# Difference between revisions of "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 $\theta = \boxed{\textbf{(B) } 15^\circ}$.

## Solution 2

Draw $BO$. Let $y = \angle BAO$. Since $AB = OD = BO$, triangle $ABO$ is isosceles, so $\angle BOA = \angle BAO = y$. Angle $\angle EBO$ is exterior to triangle $ABO$, so $\angle EBO = \angle BAO + \angle BOA = y + y = 2y$.

Triangle $BEO$ is isosceles, so $\angle BEO = \angle EBO = 2y$. Then $\angle EOD$ is external to triangle $AEO$, so $\angle EOD = \angle EAO + \angle AEO = y + 2y = 3y$. But $\angle EOD = 45^\circ$, so $\angle BAO = y = 45^\circ/3 = \boxed{15^\circ}$. That means the answer is $\boxed{\textbf{(B) } 15^\circ}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 