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

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(→‎Solution)
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<math> \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} </math>. The area of  
 
<math> \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} </math>. The area of  
  
−
==Solution==
+
==Solution 1==
 
http://img443.imageshack.us/img443/8034/circlenc1.png
 
http://img443.imageshack.us/img443/8034/circlenc1.png
  
Line 21: Line 21:
 
Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is
 
Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is
 
<math>\frac{\frac{\sqrt{2}}{9}}{\frac{\sqrt{2}}{3}}</math> = <math>\frac{1}{3} \Rightarrow C</math>
 
<math>\frac{\frac{\sqrt{2}}{9}}{\frac{\sqrt{2}}{3}}</math> = <math>\frac{1}{3} \Rightarrow C</math>
 +
==Solution 2==
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 +
Since <math>\triangle DCE</math> and <math>\triangle ABD</math> share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from <math>C</math> to <math>DE</math>.
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 +
<asy>
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import graph;
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import olympiad;
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pair O,A,B,C,D,E,F;
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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);
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draw(Circle((0,0),15));
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draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);
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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);
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markscalefactor=0.2;
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draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue);
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</asy>
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<math>OD=r, OC=\frac{1}{3}r</math>.
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 +
Since <math>m\angle DCO=m\angle DFC=90^\circ</math>, then <math>\triangle DCO\cong \triangle DFC</math>. So the ratio of the two altitudes is <math>\frac{DC}{CF}=\frac{DO}{OC}=\frac{1}{3}\Rightarrow \text{(C)}</math>
  
 
==See also==
 
==See also==

Revision as of 19:58, 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); 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 $\frac{DC}{CF}=\frac{DO}{OC}=\frac{1}{3}\Rightarrow \text{(C)}$

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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