1986 AHSME Problems/Problem 27

Problem

In the adjoining figure, $AB$ is a diameter of the circle, $CD$ is a chord parallel to $AB$ , and $AC$ intersects $BD$ at $E$ , with $\angle AED = \alpha$ . The ratio of the area of $\triangle CDE$ to that of $\triangle ABE$ 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$",E,N); [/asy]$

$\textbf{(A)}\ \cos\ \alpha\qquad \textbf{(B)}\ \sin\ \alpha\qquad \textbf{(C)}\ \cos^2\alpha\qquad \textbf{(D)}\ \sin^2\alpha\qquad \textbf{(E)}\ 1-\sin\ \alpha$

Solution

$ABE$ and $DCE$ are similar isosceles triangles. It remains to find the square of the ratio of their sides. Draw in $AD$ . Because $AB$ is a diameter, $\angle ADB=\angle ADE=90^{\circ}$ . Thus, $\frac{DE}{AE}=\cos\alpha$ So $\frac{DE^2}{AE^2}=\cos^2\alpha$ The answer is thus $\fbox{(C)}$ .

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
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1986 AHSME Problems/Problem 27

Problem

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