I had to read both solutions and stare for a long time before I realized:
Take the lozenge shape and cut it in half along the long axis. Put each half in the chord of the smaller circles. Now the remaining shape is a right triangle with short sides of length 2r. πr² - 2r² = (π-2)r²
Fun post and neat problem. Great that your kids are doing so well in math. I did manage to see the overlap dark section is half the total dark area, but then got stuck trying to find its area in a geometric way. Really appreciated Chris's approach once I read it.
My first thought was to write out the equations of the semicircles and then solve integrals for the areas. My second thought was to draw it carefully on cardstock, weigh it, then cut away the light or dark segments and re-weigh it.
I'm still struggling a bit with the "semicircle overlap is half the total dark area" logic.
I had to read both solutions and stare for a long time before I realized:
Take the lozenge shape and cut it in half along the long axis. Put each half in the chord of the smaller circles. Now the remaining shape is a right triangle with short sides of length 2r. πr² - 2r² = (π-2)r²
I explained that method to SteelyKid, who approves.
[sorry, it would be clearer if that 2r² were written as (2r)²/2]
Fun post and neat problem. Great that your kids are doing so well in math. I did manage to see the overlap dark section is half the total dark area, but then got stuck trying to find its area in a geometric way. Really appreciated Chris's approach once I read it.
My first thought was to write out the equations of the semicircles and then solve integrals for the areas. My second thought was to draw it carefully on cardstock, weigh it, then cut away the light or dark segments and re-weigh it.
I'm still struggling a bit with the "semicircle overlap is half the total dark area" logic.
I had to do a double take to follow the "semicircle overlap is half the total dark area" logic, but that is elegant.