $=3$. Therefore, $N$ is a fixed point. The distance from point $A$ to the moving line $l$ is maximized when and only when $P Q \perp A N$, and the maximum value is
$$
\sqrt{A M^{... | \sqrt{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,770 |
15. In $\triangle A B C$, $\angle C=30^{\circ}, B M$ is a median, $A C=2 a$. If $\triangle A B C$ is folded along $B M$, then the overlapping area of the two smaller $\triangle A B M$ and $\triangle B C M$ is exactly $\frac{1}{4}$ of the area of $\triangle A B C$. Try to find the area of $\triangle A B C$ (expressed in... | 15. Based on the size relationship between $\angle A B M$ and $\angle C B M$, there are three scenarios after folding.
(1) $\angle A B M > \angle C B M$.
Suppose $B C$ is folded to $B C^{\prime}$ (as shown in Figure 10), intersecting $A C$ at point $D$. Then
$$
\begin{array}{c}
S_{\triangle B D M} = \frac{1}{4} S_{\tri... | \frac{\sqrt{3}}{2} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,771 |
16. The integer $a$ makes the system of equations about $x, y$
$$
\left\{\begin{array}{l}
x-2 y=3 a-b, \\
x y=b^{2}-2 a^{2}+3 b+4
\end{array}\right.
$$
have real solutions for every real number $b$. Find the value of the integer $a$. | 16. Substituting $x=3 a-b+2 y$ into $x y=b^{2}-2 a^{2}+$ $3 b+4$, we get
$$
\begin{array}{l}
2 y^{2}+(3 a-b) y-\left(b^{2}-2 a^{2}+3 b+4\right)=0 . \\
\Delta=(3 a-b)^{2}+8\left(b^{2}-2 a^{2}+3 b+4\right) \\
=9 b^{2}-6(a-4) b-7 a^{2}+32 \\
=(3 b-a+4)^{2}-\left(8 a^{2}-8 a-16\right) \geqslant 0 .
\end{array}
$$
Since fo... | -1,0,1,2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,772 |
3. In a regular tetrahedron $A B C D$ with edge length 1, $M$ and $N$ are the midpoints of $A D$ and $B C$ respectively. Then the distance between the skew lines $A B$ and $M N$ is ( ).
(A) $\frac{1}{4}$
(B) $\frac{1}{12}$
(C) $\frac{\sqrt{2}}{4}$
(D) $\frac{\sqrt{3}}{8}$ | 3.C.
As shown in Figure 1, take the midpoints $E$ and $F$ of $AB$ and $CD$ respectively, and connect $EN$, $NF$, $FM$, $ME$, $EF$, $FB$, $FA$, and the intersection of $EF$ and $MN$ at point $O$. Clearly, quadrilateral $ENFM$ is a planar quadrilateral and is a rhombus, so $MN \perp EF$. Also, $EF$ is the median of isos... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,775 |
4. In the arithmetic sequence $\left\{a_{n}\right\}$, $a_{1}=-5$, the average of the first 11 terms is 5. If one term is removed, the average of the remaining 10 terms is 4, then the term removed is ( ).
(A) $a_{8}$
(B) $a_{9}$
(C) $a_{10}$
(D) $a_{11}$. | 4.D.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,776 |
5. Given that $S-A B C$ is a tetrahedron with three edges mutually perpendicular, and $O$ is a point within the base $A B C$. If $\angle O S A=$ $\alpha, \angle O S B=\beta, \angle O S C=\gamma$, then the range of $\tan \alpha \cdot \tan \beta \cdot$ $\tan \gamma$ is ( ).
(A) $[2 \sqrt{2},+\infty)$
(B) $(0,2 \sqrt{2})$... | 5.A.
Through point $O$, construct planes parallel to $S A$, $S B$, and $S C$, forming a rectangular parallelepiped with $O S$ as its diagonal. Then, $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$.
Thus, $\sin ^{2} \alpha=\cos ^{2} \beta+\cos ^{2} \gamma \geqslant 2 \cos \gamma \cdot \cos \beta$.
Similarly, $\si... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,777 |
Example 6 As shown in Figure 5, in Rt $\triangle C A B$, $\angle A=$ $90^{\circ}, \angle B 、 \angle C$ are bisected and intersect at $F$, and intersect the opposite sides at points $D 、 E$. Find $S_{\text {quadrilateral } B C D E}: S_{\triangle B F C}$. | Solution: Let $A B=c, A C=b, B C=a$. By the property of the internal angle bisector, we have
$$
\frac{A E}{E B}=\frac{b}{a} \text {. }
$$
Then $\frac{A E}{c}=\frac{b}{a+b}$, i.e., $A E=\frac{b c}{a+b}$.
Thus, $B E=c-A E=\frac{a c}{a+b}$.
Therefore, $S_{\triangle C B E}=\frac{1}{2} B E \cdot b=\frac{a b c}{2(a+b)}$.
An... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,778 |
6. Let set $A=\{1,2,3,4,5,6\}$, and a one-to-one mapping $f: A \rightarrow A$ satisfies that for any $x \in A$, $f(f(f(x)))$ $=x$. Then the number of mappings $f$ that satisfy the above condition is ( ).
(A) 40
(B) 41
(C) 80
(D) 81 | 6.D.
If there exists $x \in A$ such that $f(f(x))=x, f(x) \neq x$, then $f(f(f(x)))=f(x) \neq x$, which contradicts the given condition. Therefore, for any $x \in A$, either $f(x)=x$, or $f(x)=x_{1}, f(x_{1})=x_{2}, f(x_{2})=x$, and $x, x_{1}, x_{2}$ are distinct. Hence, there are only the following three scenarios:
(... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,779 |
8. The volume of the inscribed sphere in a tetrahedron with all edges equal to 1 is $\qquad$ . | 8. $\frac{\sqrt{6} \pi}{216}$ | \frac{\sqrt{6} \pi}{216} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,781 |
9. Given point $P(2,1)$, draw a line $l$ through $P$ that intersects the positive $x$-axis and $y$-axis at points $A$ and $B$ respectively. The equation of the line $l$ that minimizes the perimeter of $\triangle A O B$ (where $O$ is the origin) is $\qquad$
Translate the above text into English, please retain the origi... | $$
9.3 x+4 y-10=0 \text {. }
$$
Let $\angle B A O=\theta, t=$ $\tan \frac{\theta}{2}$. Denote the perimeter of $\triangle A B O$ as $s$, then
$$
\begin{aligned}
s= & (2+\cot \theta)+ \\
& (1+2 \tan \theta)+ \\
& \left(\frac{1}{\sin \theta}+\frac{2}{\cos \theta}\right) \\
= & 6+\frac{1-t}{t}+\frac{4 t}{1-t} \geqslant 1... | 3 x+4 y-10=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,782 |
10. Given the function
$$
f(x)=\log _{2}\left[a x^{2}+(a+2) x+(a+2)\right] \text {. }
$$
If $f(x)$ has a maximum or minimum value, then the range of values for $a$ is $\qquad$ . | 10. $(-2,0) \cup\left(\frac{2}{3},+\infty\right)$.
Since $f(x)$ has a maximum or minimum value, $a$ satisfies $\left\{\begin{array}{l}a>0, \\ \Delta \geq 0 .\end{array}\right.$ Solving this, we get $-2 < a < 0$ or $a > \frac{2}{3}$. | (-2,0) \cup\left(\frac{2}{3},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,783 |
11. Let vectors $a$ and $b$ satisfy $|a|=1$, $|b|=2$, and the angle between $a$ and $b$ is $60^{\circ}$. If the angle between the vectors $7a+2tb$ and $ta+b$ is obtuse, then the range of the real number $t$ is $\qquad$ | 11. $\left(-7,-\frac{\sqrt{14}}{2}\right) \cup\left(-\frac{\sqrt{14}}{2},-\frac{1}{2}\right)$.
From $(7 a+2 t b) \cdot(t a+b)<0$, we get
$-7<t<-\frac{1}{2}$.
When $7 a+2 t b$ is collinear with $t a+b$, we have $\frac{7}{t}=\frac{2 t}{1}, t= \pm \frac{\sqrt{14}}{2}$.
