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3. R1 In $\triangle A B C$, the incircle touches the hypotenuse $A B$ at point $P, A P=7, B P=5$. Then the radius of the incircle is equal to ( ).
$(\Lambda) 6$
(B) $\sqrt{35}$
(C) $-6+\sqrt{71}$
(I))2 | 3.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | not found | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,596 |
Example 8 As shown in Figure $8, \triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ have three sides $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$ respectively, and $\angle B=\angle B^{\prime}$, $\angle A+\angle A^{\prime}=180^{\circ}$. Try to prove: $a a^{\prime}=b b^{\prime}+c c^{\prime}$. | Analysis: Since $\angle B=\angle B B^{\prime}, \angle A+\angle \angle A^{\prime}=$ $180^{\circ}$, by analogy, we think of Ptolemy's theorem and construct a cyclic quadrilateral for the proof.
Proof: Construct the circumcircle of $\triangle A B C$, and draw $C D / / A B$ intersecting the circle at $D$. Connect $A D$ an... | a a^{\prime}=b b^{\prime}+c c^{\prime} | Geometry | proof | Yes | Yes | cn_contest | false | 710,599 |
Example 1 The elements of set $A$ are all integers, the smallest of which is 1, and the largest is 100. Except for 1, each element is equal to the sum of two numbers (which can be the same) in set $A$. Find the minimum number of elements in set $A$.
| Solution: Construct a set with as many elements as possible to satisfy the conditions, such as $\{1,2,3,5,10,20,25,50, 9\}$.
Extend $\left\{1,2, x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, 100\right\}$ to also satisfy the conditions, then $x_{1} \leqslant 4, x_{2} \leqslant 8, x_{3} \leqslant 16, x_{4} \leqslant 32, x_{5}$ $\l... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,610 |
2. The incircle of $\triangle A B C$ touches sides $A B, B C, C A$ at $D, E, F$ respectively. The incircle of $\triangle D E F$ touches sides $D E, E F, F D$ at $P, Q, R$ respectively. Given that $\triangle A B C \backsim \triangle P Q R$, and the perimeter of $\triangle P Q R$ is 3. Then the area of $\triangle A B C$ ... | $2.4 \sqrt{3}$ | 2.4 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,611 |
II. (Full marks 15 points) $f(n)$ is a strictly increasing function defined on $\mathbf{N}$ and taking integer values (if for any $x_{1}, x_{2} \in A$, when $x_{1}<x_{2}$, we have $f\left(x_{1}\right)<f\left(x_{2}\right)$, then $f(x)$ is called a strictly increasing function on $A$). When $m, n$ are coprime, $f(m n)=f(... | II. H problem, $j(n)$ is defined on $\mathbf{N}$ with integer values and is a strictly increasing function.
$\because f(19)=f(1 \times 19)=f(1) \cdot f(19)$, but $f(19)=$ $19 \neq 0$,
$$
\therefore f(1)=1 \text {. }
$$
At this point, by $f(1)<f(2)<\cdots<f(19), f(1)=1$, $f(19)=19$, and $f(2), f(3), \cdots, f(18)$ are ... | 1862 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,615 |
)As shown in Fig. 1, in the quadrilateral pyramid $S-ABCD$, the side length of the square base $ABCD$ is $a$, and the side edges $SA=SB=SC=SD=2a$, $M$ is the midpoint of edge $SA$, and $N$ is the midpoint of edge $SC$. Find the cosine value of the angle formed by the skew lines $DM$ and $BN$. | Three, as shown in Figure 1, extend $B N$ to $P$ such that $B N = N P$, and extend $D M$ to $Q$ such that $D M = M Q$. Then $B C P S$ and $A D S Q$ are both parallelograms.
$\therefore S P \| B C$ II $A D \xlongequal{\|} Q S$.
Therefore, $Q$, $S$, and $P$ are collinear. $P Q = 2 B C = 2 A D$.
Since $A B = B C = C D = D... | \frac{1}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,616 |
Four, (Full marks 15 points) If $a<b<c<d$, prove that for any real number $t \neq -1$, the equation in $x$
$$
(x-a)(x-c)+t(x-b)(x-d)=0
$$
has two distinct real roots. | $$
\begin{array}{l}
\text { IV. The original equation is rearranged as } \\
(1+t) x^{2}-[(a+c)+(b+d) t] x+(a c+ \\
b d t)=0 \text {. } \\
\because t \neq-1 \text {, } \\
\therefore t+1 \neq 0 \text {. } \\
\end{array}
$$
Equation (1) is a quadratic equation in $x$ for any $t \neq-1$. The discriminant of equation (1) i... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,617 |
Five. (Full marks 15 points) There are 1998 athletes with numbers from 1 to 1998. Select some of these athletes to participate in the honor guard, but ensure that among the remaining athletes, no one's number is equal to the product of the numbers of any two other athletes. How many athletes can be selected for the hon... | Five, the selectable numbers are $2,3,4, \cdots, 43,44$, these 43 athletes can be chosen as flag bearers to meet the requirements of the problem. The reason is as follows:
After selecting these 43 athletes as flag bearers, the product of any two numbers among the remaining athletes' numbers (excluding number 1) will b... | 43 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,618 |
1. Choose the correct answer (7 points per question, total 42 points)
1. The sum of 99 consecutive natural numbers equals $a b c d$. If $a, b, c, d$ are all prime, then the minimum value of $a+b+c+d$ is ( ).
(A) 63
(B) 70
(C) 86
(D) 97 | -1 (B).
$$
\begin{array}{l}
\begin{aligned}
a b c d & =x+(x+1)+(x+2)+\cdots+(x+98) \\
& =99 x+(1+2+\cdots+98)=99 x+99 \times 49 \\
& =99 \times(x+49)=3 \times 3 \times 11 \times(x+49)
\end{aligned} \\
\begin{aligned}
\therefore x & =4,
\end{aligned} \\
\text { At this point } a+b+c+d=3+3+11+53=70 .
\end{array}
$$ | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 710,619 |
2. In isosceles $\triangle A B C$, the vertex angle $\angle A=108^{\circ}, B C$ $=m, A B=A C=n$, let $x=\frac{m+n}{m-n}, y=$ $\frac{(m+n)^{2}}{m n}, z=\frac{m^{3}}{n^{3}}$. Then the size relationship of $x 、 y 、 z$ is ( ).
(A) $x>y>z$
(B) $x<y<z$
(C) $x=y \neq z$
(D) $x=y=z$ | 2. (D).
Take $D$ on $BC$ such that $DC = AC = n$. It is easy to get
$$
\begin{array}{l}
\triangle D B A \sim \triangle A B \\
\Rightarrow (m-n): n = n: m \\
\Rightarrow n^{2} = m(m-n) \\
\Rightarrow m^{2} = n(m+n) \\
\Rightarrow \frac{m+n}{m-n} = \frac{m^{3}}{n^{3}} . \\
\text { Also } \frac{(m+n)^{2}}{m n} = \frac{[... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,620 |
Example 2 For a finite set $A$, there exists a function $f: N \rightarrow$ $A$, with the following property: if $i, j \in N$, and $|i-j|$ is a prime number, then $f(i) \neq f(j)$. How many elements must the set $A$ have at least? | Solution: Since the absolute value of the difference between any two numbers among $1,3,6,8$ is a prime number, according to the problem, $f(1)$, $f(3)$, $f(6)$, and $f(8)$ are four distinct elements in $A$, thus $|A| \geqslant 4$.
On the other hand, if we let $A=\{0,1,2,3\}$, and the mapping $f: N \rightarrow A$ is d... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,621 |
3. Real numbers $a, b, c$ are pairwise distinct, and the coordinates of the three points are respectively: $A(a+b, c), B(b+c, a), C(c+a, b)$. Then the positional relationship of these three points is ( ).
(A) form an obtuse triangle
(B) form a right triangle
(C) form an equilateral triangle
(D) are collinear | 3. (D).
Let $a>b>c$, it is easy to get $AB=\sqrt{2}(a-c), AC=$ $\sqrt{2}(b-c), BC=\sqrt{2}(a-b)$. And $AB-AC=\sqrt{2}(a-$ b) $=BC, \therefore A, B, C$ are collinear. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,622 |
4. $a$, $b$, $c$ represent the three sides of a triangle, the nature of the roots of the equation $b^{2} x^{2} + \left(b^{2}+c^{2}-a^{2}\right) x+c^{2}=0$ is ( ).
(A) No real roots
(B) Two equal roots
(C) Two unequal roots
(D) Cannot be determined | 4. (A).
$$
\begin{array}{c}
\Delta=\left(b^{2}+c^{2}-a^{2}\right)^{2}-4 b^{2} c^{2}=4 b^{2} c^{2} . \\
{\left[\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}-1\right]=4 b^{2} c^{2}\left(\cos ^{2} A-1\right)<0 .}
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,623 |
5. On the plane, there are two points $P$ and $Q$. The number of triangles that can be drawn with $P$ as the circumcenter and $Q$ as the incenter is ( ).
(A) Only 1 can be drawn
(B) 2 can be drawn
(C) At most 3 can be drawn
(D) Infinitely many can be drawn | 5.(D).
