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Difference between revisions of "1987 AHSME Problems/Problem 30"

(Created page with "==Problem== In the figure, <math>\triangle ABC</math> has <math>\angle A =45^{\circ}</math> and <math>\angle B =30^{\circ}</math>. A line <math>DE</math>, with <math>D</math> on...")
 
(Added a solution with explanation)
 
(One intermediate revision by one other user not shown)
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In the figure, <math>\triangle ABC</math> has <math>\angle A =45^{\circ}</math> and <math>\angle B =30^{\circ}</math>. A line <math>DE</math>, with <math>D</math> on <math>AB</math>  
 
In the figure, <math>\triangle ABC</math> has <math>\angle A =45^{\circ}</math> and <math>\angle B =30^{\circ}</math>. A line <math>DE</math>, with <math>D</math> on <math>AB</math>  
 
and <math>\angle ADE =60^{\circ}</math>, divides <math>\triangle ABC</math> into two pieces of equal area.  
 
and <math>\angle ADE =60^{\circ}</math>, divides <math>\triangle ABC</math> into two pieces of equal area.  
−
(Note: the figure may not be accurate; perhaps <math>E</math> is on <math>CB</math> instead of <math>AC</math>.)
+
(Note: the figure may not be accurate; perhaps <math>E</math> is on <math>CB</math> instead of <math>AC.) </math>
 
The ratio <math>\frac{AD}{AB}</math> is
 
The ratio <math>\frac{AD}{AB}</math> is
  
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\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>.
  
 
== See also ==
 
== See also ==

Latest revision as of 12:17, 31 March 2018

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 $\boxed{E}$.

See also

1987 AHSME (Problems • Answer Key • Resources)
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