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 2014-08-30 02:43:17