2005 AIME II 第 14 题

Question

在三角形

ABC

中，

AB = 13

、

BC = 15

、

CA = 14

。点

D

在

\overline{BC}

上，且

CD = 6

。点

E

在

\overline{BC}

上，使得

\angle BAE \cong \angle CAD

。已知

BE = \frac{p}{q}

，其中

p

和

q

是互质正整数，求

q

。

In triangle

ABC,

AB = 13,

BC = 15,

and

CA = 14.

Point

D

is on

\overline{BC}

with

CD = 6.

Point

E

is on

\overline{BC}

such that

\angle BAE \cong \angle CAD.

Given that

BE = \frac{p}{q},

where

p

and

q

are relatively prime positive integers, find

q.

答案

答案：463

知识点：面积比三角学比与比例

难度评级：3060

小提示：

比较

\frac{BD}{DC}

与

\frac{BE}{EC}

：用

\frac{1}{2}xy\sin\theta

计算以

A

为顶点的三角形面积，并把每个比值写成面积之比。

Compare

\frac{BD}{DC}

and

\frac{BE}{EC}

by writing each as a ratio of triangle areas

\frac{1}{2}xy\sin\theta

at vertex

A

大提示：

相等的角会配对：两个比值相乘得到

\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}

。

The equal angles pair up: multiplying the two ratios gives

\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}

解答：

线段

AD

将对边分成的比满足

\begin{aligned} \frac{BD}{DC} &= \frac{[ABD]}{[ACD]} \\ &= \small \frac{\frac{1}{2} AB \cdot AD \sin\angle BAD}{\frac{1}{2} AC \cdot AD \sin\angle CAD} \\ &= \frac{AB \sin\angle BAD}{AC \sin\angle CAD} \end{aligned}\text{，}

同理，

\frac{BE}{EC} = \frac{AB \sin\angle BAE}{AC \sin\angle CAE}

。

因为

\angle BAE = \angle CAD

，也有

\angle BAD = \angle CAE

（二者都是这个公共角加上

\angle EAD

），所以两个比值相乘时所有正弦都约去：

\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}

。代入

BD = 9

、

DC = 6

、

AB = 13

和

AC = 14

，得

\frac{BE}{EC} = \frac{13^2}{14^2} \cdot \frac{6}{9} = \frac{169}{294}\text{。}

因而

BE = 15 \cdot \frac{169}{169 + 294} = \frac{2535}{463}

。由于

463

是质数，且不整除

2535 = 3 \cdot 5 \cdot 13^2

，该分数已最简，所以

q = 463

。

A cevian

AD

splits the opposite side in the ratio

\begin{aligned} \frac{BD}{DC} &= \frac{[ABD]}{[ACD]} \\ &= \small \frac{\frac{1}{2} AB \cdot AD \sin\angle BAD}{\frac{1}{2} AC \cdot AD \sin\angle CAD} \\ &= \frac{AB \sin\angle BAD}{AC \sin\angle CAD}, \end{aligned}

and similarly

\frac{BE}{EC} = \frac{AB \sin\angle BAE}{AC \sin\angle CAE}.

Since

\angle BAE = \angle CAD,

we also have

\angle BAD = \angle CAE

(each is that common angle plus

\angle EAD

), so multiplying the two ratios cancels all the sines:

\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}.

With

BD = 9,

DC = 6,

AB = 13,

and

AC = 14,

this gives

\frac{BE}{EC} = \frac{13^2}{14^2} \cdot \frac{6}{9} = \frac{169}{294}.

Hence

BE = 15 \cdot \frac{169}{169 + 294} = \frac{2535}{463}.

Since

463

is prime and does not divide

2535 = 3 \cdot 5 \cdot 13^2,

the fraction is in lowest terms, and

q = 463.

2005 AIME II 第 14 题

答案：463

难度评级：3060

小提示：

大提示：

解答：

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