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2010 AMC 10B Problems/Problem 20: Difference between revisions

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== Solution 2 ==
== 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 <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>.
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 O's point of tangency with <math>GK</math> and draw another line connecting O to G. 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 30-60-90 triangle.  
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 O <math>r</math>.
Call the radius of circle <math>O</math> <math>r</math>.
<math>OG=2r</math>= height of equilateral triangle + height of regular hexagon + r
<math>OG=2r= \text{height of equilateral triangle} + \text{height of regular hexagon} + r</math>


The height of an equilateral triangle of sidelength 1 is <math>\frac{\sqrt{3}}{2}</math>. The height of a regular hexagon of sidelength 1 is <math>\sqrt{3}</math>. Therefore,
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>.
<math>OG=\frac{\sqrt{3}}{2} + \sqrt{3} + r</math>.


Line 38: Line 38:
<math>\frac{3\sqrt{3}}{2}=r</math>
<math>\frac{3\sqrt{3}}{2}=r</math>


The area of Circle O equals <math>\pi r^2=\frac{27}{4} \pi</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{1}{12}}{\frac{27}{4}}=81</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}}
{{MAA Notice}}

Latest revision as of 17:41, 4 May 2024

Problem

Two circles lie outside regular hexagon $ABCDEF$. The first is tangent to $\overline{AB}$, and the second is tangent to $\overline{DE}$. Both are tangent to lines $BC$ and $FA$. What is the ratio of the area of the second circle to that of the first circle?

$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 108$

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 $GBA$ while the second circle is inscribed in $GKJ$. From this, it's evident that the ratio of the blue area to the red area is equal to the ratio of the areas $\triangle GKJ$ to $\triangle GBA$

Since the ratio of areas is equal to the square of the ratio of lengths, we know our final answer is $\left(\frac{GK}{GB}\right)^2$. From the diagram, we can see that this is $9^2=\boxed{\textbf{(D)}\ 81}$

Solution 2

As above, we note that the first circle is inscribed in an equilateral triangle of side length $1$ (if we assume, WLOG, that the regular hexagon has side length $1$). The inradius of an equilateral triangle with side length $1$ is equal to $\frac{\sqrt{3}}{6}$. Therefore, the area of the first circle is $(\frac{\sqrt{3}}{6})^2 \cdot \pi =\frac{\pi}{12}$.


Call the center of the second circle $O$. Now we drop a perpendicular from $O$ to circle $O$'s point of tangency with $GK$ and draw another line connecting $O$ to $G$. Note that because triangle $BGA$ is equilateral, $\angle BGA=60^{\circ}$. $OG$ bisects $\angle BGA$, so we have a $30-60-90$ triangle.


Call the radius of circle $O$ $r$. $OG=2r= \text{height of equilateral triangle} + \text{height of regular hexagon} + r$

The height of an equilateral triangle of side length $1$ is $\frac{\sqrt{3}}{2}$. The height of a regular hexagon of side length $1$ is $\sqrt{3}$. Therefore, $OG=\frac{\sqrt{3}}{2} + \sqrt{3} + r$.

We can now set up the following equation:

$\frac{\sqrt{3}}{2} + \sqrt{3} + r=2r$

$\frac{\sqrt{3}}{2} + \sqrt{3}=r$

$\frac{3\sqrt{3}}{2}=r$

The area of Circle $O$ equals $\pi r^2=\frac{27}{4} \pi$

Therefore, the ratio of the areas is $\frac{\frac{27}{4}}{\frac{1}{12}}=\boxed{\textbf{(D)}\ 81}$

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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