Quadrilateral with sides, diagonals and area that are distinct integers

Question

Can you find a convex quadrilateral such that all its sides, diagonals and area are distinct integers? Note that a polygon is convex if all its internal angles are smaller than 180 degrees.

Good luck!

Answer 1

Assume convex quadrilateral has its sides $a,b,c,d$ where $a<b<c<d$, 2 diagonals $m,n$ where $m<n$, and area $Z$.
For convenience, I also assume that $(a,b)$, $(b,c)$, $(c,d)$ and $(d,a)$ are adjacent sides.

Then consider the following properties:

• Triangle Inequality:
  $a+b>m$, $a+m>b$ and $a+m>b$ stands. Also for $(c,d,m)$, $(a,d,n)$ and $(b,c,n)$.

• Heron Theorem:
  For any triangle $(p,q,r)$, the area $T$ could be calculated as: $T = \sqrt{s(s-p)(s-q)(s-r)}$ where $s=\frac{(p+q+r)}{2}$.
  Also could be represented as: $T = \frac{1}{4}\sqrt{(p+q+r)(-p+q+r)(p-q+r)(p+q-r)}$.

Here also assume that the area in any triangle $(a,b,m)$, $(c,d,m)$, $(a,d,n)$ and $(b,c,n)$ should be integer.

Due to $T$ is integer, the expression in the square root should has at least $2^4$ by prime factor decomposition. Also assume that each term $(p+q+r),(-p+q+r),(p-q+r),(p+q-r)$ in square root should be all even(contain at least $2^1$ for each), then $(p,q,r)$ will contains only 3 evens or 2 odds & 1 even.

Then start brute-force searching, and checked with the prime factor decomposition to ensure the square root value is rational and Triangle Inequality also holds for all $(a,b,m)$, $(c,d,m)$, $(a,d,n)$ and $(b,c,n)$, finally found one solution:

$(a,b,c,d) = (10,17,28,35)$, $(m,n) = (21,39)$, $Z=T_{(a,b,m)}+T_{(c,d,n)}=378$

I think there may exist a better derivation for this math question, too many assumption for my answer, and the final brute-force is not a pretty solution :P

Answer 2

I ran a computer search, which confirms Conifers' answer.

The program simply tries all values for the sides and the diagonals. I do not assume the sides are in any particular order, and the areas of the triangle on either side of a diagonal need not themselves be integer. Even so, Conifer's answer is the smallest, in the sense that its longest side is minimal. Here are the first few that my program produced:

Sides, diagonals: area
(10,17,28,35) 39,21: 378
(5,29,27,51) 52,30: 396
(17,39,33,55) 60,44: 1056
(33,39,52,56) 65,60: 1938
(16,25,33,60) 52,39: 714
(25,34,33,60) 65,39: 1014
(25,39,52,60) 65,56: 1764
(25,39,60,52) 63,56: 1764

I think that first answer (i.e. Conifers') is also the smallest in area, though the second one comes close.

Here is my program code (C#).

  using System;

  namespace test
  {
     class PseIntQuad {

        static void Main() {
           for (long d = 1; d<=55; d++){
              for (long c = 1; c < d; c++){
                 for (long b = 1; b < c; b++){
                    for (long a = 1; a < b; a++){
                       // up to rotation/reflection there are three orderings of the edges
                       SearchDiag(a, b, c, d);
                       SearchDiag(a, b, d, c);
                       SearchDiag(a, c, b, d);
                    }
                 }
              }
           }
        }

        private static void SearchDiag(long a, long b, long c, long d)
        {
           for (long n = 1; n < a + b && n < c + d; n++){   // n straddles ab, cd
              if (a >= b + n || b >= a + n || c >= d + n || d >= c + n) continue;
              if (n == a || n == b || n == c || n == d) continue;
              long area1 = CalcArea(a, b, n);
              if (area1 <= 0) continue;
              long area2 = CalcArea(c, d, n);
              if (area2 <= 0) continue;
              long area = area1 + area2;
              if (area % 4 != 0) continue;
              for (long m = 1; m < b + c && m < a + d; m++){
                 if (a >= d + m || b >= c + m || c >= b + m || d >= a + m) continue;
                 if (m == a || m == b || m == c || m == d || m == n) continue;
                 long area3 = CalcArea(b, c, m);
                 if (area3 <= 0) continue;
                 long area4 = CalcArea(a, d, m);
                 if (area4 <= 0) continue;
                 if (area3 + area4 == area){
                    Console.WriteLine("({0},{1},{2},{3})  {4},{5}: {6}", a, b, c, d, m, n, area/4);
                 }
              }
           }
        }

        private static long CalcArea(long a, long b, long c)
        {
           long p = (a + b + c) * (a + b - c) * (a - b + c) * (-a + b + c);
           long area = (long)(Math.Sqrt(p)+.5);
           return area * area == p ? area : -1;
        }
     }
  }