problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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Example 6 Given that the roots of the equation $x^{2}-(k+3) x+k^{2}=0$ are integers. Find the integer value of $k$ and the roots of the equation. | $$
\text { Sol: } \begin{aligned}
\Delta & =[-(k+3)]^{2}-4 k^{2} \\
& =-3 k^{2}+6 k+9 \geqslant 0 \\
\Rightarrow k^{2} & -2 k-3 \leqslant 0 \Rightarrow-1 \leqslant k \leqslant 3
\end{aligned}
$$
$\Rightarrow$ integers $k=-1,0,1,2,3$.
By the quadratic formula, $x=\frac{(k+3) \pm \sqrt{\Delta}}{2}$, so
when $k=-1$, $\Del... | k=-1,0,2,3, x=1,0,3,4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,556 |
13. As shown in Figure 5, in the circle $\odot O$ with radius $r$, $AB$ is the diameter, $C$ is the midpoint of $\overparen{AB}$, and $D$ is the one-third point of $\overparen{CB}$. Moreover, the length of $\overparen{DB}$ is twice the length of $\overparen{CD}$; connect $AD$ and extend it to intersect the tangent line... | ミ，３，As shown in Figure 6, connect $B E$.
$\because A B$ is the diameter.
$$
\begin{array}{l}
\therefore \angle A D B=90^{\circ}, \\
\text { Also } \because \overparen{B D}=60^{\circ}, \\
\therefore \angle D A B=30^{\circ} .
\end{array}
$$
Thus, $B D=\frac{1}{2} A B=r$.
Connecting $O C$, we know that $O C \perp A B$ an... | 2r | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,563 |
14. A unit spent 500,000 yuan to purchase a piece of high-tech equipment. According to the tracking survey of this model of equipment: after the equipment is put into use, if the maintenance and repair costs are averaged to the first day, the conclusion is: the maintenance and repair cost on the $x$-th day is
$$
\left[... | 14. (1) The equipment is put into use for $x$ days. The average daily loss is:
$$
\begin{aligned}
y= & \frac{1}{x}\left[500000+\left(\frac{1}{4} \times 0+500\right)+\left(\frac{1}{4} \times 1+500\right)\right. \\
& \left.+\left(\frac{1}{4} \times 2+500\right)+\cdots+\left(\frac{x-1}{4}+500\right)\right] \\
= & \frac{1}... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,564 |
15. A piece of lead wire of length $2 n$ (where $n$ is a natural number and $n \geqslant 4$) is folded into a triangle with integer side lengths. Let $(a, b, c)$ represent a triangle with side lengths $a, b, c$ such that $a \leqslant b \leqslant c$.
(1) For the cases $n=4, 5, 6$, write down all the $(a, b, c)$ that sat... | 15. (1) When $n=4$, the length of the lead wire is 8. Then the only group of $(a, b, c)$ that satisfies the condition is $(2,3,3)$;
When $n=5$, the length of the lead wire is 10. Then the groups of $(a, b, c)$ that satisfy the condition are $(2,4,4),(3,3,4)$;
When $n=6$, the length of the lead wire is 12. Then the gr... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,565 |
1. Find all positive integers $n \geqslant 2$, such that for all integers $a$ and $b$ coprime with $n, a \equiv b(\bmod n)$ if and only if $a b \equiv 1(\bmod n)$. | The given condition is equivalent to each integer $a$ satisfying $(a, n)=1$,
$$
a^{2} \equiv 1(\bmod n) .
$$
In fact, if $a \equiv b(\bmod n)$ is equivalent to $a b \equiv 1(\bmod n)$, then by $a b \equiv 1(\bmod n)$, when $b=a$, we have $a^{2} \equiv 1(\bmod n)$. Conversely, if $a^{2} \equiv 1(\bmod n)$, for any inte... | n=2,3,4,6,8,12,24 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,566 |
Example 7 When is $x$ a rational number such that the algebraic expression $9 x^{2}$ $+23 x-2$ is exactly the product of two consecutive positive even numbers?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Let the two consecutive positive even numbers be $k, k+2$. Then
$$
9 x^{2}+23 x-2=k(k+2) \text {, }
$$
which means $9 x^{2}+23 x-\left(k^{2}+2 k+2\right)=0$.
Since $x$ is a rational number, the discriminant must be a perfect square, i.e.,
$$
\begin{aligned}
\Delta & =23^{2}+4 \times 9\left(k^{2}+2 k+2\right)... | x=2, -\frac{41}{9}, -17, \frac{130}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,567 |
4. Find all triples of positive integers $(a, m, n)$ such that $a^{m}+1$ divides $(a+1)^{n}$. | Proof: First, we prove a corollary of the unique factorization theorem for positive integers:
If $u$ divides $v^{\prime}$, then $u$ divides $(u, v)^{l}$, where the integer $l \geqslant 1$.
In fact, let $u=p_{1}^{a_{1}} p_{2}^{o_{2}} \cdots p_{k}^{o_{k}}, v=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}}$. ... | (a, m, n) \mid a=1 \text{ or } m=1 \text{ or } a=2, m=3, n \geqslant 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,568 |
5. Prove: There exist infinitely many positive integers $n$, such that $p=n r$, where $p$ and $r$ are the semiperimeter and the inradius of a triangle with integer side lengths, respectively. | Proof: Let $a, b, c$ and $S$ be the side lengths and area of a triangle that satisfies the conditions. From $S = pr$ and $S^2 = p(p-a)(p-b)(p-c)$, we get
$$
\begin{aligned}
p & = n r \Leftrightarrow p^2 = n S \Leftrightarrow p^4 = n^2 S^2 \\
& \Leftrightarrow p^4 = n^2 p(p-a)(p-b)(p-c) \\
& \Leftrightarrow (2p)^3 = n^2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,569 |
6. Given a set of positive integers $A$ where the elements cannot be expressed as the sum of several distinct perfect squares. Prove: $A$ contains a finite number of elements. | Proof: There exists a positive integer $N$ such that
$$
N=a_{1}^{2}+a_{2}^{2}+\cdots+a_{m}^{2}, 2 N=b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2} \text {, }
$$
where $a, a_{2}, \cdots, a_{m}, b_{1}, b_{2}, \cdots, b_{n}$ are positive integers, and for any $\alpha, \beta, \gamma, \delta$, when $\alpha \neq \beta, \gamma \neq \d... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 712,570 |
1. Given two circles intersecting at $X$ and $Y$. Prove: There exist four points such that for any circle tangent to the two given circles at $A$ and $B$ and intersecting the line $X Y$ at $C$ and $D$, then $A C$, $A D$, $B C$, and $B D$ pass through one of these four points. | Proof: Let $\Omega$ be a circle tangent to two given circles at $A$ and $B$, and because it intersects the line $XY$, it is either externally tangent or internally tangent to both given circles. When internally tangent, there are two cases; we will only prove one of them.
As shown in Figure 1, let $CA$ intersect the c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,571 |
3. Let $O$ and $H$ be the circumcenter and orthocenter of an acute $\triangle ABC$, respectively. Prove that there exist points $D$, $E$, and $F$ on $BC$, $CA$, and $AB$ respectively, such that
$$
O D+D H=O E+E H=O F+F H,
$$
and the lines $A D$, $B E$, and $C F$ are concurrent. | Proof: Let the extension of $AH$ intersect the circumcircle of $\triangle ABC$ at $L$, and intersect $BC$ at $K$. Connect $OL$ to intersect $BC$ at $D$, and connect $HD$. Since $HK = KL$, then $HD = LD$. Therefore,
$$
OD + DH = OD + DL = OL = R,
$$
where $R$ is the circumradius of $\triangle ABC$.
Similarly, points $E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,572 |
5. Draw the tangents to the circumcircle of acute $\triangle A B C$ at $B$ and $A$, and let them intersect the tangent at $C$ at $T$ and $U$, respectively. Let $A T$ intersect $B C$ at $P$, and let $Q$ be the midpoint of $A P$. Let $B U$ intersect $A C$ at $R$, and let $S$ be the midpoint of $B R$. Prove that $\angle A... | Proof: Let $a=BC, b=CA, c=AB$. Since $\angle ABT=180^{\circ}-\angle C, \angle ACT=180^{\circ}-\angle B$, and $BT=CT$, then
$$
\begin{array}{l}
\frac{BP}{PC}=\frac{S_{\triangle ABT}}{S_{\triangle ACT}}=\frac{\frac{1}{2} AB \cdot BT \sin \left(180^{\circ}-C\right)}{\frac{1}{2} AC \cdot CT \sin \left(180^{\circ}-B\right)}... | a: b: c=\sqrt{2}: \sqrt{2}: 1 | Geometry | proof | Yes | Yes | cn_contest | false | 712,574 |
6. Let $A B C D$ be a convex quadrilateral, and $A B$ is not parallel to $C D$. If $X$ is a point inside quadrilateral $A B C D$ such that $\angle A D X = \angle B C X < 90^{\circ}, \angle D A X = \angle C B X < 90^{\circ}$, and let $Y$ be the intersection of the perpendicular bisectors of $A B$ and $C D$. Prove: $\ang... | Proof: Let $Z$ be the second intersection point of the circumcircles of $\triangle A D X$ and $\triangle B C X$, denoted as $\Gamma_{1}$ and $\Gamma_{2}$, with centers $O_{1}$ and $O_{2}$, respectively; let $W$ be the second intersection point of the circumcircles of $\triangle A B Z$ and $\triangle C D Z$, denoted as ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,575 |
For example, let $8 a$ be a real number greater than zero. It is known that there exists a unique real number $k$ such that the quadratic equation in $x$
$$
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 $a$. | Solution: Let the two prime roots of the equation be \( p \) and \( q \). By the relationship between roots and coefficients, we have
\[
\begin{array}{l}
p+q=-(k^{2}+a k), \\
p q=1999+k^{2}+a k .
