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1987 AHSME Problems/Problem 30: Difference between revisions

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\textbf{(D)}\ \frac{1}{\sqrt[3]{6}}\qquad
\textbf{(D)}\ \frac{1}{\sqrt[3]{6}}\qquad
\textbf{(E)}\ \frac{1}{\sqrt[4]{12}}  </math>
\textbf{(E)}\ \frac{1}{\sqrt[4]{12}}  </math>
== Solution ==
First we show that <math>E</math> is on <math>AC</math>, as in the given figure, by demonstrating that if <math>E = C</math>, then <math>\triangle ADE</math> has more than half the area, so <math>DE</math> is too far to the right. Specifically, assume <math>E = C</math> and drop an altitude from <math>C</math> to <math>AB</math> that meets <math>AB</math> at <math>F</math>. Without loss of generality, assume that <math>CF = 1</math>, so that <math>\frac{\text{Area} \ \triangle EAD}{\text{Area} \ \triangle EAB} = \frac{AD}{AB}</math> (as they have the same height, and area is <math>\frac{1}{2}</math> times the product of base and height), which is <math>\frac{1 + \frac{1}{\sqrt{3}}}{1 + \sqrt{3}} = \frac{1}{\sqrt{3}} > \frac{1}{2}</math>, as required. Thus we must move <math>DE</math> to the left, scaling <math>\triangle EAD</math> by a factor of <math>k</math> such that <math>\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}}</math>. Thus <math>\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}}</math>, which is answer <math>\boxed{E}</math>.
== Solution 2 (Trigonometry) ==
The proof of why point <math>E</math> lies on <math>AC</math> 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 <math> \dfrac 12ab\sin C</math>, 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 <math>AE = \dfrac{AD \cdot \sin 60^\circ}{\sin 75^\circ}</math> and <math>AC = \dfrac{AB \cdot \sin 30 ^\circ}{\sin 75^\circ}</math>. 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 <math>\boxed{E}</math>.


== See also ==
== See also ==

Latest revision as of 19:43, 1 June 2025

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 $\boxed{E}$.

See also

1987 AHSME (Problems • Answer Key • Resources)
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