Joining every vertex of a convex pentagon to every other vertex forms three nested figures:
the original outer pentagon
an inner pentagon
a five-pointed star between them
Suppose the perimeters of these three figures are all primes; show that the three primes’ sum is at least 20.
No they are not!
The problem description requires three perimeters to be prime and their total to be at least equal to 20.
One's immediate thought is might embrace an array where all defined shapes are regular, a path that I embraced.
Initially I let the inner pentagon to have sides of 1 unit which provides a prime perimeter of 5 units. Needless to say that pentagon may have a perimeter of either 2 or 3.
The following table does offer the length of segments of the respective objects coupled with a potentially likely prime perimeter.
Note that my decimals are very imprecise (and the table is awkward).
Object Ln Per(5)* Per(2)* Per(3)*
Pent(sm) 1 5 5 | 2 2 | 3 3
Star 1.618 16.18 17 | 6.472 7 | 9.708 11
Pent(bg) 2.618 13.09 13 | 5.236 5 | 7.854 7
Sum 34.27 35 |13.705 14 |20.562 21
* Based on the values of the inner pentagon.
The above brings the following to notice:
1. If 3 prime perimeters are possible then the interrelationship of the shapes cannot be regular.
2. While the floor value is set at 20, might it not in fact be lower?
3. The internal pentagon may/may not have equal side lengths but still comply with the prime perimeter requirements.
I assume I have not misread the problem and hopefully my data is just and of further assistance.
Posted by brianjn
on 2012-12-07 03:10:32