Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it Ibero-American Mathematical Olympiad

Problem:

$ABC$ is an equilateral triangle. $D$ is on the side $AB$ and $E$ is on the side $AC$ such that $DE$ touches the incircle. Show that $AD/DB + AE/EC = 1$ .

Solution

Solution:

Put $BD = x$ , $CE = y$ , $BC = a$ . Then since the two tangents from $B$ to the incircle are of equal length, and similarly the two tangents from $D$ and $E$ , we have $ED + BC = BD + CE$ , or $ED = x + y - a$ .

By the cosine law, $ED^2 = AE^2 + AD^2 - AE \cdot AD$ . Substituting and simplifying, we get $a = \dfrac{3 x y}{x + y}$ .

Hence $AD/DB = \dfrac{2y - x}{x + y}$ and $AE/EC = \dfrac{2x - y}{x + y}$ with sum $1$ .

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