Difference between revisions of "2010 AMC 10B Problems/Problem 20"
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== Problem == | == Problem == | ||
− | Two circles lie outside regular hexagon <math>ABCDEF</math>. The first is tangent to <math>\ | + | Two circles lie outside regular hexagon <math>ABCDEF</math>. The first is tangent to <math>\overline{AB}</math>, and the second is tangent to <math>\overline{DE}</math>. Both are tangent to lines <math>BC</math> and <math>FA</math>. What is the ratio of the area of the second circle to that of the first circle? |
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 108</math> | <math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 108</math> | ||
− | == Solution == | + | == Solution 1 == |
− | A good diagram | + | A good diagram is very helpful. |
<center> [[File:Amc10B 2010.gif]] </center> | <center> [[File:Amc10B 2010.gif]] </center> | ||
Line 16: | Line 16: | ||
Since the ratio of areas is equal to the square of the ratio of lengths, we know our final answer is <math>\left(\frac{GK}{GB}\right)^2</math>. | Since the ratio of areas is equal to the square of the ratio of lengths, we know our final answer is <math>\left(\frac{GK}{GB}\right)^2</math>. | ||
From the diagram, we can see that this is <math>9^2=\boxed{\textbf{(D)}\ 81}</math> | From the diagram, we can see that this is <math>9^2=\boxed{\textbf{(D)}\ 81}</math> | ||
+ | |||
+ | == Solution 2 == | ||
+ | As above, we note that the first circle is inscribed in an equilateral triangle of side length <math>1</math> (if we assume, WLOG, that the regular hexagon has side length <math>1</math>). The inradius of an equilateral triangle with side length <math>1</math> is equal to <math>\frac{\sqrt{3}}{6}</math>. Therefore, the area of the first circle is <math>(\frac{\sqrt{3}}{6})^2 \cdot \pi =\frac{\pi}{12}</math>. | ||
+ | |||
+ | |||
+ | Call the center of the second circle <math>O</math>. Now we drop a perpendicular from <math>O</math> to circle <math>O</math>'s point of tangency with <math>GK</math> and draw another line connecting <math>O</math> to <math>G</math>. Note that because triangle <math>BGA</math> is equilateral, <math>\angle BGA=60^{\circ}</math>. <math>OG</math> bisects <math>\angle BGA</math>, so we have a <math>30-60-90</math> triangle. | ||
+ | |||
+ | |||
+ | Call the radius of circle <math>O</math> <math>r</math>. | ||
+ | <math>OG=2r= \text{height of equilateral triangle} + \text{height of regular hexagon} + r</math> | ||
+ | |||
+ | The height of an equilateral triangle of side length <math>1</math> is <math>\frac{\sqrt{3}}{2}</math>. The height of a regular hexagon of side length <math>1</math> is <math>\sqrt{3}</math>. Therefore, | ||
+ | <math>OG=\frac{\sqrt{3}}{2} + \sqrt{3} + r</math>. | ||
+ | |||
+ | We can now set up the following equation: | ||
+ | |||
+ | <math>\frac{\sqrt{3}}{2} + \sqrt{3} + r=2r</math> | ||
+ | |||
+ | <math>\frac{\sqrt{3}}{2} + \sqrt{3}=r</math> | ||
+ | |||
+ | <math>\frac{3\sqrt{3}}{2}=r</math> | ||
+ | |||
+ | The area of Circle <math>O</math> equals <math>\pi r^2=\frac{27}{4} \pi</math> | ||
+ | |||
+ | Therefore, the ratio of the areas is <math>\frac{\frac{27}{4}}{\frac{1}{12}}=\boxed{\textbf{(D)}\ 81}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/FQO-0E2zUVI?t=1381 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2010|ab=B|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Latest revision as of 18:41, 4 May 2024
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.
![Amc10B 2010.gif](https://wiki-images.artofproblemsolving.com//d/d3/Amc10B_2010.gif)
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 side length (if we assume, WLOG, that the regular hexagon has side length
). The inradius of an equilateral triangle with side length
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
's point of tangency with
and draw another line connecting
to
. Note that because triangle
is equilateral,
.
bisects
, so we have a
triangle.
Call the radius of circle
.
The height of an equilateral triangle of side length is
. The height of a regular hexagon of side length
is
. Therefore,
.
We can now set up the following equation:
The area of Circle 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.