Question

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

$\frac14\cos10^\circ$

$\frac{\sqrt3}{8}$

$\frac14\cos40^\circ$

$\frac14\cos50^\circ$

$\frac18$

Answer: B

Concepts:triangle area congruence (geometry)similarity equilateral triangle

Difficulty rating: 2200

Small Hint:

Extend $AB$ to $F$ so that $AF=AC$ , creating an equilateral triangle

Big Hint:

Choose $G$ on $BF$ with $\angle BCG=20^\circ$ ; compare $\triangle BCG$ with $\triangle DCE$

Solution:

Extend $AB$ through $B$ to $F$ with $AF=AC=1.$ Then $\triangle ACF$ is equilateral. Choose $G$ on $BF$ so that $\angle BCG=20^\circ.$ The angle conditions give $\triangle FGC\cong\triangle ABC,$ while $\triangle BCG\sim\triangle DCE.$ Since $E$ is the midpoint of $BC,$ the similarity scale is $2,$ so $[BCG]=4[CDE].$ Decomposing the equilateral triangle gives $\frac{\sqrt3}{4} =2[ABC]+4[CDE].$ Dividing by $2$ yields $[ABC]+2[CDE]=\frac{\sqrt3}{8}.$

Therefore, the correct answer is B.

1979 AMC 12 Problem 30

Answer: B

Difficulty rating: 2200

Small Hint:

Big Hint:

Solution:

Problem 30 in Other Years