AIME 1994 · 第 15 题

AIME 1994 — Problem 15

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

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 $q+r+s\,$ ?

解析

Solution

Let $O_{AB}$ be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of $\overline{PA}$ and $\overline{PB}$ , and so forth. Then $O_{AB}, O_{BC}, O_{CA}$ are, respectively, the circumcenters of $\triangle PAB, PBC, PCA$ . According to the problem statement, the circumcenters of the triangles cannot lie within the interior of the respective triangles, since they are not on the paper. It follows that $\angle APB, \angle BPC, \angle CPA > 90^{\circ}$ ; the locus of each of the respective conditions for $P$ is the region inside the (semi)circles with diameters $\overline{AB}, \overline{BC}, \overline{CA}$ .

We note that the circle with diameter $AC$ covers the entire triangle because it is the circumcircle of $\triangle ABC$ , so it suffices to take the intersection of the circles about $AB, BC$ . We note that their intersection lies entirely within $\triangle ABC$ (the chord connecting the endpoints of the region is in fact the altitude of $\triangle ABC$ from $B$ ). Thus, the area of the locus of $P$ (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of $M_1, M_2 = \overline{AB}, \overline{BC}$ and note that $\triangle M_1BM_2 \sim \triangle ABC$ , we see that thse segments respectively cut a $120^{\circ}$ arc in the circle with radius $18$ and $60^{\circ}$ arc in the circle with radius $18\sqrt{3}$ .

$AIME diagram$

The diagram shows $P$ outside of the grayed locus; notice that the creases [the dotted blue] intersect within the triangle, which is against the problem conditions. The area of the locus is the sum of two segments of two circles; these segments cut out $120^{\circ}, 60^{\circ}$ angles by simple similarity relations and angle-chasing.

Hence, the answer is, using the $\frac 12 ab\sin C$ definition of triangle area, $\left[\frac{\pi}{3} \cdot 18^2 - \frac{1}{2} \cdot 18^2 \sin \frac{2\pi}{3} \right] + \left[\frac{\pi}{6} \cdot \left(18\sqrt{3}\right)^2 - \frac{1}{2} \cdot (18\sqrt{3})^2 \sin \frac{\pi}{3}\right] = 270\pi - 324\sqrt{3}$ , and $q+r+s = \boxed{597}$ .