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[[File:Calabi triangle.svg|300px|right|thumb]]

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.<ref>{{cite web |url=http://www.people.fas.harvard.edu/~sfinch/csolve/calabi.html |archive-url=https://archive.today/20121212215151/http://www.people.fas.harvard.edu/~sfinch/csolve/calabi.html |url-status=dead |archive-date=12 December 2012 |title=Outline of Proof Regarding Squares Wedged in Triangle | last = Calabi | first = Eugenio | author-link = Eugenio Calabi |date=3 Nov 1997 |access-date=3 May 2018}}</ref> It is an [[isosceles]] triangle which is [[obtuse triangle|obtuse]] with an [[irrational number|irrational]] but [[algebraic number|algebraic]] ratio between the lengths of its sides and its base.{{sfn|Stewart|2004|p=15}}

==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.<ref name="Wolfram">{{mathworld|title=Calabi's Triangle|id=CalabisTriangle}}</ref><ref>{{cite book |last1=Conway |first1=J.H. |author-link1=John Horton Conway |last2=Guy |first2=R.K. |author-link2=Richard K. Guy |title=The Book of Numbers |year=1996 |publisher=Springer-Verlag |location=New York|page=206|contribution=Calabi's Triangle|contribution-url=https://books.google.com/books?id=0--3rcO7dMYC&pg=PA206}}</ref> 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 {{math|△''ABC''}} is [[isosceles]] which has the same length of sides as {{math|''AB'' {{=}} ''AC''}}. If the ratio of the base to either leg is {{mvar|x}}, we can set that {{math|''AB'' {{=}} ''AC'' {{=}} 1, ''BC'' {{=}} {{mvar|x}}}}. Then we can consider the following three cases:
;case 1) {{math|△''ABC''}} is [[acute triangle]]:
:The condition is <math>0 < x < \sqrt{2}</math>.
:In this case {{math|{{mvar|x}} {{=}} 1}} is valid for [[equilateral triangle]].
;case 2) {{math|△''ABC''}} is [[right triangle]]:
:The condition is <math>x = \sqrt{2}</math>.
:In this case no value is valid.
;case 3) {{math|△''ABC''}} is [[obtuse triangle]]:
:The condition is <math>\sqrt{2} < x < 2</math>.
:In this case the Calabi triangle is valid for the largest positive root of <math> 2x^3 - 2x^2 - 3x + 2 = 0</math> at <math>x = 1.55138752454832039226...</math> ({{oeis|A046095}}).

{{Hidden begin|title = Example of answer|titlestyle = text-align:center;}}
[[File:Example_figure_of_Calabi_triangle_01.svg|right|220px|Example figure of Calabi triangle 01]]

Consider the case of {{math|''AB'' {{=}} ''AC'' {{=}} 1, ''BC'' {{=}} {{mvar|x}}}}. Then
:<math>0 < x < 2.</math>
Let a base angle be {{mvar|θ}} and a square be {{math|□''DEFG''}} on base {{math|''BC''}} with its side length as {{mvar|a}}.
Let {{mvar|H}} be the foot of the perpendicular drawn from the apex {{mvar|A}} to the base. Then
:<math>\begin{align}
HB &= HC = \cos\theta = \frac{x}{2}, \\
AH &= \sin\theta = \frac{x}{2}\tan\theta , \\
0 &< \theta < \frac{\pi}{2}.
\end{align}</math>
Then {{math|''HB'' {{=}} {{sfrac|{{mvar|x}}|2}}}} and {{math|''HE'' {{=}} {{sfrac|{{mvar|a}}|2}}}}, so {{math|''EB'' {{=}} {{sfrac|{{mvar|x}} - {{mvar|a}}|2}}}}.

From △DEB ∽ △AHB,
:<math>\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}</math>

=== case 1) {{math|△''ABC''}} is acute triangle ===
[[File:Example_figure_of_Calabi_triangle_02.svg|right|220px|Example figure of Calabi triangle 02]]

Let {{math|□''IJKL''}} be a square on side {{math|''AC''}} with its side length as {{mvar|b}}.
From △ABC ∽ △IBJ,
:<math>\begin{align}
& AB : IJ = BC : BJ \\
&\Leftrightarrow 1 : b = x : BJ \\
&\Leftrightarrow BJ = bx.
\end{align}</math>

From △JKC ∽ △AHC,
:<math>\begin{align}
& JK : JC = AH : AC \\
&\Leftrightarrow b : JC = \frac{x}{2}\tan\theta : 1 \\
&\Leftrightarrow JC = \frac{2b}{x\tan\theta}.
\end{align}</math>

Then 
:<math>\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}</math>

Therefore, if two squares are congruent,
:<math>\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}</math>

In this case, <math>\frac{\pi}{4} < \theta < \frac{\pi}{2}, 2\sin\theta\cdot2(\sin\theta - 1) \ne 0.</math>

Therefore <math>x = 1</math>, it means that {{math|△''ABC''}} is equilateral triangle.

