1987 AHSME Problems/Problem 30

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]$

$\textbf{(A)}\ \frac{1}{\sqrt{2}} \qquad \textbf{(B)}\ \frac{2}{2+\sqrt{2}} \qquad \textbf{(C)}\ \frac{1}{\sqrt{3}} \qquad \textbf{(D)}\ \frac{1}{\sqrt[3]{6}}\qquad \textbf{(E)}\ \frac{1}{\sqrt[4]{12}}$

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 $\frac{AD}{AB} = \frac{k(1+\frac{1}{\sqrt{3}})}{1+\sqrt{3}} = \frac{k}{\sqrt{3}} = (\frac{3}{4})^{\frac{1}{4}} \times (\frac{1}{9})^{\frac{1}{4}} = (\frac{1}{12})^{\frac{1}{4}} = \frac{1}{\sqrt[4]{12}}$ , which is answer $\boxed{E}$ .

Solution 2 (Trigonometry)

The proof of why point $E$ lies on $AC$ is the same as Solution 1.

We have: \begin{align} \text{Area} \ \triangle AED &= \text{Area} \ \triangle CDB\\ \text{Area} \ \triangle AED &= \text{Area} \ \triangle ABC - \text{Area} \ \triangle AED\\ 2\cdot\text{Area} \ \triangle AED &= \text{Area} \ \triangle ABC \end{align} Since the area of a triangle can be expressed as $\dfrac 12ab\sin C$ , we have \begin{align} 2 \cdot \dfrac 12 AE\cdot AD \cdot \sin45^\circ &= \dfrac 12 AC\cdot AB \cdot \sin45^\circ\\ 2 \cdot AE \cdot AD = AC \cdot AB \end{align}

Now, by Law of Sines, we have $AE = \dfrac{AD \cdot \sin 60^\circ}{\sin 75^\circ}$ and $AC = \dfrac{AB \cdot \sin 30 ^\circ}{\sin 75^\circ}$ . Thus, by substitution we have \begin{align} 2 \cdot AD \cdot \dfrac{AD \cdot \sin 60^\circ}{\sin 75^\circ} &= AB \cdot \dfrac{AB \cdot \sin 30 ^\circ}{\sin 75^\circ}\\ 2 \cdot AD^2 \cdot \dfrac{\sqrt{3}}2 &= AB^2 \cdot \dfrac 12\\ \dfrac{AD^2}{AB^2} &= \dfrac{1}{2\sqrt{3}}\\ \dfrac{AD}{AB} &= \dfrac{1}{\sqrt[4]{12}} \end{align} The answer is therefore $\boxed{E}$ .

1987 AHSME (Problems • Answer Key • Resources)
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1987 AHSME Problems/Problem 30

Contents

Problem

Solution

Solution 2 (Trigonometry)

See also