Place two pennies on a table each touching a third, but not each other. The centers of the three form a certain angle.
For what angle will the area of the convex hull of this shape be maximized?
Let A, B, and C be the centers of the three
pennies with the penny withe center A touching
the other two. Let 2x be the measure of angle
BAC and a the common radius of the pennies.
The area of the convex hull is
A(x) = Area(triangle ABC) +
a*Perimeter(triangle ABC) +
Pi*a^2
= 2*a^2*sin(2x) + a*[4*a + 4*a*sin(x)] +
Pi*a^2
A'(x) = 4*a^2*cos(2x) + 4*a^2*cos(x)
A"(x) = 8*a^2*sin(2x)  4*a^2*sin(x)
A'(x) = 0 implies cos(2x) + cos(x) = 0
implies 2*cos(2x)^2 + cos(x)  1 = 0
implies cos(x) = 1/2 or 1
implies 2x = 120 or 360 degrees
only 2x = 120 degrees makes sense
A"(60) = 8*a^2*sin(120)  4*a^2*sin(60)
is less than 0
Therefore, the A(x) is maximized when
angle BAC = 120 degrees.
QED

Posted by Bractals
on 20140830 02:43:17 