\end{array}
\]
Adding (1) and (2), we get \( p+q+p q=1999 \).
Thus, \((p+1)(q+1)=2^{4} \times 5^{3}\).
From (3), it is clea... | 2 \sqrt{502} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,577 |
1. Given $\left(a^{2}+1\right)\left(b^{2}+1\right)=3(2 a b-1)$. Then the value of $b$ $\left(\frac{1}{a}-a\right)$ is ( ).
(A)0
(B) 1
(C) -2
(D) -1 | -1. (D).
From the condition, we have $(a b-2)^{2}+(a-b)^{2}=0$.
Then $a b=2, a=b, a^{2}=2$.
Therefore, $b\left(\frac{1}{a}-a\right)=b \cdot \frac{1-a^{2}}{a}$
$$
=1-a^{2}=1-2=-1 \text {. }
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,578 |
2. Let positive integers $a$, $m$, $n$ satisfy $\sqrt{a^{2}-4 \sqrt{2}}=\sqrt{m}-$ $\sqrt{n}$. Then the values of such $a$, $m$, $n$ are ( ).
(A) one set
(B) two sets
(C) more than two sets
(D) do not exist
Translate the above text into English, please retain the original text's line breaks and format, and output the ... | 2. (A).
Squaring both sides of the original equation, we get
$$
a^{2}-4 \sqrt{2}=m+n-2 \sqrt{m n} \text {. }
$$
Given that $a$, $m$, and $n$ are natural numbers, thus, $a^{2}-4 \sqrt{2}$ is an irrational number. Therefore,
$$
\left\{\begin{array} { l }
{ \sqrt { m n } = \sqrt { 8 } , } \\
{ m + n = a ^ { 2 } }
\end{... | A | Number Theory | proof | Yes | Yes | cn_contest | false | 712,579 |
3. As shown in Figure 1, in $\triangle A B C$, point $D$ is on side $B C$. It is known that $A B=A D=2, A C$ $=4$, and $B D: D C=2: 3$. Then $\triangle A B C$ is (
(A) Acute triangle
(B) Right triangle
(C) Obtuse triangle
(D) Cannot be determined | 3. (B).
Let $BD=2x$, then $DC=3x$. Draw $AH \perp BD$ at $H$, then $BH=x, CH=4x$. From $2^{2}-x^{2}$. $=AB^{2}-BH^{2}=AH^{2}=$ $AC^{2}-CH^{2}=4^{2}-(4x)^{2}$, solving gives $x^{2}=\frac{4}{5}$.
At this point, $BC^{2}=(5x)^{2}=25 \times \frac{4}{5}=20=2^{2}+4^{2}=$ $AB^{2}+AC^{2}$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,580 |
4. Let $b$ take odd numbers between 2 and 16, and $c$ take any natural number. Then the number of quadratic equations $3 x^{2}+$ $(b+1) x+c=0$ that can be formed with two distinct real roots is ( ).
(A) 64
(B) 66
(C) 107
(D) infinitely many | 4. (A).
$\because$ The equation $3 x^{2}+(b+1) x+c=0$ has two distinct real roots, $\therefore \Delta=(b+1)^{2}-12 c>0 \Rightarrow c<\frac{1}{12}(b+1)^{2}$.
Given $b=3,5,7,9,11,13,15$.
When $b=3$, $c<\frac{4}{3}$, there is 1 value of $c$ that meets the condition; when $b=5$, $c<3$, there are 2 values of $c$ that meet t... | 64 | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,581 |
5. In a convex quadrilateral $A B C D$, $A C$ bisects $\angle B A D$, and $C B = C D$. Then the relationship between $\angle B$ and $\angle D$ is ( ).
(A) equal
(B) unequal
(C) equal or supplementary
(D) complementary or supplementary | $\begin{array}{l}\text { 5. (C). } \\ \text { As shown in figure } 4, \text { when } A B \\ =A D, \triangle A B C \\ \cong \triangle A D C, \\ \text { when } A B \neq A B, \\ \text { construct } \triangle A B^{\prime} C \cong \\ \triangle A I C. M, \\ \angle B = \angle D \\ \text { or } \angle B \text { and } \angle D ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,582 |
1. The maximum value of $y=2 x+\sqrt{6-x}$ is | $=.1 .12 \frac{1}{8}$.
Let $\sqrt{6-x}=t(t \geqslant 0)$, then $x=6-t^{2}$.
Thus $y=-2 t^{2}+t+12=-2\left(t-\frac{1}{4}\right)^{2}+12 \frac{1}{8}$.
Therefore, when $t=\frac{1}{4}$, i.e., $x=\frac{95}{16}$, $y_{\text {max }}=12 \frac{1}{8}$. | 12 \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,583 |
3. The square number $y^{2}$ is
the sum of the squares of 11 consecutive integers. Then the smallest value of the natural number $y$ is
$\qquad$ | 3.11.
Let the middle number of these 11 consecutive integers be \( a \). Then
\[
\begin{aligned}
y^{2}= & (a-5)^{2}+(a-4)^{2}+(a-3)^{2}+(a-2)^{2} \\
& +(a-1)^{2}+a^{2}+(a+1)^{2}+(a+2)^{2} \\
& +(a+3)^{2}+(a+4)^{2}+(a+5)^{2} \\
= & 11 a^{2}+2\left(5^{2}+4^{2}+3^{2}+2^{2}+1^{2}\right) \\
= & 11\left(a^{2}+10\right) .
\e... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,585 |
5. Given that the quadratic function $f(x)$ has a maximum value of 5 when $x=m (m>0)$, the quadratic function $g(x)$ has a minimum value of -2, and $g(m)=$ 25, and $f(x)+g(x)=x^{2}+16 x+13$. Then the expression for $f(x)$ is $\qquad$ . | $$
\begin{array}{l}
\text { 5. }-2 x^{2}+4 x+3 . \\
\because f(x)+g(x)=x^{2}+16 x+13, \\
f(m)=5, g(m)=25, m>0, \\
\therefore m^{2}+16 m+13=30 .
\end{array}
$$
Solving, we get $m=1$.
Let $f(x)=a(x-1)^{2}+5=a x^{2}-2 a x+5+a$.
Then $g(x)=\left(x^{2}+16 x+13\right)-\left(a x^{2}-2 a x+5+a\right)$
$$
=(1-a) x^{2}+(2 a+16)... | -2 x^{2}+4 x+3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,587 |
Example 9 Find the integer $k$ that satisfies the following conditions, such that the roots of the quadratic equation $(k-1) x^{2}+(k-5) x+k=0$ are all integers. | Solution: Let the two roots of the equation be $x_{1}$ and $x_{2}$. Then
$$
\begin{array}{l}
x_{1}+x_{2}=-\frac{k-5}{k-1}=-1+\frac{4}{k-1}, \\
x_{1} x_{2}=\frac{k}{k-1}=1+\frac{1}{k-1},
\end{array}
$$
and $x_{1}+x_{2}$ and $x_{1} x_{2}$ are both integers.
Thus, $\frac{4}{k-1}$ and $\frac{1}{k-1}$ are both integers.
Th... | k=0 \text{ or } 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,588 |
One, (20 points) Among the one hundred natural numbers from $1 \sim 100$, any 21 are chosen. Prove: there must exist four numbers, among which the sum of two numbers equals the sum of the other two numbers. | Among the 100 natural numbers from $1 \sim 100$, the sum of any two can only be $3,4,5, \cdots, 199$, a total of 197 possibilities.
When 21 natural numbers are randomly selected, the sum of any two among them totals $\frac{21 \times 20}{2}=210$ sum numbers. According to the pigeonhole principle, among these 210 sum nu... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 712,589 |
II. (25 points) As shown in Figure 3, $CD$ is the altitude on the hypotenuse $AB$ of the right $\triangle ABC$, and $M, N$ are points on $AC, BC$ respectively, such that $DM \perp DN$. Prove:
(1) The circumcircle $\odot O_{1}$ of $\triangle ADM$ and the circumcircle $\odot O_{2}$ of $\triangle BDN$ are externally tange... | Proof 1: From $\angle M C N + \angle M D N = 180^{\circ} \Rightarrow C, M, D, N$ are concyclic $\Rightarrow \angle C M D = \angle B N D$.
