2000 AIME I 第 14 题

Question

在三角形

ABC

中，角

B

与角

C

相等。点

P

和

Q

分别在

\overline{AC}

和

\overline{AB}

上，且

AP = PQ = QB = BC

。角

ACB

的大小是角

APQ

的

r

倍，其中

r

是正实数。求不超过

1000r

的最大整数。

In triangle

ABC,

it is given that angles

B

and

C

are congruent. Points

P

and

Q

lie on

\overline{AC}

and

\overline{AB},

respectively, so that

AP = PQ = QB = BC.

Angle

ACB

is

r

times as large as angle

APQ,

where

r

is a positive real number. Find the greatest integer that does not exceed

1000r.

答案

答案：571

知识点：等腰三角形正弦定理三角恒等式

难度评级：2990

小提示：

设

\angle A = \alpha

，并令

AP = 1

。等腰三角形

APQ

给出

\angle AQP = \alpha

和

AQ = 2\cos\alpha

。

Let

\angle A = \alpha

and

AP = 1.

Isosceles triangle

APQ

gives

\angle AQP = \alpha

and

AQ = 2\cos\alpha

大提示：

因为

BC = 1

，正弦定理给出

AB = \frac{1}{2\sin(\frac{\alpha}{2})}

，而

AQ + QB = AB

化为

\sin\frac{3\alpha}{2} = \frac{1}{2}

。

With

BC = 1,

the law of sines gives

AB = \frac{1}{2\sin(\frac{\alpha}{2})},

and

AQ + QB = AB

reduces to

\sin\frac{3\alpha}{2} = \frac{1}{2}

解答：

设

\angle A = \alpha

，并按比例缩放使

AP = PQ = QB = BC = 1

。在三角形

APQ

中，

AP = PQ

，所以

\angle AQP = \angle A = \alpha

，从而

\angle APQ = 180^\circ - 2\alpha

，且

AQ = \frac{\sin 2\alpha}{\sin \alpha} = 2\cos\alpha

。在三角形

ABC

中，

\angle B = \angle C = 90^\circ - \frac{\alpha}{2}

，所以

\begin{aligned} AB &= \frac{BC \sin C}{\sin A} \\ &= \frac{\cos(\frac{\alpha}{2})}{\sin\alpha} = \frac{1}{2\sin(\frac{\alpha}{2})} \end{aligned}\text{。}

由

AQ + QB = AB

，

\begin{aligned} 2\cos\alpha + 1 &= \frac{1}{2\sin(\frac{\alpha}{2})} \\ &\quad\Longrightarrow\quad 4\sin\tfrac{\alpha}{2}\cos\alpha \\ &\quad {}+ 2\sin\tfrac{\alpha}{2} = 1 \end{aligned}\text{。}

由积化和差公式，

4\sin\frac{\alpha}{2}\cos\alpha

= 2\sin\frac{3\alpha}{2}

- 2\sin\frac{\alpha}{2}

，所以方程化为

\sin\frac{3\alpha}{2} = \frac{1}{2}

。于是

\alpha = 20^\circ

或

\alpha = 100^\circ

，但后者会使

AQ = 2\cos\alpha

为负，故

\alpha = 20^\circ

。

此时

\angle ACB = 80^\circ

，

\angle APQ = 140^\circ

，所以

r = \frac{80}{140} = \frac{4}{7}

。于是

\lfloor 1000r \rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = 571

。

Let

\angle A = \alpha,

and scale so

AP = PQ = QB = BC = 1.

In triangle

APQ,

the equal sides

AP = PQ

give

\angle AQP = \angle A = \alpha,

so

\angle APQ = 180^\circ - 2\alpha

and, by the law of sines,

AQ = \frac{\sin 2\alpha}{\sin \alpha} = 2\cos\alpha.

In triangle

ABC,

\angle B = \angle C = 90^\circ - \frac{\alpha}{2},

so

\begin{aligned} AB &= \frac{BC \sin C}{\sin A} \\ &= \frac{\cos(\frac{\alpha}{2})}{\sin\alpha} = \frac{1}{2\sin(\frac{\alpha}{2})}. \end{aligned}

Since

AQ + QB = AB,

\begin{aligned} 2\cos\alpha + 1 &= \frac{1}{2\sin(\frac{\alpha}{2})} \\ &\quad\Longrightarrow\quad 4\sin\tfrac{\alpha}{2}\cos\alpha \\ &\quad {}+ 2\sin\tfrac{\alpha}{2} = 1. \end{aligned}

By the product-to-sum identity,

4\sin\frac{\alpha}{2}\cos\alpha

= 2\sin\frac{3\alpha}{2}

- 2\sin\frac{\alpha}{2},

so the equation collapses to

\sin\frac{3\alpha}{2} = \frac{1}{2}.

Then

\alpha = 20^\circ

or

\alpha = 100^\circ,

but the latter makes

AQ = 2\cos\alpha

negative, so

\alpha = 20^\circ.

Now

\angle ACB = 80^\circ

and

\angle APQ = 140^\circ,

so

r = \frac{80}{140} = \frac{4}{7},

and

\lfloor 1000r \rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = 571.

2000 AIME I 第 14 题

答案：571

难度评级：2990

小提示：

大提示：

解答：

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