Question

Let $\triangle PQR$ be a right triangle with $PQ = 90,$ $PR = 120,$ and $QR = 150.$ Let $C_1$ be the inscribed circle. Construct $\overline{ST},$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR},$ such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_1.$ Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_1.$ Let $C_2$ be the inscribed circle of $\triangle RST$ and $C_3$ the inscribed circle of $\triangle QUV.$ The distance between the centers of $C_2$ and $C_3$ can be written as $\sqrt{10n}.$ What is $n?$

Answer

Answer: 725

Concepts:incircle, incenter, and inradius right triangle similarity distance formula

Difficulty rating: 2560

Solution:

The right angle is at $P,$ so place $P = (0, 0),$ $Q = (0, 90),$ $R = (120, 0).$ The inradius of a right triangle is half the sum of the legs minus the hypotenuse: $r_1 = \frac{90 + 120 - 150}{2} = 30,$ so $C_1$ has center $(30, 30).$ The tangent line to $C_1$ perpendicular to $\overline{PR}$ (on the side toward $R$ ) is $x = 60,$ and the tangent perpendicular to $\overline{PQ}$ (toward $Q$ ) is $y = 60.$

Triangle $RST$ is similar to triangle $RPQ$ with ratio $\frac{RS}{RP} = \frac{60}{120} = \frac{1}{2},$ so its inradius is $15$ and its incircle $C_2$ is centered at $(60 + 15, 15) = (75, 15).$ Triangle $QUV$ is similar to triangle $QPR$ with ratio $\frac{QU}{QP} = \frac{30}{90} = \frac{1}{3},$ so its inradius is $10$ and $C_3$ is centered at $(10, 60 + 10) = (10, 70).$

The squared distance is $65^2 + 55^2$ $= 4225 + 3025$ $= 7250$ $= 10 \cdot 725,$ so $n = 725.$

2001 AIME II Problem 7

Answer: 725

Difficulty rating: 2560

Solution:

Problem 7 in Other Years