A midsquare is an orthodiagonal and equidiagonal quadrilateral.


Are there naturals $a,b,c,d,e$ such that the quadrilateral with sides $a,b,c,d$ and diagonal $e$ is a midsquare ?

$e^2$ is a solution for $$2e^4-2(a^2+c^2)e^2+(a^2-b^2)^2+(b^2-c^2)^2=0$$ Let $D=4b^2d^2-(a^2-c^2)^2$
Then $e^2=\dfrac{a^2+c^2+\sqrt{D}}2$

import math
for a in range (p,q): # a = AB 
    for b in range(a,q): # b = BC
        for c in range (p,q): # c = CD
            if a*a+c*c > b*b:
               d = (a*a+c*c-b*b)**.5 # d = DA
               if int(d)==d:
                  D = 4*b*b*d*d-(a*a-c*c)*(a*a-c*c) 
                  if D>=0:
                     if int(D**.5)==D**.5:
                        e = (.5*(a*a+c*c+D**.5))**.5
                        if int(e)==e: 

Some results with $\sqrt{D}\in\mathbb{N}$
$$\begin{array}{|c|c|c|c|}\hline \text{perimeter-plus } a+b+c+d+2e & \sqrt{D} & e^2&e\\ \hline 658.488\approx35+101+149+115+2\times 129.244& 9982 & 16704&24\sqrt{29}\\ \hline 1183.63\approx41+181+289+229+2\times 221.815& 13202 & 49202& \sqrt{2\times73\times337}\\ \hline 3738.168\approx221+481+899+791+2\times673.084& 49042 & 453042& 3\sqrt{2\times25169}\\ \hline 6105.496\approx299+925+1405+1099+2\times1188.748&762818 & 1413122& \sqrt{2\times706561}\\ \hline 7611.367\approx85+1339+1939+1405+2\times1421.684&275422& 2021184& 264\sqrt{29}\\ \hline 11665.227\approx845+1513+2525+2191+2\times2295.614& 3450034 & 5269842& 3\sqrt{2\times137\times2137}\\ \hline 11992.528\approx1015+1241+2759+2665+2\times2156.264& 656642 & 4649474& \sqrt{2\times661\times3517}\\ \hline 17069.827\approx709+2759+3995+2975+2\times3315.913 &5527858& 10995282 & 3\sqrt{2\times610849}\\ \hline 21602.357\approx1681+2381+4889+4589+2\times4031.178&5772718 &16250400& 60\sqrt{2\times37\times61}\\ \hline 37947.648\approx1321+6329+9049+6601+2\times 7323.824 &23647358 & 53638400&1360\sqrt{29}\\ \hline 44392.996\approx2509+7081+9901+7361+2\times8770.498&49518382&76921632&12\sqrt{2\times89\times3001}\\ \hline \end{array}$$

Two special cases

  1. The isosceles trapezoid midsquare $a,b,c,b,e,e$.
    It's easy to show that with $a,b,c,e$ integers, it doesn't exist: a nice geometric expression for $e$ as a function of $a$ and $c$.
  2. The kite midsquare $a,a,c,c,e,e,~a<c$.
    I don't know whether it exists or not.
    For it, I can't even find an integer $\sqrt{D}=\sqrt{4a^2c^2-(a^2-c^2)^2}$.
    20/11/23 Let $g$ be the distance between the midpoints of the diagonals (see Euler's quadrilateral theorem), we can show that $$a^2+2eg=c^2$$ So $g$ is a rational number, we can look for $e$ and $g$ naturals such that $$(e-g)^2+g^2=2a^2$$ $$(e+g)^2+g^2=2c^2$$ Example of an "almost solution" $$(1192-1043)^2+1043^2=1~110~050=2\times745^2~~~~~~~~~~$$ $$(1192+1043)^2+1043^2=6~083~074=2\times1744^2~\boxed{+2}$$ I still don't find any solution.
  • 1
    $\begingroup$ What is a pseudo-square? $\quad$ How would you attempt to prove this statement? $\endgroup$
    – Calvin Lin
    Nov 14 at 15:46
  • $\begingroup$ I wrote some python programs and didn't find any natural midsquare. $\endgroup$ Nov 14 at 15:53
  • $\begingroup$ Can you post these into the statement itself, not just as comments. MSE asks the poster to show their work to prevent closing of the statement, so this would be helpful. $\endgroup$
    – Calvin Lin
    Nov 14 at 16:11
  • 1
    $\begingroup$ Let O be the intersection of the diagonals. We could name $x=OA=\dfrac{a^2-b^2+e^2}{2e}$, $y=OB$, $z=OC$, $t=OD$, yes they would be rational, if $ABCD$ integer exists. $\endgroup$ Nov 14 at 20:28


You must log in to answer this question.