Problems & Puzzles: Problems

Problem 65. Prime convex polygons

Dmitry Kamenetsky sent the following nice problem as a follow-up to Problem 64:

Q1. Can you find the smallest (in area) convex pentagon, such that all its sides, diagonals and area are distinct integers? Can you find higher order polygons with such properties?

BTW, Dmitry has submitted a solution for Q1, but he is not totally sure that this is the solution with the smallest area. I will show it next week.

Q2. Can you find a convex quadrilateral, such that all its sides and diagonals are distinct primes? Can you find higher order polygons with such properties?

Contribution came from Emmanuel Vantieghem. I also promissed to post the pentagon solution sent by Dmitry some weeks ago when he suggested this follow-up problem.

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Dmitry  wrote on Feb 23, 2016:

I found the smallest convex pentagon with distinct integer sides, diagonals and area (See drawing attached). Area 9870. I didn't think that these even existed. I have searched for all sides <= 200. Now I will start looking for a hexagon. .

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Emmanuel wrote on March 15, 2016:

Q1. This is my champion with area  9870 : the pentagon  ABCDE  with
A(0,0), B(855/13, -700/13), C(195,0), D(330/13, 560/13), E(-105/13, 252/13).
and with lengths AB = 85, BC = 140, CD = 175, DE = 41, EA = 21, AC = 195, AD = 50, BD = 105, BE = 104, CE = 204.

Q2. I used an identity that I derived from the Cayley-Menger determinant to prove the following statement :

It is impossible to have the six quantities  a, b, c, d, p, q  odd.

Proof :
If you allow me to write  a^2  as  X, b^2  as  Y, c^2  as  Z, d^2  as  T, p^2  as  P, q^2  as  Q,
then the 'theorem' of Cayley-Menger states that  a, b, c, d, p, q stem from a quadrilateral iff  :
XZ(X+Z)+YT(Y+T) - YZT - XZT - XYT - XYT+(Y-Z)(X-T)P - (X-Y)(Z-T)Q - (X+Y+Z+T-P-Q)PQ = 0.  (*)
If all of the variables  a,  ..., q  are odd, then  X, ..., Q  are all congruent to 1 mod  4.The left side of (*) is then -2 mod 4, which is not zero, QED.

It is possible to have  5  lengths odd and one even :
take  (a,b,c,d,p,q) = (25,13,15,21,27,24)

Since there is no triangle with one side  2  and other sides odd, we can conclude that six primes never can happen.

Thus, if you take more than  5  points, there are subquadrilaterals whose mutual differences cannot all be odd !

As a consequence, the lengths
a = 1051, b = 1303, c = 1181, d = 593, p = 1381, q = 463
given by Dimitri (in problem64) cannot correspond to a quadrilateral since they do not satisfy the relation (*).

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On Feb 14, 2017 Dmitry Kamenetsky wrote:

Looks like Luca Petrone found 3 more solutions to Problem 65 here:
They are:

For n=6, we have the 6-gon with sides 47, 663, 264, 169, 105, 1020, diagonals 700, 884, 975, 855, 952, 1001, 425, 520, 272, and area 196950.

For n=7, we have the 7-gon with sides 235, 1320, 1360, 2340, 612, 525, 5100, diagonals 1547, 2805, 4557, 4875, 2600, 4420, 4760, 5005, 3500, 3952, 4301, 2880, 3315, 1131, and area 5695998.

For n=8, we have the 8-gon with sides 1547, 612, 525, 5355, 235, 4275, 845, 1360, diagonals 2125, 2600, 5365, 5304, 2163, 1131, 5520, 5525, 3500, 2805, 5460, 5491, 3952, 3315, 5408, 4301, 3720, 4420, 4875, 4760, and area 12473070.

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