(1) The circumcenters of $\triangle A D M$ and $\triangle B D N$ are $O_{1}$ and $O_{2}$. Draw the auxiliary lines as shown in Figure 7. By the properties of circumcenters, we easily... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,590 |
1. Given $f(x)=\frac{\sqrt{1-x^{2}}}{|x+2|-2}, g(x)=$ $\lg \left(\sqrt{1+x^{2}}-x\right)$. Then the product function $F(x)=f(x) g(x)$ on the common domain is ()
(A) is an odd function but not an even function
(C) is both an odd function and an even function
(I)) is neither an odd function nor an even function | $-1 .(\mathrm{B})$.
First, find the domain of the two functions. For $f(x)$, we have
$$
\left\{\begin{array}{l}
1-x^{2} \geqslant 0, \\
|x+2|-2 \neq 0 .
\end{array}\right.
$$
This gives $\left\{\begin{array}{l}-1 \leqslant x \leqslant 1, \\ x \neq 0 \text { and } x \neq-4 .\end{array}\right.$
Thus, the domain is $(-1,... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,592 |
2. Given $\sin \alpha=x, \cos \beta=y, \cos (\alpha+\beta)=-\frac{11}{14}$, and $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$. Then the functional relationship between $y$ and $x$ is ( ).
(A) $y=-\frac{11}{14} \sqrt{1-x^{2}}+\frac{5 \sqrt{3}}{14} x\left(\frac{11}{14}<x<1\right)$
(B) $y=-\frac{11}{14} \sqrt{1-x^{2}}+... | 2. (A).
From $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$ we know
$$
\cos \alpha=\sqrt{1-x^{2}}, \sin (\alpha+\beta)=\frac{5 \sqrt{3}}{14} .
$$
We have
$$
\begin{aligned}
y & =\cos \beta=\cos [(\alpha+\beta)-\alpha] \\
& =\cos (\alpha+\beta) \cdot \cos \alpha+\sin (\alpha+\beta) \cdot \sin \alpha \\
& =-\frac{11}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,593 |
3. Given the three sides of $\triangle A B C$ are $a, b, c$. If $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ form an arithmetic sequence in order, then $\angle B($ ).
(A) must be an acute angle
(B) must be a right angle
(C) must be an obtuse angle
(1)) cannot be determined, depends on the values of $a, b, c$ | 3. (A).
From $\frac{1}{a} 、 \frac{1}{b} 、 \frac{1}{c}$ forming an arithmetic sequence, we know that
$$
\frac{1}{a} \geqslant \frac{1}{b} \geqslant \frac{1}{c} \text { or } \frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c} \text {, }
$$
which means $a \leqslant b \leqslant c$ or $a \geqslant b \geqslant c$.
By t... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,594 |
4. As shown in Figure 1, there is a unit cube container $A B C D-A_{1} B_{1} C_{1} D_{1}$ with a volume of 1. There are holes at the midpoints of edges $A B$, $B B_{1}$, and diagonal $B C_{1}$, denoted as $E$, $F$, and $G$ respectively. If this container can be placed in any orientation, then the maximum volume of wate... | 4. (C).
Using the reverse deduction method: If taking (U), it would be equivalent to removing the volume of the small triangular prism $E-B G$ $=\frac{1}{3} \times \frac{1}{4} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{48}$ (Figure 5). However, since it is a liquid, the liquid surface will form a quadrilateral $E ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,595 |
5. Given parallel lines $x+y=2 a-1$ and the family of concentric circles $x^{2}+y^{2}=a^{2}+2 a-3$ intersect at $\left(x_{0}, y_{0}\right)$. When $x_{0} y_{0}$ takes the minimum value, the value of $a$ is ( ).
(A) 1
(B) -3
(C) $2-\frac{\sqrt{2}}{2}$
(D) $2+\frac{\sqrt{2}}{2}$ | 5. (C).
From the inequality $(x+y)^{2} \leqslant 2\left(x^{2}+y^{2}\right)$, we get
$$
(2 a-1)^{2} \leqslant 2\left(a^{2}+2 a-3\right),
$$
which simplifies to $2 a^{2}-8 a+7 \leqslant 0$.
Solving this, we get $2-\frac{\sqrt{2}}{2} \leqslant a \leqslant 2+\frac{\sqrt{2}}{2}$.
When $x=y$, $a$ can take the endpoints.
$$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,596 |
6. If $a, b$ are non-zero integers, and $(a, b)$ satisfies the equation $x^{2}-9 x y+y^{2}-9=0$, then $(a, b)$ is called a non-zero integer solution of the equation. Among the following statements about the non-zero integer solutions of the equation, the false statement is ( ).
(A) Non-zero integer solutions do not exi... | 6. (D).
Starting from the simplest case to find non-zero integer solutions. Clearly, when $x=0$, $y= \pm 3$ is an integer solution. Substituting $y=3$, we get the solutions $(0,3)$, $(27,3)$. Clearly, $(3,27)$ is also a solution.
Substituting $y=27$, we get two more solutions $(3,27)$, $(240, 27)$. Clearly, $(27,240)... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,597 |
1. Given $\sin x+\sin y=\frac{1}{3}$. Then the maximum value of $\sin y-\cos ^{2} x$ is . . $\qquad$ | $$
\text { Ni. } 1 \cdot \frac{4}{9} \text {. }
$$
From $\sin y=\frac{1}{3}-\sin x$ belonging to $[-1,1]$, we know
$$
-\frac{2}{3} \leqslant \sin x \leqslant \frac{4}{3} \text {. }
$$
Thus, $-\frac{2}{3} \leqslant \sin x \leqslant 1$.
$$
\begin{aligned}
\text { Also, } & \sin y-\cos ^{2} x \\
& =\left(\frac{1}{3}-\si... | \frac{4}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,598 |
Example 10 Given that $b$ and $c$ are integers, the equation $5 x^{2} + b x + c = 0$ has two roots that are greater than -1 and less than 0. Find the values of $b$ and $c$. | Solution: According to the graph of the quadratic function $y=5 x^{2}+b x+c$ and the given conditions:
When $x=0$, $5 x^{2}+b x+c>0$, we have $c>0$;
When $x=-1$, $5 x^{2}+b x+c>0$, we have $b^{2} \geqslant 20 c, c<5$.
If $c=1$, then from (2) and (4) we get $0<b<$ ú H. $b^{2} \equiv$ 20 , we get $b=5$;
If $c=2$, then $... | b=5, c=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,599 |
4. From the vertex of the parabola $y=x^{2}$, draw two perpendicular chords $O A$ and $O B$, and construct $O M \perp A B$. Then the equation of the locus of point $M$ is | 4. $x^{2}+y^{2}-y=0$.
Given that $A$ and $B$ are on the parabola and $O A \perp O B$, we can set the coordinates of $A$ and $B$ as $\left(m, m^{2}\right),\left(-\frac{1}{m}, \frac{1}{m^{2}}\right)(m \neq 0)$. Then the equation of $A B$ is
$$
\frac{y-m^{2}}{x-m}=\frac{m^{2}-\frac{1}{m^{2}}}{m+\frac{1}{m}}=\frac{m^{2}-1... | x^{2}+y^{2}-y=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,602 |
5. Given $a_{n}=6^{n}+8^{n}$. Then $a_{84} \equiv$ $\qquad$ $(\bmod 49)$ | 5.2.
$$
\begin{aligned}
a_{84}= & (7-1)^{84}+(7 \\
& +1)^{84} \\
= & 2\left(\mathrm{C}_{84}^{0} \cdot 7^{84}+\right. \\
& \mathrm{C}_{84}^{2} \cdot 7^{82}+\cdots \\
& \left.+\mathrm{C}_{84}^{82} \cdot 7^{2}+\mathrm{C}_{84}^{84}\right) \\
= & 45 \times M+2 . \\
a_{84}= & 2(\text { modulo } 49) .
\end{aligned}
$$ | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,603 |
6. The number of planes for which the ratio of distances to the 4 vertices of a regular tetrahedron is $1: 1: 1: 2$ is $\qquad$. | 6.32
Let the vertices of a regular tetrahedron be $A, B, C, D$. The plane $\alpha$ that divides the distances to these four points in the ratio $1: 1: 1: 2$ can be of two types:
(1) $A, B, C$ are on the same side of $\alpha$, there are 2 such planes (as shown in Figure 8).
