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Difference between revisions of "2005 AMC 12A Problems/Problem 15"

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<math>(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}</math>
 
<math>(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}</math>
  
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[[Image:2005_12A_AMC-15.png]] (Upload from [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=368491#368491])
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[[Image:2005_12A_AMC-15.png]]  
  
 
== Solution ==
 
== Solution ==
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Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac{1}{3}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r</math>.  
 
Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac{1}{3}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r</math>.  
  
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Now we have to find <math>CF</math>. Notice <math>\triangle OCD \similar \triangle OFC</math>, so we can write the [[proportion]]:
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Now we have to find <math>CF</math>. Notice <math>\triangle OCD \sim \triangle OFC</math>, so we can write the [[proportion]]:
  
 
<div style="text-align:center;"><math>\frac{OF}{OC} = \frac{OC}{OD}</math><br /><math>\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}</math><br /><math>OF = \frac 19r</math></div>
 
<div style="text-align:center;"><math>\frac{OF}{OC} = \frac{OC}{OD}</math><br /><math>\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}</math><br /><math>OF = \frac 19r</math></div>
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{{AMC12 box|year=2005|num-b=14|num-a=16|ab=A}}
 
{{AMC12 box|year=2005|num-b=14|num-a=16|ab=A}}
  
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 14:19, 23 September 2007

Problem

Let $\overline{AB}$ be a diameter of a circle and $C$ be a point on $\overline{AB}$ with $2 \cdot AC = BC$. Let $D$ and $E$ be points on the circle such that $\overline{DC} \perp \overline{AB}$ and $\overline{DE}$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$?

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

2005 12A AMC-15.png

Solution

Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or $\frac{CD}{CF}$ ($F$ is the foot of the perpendicular from $C$ to $DE$).

Call the radius $r$. Then $AC = \frac 13(2r) = \frac 23r$, $CO = \frac 13r$. Using the Pythagorean Theorem in $\triangle OCD$, we get $\frac{1}{3}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r$.

Now we have to find $CF$. Notice $\triangle OCD \sim \triangle OFC$, so we can write the proportion:

$\frac{OF}{OC} = \frac{OC}{OD}$
$\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}$
$OF = \frac 19r$

By the Pythagorean Theorem in $\triangle OFC$, we have $\left(\frac{1}{9}r\right)^2 + CF^2 = \left(\frac{1}{3}r\right)^2 \Longrightarrow CF = \sqrt{\frac{8}{81}r^2} = \frac{2\sqrt{2}}{9}r$.

Our answer is $\frac{CD}{CF} = \frac{\frac{2\sqrt{2}}{3}r}{\frac{2\sqrt{2}}{9}r} = \frac 13 \Longrightarrow \mathrm{(C)}$.

See also

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 12 Problems and Solutions
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