Sindbad~EG File Manager
\documentclass{article}
\usepackage{amsfonts}
\def\floor#1{{\lfloor#1\rfloor}}
\def\ceiling#1{{\lceil#1\rceil}}
\def\Q{{\mathbb Q}}
\def\Z{{\mathbb Z}}
\def\Re{{\rm Re}}
\def\Im{{\rm Im}}
\def\A{{\mathcal A}}
\def\d{{\bf d}}
\def\S{{\bf S}}
\def\p{{\mathfrak P}}
\def\binomial#1#2{{#1 \choose #2}}
\def\norm{{\mathcal N}}
\def\onehalf{\frac 1 2}
\def\mod{\mbox{\ mod\ }}
\def\sqN{{\sqrt{\frac N 2}}}
\def\vector#1#2#3{\left(
\begin{array}{l}
#1\\
#2\\
#3
\end{array}\right)}
\def\matrix#1#2#3#4#5#6#7#8#9{{
\left( \begin{array}{ccc}
#1 & #2 & #3 \\
#4 & #5 & #6 \\
#7 & #8 & #9
\end{array} \right)}}
\def\determinant#1#2#3#4#5#6#7#8#9{{
\left\vert \begin{array}{ccc}
#1 & #2 & #3 \\
#4 & #5 & #6 \\
#7 & #8 & #9
\end{array} \right\vert}}
\def\dettwo#1#2#3#4{{
\left\vert \begin{array}{ccc}
#1 & #2 \\
#3 & #4
\end{array} \right\vert}}
% Set left margin - The default is 1 inch, so the following
% command sets a 1.25-inch left margin.
\setlength{\oddsidemargin}{0.25in}
% Set width of the text - What is left will be the right margin.
% In this case, right margin is 8.5in - 1.25in - 6in = 1.25in.
\setlength{\textwidth}{6in}
% Set top margin - The default is 1 inch, so the following
% command sets a 0.75-inch top margin.
\setlength{\topmargin}{-0.25in}
% Set height of the text - What is left will be the bottom margin.
% In this case, bottom margin is 11in - 0.75in - 9.5in = 0.75in
\setlength{\textheight}{8in}
\usepackage{pstricks}
% Set the beginning of a LaTeX document
\begin{document}
% I start by defining macros to produce pictures of tilings in PSTricks.
\psset{unit=3cm}
% so I'll make all my tilings fit in one unit width
\def\ThreeTiling{%
\pspicture(1, 1)
\qline(0,0)(1,0) % base
\qline(0,0)(0.5,0.866025404) % left side
\qline(0.5,0.866025404)(1,0) % right side
\qline(0,0)(0.5,0.288675135) % lower left to center
\qline(0.50,0.866025404)(0.50,0.288675135) % top to center
\qline(0.50,0.288675135)(1,0) % center to lower right
\endpspicture}
\def\ThreeTilingA{%
\pspicture(2, 1)
\qline(0,0)(2,0) % base
\qline(0,0)(0.5,0.866025404) % left side
\qline(0.5,0.866025404)(2,0) % right side
\qline(0,0)(1,0.577350269) % lower left to upper right
\qline(1,0.577350269)(1,0) % vertical
\endpspicture}
\def\FourTilingA{%
\psset{unit=2cm}
\pspicture(2,1.2)
\qline(0,0)(1.73205081,0) % base
\qline(1.73205081,0)(1.73205081,1) % right side
\qline(0.866025404,0.5)( 0.866025404,0) % center vertical
\qline( 0.866025404,0.5)(1.73205081,0) % downslope diagonal
\qline( 0,0)(1.73205081,1) % top
\qline( 0.866025404,0.5)(1.73205081,0.5) %horizontal
\endpspicture
\psset{unit=3cm}
}
\def\FourTilingB{%
\psset{unit=2cm}
\pspicture(2,1.2)
\qline(0,0)(1.73205081,0) % base
\qline(1.73205081,0)(1.73205081,1) % right side
\qline(0.866025404,0.5)( 0.866025404,0) % center vertical
\qline( 0.866025404,0.5)(1.73205081,0) % horizontal
\qline( 0,0)(1.73205081,1) % top
\qline(1.29903811,0.25)(1.73205081,1)
\endpspicture
\psset{unit=3cm}
}
\def\FourTilingC{%
\psset{unit=2cm}
\pspicture(2,1.2)
\qline(0,0)(1.73205081,0) % base
\qline(1.73205081,0)(1.73205081,1) % right side
\qline(0.866025404,0.5)( 0.866025404,0) % center vertical
\qline( 0.866025404,0.5)(1.73205081,0) % downslope diagonal
\qline( 0,0)(1.73205081,1) % top
\qline(1.73205081,0)(1.29903811,0.75)
\endpspicture
\psset{unit=3cm}
}
\def\FourTiling{%
\pspicture(1,1)
\qline(0,0)(1,0) % base
\qline(0,0)(0.5,0.866025404) % left side
\qline(0.5,0.866025404)(1,0) % right side
\qline(0.25,0.433012602)(0.75,0.433012602) %horizontal halfway up
\qline(0.25,0.433012602)(0.5,0) %left to bottom
\qline(0.5,0)(0.75,0.433012602) % bottom to right
\endpspicture}
\def\FiveTiling{%
\psset{unit=2cm}
\pspicture(2.5,1.2)
\qline(0,0)(2.5,0) % base
\qline(0,0)(2,1) % left side
\qline(2,1)(2.5,0) % right side
\qline(2,1)(2,0) % vertical
\qline(1,0.5)(2,0.5) % horizontal
\qline(2,0.5)(1,0) % diagonal
\qline(1,0)(1,0.5) % second vertical
\psset{unit=3cm}
\endpspicture}
\def\FiveTilingB{%
\psset{unit=2cm}
\pspicture(2.5,1.2)
\qline(0,0)(2.5,0) % base
\qline(0,0)(2,1) % left side
\qline(2,1)(2.5,0) % right side
\qline(2,1)(2,0) % vertical
\qline(1,0.5)(2,0.5) % horizontal
\qline(1,0.5)(2,0) % diagonal
\qline(1,0)(1,0.5) % second vertical
\psset{unit=3cm}
\endpspicture}
\def\SixTiling{%
\pspicture(1, 1)
\qline(0,0)(1,0) % base
\qline(0,0)(0.5,0.866025404) % left side
\qline(0.5,0.866025404)(1,0) % right side
\qline(0,0)(0.75, 0.433012702) % lower left to right, (3/4, sqrt(3)/4)
\qline(0.50,0.866025404)(0.50,0) % top to bottom
\qline(0.25, 0.433012702)(1,0) % left to lower right
\endpspicture}
\def\SixTilingA{%
\psset{unit=2cm}
\pspicture(1, 1)
\qline(0,0)( 3.46410162,0) % base (2 sqrt 3, 0)
\qline(0,0)(1.73205081,1) % left side (sqrt 3, 0)
\qline(1.73205081,1)( 3.46410162,0) % right side
\qline(1.73205081,1)(1.73205081,0) % top to bottom (sqrt 3,1) to (sqrt 3,0)
\qline( 0.866025404,0.5)(1.15470054,0) % (sqrt(3)/2, 1/2) to ((2/3) sqrt(3), 0)
\qline(1.15470054,0)(1.73205081,1)
\qline(1.73205081,1)(2.30940108,0) % top to ((4/3) sqrt 3, 0)
\qline(2.30940108,0)(2.59807621,0.5) % ((4/3) sqrt 3, 0) to ((3/2) sqrt 3, 1/2)
\endpspicture}
\def\EightTiling{%
\pspicture(1,1)
\qline(0,0)(1,0) % base
\qline(0,0)(0.5,0.866025404) % left side
\qline(0.5,0.866025404)(1,0) % right side
\qline(0.25,0.433012602)(0.75,0.433012602) %horizontal halfway up
\qline(0.25,0.433012602)(0.25,0) % vertical left
\qline(0.75,0.433012602)(0.75,0) % vertical right
\qline(0.5,0.866025404)(0.5,0) % vertical center
\qline(0.25,0.433012602)(0.5,0) %left to bottom
\qline(0.5,0)(0.75,0.433012602) % bottom to right
\endpspicture}
\def\NineTiling{%
\pspicture(1,1)
\qline(0,0)(1,0) % base
\qline(0,0)(0.5,0.866025404) % left side
\qline(0.5,0.866025404)(1,0) % right side
\qline(0.167777,0.288675135)(0.833333,0.288675135) %horizontal one-third up
\qline(0.333333,0.577350269)(0.6777777, 0.577350269) % horizontal two-thirds up
\qline(0.1677777,0.288675135)(0.333333,0)
\qline(0.333333,0.577350269)(0.6777777,0)
\qline(0.333333,0)(0.677777,0.577350269)
\qline(0.677777,0)(0.833333,0.288675135)
\endpspicture}
\def\NineTilingA{%
\psset{unit=1cm}
\pspicture(6,3.3)
\qline(0,0)(6,0) % base
\qline(0,0)(6,3) % left side
\qline(6,0)(6,3) % right side
\qline(2,0)(2,1) % vertical
\qline(4,0)(4,2)
\qline(5,0)(5,2)
\qline(4,2)(6,2) % horizontal
\qline(2,1)(4,1)
\qline(2,0)(4,1) % diagonal
\qline(4,0)(5,2)
\qline(5,0)(6,2)
\endpspicture
\psset{unit=3cm}
}
\def\TwelveTiling{%
\pspicture(1,1)
\qline(0,0)(1,0) % base
\qline(0,0)(0.5,0.866025404) % left side
\qline(0.5,0.866025404)(1,0) % right side
\qline(0.25,0.433012602)(0.75,0.433012602) %horizontal halfway up
\qline(0.25,0.433012602)(0.5,0) %left to bottom
\qline(0.5,0)(0.75,0.433012602) % bottom to right
% that finishes a 4-tiling. Now divide the 4 triangles into 3 equilaterals each
\qline(0,0)(0.25,0.144337567)
\qline(0.25,0.144337567)(0.25,0.433012602)
\qline(0.25,0.144337567)(0.5,0)
\qline(0.5,0)(0.75,0.144337567)
\qline(0.75,0.144337567)(0.75,0.433012602)
\qline(0.75,0.144337567)(1,0)
\qline(0.5,0.288675035)(0.5,0)
\qline(0.5,0.288675035)(0.25,0.433012602)
\qline(0.5,0.288675035)(0.75,0.433012602)
\qline(0.5,0.577350169)(0.5,0.866025404)
\qline(0.5,0.577350169)(0.25,0.433012602)
\qline(0.5,0.577350169)(0.75,0.433012602)
\endpspicture}
\def\ThirteenTiling{%
% sqrt(3)/ 3 = 0.577350269
% 2 sqrt(3)/ 3 = 1.15470054
% sqrt(3) = 1.7325081
% sqrt(3) + (2/3) = 2.39871747
% sqrt(3) + (1/3) = 2.06538414
\pspicture(2.9,1.2)
\qline(0,0)(2.39871747,0) % base
\qline(0,0)(1.7325081,1) % left side
\qline(1.7325081,1)(2.39871747,0) % right side
\qline(1.7325081,0)(1.7325081,1) % center vertical
\qline(2.06538414,0)(2.06538414,0.5) %vertical on right
\qline(0.577350269,0)(0.577350269,0.333333333) % left vertical
\qline(1.15470054,0)(1.15470054, 0.666666667) % second vertical
\qline(1.15470054,0.67777777)(1.7325081,0.67777777) % top horizontal on left side
\qline(0.577350269,0.33333333)(1.7325081,0.33333333) % middle horizontal on left
\qline(1.7325081,0.5)(2.06538414,0.5) % middle horizontal on right
\qline(1.7325081,0.666666667)(0.577350269,0) % top diagonal on left
\qline(1.7325081,0.333333333)(1.15470054,0) % lower diagonal on left
\qline(1.7325081,0.5)(2.06538414,0)
\endpspicture}
\def\SixteenTiling{%
\pspicture(1,1)
\qline(0,0)(1,0) % base
\qline(0,0)(0.5,0.866025404) % left side
\qline(0.5,0.866025404)(1,0) % right side
\qline(0.125,0.216506351)(0.875,0.216506351) %horizontal one-quarter
\qline(0.25,0.433012702)(0.75, 0.433012702) % horizontal halfway up
\qline(0.375,0.649519052)(0.625,0.649519052 ) % horizontal three-quarters up
% next three downward diagonals
\qline(0.125,0.216506351)(0.25,0)
\qline(0.25,0.433012702)(0.5,0)
\qline(0.375,0.649519052)(0.75,0)
% next three upward diagonals
\qline(0.25,0)(0.625,0.649519052 )
\qline(0.5,0)(0.75, 0.433012702)
\qline(0.75,0)(0.875,0.216506351)
\endpspicture}
% sqrt 3 = 1.73205081
% 2 sqrt 3 = 3.46410162
% 3 sqrt 3 = 5.19615242
% 4 sqrt 3 = 6.92820323
% 5 sqrt 3 = 8.66025404
% 6 sqrt 3 = 10.3923048
\def\TwentySevenTiling{%
\psset{unit=0.5cm}
\pspicture(10.3923048,9.4)
\qline(0,0)(10.3923048,0) % base
\qline(0,0)(5.19615242,9) % left side
\qline(5.19615242,9)(10.3923048,0) % right side
\qline(0,0)(1.73205081,1) % three tiles along the bottom
\qline(1.73205081,1)(3.46410162,0)
\qline(3.46410162,0)(5.19615242,1)
\qline(5.19615242,1)(6.92820323,0)
\qline(6.92820323,0)(8.66025404,1)
\qline(8.66025404,1)(10.3923048,0)
\qline(1.73205081,1)(8.66025404,1) % horizontal above the bottom three tiles
\qline(1.73205081,1)(1.73205081,3) % three vertical lines
\qline(5.19615242,1)(5.19615242,3)
\qline(8.66025404,1)(8.66025404,3)
\qline(1.73205081,3)(3.46410162,4) % finish the two bottom hexagons
\qline(3.46410162,4)(5.19615242,3)
