2019 AIME I 第 13 题

Question

三角形

ABC

的边长为

AB = 4

、

BC = 5

、

CA = 6

。点

D

和

E

在射线

AB

上，且

AB \lt AD \lt AE

。点

F \neq C

是

\triangle ACD

与

\triangle EBC

的外接圆的一个交点，并满足

DF = 2

、

EF = 7

。则

BE

可以表示为

\frac{a + b\sqrt{c}}{d}

，其中

a

、

b

、

c

、

d

为正整数，

a

与

d

互质，且

c

不被任何质数的平方整除。求

a + b + c + d

。

Triangle

ABC

has side lengths

AB = 4,

BC = 5,

and

CA = 6.

Points

D

and

E

are on ray

AB

with

AB \lt AD \lt AE.

The point

F \neq C

is a point of intersection of the circumcircles of

\triangle ACD

and

\triangle EBC

satisfying

DF = 2

and

EF = 7.

Then

BE

can be expressed as

\frac{a + b\sqrt{c}}{d},

where

a,

b,

c,

and

d

are positive integers such that

a

and

d

are relatively prime, and

c

is not divisible by the square of any prime. Find

a + b + c + d.

答案

答案：32

知识点：圆内接四边形根轴圆幂余弦定理

难度评级：3370

小提示：

圆内接四边形给出

\angle FDA = \angle FCA

和

\angle FEB = \angle FCB

，所以

\angle DFE = \angle ACB

；接着用余弦定理求

DE

The cyclic quadrilaterals give

\angle FDA = \angle FCA

and

\angle FEB = \angle FCB,

so

\angle DFE = \angle ACB;

now the law of cosines finds

DE

大提示：

直线

CF

是根轴：若它与直线

AB

交于

G

，幂相等给出

GA \cdot GD = GB \cdot GE

。为定位

G

，使用角

\angle GCA

，这个角可由三角形

DEF

求出

Line

CF

is the radical axis: where it crosses line

AB

at

G,

the powers give

GA \cdot GD = GB \cdot GE.

Locate

G

using the angle

\angle GCA

found from triangle

DEF.

解答：

点

D

、

E

位于

B

以外的射线

AB

上，而点

F

在直线

AB

的另一侧（相对于

C

而言）。由于

ACFD

与

BCFE

为圆内接四边形，圆周角给出

\angle FDA = \angle FCA

和

\angle FEB = \angle FCB

。记

\alpha = \angle FCA

、

\beta = \angle FCB

，则三角形

DEF

有

\angle FDE = 180^\circ - \alpha

、

\angle FED = \beta

，所以

\angle DFE = \alpha - \beta = \angle ACB

。在三角形

ABC

中，

\cos \angle ACB = \frac{25 + 36 - 16}{2 \cdot 5 \cdot 6} = \frac{3}{4}

，所以在三角形

DFE

中用余弦定理得到

\begin{aligned} DE^2 &= 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \tfrac{3}{4} \\ &= 32, \\ DE &= 4\sqrt{2} \end{aligned}\text{。}

在三角形

DFE

中，

\cos \angle FDE = \frac{4 + 32 - 49}{2 \cdot 2 \cdot 4\sqrt{2}} = -\frac{13\sqrt{2}}{32}

，所以

\alpha

为锐角，且

\cos\alpha = \frac{13\sqrt{2}}{32}

、

\sin\alpha = \sqrt{1 - \frac{338}{1024}} = \frac{7\sqrt{14}}{32}

。设

G

为直线

CF

与直线

AB

的交点。在三角形

ACG

中，

\angle GAC = \angle BAC

，且

\cos \angle BAC = \frac{16 + 36 - 25}{2 \cdot 4 \cdot 6} = \frac{9}{16}

、

\sin \angle BAC = \frac{5\sqrt{7}}{16}

，又

\angle ACG = \alpha

，所以

\sin(\angle BAC + \alpha)

= \frac{5\sqrt{7}}{16} \cdot \frac{13\sqrt{2}}{32}

+ \frac{9}{16} \cdot \frac{7\sqrt{14}}{32}

= \frac{\sqrt{14}}{4}

，并且

\begin{aligned} AG &= \frac{AC \sin \alpha}{\sin(\angle BAC + \alpha)} \\ &= \frac{6 \cdot \frac{7\sqrt{14}}{32}}{\frac{\sqrt{14}}{4}} \\ &= \frac{21}{4} \end{aligned}\text{。}

