Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it

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

Solution

Put $\mathrm{BD}=\mathrm{x}, \mathrm{CE}=\mathrm{y}, \mathrm{BC}=\mathrm{a}$ . Then since the two tangents from $\mathrm{B}$ to the incircle are of equal length, and similarly the two tangents from $\mathrm{D}$ and $\mathrm{E}$ , we have $\mathrm{ED}+\mathrm{BC}=\mathrm{BD}+\mathrm{CE}$ , or $\mathrm{ED}=$ $x+y-a$ . By the cosine law, $E D^{2}=A E^{2}+A D^{2}-A E \cdot AD$ . Substituting and simplifying, we get $a=3 x y /(x+y)$ . Hence $A D / D B=(2 y-x) /(x+y)$ and $A E / E C=(2 x-y) /(x+y)$ with sum 1.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.