1979 AHSME Problems/Problem 12

Problem 12

$[asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\circ$",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]$

In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$ . Point $A$ lies on the extension of $DC$ past $C$ ; point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle. If length $AB$ equals length $OD$ , and the measure of $\measuredangle EOD$ is $45^\circ$ , then the measure of $\measuredangle BAO$ is

$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$

Solution

Solution by e_power_pi_times_i

Because $AB = OD$ , triangles $ABO$ and $BOE$ are isosceles. Denote $\measuredangle BAO = \measuredangle AOB = \theta$ . Then $\measuredangle ABO = 180^\circ-2\theta$ , and $\measuredangle EBO = \measuredangle OEB = 2\theta$ , so $\measuredangle BOE = 180^\circ-4\theta$ . Notice that $\measuredangle AOB + \measuredangle BOE + 45^\circ = 180^\circ$ . Therefore $\theta+180-4\theta = 135^\circ$ , and $\theta = \boxed{\textbf{(B) } 15^\circ}$ .

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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1979 AHSME Problems/Problem 12

Problem 12

Solution

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