=== case 2) {{math|△''ABC''}} is right triangle ===
[[File:Example_figure_of_Calabi_triangle_03.svg|right|400px|Example figure of Calabi triangle 03]]

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

Then no value is valid.
{{Clear}}

=== case 3) {{math|△''ABC''}} is obtuse triangle ===
[[File:Example_figure_of_Calabi_triangle_04.svg|right|475px|Example figure of Calabi triangle 04]]

Let {{math|□''IJKA''}} be a square on base {{math|''AC''}} with its side length as {{mvar|b}}.

From △AHC ∽ △JKC,
:<math>\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}</math>

Therefore, if two squares are congruent,
:<math>\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}</math>

In this case,
:<math>\tan\theta = \frac{\sqrt{(2 + x)(2 - x)}}{x}.</math>

So, we can input the value of {{math|tan''θ''}},
:<math>\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}</math>

In this case, <math>\sqrt{2} < x < 2</math>, we can get the following equation:
:<math>2x^3 - 2x^2 - 3x + 2 = 0.</math>

{{Hidden end}}

== Root of Calabi's equation ==
If {{mvar|x}} is the largest positive [[zero of a function|root]] of Calabi's equation:
: <math> 2x^3 - 2x^2 - 3x + 2 = 0 , \sqrt{2} < x < 2</math>
we can calculate the value of {{mvar|x}} by following methods.

=== Newton's method ===
We can set the function <math>f : \mathbb{R} \rarr \mathbb{R}</math> as follows:
: <math>\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}</math>
The function {{mvar|f}} is [[Continuous function|continuous]] and [[Differentiable function|differentiable]] on <math>\mathbb{R}</math> and
: <math>\begin{align}
f(\sqrt{2}) &= \sqrt{2} - 2 < 0, \\
f(2)        &= 4 > 0, \\
f'(x)       &> 0 , \forall x \in [\sqrt{2}, 2].
\end{align}</math>
Then {{mvar|f}} is [[Monotonic function|monotonically increasing function]] and by [[Intermediate value theorem]], the Calabi's equation {{math|{{mvar|f}}({{mvar|x}}) {{=}} 0}} has unique solution in open interval <math>\sqrt{2} < x < 2</math>.

The value of {{mvar|x}} is calculated by [[Newton's method]] as follows:
: <math>\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}</math>
{| class="wikitable mw-collapsible mw-collapsed" style="text-align:right; font-size:smaller; width:100%"
|+ Newton's method for the root of Calabi's equation
! NO !! itaration value
|-
! {{mvar|x}}{{sub|0}}
| '''1.'''41421356237309504880168872420969807856967187537694...
|-
! {{mvar|x}}{{sub|1}}
| '''1.5'''8943369375323596617308283187888791370090306159374...
|-
! {{mvar|x}}{{sub|2}}
| '''1.55'''324943049375428807267665439782489231871295592784...
|-
! {{mvar|x}}{{sub|3}}
| '''1.5513'''9234383942912142613029570413117306471589987689...
|-
! {{mvar|x}}{{sub|4}}
| '''1.5513875245'''8074244056538641010106649611908076010328...
|-
! {{mvar|x}}{{sub|5}}
| '''1.55138752454832039226'''341994813293555945836732015691...
|-
! {{mvar|x}}{{sub|6}}
| '''1.55138752454832039226195251026462381516359'''470986821...
|-
! {{mvar|x}}{{sub|7}}
| '''1.55138752454832039226195251026462381516359170380388'''...
|}

=== Cardano's method ===
The value of {{mvar|x}} can expressed with [[complex number]]s by using [[Cardano's method]]:
: <math> x = {1 \over 3} \Bigg(1 + \sqrt[3]{-23 + 3i \sqrt{237} \over 4} + \sqrt[3]{-23 - 3i \sqrt{237} \over 4} \Bigg) .</math><ref name="Wolfram" />{{sfn|Stewart|2004|pp=7-10}}{{efn|If we set the [[polar form]] of complex number, we can calculate the value of {{mvar|x}} as follows:
: <math>\begin{align}
\alpha           &= re^{i\varphi}=r(\cos\varphi +i\sin\varphi)=\frac{-23 + 3i \sqrt{237}}{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}}, \\
\sqrt[3]{\alpha} &= \sqrt[3]{r}e^{\frac{i\varphi}{3}}=\sqrt[3]{r}\Big(\cos\Big(\frac{\varphi}{3}\Big) +i\sin\Big(\frac{\varphi}{3}\Big)\Big), \\
\sqrt[3]{\alpha} + \sqrt[3]{\bar{\alpha}} &= 2 \sqrt[3]{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 + \sqrt[3]{\alpha} + \sqrt[3]{\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}</math>
Then this Cardano's method is equivalent as [[#Viète's method|Viète's method]].}}

