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Difference between revisions of "1979 AHSME Problems/Problem 30"

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Problem 30 of the 1979 AHSME does not exist.
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== Problem ==
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<asy>
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size(200);
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import cse5;
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pathpen=black;
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anglefontpen=black;
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pointpen=black;
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anglepen=black;
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dotfactor=3;
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pair A=(0,0),B=(0.5,0.5*sqrt(3)),C=(3,0),D=(1.7,0),EE;
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EE=(B+C)/2;
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D(MP("$A$",A,W)--MP("$B$",B,N)--MP("$C$",C,E)--cycle);
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D(MP("$E$",EE,N)--MP("$D$",D,S));
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D(D);D(EE);
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MA("80^\circ",8,D,EE,C,0.1);
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MA("20^\circ",8,EE,C,D,0.3,2,shift(1,3)*C);
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draw(arc(shift(-0.1,0.05)*C,0.25,100,180),arrow =ArcArrow());
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MA("100^\circ",8,A,B,C,0.1,0);
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MA("60^\circ",8,C,A,B,0.1,0);
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//Credit to TheMaskedMagician for the diagram
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</asy>
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In <math>\triangle ABC</math>, <math>E</math> is the midpoint of side <math>BC</math> and <math>D</math> is on side <math>AC</math>.
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If the length of <math>AC</math> is <math>1</math> and <math>\measuredangle BAC = 60^\circ, \measuredangle ABC = 100^\circ, \measuredangle ACB = 20^\circ</math> and
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<math>\measuredangle DEC = 80^\circ</math>, then the area of <math>\triangle ABC</math> plus twice the area of <math>\triangle CDE</math> equals
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<math> \textbf{(A) }\frac{1}{4}\cos 10^\circ\qquad
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\textbf{(B) }\frac{\sqrt{3}}{8}\qquad
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\textbf{(C) }\frac{1}{4}\cos 40^\circ\qquad
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\textbf{(D) }\frac{1}{4}\cos 50^\circ\qquad
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\textbf{(E) }\frac{1}{8} </math>
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== Solution ==
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Let <math>F</math> be the point on the extension of side <math>AB</math> past <math>B</math> for which <math>AF=1</math>. Since <math>AF=AC</math> and <math>\measuredangle FAC = 60^\circ</math>,<math>\triangle ACF</math> is equilateral. Let <math>G</math> be the point on line segment <math>BF</math> for which <math>\measuredangle BCG=20^\circ</math>. Then <math>\triangle BCG</math> is similar to <math>\triangle DCE</math> and <math>BC=2(EC)</math>. Also <math>\triangle FGC</math> is congruent to <math>\triangle ABC</math>. Therefore, <math>[\triangle ACF]  = ([\triangle ABC] + [\triangle GCF]) + [\triangle BCG]</math>. Plugging in the values that we know and then dividing by 2 results in an answer of <math>\boxed{B) \frac{\sqrt{3}}{8}.}</math>
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This solution is from the solution manual but was typed here by alpha_2.

Latest revision as of 14:49, 30 November 2020

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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