Question

The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P,$ and its radius is $21.$ Given that $AP = 23$ and $PB = 27,$ find the perimeter of the triangle.

Answer

Solution:

Tangent segments from a point are equal, so the tangent lengths from $A,$ $B,$ $C$ are $23,$ $27,$ and some $z.$ Then the sides are $50,$ $23 + z,$ $27 + z,$ the semiperimeter is $s = 50 + z,$ and Heron's formula gives area $\sqrt{(50 + z) \cdot z \cdot 23 \cdot 27}.$

The area also equals $rs = 21(50 + z).$ Squaring $21(50 + z) = \sqrt{621 z (50 + z)}$ and dividing by $50 + z$ gives $441(50 + z) = 621 z,$ so $180 z = 22050$ and $z = \frac{245}{2}.$

The perimeter is $2s = 2\left(50 + \frac{245}{2}\right) = 345.$

1999 AIME Problem 12

Solution:

Problem 12 in Other Years