2016/17 AHSMC Part 1 Problem 10

Problem 10 of 16

Let $D$ be an arbitrary point on the side $BC$ of the equilateral triangle $ABC$ . Points $E$ and $F$ are on $AB$ and $AC$ respectively so that $DE \perp AB$ and $DF \perp AC$ and $E_1$ , $F_1$ are points on $BC$ such that $EE_1 \perp BC$ and $FF_1 \perp BC$ . If $E_1F_1$ has length $\dfrac{1}{2}$ then the length of $BC$ is

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Solution

Let $BC = a$ . We have $DE_1 = DE \cos 30^\circ = \dfrac{\sqrt{3}}{2}\, DE$ and $DF_1 = DF \cos 30^\circ = \dfrac{\sqrt{3}}{2}\, DF$ , hence $E_1F_1 = \dfrac{\sqrt{3}}{2}(DE+DF)$ . On the other hand $DE \cdot AB + DF \cdot AC = 2\,\mathrm{Area}(ABC)$ that is, $a(DE+DF) = a^2 \dfrac{\sqrt{3}}{2}$ and hence $DE+DF = a \dfrac{\sqrt{3}}{2}$ . Therefore, $E_1F_1 = \dfrac{3a}{4}$ . Since $E_1F_1 = \dfrac{1}{2}$ we must have $a = \dfrac{2}{3}$ . The answer is (A).

University of Alberta

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