# Difference between revisions of "2005 AMC 10A Problems/Problem 23"

## Problem

Let $AB$ be a diameter of a circle and let $C$ 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 $\triangle ABD$?

$[asy] unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("E",E,SSE); label("B",B,E); label("A",A,W); label("D",D,NNW); label("C",C,SW); draw(rightanglemark(D,C,B,2));[/asy]$

$\mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3}$

## Solution 1

Let us assume that the diameter is of length $1$.

$AC$ is $\frac{1}{3}$ of diameter and $CO$ is $\frac{1}{2}-\frac{1}{3} = \frac{1}{6}$.

$OD$ is the radius of the circle, so using the Pythagorean theorem height $CD$ of $\triangle AOC$ is $\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2} = \frac{\sqrt{2}}{3}$. This is also the height of the $\triangle ABD$.

Area of the $\triangle DCO$ is $\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}$ = $\frac{\sqrt{2}}{36}$.

The height of $\triangle DCE$ can be found using the area of $\triangle DCO$ and $DO$ as base.

Hence the height of $\triangle DCE$ is $\frac{\frac{\sqrt{2}}{36}}{\frac{1}{2}\cdot\frac{1}{2}}$ = $\frac{\sqrt{2}}{9}$.

The diameter is the base for both the triangles $\triangle DCE$ and $\triangle ABD$.

Hence, the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$ is $\frac{\frac{\sqrt{2}}{9}}{\frac{\sqrt{2}}{3}}$ = $\frac{1}{3} \Rightarrow C$

## Solution 2

Since $\triangle DCE$ and $\triangle ABD$ share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from $C$ to $DE$.

$[asy] import graph; import olympiad; pair O,A,B,C,D,E,F; O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); draw(Circle((0,0),15)); draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B); label("A",A,NE);label("B",B,W);label("C",C,SE);label("D",D,NE);label("E",E,SW);label("O",O,SW);label("F",F,NW); markscalefactor=0.2; draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue); [/asy]$ $OD=r, OC=\frac{1}{3}r$.

Since $m\angle DCO=m\angle DFC=90^\circ$, then $\triangle DCO\cong \triangle DFC$. So the ratio of the two altitudes is $\frac{CF}{DC}=\frac{OC}{DO}=\frac{1}{3}\Rightarrow \text{(C)}$

## Solution 3

Say the center of the circle is point $O$; Without loss of generality, assume $AC=2$, so $CB=4$ and the diameter and radius are $6$ and $3$, respectively. Therefore, $CO=1$, and $DO=3$. The area of $\triangle DCE$ can be expressed as $\frac{1}{2}(CD)(6)\text{sin }(CDE).$ $\frac{1}{2}(CD)(6)$ happens to be the area of $\triangle ABD$. Furthermore, $\text{sin } CDE = \frac{CO}{DO},$ or $\frac{1}{3}.$ Therefore, the ratio is $\frac{1}{3}.$