Draw a circle $\odot P$ such that $Q$ is inside $\odot P$; take any point $A$ on $\odot P$, and let the ray $A Q$ intersect $\odot P$ at $D$; with $D$ as the center and $D Q$ as the radius, draw an arc intersecting $\odot P$ at $B$ and $C$, $\triangle A B C$ is the desired triangle. Since $\odot P$ can be of an... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,624 |
6. The graph of the inverse proportion function $y=\frac{k}{x}$ is an axially symmetric figure, one of its axes of symmetry is the graph of which of the following linear functions? ( ).
(A) $y=-k x$
(B) $y=|k| x$
(C) $y=\frac{-k}{|k|} x$
(D) $y=k x$ | 6. (C).
When $k>0$, the axis of symmetry is $y=-x$; when $k>0$, the axis of symmetry is $y=x$. Therefore, it is $y=-\frac{k}{|k|} x$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,625 |
1. As shown in Figure 1, in trapezoid $ABCD$, $DC \parallel AB$, $DC : AB = 1 : 2$, $MN \parallel BD$ and bisects $AC$. If the area of trapezoid $ABCD$ is $\overline{ab}$, and $S_{\triangle AMN} = \overline{ba}$, then $\bar{a} + \bar{b} + \overline{ba} =$ | ニ、1.99.
$\triangle A M N \backsim \triangle A B D$, similarity ratio $=\frac{\frac{1}{2} A C}{\frac{2}{3} A C}=\frac{3}{4}$,
$$
\begin{array}{l}
S_{\triangle A M N}=\frac{9}{16} S_{\triangle A B D}=\frac{9}{16} \times \frac{2}{3} S_{A B C D}=\frac{3}{8} S_{A B C D} \Rightarrow(10 b \\
+a)=\frac{3}{8}(10 a+b) \Rightarro... | 99 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,626 |
One. (Full marks 20 points) As shown in Figure $3, \angle C A B=\angle A B D=$ $90^{\circ}, A B=A C+B D . A D$ intersects $B C$ at $P$, and a circle $\odot P$ is constructed to be tangent to $A B$. Determine whether the circle $\odot O$ with $A B$ as its diameter intersects, is internally tangent to, or contains $\odot... | -、$\odot O$ is internally tangent to $\odot P$.
Suppose $\odot P$ is tangent to $AB$ at $E$, and connect $PE$. Then $PE \perp AB$. Let $AC = a, BD = b (a \leqslant b)$. Then $AB = a + b$. Also, let the radius of $\odot O$ be $R$, and the radius of $\odot P$ be $r$.
$$
\begin{array}{l}
\text{Clearly, } R = \frac{1}{2}(a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,630 |
II. (Full marks 25 points) The smaller bases and the slant sides of two right trapezoids are both equal to 1, and the two smaller interior angles are complementary. Let their areas be $S_{1}$ and $S_{2}$, respectively. Prove that: $1<S_{1}+S_{2}<2$.
---
The smaller bases and the slant sides of two right trapezoids ar... | Two right trapezoids can be combined to form a rectangle with a square removed, or a large trapezoid with an isosceles right triangle removed (as shown in Figures 8 and 9). Using the former to prove this, we have:
$$
\begin{aligned}
S_{1}+S_{2}=S_{\text {trapezoidABCD }}- \\
S_{\text {squarePFDE }}=A B \cdot B C-1 .
\e... | 1<S_{1}+S_{2}<2 | Geometry | proof | Yes | Yes | cn_contest | false | 710,631 |
Example 1 In $\triangle A B C$, $A B=A C, \angle A=$ $80^{\circ}, D$ is a point inside the shape, and $\angle D A B=\angle I B A=$ $10^{\circ}$. Find the degree measure of $\angle A C D$.
---
Note: The symbol $\mathrm{I}$ in the original text seems to be a typo or a misprint, and it should likely be $D$ for consisten... | Solution 1: As shown in Fig. $\mathrm{i}$, let $E$ be the point symmetric to $D$ with respect to the perpendicular bisector of $BC$. It is easy to see that $\triangle ADE$ is an equilateral triangle.
From the given information, we have $DA = DB$. Therefore, $EA = EC = ED$, which means $E$ is the circumcenter of $\tria... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,632 |
Example 2 Given in $\triangle A B C$, $A B=A C$, $\angle A=20^{\circ}, D 、 E$ are points on $A C 、 A B$ respectively, $\angle D B C=60^{\circ}, \angle E C B=50^{\circ}$. Find the degree measure of $\angle B D E$. | Solution: As shown in Figure 5, let $E^{\prime}$ be the symmetric point of $E$ with respect to $AC$. Connect $E^{\prime}A$, $E^{\prime}B$, $E^{\prime}C$, $E^{\prime}D$, and $E^{\prime}E$. It is known that $\triangle E^{\prime}EC$ is an equilateral triangle, $BE^{\prime}$ is the perpendicular bisector of $EC$, and $D$ i... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,633 |
Example 4 A sequence of integers is defined as follows:
$$
\begin{array}{l}
a_{0}=0, a_{1}=1, \\
a_{n}=2 a_{n-1}+a_{n-2} \quad(n \geqslant 2, \cdots) .
\end{array}
$$
Prove: $2^{k} \mid a_{n}$ if and only if $2^{k} \mid n$.
(29th IMO Shortlist Problem)
Analysis: The previous example constructed a recurrence relation u... | Proof: From conditions (1) and (2), we have
$$
a_{n}=\frac{1}{2 \sqrt{2}}\left[(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}\right] \text {. }
$$
Let \( n=2^{k}(2 l+1) \) (where \( l=0,1,2, \cdots \); \( k=0,1,2, \cdots \)). Then
$$
\begin{array}{l}
a_{n}=\frac{(1+\sqrt{2})^{2^{k}}-(1-\sqrt{2})^{2^{k}}}{2 \sqrt{2}} . \\
{\left[(1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,634 |
1. Given $\operatorname{ctg} \theta=\sqrt[3]{7}\left(0^{\circ}<\theta<90^{\circ}\right)$. Then, $\frac{\sin ^{2} \theta+\sin \theta \cos \theta+2 \cos ^{2} \theta}{\sin ^{2} \theta+\sin \theta \cos \theta+\cos ^{2} \theta}=$ $\qquad$ . | \begin{aligned} \text { 2.1. } & \frac{13-\sqrt[3]{49}}{6} . \\ \text { Original expression } & =1+\frac{\cos ^{2} \theta}{\sin ^{2} \theta+\sin \theta \cos \theta+\cos ^{2} \theta} \\ & =1+\frac{\operatorname{ctg}^{2} \theta}{1+\operatorname{ctg} \theta+\operatorname{ctg}^{2} \theta} \\ & =1+\frac{\sqrt[3]{49}(\sqrt[3... | \frac{13-\sqrt[3]{49}}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,635 |
3. Given points $A(1,2)$ and $B(3,4)$, there is a point $P$ on the coordinate axis, and $P A + P B$ is minimized. Then the coordinates of point $P$ are $\qquad$. | 3. $\left(0, \frac{5}{2}\right)$.
Using knowledge of axis symmetry, ruler and compass constructions, and linear functions, we know that the minimum value of $P A + P l 3$ occurs when $P$ is on the $y$-axis, which is smaller than when $P$ is on the $x$-axis. Thus, the coordinates of $P$ are $\left(0, \frac{5}{2}\right)... | \left(0, \frac{5}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,637 |
(1) (Full marks 20 points) Given the equation about $x$
$$
\left(m^{2}-1\right) x^{2}-3(3 m-1) x+18=0
$$
has two positive integer roots ($m$ is an integer). The three sides $a$, $b$, $c$ of $\triangle A B C$ satisfy $c=2 \sqrt{3}$, $m^{2}+a^{2} m-8 a=0$, $m^{2}+b^{2} m-8 b=0$. Find:
(1) the value of $m$;
(2) the area ... | (1) The equation has two real roots, then $m^{2}-1 \neq 0$. Solving the equation gives $x_{1}=\frac{6}{m+1}, x_{2}=\frac{3}{m-1}$.
From the problem, we have
$\left\{\begin{array}{l}m+1=1,2,3,6, \\ m-1=1,3,\end{array}\right.$
which means $\left\{\begin{array}{l}m=0,1,2,5, \\ m=2,4 .\end{array}\right.$
Therefore, $m=2$.
... | 1 \text{ or } \sqrt{9+12 \sqrt{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,639 |
II. (Full marks 25 points) In $\triangle ABC$, the altitude $AD$ is inside the triangle. Lines through the incenters $I_{1}, I_{2}$ of $\triangle ABD$ and $\triangle ACD$ intersect $AB$ and $AC$ at $E$ and $F$, respectively.
(1) If $\angle BAC=90^{\circ}$, then $AE=AF$;
(2) If $AE=AF$, does $\angle BAC=90^{\circ}$ also... | II. As shown in Figure 3, connect $D I_{1}$ and $D I_{2}$, and extend $D I_{1}$ to intersect $A B$ at $G$.
$\because \angle B A C=90^{\circ}$,
$A D \perp B C$ at $D$,
$\therefore$ Rt $\triangle \Lambda B D$
$\sim R t \triangle C A D$.
