Calabi triangle

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.^[3]^[4] Thus, the Calabi triangle may be defined as a triangle that is not equilateral and has three placements for its largest square.

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Shape

Summarize

Perspective

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}}

Then $HB = .mw-parser-output .sfrac{white-space:nowrap}.mw-parser-output .sfrac.tion,.mw-parser-output .sfrac .tion{display:inline-block;vertical-align:-0.5em;font-size:85%;text-align:center}.mw-parser-output .sfrac .num{display:block;line-height:1em;margin:0.0em 0.1em;border-bottom:1px solid}.mw-parser-output .sfrac .den{display:block;line-height:1em;margin:0.1em 0.1em}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);clip-path:polygon(0px 0px,0px 0px,0px 0px);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}⁠x/2⁠$ 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

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,

{\displaystyle {\begin{aligned}&AH:HC=JK:KC\\&\Leftrightarrow \sin \theta

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.

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Root of Calabi's equation

Summarize

Perspective

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}}

More information NO, itaration value ...

Newton's method for the root of Calabi's equation
NO	itaration value
$x$ ₀	1.41421356237309504880168872420969807856967187537694...
$x$ ₁	1.58943369375323596617308283187888791370090306159374...
$x$ ₂	1.55324943049375428807267665439782489231871295592784...
$x$ ₃	1.55139234383942912142613029570413117306471589987689...
$x$ ₄	1.55138752458074244056538641010106649611908076010328...
$x$ ₅	1.55138752454832039226341994813293555945836732015691...
$x$ ₆	1.55138752454832039226195251026462381516359470986821...
$x$ ₇	1.55138752454832039226195251026462381516359170380388...

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 )}.

^[3]^[5]^[a]

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}}

^[2]

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

.^[3]^[6]^[7]^[b]

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}}

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Definition