All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Ellipses (Posted on 2004-05-17) Difficulty: 5 of 5
What is the area of the smallest ellipse that can be circumscribed around a 3-4-5 triangle?

What is the area of the largest ellipse that can be inscribed in a 3-4-5 triangle?

No Solution Yet Submitted by SilverKnight    
Rating: 3.8333 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution General Solution | Comment 6 of 12 |

Let T be any triangle represented on a coordinate plane by letting point A be (0,0), B be (a,0) and C be (b,c).

The transformation (x + ((a-2b)/(2c)) * y, y) = (s,t) transforms the triangle T to an isosceles triangle T". 
The transformation (s, ((a*sqrt(3))/(2c)) * t) = (u,v) transforms the isosceles triangle T" to an equilateral triangle T'. 
The composite transformation T to T' is (x + ((a-2b)/(2c)) * y, (a*sqrt(3))/(2c)) * y) = (u,v). 
All three transformations preserve area ratios, that is the ratio of two areas before a transformation will be the same as the ratio of the areas they map to after the tramsformation.

Triangle T' is an equilateral triangle with side length of a, vertecies at A'(0,0), B'(a,0), C'( a/2, (a*sqrt(3))/2 ) and center ( a/2, a/(2*sqrt(3)) ). 

The minimum circumellipse of T' (Ec') is the circle ( u - a/2 )^2 + ( v - a/(2*sqrt(3)) )^2 = ( a/sqrt(3) )^2. 
The maximum inellipse (Ei') is the circle ( u - a/2 )^2 + ( v - a/(2*sqrt(3)) )^2 = ( a/(2*sqrt(3)) )^2.  Both the circumellipse and inellipse have the same center.

Since area ratios are preserved, the equation T/Ec=T'/Ec' can be used to calculate the area of the circumellipse for the original triangle T. 
T = 1/2 * a * c.  T' = a^2*sqrt(3)/4.  Ec' = pi*a^2/3
(1/2 * a * c)/(Ec) = (a^2*sqrt(3)/4)/(pi*a^2/3)
Ec = 2*pi*a*c/(3*sqrt(3))

Similarily for the inellipse, Ei = pi*a*c/(6*sqrt(3))

The transformation can be inverted by substituting x + ((a-2b)/(2c)) * y for u and  (a*sqrt(3))/(2c)) * y for v.  Substituting those into the equations for the ellipses and simplifying yields:
Circumellipse: c^2*x^2 + (a*c-2b*c)*xy + (a^2-a*b+b^2)*y^2 - a*c^2*x + (a*b*c-a^2*c)*y = 0
Inellpise: c^2*x^2 + (a*c-2b*c)*xy + (a^2-a*b+b^2)*y^2 - a*c^2*x + (a*b*c-a^2*c)*y + (a^2*c^2)/4 = 0
Both have the same center: ((a+b)/3, c/3)

The areas of the ellipses can be calculated directly from the standard form equations by using the formula (Area)^2 = 4*pi^2*(4A*C*F+B*D*E-A*E^2-C*D^2-B^2*F)^2/(4*A*C-B^2)^3.  After some laborous algebra, the areas work out to be the same as above.

For the 3-4-5 triangle mentioned in the problem, substitute a=4, b=0, c=3 to get circumellipse area = Ec = 2*pi*4*3/(3*sqrt(3)) = 8*pi/sqrt(3) and inellipse area = Ei = pi*4*3/(6*sqrt(3)) = 2*pi/sqrt(3).  The equation of the circumellipse is 9x^2+12xy+16y^2-36x-48y=0 and the equation of the inellipse is 9x^2+12xy+16y^2-36x-48y+36=0.  Both ellipses have center (4/3, 1).

  Posted by Brian Smith on 2004-05-18 14:00:29
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information