The New Sudoku Players' Forum

[StrmCkr]

the area of a partial lune is equal to the area of a kite.

see attached picture:

the out side square is set at sqrt(2)/2 " and the internal squares are set at 1/2" and 90 degree offset

the internal inscribed circle has a diameter = sqrt(2/2)"

all arcs have a radius of 1/2" from the 8 square points.

4 internal quadrant squares are halfed by right angle triangles for a mid line and then the intersecting 90 degree off set line from the other square is mirrored {repeat for all 4 quadrents}

then form the internal middle square from those segments.

from there it is a easy process of identifying the identical parts and manipulating them to prove the kite = the partial lune.

anyone have a better way to write this out mathematically instead of a "picto" gram solution.

[blue]

Using a different picture (below) the equality would follow, if it could be shown that the darkened regions have the same areas as the corresponding green regions.

For the "partial lune" side, it should be obvious (given that the two arcs have the same radius).
For the kite side, it would follow after showing that the distance from A to B is the same as the distance from B to C ... d(A,B) = d(B,C).

The 2nd part is can be verified, based on the point coordinates shown.
The area(s) can then be "read off" as:

Area = d(A,D) * d(D,C) = (sqrt(2) - 1) / 16

kpl.jpg (98.77 KiB) Viewed 1213 times

[StrmCkr]

thanks blue for confirming what i thought.

basically the same thing I did with out adding pieces, mind you adding the parts you did is something I originally did as it left the units in a perfect square state rather then fraction.