(1) $\frac{A A_{1}}{A_{1} D}=\frac{B B_{1}}{B... | 32 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,604 |
Three. (20 points) Given that the function $f(x)$ is defined on $(-1,1)$. $f\left(\frac{1}{2}\right)=-1$, and satisfies for $x, y \in(-1,1)$,
$$
f(x)+f(y)=f\left(\frac{x+y}{1+x y}\right) .
$$
(1) Prove: $f(x)$ is an odd function on $(-1,1)$;
(2) For the sequence $x_{1}=\frac{1}{2}, x_{n+1}=\frac{2}{1+\frac{x_{n}}{x_{n}... | (1) For $0 \in(-1,1)$, we have
$$
f(0)+f(0)=f\left(\frac{0+0}{1+0}\right)=f(0) .
$$
Thus, $f(0)=0$.
For $x \in(-1,1)$, we have
$$
f(x)+f(-x)=f\left(\frac{x-x}{1-x^{2}}\right)=f(0)=0 .
$$
This implies $f(-x)=-f(x)$.
Therefore, $f(x)$ is an odd function on $(-1,1)$.
(2) Taking $x=y$, we have
$$
2 f(x)=f\left(\frac{2 x}... | 1+f\left(\frac{1}{5}\right)+f\left(\frac{1}{11}\right)+\cdots+f\left(\frac{1}{n^{2}+3 n+1}\right)+f\left(\frac{1}{n+2}\right)=0 | Algebra | proof | Yes | Yes | cn_contest | false | 712,605 |
Four, (20 points) A basketball on the court casts an elliptical shadow under the illumination of a point light source. Discuss whether the point of contact between the basketball and the ground is a focus of the elliptical shadow, and provide reasons.
---
A basketball on the court casts an elliptical shadow under the... | Four, the answer is affirmative. The proof is as follows.
As shown in Figure 10, let the sphere $O$ touch the ground $M$ at $F$, and $S$ be the light source. The conical surface formed by the light source tangent to the sphere intersects the sphere at $\odot O_{1}$. Let the angle between the generatrix of the conical s... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,606 |
One, (50 points) As shown in Figure 4, extend the sides of the convex pentagon $A B C D E$ outward to form five triangles:
$\triangle F A B, \triangle G B C, \triangle H C D, \triangle K D E, \triangle L E A$.
Prove: The five intersection points of the circumcircles of these five triangles, $A_{1}, B_{1}, C_{1}, D_{1},... | Since the five circles are of equal status, it is sufficient to prove that points $A_{1} 、 B_{1} 、 D_{1} 、 E_{1}$ are concyclic. By analogy, we can also prove that points $A_{1} 、 B_{1} 、 C_{1} 、 E_{1}$ are concyclic, thus proving that the five points are concyclic. To prove that points $A_{1} 、 B_{1} 、 D_{1} 、 E_{1}$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,608 |
II. (50 points) The player numbers of a sports team are uniquely selected from the positive integers 1 to 100. If the number of any player is neither the sum of the numbers of any other two players nor twice the number of another player, what is the maximum number of players this sports team can have? | Second, all odd numbers $1, 3, \cdots, 99$ totaling 50 can be the numbers of the team members. Below is the proof that it is impossible to increase further. If the sports team
has 51 members, from smallest to largest, denoted as
$$
a_{1}, a_{2}, \cdots, a_{50}, a_{51} \text {. }
$$
Take the differences
$$
a_{51}-a_{1... | 50 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,609 |
Example 11 Try to find all positive integers $a$ such that the equation $a x^{2}+2(2 a-1) x+4(a-3)=0$ has at least one integer solution. | Solution: Since the parameter $a$ is linear, we can express $a$ in terms of $x$, i.e.,
$$
a=\frac{2(x+6)}{(x+2)^{2}}.
$$
Since $a$ is a positive integer, then $\frac{2(x+6)}{(x+2)^{2}} \geqslant 1$.
Solving this, we get $-4 \leqslant x \leqslant 2$ and $x \neq -2$.
Thus, $x=-4,-3,-1,0,1,2$.
Substituting these into (1)... | a=1,3,6,10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,610 |
Three. (50 points) In a convex $n$-sided rose garden $(n \geqslant 4)$, there is 1 red rose planted at each of the $n$ vertices. There is a straight path between every two red roses, and these paths do not have the situation of "three lines intersecting at one point."
--- They divide the garden into many non-overlappi... | Three, (1) Solution 1: The straight paths in the rose garden form a convex $n$-sided polygon and its diagonals $\left(k_{n}\right)$. It is easy to know that there are $C_{n}^{2}-n$ diagonals in the graph. Furthermore, since the diagonals do not have "three lines intersecting at one point", the number of intersection po... | 99 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,611 |
Let $a, b, c \in \mathbf{R}^{+}$. Try to prove:
$$
\sqrt[3]{\left(\frac{a}{b+c}\right)^{2}}+\sqrt[3]{\left(\frac{b}{c+a}\right)^{2}}+\sqrt[3]{\left(\frac{c}{a+b}\right)^{2}} \geqslant \frac{3}{\sqrt[3]{4}}
$$ | Proof: By the three-variable mean inequality, we have
$$
\begin{array}{l}
2 a(b+c)^{2}=2 a(b+c)(b+c) \\
\leqslant\left[\frac{2 a+(b+c)+(b+c)}{3}\right]^{3} \\
=\frac{8(a+b+c)^{3}}{27} . \\
\text { Then } \sqrt[3]{\left(\frac{a}{b+c}\right)^{2}} \geqslant \sqrt[3]{\frac{27 a^{3}}{4(a+b+c)^{3}}} \\
=\frac{3}{\sqrt[3]{4}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,612 |
$=6$ integer solutions.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: Let the two integer roots of the equation be $x_{1}$ and $x_{2}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-(10 a+b), \\
x_{1} x_{2}=10 b+a .
\end{array}\right.
$$
We have $10 x_{1}+10 x_{2}+x_{1} x_{2}=-99 a(1 \leqslant a \leqslant 9)$.
Thus $\left(x_{1}+10\right)\left(x_{2}+10\right)=100-99 a$.
By the p... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,613 |
High 107 Let the three sides of $\triangle ABC$ be $a, b, c$, and the corresponding angles in radians be $\alpha, \beta, \gamma, n$ be a natural number. Prove:
$$
\left(\frac{\pi}{3}\right)^{n} \leqslant \frac{a \alpha^{n}+b \beta^{n}+c \gamma^{n}}{a+b+c}<\frac{\pi^{n}}{2} .
$$ | Proof: $\because(\beta-\gamma)\left(b \beta^{n-1}-c \gamma^{n-1}\right)+(\gamma-\alpha)$.
$$
\begin{array}{l}
\left(c \gamma^{n-1}-a \alpha^{n-1}\right)+(\alpha-\beta)\left(a \alpha^{n-1}-b \beta^{n-1}\right) \geqslant 0, \\
\therefore 2\left(a \alpha^{n}+b \beta^{n}+c \gamma^{n}\right) \\
\geqslant a \alpha^{n-1}(\bet... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,614 |
108 Given $a, b, c \in \mathbf{R}^{+}, a b c=1$. Prove:
$$
(a+b)(b+c)(c+a) \geqslant 4(a+b+c-1) .
$$ | Proof: Without loss of generality, let $a \geqslant 1$. Equation (1) can be rewritten as
$$
\begin{array}{l}
a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)+6 \\
\geqslant 4(a+b+c),
\end{array}
$$
which is equivalent to $\left(a^{2}-1\right)(b+c)+\left(b^{2}-1\right)(a+c)+\left(c^{2}-1\right)(a+b)+6 \geqslant 4 a+3(b+c)$.
Since $(a+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,615 |
Example 1 Using the two vertices of a triangle and $n(n \geqslant 1)$ points inside it as vertices, construct all non-overlapping small triangles. Find the number of these small triangles. | Solution: Since each of the $n$ points inside the original triangle is a common vertex of several small triangles, and the sum of the angles at that vertex in these small triangles is $360^{\circ}$, the total sum of the interior angles of all the small triangles is $360^{\circ} \cdot n + 180^{\circ}$. Therefore, the nu... | 2n + 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,616 |
Example 2 If a wire of length 20 is used to form a triangle with all three sides of integer lengths, then the number of non-congruent triangles is ( ).
(A) 5
(B) 6
(C) 8
(D) 10
(1998, Beijing Junior High School Mathematics Competition Preliminary) | Solution: Let the three sides of the triangle be $a, b, c$. And let $a \leqslant b \leqslant c$, then $a+b>c$. Therefore,
$$
2 c<a+b+c=20 \leqslant 3 c \text {. }
$$
We have $6 \frac{2}{3} \leqslant c \leqslant 10$.
Thus, $c=7,8,9$.
(1) When $c=7$, $(a, b)=(6,7)$;
(2) When $c=8$,
$$
(a, b)=(4,8),(5,7),(6,6) \text {; }... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,617 |
1. Inside a triangle (not on the sides), there are 3 points. Together with the three vertices of the original triangle, there are 6 points in total. Using these 6 points as vertices, construct all non-overlapping triangles. If no three points among these 6 points are collinear, the number of triangles constructed is $n... | (A)
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 712,618 |
2. In the convex quadrilateral $ABCD$, $AB=CD$, $AC$ is a diagonal, $\angle DAC>\angle BCA$, and $\angle DAC$ and $\angle BCA$ are supplementary, $\angle BAC>\angle ACD$, and $\angle BAC$ and $\angle ACD$ are complementary. Then $\angle B=$ | $2.45^{\circ}$.
