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. It is an isosceles triangle which is obtuse with an irrational but algebraic ratio between the lengths of its sides and its base.

Definition

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. Thus, the Calabi triangle may be defined as a triangle that is not equilateral and has three placements for its largest square.

Shape

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 $0<x<{\sqrt {2}}$ .; In this case $x = 1$ is valid for equilateral triangle.
case 2) $△ ABC$ is right triangle: The condition is $x={\sqrt {2}}$ .; In this case no value is valid.
case 3) $△ ABC$ is obtuse triangle: The condition is ${\sqrt {2}}<x<2$ .; In this case the Calabi triangle is valid for the largest positive root of $2x^{3}-2x^{2}-3x+2=0$ at $x=1.55138752454832039226...$ (OEIS: A046095).

Example of answer

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

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

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

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From △DEB ∽ △AHB,

{\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

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

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

From △JKC ∽ △AHC,

{\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

{\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,

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

Therefore $x=1$ , it means that $△ ABC$ is equilateral triangle.

case 2) $△ ABC$ is right triangle

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

Then no value is valid.

case 3) $△ ABC$ is obtuse triangle

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

From △AHC ∽ △JKC,

{\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,

{\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,

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

So, we can input the value of $tan θ$ ,

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

2x^{3}-2x^{2}-3x+2=0.

Root of Calabi's equation

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

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

we can calculate the value of $x$ by following methods.

Newton's method

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

{\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 $\mathbb {R}$ and

{\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 ${\sqrt {2}}<x<2$ .

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

{\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}}

Cardano's method

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

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

Viète's method

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

{\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}}

Lagrange's method

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, ...] =

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 }}}}}}}}}}}}}}}}}}}}}}}}}}}}

.

base angle and apex angle

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

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

Footnotes

Notes

Citations

References

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

External links

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