Difference between revisions of "1986 AHSME Problems/Problem 27"
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==Solution== | ==Solution== | ||
| + | <math>ABE</math> and <math>DCE</math> are similar isosceles triangles. It remains to find the square of the ratios of their sides. Draw in <math>AD</math>. Because <math>AB</math> is a diameter, <math>\angle ADB=\angle ADE=90^{\circ}</math>. Thus, <cmath>\frac{DE}{AE}=\cos\alpha</cmath> So <cmath>\frac{DE^2}{AE^2}=\cos^2\alpha</cmath> The answer is thus <math>\fbox{(C)}</math>. | ||
== See also == | == See also == | ||
Revision as of 14:51, 2 August 2016
Problem
In the adjoining figure, 
is a diameter of the circle, 
is a chord parallel to ,
and 
intersects 
at 
, with 
. The ratio of the area of 
to that of 
is
![[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair A=(-1,0), B=(1,0), E=(0,-.4), C=(.6,-.8), D=(-.6,-.8), E=(0,-.8/(1.6)); draw(unitcircle); draw(A--B--D--C--A); draw(Arc(E,.2,155,205)); label("$A$",A,W); label("$B$",B,C); label("$C$",C,C); label("$D$",D,W); label("$\alpha$",E-(.2,0),W); label("$E$",E,N); [/asy]](http://latex.artofproblemsolving.com/5/2/8/5287124b7c64cb0da4d33b2a32c979894f04d5d7.png)
Solution
and 
are similar isosceles triangles. It remains to find the square of the ratios of their sides. Draw in 
. Because is a diameter, 
. Thus, ![\[\frac{DE}{AE}=\cos\alpha\]](http://latex.artofproblemsolving.com/e/c/9/ec94346de7816a6e8d4ebb4cfb038c6a5575bf42.png)
So ![\[\frac{DE^2}{AE^2}=\cos^2\alpha\]](http://latex.artofproblemsolving.com/1/0/9/109020d930f188ac7d6b98d14416890eb0375715.png)
The answer is thus 
.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 26 |
Followed by Problem 28 | |
| 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. 