When $t=-\frac{\sqrt{14}}{2}$, $7 a+2 t b$ and $t a+... | \left(-7,-\frac{\sqrt{14}}{2}\right) \cup\left(-\frac{\sqrt{14}}{2},-\frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,784 |
12. Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with left and right foci $F_{1}$ and $F_{2}$, respectively, a line $l$ passing through the right focus $F_{2}$ intersects the ellipse at points $P$ and $Q$. Then the maximum value of the area of the incircle of $\triangle F_{1} P Q$ is $\qquad$ | 12. $\frac{9 \pi}{16}$.
Let the equation of line $l$ be $x=m y+1, P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$.
From $\left\{\begin{array}{l}x=m y+1, \\ \frac{x^{2}}{4}+\frac{y^{2}}{3}=1,\end{array}\right.$ we get
$\left(3 m^{2}+4\right) y^{2}+6 m y-9=0$.
Thus, $y_{1}+y_{2}=\frac{-6 m}{3 m^{2}+4}, y_{1} y_{... | \frac{9 \pi}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,785 |
Three, (15 points) Let the line $l: y=x+c$ intersect the ellipse $\frac{x^{2}}{2}+y^{2}=1$ at two points $A$ and $B$ (which can coincide), and intersect the circle $(x-2)^{2}+(y+2)^{2}=4$ at two points $C$ and $D$ (which can coincide). Find the maximum value of $9|A B|^{2}+|C D|^{2}$. | Three, the chord length formula for the line $l: y=x+c$ intersecting a curve is $\sqrt{2\left(x_{1}+x_{2}\right)^{2}-8 x_{1} x_{2}}$.
Substituting the line equation into the equations of the ellipse and the circle, we get
$$
\begin{array}{l}
\frac{3}{2} x^{2}+2 c x+c^{2}-1=0, \\
2 x^{2}+2 c x+4+4 c+c^{2}=0 .
\end{array... | 256 \sqrt{2}-336 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,786 |
Four, (15 points) Prove: Given any 7 real numbers, there must exist two real numbers $x, y$ satisfying $0 \leqslant \frac{x-y}{1+x y}<\frac{\sqrt{3}}{3}$.
| Let the seven real numbers be $\tan \theta_{1}, \tan \theta_{2}, \cdots$, $\tan \theta_{7}$, and assume without loss of generality that $-\frac{\pi}{2}<\theta_{1} \leqslant \theta_{2} \leqslant \cdots \leqslant \theta_{7}<\frac{\pi}{2}$.
Divide the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ into 6 subinterva... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,787 |
Five. (15 points) Let $\left\{a_{n}\right\}$ be an integer sequence, and it satisfies: for any $n\left(n \in \mathbf{N}_{+}\right)$, we have
$$
(n-1) a_{n+1}=(n+1) a_{n}-2(n-1) \text {, }
$$
and $2008 \mid a_{2000}$. Find the smallest positive integer $n(n \geqslant 2)$, such that $2008 \mid a_{n}$. | When $n=1$, we have $a_{1}=0$.
When $n \geqslant 2$, the original equation transforms to
$$
a_{n+1}=\frac{n+1}{n-1} a_{n}-2 \text{. }
$$
Let $b_{n}=\frac{a_{n}}{n-1}$, then $n b_{n+1}=(n+1) b_{n}-2$.
Thus, for $n \geqslant 2$, we have
$$
b_{n+1}-2=\frac{n+1}{n}\left(b_{n}-2\right) \text{. }
$$
From equation (2), we k... | 501 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,788 |
Example 7 As shown in Figure 6, in $\triangle A B C$, $A B=A C, D$ is a point on $B C$, $E$ is a point on side $A D$, and it satisfies $\angle B E D=$ $2 \angle C E D=\angle B A C$. Prove: $B D=2 C D$. | Proof: Note that
$$
\begin{array}{l}
\angle B E D=\angle A B E+\angle B A E, \\
\angle B A C=\angle C A E+\angle B A E .
\end{array}
$$
Since $\angle B E D=\angle B A C$, it follows that $\angle A B E=\angle C A E$.
Thus, $\frac{B D}{D C}=\frac{S_{\triangle A B E}}{S_{\triangle A C E}}$
$$
=\frac{\frac{1}{2} A B \cdot... | B D=2 C D | Geometry | proof | Yes | Yes | cn_contest | false | 719,789 |
Six. (15 points) Answer the following two questions and provide examples or proofs.
(1) For any positive integer $n$, does there exist $n$ points in the plane that are not all on the same line, such that the distance between any two points is a positive integer?
(2) Does there exist an infinite set of points $M$ in the... | (1) Existence.
For any $n\left(n \in \mathbf{N}_{+}\right)$, take $n$ distinct prime numbers $p_{1}, p_{2}, \cdots, p_{n}$. Let
$a_{i}=p_{1} p_{2} \cdots p_{i}, b_{i}=p_{i+1} p_{i+2} \cdots p_{n}(i=1$, $2, \cdots, n-1)$,
obviously, $a_{i} 、 b_{i} \in \mathbf{N}_{+}$.
Let $m=p_{1} p_{2} \cdots p_{n}$, then, $m=a_{i} b_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,790 |
1. $f$ is a mapping from the set $M=\{a, b, c, d\}$ to $N=\{0, 1, 2\}$, and
$$
f(a)+f(b)+f(c)+f(d)=4.
$$
Then the number of different mappings is $($ .
(A) 13
(B) 19
(C) 21
(D) 23 | -1.B.
(1) When the image set is $\{1\}$, we have $f(a)=f(b)=$ $f(c)=f(d)=1$, there is 1 such mapping;
(2) When the image set is $\{0,2\}$, $f(a) 、 f(b)$ 、 $f(c) 、 f(d)$ have 2 zeros and 2 twos. Enumerating, we find there are 6 such mappings;
(3) When the image set is $\{0,1,2\}$, $f(a) 、 f(b)$ 、 $f(c) 、 f(d)$ have 1 ze... | 19 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,791 |
2. Given $\frac{\cos x+\sin x}{\cos x-\sin x}=2007$. Then
$$
\csc \left(\frac{\pi}{2}+2 x\right)+\tan (3 \pi+2 x)
$$
is equal to $(\quad$.
(A) 2005
(B) 2006
(C) 2007
(D) 2008 | 2.C.
$$
\begin{array}{l}
\text { Given } \frac{\cos x+\sin x}{\cos x-\sin x}=2007, \text { we know that } \\
\frac{1+\tan x}{1-\tan x}=\tan \left(x+\frac{\pi}{4}\right)=2007 . \\
\text { Therefore, } \csc \left(\frac{\pi}{2}+2 x\right)+\tan (3 \pi+2 x) \\
=\frac{1}{\cos 2 x}+\frac{\sin 2 x}{\cos 2 x}=\frac{1+\sin 2 x}{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,792 |
3. Given the function $f(x)$ satisfies: for any $x, y \in \mathbf{R}$, we have
$$
f\left(x+y^{2}\right)=f(x)+2 f^{2}(y) \text { and } f(1) \neq 0 \text {. }
$$
Then the value of $f(2007)$ is $(\quad)$.
(A) 1002.5
(B) 1003
(C) 1003.5
(D) 1004 | 3. C.
Let $x=y=0$, we get $f(0)=0$.
Let $x=0, y=1$, we get $f(1)=\frac{1}{2}$.
Let $y=1$, we get $f(x+1)-f(x)=\frac{1}{2}$.
Therefore, $\{f(n)\}$ is an arithmetic sequence with the first term $f(1)=\frac{1}{2}$ and common difference $\frac{1}{2}$. Thus,
$$
f(2007)=f(1)+\frac{1}{2}(2007-1)=1003.5
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,793 |
4. Let $a=\frac{1}{2} \cos 6^{\circ}-\frac{\sqrt{3}}{2} \sin 6^{\circ}$,
$$
b=\frac{2 \tan 13^{\circ}}{1-\tan ^{2} 13^{\circ}}, c=\sqrt{\frac{1-\cos 50^{\circ}}{2}} \text {. }
$$
Then we have ().