Since $I_{1}$ and $I_{2}$ are the incenters of $\triangle A B D$ and $\triangle C A D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,640 |
Three. (Full marks 25 points) A steel plant uses 2 tons of raw material $A$ and 5 tons of raw material $B$ to produce 1 ton of type 甲 steel pipe, and 3 tons of raw material $A$ and 1 ton of raw material $B$ to produce 1 ton of type 乙 steel pipe. The unit profit (unit: ten thousand yuan/ton) from selling these two types... | Three, let the weekly production of two types of steel pipes, A and B, be $x$ tons and $y$ tons, respectively. According to the problem, we have:
$$
\left\{\begin{array}{l}
2 x+5 y \leqslant 10, \\
4 x+3 y \leqslant 12, \\
0 \leqslant x \leqslant 2.5, \\
0 \leqslant y \leqslant 1.5 .
\end{array}\right.
$$
The weekly t... | 11.5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,641 |
1. Given the triangle connecting $A(-3,0), B(0,-3)$, $C\left(\frac{15}{7}, \frac{24}{7}\right)$ and the circle $x^{2}+y^{2}=R^{2}(R>$ 0 ) always have common points. Then the range of the circle radius $R$ is
(A) $\left.\left(0, \frac{3 \sqrt{10}}{10}\right) \cup \frac{3 \sqrt{ }}{7},+\infty\right)$
(B) $\left(\frac{3 \... | -1 (B).
As shown in Figure 11, from the given information, the equations of the lines on which the three sides lie are
$$
\begin{array}{l}
A B: y=-x-3, \\
A C: y=\frac{2}{3}(x+3), \\
B C: y=3(x-1) .
\end{array}
$$
Further, the distances from the center of the circle to the three vertices and the three sides can be cal... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,642 |
2. As shown in Figure 1, in the rectangular prism, $A B$, $D B_{1}$, $A_{1} D_{1}$, and $C C_{1}$ are pairwise skew lines. Now, draw a pair of intersecting lines $l_{1}$ and $l_{2}$, such that $l_{1}$ intersects both $A B$ and $D B_{1}$, and $l_{2}$ intersects both $A_{1} D_{1}$ and $C C_{1}$. The number of such pairs ... | 2. (D).
Solution 1: As shown in Figure 12, draw a plane $\alpha$ through $AB$ intersecting $A_{1} D_{1}$ at $E$, $C C_{1}$ at $F$, and $D B_{1}$ at $P$ (think about how to accurately locate point $P$), and connect $E F$ as $l_{2}$. Then $E F$, point $P$, and $AB$ are all on plane $\alpha$. In $\alpha$, through $P$, in... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,643 |
3. As shown in Figure 2, from a point $A$ outside the circle $\odot O$, draw two tangents $A B$ and $A C$, and connect $B C$. Then the relationship between the perimeter $p$ of $\triangle A B C$ and the length $l$ of the minor arc $\overparen{B C}$ is ( ).
(A) $p>2 l$
(B) $p=2 l$
(C) $p<2 l$
(D) Cannot be determined, d... | 3. (A).
Let the radius of $\odot \bigcirc$ be $R$, connect $O A$ intersecting $B C$ at $D$, and intersecting the obvious area relationship of the arc (Fig. 14)
$S_{\triangle A O C}>S_{\text {segment } DE}$,
we get $\frac{1}{2} R^{2} \operatorname{tg} \theta>\frac{1}{2} R^{2} \theta$,
thus $\theta<\operatorname{tg} \t... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,644 |
Example 5 Proof: For any positive integer $n$, the inequality $(2 n+1)^{n} \geqslant(2 n)^{n}+(2 n-1)^{n}$ holds.
(21st All-Soviet Union Mathematical Competition) | Proof: By the binomial theorem, we have
$$
\begin{array}{l}
(1+2 n)^{n}-(2 n-1)^{n} \\
=2\left[(2 n)^{n-1} \mathrm{C}_{n}^{1}+(2 n)^{n-3} \mathrm{C}_{n}^{3}+\cdots\right] \\
\quad \geqslant(2 n)^{n} .
\end{array}
$$
Thus, the problem is proved. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,645 |
4. Given that $x$ is an acute angle. Then $\sin ^{3} x+\cos ^{3} x=\frac{\sqrt{2}}{2}$ is a ( ) for $x=\frac{\pi}{4$.
(A) sufficient but not necessary condition
(B) necessary but not sufficient condition
(C) sufficient and necessary condition
(D) neither sufficient nor necessary condition | 4. (C).
Solution 1: Necessity.
Take $x=\frac{\pi}{4}$, then we have
$$
\left(\sin \frac{\pi}{4}\right)^{3}+\left(\cos \frac{\pi}{4}\right)^{3}=2\left(\frac{\sqrt{2}}{2}\right)^{3}=\frac{\sqrt{2}}{2} \text {. }
$$
Sufficiency.
By the three-dimensional mean inequality, we have
$$
\begin{array}{l}
\sqrt{2} \sin ^{3} x+\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,646 |
5. As shown in Figure 3, rectangle $B_{1} B C C_{1}$ is formed by joining square $B_{1} B A A_{1}$ and $A_{1} A C C_{1}$. Now, fold the figure along $A_{1} A$ to form a right dihedral angle, as shown in Figure 4. Take a point $P$ on line segment $B_{1} C$, and let the distance from $P$ to $A B$ be $P E$, and the distan... | 5. (B).
As shown in Figure 15, if the shape is completed into a cube, then $B_{1} C$ is a body diagonal. Rotating around the body diagonal by $120^{\circ}$, $A_{1}$ becomes $B$, and $C_{1}$ becomes $A$. Therefore, the distance from point $P$ to $A_{1} C_{1}$, $P F$, becomes the distance from $P$ to $A B$, $P E$. Hence... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,647 |
6. For $b>a>0$, take points $A_{k}\left(x_{k}\right.$, $\left.y_{k}\right)(k=1,2, \cdots, n)$ in the first quadrant, such that $a, x_{1}, x_{2}, \cdots, x_{n}, b$ form an arithmetic sequence, and $a, y_{1}, y_{2}, \cdots, y_{n}, b$ form a geometric sequence. Then the relationship between the points $A_{1}, A_{2}, \cdot... | 6. (C).
From the general terms of arithmetic and geometric sequences, we have
$$
x_{k}=a+\frac{b-a}{n+1} \cdot k, \quad y_{k}=a\left(\frac{b}{a}\right)^{\frac{k}{n+1}}
$$
Thus,
$$
\begin{aligned}
x_{k} & =\frac{k b+(n+1-k) a}{n+1} \\
& =a \frac{k\left(\frac{b}{a}\right)+(n+1-k)}{n+1} \\
& >a \sqrt[n+1]{\left(\frac{b}... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,648 |
1. The solution set of the equation $\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$ is $\qquad$ . | Let $y=\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$, then we have
$$
\left\{\begin{array}{l}
5^{y}=3^{x}+4^{x}, \\
4^{y}=5^{x}-3^{x}
\end{array}\right.
$$
(1) + (2) gives $5^{y}+4^{y}=5^{x}+4^{x}$.
However, the function $f(x)=5^{x}+4^{x}$ is an increasing function, so from (3) we have
$$
f(y)=f(... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,649 |
2. In the Cartesian coordinate system, the coordinates of the three vertices of the right triangle $Rt \triangle OAB$ are $O(0,0), A(1,0)$, and $B(1,2)$. Take any point $C(x, 2x) (0<x<1)$ on the hypotenuse, and draw $CD \perp OA$ at $D$, and $CE \perp AB$ at $E$. Let the area of $\triangle OCD$ be $S_{1}(x)$, the area ... | $2 \cdot \frac{1}{9}$.
From the area formula, it is easy to get
$$
\begin{array}{l}
S_{1}(x)=\frac{1}{2} O D \cdot D C \\
=x^{2}(0<x<1), \\
S_{2}(x)=D C \cdot D A \\
=2 x(1-x) \\
(0<x<1), \\
S_{3}(x)=\frac{1}{2} C E \cdot E B \\
=(1-x)^{2}(0<x<1) .
\end{array}
$$
With the help of the figure (Figure 16), it is convenie... | \frac{4}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,650 |
3. The number of different sequences $a_{1}, a_{2}, \cdots, a_{100}$ that satisfy $\left(a_{n}-1\right)\left(a_{n+1}+1\right)=0(n=1,2$, $\cdots, 99)$ is $\qquad$ . | 3. Infinity.
For $\left(a_{1}-1\right)\left(a_{2}+1\right)=0$, we can take $a_{1}=1$, or $a_{2}=-1$, but it is not necessary to take $a_{1}=1$ and $a_{2}=-1$. Therefore, the sequence $a_{1}=1, a_{2}$ can be any value,
$$
a_{3}=a_{4}=\cdots=a_{100}=-1 \text {, }
$$
satisfies the condition, hence there are infinitely ma... | Infinity | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,651 |
4. Given that the pure imaginary numbers $x_{1}, x_{2}, \cdots, x_{1999}$ have a modulus of 1. Then the remainder when $x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{1998} x_{1999}+x_{1999} x_{1}$ is divided by 4 is $\qquad$ | 4.1.
Obviously, if $x_{k}$ takes $i$ or $-i$, we have
$$
\begin{array}{l}
\left(i+x_{k}\right)\left(i-x_{k+1}\right) \\
=\left\{\begin{array}{l}
0, \text{ when } x_{k}=-i \text{ or } x_{k+1}=i; \\
-4, \text{ when } x_{k}=i \text{ and } x_{k+1}=-i
\end{array}\right.
