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1994 AIME Problems/Problem 15

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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\,?

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}.

pair project(pair X, pair Y, real r){return X+r*(Y-X);}path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a... 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..., and q+r+s = \boxed{597}.

See also

1994 AIME (ProblemsResources)
Preceded by
Problem 14
Followed by
Last question
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