Something to lighten the mood

I think it's time for a mathematical joke! 😉
 

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see image - it was great fun, thanks! 😎
Nice approach.
You start with cube edge =1 then go at finding triangle edge.
I did the other way starting with triangle edge=1.
I used the trick of the isocele triangle seen in the previous puzzle ( the puzzle asking for the angle that happened to be 120°). Because of the 15° angle of the isocele triangle, it's complement happens to come in tan(75°).
So I found simple formulas but containing a not so nice tan() term.
Then matching your formulas with mines, I discovered that tan(75°) is 2 + sqrt(3).
Then I reworked my formulas that became quite simple, but written different than yours.

Now, new mind bender:
Prouve tan(75°) = 2 + sqrt(3)
 
About puzzle in post #1120
Two circles in a rectangle.

I think I found a way for the top right red area.
triangle area - ( half circle area - segment area )
Giving
5² - Pi.5²/2 + Segment area
Segment angle is: a = 180° - 2*atan(0.5)
Segment area = ( a* Pi/360 - sin(a)/2)*5²

This is too horrible, I give up.
 
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