Difference between revisions of "2010 AMC 10B Problems/Problem 20"

Revision as of 18:28, 5 September 2022

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 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 $\frac{\sqrt{3}}{6}$ . 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 sidelength 1 is $\frac{\sqrt{3}}{2}$ . The height of a regular hexagon of sidelength 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

Revision as of 18:27, 5 September 2022 (view source) Nathanj (talk \| contribs) m (Edited parentheses) ← Older edit		Revision as of 18:28, 5 September 2022 (view source) Nathanj (talk \| contribs) m Newer edit →
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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 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>.

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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