Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Find the answer

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

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Note that $\angle B X I=\angle B Y I=90$ , which means that $A B$ and $B C$ are tangent to the incircle of $A B C$ at $X$ and $Y$ respectively. So $B X=B Y=\frac{A B+B C-A C}{2}$ , which means that $\frac{2}{5}=\frac{[B X I Y]}{[A B C]}=\frac{A B+B C-A C}{A B+B C+A C}$ . The smallest perimeter is achieved when $A B=A C=3$ and $B C=4$ . The area of this triangle $A B C$ is $2 \sqrt{5}$ .

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