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

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

Revision as of 12:29, 24 December 2015

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 $\triangle ABD$?

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

Solution 1

Circlenc1.png

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 $\frac{1}{3}.$

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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All AMC 10 Problems and Solutions

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