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Anonymous 1123
I see Calabi's Triangle at [here](https://mathworld.wolfram.com/CalabisTriangle.html?fbclid=IwAR288tVqTHVQgGwFMScUCt7jKDRPz6PyPSPMnTxkSUwfsNLDytSVsh7hDvE), and this is not my code. I found at [here](https://www.facebook.com/groups/TikZandAsymptote/?epa=SEARCH_BOX)
\def\a{5} \def\b{3} \pgfmathsetmacro\h{sqrt(\b^2-\a^2/4)}
\path (0:0) coordinate (O)
(180:\a/2) coordinate (A)
(0:\a/2) coordinate (B)
(90:\h) coordinate (C)
(0:\h) coordinate (I)
(I)+(C) coordinate (H)
(intersection of A--H and B--C) coordinate (P)
(barycentric cs:P=1,I=1,H=-1)coordinate (PQ)
(intersection of P--PQ and A--B) coordinate (Q)
(barycentric cs:O=-2,Q=1) coordinate (R)
([rotate around={90:(R)}]Q) coordinate (S)
([rotate around={90:(B)}]C)coordinate(J)
(intersection of C--J and A--B) coordinate (M)
(barycentric cs:M=1,B=1,J=-1)coordinate (MN)
(intersection of M--MN and C--B) coordinate(N)
([rotate around={90:(M)}]N)coordinate(L)
([rotate around={-90:(A)}]C)coordinate(K)
(intersection of C--K and A--B) coordinate(Y)
(barycentric cs:Y=1,A=1,K=-1)coordinate (YZ)
(intersection of Y--YZ and C--A) coordinate (Z)
([rotate around={-90:(Y)}]Z)coordinate(X);
\draw[densely dotted,very thin]
(O)--(C)--(H)--(I) (A)--(H)
\draw (A)--(B)--(C)--cycle

![ScreenHunter 825.png](/image?hash=1a0cd77af1a7ce5e55ffd2936c8e7fbc1d180fa0b764abea243248d53c097eae)
How to draw Calabi's Triangle with another way?
Top Answer
I do not understand the code that you posted (and which comes from a website in which I do not understand the language unfortunately), but if you are willing to just take the edge length ratio from [your first link](https://mathworld.wolfram.com/CalabisTriangle.html?fbclid=IwAR288tVqTHVQgGwFMScUCt7jKDRPz6PyPSPMnTxkSUwfsNLDytSVsh7hDvE), then it is rather easy to draw the thing. The inscribed squares can be constructed by computing the intersections of some 45 degree lines with the opposite edges. 

\begin{tikzpicture}[line cap=round,line join=round,scale=5,
	declare function={q=1.55139;phi=acos(q/2);}]
  \draw[thick] (-1/2,0) coordinate (A) -- 
   (1/2,0) coordinate (B) -- (0,{sin(phi)/q}) coordinate (C) -- cycle;
  \draw  ($(C)+(180+phi+45:0.5)$) coordinate (auxA) 
  (intersection of C--auxA and A--B) coordinate (p1ll) 
  -- ($(A)!(p1ll)!(C)$) -- (C)
  let \p1=($(p1ll)-(C)$),\n1={veclen(\x1,\y1)/sqrt(2)} in
  -- ++ (270+phi:\n1) -- cycle
  ($(C)+(-phi-45:0.5)$) coordinate (auxB) 
  (intersection of C--auxB and A--B) coordinate (p2ll) 
  -- ($(B)!(p2ll)!(C)$) -- (C)
  -- ++ (-phi-90:\n1) -- cycle
  (-\n1/2,0) rectangle ++ (\n1,\n1);
![Screen Shot 2020-09-07 at 12.34.59 PM.png](/image?hash=ee3f5c08aff1eb6700a87ac34a5120f1790f284367d8dc568087f8782a371c36)

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