Question

Given a point $P$ on a triangular piece of paper $ABC,$ consider the creases that are formed in the paper when $A,$ $B,$ and $C$ are folded onto $P.$ Let us call $P$ a fold point of $\triangle ABC$ if these creases, which number three unless $P$ is one of the vertices, do not intersect. Suppose that $AB=36,$ $AC=72,$ and $\angle B=90^\circ.$ Then the area of the set of all fold points of $\triangle ABC$ can be written in the form $q\pi-r\sqrt s,$ where $q,$ $r,$ and $s$ are positive integers and $s$ is not divisible by the square of any prime. What is $q+r+s?$

Answer

Answer: 597

Concepts:paper folding perpendicular bisector area decomposition

Difficulty rating: 2840

Small Hint:

Two fold creases meet at the circumcenter of the triangle formed by $P$ and the corresponding two vertices

Big Hint:

The fold-point locus is the intersection of the disks with diameters $AB$ and $BC$

Solution:

The creases for two vertices meet at the circumcenter of the triangle formed with $P.$ This intersection lies off the paper exactly when the angle at $P$ is obtuse, so the fold-point locus is the intersection of the three diameter disks for $AB,$ $BC,$ and $CA.$ The $CA$ disk contains the entire right triangle, leaving the intersection of the $AB$ and $BC$ disks.

Here $BC=\sqrt{72^2-36^2}=36\sqrt3.$ The two relevant radii are $18$ and $18\sqrt3,$ and their lens consists of circular segments with central angles $120^\circ$ and $60^\circ.$ Its area is $\begin{aligned}&\left(\frac\pi3\cdot18^2\right.\\&\qquad\left.-\frac12\cdot18^2\sin120^\circ\right)\\&+\left(\frac\pi6(18\sqrt3)^2\right.\\&\qquad\left.-\frac12(18\sqrt3)^2\sin60^\circ\right)\\&=270\pi-324\sqrt3.\end{aligned}$ Thus $q+r+s$ equals $270+324+3=597.$

1994 AIME Problem 15

Answer: 597

Difficulty rating: 2840

Small Hint:

Big Hint:

Solution:

Problem 15 in Other Years