In a 8½x11 sheet of paper I drew two equal nonoverlapping circles  both completely inside the paper, of course.
What's the largest portion of the paper I could cover with the circles?
What would be the answer if I drew THREE equal circles?
First, I will call the width W (the 8.5" sides), and the length L (the 11" sides). Second I will orient the paper in "portrait" fashion… L sides are vertical, W sides are horizontal.
For 2 circles:
I haven’t proven this, but I believe the largest circles would occur when you have the two circles "jammed" into two opposite corners of the paper. So here’s how you find the largest radius, R :
The best approach I had was to sketch out my idea, marking the centers of the two circles. Then I drew a rectangle (with sides parallel to W and L) that had two opposite vertices at the two centers of the circles. Notice that the diagonal of this rectangle is 2R. Also, the width is W2R, since the center of each circle is R away from the L sides. And the length of the rectangle is L2R, since the center of each circle is R away from the W sides.
Ok, so (2R)^2 = (W2R)^2 + (L2R)^2
4R^2 = W^2 – 4WR + 4R^2 + L^2 – 4LR + 4R^2
0 = 4R^2 – 4(W+L)R + (W^2 + L^2)
Ok, well solving for the roots we get
R = [4(W+L) +/ sqrt((4(W+L))^2 – 4*4*(W^2+L^2)]/(2*4)
R = [4(W+L) +/ sqrt(4^2(W+L)^2 – 4^2*(W^2+L^2)]/8
R = [4(W+L) +/ 4*sqrt((W+L)^2 – (W^2+L^2)]/8
R = [(W+L) +/ sqrt((W+L)^2 – (W^2+L^2)]/2
R = [(W+L) +/ sqrt(W^2 + 2WL + L^2 – W^2 – L^2)]/2
R = [(W+L) +/ sqrt(2WL)]/2
So R = (19 – sqrt(187))/2 = 2.91 in
So the largest portion of the paper you could cover is
A = 2*pi*R^2 = 53.30 in^2
This is about 57% of the paper.
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For 3 circles:
Again, I haven’t proven this, but my intuition tells me that I would want to jam two of the circles into two adjacent corners (two ends of the L, not the ends of W), and then the third circle would be tangent to the midpoint of the L side. And here’s how to find the largest radius, r :
The best approach I had was to sketch out my idea, marking the centers of the three circles. I will call the circle in the bottom left corner A, the circle in the top left corner B, and the circle tangent to the middle of the right side C.
Then I drew a rectangle (with sides parallel to W and L) that had two opposite vertices at the two centers of circles A and C. Notice that the diagonal of this rectangle is 2r. Also, the width is W2r, since the center of each circle is r away from the L sides. And the length of the rectangle is L/2r, since the center of circle A is r away from the W side and the center of circle C is halfway between the two W sides.
Ok, so (2r)^2 = (W2r)^2 + (L/2r)^2
(2r)^2 = (W2r)^2 + [(L2r)/2]^2
4r^2 = W^2 – 4Wr + 4r^2 + (L^2 – 4Lr + 4r^2)/4
0 = W^2 – 4Wr + L^2/4 – Lr + r^2
0 = r^2 – (4W+L)r + (W^2 + L^2/4)
Ok, well solving for the roots we get
r = [(4W+L) +/ sqrt((4W+L)^2 – 4*1*(W^2+L^2/4)]/2
r = [(4W+L) +/ sqrt(16W^2 + 8WL +L^2 – 4W^2 – L^2)]/2
r = [(4W+L) +/ sqrt(12W^2 + 8WL)]/2
So r = (45 – sqrt(1615))/2 = 2.41 in
So the largest portion of the paper you could cover is
A = 3*pi*r^2 = 54.58 in^2
This is about 58.37% of the paper.

Posted by nikki
on 20041021 17:54:46 