Extend $B C$ to $E$, such that $C E=A D$.
Connect $A E$, then $\triangle D A C \cong$ $\triangle E C A$. We have $\angle D=\angle E$, $\angle A C D=\angle E A C, A E=C D$. $\because A B=C D$,
$\therefore A B=A E, \angle B=\angle E=45^{\circ}$. | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,619 |
3. $m, n$ are positive integers. If $\frac{2000}{2001}<\frac{n}{m}<\frac{2001}{2002}$, then the fraction $\frac{n}{m}=$ $\qquad$ when $m$ is the smallest. | 3. $\frac{4001}{4003}$.
First, give the general solution formula: $m, n, a, b, c, d$ are all positive integers, and $\frac{a}{b}m a, m c>n d$. We can set
$\left\{\begin{array}{l}b n-a m=x, \\ c m-d n=y .\end{array}(x, y\right.$ are positive integers $)$
Solving, we get $\left\{\begin{array}{l}m=\frac{d x+b y}{b c-a d}... | \frac{4001}{4003} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,620 |
4. Through the vertex of isosceles $\triangle A B C$, draw a line intersecting the extension of the opposite side at $D$. If the resulting triangles are all isosceles, then $\triangle A B C$ has $\qquad$ such configurations. | 4.5.
Draw a straight line from the base angle to intersect with the extension of the opposite side, there are 5 $\triangle A B C$ that meet the conditions, with base angles of $45^{\circ} 、 72^{\circ} 、 36^{\circ} 、 \frac{180^{\circ}}{7}$ 、 $\frac{2 \times 180^{\circ}}{7}$.
Draw a straight line from the vertex angle t... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,621 |
(20 points) On a certain day in a certain month in 2001 at midnight, teachers and students of a certain school were watching the live broadcast from Moscow of the announcement of the host city for the 2008 Summer Olympic Games. Beijing won the bid. In their excitement, they noticed that the number of people present was... | Let the number of teachers be $a$, the number of students be $b$, the number of months be $c$, the number of days be $d$, and the number of girls be $e$. From the problem, we have
$$
\left\{\begin{array}{l}
a+b=d, \\
c+p=b . \\
b+c+d-a \leqslant 30, \\
au$, so $e$ is even.
Substituting (1) into (1), we get $2 b+c \leqs... | 29 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,622 |
Three. (25 points) Find real numbers $a, b$, such that the function $y_1 = x^2 + ax + b$, $y_2 = x^2 + bx + a$ and the x-axis have four intersection points, where the distances between adjacent points are equal.
---
Translate the above text into English, please retain the original text's line breaks and format, and o... | Three, from the problem, we know $a \neq b$. Without loss of generality, let $a > b$, and set $y_{1}=y_{2}$, then $x^{2}+a x+b=x^{2}+b x+u$.
We get $x=1$, so $y_{1}$ and $y_{2}$ intersect only at the point $(1, a+b+1)$.
The graphs of $y_{1}$ and $y_{2}$ that meet the problem's conditions have 4 forms:
Figure 5
Without... | a=0, b=-4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,624 |
1. Function $f(x)=x^{a}, x \in(-1,0) \cup(0$, 1). If the inequality $f(x)>|x|$ holds, then under the condition $\alpha \in$ $\left\{-2,-1,-\frac{2}{3}, 0, \frac{1}{3}, \frac{2}{3}, 1,2\right\}$, the number of values $\alpha$ can take is ( ).
(A) 4
(B) 3
(C) 2
(D) 1 | $-1 . \mathrm{A}$.
Let $I=(0,1) \cup(-1,0)$. To make $f(x)=x^{a}>|x|$, $x^{a}$ should be greater than 0 on $I$, knowing that $a=-1, 1, \frac{1}{3}$ obviously do not meet the requirements.
When $a=0$, $f(x)=1,0<|x|$ holds;
When $\alpha=2$, on $I$, $f(x)>1,0<|x|$;
When $a=\frac{2}{3}$, $f(x)=|x|^{\frac{2}{3}}$, by $\fr... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 712,625 |
2. The 12 medians of the 4 equilateral triangular faces of a regular tetrahedron can form \( n \) different acute angles, then the value of \( n \) is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 2. B.
The cosine values of the angles formed by the medians on the surface of a regular tetrahedron's equilateral triangles are $\frac{1}{6}$, $\frac{1}{3}$, $\frac{5}{6}$, $\frac{2}{3}$, and $\frac{1}{2}$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,626 |
3. When $a$ and $b$ are both rational numbers, the point $P(a, b)$ is called a rational point. Let $A(\sqrt{1998}, 0), B(0$, $\sqrt{2000})$, then on the line $A B$ ( .
(A) there are no rational points
(B) there is only one rational point
(C) there are only two rational points
(D) there are infinitely many rational poin... | 3. A.
Let the rational point $P(a, b)$ be on the line $AB$, then
$$
\begin{array}{l}
b=-\frac{\sqrt{2000}}{\sqrt{1998}}(a-\sqrt{1998}), \\
b^{2}=\frac{2000}{1998}\left(a^{2}+1998-2 a \sqrt{1998}\right) .
\end{array}
$$
When $a$ is a non-zero rational number, $b^{2}$ is an irrational number, which contradicts the fact... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 712,627 |
4. In the complex plane, the maximum internal angle of the triangle formed by the points corresponding to the complex numbers $\frac{1}{\mathrm{i}}$, $\frac{2}{\mathrm{i}-1}$, and $(\mathrm{i}-1)^{3}$ is equal to ( ).
(A) $\pi-\arccos \frac{2}{\sqrt{13}}$
(B) $-\arccos \frac{2}{\sqrt{13}}$
(C) $45^{\circ}$
(D) $120^{\c... | $$
\begin{array}{l}
\text { 4. A. } \\
\text { Given } \frac{1}{i}=-i, \frac{2}{i-1}= \\
-1-i, (i-1)^{3}=2+2 i
\end{array}
$$
we know that the points corresponding to them are $A$,
$B$, and $C$ as shown in Figure 2. The largest interior angle of $\triangle ABC$ is $\angle A$.
$$
\begin{aligned}
& \text { Given } \arc... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,628 |
2. The number of triangles with unequal integer sides and a perimeter less than 13 is $\qquad$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | (3)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 3 | Number Theory | proof | Yes | Yes | cn_contest | false | 712,629 |
5. If the infinite decreasing geometric sequence $\left\{a_{n}\right\}$ satisfies
$$
\lim _{n \rightarrow \infty} \frac{a_{1}+a_{4}+a_{7}+\cdots+a_{3 n-2}}{a_{1}+a_{2}+\cdots+a_{n}}=\frac{3}{4},
$$
then, the common ratio $q$ of the sequence is ().
(A) $\frac{\sqrt{21}-3}{6}$
(B) $\frac{-3-\sqrt{21}}{6}$
(C) $\frac{\sq... | 5. A.
$$
\begin{array}{l}
\lim _{n \rightarrow \infty} \frac{a_{1}+a_{4}+\cdots+a_{3 n-2}}{a_{1}+a_{2}+\cdots+a_{n}}=\frac{a_{1}}{1-q^{3}} \cdot \frac{1-q}{a_{1}} \\
=\frac{1}{q^{2}+q+1} . \\
\text { Also } \frac{1}{1+q+q^{2}}=\frac{3}{4},
\end{array}
$$
We have $3 q^{2}+3 q-1=0$.
Then $q=\frac{-3 \pm \sqrt{21}}{6}$, ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,630 |
6. At the bottom of a 15 cm long, uniformly thick cylindrical candle, a thin metal sheet (negligible volume) is fixed, allowing the candle to float vertically in water with 1 cm of its top end above the water surface. Given that the density of the candle is 0.85 g/cm³. Now, if the candle is lit, the total length burned... | 6.C.
Let the cross-sectional area of the candle be $\mathrm{S}$ square centimeters, and the weight of the metal plate be $a$ grams. According to the problem, we have $15 \times S \times 0.85 + a = 14 \times S \times 1$.
Suppose that after the 1 centimeter of the candle above the water is burned, the candle rises $x$ ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,631 |
1. The solution to the inequality $\lg x-\frac{1}{\lg x} \leqslant 0$ is | $$
\text { 2.1. }\left(0, \frac{1}{10}\right] \cup(1,10] \text {. }
$$
From the original inequality, we have $\frac{(\lg x+1)(\lg x-1)}{\lg x} \leqslant 0$.
Then $0<\lg x \leqslant \frac{1}{10}$ or $1<\lg x \leqslant 10$.