\qline(5.19615242,3)(6.92820323,4)
\qline(6.92820323,4)(8.66025404,3) % OK, two hexagons done
\qline(3.46410162,4)(3.46410162,6) % left vertical of top hexagon
\qline(3.46410162,6)(5.19615242,7) % top left of top hexagon
\qline(5.19615242,7)(6.92820323,6) % top right of top hexagon
\qline(6.92820323,6)(6.92820323,4) % right vertical of top hexagon
\qline(5.19615242,7)(5.19615242,9) % uppermost vertical
\qline(1.73205081,1)(5.19615242,7)
\qline(5.19615242,7)(8.66025404,1)
\qline(3.46410162,4)(6.92820323,4)
\qline(3.46410162,4)(5.19615242,1)
\qline(5.19615242,1)(6.92820323,4)
\qline(3.46410162,2)(1.73205081,1) % radii of lower left triangle
\qline(3.46410162,2)(3.46410162,4)
\qline(3.46410162,2)(5.19615242,1)
\qline(6.92820323,2)(5.19615242,1) % radii of lower right triangle
\qline(6.92820323,2)(6.92820323,4)
\qline(6.92820323,2)(8.66025404,1)
\qline(5.19615242,5)(3.46410162,4) % radii of top triangle
\qline(5.19615242,5)(6.92820323,4)
\qline(5.19615242,5)(5.19615242,7)
\endpspicture}
\def\TwelveTilingA{% triple-square case, m=2
\pspicture(1,1)
\psset{unit=0.005cm}
\qline(409.5,10)(609.25,125.3257)
\qline(609.25,125.3257)(675.8334,10)
\qline(675.8334,10)(409.5,10)
\qline(809,240.6514)(809,10)
\qline(809,10)(675.8334,10)
\qline(675.8334,10)(809,240.6514)
\qline(809,240.6514)(609.25,125.3257)
\qline(609.25,125.3257)(675.8334,10)
\qline(675.8334,10)(809,240.6514)
\qline(809,240.6514)(609.25,125.3257)
\qline(609.25,125.3257)(542.6667,240.6514)
\qline(542.6667,240.6514)(809,240.6514)
\qline(409.5,10)(409.5,240.6514)
\qline(409.5,240.6514)(542.6667,240.6514)
\qline(542.6667,240.6514)(409.5,10)
\qline(409.5,10)(609.25,125.3257)
\qline(609.25,125.3257)(542.6667,240.6514)
\qline(542.6667,240.6514)(409.5,10)
\qline(9.999969,10)(209.75,125.3257)
\qline(209.75,125.3257)(276.3333,10)
\qline(276.3333,10)(9.999969,10)
\qline(409.5,240.6514)(409.5,10)
\qline(409.5,10)(276.3333,10)
\qline(276.3333,10)(409.5,240.6514)
\qline(409.5,240.6514)(209.75,125.3257)
\qline(209.75,125.3257)(276.3333,10)
\qline(276.3333,10)(409.5,240.6514)
\qline(409.5,240.6514)(609.25,355.9771)
\qline(609.25,355.9771)(675.8334,240.6514)
\qline(675.8334,240.6514)(409.5,240.6514)
\qline(809,471.3029)(809,240.6514)
\qline(809,240.6514)(675.8334,240.6514)
\qline(675.8334,240.6514)(809,471.3029)
\qline(809,471.3029)(609.25,355.9771)
\qline(609.25,355.9771)(675.8334,240.6514)
\qline(675.8334,240.6514)(809,471.3029)
\endpspicture}
\def\TwelveTilingB{% triple-square case, m=2
\pspicture(1,1)
\psset{unit=0.005cm}
\qline(276.3333,10)(476.0833,125.3257)
\qline(476.0833,125.3257)(542.6667,10)
\qline(542.6667,10)(276.3333,10)
\qline(675.8333,240.6514)(675.8333,10)
\qline(675.8333,10)(542.6667,10)
\qline(542.6667,10)(675.8333,240.6514)
\qline(675.8333,240.6514)(476.0833,125.3257)
\qline(476.0833,125.3257)(542.6667,10)
\qline(542.6667,10)(675.8333,240.6514)
\qline(9.999969,10)(276.3333,10)
\qline(276.3333,10)(209.75,125.3257)
\qline(209.75,125.3257)(9.999969,10)
\qline(409.5,240.6514)(276.3333,10)
\qline(276.3333,10)(209.75,125.3257)
\qline(209.75,125.3257)(409.5,240.6514)
\qline(409.5,240.6514)(276.3333,10)
\qline(276.3333,10)(476.0833,125.3257)
\qline(476.0833,125.3257)(409.5,240.6514)
\qline(409.5,240.6514)(675.8334,240.6514)
\qline(675.8334,240.6514)(609.2501,355.9771)
\qline(609.2501,355.9771)(409.5,240.6514)
\qline(809,471.3029)(675.8334,240.6514)
\qline(675.8334,240.6514)(609.2501,355.9771)
\qline(609.2501,355.9771)(809,471.3029)
\qline(675.8333,10)(809,10)
\qline(809,10)(675.8333,240.6514)
\qline(675.8333,240.6514)(675.8333,10)
\qline(809,240.6514)(809,10)
\qline(809,10)(675.8333,240.6514)
\qline(675.8333,240.6514)(809,240.6514)
\qline(675.8333,240.6514)(809,240.6514)
\qline(809,240.6514)(809,471.3029)
\qline(809,471.3029)(675.8333,240.6514)
\endpspicture}
\def\TwentySevenTilingA{% triple-square case, m=3
\pspicture(1.0,1.0)
\psset{unit=0.005cm}
\qline(542.6667,10)(675.8334,86.88382)
\qline(675.8334,86.88382)(720.2222,10)
\qline(720.2222,10)(542.6667,10)
\qline(809,163.7676)(809,10)
\qline(809,10)(720.2222,10)
\qline(720.2222,10)(809,163.7676)
\qline(809,163.7676)(675.8334,86.88382)
\qline(675.8334,86.88382)(720.2222,10)
\qline(720.2222,10)(809,163.7676)
\qline(809,163.7676)(675.8334,86.88382)
\qline(675.8334,86.88382)(631.4445,163.7676)
\qline(631.4445,163.7676)(809,163.7676)
\qline(542.6667,10)(542.6667,163.7676)
\qline(542.6667,163.7676)(631.4445,163.7676)
\qline(631.4445,163.7676)(542.6667,10)
\qline(542.6667,10)(675.8334,86.88382)
\qline(675.8334,86.88382)(631.4445,163.7676)
\qline(631.4445,163.7676)(542.6667,10)
\qline(276.3333,10)(409.5,86.88382)
\qline(409.5,86.88382)(453.8889,10)
\qline(453.8889,10)(276.3333,10)
\qline(542.6667,163.7676)(542.6667,10)
\qline(542.6667,10)(453.8889,10)
\qline(453.8889,10)(542.6667,163.7676)
\qline(542.6667,163.7676)(409.5,86.88382)
\qline(409.5,86.88382)(453.8889,10)
\qline(453.8889,10)(542.6667,163.7676)
\qline(542.6667,163.7676)(409.5,86.88382)
\qline(409.5,86.88382)(365.1111,163.7676)
\qline(365.1111,163.7676)(542.6667,163.7676)
\qline(276.3333,10)(276.3333,163.7676)
\qline(276.3333,163.7676)(365.1111,163.7676)
\qline(365.1111,163.7676)(276.3333,10)
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\qline(720.2222,317.5352)(809,471.3029)
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\def\TwentySevenTilingB{% triple-square case, m=3
\pspicture(1.0,1.0)
\psset{unit=0.005cm}
\qline(453.8889,10)(587.0555,86.88382)
\qline(587.0555,86.88382)(631.4445,10)
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\qline(809,317.5352)(809,471.3029)
\qline(809,471.3029)(720.2222,317.5352)
\endpspicture}
\def\FortyEightTiling{%hexagonal case, m=4
\pspicture(2.0,2.0)
\psset{unit=0.01cm}
\qline(10,10)(148.75,10)
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\qline(426.25,170.2147)(426.25,90.10736)
\qline(426.25,90.10736)(495.625,50.05371)
\endpspicture}
\def\OneHundredTwentyFiveTiling{%hexagonal case, m=4
\pspicture(2.0,2.0)
\psset{unit=0.01cm}
\qline(10,10)(121,10)
\qline(121,10)(65.5,42.04297)
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\qline(121,10)(232,10)
\qline(232,10)(176.5,42.04297)
\qline(176.5,42.04297)(121,10)
\qline(232,10)(343,10)
\qline(343,10)(287.5,42.04297)
\qline(287.5,42.04297)(232,10)
\qline(343,10)(454,10)
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\qline(398.5,42.04297)(343,10)
\qline(454,10)(565,10)
\qline(565,10)(509.5,42.04297)
\qline(509.5,42.04297)(454,10)
\qline(10,10)(65.5,42.04297)
\qline(65.5,42.04297)(65.5,106.1288)
\qline(65.5,106.1288)(10,10)
\qline(65.5,106.1288)(121,138.1718)
\qline(121,138.1718)(121,202.2576)
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\qline(121,202.2577)(176.5,234.3006)
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\title{Tiling triangle $ABC$ with \\ congruent triangles similar to $ABC$ } % Enter your title between curly braces
\author{Michael Beeson} % Enter your name between curly braces
\date{\today} % Enter your date or \today between curly braces
\maketitle
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{corollary}{Corollary}
\newtheorem{definition}{Definition}
\newtheorem{conjecture}{Conjecture}
\abstract
We investigate the problem of cutting a triangle $ABC$ into $N$ congruent triangles (the ``tiles''), each of which
is similar to $ABC$. The more general problem when the tile is not similar to $ABC$ is not treated in this paper;
see \cite{beeson}.
We give a complete
characterization of the numbers $N$ for which some triangle $ABC$ can be tiled by $N$ tiles similar to $ABC$, and
also a complete characterization of the numbers $N$ for which a particular triangle $ABC$ can so tiled. It has
long been known that there is always a ``quadratic tiling'' when $N$ is a square. We show that unless $ABC$
is a right triangle, $N$ must be a square. On the other hand, if $ABC$ is a right triangle, there are two
more possibilities: $N$ can be a sum of two squares $e^2 + f^2$ if the tangent of one of the angles is the
rational number $e/f$, or in case $ABC$ is a 30-60-90 triangle, $N$ can be three times a square.
\smallskip
The key idea is that
the similarity factor $\sqrt N$ is an eigenvalue of a certain matrix. The proofs we give involve only undergraduate
level linear algebra.
\section{Examples of Tilings}
We consider the problem of cutting a triangle into $N$ congruent triangles. Figures 1
through 4 show that,
at least for certain triangles, this can be done with $N = 3$, 4, 5, 6, 9, and 16. Such a configuration
is called an $N$-tiling.
\begin{figure}[h] % Fig. 1
\label{figure:3-tilings}
\caption{Two 3-tilings }
\hskip 2.5cm
\ThreeTilingA
\hskip 1cm
\ThreeTiling
\end{figure}
\begin{figure}[h] % Fig. 2
\label{figure:nsquared}
\caption{A 4-tiling, a 9-tiling, and a 16-tiling}
\hskip 2cm
\FourTiling
\hskip 1cm
\NineTiling
\hskip 1cm
\SixteenTiling
\end{figure}
\begin{figure}[h] % Fig. 3
\label{figure:4-tilings}
\caption{Three 4-tilings}
\hskip 1.7cm
\FourTilingA
\FourTilingB
\FourTilingC
\end{figure}
The method illustrated for $N=4$ ,$9$, and $16$ clearly generalizes to any perfect square $N$.
While the exhibited 3-tiling, 6-tiling, and 5-tiling clearly depend on the exactly angles of the triangle,
{\em any} triangle can be decomposed into $n^2$ congruent triangles by drawing $n-1$ lines, parallel to each edge and
dividing the other two edges into $n$ equal parts. Moreover, the large (tiled) triangle is similar to the
small triangle (the ``tile''). We call such a tiling a {\em quadradtic tiling.}
It follows that if we have a tiling of a triangle $ABC$ into $N$ congruent triangles, and $m$ is any
integer, we can tile $ABC$ into $Nm^2$ triangles by subdividing the first tiling, replacing each of the $N$ triangles by $m^2$ smaller ones.
