2012 AMC 12A Problem 18

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.