Hence, the solution set of the original inequality is $\left(0, \frac{1}{10}\right] \cup(1,10]$. | \left(0, \frac{1}{10}\right] \cup(1,10] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,632 |
2. The smallest positive integer $x$ that satisfies the equation $\tan 19 x^{\circ}=\frac{\cos 99^{\circ}+\sin 99^{\circ}}{\cos 99^{\circ}-\sin 99^{\circ}}$ is $x=$ $\qquad$ . | 2.36.
The right side of the original equation $=\frac{1+\tan 99^{\circ}}{1-\tan 99^{\circ}}=\frac{\tan 45^{\circ}+\tan 99^{\circ}}{1-\tan 45^{\circ} \cdot \tan 99^{\circ}}$ $=\tan 144^{\circ}$.
Then $19 x=144+180 k(k \in \mathbf{N})$.
So $x=\frac{180 k+144}{19}=9 k+7+\frac{9 k+11}{19}$.
We know $19 \mid 9 k+11$. Thus... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,633 |
3. In a regular tetrahedron with edge length $a$, a cylinder is embedded such that its height is exactly half the height of the tetrahedron, and its top base is tangent to the tetrahedron's side faces. Then the surface area of this cylinder is $\qquad$ . | 3. $\frac{\pi(2 \sqrt{2}+1)}{24} a^{2}$.
As shown in Figure 3, let the regular tetrahedron be $P-ABC$, and $O$ be the centroid of the base $\triangle ABC$. $E$ is the midpoint of the height $PO$, and $AD$ is the median of the equilateral $\triangle ABC$. $F$ is the midpoint of $PD$. It is easy to see that $EF \paralle... | \frac{\pi(2 \sqrt{2}+1)}{24} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,634 |
4. $M$ is a moving point on the circle $x^{2}+y^{2}-6 x-8 y=0$, $O$ is the origin, and $N$ is a point on the ray $O M$. If $|O M| \cdot|O N|=150$, then the equation of the locus of point $N$ is $\qquad$ . | $4 \cdot 3 x+4 y-75=0$.
This problem can be solved using the special point method.
As shown in Figure 4, it is easy to know that the radius of the given circle is $r=5$, and the center of the circle is point $O_{1}(3,4)$. Let the ray $O O_{1}$ intersect the circle at $M_{1}$, with $\left|O M_{1}\right|=$ 10. Take poin... | 3 x+4 y-75=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,635 |
5. In Pascal's Triangle, each number is the sum of the two numbers directly above it. The first few rows of this triangle are as follows:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Row 0 & \multicolumn{6}{|c|}{1} \\
\hline Row 1 & & & 1 & & 1 & \\
\hline Row 2 & & 1 & & 2 & & 1 \\
\hline Row 3 & & & 3 & & 3 & 1 \\
\hline ... | 5. Line 62.
In Pascal's Triangle, its $n$-th row consists of binomial coefficients $C_{n}^{k}$ $(k=0,1, \cdots, n)$. If the ratio of three consecutive terms in the $n$-th row of Pascal's Triangle is $3: 4: 5$, then there exists a positive integer $k$ such that
$$
\begin{array}{l}
\frac{3}{4}=\frac{C_{n}^{k-1}}{C_{n}^{... | 62 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,636 |
6. Given the following conditions for the sequence $\left\{a_{n}\right\}$:
(1) $\left\{2^{a_{n}}\right\}$ is a geometric sequence;
(2) The sum of the first $n$ terms $S_{n}=n^{2}+1$;
(3) $a_{1}>0$, and $a_{k}=\frac{2}{k-1}\left(a_{1}+a_{2}+\cdots+\right.$ $\left.a_{k-1}\right), k \geqslant 2$.
The condition that makes ... | 6. (1)(3).
First, consider (1). Let $b_{n}=2^{a_{n}}, n \geqslant 1, n \in \mathbf{N}$, then $a_{n}=\log _{2} b_{n}$. Since $\left\{2^{a_{n}}\right\}$ is a geometric sequence, we have
$2^{a_{n}}=b_{1} \times q^{n-1}\left(b_{1}, q\right.$ are positive constants $)$,
thus, $a_{n}=\log _{2} b_{n}=\log _{2} b_{1}+(n-1) \l... | (1)(3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,637 |
Three. (20 points) Given point $A(\sqrt{5}, 0)$ and the curve $y=$ $\sqrt{\frac{x^{2}}{4}-1}(2 \leqslant x \leqslant 2 \sqrt{5})$ with points $P_{1} 、 P_{2}, \cdots$ 、 $P_{n}$. If $\left|P_{1} A\right| 、\left|P_{2} A\right| 、 \cdots 、\left|P_{n} A\right|$ form an arithmetic sequence with common difference $d \in\left(\... | The given curve is a segment of the following hyperbola, i.e.,
$$
\frac{1}{4} x^{2}-y^{2}=1(2 \leqslant x \leqslant 2 \sqrt{5}, y \geqslant 0) \text {. }
$$
$A(\sqrt{5}, 0)$ is its right focus, as shown in Figure 5 (where the line $l$ is the right directrix $x=\frac{4}{\sqrt{5}}$, and the point $P(2 \sqrt{5}, 2)$, with... | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,638 |
3. Draw 6 chords in a circle, dividing the circle into \( n \) plane parts. The largest \( n \) is ().
(A) 18
(B) 22
(C) 24
(D) 32 | (B)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,640 |
Five. (20 points) Given $f(x)=a x^{3}+b x^{2}+c x +d$, and $f(1) 、 f(0) 、 f(-1) 、 f(2)$ are all integers. Prove: When $x$ is an integer, $f(x)$ is also an integer. | Given $f(0)=d, f(1)=a+b+c+d$, $f(-1)=-a+b-c+d, f(2)=8a+4b+2c+d$.
$\because f(0) 、 f(1)$ are integers,
$\therefore d 、 a+b+c$ are integers.
Also, $\because f(-1)=2b+d-(a+b+c)$ is an integer.
$\therefore 2b$ is an integer.
$\because f(2)=6a+2b+d+2(a+b+c)$ is an integer,
$\therefore 6a+2b$ is an integer, and $6a$ is also ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,641 |
一、(50 points) Given that the incircle $\odot I$ of $\triangle A B C$ trisects the median $A M$. Prove: The ratio of the three sides of $\triangle A B C$ is $5: 10: 13$.
---
Note: The translation maintains the original format and line breaks as requested. | As shown in Figure 7, let
the incircle $\odot I$ of $\triangle ABC$ touch the three sides at $D, E, F$,
$AM$ intersects $\odot I$ at $P, Q$, and connect $IA, IM, IC, IE, IP$.
Draw $IN \perp PQ$ at $N$, then $PN = QN$.
Also, $AP = MQ$, then $AN = NM$, so $IA = IM$.
Also, $\angle ACI = \angle MCI$, in $\triangle ACI$ a... | 5: 10: 13 | Geometry | proof | Yes | Yes | cn_contest | false | 712,642 |
II. (50 points) There are three types of stocks. The sum of the number of shares of the first two types equals the number of shares of the third type. The total value of the second type of stock is four times that of the first type. The total value of the first and second types of stocks equals the total value of the t... | Let the number of shares of the first, second, and third types of stocks be $k$, $m$, and $n$, respectively, and the value of each share be $x$, $y$, and $z$ (yuan). The problem requires finding the value of:
$$
f = \frac{k}{k+m+n} \times 100\%
$$
From the given information,
$$
\begin{array}{l}
k+m=n, \quad k x+m y=n z... | 12.5\% \leqslant f \leqslant 15\% | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,643 |
$$
\begin{array}{c}
\text { Three. (50 points) Given } x_{i} \in \mathbf{R}, a_{i} 、 b_{i} \in \mathbf{Z}^{+}(i= \\
1,2, \cdots, n), \text { let } \\
a=\frac{a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}}{a_{1}+a_{2}+\cdots+a_{n}}, \\
b=\frac{b_{1} x_{1}+b_{2} x_{2}+\cdots+b_{n} x_{n}}{b_{1}+b_{2}+\cdots+b_{n}} .
\end{arr... | $$
\begin{array}{l}
|a-b| \leqslant\left|a-x_{p}\right| . \\
\left|a-x_{p}\right| \leqslant\left|x_{q}-x_{p}\right| .
\end{array}
$$
Since $|a-b|=\left|a-\frac{b_{1} x_{1}+b_{2} \cdot x_{2}+\cdots+b_{n} \cdot x_{n}}{b_{1}+b_{2}+\cdots+b_{n}}\right|$
$$
\begin{array}{l}
=\frac{\mid b_{1}\left(a-x_{1}\right)+b_{2}\left(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,644 |
Let $P_{n}=(1+1)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{7}\right) \cdots$ $\left(1+\frac{1}{3 n-2}\right)$. Find the greatest integer part of $P_{2000}$. | $$
\begin{array}{l}
\text{First, prove the following inequality:} \\
\sqrt[3]{\frac{7 n+1}{7 n-6}}1) .
