Problems & Puzzles: Puzzles

Puzzle 1155 Equilateral triangle with an interior point such that... G. L. Honaker, Jr. sent the following nice puzzle It is well-known that if an equilateral triangle has sides of length 112, then it contains an interior point at integer distances 57, 65, and 73 from its vertices. (*) Q: Is it possible to construct an equilateral triangle of integer side-length x, such that an integer point has three prime distances from its vertices? Send x and the three primes distances and a graph representing them?   By my side I have found the work of  Ignacio Larrosa Cañestro over the same issue, that could be of some interest for some of you in this puzzle. 1st Link 2nd Link 3rd Link In this last pdf reference you can find an x value, 3192, that has two interior points which have two sets of two prime integer distances, each ; 1423 1885 2347 and 523 2707 3145 So, perhaps a solution with three prime distances is possible... __________________ (*) Wells, D. The Penguin Dictionary of Curious and Interesting Numbers London: Penguin Group. (1987), page 119. BTW "This is the smallest possible side length of an equilateral triangle that contains a point at integer distances from the vertices" See Ref 4

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During the week 23-30 Dec, 2023 contributions came from Giorgos Kalgeropoulos, Emmanuel Vantiegehm

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First of all I would like to point out that in the given example the point with distances {1423, 1885, 2347} is interior BUT the point with distances {523, 2707, 3145} is exterior.
 

I could not find a point with 3 prime distances but I found 3 more interior points with 2 prime distances:

 

     a                b              c             integer-side x

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  28753       71257      78792              97955
 

  10867     101503    102085             111048
 

135391     174859    233016             303875

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Emmanuel wrote:

 I examined, for every  x < 64000  if there is an interior point with the smallest two distances prime.
 I found none !
 (The suggested solution  {523, 2707, 3145, 3192} (with primes 523, 2707) concerns an exterior point ;

   from the  96  solutions given in 3rdLink, only 67 give interior points)

Here es a  list all the 29 solutions that give exterior points?

(I just give  {a, b, c, x} :

 

{43, 248, 285, 287}
{152, 365, 497, 507}
{217, 425, 608, 633}
{323, 392, 645, 713}
{57, 673, 715, 728}
{245, 632, 817, 873}
{285, 1067, 1288, 1343}
{551, 1055, 1519, 1584}
{553, 1105, 1488, 1657}
{824, 915, 1591, 1729}
{811, 1029, 1744, 1805}
{559, 1415, 1896, 1939}
{392, 1653, 1987, 2015}
{721, 1480, 2131, 2139}
{715, 1737, 2353, 2408}
{855, 2197, 2863, 3032}
{1267, 1857, 2840, 3113}
{523, 2707, 3145, 3192}
{936, 2431, 3059, 3365}
{731, 2755, 3399, 3416}
{1083, 2408, 3233, 3475}
{1769, 1899, 3379, 3640}
{1501, 2216, 3465, 3679}
{1568, 2197, 3515, 3723}
{903, 3035, 3608, 3937}
{2131, 2261, 4120, 4329}
{1987, 2728, 4305, 4693}
{2291, 2735, 4776, 4921}
{1256, 3741, 4805, 4921}

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Related to the Giorgos solution ( 28753, 71257, 78792; 97955) G.L. Honaker, Jr. notices, on Dec 30, that it is an "almost" solution because 78921... is prime!

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On Jan 18, 2025 Alessandro Casini wrote:

I don't have any solution yet, but I would like to put down on paper some properties that they must possess. Let (p1, p2, p3) be a solution with p1 < p2 < p3.

• p3 <= p1 + p2, since the inner radicand in the 3rd link's formula must be no negative. The equal sign is only possible if p1 = 2.
• p = 2 can't be part of a solution, so the three primes are all odd. Indeed, if p1 = 2, it follows from the previous point that the only possible triplet is (2, p, p+2), with p a twin prime. In that case, the formula gives d^2 = p^2 + 2p + 4, but the latter is always equal to 3 mod 4, so it can't be a perfect square.
• p1 = 3 is impossible too. In fact, as previously, we have (3, p, p+2) and consequentially the inner radicand (which I will call the main discriminant) is equal to 3*5*(2p - 1)*(2p + 5). To be able to be a perfect square, it must contain a further factor 5, therefore one between 2p + 5 and 2p - 1 is a multiple of 5. In the first case, p must be 5 and (3, 5, 7) is actually a solution, though exterior. In the latter, 5 | 2p - 1 = 2*(p + 2) - 5,  so p + 2 must be 5, since it's prime, but that's impossible.
  This reasoning can also be applied to other prime numbers (as p1 = 5), but it doesn't work for everyone. Let me point out that this only takes in account the innermost square root.
• From the implicit relation between the three distances and the triangle side-length, p1^2 + p2^2 +p3^2 + d^2 must be a multiple of 3. Since p^2 = 1 mod 3 (p = 3 is excluded), 3 divides the side-length d.
• I noticed that if a triplet of primes (p1, p2, p3) makes the main discriminant a perfect square, then the differences p3 - p2 and p2 - p1 are divisible by 3 and in particular p1 = p2 = p3 = 1 mod 3, excluding for some reason the case 2 mod 3. Furthermore, this no longer applies if the numbers are simply odd ones not multiples of 3. I have no idea or proof of this fact, and for the moment I have no counterexamples, so I can only make my own conjecture: if the expression 3*(a+b+c)*(a+b-c)*(a-b+c)*(-a+b+c) is a perfect integer square, with (a, b, c) primes, then a = b = c = 1 mod 3. Proving this would undoubtedly narrow down the cases to be tested, only taking in account the main discriminant.

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