(A) $ab>c$
(D) $b<c<a$ | 4.A.
Since $a=\frac{1}{2} \cos 6^{\circ}-\frac{\sqrt{3}}{2} \sin 6^{\circ}=\sin 24^{\circ}$,
$$
\begin{aligned}
b & =\frac{2 \tan 13^{\circ}}{1-\tan ^{2} 13^{\circ}}=\tan 26^{\circ}, \\
c & =\sqrt{\frac{1-\cos 50^{\circ}}{2}}=\sin 25^{\circ},
\end{aligned}
$$
Therefore, $a<c$.
Also, for $x \in\left(0, \frac{\pi}{2}\r... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,794 |
5. The function $f(x)=x^{2}-2 a x+a$ has a minimum value on the interval $(-\infty, 1)$. Then the function $g(x)=\frac{f(x)}{x}$ on the interval $(1,+\infty)$ must $(\quad)$.
(A) have a minimum value
(B) have a maximum value
(C) be a decreasing function
(D) be an increasing function | 5.D.
Given that $f(x)=x^{2}-2 a x+a$ has a minimum value in the interval $(-\infty, 1)$, we get $a<1$.
Thus, $g(x)=\frac{f(x)}{x}=x+\frac{a}{x}-2 a$.
When $a \leqslant 0$, $g(x)=x+\frac{a}{x}-2 a$ is definitely an increasing function on $(1, +\infty)$;
When $0<a<1$, $g(x)$ is decreasing on $(0, \sqrt{a})$ and increas... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,795 |
6. The minimum value of the function $f(x)=\sum_{n=1}^{19}|x-n|$ is ( ).
(A) 90
(B) 171
(C) 190
(D) 45 | 6.A.
$$
\begin{array}{l}
f(x)=\sum_{n=1}^{19}|x-n| \\
=\sum_{n=1}^{9}|x-n|+|x-10|+\sum_{n=11}^{19}|x-n| \\
\geqslant\left|\sum_{n=1}^{9}(x-n)\right|+\left|\sum_{n=11}^{19}(n-x)\right|+|x-10| \\
\geqslant\left|\sum_{n=1}^{9}(x-n)+\sum_{n=11}^{19}(n-x)\right|=90,
\end{array}
$$
The equality holds if and only if \( x=10 ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,796 |
7. Define an operation of sets $A$ and $B$: $A * B=$ $\left\{x \mid x=x_{1}+x_{2}\right.$, where, $\left.x_{1} \in A, x_{2} \in B\right\}$. If $A$ $=\{1,2,3\}, B=\{1,2\}$, then the sum of all elements in $A * B$ is $\qquad$ | 二、7.14.
$A * B$ 中元素为 $2,3,4,5$, 故其所有元素数字之和为 14 .
---
Second, 7.14.
The elements of $A * B$ are $2,3,4,5$, so the sum of all the digits of its elements is 14. | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,797 |
8. Given the condition $p:|4 x-3|<1$;
condition $q: x^{2}-(2 a+1) x+a(a+1) \leqslant 0$.
If $p$ is a necessary but not sufficient condition for $q$, then the range of real number $a$ is $\qquad$. | 8. $[0,0.5]$.
From $p$ being a necessary but not sufficient condition for $q$, we know that $q$ is a necessary but not sufficient condition for $p$.
Given $p: \frac{1}{2}<x<1, q: a \leqslant x \leqslant a+1$.
Therefore, $a+1 \geqslant 1$ and $a \leqslant \frac{1}{2}$.
Thus, $0 \leqslant a \leqslant \frac{1}{2}$. | 0 \leqslant a \leqslant \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,798 |
9. The sequence $1,1,2,1,1,3,1,1,1,4,1,1,1,1$, $5, \cdots, \underbrace{1,1, \cdots, 1}_{n-1 \uparrow}, n, \cdots$ has the sum of its first 2007 terms as | 9.3898 .
In the sequence, from number 1 to $n$ there are $n+\frac{n(n-1)}{2}$ terms, thus, from number 1 to 62 there are $62+\frac{62 \times(62-1)}{2}$ $=1953$ terms, followed by 54 ones.
Therefore, the sum of the first 2007 terms is
$$
\begin{array}{l}
(1+2+\cdots+62)+(1+2+\cdots+61)+54 \\
=3898
\end{array}
$$ | 3898 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,799 |
Example 8 As shown in Figure 7, draw two tangents from point $P$ to $\odot O$, with points of tangency at $A$ and $B$. Draw an arbitrary secant line $P N M$ through $P$ intersecting $A B$ at $Q$, where $M$ and $N$ are on $\odot O$. Prove:
$$
\frac{P N}{P M}=\frac{Q N}{Q M} .
$$ | Analysis: Since there is no similar relationship in the problem, and the line segments to be proved are all on the same line, we can consider the area method.
Proof: As shown in Figure 7, connect $A M, A N, B M, B N$.
Notice that
$$
\begin{array}{l}
\frac{P N}{P M}=\frac{S_{\triangle S B V}}{S_{\triangle P B M}}=\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,800 |
10. The function $f(x)$ defined on $\mathbf{R}$ satisfies
$$
f\left(f(x)-x^{2}+x\right)=f(x)-x^{2}+x \text {. }
$$
If there is exactly one real number $x_{0}$ such that $f\left(x_{0}\right)=$ $x_{0}$, then $f(x)=$ $\qquad$ | 10. $x^{2}-x+1$.
Since there is only one real number $x_{0}$ such that $f\left(x_{0}\right) = x_{0}$, it follows that $f(x) - x^{2} + x = x_{0}$ always holds.
Let $x = x_{0}$, we get $f\left(x_{0}\right) - x_{0}^{2} + x_{0} = x_{0}$.
Solving this, we get $x_{0} = 0$ or 1.
When $x_{0} = 0$, $f(x) = x^{2} - x$, in this ... | x^{2} - x + 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,801 |
12. Given that the equation $x^{3}+3 x^{2}-x+a$ $=0$ has three real roots that form an arithmetic sequence. Then the real number $a=$ $\qquad$ | 12. -3 .
Let these three roots be $b-d$, $b$, and $b+d$. Then
$$
\begin{array}{l}
x^{3}+3 x^{2}-x+a \\
=(x-b+d)(x-b)(x-b-d),
\end{array}
$$
i.e., $3 x^{2}-x+a$
$$
=-3 b x^{2}+\left(3 b^{2}-d^{2}\right) x-b^{3}+b d^{2} \text {. }
$$
Comparing coefficients, we get
$$
-3 b=3,3 b^{2}-d^{2}=-1,-b^{3}+b d^{2}=a \text {. }... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,803 |
Three. (20 points) As shown in Figure 1, line $l$ passes through the endpoint $A$ of line segment $AB$ and is perpendicular to $AB$. Points $E$ and $F$ are any two points on $l$, and $EB \perp BF$. $O$ is the midpoint of $AB$. Draw $EP \parallel AB$, connect $FO$ and extend it to intersect $EP$ at point $P$, then conne... | Three, as shown in Figure 2, extend $E P$ and $F B$ to intersect at point $C$.
Since $O$ is the midpoint of $A B$ and $E P \parallel A B$, we have
$$
E P = P C.
$$
Also, since $E B \perp B F$, $P B$ is the median to the hypotenuse $E C$ of $\mathrm{Rt} \triangle E B C$.
Thus, $P B = P C$, which means
$\angle P C B = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,804 |
Four. (20 points) Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms $S_{n}=2 a_{n}-3 \times 2^{n}+4\left(n \in \mathbf{N}_{+}\right)$.
(1) Find the general term formula for the sequence $a_{n}$;
(2) Let $T_{n}$ be the sum of the first $n$ terms of the sequence $\left\{S_{n}-4\right\}$, find ... | (1) $S_{n+1}-S_{n}=2\left(a_{n+1}-a_{n}\right)-3\left(2^{n+1}-2^{n}\right)$, i.e., $a_{n+1}=2 a_{n}+3 \times 2^{n}$.
