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 $BI?$

$15$

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

$3\sqrt{26}$

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

$9\sqrt{3}$

Answer: A

Concepts:incircle, incenter, and inradius Heron’s Formula Pythagorean Theorem

Difficulty rating: 1980

Solution:

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.

2012 AMC 12A Problem 18

Answer: A

Difficulty rating: 1980

Solution:

Problem 18 in Other Years