\end{array}
$$
Both are multiples of 4, so the sum i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,652 |
5. There are 10 people with distinct heights standing in a row, and each person is either taller or shorter than the person in front of them. The number of arrangements that meet this condition is $\qquad$ (answer with a number).
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a repetition of the instr... | 5.512.
According to the problem, the person standing in the last position must be the tallest or the shortest. Otherwise, there would be people both taller and shorter than the 10th person, which contradicts the given conditions. Therefore, there are two ways to arrange the 10th position.
Let the number of arrangemen... | 512 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,653 |
6. Using $1,2, \cdots, n$ to form an $n$-digit number without repeating digits, where 2 cannot be adjacent to 1 or 3, a total of 2400 different $n$-digit numbers are obtained. Then $n=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation ... | 6.7 .
Obviously, $n \geqslant 4$, otherwise 2 must be adjacent to 1 or 3.
When 2 is adjacent to 1, there are $2 \cdot(n-1)$! arrangements, and when 2 is adjacent to 3, there are also $2 \cdot(n-1)$! arrangements. When 2 is adjacent to both 1 and 3, there are $2 \cdot(n-2)$! arrangements. Therefore, the number of arran... | 7 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 710,654 |
Three. (Full marks 20 points) Given the parabola $S: y=a x^{2} +b x+c(a \neq 0)$ and the quadratic curve $a x^{2}+y^{2}+b x+a y+c+a=0$ intersect at 4 distinct points. From sketches 7 and 8, it can be seen that the following three conclusions are true. Please provide proofs.
(1) Among the 4 intersection points of the tw... | (1) Solve the system of equations
$$
\left\{\begin{array}{l}
y=a x^{2}+b x+c, \\
a x^{2}+y^{2}+b x+a y+c+a=0 .
\end{array}\right.
$$
Eliminating $x$, we get
$$
y^{2}+(1+a) y+a=0 \text {. }
$$
Solving for $y$, we get $y_{1}=-1, y_{2}=-a$.
Thus, among the 4 intersection points of the two curves, at least two have negat... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,655 |
Example 6 Let $a, b \in \mathbf{R}^{+}$, and $\frac{1}{a}+\frac{1}{b}=1$. Try to prove that for each $n \in \mathbf{N}$, we have
$$
(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1} .
$$
(1988, National High School Mathematics Competition)
Analysis: This problem is generally solved using mathematical induction. However, ... | Proof 1: From $1=\frac{1}{a}+\frac{1}{b} \geqslant \frac{2}{\sqrt{a b}}$
$$
\Rightarrow \sqrt{a b} \geqslant 2 \text {. }
$$
Directly applying the binomial theorem to the left side of the desired inequality, we have
$$
\begin{aligned}
& (a+b)^{n}-a^{n}-b^{n} \\
= & \mathrm{C}_{n}^{1} a^{n-1} b+\mathrm{C}^{2} a^{n}{ }^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,656 |
Four. (Full marks 20 points) As shown in Figure 8, point $O$ represents the sun, $\triangle A B C$ represents a triangular sunshade, and points $A$ and $B$ are two fixed points on the ground in the north-south direction. The sunlight $O C D$ from the due west direction casts the shadow of the sunshade onto the ground, ... | (1) As shown in Figure 17, draw the perpendicular line $O H$ from $O$ to the ground, connect $H D$ to intersect $A B$ at $E$, and connect $C E$. Then $H D$ is the projection of the oblique line $O D$ on the ground, and $\angle C D E = \theta$. Given that $A B$ is in the north-south direction and $C D$ is in the east-we... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,657 |
Five. (Full Score 20 Points) From Sketch 9, it is evident that the equation $\log _{5}(1+\sqrt{x})$ $=\log _{16} x$ has and only has one solution. Please prove this conclusion.
---
From Sketch 9, it is evident that the equation $\log _{5}(1+\sqrt{x})$ $=\log _{16} x$ has and only has one solution. Please prove this c... | When $x=16$, substituting it into the equation shows that it has a solution. To prove the uniqueness of the solution, it is only necessary to prove that $\log _{5}(1+\sqrt{x})$ and $\log _{16} x$ are not equal in the intervals $(0,16)$ and $(16,+\infty)$.
(1) When $0 < x < 16$, we have $\frac{\sqrt{x}}{4} < 1$, thus
$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,658 |
One, (Full marks 50 points) Given a right trapezoid $A B C D$ with a point $E$ on the right-angle leg $A B$, such that $\triangle C D E$ is an equilateral triangle, and $\frac{A E}{C D}=\frac{\sqrt{6}}{4}$. Find the value of $\frac{A E}{A B}$. | Solution 1: As shown in Figure 19, draw $D F \perp C E$ at $F$, and through $F$ draw a perpendicular to $B C$ intersecting $B C$ at $G$ and $A D$ at $H$. We get $F$ as the midpoint of $C E$, and $G$ as the midpoint of $B C$, thus
$$
\begin{array}{l}
F G=\frac{1}{2} B E \\
=\frac{1}{2}(A B-A E) .
\end{array}
$$
Then $H... | \frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,659 |
II. (Full marks 50 points) There are $n$ points on a circle $(n \geqslant 2)$, which are connected by chords, with no three chords intersecting at the same point inside the circle. Let $a_{n}$ denote the number of non-overlapping regions formed inside the circle.
(1) Directly draw to find $a_{2}, a_{3}, a_{4}, a_{5}, a... | (1) From Figure 21, we can see that $a_{2}=2, a_{3}=4, a_{4}=8, a_{5}=16, a_{6}=31$.
From the first 4 values, we might guess that $a_{n}=2^{n-1}$, but $a_{6}$ disproves this guess.
(2) Generally, we divide the required area into two parts: one part consists of $n$ segments, and the other part is the region inside the p... | a_{n}=\mathrm{C}_{n}^{4}+\mathrm{C}_{n-1}^{2}+n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,660 |
Three. (Full marks 50 points) Prove that the sequence
$$
a_{n}=3^{n} \cos \left(n \arccos \frac{1}{3}\right)(n=1,2 \cdots)
$$
is such that each term is an integer, but none of them are multiples of 3. | Three, Proof 1: Let $\theta=\arccos \frac{1}{3}$, then
$$
\cos \theta=\frac{1}{3} \text {, and } a_{n}=3^{n} \cos n \theta \text {. }
$$
(1) When $n=1,2$, we have
$$
\begin{array}{l}
a_{1}=3 \cos \theta=1, \\
a_{2}=3^2 \cos 2 \theta=3^{2}\left(2 \cos ^{2} \theta-1\right)=-7
\end{array}
$$
Assume $a_{k-1}, a_{k}$ are in... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,661 |
75. There is a wonderful positive integer $A$, which is composed of only two different digits. However, $A^{2}$ contains exactly the 10 digits 0, 1, 2, ..., 9. Try to find the positive integer $A$. | Since $A^{2}$ contains exactly the digits $0,1,2, \cdots, 9$, we have
$$
\begin{array}{l}
1023456789 \leqslant A^{2} \leqslant 9876543210 . \\
\therefore 31992 \leqslant A \leqslant 99380 .
\end{array}
$$
Therefore, $A$ is a five-digit number, and the unit digit of $A$ is not 0.
$$
\text { Also, } \because A^{2} \equi... | 67677, 97779 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,662 |
76. On the sides $BC$ and $CD$ of the square $ABCD$, there are points $M$ and $N$ respectively, such that $\angle MAN = 45^{\circ}$. Prove:
$$
\frac{AM}{AN} = \sqrt{\frac{AB + BM}{AD + DN}}.
$$ | Proof: For convenience, let $A B=1, B M=m, D N=n$. Clearly, we have
$$
\begin{aligned}
A M & =\sqrt{A B^{2}+B M^{2}} \\
& =\sqrt{1+m^{2}}, \\
A N & =\sqrt{A D^{2}+D N^{2}} \\
& =\sqrt{1+n^{2}} .
\end{aligned}
$$
The conclusion of this problem can be rewritten as
$$
\frac{1+m^{2}}{1+n^{2}}=\frac{1+m}{1+n}
$$
Construct... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,663 |
75. Given $a, b, c$ are the lengths of the sides of a triangle. Prove:
$$
\frac{23}{16}<\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}<\frac{25}{16} .
$$ | Proof: $\because 2\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)-3$
$$
\begin{array}{l}
=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a} \\
=\frac{a-b}{a+b}-\frac{a-b}{b+c}+\frac{c-a}{c+a}-\frac{c-a}{b+c} \\
=\frac{(a-b)(c-a)}{(a+b)(b+c)}+\frac{(c-a)(b-a)}{(c+a)(b+c)} \\
=-\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,664 |
Given $a \geqslant b \geqslant c>0$, and $a+b+c=3$. Prove:
$$
\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \geqslant 3+Q
$$
where $Q=|(a-1)(b-1)(c-1)|$. | $$
\begin{array}{l}
\geqslant 2 a+2 b+2 c=6, \\
a+b+c \geqslant 3 \sqrt[3]{a b c}, \\
\therefore a b c \leqslant 1 . \\
\text { From (2) we get } \\
\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \\
\geqslant 6+a b c-1-a b-b c-a c=3+Q .
