Question

Let $AB$ be a diameter of a circle and $C$ be a point on $AB$ with $2 \cdot AC = BC.$ Let $D$ and $E$ be points on the circle such that $DC \perp AB$ and $DE$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD?$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

Solution:

Let $O$ be the center. Since $2 \cdot AC = BC,$ we have $AC = \dfrac{AB}{3},$ and $AO = \dfrac{AB}{2},$ so $CO = AO - AC = \dfrac{AB}{2} - \dfrac{AB}{3} = \dfrac{AB}{6}.$

Triangles $DCO$ and $DAB$ share the same altitude from $D$ to line $AB,$ so $\dfrac{[DCO]}{[DAB]} = \dfrac{CO}{AB} = \dfrac{1}{6}.$

Because $O$ is the midpoint of $DE,$ triangles $DCO$ and $ECO$ have equal areas, so $[DCE] = 2\,[DCO] = \dfrac{2}{6}[DAB] = \dfrac{1}{3}[DAB].$

Thus, the correct answer is C.

2005 AMC 12A Problem 15

Solution:

Problem 15 in Other Years