2005 AMC 12A Problem 15

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}

Let

O

be the center. Since

AC = \dfrac{AB}{3}

and

AO = \dfrac{AB}{2},

find

CO

\triangle DCO

and

\triangle ECO

have equal areas because

O

is the midpoint of

DE

Let

O

be the center. Since

2 \cdot AC = BC,

we have

AC = \dfrac{AB}{3},

and

AO = \dfrac{AB}{2},

so

\begin{aligned} &CO = AO - AC \\ &= \dfrac{AB}{2} - \dfrac{AB}{3} \\ &= \dfrac{AB}{6}. \end{aligned}

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

\begin{aligned} &[DCE] = 2\,[DCO] \\ &= \dfrac{2}{6}[DAB] = \dfrac{1}{3}[DAB]. \end{aligned}

Thus, the correct answer is C.

Problem 15 in other years