\end{array}
$$
From (2) we get
$$
\begin{array}{l}
\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \\
\ge... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,665 |
Example 7 The sequence $\left\{a_{n}\right\}(n \geqslant 0), a_{n}=[n \sqrt{2}]$. Prove: $\left\{a_{n}\right\}$ contains infinitely many perfect squares.
(30th IMO Preliminary Problem)
Analysis: The proof process of Example 4 inspires us: the expansion of $(1+$ $\sqrt{2})^{n}$ can be expressed in the form $x \cdot \sqr... | Prove: For $m$ being a positive integer, by the binomial theorem we have
$$
\begin{array}{l}
(\sqrt{2}+1)^{m}=x_{m} \cdot \sqrt{2}+y_{m}, \\
(\sqrt{2}-1)^{m}=x_{m} \cdot \sqrt{2}-y_{m}. \\
\left(x_{m}, y_{m} \in \mathbf{N}\right)
\end{array}
$$
Multiplying the two equations, we get
$$
\begin{array}{l}
2 x_{m}^{2}-y_{m... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,666 |
Example 8 Let the sequence $g(n)$ be defined as follows:
$$
\begin{array}{l}
g(1)=0, g(2)=1, \\
g(n+2)=g(n+1)+g(n)+1 . \\
(n \geqslant 1) \quad .
\end{array}
$$
If $n$ is a prime number greater than 5, prove that:
$$
n \mid g(n)(g(n)+1) \text {. }
$$
(29th IMO Shortlist Problem)
Analysis: Recalling the structure of Ex... | Prove: Let $f(n)=g(n)+1$. Then
$$
\begin{array}{l}
f(1)=1, f(2)=2, \\
f(n+2)=f(n+1)+f(n) .
\end{array}
$$
From (1) and (2), it is easy to deduce that
$$
\begin{aligned}
f(n)= & \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right) \\
= & \frac{1}{2^{n}}\left(\ma... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,667 |
Example 9 The sequence $a_{0}, a_{1}, a_{2}, \cdots$ satisfies $a_{0}=1, a_{1}=1, a_{n+1}=2 a_{n}+(a-1) a_{n-1}(n \in \mathbf{N}), a \in \mathbf{N}$ is a parameter. Let $p_{0}>2$ be a given prime. Find the smallest value of $a$ that satisfies the following two conditions.
(i) If $p$ is a prime and $p \leqslant p_{0}$, ... | Given the conditions of the problem,
$$
a_{n}=\frac{1}{2 \sqrt{a}}\left[(1+\sqrt{a})^{n}-(1-\sqrt{a})^{n}\right] \text {. }
$$
Noting that $p$ is a prime number, then
$$
p \mid \mathrm{C}_{p}^{k}(k=1,2, \cdots, p-1) .
$$
By the binomial theorem, we have
$$
\begin{aligned}
a_{p} & =\frac{1}{2 \sqrt{a}}\left[(1+\sqrt{a... | c | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,668 |
Example 1 Given a quadratic equation $x^{2} + m x - m + 1 = 0$ ($m$ is an integer) has two distinct positive integer roots. Find the value of $m$.
(1996, Sichuan Province Math Competition)
Analysis: For a quadratic equation to have integer roots, it must first satisfy that the discriminant is a perfect square. If we fo... | Solution: Let the two positive integer roots of the equation be $\alpha, \beta$, then
$$
\begin{array}{l}
\alpha+\beta=-m, \alpha \beta=-m+1 . \\
\therefore \alpha \beta-\alpha-\beta=1 .
\end{array}
$$
Thus, $\alpha \beta-\alpha-\beta+1=2$,
which means $(\alpha-1)(\beta-1)=2$.
Therefore, we have
$$
\begin{array}{l}
\l... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,670 |
Example 2 The sum of two positive integers is 1997 less than their product, and one of them is a perfect square. Then the larger number and the smaller (1997, Tianjin Junior High School Mathematics Competition) | Solution: Let the two positive integers be $\alpha, \beta$, and $\alpha>\beta$. Then
$$
\begin{array}{l}
\alpha+\beta+1997=\alpha \beta . \\
\therefore \alpha \beta-\alpha-\beta+1=1998,
\end{array}
$$
which means
$$
\begin{array}{l}
(\alpha-1)(\beta-1) \\
=1998=999 \times 2=333 \times 3 \times 2 .
\end{array}
$$
Sinc... | 663 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,671 |
Example 3 Given $x y z=1, x+y+z=2, x^{2}+$ $y^{2}+z^{2}=3$. Then
$$
\frac{1}{x y+z-1}+\frac{1}{y z+x-1}+\frac{1}{z x+y-1}
$$
is ( ).
(A) 1
(B) $-\frac{1}{2}$
(C) 2
(D) $-\frac{2}{3}$
(1997, Taiyuan Junior High School Mathematics Competition) | Given: $\because x+y+z=2$,
$$
\begin{aligned}
\therefore & x y+z-1 \\
& =x y+(2-x-y)-1 \\
& =(x-1)(y-1) .
\end{aligned}
$$
Similarly, we have
$$
\begin{array}{l}
y z+x-1=(y-1)(z-1), \\
z x+y-1=(z-1)(x-1) .
\end{array}
$$
Since $x y z=1, x^{2}+y^{2}+z^{2}=3$, then
$$
\begin{array}{l}
x y+y z+z x \\
=\frac{1}{2}(x+y+z)... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,672 |
Example 3 Given in $\triangle A B C$, $A B=A C$, $\angle B A C=100^{\circ}, P$ is a point on the angle bisector of $\angle C$, $\angle P B C=10^{\circ}$. Find the degree measure of $\angle A P B$. | Solution: As shown in Figure 6, take a point $D$ on the extension of $CA$ such that $CD = CB$, and connect $DP$ and $DB$. Let the intersection of $DB$ and $CP$ be $E$. It is easy to see that $B$ and $D$ are symmetric with respect to $PC$. Therefore, $\triangle PBD$ is an equilateral triangle, and $\angle PDA = \angle P... | 70^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,673 |
Example 4 Given $u+v=96$, and the quadratic equation $x^{2} + u x + v = 0$ has integer roots, then its largest root is $\qquad$
(1996, Anhui Province Partial Areas Junior High School Mathematics League) | Solution: Let the two integer roots of the equation be $x_{1}$ and $x_{2}$, then we have
$$
\begin{array}{l}
x_{1}+x_{2}=-u, x x_{2}=v . \\
\because u+v=96, \\
\therefore x_{1} x_{2}-\left(x_{1}+x_{2}\right)=v+u=96 .
\end{array}
$$
Thus, $x_{1} x_{2}-x_{1}-x_{2}+1=97$,
which means $\left(x_{1}-1\right)\left(x_{2}-1\ri... | 98 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,674 |
Example 5 If the two legs of a right triangle are integers, and are roots of the equation $m x^{2}-2 x-m+1=0$ (where $m$ is an integer), does such a triangle exist? If it exists, find the lengths of the three sides of all such triangles; if not, explain the reason.
(1997, Hubei Huanggang City Junior High School Mathema... | Solution: Let the two legs of the right triangle be $\alpha, \beta$.
$\because \alpha 、 \beta$ are integers,
$$
\therefore \alpha \geqslant 1, \beta \geqslant 1 \text {. }
$$
$\because \alpha$ 、 $\beta$ are the roots of the equation $m x^{2}-2 x-m+1=0$ ( $m$ is an integer),
$$
\begin{array}{l}
\therefore \alpha+\beta=\... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,675 |
Example 1 The number of proper subsets of the set $\left\{x \left\lvert\,-1 \leqslant \log _{\frac{1}{x}} 10<-\frac{1}{2}\right., 1<\right.$ $x \in \mathbf{N}\}$ is $\qquad$
(1996, National High School Mathematics Competition) | $$
\begin{array}{l}
\text { Solution: }\left\{x \left\lvert\,-1 \leqslant \log _{x}^{1} 10<-\frac{1}{2}\right., 1<x \in \mathbf{N}\right. \\
=\left\{x \left\lvert\,-1 \leqslant \frac{1}{\lg \frac{1}{x}}<-\frac{1}{2}\right., 1<x \in \mathbf{N}\right\} \\
=\{x \mid 1 \leqslant \lg x<2,1<x \in \mathbf{N}\} \\
=\{x \mid 10... | 2^{90}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,676 |
Example 2 Let $S$ be a set with 6 elements. In how many ways can two (not necessarily distinct) subsets of $S$ be chosen so that the union of the two subsets is $S$? The order of selection does not matter. For example, the pair of subsets $\{a, c\}, \{b, c, d, e, f\}$ and the pair of subsets $\{b, c, d, e, f\}, \{a, c\... | Solution: Let $S=A \cup B$, and without loss of generality, assume $|A| \leqslant|B|$ (denote the number of elements in set $S$ as $|S|$).
If $|A|=0$, then $A=\varnothing, B=S$, there is only 1 way to choose;
If $|A|=1$, then $|B|=6,5$. At this time, there are $\mathrm{C}_{6}^{1} \mathrm{C}_{6}^{6} +\mathrm{C}_{6}^{1... | 365 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,677 |
Example 3 Let $S$ be a subset of $\{1,2,3, \cdots, 50\}$ with the following property: the sum of any two distinct elements of $S$ is not divisible by 7. Then, the maximum number of elements in $S$ is ( ).
