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

Revision as of 21:31, 5 February 2010

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 $\triangle ABD$ ?

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

Solution 1

http://img443.imageshack.us/img443/8034/circlenc1.png

$AC$ 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 $\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2$ (Error compiling LaTeX. Unknown error_msg) = $\frac{\sqrt{2}}{3}$ . 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 $\frac{DC}{CF}=\frac{DO}{OC}=\frac{1}{3}\Rightarrow \text{(C)}$

2005 AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2005 AMC 10A Problems/Problem 23"

Revision as of 21:31, 5 February 2010

Contents

Problem

Solution 1

Solution 2

See also

@@ Line 31: / Line 31: @@
 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--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;