2013 AMC 12B Problem 19

Problem 19 of 25HarderGeometry

In triangle

ABC,

AB = 13,

BC = 14,

and

CA = 15.

Distinct points

D,

E,

and

F

lie on segments

BC,

CA,

and

DE,

respectively, such that

AD \perp BC,

DE \perp AC,

and

AF \perp BF.

The length of segment

DF

can be written as

\dfrac{m}{n},

where

m

and

n

are relatively prime positive integers. What is

m + n?

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Solution

The altitude from

A

BC

gives

BD = 5,

CD = 9,

AD = 12.

Because

DE \perp AC,

triangle

AED \sim ADC,

giving

DE = \tfrac{36}{5}

and

AE = \tfrac{48}{5}.

Since

\angle AFB = \angle ADB = 90^\circ,

quadrilateral

ABDF

is cyclic, so

\angle ABD = \angle AFE,

making right triangles

ABD

and

AFE

similar:

\dfrac{FE}{5} = \dfrac{\frac{48}{5}}{12},

FE = 4.

Hence

DF = DE - FE

= \tfrac{36}{5} - 4

= \tfrac{16}{5},

and

m + n = 21.

Thus, the correct answer is B.

Tagged: cyclic quadrilateral · similarity · altitude