Pearl
THIS CHANGES EVERYTHING
You're the only 50 year old I know that would take off his shirt at a friend's house party. Much love.
just saw this, and saw the photo recently, surprisingly not disappointed
You're the only 50 year old I know that would take off his shirt at a friend's house party. Much love.
But that pi is messing up the optic....Actually today is Pi Day....3/14 or 3.14🙂
It's much more than that - it's the once-in-our-lifetime "Rounded Pi Day". Pi to the first five decimal places is 3.14159. If you round that, you get 3.1416, which is today's date. So in honor of this special Rounded Pi Day, here is a problem to solve: prove that the red portion has the same area as the blue portion in the diagram shown here ...
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ok - here is what i'm thinking -
the only thing you need to know is the area of a circle = πr^2 (π times the radius times the radius)
the are of the slice of "pie" is πr^2/4 (1/4 of the whole)
the area of the red is the whole slice removing the area of the two inscribed half circles then add the blue back in (cause it was taken out twice) so two half circles make a whole, but the radius is half that of the big slice π(r/2)^2
area of red = πr^2/4 - π(r/2)^2 + area of blue
πr^2/4 - π(r/2)^2 = 0
area of red = area of blue
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on the same note, last year
3/14/15 9:26:53.589.....
Yeah, that's probably quicker than how I did it. I looked at the quarter circle as having area πr^2/4. The two smaller half circles have a radius equal to r/2, which means each of them have an area of 1/2 πr^2/4, or together they have πr^2/4. So the area of the two semi circles is equal to the area of the larger quarter circle. That means that the overlap between them (blue) must be equal to the portion not included (red.) Putting that in equation form collapses to exactly yours, just cosmetically different:
πr^2/4 = (1/2*π(r/2)^2 + 1/2*π(r/2)^2 - blue) + red
or
0 = -blue + red
or
blue = red.