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Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

In $\triangle ABC$ , the incircle centered at $I$ touches sides $AB$ and $BC$ at $X$ and $Y$ , respectively. Additionally, the area of quadrilateral $BXIY$ is $\frac{2}{5}$ of the area of $ABC$ . Let $p$ be the smallest possible perimeter of a $\triangle ABC$ that meets these conditions and has integer side lengths. Find the smallest possible area of such a triangle with perimeter $p$ .

Solution

Solution:

Note that $\angle BXI = \angle BYI = 90^\circ$ , which means that $AB$ and $BC$ are tangent to the incircle of $ABC$ at $X$ and $Y$ respectively. So $BX = BY = \frac{AB + BC - AC}{2}$ , which means that $\frac{2}{5} = \frac{[BXIY]}{[ABC]} = \frac{AB + BC - AC}{AB + BC + AC}$ . The smallest perimeter is achieved when $AB = AC = 3$ and $BC = 4$ . The area of this triangle $ABC$ is $2 \sqrt{5}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.