2005 AMC 10A Problem 23

Attempt Problem 23 of the 2005 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10A solutions, or check the answer key.

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23.

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}

Answer: C

Concepts:area ratio circle triangle area

Difficulty rating: 2010

Small hint:

Let

O

be the center; from

2 \cdot AC = BC

locate

C

using

AC = \frac{1}{3}AB

Big hint:

Since

O

is the midpoint of

DE,

[\triangle DCE] = 2\,[\triangle DCO],

and

\triangle DCO

shares the altitude from

D

with

\triangle DAB

Solution:

Let

O

be the center. From

2 \cdot AC = BC

and

AC + BC = AB,

we get

AC = \dfrac{AB}{3},

so

CO = \dfrac{AB}{2} - \dfrac{AB}{3} = \dfrac{AB}{6}.

Triangles

DCO

and

DAB

share the apex

D

with bases

CO

and

AB

on the same line, so

[\triangle DCO] = \dfrac{CO}{AB}[\triangle DAB]

= \dfrac{1}{6}[\triangle DAB].

Because

O

is the midpoint of

DE,

[\triangle DCE] = 2\,[\triangle DCO]

= \dfrac{1}{3}[\triangle DAB].

Thus, the correct answer is C.