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 is isosceles which has the same length of sides as . If the ratio of the base to either leg is , we can set that . Then we can consider the following three cases:
;case 1) is acute triangle:
:The condition is $0 < x < \sqrt$.
:In this case is valid for equilateral triangle.
;case 2) is right triangle:
:The condition is $x = \sqrt$.
:In this case no value is valid.
;case 3) is obtuse triangle:
:The condition is $\sqrt < x < 2$.
:In this case the Calabi triangle is valid for the largest positive root of $2x^3 - 2x^2 - 3x + 2 = 0$ at $x = 1.55138752454832039226...$ ().

Consider the case of . Then
:$0 < x < 2.$
Let a base angle be and a square be on base with its side length as .
Let be the foot of the perpendicular drawn from the apex to the base. Then
:$\begin HB &= HC = \cos\theta = \frac, \\ AH &= \sin\theta = \frac\tan\theta , \\ 0 &< \theta < \frac. \end$
Then and , so .

From △DEB ∽ △AHB,
:$\begin & EB : DE = HB : AH \\ &\Leftrightarrow \bigg(\frac\bigg) : a = \cos \theta : \sin \theta = 1 : \tan \theta \\ &\Leftrightarrow a = \bigg(\frac\bigg)\tan \theta \\ &\Leftrightarrow a = \frac. \\ \end$

case 1) is acute triangle

Let be a square on side with its side length as .
From △ABC ∽ △IBJ,
:$\begin & AB : IJ = BC : BJ \\ &\Leftrightarrow 1 : b = x : BJ \\ &\Leftrightarrow BJ = bx. \end$

From △JKC ∽ △AHC,
:$\begin & JK : JC = AH : AC \\ &\Leftrightarrow b : JC = \frac\tan\theta : 1 \\ &\Leftrightarrow JC = \frac. \end$

Then
:$\begin &x = BC = BJ + JC = bx + \frac \\ &\Leftrightarrow x = b\frac \\ &\Leftrightarrow b = \frac. \end$

Therefore, if two squares are congruent,
:$\begin &a = b \\ &\Leftrightarrow \frac = \frac \\ &\Leftrightarrow x\tan\theta\cdot(x^2 \tan\theta + 2) = x^2 \tan\theta(\tan\theta + 2) \\ &\Leftrightarrow x\tan\theta\cdot(x(\tan\theta + 2) - (x^2 \tan\theta + 2)) = 0 \\ &\Leftrightarrow x\tan\theta\cdot(x\tan\theta - 2)\cdot(x - 1) = 0 \\ &\Leftrightarrow 2\sin\theta\cdot2(\sin\theta - 1)\cdot(x - 1) = 0. \end$

In this case, $\frac < \theta < \frac, 2\sin\theta\cdot2(\sin\theta - 1) \ne 0.$

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

case 2) is right triangle

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

Then no value is valid.

case 3) is obtuse triangle

Let be a square on base with its side length as .

From △AHC ∽ △JKC,
:$\begin & AH : HC = JK : KC \\ &\Leftrightarrow \sin\theta : \cos\theta = b : (1-b) \\ &\Leftrightarrow b\cos\theta = (1-b)\sin\theta \\ &\Leftrightarrow b = (1-b)\tan\theta \\ &\Leftrightarrow b = \frac. \end$

Therefore, if two squares are congruent,
:$\begin &a = b \\ &\Leftrightarrow \frac = \frac \\ &\Leftrightarrow \frac = \frac \\ &\Leftrightarrow x(\tan\theta + 1) = \tan\theta + 2 \\ &\Leftrightarrow (x - 1)\tan\theta = 2 - x. \end$

In this case,
:$\tan\theta = \frac.$

So, we can input the value of ,
:$\begin &(x - 1)\tan\theta = 2 - x \\ &\Leftrightarrow (x - 1)\frac = 2 - x \\ &\Leftrightarrow (2 - x)\cdot((x - 1)^2 (2 + x) - x^2 (2 - x)) = 0 \\ &\Leftrightarrow (2 - x)\cdot(2x^3 - 2x^2 - 3x + 2) = 0. \end$

In this case, $\sqrt < x < 2$, we can get the following equation:
:$2x^3 - 2x^2 - 3x + 2 = 0.$

Root of Calabi's equation

If is the largest positive root of Calabi's equation:
: $2x^3 - 2x^2 - 3x + 2 = 0 , \sqrt < x < 2$
we can calculate the value of by following methods.

Newton's method

We can set the function $f : \mathbb \rarr \mathbb$ as follows:
: $\begin f(x) &= 2x^3 - 2x^2 -3x + 2, \\ f'(x)&= 6x^2 - 4x - 3 = 6\bigg(x - \frac\bigg)^2 - \frac. \end$
The function is continuous and differentiable on $\mathbb$ and
: \begin
f(\sqrt) &= \sqrt - 2 < 0, \\
f(2) &= 4 > 0, \\
f'(x) &> 0 , \forall x \in sqrt, 2
\end
Then is monotonically increasing function and by Intermediate value theorem, the Calabi's equation has unique solution in open interval $\sqrt < x < 2$.

The value of is calculated by Newton's method as follows:
: $\begin x_0 &= \sqrt, \\ x_ &= x_n - \frac = \frac. \end$

Cardano's method

The value of can expressed with complex numbers by using Cardano's method:
: x = \Bigg(1 + \sqrt + \sqrt \Bigg) .

Viète's method

The value of can also be expressed without complex numbers by using Viète's method:
: $\begin x &= \bigg(1 + \sqrt \cos\!\bigg( \cos^\!\!\bigg(\!- \bigg) \bigg) \bigg) \\ &= 1.55138752454832039226195251026462381516359170380389\cdots . \end$

Lagrange's method

The value of has continued fraction representation by Lagrange

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