2017 AMC 12A Problem 24

Question

Quadrilateral

ABCD

is inscribed in circle

O

and has sides

AB=3,

BC=2,

CD=6,

and

DA=8.

Let

X

and

Y

be points on

BD

such that

\dfrac{DX}{BD}=\dfrac{1}{4}

and

\dfrac{BY}{BD}=\dfrac{11}{36}.

Let

E

be the intersection of line

AX

and the line through

Y

parallel to

AD.

Let

F

be the intersection of line

CX

and the line through

E

parallel to

AC.

Let

G

be the point on circle

O

other than

C

that lies on line

CX.

What is

XF\cdot XG?

17

\dfrac{59-5\sqrt2}{3}

\dfrac{91-12\sqrt3}{4}

\dfrac{67-10\sqrt2}{3}

18

Answer: A

Concepts:similarity power of a point law of cosines

Difficulty rating: 2520

Small hint:

The parallels give

\triangle XEY\sim\triangle XAD

and

\triangle XEF\sim\triangle XAC

Big hint:

Combine those ratios with Power of a Point to get

XF\cdot XG=XB\cdot XY;

find

BD^2

via Law of Cosines

Solution:

Because

YE\parallel AD

and

EF\parallel AC,

we get

\triangle XEY\sim\triangle XAD

and

\triangle XEF\sim\triangle XAC,

giving

\dfrac{XY}{XE}=\dfrac{XD}{XA}

and

\dfrac{XF}{XE}=\dfrac{XC}{XA}.

Hence

\dfrac{XC}{XD}=\dfrac{XF}{XY},

so

XF\cdot XD=XC\cdot XY.

Power of a Point at

X

gives

XC\cdot XG=XD\cdot XB,

and combining yields

XF\cdot XG=XB\cdot XY.

With

d=BD,

DX=\dfrac14 d

and

BY=\dfrac{11}{36}d,

so

\begin{aligned} XF\cdot XG &=\left(d-\tfrac14 d\right) \\ &\quad {}\cdot\left(d-\tfrac14 d-\tfrac{11}{36}d\right) \\ &=\dfrac34 d\cdot\dfrac49 d=\dfrac{d^2}{3}. \end{aligned}

Since

ABCD

is cyclic,

\angle BAD

and

\angle BCD

are supplementary. The Law of Cosines on

\triangle ABD

and

\triangle CBD