Calabi triangle

The Calabi triangle is a special triangle found by Eugenio Calabi and defined by its property of having three different placements for the largest square that it contains.^[1] It is an isosceles triangle which is obtuse with an irrational but algebraic ratio between the lengths of its sides and its base.^[2]

Consider the largest square that can be placed in an arbitrary triangle. It may be that such a square could be positioned in the triangle in more than one way. If the largest such square can be positioned in three different ways, then the triangle is either an equilateral triangle or the Calabi triangle.^[3][4] Thus, the Calabi triangle may be defined as a triangle that is not equilateral and has three placements for its largest square.

The triangle △ABC is isosceles which has the same length of sides as AB = AC. If the ratio of the base to either leg is x, we can set that AB = AC = 1, BC = x. Then we can consider the following three cases:

case 1) △ABC is acute triangle

The condition is ${\displaystyle 0<x<{\sqrt {2}}}$.

In this case x = 1 is valid for equilateral triangle.

case 2) △ABC is right triangle

The condition is ${\displaystyle x={\sqrt {2}}}$.

In this case no value is valid.

case 3) △ABC is obtuse triangle

The condition is ${\displaystyle {\sqrt {2}}<x<2}$.

In this case the Calabi triangle is valid for the largest positive root of ${\displaystyle 2x^{3}-2x^{2}-3x+2=0}$ at ${\displaystyle x=1.55138752454832039226...}$ (OEIS: A046095).

Example of answer

Consider the case of AB = AC = 1, BC = x. Then

${\displaystyle 0<x<2.}$

Let a base angle be θ and a square be □DEFG on base BC with its side length as a. Let H be the foot of the perpendicular drawn from the apex A to the base. Then

${\displaystyle {\begin{aligned}HB&=HC=\cos \theta ={\frac {x}{2}},\\AH&=\sin \theta ={\frac {x}{2}}\tan \theta ,\\0&<\theta <{\frac {\pi }{2}}.\end{aligned}}}$

Then HB = ⁠x/2⁠ and HE = ⁠a/2⁠, so EB = ⁠x - a/2⁠.

From △DEB ∽ △AHB,

${\displaystyle {\begin{aligned}&EB:DE=HB:AH\\&\Leftrightarrow {\bigg (}{\frac {x-a}{2}}{\bigg )}:a=\cos \theta :\sin \theta =1:\tan \theta \\&\Leftrightarrow a={\bigg (}{\frac {x-a}{2}}{\bigg )}\tan \theta \\&\Leftrightarrow a={\frac {x\tan \theta }{\tan \theta +2}}.\\\end{aligned}}}$

case 1) △ABC is acute triangle

[edit]

Let □IJKL be a square on side AC with its side length as b. From △ABC ∽ △IBJ,

${\displaystyle {\begin{aligned}&AB:IJ=BC:BJ\\&\Leftrightarrow 1:b=x:BJ\\&\Leftrightarrow BJ=bx.\end{aligned}}}$

From △JKC ∽ △AHC,

${\displaystyle {\begin{aligned}&JK:JC=AH:AC\\&\Leftrightarrow b:JC={\frac {x}{2}}\tan \theta :1\\&\Leftrightarrow JC={\frac {2b}{x\tan \theta }}.\end{aligned}}}$

Then

${\displaystyle {\begin{aligned}&x=BC=BJ+JC=bx+{\frac {2b}{x\tan \theta }}\\&\Leftrightarrow x=b{\frac {x^{2}\tan \theta +2}{x\tan \theta }}\\&\Leftrightarrow b={\frac {x^{2}\tan \theta }{x^{2}\tan \theta +2}}.\end{aligned}}}$

Therefore, if two squares are congruent,

${\displaystyle {\begin{aligned}&a=b\\&\Leftrightarrow {\frac {x\tan \theta }{\tan \theta +2}}={\frac {x^{2}\tan \theta }{x^{2}\tan \theta +2}}\\&\Leftrightarrow x\tan \theta \cdot (x^{2}\tan \theta +2)=x^{2}\tan \theta (\tan \theta +2)\\&\Leftrightarrow x\tan \theta \cdot (x(\tan \theta +2)-(x^{2}\tan \theta +2))=0\\&\Leftrightarrow x\tan \theta \cdot (x\tan \theta -2)\cdot (x-1)=0\\&\Leftrightarrow 2\sin \theta \cdot 2(\sin \theta -1)\cdot (x-1)=0.\end{aligned}}}$

In this case, ${\displaystyle {\frac {\pi }{4}}<\theta <{\frac {\pi }{2}},2\sin \theta \cdot 2(\sin \theta -1)\neq 0.}$

Therefore ${\displaystyle x=1}$, it means that △ABC is equilateral triangle.

case 2) △ABC is right triangle

[edit]

In this case, ${\displaystyle x={\sqrt {2}},\tan \theta =1}$, so ${\displaystyle a={\frac {\sqrt {2}}{3}},b={\frac {1}{2}}.}$

Then no value is valid.

case 3) △ABC is obtuse triangle

[edit]

Let □IJKA be a square on base AC with its side length as b.

