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

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==Problem==
 
==Problem==
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In this diagram <math>AB</math> and <math>AC</math> are the equal sides of an isosceles <math>\triangle ABC</math>, in which is inscribed equilateral <math>\triangle DEF</math>.
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Designate <math>\angle BFD</math> by <math>a</math>, <math>\angle ADE</math> by <math>b</math>, and <math>\angle FEC</math> by <math>c</math>. Then:
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<asy>
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size(150);
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defaultpen(linewidth(0.8)+fontsize(10));
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pair A=(5,12),B=origin,C=(10,0),D=(5/3,4),E=(10-5*.45,12*.45),F=(6,0);
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draw(A--B--C--cycle^^D--E--F--cycle);
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draw(anglemark(E,D,A,1,45)^^anglemark(F,E,C,1,45)^^anglemark(D,F,B,1,45));
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label("$b$",(D.x+.2,D.y+.25),dir(30));
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label("$c$",(E.x,E.y-.4),S);
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label("$a$",(F.x-.4,F.y+.1),dir(150));
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label("$A$",A,N);
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label("$B$",B,S);
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label("$C$",C,S);
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label("$D$",D,dir(150));
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label("$E$",E,dir(60));
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label("$F$",F,S);</asy>
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<math>\textbf{(A)}\ b=\frac{a+c}{2}\qquad
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\textbf{(B)}\ b=\frac{a-c}{2}\qquad
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\textbf{(C)}\ a=\frac{b-c}{2} \qquad
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\textbf{(D)}\ a=\frac{b+c}{2}\qquad
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\textbf{(E)}\ \text{none of these}    </math>
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==Solution==
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Since <math>\triangle DEF</math> is an [[equilateral triangle]], all of the angles are <math>60^{\circ}</math>.
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The angles in a line add up to <math>180^{\circ}</math>, so
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<cmath>\angle FDB = 120 - b</cmath>
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<cmath>\angle EFC = 120 - a</cmath>
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The angles in a triangle add up to <math>180^{\circ}</math>, so
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<cmath>\angle ABC = 60 + b - a</cmath>
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<cmath>\angle ACB = 60 - c + a</cmath>
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Since <math>\triangle ABC</math> is [[isosceles triangle|isosceles]] and <math>AB = AC</math>, by Base-Angle Theorem,
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<cmath>60 + b - a = 60 - c + a</cmath>
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<cmath>b + c = 2a</cmath>
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<cmath>a = \frac{b+c}{2}</cmath>
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The answer is <math>\boxed{\textbf{(D)}}</math>.
  
 
==See Also==
 
==See Also==
−
{{AHSME 40p box|year=1960 |before=[[Problem 37]]|after=[[Problem 39]]}}
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{{AHSME 40p box|year=1960|num-b=37|num-a=39}}
 +
 
 +
[[Category:Introductory Geometry Problems]]

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 (Problems • Answer Key • Resources)
Preceded by
Problem 37 		Followed by
Problem 39
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions
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