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Difference between revisions of "2007 UNCO Math Contest II Problems/Problem 9"

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Let <math>ABC</math> be the equilateral triangle and let <math>A_1B_1</math> be the internal common tangent of the two largest circles, with the points <math>A_1</math> and <math>B_1</math> laying on the sides <math>AB</math> and <math>BC</math>, respectively.
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Let <math>L</math> be the point of tangency of the two circles.
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Let <math>OM</math> be the inscribed circle's radius, with <math>O</math> being its center and <math>M</math> being the midpoint of <math>AB</math>. Let <math>BH</math> be the triangle's altitude. We can calculate <math>OM</math> using Pythagoras' Theorem:
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<math>OM^2 = BO^2 - BM^2</math>
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<math>OM^2 = \left(\frac{2}{3} \times BH \right)^2 - \left(\frac{1}{2} \times AB \right)^2</math>
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<math>OM^2 = \left( \frac{2 \sqrt{3}}{3} \right)^2 - 1^2</math>
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<math>OM = \frac{\sqrt3}{3}</math>
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<math>\Rightarrow OM = \frac{BH}{3}</math>
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From there we have <math>BL = OM = \frac{BH}{3}</math>. This means the scale factor of the similar triangles <math>ABC</math> and <math>A_1B_1B</math> is <math>\frac {1}{3}</math> (because the ratio of their altitudes <math>\frac{BL}{BH}</math> is equal to <math>\frac{1}{3}</math> .)
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Since the smaller circle is inscribed in <math>A_1B_1B</math>, its radius is 3 times smaller than <math>OM</math>, i.e. equal to <math>\frac{\sqrt3}{9}</math> . Repeating this trend infinitely and summing all the areas gives us:
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<math>Area = \pi \left(\frac{\sqrt3}{3}\right)^2 + \pi \left(\frac{\sqrt3}{9}\right)^2 + \pi \left(\frac{\sqrt3}{27}\right)^2 +\pi \left(\frac{\sqrt3}{81}\right)^2 + ...</math>
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<math>Area = \pi \left(\frac{1}{3} + \frac{1}{3^3} + \frac{1}{3^5} + \frac{1}{3^7} + ... \right)</math>
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We have the geometric progression <math>B_n = \frac{1}{3^{2n-1}}</math> with a scale factor of <math>\frac{1}{3}</math> and a common ratio of <math>\frac{1}{9}</math> :
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<math>Area = \pi \times \frac{\frac{1}{3}}{1 - \frac{1}{9}}</math>
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<math>Area = \frac{3}{8} \pi</math>
 
== See Also ==
 
== See Also ==
 
{{UNCO Math Contest box|n=II|year=2007|num-b=8|num-a=10}}
 
{{UNCO Math Contest box|n=II|year=2007|num-b=8|num-a=10}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 17:38, 29 January 2018

Problem

A circle is inscribed in an equilateral triangle whose side length is $2$. Then another circle is inscribed externally tangent to the first circle but inside the triangle as shown. And then another, and another. If this process continues forever what is the total area of all the circles? Express your answer as an exact multiple of $\pi$ (and not as a decimal approximation).

[asy] path T=polygon(3); draw(unitcircle,black); draw(scale(2)*T,black); draw(shift(2/sqrt(3),-2/3)*scale(1/3)*unitcircle,black); draw(shift(2/sqrt(3)/3,-2/9)*shift(2/sqrt(3),-2/3)*scale(1/9)*unitcircle,black); [/asy]


Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Let $ABC$ be the equilateral triangle and let $A_1B_1$ be the internal common tangent of the two largest circles, with the points $A_1$ and $B_1$ laying on the sides $AB$ and $BC$, respectively.

Let $L$ be the point of tangency of the two circles.

Let $OM$ be the inscribed circle's radius, with $O$ being its center and $M$ being the midpoint of $AB$. Let $BH$ be the triangle's altitude. We can calculate $OM$ using Pythagoras' Theorem:


$OM^2 = BO^2 - BM^2$


$OM^2 = \left(\frac{2}{3} \times BH \right)^2 - \left(\frac{1}{2} \times AB \right)^2$


$OM^2 = \left( \frac{2 \sqrt{3}}{3} \right)^2 - 1^2$


$OM = \frac{\sqrt3}{3}$


$\Rightarrow OM = \frac{BH}{3}$


From there we have $BL = OM = \frac{BH}{3}$. This means the scale factor of the similar triangles $ABC$ and $A_1B_1B$ is $\frac {1}{3}$ (because the ratio of their altitudes $\frac{BL}{BH}$ is equal to $\frac{1}{3}$ .)

Since the smaller circle is inscribed in $A_1B_1B$, its radius is 3 times smaller than $OM$, i.e. equal to $\frac{\sqrt3}{9}$ . Repeating this trend infinitely and summing all the areas gives us:


$Area = \pi \left(\frac{\sqrt3}{3}\right)^2 + \pi \left(\frac{\sqrt3}{9}\right)^2 + \pi \left(\frac{\sqrt3}{27}\right)^2 +\pi \left(\frac{\sqrt3}{81}\right)^2 + ...$


$Area = \pi \left(\frac{1}{3} + \frac{1}{3^3} + \frac{1}{3^5} + \frac{1}{3^7} + ... \right)$


We have the geometric progression $B_n = \frac{1}{3^{2n-1}}$ with a scale factor of $\frac{1}{3}$ and a common ratio of $\frac{1}{9}$ :


$Area = \pi \times \frac{\frac{1}{3}}{1 - \frac{1}{9}}$


$Area = \frac{3}{8} \pi$

See Also

2007 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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