Dividing both sides by $2^{n+1}$ gives $\frac{a_{n+1}}{2^{n+1}}=\frac{a_{n}}{2^{n}}+\frac{3}{2}$, thus $\frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=\frac{3}{2}$.
From $a_{1}=2 a_{1}-3 \times 2^{1}+4$, we... | T_{n}=14-(14-6 n) 2^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,805 |
Five. (20 points) Define an odd function $f(x)$ on $[-1,1]$ that satisfies $f(1)=1$, and for $a, b \in[-1,1], a+b \neq 0$, we have $\frac{f(a)+f(b)}{a+b}>0$.
(1) Prove: $f(x)$ is an increasing function on $[-1,1]$;
(2) Prove: When $\frac{1}{3} \leqslant x \leqslant 1$, $f(x) \leqslant 3 x$;
(3) If $f(x) \leqslant m^{2}... | (1) For any $x_{1}, x_{2} \in[-1,1]$, and $x_{1}<x_{2}$,
$$
x_{1}+\left(-x_{2}\right)<0 \text {, }
$$
Therefore, $f\left(x_{1}\right)-f\left(x_{2}\right)<0$.
Thus, $f(x)$ is an increasing function on $[-1,1]$.
(2) From (1), when $\frac{1}{3} \leqslant x \leqslant 1$,
$$
f(x)_{\max }=f(1)=1, (3 x)_{\min }=1 \text {. }
... | m \leqslant-2 \text{ or } m \geqslant 2 \text{ or } m=0 | Algebra | proof | Yes | Yes | cn_contest | false | 719,806 |
Six. (20 points) In the sequence $\left\{a_{n}\right\}$,
$$
a_{n}=\cos 2^{n} \theta(n \in \mathbf{N}) \text {. }
$$
(1) If the set $A=\left\{\theta \mid a_{0}>0, a_{1}<0, \theta \in\right.$ $\mathbf{R}\}, B=[0,2 \pi]$, find $A \cap B$;
(2) Find all real numbers $\theta$, such that every term of the sequence $\left\{a_{... | Let $a_{0}>0$, i.e., $\cos \theta>0$.
Solving gives $2 k \pi-\frac{\pi}{2}<\theta<2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z})$.
Since $a_{n+1}=2 a_{n}^{2}-1<0$, we have $-1 \leqslant 2 a_{n}^{2}-1<0$, which implies $0 \leqslant a_{n}^{2}<\frac{1}{2}$.
Thus, $-\frac{\sqrt{2}}{2}<a_{n}<\frac{\sqrt{2}}{2}$.
Since $a_{n}<0$, w... | \left\{\theta \left\lvert\, \theta=2 k \pi \pm \frac{2 \pi}{3}(k \in \mathbf{Z})\right.\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,807 |
1. (i) (Grade 11) $a, b, c$ are real numbers, and the sets are
$$
\begin{array}{l}
A=\left\{x \mid x=a^{2}-4 a\right\}, \\
B=\left\{y \mid y=b^{2}+2 b-3\right\}, \\
C=\left\{z \mid z=\sqrt{c^{2}+6 c+10}-5\right\}
\end{array}
$$
The relationship between them is ( ).
(A) $A=B=C$
(B) $A \cap B \cap C=\varnothing$
(C) $A ... | (i) A.
$$
A=B=C=\{t \mid t \geqslant-4\} \text {. }
$$
(ii) D.
From $2 a^{2}-b^{2}=a^{2}+2 b^{2}$, we get $a^{2}=3 b^{2}$.
$$
e^{2}=\frac{2 a^{2}-b^{2}}{2 a^{2}}=\frac{5 b^{2}}{6 b^{2}}=\frac{5}{6}, e=\frac{\sqrt{30}}{6} .
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,808 |
2. (i) (Grade 11) The range of the function $y=\frac{2 \sin x \cdot \cos x-3}{\sin x+\cos x-2}$ is ( ).
(A) $\left[\frac{1}{2}, 3\right]$
(B) $[2-\sqrt{2}, 2+\sqrt{2}]$
(C) $[\sqrt{2}-1, \sqrt{2}+1]$
(D) $\left[1-\frac{\sqrt{2}}{2}, 1+\frac{\sqrt{2}}{2}\right]$
(ii) (Grade 12) Given that $a, b, c, d$ are distinct posit... | 2. (i)B.
Since $2 \sin x \cdot \cos x - 3 = (\sin x + \cos x)^2 - 4$, therefore, $y = (\sin x + \cos x) + 2$.
Also, $-\sqrt{2} \leq \sin x + \cos x \leq \sqrt{2}$, then
$2 - \sqrt{2} \leq y \leq 2 + \sqrt{2}$.
(ii) B.
$$
\begin{array}{l}
z^{2} - x^{2} = a c + b d + 2 \sqrt{a b c d} - (a c + b d) \\
= 2 \sqrt{a b c d} ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,809 |
3. (i) (Grade 11) In $\triangle A B C$, the necessary and sufficient condition for
$$
\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=\cos ^{2} \frac{B}{2}
$$
to hold is ( ).
(A) $a c=b^{2}$
(B) $a^{2}+c^{2}=b^{2}$
(C) $a+c=2 b$
(D) $\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$
(ii) (Grade 12) In square $A B... | 3. (i) C.
$$
\begin{array}{l}
\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=\cos ^{2} \frac{B}{2} \\
\Leftrightarrow \frac{1-\cos A}{2}+\frac{1-\cos C}{2}=\cos B \\
\Leftrightarrow 1-\cos B=\frac{1}{2}(\cos A+\cos C) \\
\Leftrightarrow 2 \sin ^{2} \frac{B}{2}=\cos \frac{A+C}{2} \cdot \cos \frac{A-C}... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,810 |
Example 9 As shown in Figure 8, let $I$ be the incenter of $\triangle ABC$, $B_{1}$ and $C_{1}$ be the midpoints of sides $AB$ and $AC$ respectively, the extension of $C_{1}I$ intersects $AB$ at point $B_{2}$, and the extension of $B_{1}I$ intersects the extension of $AC$ at point $C_{2}$. Given that $S_{\triangle AB_{... | Solution: As shown in Figure 8, connect $A I$. Let $A C=b, A B=c$, $B C=a, A B_{2}=c^{\prime}, A C_{2}=b^{\prime}$, and the inradius of the incircle be $r$.
From $\frac{b c}{2} \sin A=S_{\triangle A B C}=S_{\triangle A B_{2} C_{2}}=\frac{b^{\prime} c^{\prime}}{2} \sin A$, we get $b c=b^{\prime} c^{\prime}$.
Notice that... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,811 |
4. If $\alpha \_\beta$ are acute angles, and
$$
\begin{array}{l}
3 \sin ^{2} \alpha+2 \sin ^{2} \beta=1, \\
3 \sin 2 \alpha-2 \sin 2 \beta=0,
\end{array}
$$
then $\sin (\alpha+2 \beta)=(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{3}}{2}$
(D) 1 | 4.D.
$$
\begin{array}{l}
\text { Given } \cos (\alpha+2 \beta)=\cos \alpha \cdot \cos 2 \beta-\sin \alpha \cdot \sin 2 \beta \\
=\cos \alpha \cdot 3 \sin ^{2} \alpha-\sin \alpha \cdot 3 \sin \alpha \cdot \cos \alpha=0,
\end{array}
$$
and $0<\alpha+2 \beta<\frac{3 \pi}{2}$, we have $\alpha+2 \beta=\frac{\pi}{2}$. There... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,812 |
5. There are 7 points on a plane, with no three points collinear. Using these points as vertices to form triangles, such that any two triangles have at most one common vertex and no common edges. Then the maximum number of triangles that can be formed is ( ).
(A) 3
(B) 7
(C) 21
(D) 35 | 5.B.
On the one hand, 7 points can connect to form $\frac{7(7-1)}{2}=21$ edges, and each edge can belong to at most one triangle, so 21 edges can form at most 7 triangles; on the other hand, with 1, 2, $\cdots$, 7 as vertices, 7 triangles that meet the conditions can actually be constructed. For example, by rotating t... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,813 |
6. The unit digit of $\left[(1+\sqrt{2})^{10}\right]$ is $(\quad)$ (where, $[x]$ denotes the greatest integer not exceeding $x$).