直线

CF

是两圆的根轴，所以

GA \cdot GD = GB \cdot GE

。设

x = BD

，则

BE = x + DE = x + 4\sqrt{2}

：

\begin{aligned} &\frac{21}{4}\left(x - \frac{5}{4}\right) \\ &= \frac{5}{4}\left(x - \frac{5}{4} + 4\sqrt{2}\right) \end{aligned}\text{，}

因为

GD = 4 + x - \frac{21}{4}

，且

GB = \frac{21}{4} - 4

。这给出

16\left(x - \frac{5}{4}\right) = 20\sqrt{2}

，所以

x = \frac{5 + 5\sqrt{2}}{4}

，并且

BE = \frac{5 + 21\sqrt{2}}{4}

。因此

a + b + c + d = 5 + 21 + 2 + 4

= 32

。

Points

D,

E

lie beyond

B

on ray

AB,

and

F

lies on the opposite side of line

AB

from

C.

Since

ACFD

and

BCFE

are cyclic, the inscribed angles give

\angle FDA = \angle FCA

and

\angle FEB = \angle FCB.

Writing

\alpha = \angle FCA

and

\beta = \angle FCB,

triangle

DEF

has angles

\angle FDE = 180^\circ - \alpha

and

\angle FED = \beta,

so

\angle DFE = \alpha - \beta = \angle ACB.

From triangle

ABC,

\cos \angle ACB = \frac{25 + 36 - 16}{2 \cdot 5 \cdot 6} = \frac{3}{4},

so the law of cosines in triangle

DFE

gives

\begin{aligned} DE^2 &= 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \tfrac{3}{4} \\ &= 32, \\ DE &= 4\sqrt{2}. \end{aligned}

In triangle

DFE,

\cos \angle FDE = \frac{4 + 32 - 49}{2 \cdot 2 \cdot 4\sqrt{2}} = -\frac{13\sqrt{2}}{32},

so

\alpha

is acute with

\cos\alpha = \frac{13\sqrt{2}}{32}

and

\sin\alpha = \sqrt{1 - \frac{338}{1024}} = \frac{7\sqrt{14}}{32}.

Let

G

be the intersection of line

CF

with line

AB.

In triangle

ACG,

\angle GAC = \angle BAC

has

\cos \angle BAC = \frac{16 + 36 - 25}{2 \cdot 4 \cdot 6} = \frac{9}{16},

\sin \angle BAC = \frac{5\sqrt{7}}{16},

and

\angle ACG = \alpha,

so

\sin(\angle BAC + \alpha)

= \frac{5\sqrt{7}}{16} \cdot \frac{13\sqrt{2}}{32}

+ \frac{9}{16} \cdot \frac{7\sqrt{14}}{32}

= \frac{\sqrt{14}}{4}

and

\begin{aligned} AG &= \frac{AC \sin \alpha}{\sin(\angle BAC + \alpha)} \\ &= \frac{6 \cdot \frac{7\sqrt{14}}{32}}{\frac{\sqrt{14}}{4}} \\ &= \frac{21}{4}. \end{aligned}

Line

CF

is the radical axis of the two circles, so

GA \cdot GD = GB \cdot GE.

With

x = BD

and

BE = x + DE = x + 4\sqrt{2}:

\begin{aligned} &\frac{21}{4}\left(x - \frac{5}{4}\right) \\ &= \frac{5}{4}\left(x - \frac{5}{4} + 4\sqrt{2}\right), \end{aligned}

since

GD = 4 + x - \frac{21}{4}

and

GB = \frac{21}{4} - 4.

This gives

16\left(x - \frac{5}{4}\right) = 20\sqrt{2},

so

x = \frac{5 + 5\sqrt{2}}{4}

and

BE = \frac{5 + 21\sqrt{2}}{4}.

Therefore

a + b + c + d = 5 + 21 + 2 + 4

= 32.

2019 AIME I 第 13 题

答案：32

难度评级：3370

小提示：

大提示：

解答：

其他年份的第 13 题