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.Template:Sfn

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 Template:Math is isosceles which has the same length of sides as Template:Math. If the ratio of the base to either leg is Template:Mvar, we can set that Template:Math. Then we can consider the following three cases:

case 1) Template:Math is acute triangle

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

In this case Template:Math is valid for equilateral triangle.

case 2) Template:Math is right triangle

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

In this case no value is valid.

case 3) Template:Math 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 $x=1.55138752454832039226...$ (Template:Oeis).

Example figure of Calabi triangle 01

Consider the case of Template:Math. Then

$0<x<2.$

Let a base angle be Template:Mvar and a square be Template:Math on base Template:Math with its side length as Template:Mvar. Let Template:Mvar be the foot of the perpendicular drawn from the apex Template:Mvar to the base. Then

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

Then Template:Math and Template:Math, so Template:Math.

From △DEB ∽ △AHB,

$\begin{array}{cc}& EB:DE=HB:AH\\ & \iff \left(\frac{x-a}{2}\right):a=\cos \theta :\sin \theta =1:\tan \theta \\ & \iff a=\left(\frac{x-a}{2}\right)\tan \theta \\ & \iff a=\frac{x\tan \theta }{\tan \theta +2}.\\ \end{array}$

case 1) Template:Math is acute triangle

Example figure of Calabi triangle 02

Let Template:Math be a square on side Template:Math with its side length as Template:Mvar. From △ABC ∽ △IBJ,

$\begin{array}{cc}& AB:IJ=BC:BJ\\ & \iff 1:b=x:BJ\\ & \iff BJ=bx.\end{array}$

From △JKC ∽ △AHC,

$\begin{array}{cc}& JK:JC=AH:AC\\ & \iff b:JC=\frac{x}{2}\tan \theta :1\\ & \iff JC=\frac{2b}{x\tan \theta }.\end{array}$

Then

$\begin{array}{cc}& x=BC=BJ+JC=bx+\frac{2b}{x\tan \theta }\\ & \iff x=b\frac{x^2\tan \theta +2}{x\tan \theta }\\ & \iff b=\frac{x^2\tan \theta }{x^2\tan \theta +2}.\end{array}$

Therefore, if two squares are congruent,

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

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

Therefore $x=1$, it means that Template:Math is equilateral triangle.

case 2) Template:Math is right triangle

Example figure of Calabi triangle 03

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

Then no value is valid. Template:Clear

case 3) Template:Math is obtuse triangle

Example figure of Calabi triangle 04

Let Template:Math be a square on base Template:Math with its side length as Template:Mvar.

From △AHC ∽ △JKC,

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

Therefore, if two squares are congruent,

$\begin{array}{cc}& a=b\\ & \iff \frac{x\tan \theta }{\tan \theta +2}=\frac{\tan \theta }{1+\tan \theta }\\ & \iff \frac{x}{\tan \theta +2}=\frac{1}{1+\tan \theta }\\ & \iff x(\tan \theta +1)=\tan \theta +2\\ & \iff (x-1)\tan \theta =2-x.\end{array}$

In this case,

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

So, we can input the value of Template:Math,

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

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 Template:Mvar 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 Template:Mvar by following methods.

Newton's method

We can set the function $f:ℝ\to ℝ$ as follows:

$\begin{array}{cc}f(x)& =2x^3-2x^2-3x+2,\\ f^{\prime }(x)& =6x^2-4x-3=6(x-\frac{1}{3}{)}^2-\frac{11}{3}.\end{array}$

The function Template:Mvar is continuous and differentiable on $ℝ$ and

$\begin{array}{cc}f(\sqrt{2})& =\sqrt{2}-2<0,\\ f(2)& =4>0,\\ f^{\prime }(x)& >0,\mathrm{\forall }x\in [\sqrt{2},2].\end{array}$

Then Template:Mvar is monotonically increasing function and by Intermediate value theorem, the Calabi's equation Template:Math has unique solution in open interval $\sqrt{2}<x<2$.

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

$\begin{array}{cc}{x}_0& =\sqrt{2},\\ x_{n+1}& =x_n-\frac{f(x_n)}{f^{\prime }(x_n)}=\frac{4x_n^3-2x_n^2-2}{6x_n^2-4x_n-3}.\end{array}$

Newton's method for the root of Calabi's equation

NO	itaration value
Template:MvarTemplate:Sub	1.41421356237309504880168872420969807856967187537694...
Template:MvarTemplate:Sub	1.58943369375323596617308283187888791370090306159374...
Template:MvarTemplate:Sub	1.55324943049375428807267665439782489231871295592784...
Template:MvarTemplate:Sub	1.55139234383942912142613029570413117306471589987689...
Template:MvarTemplate:Sub	1.55138752458074244056538641010106649611908076010328...
Template:MvarTemplate:Sub	1.55138752454832039226341994813293555945836732015691...
Template:MvarTemplate:Sub	1.55138752454832039226195251026462381516359470986821...
Template:MvarTemplate:Sub	1.55138752454832039226195251026462381516359170380388...

Cardano's method

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

$x=\frac{1}{3}(1+\sqrt[3]{\frac{-23+3i\sqrt{237}}{4}}+\sqrt[3]{\frac{-23-3i\sqrt{237}}{4}}).$Template:SfnTemplate:Efn

Viète's method

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

$\begin{array}{cc}x& =\frac{1}{3}(1+\sqrt{22}\cos (\frac{1}{3}{\cos }^{-1}(-\frac{23}{11\sqrt{22}})\left)\right)\\ & =1.55138752454832039226195251026462381516359170380389\cdots .\end{array}$Template:Sfn

Lagrange's method

The value of Template:Mvar 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, ...] =

$1+\frac{1}{1+\frac{1}{1+\frac{1}{4+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{5+\frac{1}{2+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{1+\frac{1}{390+\cdots }}}}}}}}}}}}}}$.Template:Efn

base angle and apex angle

The Calabi triangle is obtuse with base angle Template:Mvar and apex angle Template:Mvar as follows:

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

See also

• Biggest little polygon
• Cubic equation
• Inscribed square in a triangle

References

• Template:Citation - Galois Theory Errata.

External links

• Template:Mathworld