(A) 1
(B) 3
(C) 5
(D) 7 | 6.C.
Notice that $(1+\sqrt{2})^{10}=(3+2 \sqrt{2})^{5}$.
Let $a=3+2 \sqrt{2}, b=3-2 \sqrt{2}, S_{k}=a^{k}+b^{k}$.
Since $a$ and $b$ are the roots of the equation $x^{2}-6 x+1=0$, then
$$
\begin{array}{l}
a^{n}=6 a^{n-1}-a^{n-2}, b^{n}=6 b^{n-1}-b^{n-2}, \\
S_{n}=6 S_{n-1}-S_{n-2} .
\end{array}
$$
Note that $S_{0}=2, ... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,814 |
7. (i) (Grade 11) Let $A=\{2,4,7,8,13,15\}$. If a non-empty set $M$ satisfies that adding 4 to each element of $M$ results in a subset of $A$, and subtracting 4 from each element of $M$ also results in a subset of $A$, then $M=$ $\qquad$
(ii) (Grade 12) Randomly fill the numbers $1,2, \cdots, 9$ into the nine cells of ... | (ii) $\frac{1}{14}$.
In $1,2, \cdots, 9$, there are 5 odd numbers, which must occupy exactly one row and one column. There are 9 ways to choose any row and column, so $P=\frac{9 \times 5! \times 4!}{9!}=\frac{1}{14}$.
(ii) $\{11\}$.
$$
\begin{array}{l}
\text { Let } A_{1}=\{x \mid x=a-4, a \in A\} \\
=\{-2,0,3,4,9,11... | M=\{11\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,815 |
8. (i) (Grade 11) Given the function
$$
f(x)=2 \cos \left(\frac{k}{4} x+\frac{\pi}{3}\right)
$$
the smallest positive period is no greater than 2. Then the smallest positive integer value of $k$ is $\qquad$ .
(ii) (Grade 12) The line $y=k x-2$ intersects the parabola $y^{2}$ $=8 x$ at points $A$ and $B$. If the x-coor... | 8. (i) 13.
From the period $T=\frac{8 \pi}{k} \leqslant 2$, we get $k \geqslant 4 \pi > 12.56$.
Therefore, the positive integer $k \geqslant 13$.
(ii) $2 \sqrt{15}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Since $y=\frac{k y^{2}}{8}-2$, i.e., $k y^{2}-8 y-16=0$, then
$$
\begin{array}{l}
y_{1}+y_... | 2 \sqrt{15} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,816 |
9. (i) (Grade 10) $\frac{\cos 10^{\circ}}{\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}}=$ $\qquad$ .
(ii) (Grade 11) Using the digits $1,2,3,4,5,6$ to number the six faces of a cube. If a numbered cube can be rotated to match the numbering of another cube, they are considered the same numbering method. Therefore, the numbe... | 9. (i) $\sqrt{2}$.
Since $\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}$
$=\cos 40^{\circ} \sqrt{1-\cos 80^{\circ}}$
$=\cos 40^{\circ} \cdot \sqrt{2} \sin 40^{\circ}=\frac{\sqrt{2}}{2} \sin 80^{\circ}=\frac{\sqrt{2}}{2} \cos 10^{\circ}$,
therefore, $\frac{\cos 10^{\circ}}{\cos 40^{\circ} \sqrt{1-\sin 10^{\circ}}}=\sqrt{2}$... | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,817 |
10. If $\frac{1^{2}+3^{2}+\cdots+(2 n-1)^{2}}{2^{2}+4^{2}+\cdots+(2 n)^{2}}=\frac{13}{14}$, then
$$
n=
$$ | 10.20.
From $\sum_{k=1}^{2 n} k^{2}=\frac{27}{14} \sum_{k=1}^{n}(2 k)^{2}$, we get
$$
\begin{array}{l}
\frac{2 n(2 n+1)(4 n+1)}{6} \\
=\frac{27}{14} \times 4 \times \frac{n(n+1)(2 n+1)}{6},
\end{array}
$$
which means $4 n+1=\frac{27}{7}(n+1)$.
Therefore, $n=20$. | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,818 |
11. The sequence $\left\{a_{n}\right\}$ satisfies -
$$
\begin{array}{l}
a_{1}=8, a_{2}=26, \\
a_{n}=a_{n-1}+a_{n-2}+a_{n-1} a_{n-2} .
\end{array}
$$
Then $a_{10}=$ | 11. $3^{144}-1$
According to the condition, we have $a_{n}+1=\left(a_{n-1}+1\right)\left(a_{n-2}+1\right)$ and $\left\{a_{n}\right\}$ is a strictly increasing sequence of positive integers.
$$
\begin{array}{l}
\text { and } \log _{3}\left(a_{n}+1\right) \\
=\log _{3}\left(a_{n-1}+1\right)+\log _{3}\left(a_{n-2}+1\righ... | 3^{144}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,819 |
12. Let the function $f(x): \mathbf{R} \rightarrow \mathbf{R}$, and it satisfies for any $x, y \in \mathbf{R}$,
$$
\begin{array}{l}
f(x) f(y) \\
=f(2 x y+3)+3 f(x+y)-3 f(x)+6 x .
\end{array}
$$
Then $f(x)=$ | 12. $f(x)=2 x+3$.
Swapping $x$, we get
$$
-3 f(x)+6 x=-3 f(y)+6 y \text {. }
$$
Thus, $f(x)-2 x$ is a constant.
Let $f(x)=2 x+a$.
$$
\begin{aligned}
\text { By } & (2 x+a)(2 y+a) \\
= & 2(2 x y+3)+a+3(2 x+2 y+a)- \\
& 3(2 x+a)+6 x,
\end{aligned}
$$
we get
$$
\begin{array}{l}
2(x+y)(3-a)=a^{2}-a-6 \\
=(a-3)(a+2),
\en... | f(x)=2x+3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,820 |
13. Find the smallest positive number $a$ such that there exists a positive number $b$, for which the inequality
$$
\sqrt{1-x}+\sqrt{1+x} \leqslant 2-b x^{a}
$$
holds for all $x \in [0,1]$.
For the value of $a$ found, determine the largest positive number $b$ that satisfies the above inequality. | $$
\begin{array}{l}
\sqrt{1-x}+\sqrt{1+x}-2 \\
=\frac{(\sqrt{1-x}+\sqrt{1+x}-2)(\sqrt{1-x}+\sqrt{1+x}+2)}{\sqrt{1-x}+\sqrt{1+x}+2} \\
=\frac{2\left(\sqrt{1-x^{2}}-1\right)}{\sqrt{1-x}+\sqrt{1+x}+2} \\
=\frac{-2 x^{2}}{(\sqrt{1-x}+\sqrt{1+x}+2)\left(\sqrt{1-x^{2}}+1\right)} .
