Difference between revisions of "2005 Alabama ARML TST Problems/Problem 3"
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==Problem== | ==Problem== | ||
− | The difference between the | + | The difference between the [[area]]s of the [[circumcircle]] and [[incircle]] of an [[equilateral triangle]] is <math>300\pi</math> square units. Find the number of units in the [[length]] of a [[edge | side]] of the [[triangle]]. |
==Solution== | ==Solution== | ||
+ | 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>.) | ||
− | + | 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== | |
− | + | {{ARML box|year=2005|state=Alabama|num-b=2|num-a=4}} | |
− | + | ||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 19:38, 6 February 2010
Problem
The difference between the areas of the circumcircle and incircle of an equilateral triangle is square units. Find the number of units in the length of a side of the triangle.
Solution
Let be the radius of the circumcircle and let
be the radius of the incircle. Then we have
. If the center of these two circles is
, the vertices are
and
, and
is the midpoint of side
, triangle
is a
right triangle, and its hypotenuse has length
and its shorter leg has length
. Thus
. (There are many other arguments to get to this conclusion; for instance,
is also the centroid of the triangle and
is a median, so
trisects
and
.)
Then so
and the side length of the triangle is equal to
.
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 |