1982 AMC 12 Problem 24

Question

In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If

AG=2,

GF=13,

FC=1

and

HJ=7,

then

DE

equals

2\sqrt{22}

7\sqrt3

9

10

13

All three sides of the triangle have length

16

Apply power of a point from

A,

B,

and

C

to relate the two base segments and

DE

First, power of

A

gives

AH(AH+7)=AG\cdot AF.

The latter product is

2\cdot15=30,

so

AH=3,

BJ=6,

and

BH=13.

Let

BD=u,

DE=x,

EC=v.

From the secants at

B

and

C,

\begin{aligned} u(u+x)&=BH\cdot BJ\\ &=13\cdot6=78,\\ v(v+x)&=CF\cdot CG\\ &=1\cdot14=14. \end{aligned}

Also

u+x+v=16.

Subtracting the power equations and using the sum gives

u-v=4.

Hence

u=10-\frac{x}{2}

and

v=6-\frac{x}{2}.

Substitution into

v(v+x)=14

gives

x^2=88,

so

DE=2\sqrt{22}.

Therefore, the correct answer is A.

Problem 24 in other years