Hence the set of $N$ for which an $N$-tiling of some triangle exists is closed under multiplication by squares.
Let $N$ be of the form $n^2 + m^2$. Let triangle $T$ be a right triangle with perpendicular sides $n$ and $m$, say with $n \ge m$.
Let $ABD$ be a right triangle with base $AD$ of length $m^2$, the right angle at $D$ and altitude $mn$, so side $BD$ has length $mn$. Then $ABD$ can be decomposed into $m$ triangles congruent to $T$, arranged with their short sides
(of length $m$)
parallel to the base $AD$. Now, extend $AD$ to point $C$, located $n^2$ past $D$. Triangle $ADC$ can be tiled
with $n^2$ copies of $T$, arranged with their long sides parallel to the base. The result is a tiling of
triangle $ABC$ by $n^2 + m^2$ copies of $T$. The first 5-tiling exhibited in Fig. 3 is the simplest
example, where $n=2$ and $m=1$. The case $N = 13 = 3^2 + 2^2$ is illustrated in Fig. 5. We call these tilings ``biquadratic.''
More generally, a {\em biquadratic tiling} of triangle $ABC$ is one in which $ABC$ has a right angle at $C$, and can be divided by an altitude
from $C$ to $AB$ into two triangles, each similar to $ABC$, which can be tiled respectively by $n^2$ and $m^2$ copies of a triangle similar to $ABC$.
The second 5-tiling shows that this can be sometimes be done more generally than by combining two quadratic tilings.
\begin{figure}[h] % Fig. 4
\label{figure:5-tilings}
\caption{Two 5-tilings}
\hskip 2cm
\FiveTiling
\hskip 1 cm
\FiveTilingB
\end{figure}
\begin{figure}[h] % Fig. 5
\label{figure:13-tilings}
\caption{A 13-tiling}
\hskip 2cm \ThirteenTiling
\end{figure}
If the original triangle $ABC$ is chosen to be isosceles,
then each of the $n^2$ triangles can be divided in half by an altitude; hence any isosceles triangle can be decomposed into $2n^2$ congruent
triangles. If the original triangle is equilateral, then it can be first decomposed into $n^2$ equilateral triangles, and then these
triangles can be decomposed into 3 or 6 triangles each, showing that any equilateral triangle can be decomposed into $3n^2$ or $6n^2$
congruent triangles. These tilings are neither quadratic nor biquadratic.
For example we can 12-tile an equilateral triangle in two different ways, starting with a 3-tiling and then subdividing each triangle
into 4 triangles (``subdividing by 4''), or starting with a 4-tiling and then subdividing by 3.
\begin{figure}[h] % Fig. 6
\label{figure:6-8-12-tilings}
\caption{A 6-tiling, an 8-tiling, and a 12-tiling}
\hskip 2cm
\SixTiling
\hskip 1cm
\EightTiling
\hskip 1cm
\TwelveTiling
\end{figure}
Examples like these led us to the following definitions:
A tiling $E$ of triangle $ABC$ (with tile $T_2$ is a {\em subtiling} of another tiling $F$ of $ABC$ (with tile $T$), if $T$ can be tiled by the tile $T_2$ and the tiling $E$ is obtained by tiling each copy of $T$ in $F$ with triangle $T_2$. It is not required that the {\em same}
tiling be used for each copy of $T$. For example, we could take $F$ to be one of the two five-tilings, and then tile each of the tiles
in that tiling by one of its two five-tilings. In this way we can obtain 32 different 25-tilings, none of them quadratic.
A tiling of $ABC$ is called {\em composite} if it is a subtiling of some tiling into fewer triangles. It is called {\em prime} if it is
not composite. Note that a quadratic $N^2$ tiling is prime if and only if $N$ is a prime number.
The examples above do not exhaust all possible tilings, even when $N$ is a square. For example, Fig. 7 shows a 9-tiling that is not produced by
those methods:
\begin{figure}[h] % Fig. 7
\label{figure:9-tilings}
\caption{Another 9-tiling}
\hskip 4cm
\NineTilingA
\end{figure}
There is another family of $N$-tilings, in which $N$ is of the form $3m^2$, and both the tile and the tiled triangle are 30-60-90 triangles.
The case $m=1$ is given in Fig. 1; the case $m=2$ makes $N=12$. There are two ways to 12-tile a 30-60-90 triangle with 30-60-90 triangle.
One is to first quadratically 4-tile it, and then subtile the four triangles with the 3-tiling of Figure 1. This produces the first
12-tiling in Fig. 8. Somewhat surprisingly, there is another way to tile the same triangle with the same 12 tiles, also shown in Fig. 8; the second
tiling is prime.
The next member of this family is $m=3$, which makes $N=27$. Two 27-tilings are shown in Fig. 9; the first obtained by subtiling a quadratic tiling,
and the second one prime. Similarly, there are two 48-tilings (not shown).
\begin{figure}[h] % Fig. 8
\caption{Two 12-tilings }
\hskip 3cm
\TwelveTilingA
\hskip 2cm
\TwelveTilingB
\end{figure}
\begin{figure}[h] % Fig. 9
\caption{Two 27-tilings }
\hskip 3cm
\TwentySevenTilingA
\hskip 2cm
\TwentySevenTilingB
\end{figure}
Until October 12, 2008, no examples were known of more complicated tilings than those illustrated above. Then we found the
beautiful 27-tiling shown in Fig. 10. This tiling is one of a family of $3k^2$ tilings (the case $k=3$).
The next case is a 48-tiling, made from six hexagons (each containing 6 tiles) bordered by
4 tiles on each of 3 sides. In general one can arrange $1+2+\ldots+k$ hexagons in bowling-pin fashion, and add $k+1$ tiles
on each of three sides, for a total number of tiles of $6(1+2+\ldots+k)+3(k+1) = 3k(k+1) + 3(k+1) = 3(k+1)^2$.
Figure 11 shows more members of this family.
\begin{figure}[h] % Fig. 10
\caption{A prime 27-tiling}
\hskip 5cm
\TwentySevenTiling
\end{figure}
\begin{figure}[h] % Fig. 11
\caption{$3m^2$ tilings for $m=4$ and $m=5$}
\hskip 1.5cm
\FortyEightTiling
\hskip 0.5cm
\OneHundredTwentyFiveTiling
\end{figure}
\section{Previous work}
The examples given in Figures 1 through 6 are well-known. They have been discussed,
in particular, in connection with ``rep-tiles'' \cite{golomb}. A ``rep-tile'' is a set of points $X$ in the plane (not necessarily just a triangle) that
can be dissected into $N$ congruent sets, each of which is similar to $S$.
An $N$-tiling in which the tiled triangle $ABC$ is similar to the
triangle $T$ used as the tile is a special case of this situation. That is the case, for example, for the $n^2$ family and the $n^2 + m^2$ family, but not for the 3-tiling, 6-tiling, or the 12-tiling exhibited above. Thus the concepts of an $N$-tiling and rep-tiles overlap, but neither subsumes the other. The paper \cite{goldberg} also contains a diagram showing the $n^2$ family of tilings, but the problem considered there is different: one
is allowed to cut $N$ copies of the tile first, before assembling the pieces into a large figure, but the large figure must be similar to the original
tile. The two books \cite{boltyanski} and \cite{boltyanski2} have tantalizing titles, but deal with other problems.
Only after completing the work in this paper did I encounter
Soifer's book \cite{soifer}, when the second edition came out, although the first edition had been out for 19 years.
The book contains the observation that
if the tile $T$ is similar to the tiled triangle then $\sqrt N$ is an eigenvalue of a certain matrix, so that observation
is, as it turns out, not new. The book, however, does not contain any examples of tilings beyond the quadratic tilings,
though it gives an indication
that at least the biquadratic tilings were known, since it says that the 1989 Russian Mathematical Olympiad contained the problem to show that
if $N$ is a sum of two squares then there is a triangle that can be $N$-tiled. Soifer states (p. 48) an open problem about triangle
tiling, and says that Paul Erd\"os offered a \$25 prize for the first solution. He does not state where or when Erd\"os mentioned these
problems. The problem statement is: Find all positive integers $N$ such that at least one triangle can be cut into $N$
triangles congruent to each other. This is Soifer's ``Problem 6.7.''
Soifer also states some related problems. His ``Problem 6.5'' is: For each triangle $ABC$, find all positive integers $N$ such that
$T$ can be cut into $N$ triangles congruent to each other, and the number of distinct partitions of $T$ into $N$ congruent
triangles. Soifer says that his Problem 6.5 is ``open and very difficult.''
Soifer's ``Problem 6.6'' is also a \$25 Erd\"os problem: Find (and classify) all triangles that can only be cut into $n^2$
congruent triangles for any integer $n$.
Soifer claims, without publishing a proof, that if the sides and angles of $ABC$ are integrally independent, then $ABC$
admits only quadratic tilings. He proves that the perfect squares are exactly the $N$ for
which {\em every} triangle $ABC$ can be $N$-tiled by some triangle.
Dima Fan-Der-Flaas informed me that the problem of finding an $N$-tiling of some triangle when $N=1989$ was posed on the
Russian Mathematical Olympiad in 1989; it was solved by a few students, who had to discover what we call here the ``biquadratic
tilings'', and realize that 1989 is a sum of two squares and the relevance of that fact. I would like to thank Dima for his
careful reading of parts of some drafts of this paper.
\section{The impossibility of certain tilings}
The elementary constructions just described suffice to produce $N$-tilings when $N$ has
one of the forms $n^2$, $n^2 + m^2$, $2n^2$, $3n^2$, or $6n^2$. Of the tilings we have exhibited,
many have the tile similar to the tiled triangle $ABC$; the others have $ABC$ equilateral or at least
isosoceles. In this paper we deal only with the case when the tile is similar to $ABC$; in which
case it turns out that the only possible forms of $N$ are $n^2$, $n^2+m^2$, and $3n^2$. Our main theorem implies that
if $N$ is not of one of these forms, then there is no $N$-tiling of any triangle $ABC$ by a tile similar to $ABC$;
but it also shows that the second and third form can only apply when $ABC$ is a right triangle.
\section{Definitions, notation, and some simple lemmas}
We give a mathematically precise definition of ``tiling'' and fix some terminology and notation.
Given a triangle $T$ and a larger triangle $ABC$, a ``tiling'' of triangle $ABC$ by triangle $T$ is
a list of triangles $T_1,\ldots,T_n$ congruent to $T$, whose interiors are disjoint, and the closure of whose union is triangle $ABC$. A ``strict
vertex'' of the tiling is a vertex of one of the $T_i$ that does not lie on the interior of an edge of another $T_j$.
A ``strict tiling'' is one in which no $T_i$ has a vertex lying on the interior of an edge of another $T_j$, i.e. every vertex is strict.
For example, the biquadratic tilings (illustrated above for $N=5$ and $N=13$) are not strict, but all the other tilings shown above are strict.
The letter ``$N$'' will always be used for the number of triangles used in the tiling. An $N$-tiling of $ABC$ is a tiling that uses
$N$ copies of some triangle $T$.
Let $a$, $b$, and $c$ be the sides of triangle $ABC$, and angles $\alpha$, $\beta$, and $\gamma$ be the angles
opposite sides $a$, $b$, and $c$, i.e. the interior angles at vertices $A$, $B$, and $C$. An {\em interior vertex}
in a tiling of $ABC$ is a vertex of one of $T_i$ that does not lie on the boundary of $ABC$. A {\em boundary vertex}
is a vertex of one of the $T_i$ that lies on the boundary of $ABC$.
By the law of sines we have
$$ \frac a {\sin \alpha} = \frac b {\sin \beta} = \frac c {\sin \gamma} $$
Up to similarity then we may assume
\begin{eqnarray*}
a &=& \sin \alpha \\
b &=& \sin \beta \\
c &=& \sin \gamma
\end{eqnarray*}
Since $\gamma = \pi - (\alpha + \beta)$ we have $\sin(\gamma) = \sin(\alpha + \beta)$, so
$$ p \sin \alpha + q \sin \beta + r \sin(\alpha + \beta) = 0.$$
The meanings of all these symbols will be fixed throughout the rest of the paper.
A non-strict vertex $V$ is one that lies on an edge of $T_j$, with
$T_j$ on one side of the edge and (more than one) $T_i$ having vertex $V$ on the other side.
Consider the maximal line segment $S$ extending this edge which is contained in the union of the edges of the tiling.
This is defined to be the {\em maximal segment} of $V$.
Since there are triangles on each side of $S$, there are triangles on each side of $S$ at every point of $S$ (since
$S$ cannot extend beyond the boundary of $ABC$). Hence the length of $S$ is a sum of lengths of sides of triangles $T_i$
in two different ways (though the summands may possibly be the same numbers in a different order). Let us assume
for the moment that the summands are not the same numbers. Then
it follows that some linear relation of the form
$$ pa + qb + rc = 0$$
holds, with $p$, $q$, and $r$ integers not all zero (one of which must of course be negative), and the sum of the absolute values
of $p$, $q$, and $r$ is less than or equal to $N$, since there are no more than $N$ triangles.
