2006 AMC 10A Problems/Problem 16

Revision as of 21:11, 15 September 2007 by Azjps (talk | contribs) (solution)

Problem

2006 AMC10A-16.png

A circle of radius 1 is tangent to a circle of radius 2. The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $\overline{AB}$ and $\overline{AC}$ are congruent. What is the area of $\triangle ABC$?

$\mathrm{(A) \ } \frac{35}{2}\qquad\mathrm{(B) \ } 15\sqrt{2}\qquad\mathrm{(C) \ } \frac{64}{3}\qquad\mathrm{(D) \ } 16\sqrt{2}\qquad\mathrm{(E) \ } 24\qquad$

Solution

2006 AMC10A-16a.png

Note that $\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC \displaystyle$. Using the first pair of similar triangles, we write the proportion:

$\frac{AO_1}{AO_2} = \frac{DO_1}{DO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3$

By the Pythagorean Theorem we have that $AD = \sqrt{3^2-1^2} = \sqrt{8}$.

Now using $\triangle ADO_1 \sim \triangle AFC$,

$\frac{AD}{AF} = \frac{DO_1}{CF} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{CF} \Longrightarrow CF = 2\sqrt{2}$

The area of the triangle is $\frac{1}{2}bh = \frac{1}{2}\left(2\cdot 2\sqrt{2}\right)\left(8\right) = 16\sqrt{2}\ \mathrm{(D)}$.

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions
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