\end{array}
$$
To make $\sqrt{1-x}+\sqrt{1+... | a=2, b=\frac{1}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,821 |
Example 1 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=\frac{1}{2}, a_{n}=a_{n-1}+\frac{1}{n^{2}} a_{n-1}^{2} \text {. }
$$
Prove: $\frac{n+1}{n+2}<a_{n}<n$. | It is known that $a_{n}>a_{n-1}>0$.
$$
\begin{array}{l}
\text { Also, } a_{n}=a_{n-1}+\frac{1}{n^{2}} a_{n-1}^{2} . \\
\frac{n^{2}}{n^{2}+n-1} a_{n}. \text{ Therefore, }
$$
a_{n}>a_{n-1}+\frac{1}{n^{2}} a_{n-1} \cdot \frac{n^{2}}{n^{2}+n-1} a_{n},
$$
which means $\frac{1}{a_{n-1}}-\frac{1}{a_{n}}>\frac{1}{n^{2}+n-1}>\... | \frac{n+1}{n+2}<a_{n}<n | Inequalities | proof | Yes | Yes | cn_contest | false | 719,822 |
14. As shown in Figure 1, quadrilaterals $ABCD$ and $BCFE$ are both right trapezoids, where $DC \parallel AB$, $CF \parallel BE$, $BC \perp AB$, $EF \perp BE$, and points $A$, $E$, $D$, $F$ lie on the same line. $P$ is a point inside the trapezoid such that $AP = AB$ and $DP = DC$. Prove that $\angle APE = \angle DPF$. | 14. As shown in Figure 4, with point $A$ as the center and $A B$ as the radius, draw $\odot A$. Since $A P=A B$, point $P$ lies on $\odot A$. With point $D$ as the center and $D C$ as the radius, draw $\odot D$. Since $D P=D C$, point $P$ lies on $\odot D$. Let the other intersection point of the two circles be $Q$, th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,823 |
15. The sequence of positive integers $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{2}=15, \\
2+\frac{4}{a_{n}+1}<\frac{a_{n}}{a_{n}-4 n+2}+\frac{a_{n}}{a_{n+1}-4 n-2} \\
<2+\frac{4}{a_{n}-1} .
\end{array}
$$
Calculate: $S_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}$. | 15. First, find the general term $a_{n}$.
When $n=1$, the given condition becomes
$$
2+\frac{4}{a_{1}+1}1, \\
\frac{2}{a_{1}-2}+\frac{a_{1}}{9}-\frac{4}{a_{1}-1}0, \text{ i.e., } \left(a_{1}-10\right)\left(a_{1}^{2}-11\right)>2.
$$
Thus, $a_{1} \leqslant 3$ or $a_{1} \geqslant 11$.
From equation (3), we have $a_{1}^{3... | \frac{n}{2 n+1} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,824 |
1. In the Cartesian coordinate system $x O y$, the line $l: y=$ $-\frac{4}{3} x+4$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively. If the line $l$ is rotated counterclockwise by $90^{\circ}$ around point $O$, it intersects the $y$-axis at point $A^{\prime}$ and the line $l$ at point $C$. T... | $-1 . B$.
As shown in Figure 3, let the line $l$ rotate around point $O$ and intersect the $x$-axis at $B'$. It is easy to know that $O A=3$, $O B=4$, so $O A'=3$, $O B'=4$. Therefore, $A' B=1$.
By the Pythagorean theorem, we have
$$
A B=5 \text {. }
$$
By the properties of rotation, we know $\angle A' C B=90^{\circ}$... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,825 |
2. In $\triangle A B C$, $A B=A C=a, B C=b$, $\angle A=36^{\circ}$. Let $m=\frac{a+b}{a-b}, n=\frac{(a+b)^{2}}{a b}, p=\frac{a^{3}}{b^{3}}$. Then the size relationship of $m, n, p$ is $(\quad$.
(A) $m>n>p$
(B) $n>p>m$
(C) $p>m>n$
(D) $m=n=p$ | 2.D.
From the problem, we know $\angle A B C=\angle A C B=72^{\circ}$. As shown in Figure 4, construct $B D$ to bisect $\angle A B C$. Then
$$
\begin{array}{l}
\angle A B D=\angle D B C=36^{\circ}, \\
\angle B D C=72^{\circ} .
\end{array}
$$
Therefore, $\triangle B C D \backsim \triangle A B C$
$$
\begin{aligned}
& \... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,826 |
3. In $\triangle A B C$, $B D$ bisects $\angle A B C$ and intersects $A C$ at point $D, C E$ bisects $\angle A C B$ and intersects $A B$ at point $E$. If $B E+$ $C D=B C$, then the degree measure of $\angle A$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $90^{\circ}$ | 3.C.
As shown in Figure 5, let $B D$ and $C E$ intersect at point $O$. Take a point $F$ on $B C$ such that $B F = B E$, and connect $O F$.
From the given information, we know $\angle 1 = \angle 2$, and $B O = B O$, therefore,
$$
\begin{array}{l}
\triangle B E O \cong \triangle B F O, \\
\angle E O B = \angle F O B.
\... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,827 |
4. For the convex quadrilateral $A B C D$, the four vertices satisfy: the product of the distances from each vertex to the other three vertices is equal. Then the quadrilateral $A B C D$ is ( ).
(A) Isosceles trapezoid
(B) Rhombus
(C) Rectangle
(D) Square | 4.C.
From the problem, we have
$$
\begin{array}{l}
A B \cdot A C \cdot A D=B A \cdot B C \cdot B D \\
=C A \cdot C B \cdot C D=D A \cdot D B \cdot D C . \\
\text { Then }\left\{\begin{array}{l}
A C \cdot A D=B C \cdot B D, \\
A B \cdot B D=A C \cdot C D, \\
C A \cdot C B=D A \cdot D B .
\end{array}\right.
\end{array}
... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,828 |
5. If a pair of different natural numbers $a, b$ satisfies that their arithmetic mean $A=\frac{a+b}{2}$ and geometric mean $G=\sqrt{a b}$ are both two-digit numbers, and one of $A$ and $G$ can be obtained by swapping the tens and units digits of the other, then such a pair of natural numbers is called a "good pair". Ho... | 5.A.
From the problem, we have $\left\{\begin{array}{l}a+b=2 A, \\ a b=G^{2} .\end{array}\right.$
Then $a, b$ are the roots of the quadratic equation $x^{2}-$ $2 A x+G^{2}=0$.
Solving for $a, b$ gives $a, b=A \pm \sqrt{A^{2}-G^{2}}$.
Thus, $\sqrt{A^{2}-G^{2}}$ must be a natural number.
Let $A=10 p+q(1 \leqslant p, q \... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,829 |
6. As shown in Figure $1, P$ is a point on the arc $\overparen{A B}$ of the circumcircle of the equilateral $\triangle A B C$. Then the relationship between $B C$ and $P C$ is ( ).
(A) $B C \leqslant \frac{\sqrt{3}}{2} P C$
(B) $B C \leqslant \frac{\sqrt{3}}{3} P C$
(C) $B C \geqslant \frac{\sqrt{3}}{2} P C$
(D) $B C \... | 6. C.
First, prove that in $\triangle ABC$, if $\angle A=60^{\circ}$, then $BC^{2}=AB^{2}+AC^{2}-AB \cdot AC$.
As shown in Figure 6, draw $CD \perp AB$, then $AD=\frac{1}{2} AC$,
$$
CD=\frac{\sqrt{3}}{2} AC .
$$
Thus, $BC^{2}=BD^{2}+CD^{2}$
$=(AB-AD)^{2}+CD^{2}$
$=\left(AB-\frac{1}{2} AC\right)^{2}+\left(\frac{\sqrt... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,830 |
1. For the system of equations $\left\{\begin{array}{l}x^{2}+2 a y=5, \\ y-x=6 a\end{array}\right.$ to have positive integer solutions, the value of $a$ is . $\qquad$ | 2.1. $\frac{1}{2}$ or $\frac{1}{6}$.
From the second equation, we get $a=\frac{y-x}{6}$, substituting into the first equation and rearranging yields
$$
y^{2}-x y+3 x^{2}-15=0 \text {. }
$$
If this equation has positive integer solutions, then $\Delta=60-11 x^{2}$ must be a perfect square.
It is easy to see that $x=1,2... | \frac{1}{2} \text{ or } \frac{1}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,831 |
2. The incenter of $\triangle A B C$ is $I$, and the angle bisector of $\angle B$ intersects $A C$ at point $P$. If $A P+A B=B C$, and $A B=3, B C=$ 5, then the value of $A I$ is $\qquad$ . | 2.2.
As shown in Figure 7, on segment $B C$, take $B A^{\prime}=B A$, extend $A I$ to intersect $B C$ at $Q$, and connect $P A^{\prime}$.
$$
\begin{array}{l}
\text { Given } B A^{\prime}=B A, \\
\angle A B P=\angle A^{\prime} B P, \\
B P=B P,
\end{array}
$$
we have $\triangle A B P \cong \triangle A^{\prime} B P$.
