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.
Logging you in...
nainankovoor 17p · 352 weeks ago
My way was to divide the curvilinear 'square' into four congruent figures, each a 'right triangle' with an arc-hypotenuse.
Each such triangle in turn is included in a sector of a unit circle with angle π/6. The circular sector has area π/12; removing the 'right triangle' on its curved circumference leaves a figure that may be split into two congruent acute-angled triangles. The combined area of these triangles is that of a parallelogram with base (sin π/3 - 1/2) and altitude 1/2.
Thus the area of the curvilinear 'square' is 4 (π/12 - (sin π/3 - 1/2)/2), which works out to π/3 - 2 sin π/3 plus 1.
All in all, well within the capabilities of an Indian ninth-grader (who might even solve it in seven minutes if he keeps his cool)!
nainankovoor 17p · 352 weeks ago
There are echoes of the Inclusion-Exclusion Principle in your approach which would be more pronounced in the 3-D case. My approach seems to generalise easily, to 3-D at least.