All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes
3 cents (Posted on 2014-08-29) Difficulty: 3 of 5
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?

No Solution Yet Submitted by Jer    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 3

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) +

       = 2*a^2*sin(2x) + a*[4*a + 4*a*sin(x)] +

  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.


  Posted by Bractals on 2014-08-30 02:43:17
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information