2012 AMC 12A Problem 18

Question

Triangle

ABC

has

AB = 27,

AC = 26,

and

BC = 25.

Let

I

denote the intersection of the internal angle bisectors of

\triangle ABC.

What is

BI?

15

5 + \sqrt{26} + 3\sqrt{3}

3\sqrt{26}

\dfrac{2}{3}\sqrt{546}

9\sqrt{3}

Drop perpendiculars from

I

to the sides; the tangent segment from

B

has length

s - b

Find the inradius from area

= rs

and Heron’s formula, then use the right triangle with legs

r

and

BD

Let

D

be the foot of the perpendicular from the incenter

I

to

BC.

The tangent length

BD = s - AC,

where

s = \tfrac12(25 + 26 + 27) = 39,

so

BD = 39 - 26 = 13.

By Heron’s formula the area is

\sqrt{39 \cdot 14 \cdot 13 \cdot 12},

and the inradius satisfies

r^2 = \dfrac{(s-a)(s-b)(s-c)}{s}

= \dfrac{14 \cdot 13 \cdot 12}{39}

= 56.

In right triangle

BDI,

BI^2 = r^2 + BD^2

= 56 + 169

= 225,

so

BI = 15.

Thus, the correct answer is A.

Problem 18 in other years