From △AHC ∽ △JKC,

${\displaystyle {\begin{aligned}&AH:HC=JK:KC\\&\Leftrightarrow \sin \theta :\cos \theta =b:(1-b)\\&\Leftrightarrow b\cos \theta =(1-b)\sin \theta \\&\Leftrightarrow b=(1-b)\tan \theta \\&\Leftrightarrow b={\frac {\tan \theta }{1+\tan \theta }}.\end{aligned}}}$

Therefore, if two squares are congruent,

${\displaystyle {\begin{aligned}&a=b\\&\Leftrightarrow {\frac {x\tan \theta }{\tan \theta +2}}={\frac {\tan \theta }{1+\tan \theta }}\\&\Leftrightarrow {\frac {x}{\tan \theta +2}}={\frac {1}{1+\tan \theta }}\\&\Leftrightarrow x(\tan \theta +1)=\tan \theta +2\\&\Leftrightarrow (x-1)\tan \theta =2-x.\end{aligned}}}$

In this case,

${\displaystyle \tan \theta ={\frac {\sqrt {(2+x)(2-x)}}{x}}.}$

So, we can input the value of tanθ,

${\displaystyle {\begin{aligned}&(x-1)\tan \theta =2-x\\&\Leftrightarrow (x-1){\frac {\sqrt {(2+x)(2-x)}}{x}}=2-x\\&\Leftrightarrow (2-x)\cdot ((x-1)^{2}(2+x)-x^{2}(2-x))=0\\&\Leftrightarrow (2-x)\cdot (2x^{3}-2x^{2}-3x+2)=0.\end{aligned}}}$

In this case, ${\displaystyle {\sqrt {2}}<x<2}$, we can get the following equation:

${\displaystyle 2x^{3}-2x^{2}-3x+2=0.}$

If x is the largest positive root of Calabi's equation:

${\displaystyle 2x^{3}-2x^{2}-3x+2=0,{\sqrt {2}}<x<2}$

we can calculate the value of x by following methods.

We can set the function ${\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }$ as follows:

${\displaystyle {\begin{aligned}f(x)&=2x^{3}-2x^{2}-3x+2,\\f'(x)&=6x^{2}-4x-3=6{\bigg (}x-{\frac {1}{3}}{\bigg )}^{2}-{\frac {11}{3}}.\end{aligned}}}$

The function f is continuous and differentiable on ${\displaystyle \mathbb {R} }$ and

${\displaystyle {\begin{aligned}f({\sqrt {2}})&={\sqrt {2}}-2<0,\\f(2)&=4>0,\\f'(x)&>0,\forall x\in [{\sqrt {2}},2].\end{aligned}}}$

Then f is monotonically increasing function and by Intermediate value theorem, the Calabi's equation f(x) = 0 has unique solution in open interval ${\displaystyle {\sqrt {2}}<x<2}$.

The value of x is calculated by Newton's method as follows:

${\displaystyle {\begin{aligned}x_{0}&={\sqrt {2}},\\x_{n+1}&=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}={\frac {4x_{n}^{3}-2x_{n}^{2}-2}{6x_{n}^{2}-4x_{n}-3}}.\end{aligned}}}$

The value of x can expressed with complex numbers by using Cardano's method:

${\displaystyle x={1 \over 3}{\Bigg (}1+{\sqrt[{3}]{-23+3i{\sqrt {237}} \over 4}}+{\sqrt[{3}]{-23-3i{\sqrt {237}} \over 4}}{\Bigg )}.}$^[3][5][a]

The value of x can also be expressed without complex numbers by using Viète's method:

${\displaystyle {\begin{aligned}x&={1 \over 3}{\bigg (}1+{\sqrt {22}}\cos \!{\bigg (}{1 \over 3}\cos ^{-1}\!\!{\bigg (}\!-{23 \over 11{\sqrt {22}}}{\bigg )}{\bigg )}{\bigg )}\\&=1.55138752454832039226195251026462381516359170380389\cdots .\end{aligned}}}$^[2]

The value of x has continued fraction representation by Lagrange's method as follows:
[1, 1, 1, 4, 2, 1, 2, 1, 5, 2, 1, 3, 1, 1, 390, ...] =

${\displaystyle 1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{4+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{5+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{3+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{390+\cdots }}}}}}}}}}}}}}}}}}}}}}}}}}}}}$.^[3][6][7][b]

The Calabi triangle is obtuse with base angle θ and apex angle ψ as follows:

${\displaystyle {\begin{aligned}\theta &=\cos ^{-1}(x/2)\\&=39.13202614232587442003651601935656349795831966723206\cdots ^{\circ },\\\psi &=180-2\theta \\&=101.73594771534825115992696796128687300408336066553587\cdots ^{\circ }.\\\end{aligned}}}$

• Stewart, Ian (2004), Galois Theory (3rd ed.), Chapman and Hall/CRC, ISBN 978-1-58488-393-7 - Galois Theory Errata.

• Weisstein, Eric W. "Calabi's Triangle". MathWorld.