If $S$ is a maximal segment containing a non-strict vertex, then there will be integers $n$ and $m$
such that $n$ triangles have a side contained in $S$ and lie on one side of $S$, and $m$ triangles
have a side in $S$ and lie on the other side of $S$. In that case we say $S$ is of type $m:n$.
For example, Fig. 3 shows a 5-tiling with a maximal segment of type $1:2$. This definition does
not require that the lengths of the subdivisions of the maximal segment all be the same (as they are in Fig. 3).
A {\em quadratic tiling} is one in which $N$ is a perfect square, say $N = m^2$, and the tiling is produced by drawing
$m-1$ equally spaced lines parallel to each side, dividing each edge into $m$ equal segments. In such a tiling,
the tile $T$ is similar to the large triangle $ABC$. An {\em angle relation} is an equation
$$p \alpha + q\beta + r \gamma = 2\pi$$
where $p$, $q$, and $r$ are non-negative integers, not all equal. (Since we always have $\alpha + \beta + \gamma = \pi$,
we do not count that equation or its multiples as an angle relation.)
A {\em split vertex} occurs when two copies of the tile in a triangle share one of the vertices of the large triangle.
The following lemma is simple and fundamental:
\begin{lemma} \label{lemma:rightangle}
If, in a tiling, $P$ is a boundary vertex (or a non-strict interior vertex)
and only one interior edge emanates from $P$,
then both angles at $P$ are right angles and $\gamma = \pi/2$.
\end{lemma}
\noindent{\em Proof.}
If the two angles at $P$ are different, then their sum is less than $\pi$, since the sum of all three angles
is $\pi$. Therefore the two angles are the same. But $2\alpha \le \alpha + \beta < \pi$ and $2\beta \le \beta + \gamma < \pi$.
Therefore both angles are $\gamma$. But then $2\gamma = \pi$, so
$\gamma = \pi/2$.
%The following result we call ``Euler's equation'', because it is related to Euler's famous formula for the numbers of vertices, edges, and
%faces of a polygon, although we have not derived it that way here.
%
%\begin{lemma}
%Let a tiling contain $N_b$ boundary vertices, $N_s$ strict interior vertices, and $N_n$ non-strict interior vertices. We then have
%\begin{eqnarray} \label{eq:euler}
% N-1 &=& N_b + N_n + 2 N_s
%\end{eqnarray}
%\end{lemma}
%
%\noindent {\em Proof}.
%We count vertices. At each strict interior vertex, the sum of the angles of the tiles sharing that vertex is $2\pi$.
%At each non-strict interior vertex and at each boundary vertex (that is, vertex lying on the boundary of triangle $ABC$ but not
%equal to $A$, $B$, or $C$), the angle sum is $\pi$. The total angle sum of all $N$ copies of the tile is $N\pi$,
%which must be accounted for by the $\pi$ at the vertices of $ABC$, plus the contributions at the other vertices.
The following lemma identifies those relatively few rational multiples of $\pi$ that have rational tangents or
whose sine and cosine satisfy a polynomial of low degree over $\Q$.
\begin{lemma} \label{lemma:euler}
Let $\zeta = e^{i\theta}$ be algebraic
of degree $d$ over $\Q$, where
$\theta$ is a rational multiple of $\pi$, say $\theta = 2m \pi/n$, where $m$ and $n$ have no common factor.
\smallskip
Then $d = \varphi(n)$, where $\varphi$ is the Euler totient function.
In particular if $d = 4$, which is the case when $\tan \theta$ is rational and $\sin \theta$ is not,
then $n$ is 5, 8, 10, or 12; and if $d=8$ then $n$ is 15, 16, 20, 24, or 30.
\end{lemma}
\noindent{\em Remark}. For example, if $\theta = \pi/6$, we have $\sin \theta = 1 / 2$, which
is of degree 1 over $\Q$. Since $\cos \theta = \sqrt 3 / 2$, the number
$\zeta = e^{i\theta}$ is in $\Q(i,\sqrt{3})$, which is of degree 4 over $\Q$. The number $\zeta$ is a
12-th root of unity, i.e. $n$ in the theorem is 12 in this case; so the minimal polynomial of $\zeta$ is
of degree $\varphi(12) = 4$. This example shows that the theorem is best possible.
\smallskip
\noindent{\em Remark}. The hypothesis that $\theta$ is a rational multiple of $\pi$ cannot be dropped.
For example, $x^4-2x^3+x^2-2x+1$ has two roots on the unit circle and two off the unit circle.
\smallskip
\noindent{\em Proof}.
Let $f$ be a polynomial with rational coefficients of degree $d$ satisfied by $\zeta$.
Since $\zeta = e^{i 2m\pi/n}$,
$\zeta$ is an $n$-th root of unity, so its minimal polynomial has degree $d = \varphi(n)$,
where $\varphi$ is the Euler totient function. Therefore $\varphi(n) \le d$. If $\tan \theta$
is rational and $\sin \theta$ is not, then $\sin \theta$ has degree 2 over $\Q$, so $\zeta$
has degree 2 over $Q(i)$, so $\zeta$ has degree 4 over $\Q$. The stated values
of $n$ for the cases $d=4$ and $d=8$ follow from the well-known formula for $\varphi(n)$. That completes the
proof of (ii) assuming (i).
\begin{corollary} \label{corollary:euler}
If $\sin \theta$ or $\cos \theta$ is rational, and $\theta < \pi$ is a rational multiple of $\pi$,
then $\theta$ is a multiple of $2\pi/n$ where $n$ is 5, 4, 8, 10, or 12.
\end{corollary}
\noindent{\em Proof}. Let $\zeta = \cos \theta + i \sin \theta = e^{i\theta}$. Under the stated hypotheses,
the degree of $\Q(\zeta)$ over $\Q$ is 2 or 4. Hence, by the lemma, $\theta$ is a multiple of $2\pi/n$,
where $n = 5$, 8, 10, or 12 (if the degree is 4) or $n = 3$ or $6$ (if the degree is 3). But the cases
3 and 6 are superfluous, since then $\theta$ is already a multiple of $2\pi/12$.
%\begin{lemma} \label{lemma:euler2}
%Let $\zeta = e^{i\theta}$ be algebraic
%of degree 2 over $\Q$. Then $\theta$ is a rational multiple of $\pi$.
%\end{lemma}
%
%\noindent{\em Proof}.
%Since $\theta$ is real, $\zeta$ is a root of the polynomial
%\begin{eqnarray*}
% (x-\zeta)(x-\bar \zeta) &=& (x-e^{i\theta})(x-e^{-i\theta}) \\
% && x^2 - 2 (\cos \theta) x + 1
%\end{eqnarray*}
%If $\cos \theta$ is rational, then this is the minimal polynomial of $\zeta$, and this must be the
% case if $\zeta$ is of degree 2, for then $\zeta^2 is some linear function of $\zeta$, say $\zeta^2 = p\zeta + q$,
% and we have $0 = \zeta^2 - 2(\cos \theta) \zeta + 1 = (p - 2\cos\theta) \zeta + q + 1$; and since $\zeta is not real
% we must have $p-2\cos \theta = 0$, so $\cos \theta$ is rational.
\section{Quadratic and non-quadratic tilings}
In this section we give a simple sufficient condition for a tiling to be quadratic.
\begin{lemma} \label{lemma:quadratic} Suppose tile $T$ strictly tiles triangle $ABC$. If the tile $T$ is similar to the
triangle $ABC$, and there are no angle relations, then the tiling is a quadratic tiling.
\end{lemma}
\noindent{\em Proof}.
Note that since there are no angle relations, the three angles $\alpha$, $\beta$,
and $\gamma$ are pairwise unequal: for example, if $\alpha = \beta$, then the relation $\alpha + \beta + \gamma = \pi$
implies $2\alpha + \gamma = \pi$, which is an angle relation.
Since $T$ is similar to triangle $ABC$, and angle $A$ is the smallest angle of $ABC$,
angle $A = \alpha$. Then consider the copy $T_1$ of the tile that shares vertex $A$. Its two sides
lie on the sides of triangle $ABC$. We can relabel the vertices $B$ and $C$ if necessary so that
the angle of $T_1$ at its vertex $P_1$ on side $AB$ is $\beta$, and its angle at its vertex $Q_1$ on side $AC$
is $\gamma$.
There must be exactly three copies of the tile meeting at $P_1$, and the three angles
at $P_1$ are (in some order) $\alpha$, $\beta$, and $\gamma$, because any other vertex behavior gives rise to an
angle relation. Let the tiles meeting at $P_1$ be $T_1$, $T_2$, and $T_3$, numbered so that $T_2$ and $T_1$ share
a side. That shared side is $a$, since it is opposite angle $A$ in $T_1$. Then $T_2$ does not have angle $\alpha$
at $P_1$, since the $\alpha$ vertex of $T_2$ has to be opposite side $P_1Q_1$. $T_2$ does not have angle $\beta$
at $P_1$, since $T_1$ has angle $\beta$ there, and only one $\beta$ can occur at $P_1$. Therefore $T_2$ has angle
$\gamma$ at $P_1$. Therefore $T_3$ has angle $\alpha$ at $P_1$. Since the tiling is strict, the angle of $T_3$
at its second vertex $P_2$ on side $AB$ must be $\beta$; otherwise the shared sides of $T_2$ and $T_3$ will have
different length, since the length of that side of $T_2$ is $b$. But now, we are in the same situation with $T_2$
as we originally were with $T_1$: the two angles along side $AB$ are $\alpha$ and $\beta$ (in that order). We
can argue as before that the three triangles $T_3$, $T_4$, and $T_5$ meeting at $P_2$ have angles $\beta$,$\gamma$,
and $\alpha$ at $P_2$, in that order. Continuing down side $AB$ in this fashion, we eventually reach a tile
$T_{2m-1}$ that has $B$ for a vertex; there will be $m$ copies of the tile sharing a side with $AB$; there will
be $m-1$ vertices $P_,\ldots, P_{m-1}$ along $AB$, each shared by three triangles; the number of tiles used is
$2m-1$. The third vertices of these triangles are points $Q_1,\ldots,Q_{m-1}$, lying on a line parallel to $AB$,
and the last point $Q_{m-1}$ lies on $BC$. The triangle $Q_1CQ_{m-1}$ is thus tiled by the restriction of the
original tiling to that triangle. This restricted tiling is still strict and has no angle relations. By
induction, we can assume that this restricted tiling is quadratic. Since it has $m-1$ tiles along side $Q_1Q_{m-1}$,
we have $(m-1)^2 = N-(2m-1)$. Then $N = (m-1)^2 + 2m -1 = m^2$. That completes the proof.
\smallskip
\noindent{\em Remark}.
The 5-tiling in Figure \ref{figure:5-tilings} has $T$ similar to $ABC$, but it has an angle relation $2\alpha + 2\beta = \pi$,
and it also has a non-strict vertex. It is natural to ask if the hypotheses of the lemma can be weakened by dropping
one or the other of the hypotheses. Does there exist a strict non-quadratic tiling in which $T$ is
similar to $ABC$? (Angle relations are OK.)
Does there exist a non-quadratic tiling with no angle relations in which $T$ is similar to $ABC$? (Non-strict vertices are
OK, but the angles meeting there would have to be exactly one each of $\alpha$, $\beta$, and $\gamma$.) We do not
know the answer to either of those questions. Note that
if $\gamma$ is a right angle we have an angle relation $2\gamma = \pi$.
\section{The d-matrix, and a related eigenvalue problem}
Let triangle $ABC$ be tiled by the tile $T$, whose sides are $a$, $b$, and $c$. Let the sides of $ABC$ be
$X$, $Y$, and $Z$. We assume the triangle is labeled so that angles $A$, $B$, and $C$ are listed in non-decreasing order;
hence also $X \le Y \le Z$. In case triangle $ABC$ is similar to the tile, this implies that angle $A = \alpha$,
angle $B = \beta$, and angle $C = \gamma$.
Each side $X$, $Y$, and $Z$ is a linear combination of $a$, $b$, and $c$, the coefficients specifying how many tiles
share sides of length $a$, $b$, and $c$ with $X$, $Y$, or $Z$. These nine numbers are the entries of the matrix
$\d$, such that
$$ \vector X Y Z = \d \vector a b c $$
If the triangle $ABC$ is similar to the tile, then we have
$$ \vector X Y Z = \sqrt N \vector a b c$$
because each side of $ABC$ must be $\sqrt N$ times the corresponding side of the tile $T$, in order that the
area of $ABC$ can be $N$ times the area of $T$. Therefore
$$ \d \vector a b c = \sqrt N \vector a b c.$$
That is, $\sqrt N$ is an eigenvalue of $\d$, and $(a,b,c)$ is an eigenvector for that eigenvalue.
If triangle $T$ is isosceles, then $\d$ is not (yet) uniquely defined. In that case we have either $a = b$ or
$b = c$; our convention is to ignore $b$, so that when $T$ is isosceles, the middle column of the $\d$ matrix is zero.
We will not make use of the $\d$ matrix when $T$ is equilateral, but for completeness, we define the $\d$ matrix
in that case to have non-zero entries only in the first column.
If $T$ is not isosceles, then the coefficients in the $\d$ matrix are integers between 0 and $N-1$, inclusive, assuming $N > 2$: Not all $N$ triangles
can share a side of triangle $ABC$, since if $N > 2$, there would be two adjacent vertices along that side at which
only two triangles meet; but then by Lemma \ref{lemma:rightangle}, the copy of the tile between those vertices
would have two right angles.
