1996 USAMO Problems/Problem 3

Problem

Let $ABC$ be a triangle. Prove that there is a line $l$ (in the plane of triangle $ABC$ ) such that the intersection of the interior of triangle $ABC$ and the interior of its reflection $A'B'C'$ in $l$ has area more than $\frac{2}{3}$ the area of triangle $ABC$ .

Solution

Let the triangle be $ABC$ . Assume $A$ is the largest angle. Let $AD$ be the altitude. Assume $AB \le AC$ , so that $BD \le BC/2$ . If $BD > \frac{BC}{3}$ , then reflect in $AD$ . If $B'$ is the reflection of $B'$ , then $B'D = BD$ and the intersection of the two triangles is just $ABB'$ . But $BB' = 2BD > \frac{2}{3} BC$ , so $ABB'$ has more than $\frac{2}{3}$ the area of $ABC$ .

If $BD < BC/3$ , then reflect in the angle bisector of $C$ . The reflection of $A'$ is a point on the segment $BD$ and not $D$ . (It lies on the line $BC$ because we are reflecting in the angle bisector. $A'C > DC$ because $\angle{CAD} < \angle{CDA} = 90^{\circ}$ . Finally, $A'C \le BC$ because we assumed $\angle B$ does not exceed $\angle A$ ). The intersection is just strictly larger than $AA'C$ . But $\frac{[AA'C]}{[ABC]} = \frac{CA'}{CB} > CD/CB \ge 2/3$ .

1996 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6
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1996 USAMO Problems/Problem 3

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