# 1987 AHSME Problems/Problem 30

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## Problem

In the figure, $\triangle ABC$ has $\angle A =45^{\circ}$ and $\angle B =30^{\circ}$. A line $DE$, with $D$ on $AB$ and $\angle ADE =60^{\circ}$, divides $\triangle ABC$ into two pieces of equal area. (Note: the figure may not be accurate; perhaps $E$ is on $CB$ instead of $AC.)$ The ratio $\frac{AD}{AB}$ is $[asy] size((220)); draw((0,0)--(20,0)--(7,6)--cycle); draw((6,6)--(10,-1)); label("A", (0,0), W); label("B", (20,0), E); label("C", (7,6), NE); label("D", (9.5,-1), W); label("E", (5.9, 6.1), SW); label("45^{\circ}", (2.5,.5)); label("60^{\circ}", (7.8,.5)); label("30^{\circ}", (16.5,.5)); [/asy]$ $\textbf{(A)}\ \frac{1}{\sqrt{2}} \qquad \textbf{(B)}\ \frac{2}{2+\sqrt{2}} \qquad \textbf{(C)}\ \frac{1}{\sqrt{3}} \qquad \textbf{(D)}\ \frac{1}{\sqrt{6}}\qquad \textbf{(E)}\ \frac{1}{\sqrt{12}}$

## Solution

First we show that $E$ is on $AC$, as in the given figure, by demonstrating that if $E = C$, then $\triangle ADE$ has more than half the area, so $DE$ is too far to the right. Specifically, assume $E = C$ and drop an altitude from $C$ to $AB$ that meets $AB$ at $F$. Without loss of generality, assume that $CF = 1$, so that $\frac{\text{Area} \ \triangle EAD}{\text{Area} \ \triangle EAB} = \frac{AD}{AB}$ (as they have the same height, and area is $\frac{1}{2}$ times the product of base and height), which is $\frac{1 + \frac{1}{\sqrt{3}}}{1 + \sqrt{3}} = \frac{1}{\sqrt{3}} > \frac{1}{2}$, as required. Thus we must move $DE$ to the left, scaling $\triangle EAD$ by a factor of $k$ such that $\text{Area} \ \triangle EAD = \frac{1}{2} \ \text{Area} \ \triangle CAB \implies \frac{1}{2}k^{2}(1+\frac{1}{\sqrt{3}}) = \frac{1}{4}(1+\sqrt{3}) \implies k^{2} = \frac{\sqrt{3}}{2} \implies k = (\frac{3}{4})^{\frac{1}{4}}$. Thus $\frac{AD}{AB} = \frac{k(1+\frac{1}{\sqrt{3}})}{1+\sqrt{3}} = \frac{k}{\sqrt{3}} = (\frac{3}{4})^{\frac{1}{4}} \times (\frac{1}{9})^{\frac{1}{4}} = (\frac{1}{12})^{\frac{1}{4}} = \frac{1}{\sqrt{12}}$, which is answer $\boxed{E}$.

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