Dividing a square with arcs

Feb 2020
2
0
Pittsburgh, PA
over at


We find a problem involving dividing a square into two areas and comparing the areas. A geometrical solution there is convincing (the blue and white regions have equal areas), but when I attack the same problem with a more cumbersome algebraic approach, I keep getting a different answer. Thus (see the website) :

The white region is the two semicircles minus the lens that has been counted twice. Call the side of the square s and the lens L. We have
W = 2 (½) pi (s/2)^2 - L
Since the whole square can be seen as four of those semicircles minus four of those lenses, we have
s^2 = 4 ( ½ ) pi (s/2)^2 - 4 L
Solving this last for L, we have
L = (pi/8 - 1/4) s ^2
Plugging this into the first equation, we have
W = (pi/8+ 1/4) s^2
The Blue region must be
B = S^2 - W = s^2 - (pi/8 + 1/4) s ^2 = (3/4 - pi/8)s^2
This is not equal to W. What is my error?
J
 

skipjack

Forum Staff
Dec 2006
21,482
2,472
Your first equation should be W = 2 (½) pi (s/2)^2 - 2 L.
 
Oct 2013
717
91
New York, USA
Since L was subtracted instead of 2L, the equation for W is too big by L. Taking W = (pi/8+ 1/4) s^2 and decreasing by L = (pi/8 - 1/4) s^2, makes W = (pi/8 + 1/4)s^2 - (pi/8 - 1/4)s^2. Then the pi/8 cancels out, and it becomes (1/4)s^2 - (-1/4)s^2 = (1/4)s^2 + (1/4)s^2 = (1/2)s^2.
 
Feb 2020
2
0
Pittsburgh, PA
to skipjack and EvanJ: Thanks so much ! My error seems obvious now that you've pointed it out, but I might never have seen it without your help.