Difference between revisions of "2006 AMC 10A Problems/Problem 16"
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== Problem == | == Problem == | ||
| + | [[Image:2006_AMC10A-16.png]] | ||
| − | {{ | + | A [[circle]] of [[radius]] 1 is tangent to a circle of radius 2. The sides of <math>\triangle ABC</math> are [[tangent]] to the circles as shown, and the sides <math>\overline{AB}</math> and <math>\overline{AC}</math> are [[congruent]]. What is the area of <math>\triangle ABC</math>? |
| − | + | <math>\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</math> | |
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== Solution == | == Solution == | ||
| + | [[Image:2006_AMC10A-16a.png]] | ||
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| + | Note that <math>\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC \displaystyle</math>. Using the first pair of [[similar triangles]], we write the [[proportion]]: | ||
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| + | <div style="text-align:center;"><math>\frac{AO_1}{AO_2} = \frac{DO_1}{DO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3</math></div> | ||
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| + | By the [[Pythagorean Theorem]] we have that <math>AD = \sqrt{3^2-1^2} = \sqrt{8}</math>. | ||
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| + | Now using <math>\triangle ADO_1 \sim \triangle AFC</math>, | ||
| − | {{ | + | <div style="text-align:center;"><math>\frac{AD}{AF} = \frac{DO_1}{CF} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{CF} \Longrightarrow CF = 2\sqrt{2}</math></div> |
| − | == | + | The area of the triangle is <math>\frac{1}{2}bh = \frac{1}{2}\left(2\cdot 2\sqrt{2}\right)\left(8\right) = 16\sqrt{2}\ \mathrm{(D)}</math>. |
| − | + | == See also == | |
| − | + | {{AMC10 box|year=2006|num-b=15|num-a=17|ab=A}} | |
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Revision as of 20:11, 15 September 2007
Problem
A circle of radius 1 is tangent to a circle of radius 2. The sides of 
are tangent to the circles as shown, and the sides 
and 
are congruent. What is the area of ?
Solution
Note that 
. Using the first pair of similar triangles, we write the proportion:
By the Pythagorean Theorem we have that 
.
Now using 
,
The area of the triangle is 
.
See also
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
