Manasataramgini has a problem from plane geometry embedded in a short story, which I extract here:
(I think such time requirements are puerile.) I came up with a solution that is different from the one posted by Manasataramgini, and I've been wondering if I got lucky or whether there is a method here. I haven't been able to decide, so this post.)
Manasataramgini's method is to add up the square (blue dotted lines) and the four segments of circle created by chords of π/6.
My method was to note that I know A (congruent to a segment of a circle of 2π/3) and hence B (quarter circle - A) and hence C (= 1 - 2*B - A). I know D (twice a segment of a circle of π/2), and 2*D + 4*C counts the area of interest twice and the rest of the square once, so just subtract the unit area of square from it. After that it is just arithmetic.
Given a unit square, if a point lies on the same plane as the square at not more than a unit distance simultaneously from each of the four vertices of the square then what will be: 1) the minimum distance it can reach from any side of the square; 2) what fraction of the area of the square can the point be located in.Since the short story includes the requirement of solving it in 7 minutes, I was asked to try my hand at it.
(I think such time requirements are puerile.) I came up with a solution that is different from the one posted by Manasataramgini, and I've been wondering if I got lucky or whether there is a method here. I haven't been able to decide, so this post.)
My method was to note that I know A (congruent to a segment of a circle of 2π/3) and hence B (quarter circle - A) and hence C (= 1 - 2*B - A). I know D (twice a segment of a circle of π/2), and 2*D + 4*C counts the area of interest twice and the rest of the square once, so just subtract the unit area of square from it. After that it is just arithmetic.