(A) 6
(B) 7
(C) 14
(D) 22
(E) 23
(43rd American High School Mathematics Examination) | Solution: Divide $A=\{1,2,3, \cdots, 50\}$ into 7 subsets: $A_{0}, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$, where the elements of $A_{i}$ have a remainder of $i$ when divided by 7 ($i=0,1,2,3,4,5,6$), i.e., $A_{0}=\{0,7,14,21,28,35,42,49\}$,
$$
\begin{array}{l}
A_{1}=\{1,8,15,22,29,36,43,50\}, \\
A_{2}=\{2,9,16,23,30... | E | Number Theory | MCQ | Yes | Yes | cn_contest | false | 710,678 |
Example 4 Let $M=\{1,2,3, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: if $x \in A, 15 x \notin A$, then the maximum number of elements in $A$ is $\qquad$
(1995, National High School Mathematics Competition) | Solution: Construct subset $A$ as follows:
$1995 \div 15=133$, let $A_{1}=\{134,135, \cdots$, 1995\}, then $\left|A_{1}\right|=1862$ (elements);
$133 \div 15=8$ remainder 13, let $A_{2}=\{9,10, \cdots$, 133\}, then $\left|A_{2}\right|=125$ (elements);
Let $A_{3}=\{1,2,3,4,5,6,7,8\}$, then $\left|A_{3}\right|=$ 8 (elem... | 1870 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,679 |
Example 5 Let the set $M=\{1,2,3, \cdots, 1000\}$. For any non-empty subset $X$ of $M$, let $\alpha_{X}$ denote the sum of the largest and smallest numbers in $X$. Then the arithmetic mean of all such $\alpha_{X}$ is $\qquad$
(1991, National High School Mathematics Competition) | Solution: Construct the subset $X^{\prime}=\{1001-x \mid x \in X\}$, then all non-empty subsets can be divided into two categories: $X^{\prime}=X$ and $X^{\prime} \neq X$.
When $X^{\prime}=X$, it must be that $X^{\prime}=X=M$, thus, $\alpha_{X}=1001$.
When $X^{\prime} \neq X$, let $x$ and $y$ be the maximum and minim... | 1001 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,680 |
Example 6 For any set $S$, let $|S|$ denote the number of elements in set $S$, and let $n(S)$ denote the number of subsets of set $S$, including the empty set and set $S$ itself. If $A$, $B$, and $C$ are three sets that satisfy the condition $n(A) + n(B) + n(C) = n(A \cup B \cup C)$ and $|A| = |B| = 100$, then the mini... | Solution: Since a set containing $k$ elements has $2^{k}$ subsets, from $n(A)+n(B)+n(C)=n(A \cup B \cup C)$, we get $2^{100}+2^{100}+2^{|C|}=2^{|A \cup B \cup C|}$.
$$
\therefore 1+2^{|C|-101}=2^{|A \cup B \cup C|-101}
$$
It is easy to see that $1+2^{|C|-101}$ is a number greater than 1 and a power of 2.
$$
\therefore... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 710,681 |
For example, $73 \leqslant n \in \mathbf{N}$, let $S$ be the set of all non-empty subsets of the set $\{2,3, \cdots, n\}$. For each $S_{i} \subseteq S, i=1,2, \cdots, 2^{n-1}-1$, let $p_{i}$ be the product of all elements in $S_{i}$. Find $p_{1}+p_{2}+\cdots+p_{2^{n-1}-1}$. | When $n=3$, it is easy to get
$$
\begin{array}{l}
p_{1}+p_{2}+p_{3} \\
=2+3+2 \times 3=11=\frac{4!}{2}-1,
\end{array}
$$
When $n=4$, we have
$$
\begin{array}{l}
p_{1}+p_{2}+p_{5}+p_{4}+p_{5}+p_{6}+p_{7} \\
=11+4+4(2+3+2 \cdot 3)=59 \\
=\frac{5!}{2}-1 ;
\end{array}
$$
From this, we conjecture
$$
\begin{array}{l}
p_{1}... | \frac{(n+1)!}{2}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,682 |
Example 8 If $M=\left\{(x, y)|| \operatorname{tg} \pi y \mid+\sin ^{2} \pi x\right.$ $=0\}, N=\left\{(x, y) \mid x^{2}+y^{2} \leqslant 2\right\}$, then the number of elements in $M \cap N$ is ( ).
(A) 4
(B) 5
(C) 8
(D) 9
(1993, National High School Mathematics Competition) | Solution: From the condition that the sum of non-negative numbers is zero, we get
$$
\left\{\begin{array}{l}
\operatorname{sin} \pi x = 0, \\
\operatorname{tr} \pi y = 0
\end{array} \Rightarrow \left\{\begin{array}{l}
x=l, \dot{k} \in \mathbf{Z}, \\
y=i, l \in \mathbf{Z}.
\end{array}\right.\right.
$$
That is, the set ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,683 |
Example 4 In hexagon $A B C D E F$, $\angle A=\angle B$ $=\angle C=\angle D=\angle E=\angle F$, and $A B+B C=$ $11, F A-C D=3$. Then $B C+D E=$ ?
(1994, Beijing Junior High School Mathematics Competition) | Solution: As shown in Figure 7, each angle of the hexagon is equal, all being $120^{\circ}$, so each exterior angle is $60^{\circ}$. Let the lines $A B, C D$, and $E F$ intersect at $P$, $Q$, and $R$, respectively, forming four equilateral triangles, i.e., $\triangle R A F$, $\triangle R P Q$, $\triangle B P C$, and $\... | 14 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,684 |
Example 9 Let the plane region $D$ be represented by $N(D)$, which denotes the number of all integer points (i.e., points on the xoy plane where both coordinates $x$ and $y$ are integers) belonging to $D$. If $A$ represents the region enclosed by the curve $y=x^{2} (x \geqslant 0)$ and the two lines $x=10$, $y=1$ (incl... | Solution: Draw the figure in the Cartesian coordinate system, and it is easy to calculate from the figure
$$
\begin{array}{l}
N(A)=1^{2}+2^{2}+\cdots+10^{2} \\
=\frac{1}{6} \times 10(10+1)(2 \times 10+1)=385, \\
N(B)=\left(101-1^{2}\right)+\left(101-2^{2}\right)+\cdots \\
+\left(101-10^{2}\right) \\
=101 \times 10-\lef... | 1010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,685 |
Question 3: There are $n$ street lamps on one side of a road. To save electricity, $r(r \geqslant 2)$ lamps need to be turned off. To ensure lighting is not affected, it is required that there are at least $k(n \geqslant r+k(r$ -1)) lamps between any two turned-off lamps. How many different ways are there to turn off t... | Solution: Represent these $n$ blood lamps with the numbers $1,2,3, \cdots, n$, then problem 3 is transformed into selecting $r$ numbers $a_{1}, a_{2}, \cdots, a_{r}$ from these $n$ numbers, and satisfying $a_{1}<a_{2}-k<a_{3}-2k<\cdots<a_{r}-(r-1)k$ (where $k$ is a constant, $i, j=1,2 \cdots n$). How many such number g... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,686 |
1. Calculate $\frac{1}{1-\sqrt[4]{3}}+\frac{1}{1+\sqrt[4]{3}}+\frac{2}{1+\sqrt{3}}$ is ( ).
(A) 1
(B) -1
(C) 2
(D) -2 | $-1 .(\mathrm{D})$
original expression $=\frac{2}{1-\sqrt{3}}+\frac{2}{1+\sqrt{3}}=\frac{4}{1-3}=-2$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,688 |
2. The perimeter of $\triangle A B C$ is $24, M$ is the midpoint of $A B$. $M C=M A=5$. Then the area of $\triangle A B C$ is ( ).
(A) 12
(E) 16
(C) 24
(D) 30 | 2. (C)
$$
\begin{array}{l}
\because M A=M B=M C=5, \\
\therefore \angle A C B=90^{\circ} .
\end{array}
$$
Given the perimeter is 24, then $A C+B C=14, A C^{2}+B C^{2}=$ $10^{2}$
$$
\begin{array}{l}
\begin{aligned}
\therefore 2 A C \cdot B C & =(A C+B C)^{2}-\left(A C^{2}+B C^{2}\right) \\
& =14^{2}-10^{2}=4 \times 24 ... | 24 | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,689 |
3. Let $b>a$, the graphs of the linear functions $y=b x+a$ and $y$ $=a x+b$ are drawn in the Cartesian coordinate system, then there is a set of values for $a$ and $b$ such that one of the following 4 graphs is correct ( ).
| 3. (B)
The intersection point $(x, y)$ of two lines is the solution to the system of equations $\left\{\begin{array}{l}y=b x+a, \\ y=a x+b\end{array}\right.$, which is $(1, a+b)$.
(A) The x-coordinate of the intersection point is 1, a number, so (A) is incorrect.
(C) The x-coordinate of the intersection point is $2 \n... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,690 |
5. As shown in the figure, in isosceles trapezoid $ABCD$, $AB \parallel DC, AB=998, DC=1001, AD=1999$, point $P$ is on line segment $AD$. Then the number of points $P$ that satisfy the condition $\angle APC = 90^{\circ}$ is ( ).