\end{array}
$$
The inequality (1) is equivalent to
$$
\begin{array}{l}
\frac{7 n+1}{7 n-6}0 \\
\Leftrightarrow 27(n-1)^{2}+13(n-1)>0 .
\end{array}
$$
The last inequality holds for $n>1$, thus inequality (1) holds.
S... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,645 |
On a circle with 110 points, 5 points can be connected to form a convex pentagon and a pentagram. Prove: the product of the distances from any point on this circle to the five sides of the convex pentagon is equal to the product of the distances from that point to the five sides of the pentagram. | Proof: First, prove the proposition: If lines $l_{1}$ and $l_{2}$ are tangent to $\odot O$ at $A$ and $B$, and $P$ is any point on $\odot O$, $P Q \perp A B$, $P E \perp l_{1}$, $P F \perp l_{2}$, with $Q$, $E$, and $F$ being the feet of the perpendiculars, then $P Q^{2}=P E \cdot P F$.
Connect $P A$, $P B$, $Q E$, and... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,646 |
In a string composed of decimal digits, if it contains an even number of digit 0, it is considered a valid codeword. For example, 001230407986 is a valid codeword, while 209807004508 is an invalid codeword. Find the number of valid codewords of length $n$. | Let $a_{n}$ denote the number of valid codewords of length $n$. Since in a digit string of length 1, only the digit string 0 is an invalid codeword, we have $a_{1}=9$. A valid codeword of length $n$ can be obtained by adding a digit to a digit string of length $n-1$ in two ways.
The first method is to add a non-zero d... | \frac{1}{2}\left(8^{n}+10^{n}\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,647 |
Let $a, b, c \in \mathbf{R}^{+}$. Try to prove:
$$
\frac{a}{b^{2}}+\frac{b}{c^{2}}+\frac{c}{a^{2}} \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c} .
$$ | Proof: By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{a}{b^{2}}+\frac{2 b}{c^{2}}+\frac{4 c}{a^{2}} \geq 7 \cdot \sqrt[7]{b^{2}} \cdot\left(\frac{b}{c^{2}}\right)^{2} \cdot\left(\frac{c}{a^{2}}\right)^{4}=\frac{7}{a}, \\
\frac{b}{c^{2}}+\frac{2 c}{a^{2}}+\frac{4 a}{b^{2}} \geq \frac{7}{b}, \frac{c}{a^{2}}+\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,648 |
4. The distance between two points $A$ and $B$ on a plane is $a+b$, where $a, b$ are constants greater than 0. Now there are $x$ lines on the plane such that the distances from points $A$ and $B$ to this line are $a$ and $b$ respectively. Then $x$ equals ( ).
(A) infinitely many
(B) 3
(C) 2
(D) 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 | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,649 |
5. In the plane of isosceles $\triangle A B C(A B=A C \neq B C)$, there is a point $P$ such that $\triangle P A B$, $\triangle P B C$, and $\triangle P A C$ are all isosceles triangles. Then the number of points that satisfy this condition is ( ).
(A) 1
(B) 3
(C) 6
(D) 7 | (C)
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 | 712,650 |
6. Given that $\triangle A B C$ is a right triangle, $\angle C$ is the right angle, $A C$ $\neq B C$. If point $P$ is a point on the plane where $\triangle A B C$ is located, $P \neq A, B, C$, such that the triangle formed by $P, B, C$ is similar to $\triangle A B C$, then the maximum number of such points $P$ is $\qqu... | (1)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,651 |
7 Find points on the sides of a rectangle such that the lines connecting this point to the endpoints of the opposite side divide the rectangle into three mutually similar triangles. Such points ( ).
(A) do not exist
(B) must have 2
(C) must have 4
(D) the number cannot be determined
保留了源文本的换行和格式。 | (I)
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly. | not found | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,652 |
Example 1 If on the interval $[1,2]$, the function $f(x)=x^{2}+p x+q$ and $g(x)=x+\frac{1}{x^{2}}$ take the same minimum value at the same point, then the maximum value of $f(x)$ on this interval is ( ).
(A) $4+\frac{11}{2} \sqrt[3]{2}+\sqrt[3]{4}$
(B) $4-\frac{5}{2} \sqrt[3]{2}+\sqrt[3]{4}$
(C) $1-\frac{1}{2} \sqrt[3]... | Solution: $\because 1 \leqslant x \leqslant 2$,
$$
\begin{array}{l}
g(x)=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^{2}} \geqslant 3 \sqrt[3]{\frac{x}{2} \cdot \frac{x}{2} \cdot \frac{1}{x^{2}}} \\
=\frac{3}{2} \sqrt[3]{2} .
\end{array}
$$
When and only when $\frac{x}{2}=\frac{1}{x^{2}}$, i.e., $x=\sqrt[3]{2}$, $g(x)$ has the... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,654 |
Example 2 If $f(x) (x \in \mathbf{R})$ is an even function with a period of 2, and when $x \in [0,1]$, $f(x) = x^{\frac{1}{19988}}$, then the ascending order of $f\left(\frac{98}{19}\right), f\left(\frac{101}{17}\right), f\left(\frac{104}{15}\right)$ is | Solution: Given that $x \in \mathbf{R}, k \in \mathbf{Z}$,
$$
\begin{array}{l}
\because f(2 k+x)=f(x)=f(-x), \\
\begin{aligned}
\therefore f\left(\frac{98}{19}\right) & =f\left(6-\frac{16}{19}\right)=f\left(-\frac{16}{19}\right) \\
& =f\left(\frac{16}{19}\right), \\
f\left(\frac{101}{17}\right) & =f\left(6-\frac{1}{17}... | f\left(\frac{101}{17}\right)<f\left(\frac{98}{19}\right)<f\left(\frac{104}{15}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,655 |
Example 3 As shown in Figure 1, in the right trapezoid $A B C D$, $A B=7, A D=$ $2, B C=3$. If point $P$ on side $A B$ makes the triangle with vertices $P, A, D$ similar to the triangle with vertices $P, B, C$, then the number of such points $P$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | Solution: (1) If $\triangle A P D \backsim \triangle B P C$, then $\frac{A P}{7-A P}=\frac{2}{3}$. Solving this, we get $A P=\frac{14}{5}$.
(2) If $\triangle A P D \backsim \triangle B C P$, then $\frac{A P}{3}=\frac{2}{7-A P}$. Solving this, we get $A P=1$ or 6. In summary, there are 3 points $P$, so the answer is (C)... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,656 |
Example 3 If $\left(\log _{2} 3\right)^{x}-\left(\log _{5} 3\right)^{r} \geqslant\left(\log _{2} 3\right)^{-v}$ $-\left(\log _{5} 3\right)^{-\nu}$, then $(\quad$.
(A) $x-y \geqslant 0$
(B) $x+y \geqslant 0$
(C) $x-y \leqslant 0$
(D) $x+y \leqslant 0$ | Sol: $\because 0<\log _{5} 3<1<\log _{2} 3$,
$\therefore y_{1}=\left(\log _{2} 3\right)^{2}$ is an increasing function.
Also, $\because\left(\log _{5} 3\right)^{x}$ is a decreasing function,
$\therefore y_{2}=-\left(\log _{5} 3\right)^{\prime}$ is an increasing function.
Therefore, $y=y_{1}+y_{2}=\left(\log _{2} 3\righ... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 712,657 |
Example 5 Given that when $x \in[0,1]$, the inequality
$$
x^{2} \cos \theta-x(1-x)+(1-x)^{2} \sin \theta>0
$$
always holds. Try to find the range of $\theta$.
| Let $f(x)=x^{2} \cos \theta-x(1-x)$
$$
\begin{array}{l}
+(1-x)^{2} \sin \theta \\
=(\cos \theta+\sin \theta+1) x^{2}-(1+2 \sin \theta) \cdot x \\
+\sin \theta .
\end{array}
$$
From the problem, we know $\left\{\begin{array}{l}f(0)=\sin \theta>0 \\ f(1)=\cos \theta>0\end{array}\right.$.
Under conditions (1) and (2), th... | 2 k \pi+\frac{\pi}{12}<\theta<2 k \pi+\frac{5 \pi}{12}(k \in \mathbf{Z}) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,659 |
1. Let $f(x)$ be an odd function on $(-\infty,+\infty)$, $f(x+2) = -f(x)$, and when $0 \leqslant x \leqslant 1$, $f(x)=x$. Then $f(2.5)=$ ( ).
(A) 0.5
(B) -0.5
(C) 1.5
(D) -1.5 | (B)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,660 |
2. Find the range of real number $a$ such that for any real number $x$ and any $\theta \in\left[0, \frac{\pi}{2}\right]$, we have
$$
\begin{array}{l}
(x+3+2 \sin \theta \cdot \cos \theta)^{2}+(x+u \sin \theta+u \cos \theta)^{2} \\
\geqslant \frac{1}{8} .