For example, consider the 5-tiling shown in Figure \ref{figure:5-tilings}. Here the shortest side of the large
triangle consists of one $c$, so the top row of the $\d$ matrix is $0\ 0 \ 1$. The middle side of the large triangle
consists of two $c$'s, so the middle row of the $\d$ matrix is $0\ 0\ 2$. The longest side of the large triangles
consists of one $a$ and two $b$'s, so the bottom row is $1\ 2\ 0$. Thus the $\d$ matrix for this example is
$$ \matrix 0 0 1 0 0 2 1 2 0$$
and the eigenvalue equation is
$$ \matrix 0 0 1 0 0 2 1 2 0 \vector a b c = \sqrt 5 \vector a b c $$
In this example we have $\alpha = \pi/6$, $\beta = \pi/3$, and $\gamma = \pi/2$, so $a = \sin \pi/6 = 1/2$,
$b = \sin \pi/3 = \sqrt 3/2$, and $c = \sin \pi/2 = 1$. One can check the eigenvalue equation numerically with these values.
Note that the $\d$ matrix for a quadratic tiling is $\sqrt N$ times the identity. We conjecture that if $N$ is a perfect
square, say $m^2$, and $\d$ is $m$ times the identity, then the tiling is quadratic.
\section{Tilings with $T$ similar to $ABC$}
In this section, we assume triangle $ABC$ is $N$-tiled by triangle $T$ similar to $ABC$. In case $N$ is a square,
we have the quadratic tiling of $ABC$; in this section we assume $N$ is not a square.
Let the sides of $T$ be $a$, $b$, and $c$,
in non-decreasing order; these are opposite the angles $\alpha$, $\beta$, and $\gamma$ of $T$.
We start by disposing of a special case.
\begin{lemma} \label{lemma:equilateral}
Suppose $T$ and $ABC$ are both equilateral, and there is an $N$-tiling of $ABC$ by $T$. Then
$N$ is a square and the tiling is a quadratic tiling.
\end{lemma}
\noindent{\em Proof}. Since all the angles of $T$ and $ABC$ are equal, and all the sides of $T$ are equal, there is
only one way to place tile $T_1$ at vertex $B$. Along side $BC$ there must be a certain number $m$ of copies of $T$;
hence the side of $ABC$ is $mc$, where $c$ is the side of $X$. We prove by induction on $m$ that such a tiling
is a quadratic tiling using $m^2$ triangles. There are $m$ tiles that share sides with $BC$.
Call them $T_1,\ldots, T_m$. This sawtooth-like configuration requires the placement of $m-1$ copies of $T$, one
between each adjacent pair of triangles $T_1,\ldots,T_m$. Now we have identified a total of $2m-1$ triangles that participate
in the original tiling, and the remaining triangles tile the smaller equilateral triangle formed by deleting the tiles
identified so far from $ABC$. The base of this triangle is smaller than the original base $BC$ by $c$, the side of $T$.
By the induction hypothesis, the tiling of this triangle is quadratic, using $(m-1)^2$ tiles. Combining this with
the row of $2m-1$ triangles along $BC$, we have a quadratic tiling with a total of $(m-1)^2 + 2m-1 = m^2$ tiles,
completing the inductive proof.
Next we review the computation of eigenvectors by cofactors. To find an eigenvector
of the $\d$ matrix with eigenvalue $\sqrt N$, consider the matrix $X := \d- \sqrt{N} I$. An eigenvector
can be found by picking any row, and then arranging the cofactors of the elements of that row as a (column) vector.
If these cofactors do not all vanish, then the result is an eigenvector. (The reader may either verify this
or just check directly that the particular eigenvalues produced this way below are indeed eigenvectors.)
Now we take up the general case of a tiling $T$ with $ABC$ similar to $T$, when $N$ is not a square.
\begin{lemma} \label{lemma:diagonalzeroes}
Let triangle $ABC$ be $N$-tiled by tile $T$ similar to $ABC$,
and suppose $N$ is not a square. Then the diagonal entries of the $\d$ matrix are zero.
\end{lemma}
\noindent{\em Proof}.
Since the area of $ABC$ is $N$ times
the area of $T$, and $T$ is similar to $ABC$, the sides of $ABC$ are $\sqrt N$ times $a$, $b$, and $c$.
Then (as discussed in a previous section) we have the
eigenvalue equation
$$ \d \vector a b c = \sqrt{N} \vector a b c.$$
The characteristic polynomial $f(x)$ of the $\d$ matrix, the determinant of $\d - x I$, is a cubic polynomial with integer
coefficients, yet has for a zero the number $\sqrt N$. This is only possible if it factors into a quadratic factor and a
linear factor. Since $N$ is not a square, the quadratic factor must be a multiple of $\lambda^2-N$. The coefficient of
$x^3$ is $-1$, and so for some $q$ we have
$$ f(x) = (x^2-N)(q-x) $$
In general the coefficient of $x^2$ in the characteristic polynomial of any 3 by 3 matrix $\d$ is the trace of $\d$,
and the constant term is the determinant of $\d$. Hence $q$ is the trace of $\d$ and $-Nq$ is the determinant of $\d$.
Since the entries of $\d$ are non-negative integers, the trace is non-negative, so $q \ge 0$.
To avoid so many subscripts, we use separate letters for the entries in the $\d$-matrix, writing it as
$$ \d = \matrix p d e g m f h \ell r $$
Since the similarity factor between $ABC$ and $T$ is $\sqrt N$, there cannot be more than $\sqrt N$ tiles with
$a$ sides along $X$, the short side of $ABC$. That is, $p \le \sqrt N$. More formally, $a \sqrt N = X = pa + db + ec \ge pa$,
so $p \le \sqrt N$. Similarly $m \le \sqrt N$ and $r \le \sqrt N$. Since $N$ is not rational, we have strict
inequalities: $p < \sqrt N$, $r < \sqrt N$, and $r < \sqrt N$. It follows that $pm < N$, etc.
We also note that there is just one tile sharing vertex $A$, where $ABC$ has its $\alpha$ angle. That tile
must have its $b$ and $c$ sides along $AB$ and $AC$, or along $AC$ and $AB$, we don't know which. Thus either
$f$ and $\ell$ are nonzero, or $m$ and $r$ are nonzero.
Suppose, for proof by contradiction, that $q$, the trace of $\d$, is not zero. Then $q = p + m + r > 0$.
Since the three eigenvalues are distinct (because $q$ is rational and $\sqrt N$ is not), the eigenspace corresponding
to $\sqrt N$ is one-dimensional. The eigenvalue equation is
$$ (\d - \lambda I) \vector u v w = 0$$
or showing the coefficients
$$ \matrix { p - \sqrt N} d e g {m-\sqrt N} f h \ell {r - \sqrt N} \vector u v w = 0 $$
We claim that there exists an eigenvector $(u,v,w)$ whose components lie in
$\Q(\sqrt N)$. To prove this we will use the cofactor method described above.
The resulting eigenvector is $(u,v,w)$, provided all three components are nonzero, where
\begin{eqnarray*}
u &=& \dettwo d e {m-\sqrt N} f \\
v &=& -\dettwo {p-\sqrt N} e g f \\
w &=& \dettwo {p-\sqrt N} d g {m-\sqrt N}
\end{eqnarray*}
Although we have not given a proof of the cofactor method's correctness, one can
easily verify directly that the exhibited vector is indeed an eigenvector for $\sqrt N$; this
also provides a check that no algebraic mistake has been made. The fact that all three cofactors are nonzero
is really only needed to conclude directly that the eigenspace of $(u,v,w)$ is one-dimensional; but we
know that directly in our case since the eigenvalues $\sqrt N$, $-\sqrt N$, and $q$ are distinct. It therefore
suffices to check that one of the cofactors $u$, $v$, $w$ is nonzero; then the others must automatically be nonzero
because $(u,v,w)$ is a nonzero multiple of $(a,b,c)$. But we give direct proofs that all three cofactors are nonzero
anyway, as it takes only one more paragraph.
We have $u = df-em + e\sqrt N$. If $u=0$ then $e=0$ and hence $df = 0$. If $v=0$ then similarly $f=0$ and $eg=0$.
Finally if $w=0$ then $p+m=0$ and hence $p=m=0$, so $N = dg$.
Assume, for proof by contradiction, that $w=0$ . Then
\begin{eqnarray*}
(m - \sqrt N)(p - \sqrt N) &=& dg \\
mp + N -\sqrt N(p + m) &=& dg
\end{eqnarray*}
Since $\sqrt N$ is irrational this means $p+m=0$, and since $p$ and $m$ are nonnegative, that implies $p=0$
and $m=0$. Hence $dg = N$. But $d \le (a/b) \sqrt N$, with equality implying that $p=d=0$,
and $g \le (b/a)\sqrt N$ with equality implying $m=f=0$. Since $dg = N$, equality must hold in both
inequalities. Hence $d = (a/b) \sqrt N$ and $g = (b/a) \sqrt N$
and $p=e=m=f=0$. But we showed above that either $m$ and $r$ are both nonzero, or $f$ and $\ell$ are both
nonzero. That is now contradicted by $m=f=0$. This contradiction shows that $w \neq 0$.
Next we give the proof that $u \neq 0$; as remarked above, this is technically superfluous, but still it is
interesting because the proof we give is not simply an abstract argument about projecting onto the one-dimensional eigenspace.
Assume, for proof by contradiction, that $u=0$. Then $df-em + e \sqrt N$ = 0. Since
$\sqrt N$ is not rational, $e=0$ and $df-em=0$. Then $df=0$, so $d=0$ or $f=0$. If $d=0$,
then since both $d$ and $e$ are zero, side $X$ of triangle $ABC$ is composed of all $a$ sides
and $X = (p - \sqrt N)a$. But since $\sqrt N$ is the similarity factor between $T$ and $ABC$, we have
$X = \sqrt N a$. Hence $p -\sqrt N = \sqrt N$. Hence $\sqrt N = p/2$, so $N^2 = p^2/4$. Hence $4N^2 = p^2$
and $p$ is even, so $N$ is a square, contradiction. This contradiction proves $d \neq 0$. Since
$d=0$ or $f=0$, we have $f=0$. Now assume, for proof by contradiction, that $g = 0$. Then since $f=0$,
side $Y$ is composed entirely of $b$ sides of tiles, so $Y = mb$. But $Y = b \sqrt N$ since $\sqrt N$ is
the similarity factor between $T$ and $ABC$. Hence $m = \sqrt N$, contradiction. That proves $g \neq 0$.
As shown above, either $f$ and $\ell$ are both nonzero or $m$ and $r$ are both nonzero. But $f=0$. Hence
both $m$ and $r$ are nonzero. Now
\begin{eqnarray*}
X &=& pa + db \mbox{\qquad since $e=0$}\\
X &=& \sqrt N a \\
(\sqrt N - p) a &=& db \\
\frac b a &=& \frac {\sqrt N - p} d \\
Y &=& \sqrt N b \\
Y &=& ga + m b \mbox{\qquad since $f=0$}\\
\frac b a &=& \frac g {\sqrt N-m} \\
\frac ba &=& \frac {\sqrt N -p} d \ = \ \frac g {\sqrt N-m}
\end{eqnarray*}
Cross-multiplying we have
\begin{eqnarray*}
dg &=& N-(m+p)\sqrt N + mp \\
m+p &=& 0 \mbox{\qquad as the coefficient of $\sqrt N$ must be zero}\\
m &=& p \ = \ 0 \mbox{\qquad as $m$ and $p$ are nonnegative}
\end{eqnarray*}
But $m$ was proved above to be nonzero. This contradiction completes the proof that $u \neq 0$.
Now assume $v=0$. Then
$$ pf - eg +f \sqrt N = 0$$
Since $N$ is irrational we have $f=0$ and $pf = eg$, but since $f=0$ we have $eg=0$.
Hence either $e=0$ or $g=0$. Assume, for proof by contradiction, that $g=0$. Then the
middle side of $ABC$ (corresponding to the middle row) is equal to $mb$ but also to $b\sqrt N$,
so $m = \sqrt N$, contradiction. This contradiction proves $g \neq 0$. Hence $e=0$. Since either
$f$ and $\ell$ are both nonzero or $m$ and $r$ are both nonzero, and we have proved $f=0$, then
$m$ and $r$ are both nonzero. Now that we have $e=0=f$, and $m \neq 0$, we reach a contradiction
by the same computation as in the case $u=0$, shown in the series of displayed equations above.
Hence $v \neq 0$.
Thus none of the three cofactors is zero.
That completes the proof that there is an eigenvector $(u,v,w)$ for the eigenvalue $\sqrt N$
with components in $\Q(\sqrt N)$.
Since the eigenspace is one-dimensional, this eigenvector is a (not necessarily rational) multiple of $(a,b,c)$.
