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Difference between revisions of "1979 AHSME Problems/Problem 12"

(Created page with "== 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=intersection...")
 
(→‎Another Solution)
 
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==Solution==
 
==Solution==
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Solution by e_power_pi_times_i
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Because <math>AB = OD</math>, triangles <math>ABO</math> and <math>BOE</math> are isosceles. Denote <math>\measuredangle BAO = \measuredangle AOB = \theta</math>. Then <math>\measuredangle ABO = 180^\circ-2\theta</math>, and <math>\measuredangle EBO = \measuredangle OEB = 2\theta</math>, so <math>\measuredangle BOE = 180^\circ-4\theta</math>. Notice that <math>\measuredangle AOB + \measuredangle BOE + 45^\circ = 180^\circ</math>. Therefore <math>\theta+180-4\theta = 135^\circ</math>, and <math>\theta = \boxed{\textbf{(B) } 15^\circ}</math>.
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==Solution 2==
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Draw <math>BO</math>. Let <math>y = \angle BAO</math>. Since <math>AB = OD = BO</math>, triangle <math>ABO</math> is isosceles, so <math>\angle BOA = \angle BAO = y</math>. Angle <math>\angle EBO</math> is exterior to triangle <math>ABO</math>, so <math>\angle EBO = \angle BAO + \angle BOA = y + y = 2y</math>.
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Triangle <math>BEO</math> is isosceles, so <math>\angle BEO = \angle EBO = 2y</math>. Then <math>\angle EOD</math> is external to triangle <math>AEO</math>, so <math>\angle EOD = \angle EAO + \angle AEO = y + 2y = 3y</math>. But <math>\angle EOD = 45^\circ</math>, so <math>\angle BAO = y = 45^\circ/3 = \boxed{15^\circ}</math>. That means the answer is <math>\boxed{\textbf{(B) } 15^\circ}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 12:34, 17 June 2021

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 $\boxed{\textbf{(B) } 15^\circ}$.

See also

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

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