2010 AMC 10B Problems/Problem 20
Problem
Two circles lie outside regular hexagon . The first is tangent to , and the second is tangent to . Both are tangent to lines and . What is the ratio of the area of the second circle to that of the first circle?
Solution 1
A good diagram is very helpful.
The first circle is in red, the second in blue. With this diagram, we can see that the first circle is inscribed in equilateral triangle while the second circle is inscribed in . From this, it's evident that the ratio of the blue area to the red area is equal to the ratio of the areas to
Since the ratio of areas is equal to the square of the ratio of lengths, we know our final answer is . From the diagram, we can see that this is
Solution 2
As above, we note that the first circle is inscribed in an equilateral triangle of sidelength 1 (if we assume, WLOG, that the regular hexagon has sidelength 1). The inradius of an equilateral triangle with sidelength 1 is equal to . Therefore, the area of the first circle is .
Call the center of the second circle . Now we drop a perpendicular from to Circle O's point of tangency with and draw another line connecting O to G. Note that because triangle is equilateral, . bisects , so we have a 30-60-90 triangle.
Call the radius of Circle O .
= height of equilateral triangle + height of regular hexagon + r
The height of an equilateral triangle of sidelength 1 is . The height of a regular hexagon of sidelength 1 is . Therefore, .
We can now set up the following equation:
The area of Circle O equals
Therefore, the ratio of the areas is
Video Solution
https://youtu.be/FQO-0E2zUVI?t=1381
~IceMatrix
See Also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
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