Part 1:
Regular hexagon ABCDEF is surrounded by two parabolas. One contains A,B,C,D and the other contains D,E,F,A. What is the ratio between the area of the intersection of the parabola's interiors to the area of the hexagon?
Part 2:
Regular hexagon ABCDEF is surrounded by two parabolas. One contains A,B,C,D and the other contains C,D,E,F. What is the ratio between the area of the intersection of the parabola's interiors to the area of the hexagon?
Part 3:
Regular hexagon ABCDEF is surrounded by two parabolas. One contains A,B,C,D and the other contains B,C,D,E. What is the ratio between the area of the intersection of the parabola's interiors to the area of the hexagon?
The result for the problem of parabolas “hugging” the hexagon,
was that, while the overlap area is an irrational number,
and the polygon area is an irrational number,
together they ratio to a rational fraction.
Further, all three ratios, share one common denominator, 29.
So, I wondered if this was a generic feature for every regular
polygon. I.e: arrange parabolas 'hugging' a polygon and
their overlap area will in a predictable ratio to the polygon
area. Going further, maybe there is one denominator associated
with every polygon, and for that denominator one numerator
associated with every angle of rotation, and even a formula
and a sequence for these.
Note that the number of unique possible positions p (or “cases”)
for the second parabola goes up with the number of polygon sides n:
(n, p) goes like: (3,1), (4,2), (5,2), (6,3), (7,3), (8,4)...
So, I explored the analogous problem using different polygons:
The first two experiments were rewarding – A rational ratio was
again produced in each case. Even better, two ratios for
the square also followed the pattern of the hexagon, having
a common denominators with D=6. The ratios are 8/6 and 11/6.
But, the octagon did not follow the expectations at all. The results
are shown in the link.
So, why are there such some well-behaved ratios of incongruous
quantities in some polygons? Here are a few thoughts...
It may have to do with the symmetry of the angles
the parabolas and polygons have in common. The area of a polygon of
side a is A = (r p) / 2, where r is the incircle radius, and p
is perimeter. This becomes A = (na^2)/4) cot(pi/n),
or A=n/4 cot(pi/n) for a unit polygon.
More sides also make for wider parabolas, where the two fitted
points on either side of the apex are more nearly linear. At larger n
nearby parabolas overlap more, since they point in nearly
the same direction. So as n increases, both polygon area and
overlap area grow. But why they can make a rational ratio
is not clear to me.
While this puzzle was hard to solve, it just kept getting
more and more interesting.
More hugging...
