Difference between revisions of "2006 AMC 10A Problems/Problem 16"
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Hence, the area of the triangle is <cmath>\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = \boxed{16\sqrt{2}\ \mathrm{(D)}}</cmath> | Hence, the area of the triangle is <cmath>\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = \boxed{16\sqrt{2}\ \mathrm{(D)}}</cmath> | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | <asy> | ||
+ | size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); | ||
+ | real t=2^0.5; | ||
+ | D((0,0)--(4*t,0)--(2*t,8)--cycle); | ||
+ | D(CR(D((2*t,2)),2)); | ||
+ | D(CR(D((2*t,5)),1)); | ||
+ | D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); | ||
+ | pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0)); | ||
+ | D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0))); | ||
+ | MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); </asy> | ||
+ | |||
+ | Since <math>\triangle{A O_1 D} \sim \triangle{A O_2 E},</math> we have that <math>\frac{A O_1}{A O_2} = \frac{O_1 D}{O_2 E} = \frac{1}{2}</math>. | ||
+ | |||
+ | Since we know that <math>O_1 O_2 = 1 + 2 = 3,</math> the total length of <math>A O_2 = 2 \cdot 3 = 6.</math> | ||
+ | |||
+ | We also know that <math>O_2 X = 2</math>, so <math>A X = A O_2 + O_2 X = 6 + 2 = 8.</math> | ||
+ | |||
+ | Also, since <math>\triangle{ABX} \sim \triangle{A E O_2},</math> we have that | ||
== See also == | == See also == |
Revision as of 15:18, 25 December 2020
Contents
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
Let the centers of the smaller and larger circles be and , respectively. Let their tangent points to be and , respectively. We can then draw the following diagram:
We see that . Using the first pair of similar triangles, we write the proportion:
By the Pythagorean Theorem, we have .
Now using ,
Hence, the area of the triangle is
Solution 2
Since we have that .
Since we know that the total length of
We also know that , so
Also, since we have that
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.