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?
By defining a right triangle where the hypotenuse connects the radii of the two equal circles, we can us Pythagorean Theorem to form a quadratic equation where we can find the radius of each of the circles.
The long leg is (11  2R).
The short leg is (17/2  2R).
The hypotenuse is 2R.
Our equation, then, is (11  2R)^{2} + (17/2  2R)^{2} = (2R)^{2}.
Simplified and rewritten into a quadratic equation, we have:
4R^{2}  78R + 773/4 = 0.
From the quadratic we get two roots:
(39/4 + SQRT(187)/2) =~ 16.5874, and
(39/4  SQRT(187)/2) =~ 2.9126
The first requires the radius of the circle to be larger than the sheet, so we can dismiss it. Thus, the radius of each circle is the second root: (39/4  SQRT(187)/2).
The area of the two circles that cover the 8 1/2 x 11 sheet of paper is, then (2*Pi*(39/4  SQRT(187)/2))*2
=~ 36.60085 sq. units.
For three circles, two of the circles would need be placed into the corners of one side where they are both tangent to the edge along the 11 inch length, with the third circle tangent to the midpoint of the opposite edge.
Again, using Pythagorean's theorem and the quadratic formula, the radii of the three circles may be found.
The long leg is (17/2  2R).
The short leg is (11/2  R).
Again, the hypotenuse is 2R.
Using Pythagorean's theorem, we have:
(17/2  2R)^{2} + (11/2  R)^{2} = (2R)^{2}
Simplifying into a quadratic equation:
R^{2}  45R + 205/2 = 0
The roots are then:
(45/2 + SQRT(1615)/2) and (45/2  SQRT(1615)/2).
Again, the first root is too large, thus we use the second:
R = (45/2  SQRT(1615)/2) =~ 2.4065
The area of the three circles that cover the 8 1/2 inch x 11 inch sheet of paper is, then (2*Pi*((45/2  SQRT(1615)/2)))*3
=~ 45.3609 sq. inches.

Posted by Dej Mar
on 20080307 07:34:42 