Th... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,832 |
1. In $\triangle A B C$, if $A B \cdot B C<0$, then the shape of $\triangle A B C$ is ( ).
(A) acute triangle
(B) right triangle
(C) obtuse triangle
(D) cannot be determined | -1.D.
From $\boldsymbol{A B} \cdot \boldsymbol{B} \boldsymbol{C}<0$, we can only conclude that the exterior angle of $\angle A B C$ is obtuse, and we cannot determine the shape of $\triangle A B C$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,833 |
2. Let $M=\sum_{k=1}^{n} \frac{k}{k^{4}+k^{2}+1}$,
$$
N=\frac{1}{n^{2}+n+1} \sum_{k=1}^{n} k \text {. }
$$
Then the relationship between $M$ and $N$ is ( ).
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Cannot be determined, depends on the value of $n$ | 2.B.
$$
\begin{array}{l}
M=\sum_{k=1}^{n} \frac{k}{k^{4}+k^{2}+1}=\sum_{k=1}^{n} \frac{k}{\left(k^{2}+1\right)^{2}-k^{2}} \\
=\sum_{k=1}^{n} \frac{1}{2}\left(\frac{1}{k^{2}-k+1}-\frac{1}{k^{2}+k+1}\right) \\
=\frac{1}{2}\left(1-\frac{1}{n^{2}+n+1}\right)=\frac{1}{2} \cdot \frac{n^{2}+n}{n^{2}+n+1} \\
=\sum_{k=1}^{n} \f... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,834 |
5. A party has prepared $2 n$ gifts, which are evenly divided into two strings and publicly hung on the wall. Each winner can (and can only) choose one gift from the bottom of either string. If there are $2 n$ winners selecting these $2 n$ gifts, then there are $\qquad$ different ways of selection. | 5. $\mathrm{C}_{2 n}^{n}$.
Suppose the two strings of gifts from bottom to top are:
$$
a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n} \text {. }
$$
Consider the $2n$ winners as $2n$ positions. Choosing $n$ positions from $2n$ positions, there are $C_{2 n}^{n}$ ways to choose. Arranging $a_{1}, a_{2}, \cdots... | \mathrm{C}_{2 n}^{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,835 |
11. (16 points) As shown in Figure 2, given the parabola $y^{2}=2 p x$ $(p>0), A B$ is a chord passing through the focus $F$. If $A B$ makes an angle $\theta(0<\theta \leqslant \frac{\pi}{2})$ with the $x$-axis, find $\angle A O B$. | 11. When $0<\theta<\frac{\pi}{2}$, the equation of $AB$ can be written as $y=\tan \theta\left(x-\frac{p}{2}\right)$, i.e.,
$$
x=\cot \theta \cdot y+\frac{p}{2} \text {. }
$$
This result also holds for $\theta=\frac{\pi}{2}$.
Substituting equation (1) into the parabola equation yields
$$
y^{2}-2 p \cot \theta \cdot y-p... | \pi-\arctan \frac{4}{3 \sin \theta} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,836 |
12. (16 points) Find the smallest positive integer $n$ such that: for any $n$ points $A_{1}, A_{2}$, $\cdots, A_{n}$ taken on the circumference of $\odot O$, among the $\mathrm{C}_{n}^{2}$ angles $\angle A_{i} O A_{j}(1 \leqslant i<j \leqslant n)$, at least 2007 are not greater than $120^{\circ}$. | 12. First, when $n=90$, as shown in Figure 7, let $AB$ be the diameter of $\odot O$. Take 45 points near points $A$ and $B$, respectively. At this time, there are only $2 \mathrm{C}_{45}^{2}=45 \times 44=1980$ angles that do not exceed $120^{\circ}$. Therefore, $n=90$ does not satisfy the problem's requirements.
Second... | 91 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,837 |
1. Given $f(x)=3^{x-m}$ with the domain $[2,4]$, and it passes through the point $A(2,1)$. Then the range of $F(x)=\left(f^{-1}(x)\right)^{2}-$ $f^{-1}(x)$ is $(\quad)$.
(A) $[2,9]$
(B) $[2,12]$
(C) $[1,9]$
(D) $[1,12]$ | 1. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,838 |
2. Given the curve $y=x^{3}-x$, draw a tangent line to this curve from a point $A(t, 0)$ on the $x$-axis. The maximum number of tangent lines that can be drawn is $(\quad$.
(A) 1
(B) 2
(C) 3
(D) 4 | 2. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Calculus | MCQ | Yes | Yes | cn_contest | false | 719,839 |
4. In the acute triangle $\triangle A B C$, $B C=1, \angle A=$ $2 \angle B$. Then the range of $A C$ is ( ).
(A) $\left(0, \frac{\sqrt{2}}{2}\right)$
(B) $\left(\frac{1}{2}, 1\right)$
(C) $\left(\frac{1}{2}, \frac{\sqrt{3}}{3}\right)$
(D) $\left(\frac{\sqrt{3}}{3}, \frac{\sqrt{2}}{2}\right)$ | 4.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,841 |
6. A frog is at point $A$ of a regular hexagon $A B C D E F$, each time it jumps to an adjacent vertex, and stops when it reaches point $D$ or after 5 jumps. How many different ways of jumping are there?
(A) 26
(B) 28
(C) 30
(D) 32 | 6. A | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,843 |
6. Given $f(x)=x^{3}$ on $[1, b]$ satisfies
$$
\frac{f(b)-f(1)}{b-1}=f^{\prime}(t)(1<t<b) \text {. }
$$
then $\lim _{b \rightarrow 1} \frac{t-1}{b-1}=$ | 6. $\frac{1}{2}$.
$$
\begin{array}{l}
\lim _{b \rightarrow 1} \frac{t-1}{b-1}=\lim _{b \rightarrow 1} \frac{\sqrt{\frac{1}{3}\left(b^{2}+b+1\right)}-1}{b-1} \\
=\lim _{b \rightarrow 1} \frac{\sqrt{b^{2}+b+1}-\sqrt{3}}{\sqrt{3}(b-1)} \\
=\lim _{b \rightarrow 1} \frac{(b-1)(b+2)}{\sqrt{3}(b-1)\left(\sqrt{b^{2}+b+1}+\sqrt... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 719,846 |
10. In $\triangle A B C$, $\angle A, \angle B$ satisfy $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}=2$.
Then the range of $\sin A+\sin B$ is | 10. $(1, \sqrt{2}]$ | (1, \sqrt{2}] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,848 |
12. If the equation with respect to $x$
$$
x^{2}+a x+\frac{1}{x^{2}}+\frac{a}{x}+b+2=0(a, b \in \mathbf{R})
$$
has real roots, then the minimum value of $a^{2}+b^{2}$ is $\qquad$. | $12 . \frac{16}{5}$ | \frac{16}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,850 |
13. Given the equation about $x$
$$
8^{x}-m 4^{x}+\left(m^{2}-3\right) 2^{x}=0(m \in \mathbf{R}) \text {. }
$$
(1) If the equation has no real roots, find the range of $m$;
(2) If the equation has a unique real root, find the range of $m$. | 13. Let $t=2^{x}$, then the original equation is transformed into an equation about $t$:
$$
t^{2}-m t+m^{2}-3=0 \text {. }
$$
Then: $\Delta=12-3 m^{2}$.
(1) The original equation has no real roots if and only if equation (1) has no real roots or only non-positive roots.
When equation (1) has no real roots, $m2$.
When ... | m \leqslant-\sqrt{3} \text { or } m>2 \text { for no real roots, and } m=2 \text { or } -\sqrt{3}<m \leqslant \sqrt{3} \text { for a unique real root} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,851 |
14. Given $O(0,0), A(1,0), B(0,2)$. Let $P_{1}, P_{2}, P_{3}$ be the midpoints of $AB, OB, OP_{1}$ respectively. For any positive integer $n$, $P_{n+3}$ is the midpoint of the segment $P_{n} P_{n+1}$. Let $P_{n}(x_{n}, y_{n})$, and set
$$
\begin{array}{l}
a_{n}=\frac{x_{n}}{2}+x_{n+1}+x_{n+2}, \\
b_{n}=a_{n} y_{n}+y_{n... | 14. (1) From the problem, we know
$$
x_{1}=\frac{1}{2}, x_{2}=0, x_{3}=\frac{1}{4}, y_{1}=1, y_{2}=1, y_{3}=\frac{1}{2} \text {. }
$$
From $x_{n+3}=\frac{x_{n}+x_{n+1}}{2}$, we get $x_{4}=\frac{1}{4}, x_{5}=\frac{1}{8}$.
