2002 AMC 10A Problem 23

Problem 23 of 25HarderGeometry

Points

A,

B,

C,

and

D

lie on a line, in that order, with

AB=CD

and

BC=12.

Point

E

is not on the line, and

BE=CE=10.

The perimeter of

\triangle AED

is twice the perimeter of

\triangle BEC.

Find

AB.

Show solution

Solution

Let

M

be the midpoint of

BC.

Since

BE=CE,

EM\perp BC

and

EM=\sqrt{10^2-6^2}=8.

By symmetry

AE=ED;

write

AB=CD=x

and

AE=ED=y.

The perimeter condition gives

2y+(2x+12)

=2(10+10+12)=64,

so

x+y=26.

Also

y^2=EM^2+(x+6)^2

=64+(x+6)^2.

Substituting

y=26-x,

(26-x)^2=64+(x+6)^2,

which simplifies to

676-52x=100+12x,

so

64x=576

and

x=9.

Thus, the correct answer is D.

More practice

Concepts: isosceles triangle · Pythagorean Theorem · perimeter

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.