(A) 0
(B) 1
(C) 2
(D) an integer not less than 3 | 5. (C)
The midpoint $M$ of $A D$ subtends a $90^{\circ}$ angle with $B C$; also, let $N$ be a point on $A D$ such that $A N: N D=998: 1001$. From the fact that $\triangle A B N$ and $\triangle D C N$ are both isosceles triangles, it follows that $N$ subtends a $90^{\circ}$ angle with $B C$. Noting that the circle with ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,692 |
6. There are the following three propositions:
(甲) If $\alpha, \beta$ are unequal irrational numbers, then $\alpha \beta + \alpha - \beta$ is an irrational number;
(乙) If $\alpha, \beta$ are unequal irrational numbers, then $\frac{\alpha - \beta}{\alpha + \beta}$ is an irrational number;
(丙) If $\alpha, \beta$ are uneq... | 6. (A)
$$
\because \alpha \beta+\alpha-\beta=\alpha \beta+\alpha-\beta-1+1=(\alpha-1)(\beta+
$$
1) +1 , if we let $\alpha=1+\sqrt{2}, \beta=-1+\sqrt{2}$, then $\alpha \beta+\alpha-\beta$ is a rational number. $\therefore$ (A) is incorrect;
If we let $\alpha=2 \sqrt{2}, \beta=\sqrt{2}$, then $\frac{\alpha-\beta}{\alpha+... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 710,693 |
1. Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$ and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ | $$
\begin{array}{l}
\because(b-c)^{2}=4(a-b)(c-a), \\
b^{2}-2 b c+c^{2}=4 a c-4 b c+4 a b-4 a^{2}, \\
\therefore(b+c)^{2}-4 a(b+c)+4 a^{2}=0 .
\end{array}
$$
Therefore, $[2 a-(b+c)]^{2}=0$, which means $2 a=b+c$.
$$
\therefore \frac{b+c}{a}=2 \text {. }
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,694 |
Example 5 In $\triangle A B C$, $E$ is the midpoint of $B C$, and $D$ is on $A C$. If $\angle B A C=60^{\circ}, \angle A C B=20^{\circ}$, $\angle D E C=80^{\circ}, S_{\triangle A B C}+2 S_{\triangle C D E}=\sqrt{3}$, find the length of $A C$. | Solution: As shown in Figure 8, take point $F$ on the extension of $AB$ such that $AF = AC$, and connect $FC$. Take point $G$ on $AF$ such that $GF = AB$, and connect $GC$. It is known that $\triangle FAC$ is an equilateral triangle; $\triangle CFG \cong \triangle CAB$; $\triangle CGB \sim \triangle CED$. We have $S_{\... | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,695 |
3. Given that $a$ and $b$ are integers, and satisfy
$$
\left(\frac{\frac{1}{a}}{\frac{1}{a}-\frac{1}{b}}-\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}\right)\left(\frac{1}{a}-\frac{1}{b}\right) \frac{1}{\frac{1}{a^{2}}+\frac{1}{b^{2}}}=\frac{2}{3} \text {. }
$$
Then $a+b=$ $\qquad$ . | 3.3.
$$
\begin{array}{l}
\text { Left side }=\frac{a b}{a-b}=\frac{2}{3}, \\
\therefore(3 b-2)(3 a-2)=4 .
\end{array}
$$
Given that $a \neq b$ and they are integers, hence $3 b-2, 3 a-2$ can only take the values 1, 4 or -1, -4.
(1) Suppose $3 b-2=1, 3 a-2=4$.
Solving gives $b=1, a=2$. Therefore, $a+b=3$.
(2) Suppose $... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,697 |
4. In square $A B C D$, $N$ is the midpoint of $D C$, $M$ is a point on $A D$ different from $D$, and $\angle N M B=\angle M B C$. Then $\operatorname{tg} \angle A B M=$ $\qquad$ | 4. $-\frac{1}{3}$.
Extend $M N$ to intersect the extension of $B C$ at $T$, let the midpoint of $M B$ be $O$, connect $T O$, then $\triangle B A M \sim \triangle T O B$.
So, $\frac{A M}{M B}=\frac{O B}{B T}$, which means $M B^{2}=2 A M \cdot B T$.
Let $D N=1, C T=M D=k$, then $A M=2-k$. $B M=\sqrt{4+(2-k)^{2}}, B T=2+... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,698 |
One, (20 points) A class participated in an intelligence competition, with three questions: $a$, $b$, and $c$. Each question either scores full marks or 0 points, where question $a$ is worth 20 points, and questions $b$ and $c$ are each worth 25 points. The competition results show that each student got at least one qu... | Let $x_{a}, x_{b}, x_{c}$ respectively represent the number of people who answered questions $a$, $b$, and $c$ correctly, then we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{a}+x_{b}=29, \\
x_{a}+x_{c}=25, \\
x_{b}+x_{c}=20 .
\end{array}\right. \\
\therefore x_{a}+x_{b}+x_{c}=37 .
\end{array}
$$
Solving, we ge... | 42 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,699 |
II. (This question is worth 25 points) As shown in the figure, let $\triangle ABC$ be a right triangle, with point $D$ on the hypotenuse $BC$, and $BD = 4DC$. It is known that a circle passes through point $C$ and intersects $AC$ at $F$, and is tangent to $AB$ at the midpoint $G$ of $AB$. Prove that $AD \perp BF$. | II. Draw $D E \perp A C$ at $E$, then $A C=\frac{5}{4} A E$,
$$
A G=\frac{3}{2} E D \text {. }
$$
Secant-Tangent Theorem
$$
\begin{array}{l}
A G^{2}=A F \cdot A C=A F \cdot \frac{5}{4} A E . \\
\text { Therefore } \frac{25}{4} E D^{2}=A F \cdot \frac{5}{4} A E,
\end{array}
$$
which means $5 E D^{2}=A F \cdot A E$.
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,700 |
Three. (This question is worth 25 points) $a$ is a real number greater than zero. It is known that there exists a unique real number $k$ such that the quadratic equation $x^{2}+\left(k^{2}+a k\right) x+1999+k^{2}+a k=0$ has two roots that are both prime numbers. Find the value of $a$. | Three, let the two prime roots of the equation be \( p \) and \( q \). By the relationship between the roots and coefficients of a quadratic equation, we have
\[
\begin{array}{l}
p+q=-k^{2}-a k, \\
p q=1999+k^{2}+a k . \\
(1)+(2), \text { we get } p+q+p q=1999 . \\
\therefore(p+1)(q+1)=2^{4} \times 5^{3} .
\end{array}
... | 2 \sqrt{502} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,701 |
3. Given the equation $x^{2}+y^{2}-2 \sqrt{m+1} x- m y+m+1=0$ (where $m$ is a positive parameter), then the locus of the center of the circle is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. The part of the hyperbola $x^{2}-4 y^{2}=1$ in the first quadrant | x^{2}-4 y^{2}=1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,704 |
Example 6 In $\triangle A B C$, $\angle A=20^{\circ}, A B=$ $A C=a, B C=b$. Prove: $a^{3}+b^{3}=3 a^{2} b$.
| Proof: As shown in Figure 9, let $E$ be the reflection of $C$ over $AB$, and $F$ be the reflection of $B$ over $AC$. $EF$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. We have $\triangle EBM \sim \triangle ABE$. Therefore,
$$
BE^2 = BM \cdot BA.
$$
Thus, $BM = \frac{b^2}{a}$.
$$
\therefore AM = AB - BM = \fra... | a^3 + b^3 = 3a^2b | Geometry | proof | Yes | Yes | cn_contest | false | 710,706 |
5. Calculate: $: \frac{C_{11}^{11}}{1}+\frac{C_{11}^{1}}{2}+\frac{C_{11}^{2}}{3}+\cdots \cdots+\frac{C_{11}^{11}}{12}$ | 5. $\frac{1365}{4}$ | \frac{1365}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,707 |
II. (16 points) Let $n \in \mathbf{N}$, and suppose that $37.5^{n} + 26.5^{n}$ is a positive integer. Find the value of $n$. | $$
=, 37.5^{n}+26.5^{n}=\frac{1}{2^{n}}\left(75^{n}+53^{n}\right) \text {. }
$$
When $n$ is a positive even number,
$$
75^{n}+53^{n} \equiv(-1)^{n}+1^{n} \equiv 2(\bmod 4) \text {, }
$$
i.e., $75^{n}+53^{n}=4 l+2$ (here $\left.\cdot l \in \mathbf{N}\right)$.
Thus, $37.5^{n}+26.5^{n}=\frac{1}{2^{n-1}}(2 l+1)$ is not a... | n=1,3,5,7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,713 |
Three, (16 points) It is known that on the parabola $y=x^{2}$, there are three vertices $A, B$, and $C$ of a square. Find the minimum value of the area of such a square.
