Difference between revisions of "2006 AMC 10A Problems/Problem 16"
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pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0)); | 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))); | 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(" | + | 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> |
We see that <math>\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC</math>. Using the first pair of [[similar triangles]], we write the proportion: | We see that <math>\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC</math>. Using the first pair of [[similar triangles]], we write the proportion: |
Revision as of 22:19, 19 September 2023
Contents
[hide]Problem
A circle of radius is tangent to a circle of radius . The sides of are tangent to the circles as shown, and the sides and are congruent. What is the area of ?
Solution 1
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
Since we know that and we have that
This equation simplified gets us
Let
By the Pythagorean Theorem on we have that
We know that , and so we have
Simplifying, we have that
Recall that .
Therefore,
Since the height is we have the area equal to
Thus our answer is .
~mathboy282
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.