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2005 AMC 12A Problems/Problem 15: Difference between revisions

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== Solution ==
== Solution ==
'''Solution 1'''
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 <math>\frac{CD}{CF}</math> (<math>F</math> is the foot of the [[perpendicular]] from <math>C</math> to <math>DE</math>).
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 <math>\frac{CD}{CF}</math> (<math>F</math> is the foot of the [[perpendicular]] from <math>C</math> to <math>DE</math>).


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Our answer is <math>\frac{CD}{CF} = \frac{\frac{2\sqrt{2}}{3}r}{\frac{2\sqrt{2}}{9}r} = \frac 13 \Longrightarrow \mathrm{(C)}</math>.
Our answer is <math>\frac{CD}{CF} = \frac{\frac{2\sqrt{2}}{3}r}{\frac{2\sqrt{2}}{9}r} = \frac 13 \Longrightarrow \mathrm{(C)}</math>.
'''Solution 2'''
Let the centre of the circle be <math>O</math>.
Note that <math>2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB</math>.
<math>O</math> is midpoint of <math>AB \Rightarrow \frac{3}{2}AC = AO \Rightarrow CO = \frac{1}{3}AO \Rightarrow CO = \frac{1}{6} AB</math>.
<math>O</math> is midpoint of <math>DE \Rightarrow</math> Area of <math>\triangle DCE = 2 \cdot</math> Area of <math>\triangle DCO = 2 \cdot (\frac{1}{6} \cdot</math> Area of <math>\triangle ABD) = \frac{1}{3} \cdot</math> Area of <math>\triangle ABD \Longrightarrow \mathrm{(C)}</math>.


== See also ==
== See also ==

Revision as of 08:16, 5 February 2009

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 $\triangle ABD$?

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

Solution

Solution 1

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


Solution 2

Let the centre of the circle be $O$.

Note that $2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB$.

$O$ is midpoint of $AB \Rightarrow \frac{3}{2}AC = AO \Rightarrow CO = \frac{1}{3}AO \Rightarrow CO = \frac{1}{6} AB$.

$O$ is midpoint of $DE \Rightarrow$ Area of $\triangle DCE = 2 \cdot$ Area of $\triangle DCO = 2 \cdot (\frac{1}{6} \cdot$ Area of $\triangle ABD) = \frac{1}{3} \cdot$ Area of $\triangle ABD \Longrightarrow \mathrm{(C)}$.

See also

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