Difference between revisions of "1960 AHSME Problems/Problem 38"

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{{AHSME 40p box|year=1960|num-b=37|num-a=39}}
 
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[[Category:Introductory Geometry Problems]]

Latest revision as of 19:21, 17 May 2018

Problem

In this diagram $AB$ and $AC$ are the equal sides of an isosceles $\triangle ABC$, in which is inscribed equilateral $\triangle DEF$. Designate $\angle BFD$ by $a$, $\angle ADE$ by $b$, and $\angle FEC$ by $c$. Then:

[asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); pair A=(5,12),B=origin,C=(10,0),D=(5/3,4),E=(10-5*.45,12*.45),F=(6,0); draw(A--B--C--cycle^^D--E--F--cycle); draw(anglemark(E,D,A,1,45)^^anglemark(F,E,C,1,45)^^anglemark(D,F,B,1,45)); label("$b$",(D.x+.2,D.y+.25),dir(30)); label("$c$",(E.x,E.y-.4),S); label("$a$",(F.x-.4,F.y+.1),dir(150)); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$D$",D,dir(150)); label("$E$",E,dir(60)); label("$F$",F,S);[/asy]

$\textbf{(A)}\ b=\frac{a+c}{2}\qquad \textbf{(B)}\ b=\frac{a-c}{2}\qquad \textbf{(C)}\ a=\frac{b-c}{2} \qquad \textbf{(D)}\ a=\frac{b+c}{2}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Since $\triangle DEF$ is an equilateral triangle, all of the angles are $60^{\circ}$. The angles in a line add up to $180^{\circ}$, so \[\angle FDB = 120 - b\] \[\angle EFC = 120 - a\] The angles in a triangle add up to $180^{\circ}$, so \[\angle ABC = 60 + b - a\] \[\angle ACB = 60 - c + a\] Since $\triangle ABC$ is isosceles and $AB = AC$, by Base-Angle Theorem, \[60 + b - a = 60 - c + a\] \[b + c = 2a\] \[a = \frac{b+c}{2}\] The answer is $\boxed{\textbf{(D)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 37
Followed by
Problem 39
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