Belated Pi Day Puzzle

As soon as you write out the areas, the answer becomes obvious. Still, a clever puzzle.

What can I say

http://www.youtube.com/watch?v=D1n5CQe1krI

Something weird happened when I drew this in solid works.

I used 24" for large circle and 12" for the smaller ones. Saved it and then saved a copy then isolated the red and blue areas then extruded them .0625" to get some volume.

The volume were identical at 1.28" cubic inches BUT the red one was 43.45 square inches and the blue one was 42.28 square inches in area????? The masses also matched....weird!

How does that work?

I assume it is a programming glitch with Solid Works the way it measures the ever decreasing small area near the end of the red segments.

I think you're right, those pointy red cusps are likely problematic to quantize.

Round off error somewhere, I suspect. Is it actually pixel resolution issue? What do you see near the edges when you zoom way in?

I just had a cursory look at the link you gave, but on steps 19 and 20 I was surprised when the areas given did not have any relationship to the radii or diameters of the circle. Something must have gone wrong there.

The're only dealing with the area of the intersection, the rest is known....

More or less what I was going to say but CR4 is not user friendly for typing math formulas..

What in hell is wrong with ya, lad? Today is St. Paddy's Day, and that shamrock is na green, and the leaves are shaped funny, and it has na stem.

ACK! That image caused a flashback to my Army days, working at Southern European Broadcasting, part of the Armed Forces Radio and Television Service.

At the time, the Army portion of AFRTS didn't have their own unit patch, so we wore the 'General Army' Patch (I forget it's real name) Which was a blue five-lointed star inside a white clover on a red circle.

This was close enough to trigger the memory. I think I have a better understanding of those poor souls suffering from PTSD; you never know what image, or sound, or smell will be a trigger, throwing your mind bank into the hell you're trying to escape. (Fortunately, my Army experiences are all of the "Crappy job, idiot bosses" hell, not the "Saving Private Ryan" hell.)

I looked at it this way: A_r = A₁ - 2A₂ + A_b.

Then put in the formulas for A₁ and A_2, and they nicely cancelled out, leaving: A_r = A_b.

It looks like you guys figured it out, but in case you're curious here is the official solution...

https://www.theguardian.com/science/2016/mar/14/did-you-solve-it-the-pi-day-puzzle-that-will-spin-you-in-circles

GA from me!!!

Very well said too. I figured it out using what geometry I could pull out of my brain and came to a similar conclusion, but was not able to verbalize it at all, let alone as well as you did.

Thanks!

Yes - good explanation and a great aid to visualising the problem.

But a little bit of maths creeps in to explain why these particular areas are equal.

My general proof applies to the particular case, IMHO.

The only maths I can see is that areas are additive. Seems obvious now, but I remember when I was (very) young, thinking that if a square slice of bread is cut by a diagonal, the resulting area is greater than if cut into 2 rectangles.

Do that and then go weigh the bread on a set of scales and see what you get. I get a PB & J.

I don't need to, I was about 3 at the time, I know the answer now .

BTW what's a PB & J? Something bread and jam? (or jelly as you call it over there)

Peanut butter and jelly on toast. If you live in the hotter regions, then peanut magma and helly on charcoal.

I haven't looked at other posts yet (or the solution if it's out), but my take is -

Let the radius of the semicircles = R, and the quadrant 2R. Then area of quadrant = pi*R², total area of both semicircles = 2*pi*R²/2 = pi*R². If the semicircles are imagined distorted into 2 semi-quadrants, non-overlapping, and then distorted back to semicircles, any area of overlap must be subtracted from the quadrant area.