A piece of paper has the precise shape of a triangle ABC with two side lengths AB and AC being respectively 36 and 72 with ∠ABC = 90^{o}
Find the area of the set of points P inside ΔABC such that if creases are made by folding (and then unfolding) each of A, B, C to P, then the creases do not overlap.
By playing with Geometers' Sketchpad we can see that point P must lie within a semicircle with diameter BC and also within a semicircle with diameter AB. It also must lie within the semicircle with diameter AC, but, this, being the hypotenuse, encompasses the whole triangle.
The two important semicircles meet at B and at a pointcall it Dalong the hypotenuse AC, so we need to find the combined areas of two backtoback segments of circles with endpoints B and D.
The segment with arc centered at the midpoint of BC has area pi*(18*sqrt(3))^2 / 6  27*9*sqrt(3).
The segment with arc centered at the midpoint of AB has area pi*18^2 / 3  9*9*sqrt(3).
That brings the total area to
pi*18^2*5/6  324*sqrt(3)
= 270*pi  324*sqrt(3) ~= 287.045554816926

Posted by Charlie
on 20140206 16:36:27 