A letter F is composed of 6 unit squares and two rectangles of unit width as in the figure:

Find the lengths of the two rectangles such that the center of gravity is at the center of the middle square.

(In reply to maybe by Art M)

Letting y be the height of the vertical rectangle and x be the width of the horizontal rectangle, and setting the origin at the center of the top left square:

Horizontally:

(0*(4+y) + 1*2 + (3/2 + x/2)*x) / (4 + y + 2 + x) = 1

2 + 3*x/2 + x^2 / 2 = 6 + x + y

 

Vertically:

(0*(2+x) + 1*1 + 2*2 + 3*1 + (7/2 + y/2)*y) / (6+x+y) = 2

8 + 7*y/2 + y^2 / 2 = 2*(6 + x + y)

 

Setting aside Excel cells for x and y, and formula cells for 2 + 3*x/2 + x^2 / 2 and for 6 + x + y; also for 8 + 7*y/2 + y^2 / 2 and for 2*(6 + x + y); as well as the ratio within each pair, and using Solver to solve the ratios as 1, each, results in a value of 3.372281323 for each of x and y, which agrees numerically with Art M's answer.

The question is, how does one analytically solve the two simultaneous quadratics? Does it necessarily involve a quartic? In other words, I'd like to see the solution rather than just the answer.

Posted by Charlie on 2007-05-17 10:16:17