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 = 90o
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 point--call it D--along the hypotenuse AC, so we need to find the combined areas of two back-to-back 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 2014-02-06 16:36:27