保留源文本的换行和格式，直接输出翻译结果。 | Let's assume that two of the three vertices are on the right side of the $y$-axis (including the $y$-axis), and let the coordinates of points $A$, $B$, and $C$ be $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$, respectively, with the slope of $BC$ being $k (k>0)$. Then we have... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,714 |
$$
\begin{array}{l}
\text { Four. (18 points) Let } f(x) \\
=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+ \\
a_{0}, g(x)=c_{n+1} x^{n+1}+ \\
c_{n} x^{n}+\cdots+c_{0} \text { be two real }
\end{array}
$$
polynomials with non-zero coefficients, and there exists a real number $r$ such that $g(x)=$ $(x-r) f(x)$. Let $a=\max \left(... | \[
\begin{array}{l}
\quad k^{\pi} 、 \because c_{n+1} x^{n+1}+c_{n} x^{n}+\cdots+c_{n}=(x-r)\left(a_{n} x^{n} \right. \\
\left. a_{n-1} x^{n-1}+\cdots+a_{0}\right), \\
\therefore c_{n+1}=a_{n}, \\
c_{n}=a_{n-1}-r a_{n}, \\
c_{n-1}=a_{n-2}-r a_{n-1}, \\
\cdots \cdots \cdots \\
c_{1}=a_{0}-r a_{1}, \\
c_{0}=-r a_{0} . \\
... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,715 |
1. The monotonic decreasing interval of the function $\left.y=\log _{\frac{1}{5}} \right\rvert\, x-2$ ! is ( ).
(A) $(-\infty, 2)$
(B) $(-\infty,-2) \cup(2,+\infty)$
(C) $(2,+\infty)$
(D) $(0,2) \cup(2,+\infty)$ | $$
\begin{array}{l}
\text { - } 、 1 .(\mathrm{C}) \text {. } \\
\text { Let } f(x)=|x-2| \text {, its }
\end{array}
$$
graph is shown in the right figure. When $x>2$, $y=\log _{\frac{1}{3}}|x-2|$ is a decreasing function. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,716 |
Example 7 Let $P$ be a point outside the equilateral $\triangle A B C$, and $P A=3, P B=4, P C=5$. Find the side length of $\triangle A B C$. | Solution: As shown in Figure 10, construct a regular $\triangle A P D$ with $P A$ as one side, and connect $B D$. It is easy to see that $\triangle A B D \cong \triangle A C P$. Therefore, $B D = P C = 5$.
In $\triangle B P D$, given $P D = 3, P B = 4, B D = 5$, we know $\angle B P D = 90^{\circ}$. It is also known th... | \sqrt{25 - 12 \sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,717 |
2. Given non-zero real numbers $x, 3, z$ form an arithmetic sequence, $x+1, y, z$ and $x, y, z+2$ each form a geometric sequence. Then the value of $y$ is ( ).
(A) 8
(B) 16
(C) 12
(D) 20 | 2. (C).
From the given information, we can obtain the system of equations $\left\{\begin{array}{l}2 y=x+z, \\ y^{2}=z(x+1), \\ y^{2}=x(z+2) .\end{array}\right.$ From (2) - (3) we get $z=2 x$. Substituting into (1) gives $x=\frac{2}{3} y$. Therefore, $z=\frac{4}{3} y$.
Substituting back into (2) yields the value of $y$... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,718 |
3. Let $z$ be a complex number, the argument of $z+2$ is $\frac{\pi}{3}$, and the argument of $z-2$ is $\frac{5 \pi}{6}$. Then $z$ equals ( ).
(A) $-\sqrt{3}+i$
(B) $-1+\sqrt{3} i$
(C) $-\frac{\sqrt{3}}{2}+\frac{1}{2} i$
(D) $-\frac{1}{2}+\frac{\sqrt{3}}{2} i$ | 3. (B).
Let $z=x+yi$, it is easy to know that $y>0$. According to the problem, we have $\left\{\begin{array}{l}\frac{y}{x+2}=\sqrt{3}, \\ \frac{y}{x-2}=-\frac{\sqrt{3}}{3} .\end{array}\right.$
By dividing the two equations, we get $x=\cdots \therefore y=\sqrt{3}$. Thus, $z=-1+\sqrt{3} i$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,719 |
4. Given the line $\left\{\begin{array}{l}x=t \cos \alpha, \\ y=t \sin \alpha\end{array}\right.$ ( $t$ is a parameter) and the circle $\left\{\begin{array}{l}x=4+2 \cos \theta, \\ y=2 \sin \theta\end{array}\right.$ ( $\theta$ is a parameter) are tangent to each other. Then the inclination angle of the line is ( ).
(A) ... | 4. (A).
Substitute the line $\left\{\begin{array}{l}x=t \cos \alpha, \\ y=t \sin \alpha\end{array}\right.$ into the equation $(x-4)^{2}+y^{2}=4$ to get
$$
t^{2}-8 t \cos \alpha+12=0 \text {. }
$$
By the problem, $\Delta=64 \cos ^{2} \alpha-48=0$,
$$
\therefore \cos \alpha= \pm \frac{\sqrt{3}}{2} \text {. }
$$
Also, ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,720 |
5. Let $a, b, c$ be three different lines, and $\alpha, \beta, \gamma$ be two different planes. Which of the following propositions is true? ( ).
(A) If $\alpha \perp \gamma, \beta \perp \gamma$, then $\alpha / / \beta$
(B) If $a \perp b, b \perp c$, then $a / / c$ or $a \perp c$
(C) If $a \subset \alpha, b, c \subset ... | 5. (D).
(A) $\alpha$ and $\beta$ can intersect, so (A) is incorrect; (B) $a$ and $c$ can be skew and not perpendicular, so (B) is incorrect; (C) when $a \perp b, a \perp c$, $\alpha$ and $\beta$ can be parallel, (C) is incorrect. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,721 |
6.8 times of shooting, hitting 3 times, among which there are exactly 2 consecutive hits. The number of such scenarios is ( .
(A) 15
(B) 30
(C) 48
(D) 60 | 6. (B).
Considering the two consecutive hits and the single hit as two elements to be inserted into 6 spaces created by the five positions, we find that the number of qualified scenarios is $P_{6}^{2}=6 \times 5=30$. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 710,722 |
7. If the base edge length and the side edge length of a regular pyramid are equal, then the pyramid must not be ( ).
(A) Triangular pyramid
(B) Square pyramid
(C) Pentagonal pyramid
(D) Hexagonal pyramid | 7. (D).
If the base edge length of a regular hexagonal pyramid is equal to the slant height, and since the base edge length is equal to the radius of the base, then in the right triangle formed by the height of the hexagonal pyramid, the radius of the base, and the slant height, the slant height and the radius of the ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,723 |
8. The polar equation of a circle is $\rho=\sqrt{2}(\cos \theta+$ $\sin \theta$ ), the coordinates of the center of the circle are ( ).
(A) $\left(2, \frac{\pi}{4}\right)$
(B) $\left(1, \frac{\pi}{4}\right)$
(C) $\left(1, \frac{3 \pi}{4}\right)$
(D) $\left(\sqrt{2},-\frac{\pi}{4}\right)$ | 8. (B).
Transform the equation of the circle into Cartesian coordinates, we get
$$
\begin{array}{l}
x^{2}+y^{2}-\sqrt{2} x-\sqrt{2} y=0, \\
\text { i.e., }\left(x-\frac{\sqrt{2}}{2}\right)^{2}+\left(y-\frac{\sqrt{2}}{2}\right)^{2}=1 .
\end{array}
$$
The center of the circle is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2... | (B) | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,724 |
1. If the 5th term of the expansion of $\left(x \sqrt{x}-\frac{1}{x}\right)^{6}$ is $\frac{15}{2}$, then $\lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-a}\right)=$ | $\begin{array}{l}\text { II, 1.1. } \\ \because T_{5}=C_{6}^{4}(x \sqrt{x})^{2} \cdot\left(-\frac{1}{x}\right)^{4}=\frac{15}{x}, \\ \text { from } \frac{15}{x}=\frac{15}{2} \text {, we get } x^{-1}=\frac{1}{2} . \\ \therefore \lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-n}\right) \\ =\frac{\frac{1}{2}}{1-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,725 |
2. A cylindrical bucket with a base diameter of $32 \mathrm{~cm}$ is filled with an appropriate amount of water. After placing an iron ball into the water, the ball is completely submerged, and the water level rises by $9 \mathrm{~cm}$ (no water spills out). The surface area of the ball is $\qquad$ $\mathrm{cm}^{2}$. | $$
2.576 \pi \mathrm{cm}^{2} \text {. }
$$
Let the radius of the sphere be $R$. According to the problem, we have
$$
\begin{array}{l}
\frac{4}{3} \pi R^{3}=\pi \cdot 16^{2} \times 9 \\
\therefore R=12 \\
\quad S=4 \pi R^{2}=4 \pi \cdot 12^{2}=576 \pi\left(\mathrm{cm}^{2}\right)
\end{array}
$$ | 576 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,726 |
Example 1 When $n \in \mathbf{N}$, is the integer part of $(3+\sqrt{7})^{n}$ odd or even? Please prove your conclusion.
(1980, Spring Festival Cup Mathematics Competition)
Analysis: Since $(3+\sqrt{7})^{n}$ can be expressed as the sum of an integer and a pure decimal, and this integer is what we are looking for. To det... | Proof: First, we affirm that the integer part of $(3+\sqrt{7})^{n}$ is odd.
Indeed, since $0<(3-\sqrt{7})^{n}<1$, and
$$
\begin{aligned}
& (3+\sqrt{7})^{n}+(3-\sqrt{7})^{n} \\
= & 2\left(3^{n} \mathrm{C}_{n}^{0}+7 \cdot 3^{n-2} \mathrm{C}_{n}^{2}+7^{2} \cdot 3^{n \cdot 4} \mathrm{C}_{n}^{4}\right. \\
& +\cdots) \\
\tri... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,728 |
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