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Difference between revisions of "1981 AHSME Problems/Problem 27"

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The ratio of these two areas is <math>\frac{\frac{x^2\sqrt{3}}{4}}{\frac{x^2(7+4\sqrt{3})}{4}} = \frac{\sqrt{3}}{7+4\sqrt{3}} = \sqrt{3}(7-4\sqrt{3}) = 7\sqrt{3} - 12</math> <math>\fbox{(D)}</math>.
 
The ratio of these two areas is <math>\frac{\frac{x^2\sqrt{3}}{4}}{\frac{x^2(7+4\sqrt{3})}{4}} = \frac{\sqrt{3}}{7+4\sqrt{3}} = \sqrt{3}(7-4\sqrt{3}) = 7\sqrt{3} - 12</math> <math>\fbox{(D)}</math>.
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-j314andrews
  
 
== See Also ==
 
== See Also ==
 
{{AHSME box|year=1981|num-b=26|num-a=28}}
 
{{AHSME box|year=1981|num-b=26|num-a=28}}

Latest revision as of 14:34, 28 June 2025

Problem

In the adjoining figure triangle $ABC$ is inscribed in a circle. Point $D$ lies on $\stackrel{\frown}{AC}$ with $\stackrel{\frown}{DC} = 30^\circ$, and point $G$ lies on $\stackrel{\frown}{BA}$ with $\stackrel{\frown}{BG}\, > \, \stackrel{\frown}{GA}$. Side $AB$ and side $AC$ each have length equal to the length of chord $DG$, and $\angle CAB = 30^\circ$. Chord $DG$ intersects sides $AC$ and $AB$ at $E$ and $F$, respectively. The ratio of the area of $\triangle AFE$ to the area of $\triangle ABC$ is

[asy] defaultpen(linewidth(.8pt)); pair C = origin; pair A = 2.5*dir(75); pair B = A + 2.5*dir(-75); path circ =circumcircle(A,B,C); pair D = waypoint(circ,(7/12)); pair G = waypoint(circ,(1/6)); pair E = intersectionpoint(D--G,A--C); pair F = intersectionpoint(A--B,D--G); label("$A$",A,N); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,SW); label("$G$",G,NE); label("$E$",E,NW); label("$F$",F,W); label("$30^\circ$",A,12S+E,fontsize(6pt)); draw(A--B--C--cycle); draw(circ); draw(Arc(A,0.25,-75,-105)); draw(D--G); [/asy]

$\textbf{(A)}\ \dfrac {2 - \sqrt {3}}{3}\qquad \textbf{(B)}\ \dfrac {2\sqrt {3} - 3}{3}\qquad \textbf{(C)}\ 7\sqrt {3}-12\qquad \textbf{(D)}\ 3\sqrt {3}-5\qquad\\ \textbf{(E)}\ \dfrac {9-5\sqrt {3}}{3}$

Solution

Since $\triangle ABC$ is isosceles, $\angle ACB = \angle ABC = \frac{150^{\circ}}{2} = 75^{\circ}$, and $\stackrel{\frown}{AC}\ =\ \stackrel{\frown}{AB}\ = 2 \cdot 75^{\circ} = 150^{\circ}$. Thus $\stackrel{\frown}{AD}\ = 150^{\circ} - 30^{\circ} = 120^{\circ}$, and $\angle AGD = \angle ACD = \frac{120^{\circ}}{2} = 60^{\circ}$. Since $DG = AB = AC$, $\stackrel{\frown}{DG}\ = 150^{\circ}$. So $\stackrel{\frown}{AG}\ = 150^{\circ} - 120^{\circ} = 30^{\circ}$. Therefore, $\angle AEG = \angle DEC = \frac{30^{\circ} + 30^{\circ}}{2} = 30^{\circ}$, and $\triangle AFE$ is isosceles with $AF = FE$. Also $\angle AFE = 180^\circ - 2 \cdot 30^\circ = 120^\circ$ and $\angle AFG = 60^\circ$, so $\triangle AFG$ is equilateral.

Let $x = AF = FE = FG$. By the Law of Cosines, $AE^2 = x^2 + x^2 - 2x \cdot x \cdot \cos(120^\circ) = 3x^2$, so $AE = x\sqrt{3}$. $\stackrel{\frown}{AG}\ =\ \stackrel{\frown}{DC}$, $AG = DC$. Thus $\triangle DEC \cong \triangle AEG$, and $EC = 2x$. Thus $AC = AB = x(2+\sqrt{3})$. The area of $\triangle AFE$ is $\frac{1}{2} \cdot x \cdot x\sqrt{3} \cdot \sin 30^\circ = \frac{x^2\sqrt{3}}{4}$, while the area of $\triangle ABC$ is $\frac{1}{2} \cdot x(2+\sqrt{3}) \cdot x(2+\sqrt{3}) \cdot \sin 30^\circ = \frac{x^2(7+4\sqrt{3})}{4}$.

The ratio of these two areas is $\frac{\frac{x^2\sqrt{3}}{4}}{\frac{x^2(7+4\sqrt{3})}{4}} = \frac{\sqrt{3}}{7+4\sqrt{3}} = \sqrt{3}(7-4\sqrt{3}) = 7\sqrt{3} - 12$ $\fbox{(D)}$.

-j314andrews

See Also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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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