Recall that the third eigenvalue of the $\d$ matrix is the trace $q = p + m + r$. We can use the cofactor
method to find an eigenvector for this eigenvalue as well, namely
\begin{eqnarray*}
V &=& \bigg( \dettwo d e {m-q} f , - \dettwo {p-q} e g f, \dettwo {p-q} d g {m-q} \bigg) \\
&=& \vector {df-em + e(p+m+r)}
{-pf + eg + f(p+m+r)}
{pm-dg - (m+p)(m+p+r) + (m+p+r)^2} \\
&=& \vector {df+e(p+r)} {eg + f(m+r)} { pm-dg + r(m+p+r)}
\end{eqnarray*}
Technically, it is not an eigenvector until we prove that the components are not zero, but we do not need that
right now; it suffices that it satisfy the eigenvalue equation.
The eigenvalue equation $\d V = (p+m+r) V$ is
\begin{eqnarray}
\hskip-0.12in \matrix p d e g m f h \ell r \vector {df-e(p+r)} {eg + f(m+r)} { pm-dg + r(m+p+r)} \hskip-0.12in &=& \hskip-0.1in (p + m + r) \vector {df-e(p+r)} {eg + f(m+r)} { pm-dg + r(m+p+r)} \label{eq:secondeigenvalue}
\end{eqnarray}
The first component of this vector equation is
\begin{eqnarray*}
p(df - e(p+r)) + d(eg+f(m+r)) + e(pm - dg + r(m+p+r) &=& (p+m+r) (df-e(p+r))
\end{eqnarray*}
Multiplying out and cancelling like terms, and dividing by 2, we find
$$ epm + er(m+p+r) = 0.$$
We argue by cases, according to whether $e=0$ or not. We first take up
the case that $e \neq 0$.
Then $pm + r(m+p+r) = 0$. Since these terms are nonnegative, they are both zero. Hence $pm = 0$ and $r(m+p+r) = 0$.
Hence $r = 0$ or $m+p+r$ = 0. In either case $r=0$.
Writing out the third component of the eigenvalue equation, and setting $r=0$, we have
\begin{eqnarray*}
h(df-ep) + \ell (eg+fm) &=& (p+m)(pm-dg) \\
h(df-ep) + \ell (eg+fm) &=& -(p+m)dg \mbox{\qquad since $pm = 0$} \\
hdf - hep + \ell eg + \ell fm &=& -pdg - m dg \\
\end{eqnarray*}
%Recall that the determinant of $\d$ is $-N(p+m+r)$.
%Expanding the determinant of $\d$ by cofactors on the bottom row, and setting $r=0$, we have
%\begin{eqnarray*}
%-N(p+m) &=& h\dettwo p d g m - \ell \dettwo p e g f + r \dettwo p d g m \\
%&=& h(pm-dg) - \ell (pf - eg) \\
%&=& hpm - hdg - \ell pf + \ell eg \\
%&=& -hdg - \ell pf + \ell eg \mbox{\qquad since $mp = 0$} \\
%N(p+m) &=& hdg + \ell pf - \ell eg
%\end{eqnarray*}
Now we write out the second component of the eigenvalue equation (\ref{eq:secondeigenvalue}), setting $r=0$:
\begin{eqnarray*}
g(df-ep) + m(eg + fm) + f(pm -dg) &=& (p+m)(eg + fm) \\
gdf - gep + m(eg + fm) + fpm - fdg &=& peg + pfm + m(eg + fm) \\
peg&=& 0 \\
pg &=& 0 \mbox{\qquad since $e \neq 0$ }
\end{eqnarray*}
Assume, for proof by contradiction, that $m \neq 0$. Then since $mp = 0$ we have $p=0$.
The equation $N(p+m) = hdg + \ell pf + \ell eg$ becomes
$$Nm = hdg + \ell eg.$$
The third component of the eigenvalue equation becomes, with $p=0$,
\begin{eqnarray*}
hdf + \ell eg + \ell fm &=& - mdg
\end{eqnarray*}
The left side is $\ge 0$ and the right side is $\le 0$. Hence both sides are equal to zero.
Since $m \neq 0$ and $e \neq 0$, we have $dg = 0$ and $hdf = 0$ and $\ell g$ = 0 and $\ell f = 0$.
We derived above (by observing that the $b$ and $c$ sides of the tile at vertex $A$ lie on the two adjacent
sides of $ABC$) that either $f$ and $\ell$ are both nonzero or $m$ and $r$ are both nonzero. Since $r = 0$
we must have $f$ and $\ell$ both nonzero. Hence $\ell f = 0$ is a contradiction.
That contradiction completes the proof that $m=0$.
Now assume, for proof by contradiction, that $p \neq 0$.
Then since $pg = 0$ we have $g=0$. Then the equation $N(p+m) = hdg + \ell p f + \ell eg$ becomes
$Np = \ell p f$. Canceling $p$ we have $N = \ell f$. But as proved above, $\ell f \le N$,
and equality holds if and only if $AC$ is composed only of $c$ sides of tiles and $AB$ is composed
only of $b$ sides. Therefore we have $h=0$ as well as $g=m=r=0$.
Then since $h=0$ and $r=0$, the long side $AB$ of $ABC$ is composed entirely of $b$ sides of tiles.
If $T$ is isosceles, then by convention the middle column of the $\d$-matrix is zero. Since $\ell f = N$,
we now have $\ell \neq 0$, so the middle column is not zero, and $T$ is not isoceles.
Since side $AB$ is composed entirely of $b$ sides of tiles, there are equally spaced vertices $V_0 = A, V_1,\ldots, V_\ell$,
spaced $b$ apart, each one of which is one side of a tile $T_i$. Tile $T_1$, which has vertices at $A$ and $V_1$, has
its $\alpha$ angle at $V_0$. All these tiles have their $\beta$ angles in the interior of $ABC$, and their $\alpha$
and $\gamma$ angles at the $V_i$. If $\gamma > \pi/2$ then there is only one possible orientation for these tiles,
as two $\gamma$ angles will not fit at any $V_i$. In that case the angle of the last tile at vertex $B$ must be
$\gamma$, contradiction, since the angle there cannot exceed $\beta$, and $\beta \neq \gamma$ since then $T$ would
be isosceles. Hence $\gamma \le \pi/2$.
In particular, the tile that shares vertex $B$ has its $b$ side along $AB$.
Therefore the tile sharing vertex $B$ and part of side $AB$
has its $\alpha$ angle at $B$, and the angle $\beta$ at vertex $B$ splits into some number of $\alpha$ angles,
so for some number $J$, we have $\beta = J \alpha$. Somewhere along $AB$ there must occur a vertex $V_k$
at which both the tile $T_k$ and the tile $T_{k+1}$ have angle $\gamma$. There is not room at $V_k$ for
a third tile, since $2\gamma + \alpha > \alpha + \beta + \gamma = \pi$. Hence there are exactly those two
tiles at $V_k$, and we have $\gamma = \pi/2$.
Since $\gamma$ is a right angle, we must have $a^2 + b^2 = c^2$. Since $(u,v,w)$ is a multiple of $(a,b,c)$ we
also have $u^2 + v^2 = w^2$. We now compute these expressions from the formulas for $(u,v,w)$. In view of $m=g=0$ we have
\begin{eqnarray*}
u &=& ef - e\sqrt N \\
v &=& f\sqrt N - fp \\
w &=& N - p \sqrt N
\end{eqnarray*}
Squaring these equations we have
\begin{eqnarray*}
u^2 &=& e^2(f^2 + N - 2f\sqrt N)\\
v^2 &=& f^2(N+p^2 - 2p\sqrt N) \\
w^2 &=& N^2 +p^2 N - 2p N \sqrt N
\end{eqnarray*}
Setting $u^2 + v^2 = w^2$ we find
\begin{eqnarray*}
e^2 f^2 + e^2 N + f^2 N + f^2 p^2 - 2(e^2f- f^2p) \sqrt N &=& N^2 + p^2 N - 2pN\sqrt N \\
\end{eqnarray*}
Equating the coefficients of $\sqrt N$ and equating the rational parts, we have
\begin{eqnarray*}
e^2 f - f^2 p &=& pN \\
e^2f^2 + e^2 N + f^2 N + f^2 p^2 &=& N^2 + p^2 N
\end{eqnarray*}
Since $\gamma$ is a right angle, $\alpha + \beta = \pi/2$. Since $\beta = J\alpha$, we have
$\alpha = \pi/(2(J+1))$, so $\alpha$ is a rational multiple of 2. We have
$\tan \alpha = b/a = v/u$, which belongs to $\Q(\sqrt N)$. We have
$$ \cos \alpha = \frac u {u^2 + v^2}$$
which is also in $\Q(\sqrt N)$. Similarly $\sin \alpha$ belongs to $\Q(\sqrt N)$. Then $\zeta = d^{i\alpha}$
is of degree 4 over $\Q$, since $\Q(\zeta) = \Q(i,\sqrt N)$. By Lemma \ref{lemma:euler}, $4(J+1)$ is 5, 8, 10,
or 12. Since $5$ and $10$ are not divisible by 4, we have $4(J+1) = 8$ or $10$. But if $4(J+1) = 8$ then
$J=1$, while we have $J \ge 2$ since $\beta = J\alpha$. The only remaining possibility is $4(J+1) = 12$,
which makes $J = 2$. Then $\alpha = \pi/6$ and $2\beta = \alpha$, so $\beta = \pi/3$.
Then $a = \sin \alpha = 1/2$, $b = \sqrt 3 / 2$ and $c = 1$. But now $\overline{AC} = fc = f$,
and
\begin{eqnarray*}
\ell b &=&\overline{AB}\\
&=& \frac 2 {\sqrt 3} \overline{AC}\\
&=& \frac 2 {\sqrt 3} fc
\end{eqnarray*}
Now we put in $c = 1$ and $b = \sqrt 3 / 2$:
\begin{eqnarray*}
\ell \frac {\sqrt 3} 2 &=& \frac 2 {\sqrt 3} f \\
3 \ell &=& 4f
\end{eqnarray*}
We have $N=\ell f = (4/3)f^2$, so $3N = 4f^2$, so $f$ is divisible by $3$, say $f = 3k$; then $N = 3(2k)^2$
is three times a square.%\
\footnote{
Tilings of this kind actually exist, but with $e=0$ (we are now in the case $e \neq 0$).
(In the next draft I will add some figures to illustrate it.)
}
It remains to show that $p=0$; in fact we claim $p=0$ and $e=0$, so side $AC$ is
also composed entirely of $b$ sides of triangles. We have
\begin{eqnarray*}
\overline{BC} &=& \frac 1 2 \overline{AB} \\
&=& \frac 1 2 \ell b \\
&=& \frac {\ell \sqrt 3} 4 \\
&=& pa + db + ec
\end{eqnarray*}
Now we put in the values $a = 1/2$, $b = \sqrt 3 /2$, and $c = 1$.
\begin{eqnarray*}
\frac {\ell \sqrt 3} 4 &=& \frac p 2 + d \frac {\sqrt 3} 2 + e
\end{eqnarray*}
This is an equation in $\Q(\sqrt 3)$. Equating the rational parts we have $0 = p/2 + e$. Since
both $p/2$ and $e$ are nonnegative, we have $p=0$ and $e=0$, as claimed. In particular $p=0$ so the
diagonal elements are nonzero, which is the conclusion of the theorem; or we could say, in particular $e=0$,
contradicting the assumption $e\neq 0$ and completing the analysis of that case.
Therefore we may now assume $e = 0$. Remember that $r=0$ was derived only under the assumption $e \neq 0$,
so the equation $r=0$ is no longer in force.
The third component of the eigenvalue equation (\ref{eq:secondeigenvalue}) is (substituting $e=0$)
\begin{eqnarray*}
hdf + \ell f(m+r) + r(pm-dg+r(m+p+r)) &=& (p+m+r)(pm + r(m+p+r))
\end{eqnarray*}
Subtracting $r^2(m+p+r)$ from both sides we have
\begin{eqnarray}
hdf + \ell fm + \ell fr + rpm -rdg &=& (p+m+r)pm + (p+m)r(m+p+r) \nonumber \\
hdf + \ell fm + \ell fr + rpm - rdg &=& (p+m+r)(pm + pr + mr) \label{eq:thirdcomponent8}
\end{eqnarray}
To get rid of $h$ and $\ell$, we expand the determinant of the $\d$ matrix by cofactors on the bottom row.
That determinant is $-Nq = -N(p+m+r)$, so we have (remembering $e=0$)
\begin{eqnarray*}
-N(p+m+r) &=& h\dettwo p d g m - \ell \dettwo p e g f + r \dettwo p d g m \\
&=& hdf + \ell pf + rpm - rdg
\end{eqnarray*}
Adding and subtracting $\ell pf$ to the left side of (\ref{eq:thirdcomponent8}) the expression for
the determinant appears, and we have
\begin{eqnarray*}
hdf + \ell pf +rpm - rdg + \ell fm + \ell fr - \ell pf &=& (p+m+r)(pm + pr + mr) \\
-N(p+m+r) + \ell fm + \ell fr - \ell pf &=& (p+m+r)(pm +pr +mr)
\end{eqnarray*}
Moving everything to the right side we have
\begin{eqnarray*}
0 &=& (p+m+r)(pm +pr + mr) + (N-\ell f)(m+r) + (N+\ell f) p
\end{eqnarray*}
Since $\ell$ is the number of $b$ sides of tiles on the long side $\sqrt N c$ of $ABC$, we have
$\ell b \le \sqrt N c$, or $\ell \le (c/b) \sqrt N$. Since $f$ is the number of $c$ sides of tiles on $AC$,
whose length is $\sqrt N b$, we have $fc \le \sqrt N b$, or $f \le (b/c) \sqrt N$. Hence
\begin{eqnarray*}
\ell f &\le& \bigg( \frac c b \sqrt N \bigg)\bigg( \frac b c \sqrt N\bigg) \\
\ell f &\le& N
\end{eqnarray*}
Hence all the terms on the right of the previous equation are nonnegative. Hence each of them is zero.
