2014 AIME II 第 11 题

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11.

在

\triangle RED

中，

RD = 1

、

\angle DRE = 75^\circ

、

\angle RED = 45^\circ

。令

M

为线段

\overline{RD}

的中点。点

C

在边

\overline{ED}

上，且

\overline{RC} \perp \overline{EM}

。将线段

\overline{DE}

沿

E

的方向延长至点

A

，使

CA = AR

。于是

AE = \frac{a - \sqrt{b}}{c}

，其中

a

和

c

是互质正整数，

b

是正整数。求

a + b + c

。

In

\triangle RED,

RD = 1,

\angle DRE = 75^\circ

and

\angle RED = 45^\circ.

Let

M

be the midpoint of segment

\overline{RD}.

Point

C

lies on side

\overline{ED}

such that

\overline{RC} \perp \overline{EM}.

Extend segment

\overline{DE}

through

E

to point

A

such that

CA = AR.

Then

AE = \frac{a - \sqrt{b}}{c},

where

a

and

c

are relatively prime positive integers, and

b

is a positive integer. Find

a + b + c.

答案

答案：56

知识点：坐标几何正弦定理斜率

难度评级：3060

小提示：

令

D

在原点，

E

在正

x

轴上；则

R = (\cos 60^\circ, \sin 60^\circ)

，并由正弦定理得到

DE = \frac{\sqrt{3}+1}{2}

Place

D

at the origin and

E

on the positive

x

-axis; then

R = (\cos 60^\circ, \sin 60^\circ)

and the law of sines gives

DE = \frac{\sqrt{3}+1}{2}

大提示：

用垂直斜率确定

x

轴上的

C

，而

CA = AR

会给出关于

A

的

x

坐标的一次方程

Perpendicular slopes locate

C

on the

x

-axis, and

CA = AR

is a linear equation in the

x

-coordinate of

A

解答：

因为

\angle RDE = 180^\circ - 75^\circ - 45^\circ

= 60^\circ

，令

D = (0,0)

，且

E

在正

x

轴上，则

R = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)

。正弦定理给出

DE = \frac{\sin 75^\circ}{\sin 45^\circ} = \frac{\sqrt{3}+1}{2}

，并且

M = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)

。

直线

EM

的斜率为

\frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} - \frac{\sqrt{3}+1}{2}} = \frac{-\sqrt{3}}{1 + 2\sqrt{3}}

，所以直线

RC

的斜率为

\frac{1 + 2\sqrt{3}}{\sqrt{3}}

。从

R

下降

\frac{\sqrt{3}}{2}

到

x

轴，水平向左移动

\frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{6\sqrt{3} - 3}{22}

，所以

C = (c, 0)

，其中

c = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{7 - 3\sqrt{3}}{11}

。

对

A = (t, 0)

，条件

CA = AR

写作

(t - c)^2 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4}

，这是关于

t

的一次方程：

t = \frac{1 - c^2}{1 - 2c} = \frac{9 + 4\sqrt{3}}{11}

。因此

\begin{aligned} AE &= t - \frac{\sqrt{3}+1}{2} \\ &= \frac{18 + 8\sqrt{3} - 11\sqrt{3} - 11}{22} \\ &= \frac{7 - 3\sqrt{3}}{22} = \frac{7 - \sqrt{27}}{22} \end{aligned}\text{，}

所以

a + b + c = 7 + 27 + 22 = 56

。

Since

\angle RDE = 180^\circ - 75^\circ - 45^\circ

= 60^\circ,

place

D = (0,0)

with

E

on the positive

x

-axis, so

R = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right).

The law of sines gives

DE = \frac{\sin 75^\circ}{\sin 45^\circ} = \frac{\sqrt{3}+1}{2},

and

M = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right).

The slope of

EM

is

\frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} - \frac{\sqrt{3}+1}{2}} = \frac{-\sqrt{3}}{1 + 2\sqrt{3}},

so line

RC

has slope

\frac{1 + 2\sqrt{3}}{\sqrt{3}}.

Descending from

R

by

\frac{\sqrt{3}}{2}

to the

x

-axis moves us left by

\frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{6\sqrt{3} - 3}{22},

so

C = (c, 0)

with

c = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{7 - 3\sqrt{3}}{11}.

For

A = (t, 0),

the condition

CA = AR

reads

(t - c)^2 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4},

which is linear in

t:

t = \frac{1 - c^2}{1 - 2c} = \frac{9 + 4\sqrt{3}}{11}.

Then

\begin{aligned} AE &= t - \frac{\sqrt{3}+1}{2} \\ &= \frac{18 + 8\sqrt{3} - 11\sqrt{3} - 11}{22} \\ &= \frac{7 - 3\sqrt{3}}{22} = \frac{7 - \sqrt{27}}{22}, \end{aligned}

so

a + b + c = 7 + 27 + 22 = 56.