1979 AHSME Problems/Problem 30

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Problem

[asy] size(200); import cse5; pathpen=black; anglefontpen=black; pointpen=black; anglepen=black; dotfactor=3; pair A=(0,0),B=(0.5,0.5*sqrt(3)),C=(3,0),D=(1.7,0),EE; EE=(B+C)/2; D(MP("$A$",A,W)--MP("$B$",B,N)--MP("$C$",C,E)--cycle); D(MP("$E$",EE,N)--MP("$D$",D,S)); D(D);D(EE); MA("80^\circ",8,D,EE,C,0.1); MA("20^\circ",8,EE,C,D,0.3,2,shift(1,3)*C); draw(arc(shift(-0.1,0.05)*C,0.25,100,180),arrow =ArcArrow()); MA("100^\circ",8,A,B,C,0.1,0); MA("60^\circ",8,C,A,B,0.1,0); //Credit to TheMaskedMagician for the diagram [/asy]

In $\triangle ABC$, $E$ is the midpoint of side $BC$ and $D$ is on side $AC$. If the length of $AC$ is $1$ and $\measuredangle BAC = 60^\circ, \measuredangle ABC = 100^\circ, \measuredangle ACB = 20^\circ$ and $\measuredangle DEC = 80^\circ$, then the area of $\triangle ABC$ plus twice the area of $\triangle CDE$ equals

$\textbf{(A) }\frac{1}{4}\cos 10^\circ\qquad \textbf{(B) }\frac{\sqrt{3}}{8}\qquad \textbf{(C) }\frac{1}{4}\cos 40^\circ\qquad \textbf{(D) }\frac{1}{4}\cos 50^\circ\qquad \textbf{(E) }\frac{1}{8}$

Solution

Let $F$ be the point on the extension of side $AB$ past $B$ for which $AF=1$. Since $AF=AC$ and $\measuredangle FAC = 60^\circ$,$\triangle ACF$ is equilateral. Let $G$ be the point on line segment $BF$ for which $\measuredangle BCG=20^\circ$. Then $\triangle BCG$ is similar to $\triangle DCE$ and $BC=2(EC)$. Also $\triangle FGC$ is congruent to $\triangle ABC$. Therefore, $[\triangle ACF]  = ([\triangle ABC] + [\triangle GCF]) + [\triangle BCG]$. Plugging in the values that we know and then dividing by 2 results in an answer of $\boxed{B) \frac{\sqrt{3}}{8}.}$

This solution is from the solution manual but was typed here by alpha_2.

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