Therefore, $a_{1}=\frac{1}{2}, a_{2}=\frac{1}{2}, a_{3}=\frac{1}{2}$.
(2) From $a_{n+1}=\frac{x_{n... | c_{n}=(-1)^{n} \frac{1}{4^{n}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,852 |
15. As shown in Figure $1, C$ is the midpoint of the semicircular arc, $P$ is a point on the extension of the diameter $B A$, a tangent line $P D$ is drawn through $P$, with $D$ as the point of tangency, the angle bisector of $\angle B P D$ intersects $A C$ and $B C$ at points $E$ and $F$ respectively. Prove that the c... | 15. As shown in Figure 2, connect $D B$, $D A$, $D E$, $D O$, and $D F$.
Since $C$ is the midpoint of the semicircle arc and $P D$ is a tangent, we have,
$$
\begin{array}{l}
O C \perp A B, P D \perp O D . \\
\text { Therefore, } \angle D P B = \angle C O D . \\
\text { Also, } P F \text { bisects } \angle D P B, \text... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,853 |
16. Let $F_{1}$ and $F_{2}$ be the left and right foci, respectively, of the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \quad(a>0, b>0)
$$
$O$ is the origin, point $P$ is on the left branch of the hyperbola, point $M$ is on the right directrix, and it satisfies
$$
F_{1} \boldsymbol{O}=\boldsymbol{P M}, O P... | 16. (1) From $\boldsymbol{F}_{1} \boldsymbol{O}=\boldsymbol{P M}$, we know that quadrilateral $P F_{1} O M$ is a parallelogram.
Also, $\boldsymbol{O P}=\lambda\left(\frac{O F_{1}}{\left|O F_{1}\right|}+\frac{\boldsymbol{O M}}{|O M|}\right)(\lambda>0)$, so $O P$ is the angle bisector of $\angle F_{1} O M$.
Therefore, q... | y= \pm \sqrt{5} x-3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,854 |
1. $a$, $b$, $c$ are distinct positive numbers, $a^{2}+c^{2}=$ $2 b c$. Then which of the following relationships may hold? ( ).
(A) $a>b>c$
(B) $b>c>a$
(C) $b>a>c$
(D) $a>c>b$ | -1.C.
If $a>b$, then $a^{2}+c^{2}>b^{2}+c^{2}>2bc$, which does not meet the condition, so we exclude options $(\mathrm{A})$ and $(\mathrm{D})$.
From $a^{2}-c^{2}=2c(b-c)$, we know that $a-c$ and $b-c$ have the same sign, thus we exclude option (B). Moreover, when $b>a>c$, $a^{2}+c^{2}=2bc$ might hold, for example, tak... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,855 |
2. Given $f(x)=\frac{1+x}{1-x}$, let $f_{1}(x)=f(x)$, $f_{k+1}(x)=f\left(f_{k}(x)\right)(k=1,2, \cdots)$. Then $f_{2007}(x)$ $=(\quad)$.
(A) $\frac{1+x}{1-x}$
(B) $\frac{x-1}{x+1}$
(C) $x$
(D) $-\frac{1}{x}$ | 2. B.
$$
\begin{array}{l}
f_{1}(x)=\frac{1+x}{1-x}, f_{2}(x)=\frac{1+f_{1}}{1-f_{1}}=-\frac{1}{x}, \\
f_{3}(x)=\frac{1+f_{2}}{1-f_{2}}=\frac{x-1}{x+1}, f_{4}(x)=\frac{1+f_{3}}{1-f_{3}}=x . \\
\text { Therefore, } 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}(x)=x .
\e... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,856 |
Three. (20 points). Given $a=(5 \cos \alpha, 4 \sin \alpha)$, $\boldsymbol{b}=(5 \cos \beta, 4 \sin \beta)$. Find the maximum value of $|\boldsymbol{a}-\boldsymbol{b}|$.
| $$
\begin{array}{l}
\text { III. Solution 1: }|\boldsymbol{a}-\boldsymbol{b}| \\
=\sqrt{5(\cos \alpha-\cos \beta)^{2}+4(\sin \alpha-\sin \beta)^{2}} \\
\leqslant \sqrt{5(\cos \alpha-\cos \beta)^{2}+5(\sin \alpha-\sin \beta)^{2}} \\
=\sqrt{10-10(\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta)} \\
=\sqrt{10[1-... | 2 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,857 |
3. Let $\alpha$ be an acute angle, $x=\frac{\sqrt{\sin \alpha}+\sqrt{\cos \alpha}}{\sqrt{\sin \alpha+\cos \alpha}}$, $y=\sqrt{\sin \alpha+\cos \alpha}$, $z=\frac{2 \sin \alpha \cdot \cos \alpha}{\sin \alpha+\cos \alpha}$.
Then the order of magnitude of $x$, $y$, and $z$ is ( ).
(A) $x \geqslant y \geqslant z$
(B) $x \g... | 3. A.
$$
\begin{array}{l}
\frac{x}{y}=\frac{\sqrt{\sin \alpha}+\sqrt{\cos \alpha}}{\sin \alpha+\cos \alpha} \geqslant \frac{\sin \alpha+\cos \alpha}{\sin \alpha+\cos \alpha}=1, \\
z=\frac{2 \sin \alpha \cdot \cos \alpha}{\sin \alpha+\cos \alpha} \leqslant \frac{2 \sin \alpha \cdot \cos \alpha}{2 \sqrt{\sin \alpha \cdot... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,858 |
4. Use red, yellow, blue, and green to color the four small squares $A, B, C, D$ in Figure 1 (using any of the colors), such that adjacent areas (squares with a common edge) are different colors. The number of different coloring methods is ( ).
(A) 24
(B) 36
(C) 72
(D) 84 | 4.D.
Choosing two colors has $\mathrm{C}_{4}^{2}$ ways, selecting diagonals for one color has 2 ways, totaling $2 \mathrm{C}_{4}^{2}=12$ ways;
Choosing three colors has $\mathrm{C}_{4}^{3}$ ways, among which one color is repeated and has $\mathrm{C}_{3}^{1}$ ways to choose, the repeated color has 2 ways to choose dia... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,859 |
5. For a regular quadrilateral pyramid, the area ratio of a diagonal section to a side face is $\sqrt{6}: 2$. Then the angle between the side face and the base is ( ).
(A) $\frac{\pi}{3}$
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{6}$
(D) $\frac{\pi}{12}$ | 5. A.
As shown in Figure 3, let the side length of the square base be 1, the height of the pyramid be $h$, and the height of the triangular side be $l$. Then
$$
\begin{array}{c}
A C=\sqrt{2}, \\
\frac{\sqrt{6}}{2}=\frac{\sqrt{2} h}{l} . \\
\text { Hence } \sin \angle P M H=\frac{h}{l} \\
=\frac{\sqrt{3}}{2}, \angle P ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,860 |
6. The smallest element of the set of positive integers $A_{k}$ is 1, and the largest element is 2007, and the elements can be arranged in an arithmetic sequence with a common difference of $k$. Then the number of elements in the union $A_{17} \cup A_{59}$ is ( ).
(A) 119
(B) 120
(C) 151
(D) 154 | 6. C.
Let $\left|A_{k}\right|$ denote the number of elements in the set $A_{k}$, and suppose
$$
\left|A_{k}\right|=n+1 \text {. }
$$
From $2007=1+n k$, we get $n=\frac{2006}{k}$. Therefore,
$$
\begin{array}{l}
\left|A_{17}\right|=\frac{2006}{17}+1=119, \\
\left|A_{59}\right|=\frac{2006}{59}+1=35, \\
\left|A_{17} \cap... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,861 |
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