Calabi triangle

 ( Types Of Triangles ) 

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 triangle with an irrational but algebraic number 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 .

In this case is valid for equilateral triangle.

case 2) is right triangle

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

In this case no value is valid.

case 3) is obtuse triangle

The condition is .

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

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

$\backslash begin\{align\}$

HB &= HC = \cos\theta = \frac{x}{2}, \\ AH &= \sin\theta = \frac{x}{2}\tan\theta , \\ 0 &< \theta < \frac{\pi}{2}. \end{align} Then and , so .

From △DEB ∽ △AHB,

$\backslash begin\{align\}$

& 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{align}

---

case 1) is acute triangle Let be a square on side with its side length as . From △ABC ∽ △IBJ,

$\backslash begin\{align\}$

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

From △JKC ∽ △AHC,

$\backslash begin\{align\}$

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

Then

$\backslash begin\{align\}$

&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{align}

Therefore, if two squares are congruent,

$\backslash begin\{align\}$

&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\cdot2(\sin\theta - 1)\cdot(x - 1) = 0. \end{align}

In this case,

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

---

case 2) is right triangle In this case, $x\; =\; \backslash sqrt\{2\},\; \backslash tan\backslash theta\; =\; 1$, so $a\; =\; \backslash frac\{\backslash sqrt\{2\}\}\{3\},\; b\; =\; \backslash frac\{1\}\{2\}.$

Then no value is valid.

---

case 3) is obtuse triangle Let be a square on base with its side length as .

From △AHC ∽ △JKC,

$\backslash begin\{align\}$

& 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{align}

Therefore, if two squares are congruent,

$\backslash begin\{align\}$

&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{align}

In this case,

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

So, we can input the value of ,

$\backslash begin\{align\}$

&(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{align}

In this case, , 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:

we can calculate the value of by following methods.

---

Newton's method We can set the function $f\; :\; \backslash mathbb\{R\}\; \backslash rarr\; \backslash mathbb\{R\}$ as follows:

$\backslash begin\{align\}$

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{align} The function is continuous and differentiable on $\backslash mathbb\{R\}$ and

$\backslash begin\{align\}$

f(\sqrt{2}) &= \sqrt{2} - 2 < 0, \\ f(2) &= 4 > 0, \\ f'(x) &> 0 , \forall x \in \sqrt{2},. \end{align} Then is monotonically increasing function and by Intermediate value theorem, the Calabi's equation has unique solution in open interval .

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

$\backslash begin\{align\}$

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{align}

+ Newton's method for the root of Calabi's equation ! NO !! itaration value

---

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

$x\; =\; \{1\; \backslash over\; 3\}\; \backslash Bigg(1\; +\; \backslash sqrt3\{-23\; +\; 3i\; \backslash sqrt\{237\}\; \backslash over\; 4\}\; +\; \backslash sqrt3\{-23\; -\; 3i\; \backslash sqrt\{237\}\; \backslash over\; 4\}\; \backslash Bigg)\; .${4} , \\

r &= \frac{1}{4}\sqrt{(-23)^2+9 \cdot 237} = \frac{1}{4}\sqrt{2 \cdot 11^3} = \Bigg(\sqrt{\frac{11}{2}}\Bigg)^3, \\ \cos\varphi &= - \frac{23}{4}\frac{1}{r} = - \frac{23 \cdot 2 \sqrt{2}}{4 \cdot 11 \sqrt{11}} = - \frac{23}{11 \sqrt{22}}, \\ \sqrt3{\alpha} &= \sqrt3{r}e^{\frac{i\varphi}{3}}=\sqrt3{r}\Big(\cos\Big(\frac{\varphi}{3}\Big) +i\sin\Big(\frac{\varphi}{3}\Big)\Big), \\ \sqrt3{\alpha} + \sqrt3{\bar{\alpha}} &= 2 \sqrt3{r} \cos\Big(\frac{\varphi}{3}\Big) = \sqrt{22} \cos\!\bigg( {1 \over 3} \cos^{-1}\!\!\bigg(\!-{23 \over 11 \sqrt{22}} \bigg) \bigg), \\ x &= \frac{1}{3}\bigg(1 + \sqrt3{\alpha} + \sqrt3{\bar{\alpha}} \bigg) = {1 \over 3} \bigg(1 + \sqrt{22} \cos\!\bigg( {1 \over 3} \cos^{-1}\!\!\bigg(\!-{23 \over 11 \sqrt{22}} \bigg) \bigg) \bigg). \end{align}

---

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

$\backslash begin\{align\}$

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{align}

---

Lagrange's method The value of has continued fraction representation by Lagrange's method as follows:
1, =

$1\; +\; \backslash cfrac\{1\}\{1\; +$

   \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    }}}}}}}}}}}}}}. - [https://gdz.sub.uni-goettingen.de/id/PPN308899857?tify=%7B%22pages%22%3A%5B544%2C545%5D%2C%22pan%22%3A%7B%22x%22%3A0.947%2C%22y%22%3A0.64%7D%2C%22view%22%3A%22toc%22%2C%22zoom%22%3A0.354%7D Œuvres II, p.539-578.] - [https://gdz.sub.uni-goettingen.de/id/PPN308899857?tify=%7B%22pages%22%3A%5B586%2C587%5D%2C%22pan%22%3A%7B%22x%22%3A0.947%2C%22y%22%3A0.64%7D%2C%22view%22%3A%22toc%22%2C%22zoom%22%3A0.354%7D Œuvres II, p.581-652.]{k_{95}} \\
 &= \frac{10264770284430115358350493989796951584352149694}{6616509493602288551995313304988866597070326335} \\
 &= 1.5513875245483203922619525102646238151635917038038871995280071201179267425542569572957604536135484903\cdots, \\
     

\varepsilon &= \frac{1}{k_{95}(k_{95}+k_{94})} \\

           &= 1.82761... \times 10^{-91}.
     

---

base angle and apex angle The Calabi triangle is Obtuse triangle with base angle and apex angle as follows:

$\backslash begin\{align\}$

\theta &= \cos^{-1}(x/2) \\

      &= 39.13202614232587442003651601935656349795831966723206\cdots^\circ, \\
     

\psi &= 180 - 2\theta \\

      &= 101.73594771534825115992696796128687300408336066553587\cdots^\circ. \\
     

\end{align}

---

See also

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

---

Footnotes

---

Notes

---

Citations

• - Galois Theory Errata.

---

External links

Post Comment