In particular $(N+\ell f)p = 0$; but $N + \ell f > 0$, so $p=0$. Then then equation becomes
$$ (m+r)mr + (N-\ell f)(m+r) = 0$$
If $m+r = 0$ then $m=0=r$ and the lemma is proved. Hence we may assume $mr = 0 $ and $N = \ell f$.
But if $N=\ell f$ then we must have equality in the two inequalities $\ell \le (c/b) \sqrt N$ and $f \le (b/c) \sqrt N$.
This implies that side $AC$ is composed only of $c$ sides of tiles and side $AB$ is composed only of $b$ sides of tiles,
so $g=m=h=r=0$. In particular $m=r=0$. That completes the proof of the lemma.
\medskip
We pause to observe that the $\d$ matrix for a biquadratic tiling, in case $N = m^2 + n^2$, has the form
\begin{eqnarray*}
\d &=& \matrix 0 0 n 0 0 m n m 0
\end{eqnarray*}
which does satisfy the conditions above (as it must).
The hypothesis that $N$ is not a square is necessary, as shown by the 9-tiling in Figure 7. Its $\d$ matrix is
$$ \matrix 1 1 0 2 2 0 0 0 3$$
and as predicted, the determinant is zero, but the trace is not zero, and the characteristic polynomial is $-x(x-3)^2$.
Continuing with the general case of $N$ not a square,
some further conclusions can be drawn about the $\d$ matrix. We have shown that
$p=m=r= 0$. The determinant is then given by
$$ {\rm det\ } \d = dfh + eg\ell$$
Since the matrix entries are nonnegative, that means that each of these two terms must contain a zero factor.
In particular, at most four entries in the $\d$ matrix are nonzero.
The negated coefficient of $\lambda$ in the characteristic equation is (since the diagonal elements are zero)
the sum of paired products of off-diagonal elements:
\begin{eqnarray} \label{eq:lambdacoef}
N &=& dg + eh + fl
\end{eqnarray}
But at least one of these three terms will be zero, as shown above.
In view of the lemma, the $\d$ matrix becomes
\begin{equation} \label{eq:dletters}
\d = \matrix 0 d e g 0 f h \ell 0
\end{equation}
and the matrix equation
$$ \d \vector a b c =\sqrt N \vector a b c $$
becomes the three equations
\begin{eqnarray*}
db + ec &=& \sqrt N a \\
ga + fc &=& \sqrt N b \\
ha + \ell b &=& \sqrt N c
\end{eqnarray*}
\begin{lemma} \label{lemma:penultimate}
Suppose $ABC$ is $N$-tiled by tile $T$ similar to $ABC$, and $N$ is not a square. Then
$\gamma$ is a right angle.
\end{lemma}
\noindent{\em Proof}. First we note that $T$ and $ABC$ are not equilateral, by Lemma \ref{lemma:equilateral}.
Next we will prove that $T$ and $ABC$ are not isosceles with $\beta = \gamma$.
Assume, for proof by contradiction, that $\beta=\gamma$.
Then, by our definition of the $\d$ matrix,
the middle column of the $\d$ matrix is zero, i.e. $b$ is counted as $c$. Then we have $d=\ell=0$ and
\begin{equation}
\d = \matrix 0 0 e g 0 f h 0 0
\end{equation}
That implies that the short side $BC$ of triangle $ABC$ has only $c$ sides of tiles on it, and the long side $AB$ has
only $a$ sides of tiles on it. At the vertex $A$, there can only be one tile, since the angle at $A$ is the smallest
angle $\alpha$ so there can be no vertex splitting. This tile has one side of length $a$ opposite angle $A$ and another
along side $AB$. Hence $a=b$. Since $T$ is not equilateral, we must have $b < c$ and $\beta < \gamma$. This contradicts
the assumption that $\beta$ is not less than $\gamma$, and thus completes the proof by contradiction that $\beta < \gamma$.
Since the $\d$ matrix has zeroes on the diagonal, no $c$ sides of tiles occur along the longest side $AB$ of triangle $ABC$;
only $a$ and $b$ sides occur there. There are $\ell + h $
tiles along $AB$; for simplicity of notation, let $k := \ell + h$ and number those tiles $T_1,\ldots,T_k$ starting at
vertex $A$. Let $V_1,\ldots, V_{k-1}$ be the vertices of those tiles on $AB$. Tile $T_1$ must have its $\alpha$ angle at vertex $A$, and since its $c$ side is not on $AB$, it must have its $\gamma$ angle at vertex $V_1$.
Tile $T_k$ cannot have a $\gamma $ angle at $B$, since $\beta < \gamma$. Hence it
has either its $\alpha$ or its $\beta$ angle at $B$, and
its $\gamma$ angle at $V_{k-1}$ (since its $c$ side does not lie on $AB$).
Continuing towards $C$ from $B$ with $T_2, T_3,\ldots$,
and continuing towards $B$ from $C$ similarly, we must encounter an index $j$ between 1 and $k-1$ such that
$T_j$ and $T_{j+1}$ both have their $\gamma$ angles at $V_j$. At that point we know $\gamma \le \pi/2$.
Assume, for proof by contradiction, that $\gamma \neq \pi/2$. Then there is at least one
additional copy $T^\prime$ of the tile between $T_j$ and $T_{j+1}$, sharing vertex $V_j$, by Lemma \ref{lemma:rightangle}. If $T^\prime$ has
its $\gamma$ angle at $V_j$ then there are exactly those three tiles meeting at $V_j$ (else $\gamma < \pi/3$) and we have
$\gamma = \pi/3$, and hence $T$ is equilateral, which as noted above is impossible.
Hence none of the additional tiles $T^\prime$ meeting at $V_j$ have a $\gamma$ angle at $V_j$.
None of the tiles $T^\prime$ can contribute a $\beta$ angle at $V_j$ either, since $2\gamma + \beta > \pi$.
Hence there is an angle relation $2 \gamma + p \alpha = \pi$, where $p$ additional tiles contribute $\alpha$ each to the angle
sum at $V_j$, and $p > 0$. But $2 \gamma + p \alpha > \gamma + \beta + \alpha = \pi$, since $\gamma > \beta$. This contradiction
completes the proof of the lemma.
\begin{lemma} \label{lemma:fnonzero}
Suppose $ABC$ is $N$-tiled by tile $T$ similar to $ABC$, and $N$ is not a square. Then $f = \d_{12}$ is
not zero.
\end{lemma}
\noindent{\em Proof}. Suppose, for proof by contradiction, that $f=0$. Then the middle row of the $\d$ matrix
is $(g,0,0)$, which means that all the tiles along side $AC$ of triangle $ABC$ share their $a$ sides with $AC$.
At vertex $A$, where $ABC$ has its smallest angle $\alpha$, there is exactly one tile $T_1$, with angle $\alpha$ at $a$.
Hence both the side of $T_1$ opposite that angle, and the side shared with $AC$, are equal to $a$. Thus $T$ is isosceles.
In that case, by convention we have agreed to write the $\d$ matrix with zeroes in the second column, so
the $\d$ matrix has the form
\begin{equation}
\d = \matrix 0 0 e g 0 0 h 0 0
\end{equation}
Now the bottom row is $(h,0,0)$, which means that all the tiles along side $AB$ share their $a$ sides with $AB$.
In particular the tile at vertex $A$ has an $a$ side along $AB$. But we have already seen that its other two sides
are $a$. Hence the tile is equilateral, contradicting Lemma \ref{lemma:equilateral}, since $N$ is not a square. That completes
the proof.
There are six letters for coefficients in the $\d$ matrix, but for any specific tiling, at most four of those
coefficients are nonzero. We will analyze some special cases. The case corresponding to the biquadratic
tilings is $d=0$ and $g=0$. We call that the ``biquadratic case''. In the biquadratic case the $\d$ matrix
has the form
\begin{eqnarray} \label{eq:dsimple}
\d = \matrix 0 0 e 0 0 f h \ell 0
\end{eqnarray}
Equation (\ref{eq:lambdacoef}) now becomes
\begin{eqnarray} \label{eq:lambdacoef2}
eh + \ell f &=& N
\end{eqnarray}
We compute the eigenvector in the biquadratic case, using the cofactor method described above. Let
$$ X = \matrix {-\sqrt N} 0 e 0 {-\sqrt N} f h \ell {-\sqrt N} $$
Taking the cofactors of the bottom row (notice the minus sign in the second component, which comes from the
definition of ``cofactor'') we find the eigenvector
\begin{eqnarray*}
\bigg( \dettwo 0 e {-\sqrt N} f,
-\dettwo {-\sqrt N} e 0 f,
\dettwo {-\sqrt N} 0 0 {- \sqrt N} \bigg)
&=& \vector { e \sqrt N} { f \sqrt N} N
\end{eqnarray*}
Note that $e \neq 0$ and $f \neq 0$, since the first two sides of $ABC$ are given by $ec$ and $fc$.
Hence the cofactors do not vanish.
We claim that the bottom two rows of $\d - \sqrt N I$, namely $(0,-\sqrt N,f)$ and $(h,\ell,-\sqrt N)$, are linearly independent.
Indeed, suppose that for some constants $p$ and $q$ we have $p(0,-\sqrt N, f) + q(h,\ell,-\sqrt N) = 0$. From the first
component we see that $qh = 0$. From the third component we see that $pf = q \sqrt N$.
If $q$ is not zero, then $\sqrt N = pf/q$, contradicting the irrationality of $\sqrt N$. Hence $q = 0$.
Hence from the second component, $p \sqrt N = 0$. Hence $p=0$.
This proves that the bottom two rows of $\d - \sqrt N I$ are linearly independent. Hence $\d - \sqrt N I$ has rank 2;
hence the eigenspace associated with the eigenvalue $\sqrt N$ is one-dimensional. It follows that the eigenvector computed above
is a multiple of $(a,b,c)$. That is, for some constant $\mu$ we have
\begin{eqnarray} \label{eq:eigenvector-biquadratic}
\vector a b c &=& \mu \vector { e \sqrt N} { f \sqrt N} N
\end{eqnarray}
The constant $\mu$ is an arbitrary scale factor; changing $\mu$ just changes the size of the tile $T$ and the triangle $ABC$
by the same factor. We are therefore free to choose $\mu$ to suit our convenience. We choose to take $\mu = \sqrt N$; then we have
\begin{eqnarray} \label{eq:abcvalues}
\vector a b c &=& \vector e f {\sqrt N}
\end{eqnarray}
\begin{lemma} \label{lemma:gammaright-sumofsquares}
Let triangle $ABC$ be $N$-tiled by $T$, and suppose $N$ is not a square and $T$ is similar to $ABC$, and
$d=g=0$ (the biquadratic case). Then $N$ is a sum of squares, specifically $N = e^2 + f^2$ where $e$ and
$f$ are as above, and $\tan \alpha = e/f$. In particular $\tan \alpha$ is rational.
\end{lemma}
\noindent{\em Proof}.
By Lemma \ref{lemma:penultimate}, we have $\gamma = \pi/2$. By the Pythagorean theorem and (\ref{eq:abcvalues})
we see that $\gamma = \pi/2$ if and only if $e^2 + f^2 = N$. Since $\gamma = \pi/2$, we have
$\tan \alpha = a/b$, and by (\ref{eq:eigenvector-biquadratic}), $\tan \alpha = e/f$. That completes the proof of the lemma.
\begin{lemma} \label{lemma:vertex-splitting-occurs}
Suppose $ABC$ is $N$-tiled by tile $T$ similar to $ABC$, and $N$ is not a square, and $d=g=0$ (the biquadratic case).
Then the right angle of $ABC$ is split by the tiling, and the tangents of the other angles of $ABC$ are rational.
\end{lemma}
\noindent{\em Proof}. We suppose, as always, that the $\gamma$ angle of $ABC$ is at $C$, the $\beta$ angle at $B$, and
the $\alpha$ angle at $A$.
Since the $\d$ matrix has the form given in (\ref{eq:dsimple}), all the tiles along side $BC$ share their $c$ sides with $BC$
(there are $e$ of them) and all the tiles along side $AC$ share their $c$ sides with $AC$ (there are $f$ of them). Suppose,
for proof by contradiction, that the
vertex at $C$ is not split. Then a single tile shares vertex $C$, so the tile has two $c$ sides, and hence is isosceles with $b=c$.
But by (\ref{eq:abcvalues}), we have $c/b = \sqrt N/e$. Hence if $b=c$ we have $N = e^2$, contradicting the hypothesis that
$N$ is not a square. Hence the vertex $C$ is split as claimed. The tangents of the other two angles are $e/f$
and $f/e$, which are rational. This completes the proof of the lemma.
