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Difference between revisions of "2005 Alabama ARML TST Problems/Problem 3"
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Let <math>R</math> be the [[radius]] of the circumcircle and let <math>r</math> be the radius of the incircle. Then we have <math>R^2-r^2=300</math>.  If the center of these two circles is <math>O</math>, the [[vertex | vertices]] are <math>A, B</math> and <math>C</math>, and <math>M</math> is the [[midpoint]] of side <math>AB</math>, triangle <math>\triangle AMO</math> is a <math>30^\circ-60^\circ-90^\circ</math> [[right triangle]], and its [[hypotenuse]] has length <math>R</math> and its shorter leg has length <math>r</math>.  Thus <math>R = 2r</math>.  (There are many other arguments to get to this conclusion; for instance, <math>O</math> is also the [[centroid]] of the triangle and <math>COM</math> is a [[median of a triangle | median]], so <math>O</math> trisects <math>CO</math> and <math>R = CO = 2OM = 2r</math>.)
 
Let <math>R</math> be the [[radius]] of the circumcircle and let <math>r</math> be the radius of the incircle. Then we have <math>R^2-r^2=300</math>.  If the center of these two circles is <math>O</math>, the [[vertex | vertices]] are <math>A, B</math> and <math>C</math>, and <math>M</math> is the [[midpoint]] of side <math>AB</math>, triangle <math>\triangle AMO</math> is a <math>30^\circ-60^\circ-90^\circ</math> [[right triangle]], and its [[hypotenuse]] has length <math>R</math> and its shorter leg has length <math>r</math>.  Thus <math>R = 2r</math>.  (There are many other arguments to get to this conclusion; for instance, <math>O</math> is also the [[centroid]] of the triangle and <math>COM</math> is a [[median of a triangle | median]], so <math>O</math> trisects <math>CO</math> and <math>R = CO = 2OM = 2r</math>.)
  
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Then <math>4r^2 - r^2 = 300</math> so <math>r = 10</math> and the side length of the triangle is equal to <math>10\sqrt 3</math>.
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Then <math>4r^2 - r^2 = 300</math> so <math>r = 10</math> and the side length of the triangle is equal to <math>20\sqrt 3</math>.
  
 
==See Also==
 
==See Also==
 
{{ARML box|year=2005|state=Alabama|num-b=2|num-a=4}}
 
{{ARML box|year=2005|state=Alabama|num-b=2|num-a=4}}
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[[Category:Intermediate Geometry Problems]]

Latest revision as of 18:38, 6 February 2010

Problem

The difference between the areas of the circumcircle and incircle of an equilateral triangle is $300\pi$ square units. Find the number of units in the length of a side of the triangle.

Solution

Let $R$ be the radius of the circumcircle and let $r$ be the radius of the incircle. Then we have $R^2-r^2=300$. If the center of these two circles is $O$, the vertices are $A, B$ and $C$, and $M$ is the midpoint of side $AB$, triangle $\triangle AMO$ is a $30^\circ-60^\circ-90^\circ$ right triangle, and its hypotenuse has length $R$ and its shorter leg has length $r$. Thus $R = 2r$. (There are many other arguments to get to this conclusion; for instance, $O$ is also the centroid of the triangle and $COM$ is a median, so $O$ trisects $CO$ and $R = CO = 2OM = 2r$.)

Then $4r^2 - r^2 = 300$ so $r = 10$ and the side length of the triangle is equal to $20\sqrt 3$.

See Also

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 2 		Followed by:
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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