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

This is just solution to smallist circumscribing ellipse.

Use cartesian. Then triangle can be (0,0), (4,0), and (0,3).

Standard form of ellipse is (x-h)^2/a^2 + (y-k)^2/b^2 = 1.

Get an equation for each point:

1. (h/a)^2 + (k/b)^2 = 1

2. ((h-4)/a)^2 + (k/b)^2 = 1

3. (h/a)^2 + ((k-3)/b)^2 = 1

The only way for both 1 and 2 to be true is if h = 2.
The only way for both 1 and 3 to be true is if k = 3/2.

So an ellipse circumscribing these points will have general form,

((x-2)/a)^2 + ((y-3/2)/b)^2 = 1

At (0,0) we get,

4/a^2 + 9/4/b^2 = 1

16b^2 + 9a^2 = 4a^2b^2

b = 3a / ( (4a^2 - 16)^(1/2) )

A(a,b) = Area = pi*a*b

= (3* a^2 * pi/2) / ( (a^2-4) ^(1/2) )

We want to minimize this.

dA = 3*pi/2 [ 2a / ((a^2-4)*(1/2)) - a^3 / ((a^2-4)^(3/2)) ]

dA = 0

0 = [2a * (a^2-4) - a^3]/(a^2-4)^(3/2)

0 = a * (a^2 - 8)

since a must be real and nonzero

a = 8^(1/2)
substituting, b = 8^(1/2) * 3/4

So the minimum area = pi * a * b =

6 * pi.

good problem.

Posted by vectorboy on 2004-05-31 11:42:47