1990 AHSME Problems/Problem 14

Problem

$[asy] draw(circle((0,0),1),black); draw((0,1)--(cos(pi/14),-sin(pi/14))--(-cos(pi/14),-sin(pi/14))--cycle,dot); draw((-cos(pi/14),-sin(pi/14))--(0,-1/cos(3pi/7))--(cos(pi/14),-sin(pi/14)),dot); draw(arc((0,1),.25,230,310)); MP("A",(0,1),N);MP("B",(cos(pi/14),-sin(pi/14)),E);MP("C",(-cos(pi/14),-sin(pi/14)),W);MP("D",(0,-1/cos(3pi/7)),S); MP("x",(0,.8),S); [/asy]$

An acute isosceles triangle, $ABC$ , is inscribed in a circle. Through $B$ and $C$ , tangents to the circle are drawn, meeting at point $D$ . If $\angle{ABC}=\angle{ACB}=2\angle{D}$ and $x$ is the radian measure of $\angle{A}$ , then $x=$

$\text{(A) } \frac{3\pi}{7}\quad \text{(B) } \frac{4\pi}{9}\quad \text{(C) } \frac{5\pi}{11}\quad \text{(D) } \frac{6\pi}{13}\quad \text{(E) } \frac{7\pi}{15}$

Solution

We can make two equations (assume angle D is y): $y+2x=180$ and $4y+x=180$ . We find that $x=\dfrac{540}{7}$ . Now we have to convert this to radians. 360 degrees is $2\pi$ radians, so since we have $\dfrac{540}{7}$ degrees, the answer is $\dfrac{3\pi}{7}$ which is $\fbox{A}.$

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1990 AHSME Problems/Problem 14

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