Problems & Puzzles:
Problems
Problem 65. Prime convex
polygons
Dmitry Kamenetsky sent the following nice problem as a
followup 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 followup problem.
***
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.
.
***
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.
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 CayleyMenger
states that a, b, c, d, p, q stem from a quadrilateral iff :
XZ(X+Z)+YT(Y+T)  YZT  XZT  XYT  XYT+(YZ)(XT)P
 (XY)(ZT)Q  (X+Y+Z+TPQ)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
(*).
***
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 6gon 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 7gon 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 8gon 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.
***