We now turn to another important case, when $e=0$. We call this the ``triple-square case'', because it will
turn out that in this case $N$ must be three times a square. The following lemma and its proof give a complete analysis
of this case.
\begin{lemma} \label{lemma:triplesquare}
Suppose $ABC$ is not equilateral and is $N$-tiled by tile $T$ similar to $ABC$,
and $N$ is not a square, and $e=0$ (the triple-square case).
Then $\alpha = \pi/6$, $\beta = \pi/3$, and $N = 3d^2$ is three times a square.
\end{lemma}
\noindent{\em Remark}. There do exist tilings for each $N$ of the form $3d^2$ that fall under the triple square case,
as we showed in Figures 9 and 10.
\smallskip
\noindent{\em Proof}. Under the hypotheses of the lemma we have
$$ \d = \matrix 0 d 0 g 0 f h \ell 0.$$
In this matrix, $d$ and $h + \ell$ are not zero, since they represent the number of tiles along $BC$ and $AB$, respectively.
By Lemma \ref{lemma:fnonzero} we have $f \neq 0$. We have
$$ X = \d - \sqrt N I = \matrix {-\sqrt N} d 0 g {-\sqrt N} f h \ell {-\sqrt N} $$
We will prove that the bottom two rows of $X$ are linearly independent. If they are linearly
dependent, then for some $p$ and $q$, we have
\begin{eqnarray*}
0 &=& p(g, - \sqrt N,f) + q(h,\ell,\sqrt N) \\
&=& pg + pf + q\ell + qh + \sqrt N (q-p)
\end{eqnarray*}
Since $N$ is not a square, the coefficient of $\sqrt N$ is zero, so $q=p$, and
\begin{eqnarray*}
0 &=& pg + pf + q\ell + qh \\
&=& pg + pf + p\ell + ph \\
&=& p(g + f + \ell + h)
\end{eqnarray*}
Since the entries of the $\d$ matrix are non-negative, and $h+ \ell$ is strictly positive, we conclude $p = q = 0$.
That proves that the bottom two rows of $X$ are linearly independent, so $X$ has rank 2 and the eigenspace of $\sqrt N$
is one-dimensional.
We then compute the eigenvector by the cofactor method. Taking the cofactors of the bottom row, we
find the eigenvector
\begin{eqnarray*}
\bigg( \dettwo 0 e {-\sqrt N} f,
-\dettwo {-\sqrt N} 0 g f,
\dettwo {-\sqrt N} d g {- \sqrt N} \bigg)
&=& \vector { df} { f \sqrt N} {N-dg}
\end{eqnarray*}
Since $(a,b,c)$ is an eigenvector and
the eigenspace is one dimensional, $(a,b,c)$ is a multiple of this computed eigenvector. By scaling the triangle
appropriately we can assume $(a,b,c)$ is actually equal to the computed eigenvector:
$$ \vector a b c = \vector {df} {f\sqrt N} {N-dg}$$
It follows that $\sin \alpha = a/c = fd/(N-dg)$ is rational and $\tan \alpha = a/b = d/\sqrt N$ is of degree 2 over $\Q$.
According to the first row of the $\d$ matrix, the tiles along $BC$ have only $b$ sides on $BC$.
Assume, for proof by contradiction, that vertex $B$ does not split. Then there is a single tile $T_1$ at vertex $B$,
which therefore shares one side with $AB$ and one side with $BC$. Triangle $T$ is not isosceles, since then by
definition the $\d$ matrix would have zeroes in the middle column. Hence the unique $b$ side of $T_1$ must be
opposite angle $B$; but it must also lie on $BC$, which is a contradiction.
Hence vertex $B$ does split. Therefore for some integer $P$ we have $\beta = P \alpha$.
Since by Lemma \ref{lemma:penultimate}, $\gamma = \pi/2$,we have
\begin{eqnarray*}
\frac \pi 2 &=& \alpha + \beta \\
&=& \alpha + P \alpha \\
&=& (P+1)\alpha
\end{eqnarray*}
Therefore
$$ \alpha = \frac \pi {2(P+1)} = \frac {2\pi} {4(P+1)}.$$
By Lemma \ref{lemma:euler} we conclude that $4(P+1)$ is one of the numbers $n =3$, 4, 5, 8, 10, or 12 for which $\phi(n) = 4$.
Of these numbers, only 4, 1, and 12 are divisible by 4, which implies $P=2$, since the values $P=0$ and $P=1$ do not correspond
to vertex splitting. Hence $P=2$ and we have $\beta = 2\alpha$, so $\alpha + \beta = \pi/2 = 3\alpha$, and $\alpha = \pi/6$.
Hence $\tan \alpha = df/(f\sqrt N) = d/\sqrt N = 1\sqrt 3$. Hence
$N = 3d^2$.
That completes the proof of the lemma.
Now we have dealt with the biquadratic case (when $d=g=0$) and the triple-square case (when $e = 0$). It remains
to show that these are the only two possible cases, when $N$ is not a square and $T$ is similar to $ABC$. Recall that
$dfh = 0$ and $eg\ell = 0$ since $\det \d = 0$; that leaves only a few possibilities to consider. We begin by
showing that if $d =0$ then we are already in the biquadratic case.
\begin{lemma} \label{lemma:dandg}
Assume triangle $ABC$ is $N$-tiled by $T$, that $N$ is not a square, that $T$ is similar to $ABC$, and that
$d$ and $g$ are two entries in the $\d$ matrix of the tiling, in the notation used above (the ones that are zero
in the biquadratic case). Then $d = 0$ implies $g = 0$, i.e. we are in the biquadratic case as soon as $d=0$.
\end{lemma}
\noindent{\em Proof}. We have
$$ X = \d - \sqrt N I = \matrix {-\sqrt N} d e g {-\sqrt N} f h l {-\sqrt N} $$
By the cofactor method described above we compute the eigenvector
$$ \vector a b c = \vector { df + e \sqrt N} { f \sqrt N + e g} { N - dg} $$
Suppose, for proof by contradiction, that $d = 0$ and $g \neq 0$.
Then
$$ \vector a b c = \vector { e \sqrt N } { f \sqrt N + eg} N $$
By Lemma \ref{lemma:penultimate}, $\gamma$ is a right angle, so by the Pythagorean theorem, we have $a^2 + b^2 = c^2$. That is,
\begin{eqnarray*}
e^2 N + (f\sqrt N + eg)^2 &=& N^2 \\
e^2 N + f^2 N + 2egf \sqrt N + e^2 g^2 &=&N^2
\end{eqnarray*}
Since $N$ is not a square, the coefficient of $\sqrt N$ is zero; that is, $egf= 0$. By hypothesis, $g \neq 0$, so $ef = 0$.
The first row of the $\d$ matrix is $(0,d,e) = (0,0,e)$, so $e \neq 0$ because there must be some triangles on the first side of $ABC$.
Therefore $f = 0$. Then the $\d$ matrix is
$$ \matrix 0 0 e g 0 0 h \ell 0 $$
Hence all the tiles on the middle side $AC$ of $ABC$ have their $a$ side on $AC$, and all the tiles on the hypotenuse $AB$
do not have their $c$ side on $AB$. Consider the tile $T_1$ sharing vertex $A$ (there is only one, since $ABC$ has angle $\alpha$ there).
It has its $a$ side on $AC$ and does not have its $c$ side on $AB$. Hence its $b$ side is on $AB$ and its $c$ side opposite angle $A$,
which is $\alpha$. Hence $a=c$ and triangles $T$ and $ABC$ are equilateral, which is a contradiction since $\gamma = \pi/2$.
This contradiction shows that the assumption $d=0$ and $g \neq 0$ is untenable, which completes the proof of the lemma.
\begin{lemma} \label{lemma:notbiquadraticimpliestriplesquare}
Assume triangle $ABC$ is $N$-tiled by $T$, that $N$ is not a square, that $T$ is similar to $ABC$,
and that $d \neq 0$. Then $e=0$, i.e. we are in the triple-square case as soon as $d \neq 0$.
\end{lemma}
\noindent{\em Proof}. We have as in the proof of the previous lemma
$$ \d = \matrix 0 d e g 0 f h \ell 0$$
$$ X = \d - \sqrt N I = \matrix {-\sqrt N} d e g {-\sqrt N} f h \ell {-\sqrt N} $$
$$ \vector a b c = \vector {df + e \sqrt N} {eg + f \sqrt N} {N-dg} $$
By Lemma \ref{lemma:penultimate} and the Pythagorean theorem we have
\begin{eqnarray*}
0 &=& c^2 - a^2 - b^2 \\
&=& (N-dg)^2 - (df + e \sqrt N)^2 - (eg + f\sqrt N)^2 \\
&=& -2(def + egf) \sqrt N + \mbox{\ rational}
\end{eqnarray*}
Since $N$ is not a square and the entries of the $\d$ matrix are nonnegative integers, we have $def = 0$ and $egf = 0$.
Since $f \neq 0$ by Lemma \ref{lemma:fnonzero}, and $d \neq 0$ by hypothesis, we have $e = 0$.
That completes the proof of the lemma.
\section{The main theorem}
The following theorem completely answers the question, ``for which $N$ does there exist an $N$-tiling
in which the tile is similar to the tiled triangle?'' Namely, $N$ is a square, or a sum of two squares,
or three times a square.
\begin{theorem} \label{theorem:Similar}
Suppose $ABC$ is $N$-tiled by tile $T$ similar to $ABC$. Then either $N$ is a square, or $T$ is a right
triangle and exactly one of the following holds:
\smallskip
(i) $T$ is a right triangle, and $N$ is a sum of two squares, specifically
$N = e^2 + f^2$, where $\tan \alpha = e/f$, or
\smallskip
(ii) $N$ is
three times a square and $T$ is a 30-60-90 right triangle, i.e. $\alpha = \pi/6$.
\smallskip
\end{theorem}
\noindent{\em Proof}. Suppose $N$ is not a square. Then by Lemma \ref{lemma:penultimate}, $\gamma$ is a right angle.
Now consider the $\d$ matrix. By Lemma \ref{lemma:diagonalzeroes} the diagonal entries are zero, so
as stated in (\ref{eq:dletters}) the $\d$ matrix has the form
$$ \d = \matrix 0 d e g 0 f h \ell 0 $$
By Lemma \ref{lemma:dandg}, if $d=0$ then also $g=0$, i.e. we are in the
``biquadratic case''. Then by Lemma \ref{lemma:gammaright-sumofsquares}, $N = e^2 +f^2$ f and $\tan \alpha = e/f$.
If $e=0$ then by Lemma \ref{lemma:triplesquare}, $N$ is three times a square and $T$ is a 30-60-90
triangle. Finally, Lemma \ref{lemma:notbiquadraticimpliestriplesquare} shows that the cases $d=0$
and $e=0$ are exhaustive. That completes the proof that at least one of the two given alternatives holds.
We now will prove that $N$ cannot be both a sum of squares and three times a square,
since the equation $x^2 + y^2 = 3z^2$ has no integer solutions. To see that, we can assume without loss of
generality that $x$, $y$, and $z$ are not all even. Note that squares are
always congruent to 0 or 1 mod 4, so the left side is 0, 1, or 2 mod 4. Then $z^2$ must be congruent to
0 mod 4, since if not, the right side is congruent to 3 mod 4. Hence $z$ is even. But $x$ and $y$ must also be even to make the left side congruent to 0 mod 4, contradiction. Hence the equation has no solutions. Thus the alternatives in the
theorem are mutually exclusive, as claimed. That completes the proof of the theorem.
\section{Conjectures}
While the theorem characterizes the possible $N$ for which triangle $ABC$ can be $N$-tiled by tiles
similar to $ABC$, it does not completely characterize the possible tilings themselves.
Note that the 9-tiling in Figure 7 shows that not every $m^2$-tiling is a quadratic tiling, so we have not
classified all the $m^2$-tilings. Briefly we conjectured that a tiling in which the $\d$-matrix
is $m$ times the identity should be a quadratic tiling, but even that is not true. One can extend the
9-tiling in Figure 7 by adding more triangles to the right and below, producing a 25-tiling in which the
$\d$-matrix is 5 times the identity. We did show, in Lemma \ref{lemma:vertex-splitting-occurs}, that
if $N$ is not a square, then the right angle of $ABC$ is split by the tiling.
We conjecture that if $ABC$ is not a right triangle, then the only tilings of $ABC$ by a tile similar to $ABC$ are
the quadratic tilings. In case $ABC$ is a right triangle, the possibility arises that two tiles form a rectangle
with diagonal. One can then erase that diagonal and draw the other diagonal, creating another tiling. Similarly,
as in the 9-tiling in Figure 7, several tiles can form a square, and the square can be rotated. Call two tilings
``immediately equivalent'' if one is obtained from the other by re-drawing a diagonal or rotating a square. Let the
relation of ``equivalence'' between tilings be the
transitive closure of immediate equivalence. In other words, two tilings are equivalent if one can be obtained from
the other by a finite number of immediate equivalences. Then we conjecture that, for any triangle $ABC$, every
tiling of $ABC$ by tiles congruent to $ABC$ is equivalent to a quadratic tiling.
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\medskip
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