2011 AMC 12B Problem 20

Question

Triangle

ABC

has

AB=13,

BC=14,

and

AC=15.

The points

D,

E,

and

F

are the midpoints of

AB,

BC,

and

AC

respectively. Let

X\ne E

be the intersection of the circumcircles of

\triangle BDE

and

\triangle CEF.

What is

XA+XB+XC?

24

14\sqrt{3}

\dfrac{195}{8}

\dfrac{129\sqrt{7}}{14}

\dfrac{69\sqrt{2}}{4}

Answer: C

Concepts:circumcircle, circumcenter, and circumradius inscribed angle Heron’s Formula

Difficulty rating: 2220

Small hint:

Use the parallel midsegments and the Inscribed Angle Theorem to show

X

is equidistant from

A,

B,

C

Big hint:

X

is the circumcenter, so

XA+XB+XC=3R

where

R=\dfrac{abc}{4\cdot[ABC]}

Solution:

Since

DE\parallel AC

and

EF\parallel AB,

we get

\angle BDE=\angle BAC=\angle EFC.

By the Inscribed Angle Theorem,

\angle BXE=\angle BDE

and

\angle EXC=\angle EFC,

so

\angle BXE=\angle EXC.

With

BE=EC,

this forces

XB=XC.

Also

\angle BXC=2\angle BAC.

The chord formula in

\triangle BXC

gives

BC=2XB\sin(\angle BAC),

while the same formula in

\triangle ABC

gives

BC=2R\sin(\angle BAC).

Thus

XB=XC=R.

The point at distance

R

from both

B

and

C

on this side of

BC

is the circumcenter of

\triangle ABC,

so

XA=XB=XC=R.

The area of the

13

-

14

-

15

triangle is

84

by Heron’s formula, so

R=\dfrac{13\cdot14\cdot15}{4\cdot84}=\dfrac{65}{8},

and

XA+XB+XC=3R=\dfrac{195}{8}.

Thus, the correct answer is C.