Difference between revisions of "1987 AHSME Problems/Problem 30"
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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>. | ||
== See also == | == See also == | ||
Latest revision as of 12:17, 31 March 2018
Problem
In the figure, 
has 
and 
. A line 
, with 
on 
and 
, divides into two pieces of equal area.
(Note: the figure may not be accurate; perhaps 
is on 
instead of 
The ratio 
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]](http://latex.artofproblemsolving.com/1/3/5/135a648963b48464cdd989019e3de4847a416079.png)
![$\textbf{(A)}\ \frac{1}{\sqrt{2}} \qquad \textbf{(B)}\ \frac{2}{2+\sqrt{2}} \qquad \textbf{(C)}\ \frac{1}{\sqrt{3}} \qquad \textbf{(D)}\ \frac{1}{\sqrt[3]{6}}\qquad \textbf{(E)}\ \frac{1}{\sqrt[4]{12}}$](http://latex.artofproblemsolving.com/0/5/5/0552faf8dc8fb3fefe89900cae35dfc47f52ce06.png)
Solution
First we show that is on 
, as in the given figure, by demonstrating that if 
, then 
has more than half the area, so is too far to the right. Specifically, assume and drop an altitude from 
to that meets at 
. Without loss of generality, assume that 
, so that 
(as they have the same height, and area is 
times the product of base and height), which is 
, as required. Thus we must move to the left, scaling 
by a factor of 
such that 
. Thus ![$\frac{AD}{AB} = \frac{k(1+\frac{1}{\sqrt{3}})}{1+\sqrt{3}} = \frac{k}{\sqrt{3}} = (\frac{3}{4})^{\frac{1}{4}} \times (\frac{1}{9})^{\frac{1}{4}} = (\frac{1}{12})^{\frac{1}{4}} = \frac{1}{\sqrt[4]{12}}$](http://latex.artofproblemsolving.com/3/8/8/388278a98681d4d01abcd946122c3470e490774c.png)
, which is answer 
.
See also
| 1987 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