\end{array}
$$ | $\left(a \leqslant \sqrt{6}\right.$ or $\left.a \geqslant \frac{7}{2}\right)$ | a \leqslant \sqrt{6} \text{ or } a \geqslant \frac{7}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 712,661 |
4. Given that $a, b, c$ are real numbers. The functions are $f(x)=a x^{2}+b x+c, g(x)=a x+b$; when $-1 \leqslant x \leqslant 1$, $|f(x)| \leqslant 1$.
( I ) Prove: $|c| \leqslant 1$;
(II ) Prove: When $-1 \leqslant x \leqslant 1$, $|g(x)| \leqslant 2$;
(III) Suppose $a>0$, when $-1 \leqslant x \leqslant 1$, the maximum... | ( (III) \( f(x)=2 x^{2}-1 \)) | f(x)=2 x^{2}-1 | Algebra | proof | Yes | Yes | cn_contest | false | 712,663 |
5. Let $f(x)$ be an even function defined on $(-\infty,+\infty)$ with a period of 2. On the interval $[2,3]$, $f(x)=-2(x-3)^{2}+4$.
(1) When $x \in[1,2]$, find the analytical expression for $f(x)$;
(II) If the rectangle $A B C D$ has two vertices $A$ and $B$ on the $x$-axis, and $C, D$ on the graph of the function $y=f... | $\begin{array}{l}\text { ((I) } f(x)=-2(x-1)^{2}+4(1 \leqslant x \leqslant 2),(\text { II) } \\ \left.\frac{16 \sqrt{6}}{4} .\right)\end{array}$ | \frac{16 \sqrt{6}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,664 |
6. Given the function $f(x)=\frac{2 x^{2}+b x+c}{x^{2}+1}(b<0)$ with the range $[1,3]$.
( [ ) Find the values of the real numbers $b, c$;
( II ) Determine the monotonicity of the function $F(x)=\lg f(x)$ on $x \in[-1,1]$, and provide a proof;
( III ) If $t \in \mathbf{R}$, prove that:
$\left.\lg \frac{7}{5} \leqslant F... | $((1) b=-2, c=2 ;(I I) F(x)$ is a decreasing function for $x \in[-1,1]$ 5.) | b=-2, c=2; F(x) \text{ is a decreasing function for } x \in[-1,1]; \lg \frac{7}{5} \leqslant F\left(\left|t-\frac{1}{6}\right| \cdots\left|t+\frac{1}{6}\right|\right) \leqslant \lg \frac{13}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,665 |
7. Given the function $f(x)=\log _{m} \frac{x-3}{x+3}$.
(1) If the domain of $f(x)$ is $[\alpha, \beta](\beta>\alpha>0)$. Determine the monotonicity of $f(x)$ over its domain, and provide a proof;
(II) If $0<\alpha<\beta$ and the domain of $f(x)$ is $[\alpha, \beta](\beta>\alpha>0)$, does such an interval $[\alpha, \be... | (( I) $01$ is an increasing function;
(II) When $0<m<\frac{2-\sqrt{3}}{4}$, there exists
$$
\begin{array}{l}
\alpha, \beta]=\left[\frac{-(2 m-1)-\sqrt{16 m^{2}-16 m+1}}{2 m},\right. \\
\left.-\frac{(2 m-1)+\sqrt{16 m^{2}-16 m+1}}{2 m}\right] ; \\
\text { When } \frac{2-\sqrt{3}}{4} \leqslant m<1 \text {, it does not ex... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,666 |
For example, $4 n$ lines intersect each other pairwise, but no three lines intersect at the same point. How many parts will such $n$ lines divide the plane into? | Solution: Let the $k$-th line divide the plane into $a_{k}\left(a_{1}=2\right)$ parts. It is easy to see that the $(k+1)$-th line can intersect the previous $k$ lines at most at $k$ points, and these $k$ points divide the $(k+1)$-th line into $k+1$ segments (including two rays). Each segment divides the region it lies ... | \frac{1}{2} n(n+1)+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,667 |
8. A ship travels uniformly against the current from point A to point B, with the distance between A and B being $s$ (kilometers), the water speed being a constant $p$ (kilometers/hour), and the ship's maximum speed in still water being $q$ (kilometers/hour) $(q>p)$. It is known that the hourly fuel cost (in yuan) is d... | $$
\text { (( I ) } y \leq k s \frac{v^{2}}{v-p}, v \in(p, q] \text {; ( II ) } 2 p \leq q
$$
When $2 p \leq q$, the actual speed of the ship's advance should be $p$. When $2 p > q$, the actual speed of the ship's advance should be $q-p$. | 2 p \leq q \text{, the actual speed of the ship's advance should be } p. \text{ When } 2 p > q, \text{ the actual speed of the ship's advance should be } q-p. | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,668 |
Example 1 Let $A B C D E F$ be a convex hexagon satisfying $A B=B C=C D, D E=E F=F A, \angle B C D=\angle E F A=60^{\circ}$. Let $G$ and $H$ be two points inside the hexagon such that
$$
\begin{array}{l}
\angle A G B=\angle D H E=120^{\circ} \text{. Prove that: } \\
A G+G B+G H+D H+H E \geqslant C F . \\
\quad(\text{IM... | This problem can be solved by the method of symmetry. Observing, we find that the hexagon $A B C D E F$ is composed of two equilateral triangles ($\triangle A F E$ and $\triangle C B I$) and a quadrilateral BDE, which is symmetric about the line $B E$. Construct the symmetric points $C^{\prime}$ and $F^{\prime}$ of $C$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,669 |
Example 2 Let $C_{1}$ and
$C_{2}$ be concentric circles, with the radius of $C_{2}$ being twice that of $C_{1}$. Quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is inscribed in $C_{1}$, and the extension of $A_{4} A_{1}$ intersects circle $C_{2}$ at $B_{1}$, the extension of $A_{1} A_{2}$ intersects circle $C_{2}$ at $B_{2}$, ... | Analysis: This problem is related to chords on a circle. It reminds us of the generalization of Ptolemy's theorem, and it involves an inequality. Therefore, we connect $O A_{2}$, $O B_{2}$, and $O B_{3}$. Let the radius of $\odot C_{1}$ be $r$. In quadrilateral $O A_{2} B_{2} B_{3}$, by Ptolemy's inequality, we have
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,670 |
Example 4, Let $A B C D E F$ be a convex hexagon, and $A B = B C, C D = D E, E F = F A$. Prove: $\frac{B C}{B E} + \frac{D E}{D A} + \frac{F A}{F C} \geqslant \frac{3}{2}$, and determine the condition for equality. (IMO38-Preliminary Problem) | Analysis: By observation, we find that the inequality involves six line segments, making it difficult to handle with many variables.
Let $A C=a, C E$
$=b, A E=c$. This way, we can use the Ptolemy inequality in multiple quadrilaterals.
In quadrilateral $A C E F$,
$$
A C \cdot E F + C E \cdot A F \geqslant A E \cdot C F ... | \frac{B C}{B E} + \frac{D E}{D A} + \frac{F A}{F C} \geqslant \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 712,672 |
Example 5 Given that $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at $W$, and they are both internally tangent to $\odot O$. Construct the external common tangent of $\odot O_{1}$ and $\odot O_{2}$ to intersect $\odot O$ at $A B$, and construct the internal common tangent of $\odot O_{1}$ and $\odot O_{2}$ to... | At this point, the reader may have noticed that Example 5 and Example 6 are related. In Example 6, when $\odot O_{1}$ and $\odot O_{2}$ are tangent, the two internal common tangents of $\odot O_{1}$ and $\odot O_{2}$ will coincide, and $G$ and $H$ will coincide, forming $W$ in Example 5. At this moment, $A B / / C D$ i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,673 |
Example 6 Given that $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$ are internally tangent to $\odot \mathrm{O}$, construct the two internal common tangents of $\odot O_{1}$ and $\odot O_{2}$, which intersect $\odot O$ at $A$ and $B$. Construct the external common tangent of $\odot O_{1}$ and $\odot O_{2}$, with poi... | Let $G$ and $H$ be the points of tangency of the internal common tangents of $\odot O_{1}$ and $\odot O_{2}$, respectively. Let $EF$ intersect $\odot O$ at points $C$ and $D$. Denote the length of the internal common tangent of $\odot O$ and $\odot O_{2}$ as $d$. For $[A.C. \odot O_{1}.D]$ (where “[* . * . * . *]” repr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,674 |
Example 1 Given $\odot\left(A, r_{A}\right) 、 \odot\left(B, r_{B}\right)$ 、 $\odot\left(C, r_{C}\right)$ intersect each other pairwise. Prove: The three common chords obtained from the pairwise intersections of these three circles are concurrent. | Proof: As shown in the figure,
4. From the fact that the line connecting the centers of two circles is perpendicular to their common chord, the problem can be viewed as drawing perpendiculars from the intersection points $A_{1}, B_{1}, C_{1}$ to the sides $BC, CA$, and $AB$ of $\triangle ABC$, and proving that these th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,676 |
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