This is in continuation of Two circles
In a 8.5x11 sheet of paper I drew two non-overlapping ellipses of equal area, with major axis of each equalling twice the minor axis - both completely inside the paper, of course.
What's the largest portion of the paper I could cover with these ellipses?
What would be the answer if I drew THREE equal ellipses?
The stretching idea gives
for a 4.25 by 11 rectangle
r = 14 - sqrt(147.6875) ≈1.8473
for a 8.5 by 5.5 rectangle
But there is a better way of doing this with the major axis of one parallel to the other two.
Orient the paper so one corner is at (0,0) and the opposite at (8.5,11).
One ellipse is taller than it is wide and is tangent to the x- and y-axes:
4(x-r)^2 + (y-2r)^2 = 4r^2
The other is wider than it is tall and is tangent to the top of the paper at the center:
(x-4.25)^2 + 4(y-(11-r))^2 = 4r^2
What is sought is a value of r that will make these tangent. In other words the system of these two equations should have exactly one solution. Unfortunately trying to find this value gives a horrendous fourth degree polynomial. So I just graphed the ellipses with different values of r until they just touched.
The solution is between 1.892 and 1.893 so to be safe use r=1.892.
The area of each ellipse is about 22.492
or 67.475 total.
Divide this by 8.5*11 to give a proportion covered of .7217
This still doesn't rule out the possibility of tilting the ellipses. I could imagine tilting the tall ones slightly outwards, giving the wide one more room. This might allow them to be bigger. Unfortunately Geometers Sketchpad doesn't do ellipses very easily, so I will not attempt this.
Posted by Jer
on 2011-01-20 16:52:19