=== Viète's method ===
The value of {{mvar|x}} can also be expressed without [[complex number]]s by using [[Cubic equation#Trigonometric solution for three real roots|Viète's method]]:
: <math>\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}</math>{{sfn|Stewart|2004|p=15}}

=== Lagrange's method ===
The value of {{mvar|x}} has [[continued fraction]] representation by [[Joseph-Louis Lagrange|Lagrange]]'s method as follows:<br />[1, 1, 1, 4, 2, 1, 2, 1, 5, 2, 1, 3, 1, 1, 390, ...] =<br />
:<math>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    }}}}}}}}}}}}}}</math>.<ref name="Wolfram" /><ref>{{Citation|last=Joseph-Louis|first=Lagrange|author-link=Joseph-Louis Lagrange|year=1769|title=Sur la résolution des équations numériques|journal=Mémoires de l'Académie royale des Sciences et Belles-lettres de Berlin|volume=23|url=http://sites.mathdoc.fr/cgi-bin/oeitem?id=OE_LAGRANGE__2_539_0}} - [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.]</ref><ref>{{Citation|last=Joseph-Louis|first=Lagrange|author-link=Joseph-Louis Lagrange|year=1770|title=Additions au mémoire sur la résolution des équations numériques|journal=Mémoires de l'Académie royale des Sciences et Belles-lettres de Berlin|volume=24|url=http://sites.mathdoc.fr/cgi-bin/oeitem?id=OE_LAGRANGE__2_581_0}} - [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.]</ref>{{efn|text=If a [[continued fraction]] {{math|[''a''{{sub|0}}, ''a''{{sub|1}}, ''a''{{sub|2}}, ...]}} are found, with numerators {{mvar|h}}{{sub|1}}, {{mvar|h}}{{sub|2}}, ... and denominators {{mvar|k}}{{sub|1}}, {{mvar|k}}{{sub|2}}, ... then the relevant recursive relation is that of [[Gaussian bracket]]s:

:{{math|''h''{{sub|''n''}} {{=}} ''a''{{sub|''n''}}''h''{{sub|''n'' − 1}} + ''h''{{sub|''n'' − 2}}}},
:{{math|''k''{{sub|''n''}} {{=}} ''a''{{sub|''n''}}''k''{{sub|''n'' − 1}} + ''k''{{sub|''n'' − 2}}}}.

The successive convergents are given by the formula

:{{bigmath|{{sfrac|''h''{{sub|''n''}}|''k''{{sub|''n''}}}} {{=}} {{sfrac|''a''{{sub|''n''}}''h''{{sub|''n'' − 1}} + ''h''{{sub|''n'' − 2}}|''a''{{sub|''n''}}''k''{{sub|''n'' − 1}} + ''k''{{sub|''n'' − 2}}}}}}.

If the continued fraction is <br />
:[1, 1, 1, 4, 2, 1, 2, 1, 5, 2, 1, 3, 1, 1, 390, 1, 1, 2, 11, 6, 2, 1, 1, 56, 1, 4, 3, 1, 1, 6, 9, 3, 2, 1, 8, 10, 9, 25, 2, 1, 3, 1, 3, 5, 2, 35, 1, 1, 1, 41, 1, 2, 2, 1, 2, 2, 3, 1, 4, 2, 1, 1, 1, 1, 3, 1, 6, 2, 1, 4, 11, 1, 2, 2, 1, 1, 6, 3, 1, 1, 1, 1, 1, 1, 4, 1, 7, 2, 2, 2, 36, 7, 22, 1, 2, 1, ...],<ref>{{OEIS|A046096}}</ref>
we can calculate the rational approxmation of {{mvar|x}} is as follows:

<table class="wikitable mw-collapsible mw-collapsed" style="width:100%;text-align:right;font-size:smaller;">
<caption>The value of numerators <span class="texhtml mvar" style="font-style:italic;">h<sub>n</sub></span> and denominators <span class="texhtml mvar" style="font-style:italic;">k<sub>n</sub></span> of continued fraction
</caption>
<tr>
<th><span class="texhtml mvar" style="font-style:italic;">n</span></th>
<th><span class="texhtml mvar" style="font-style:italic;">a<sub>n</sub></span></th>
<th><span class="texhtml mvar" style="font-style:italic;">h<sub>n</sub></span></th>
<th><span class="texhtml mvar" style="font-style:italic;">k<sub>n</sub></span>
</th></tr>
<tr>
<th>-2
</th>
<td><br /></td>
<td>0</td>
<td>1
</td></tr>
<tr>
<th>-1
</th>
<td><br /></td>
<td>1</td>
<td>0
</td></tr>
<tr>
<th>0
</th>
<td>1</td>
<td>1</td>
<td>1
</td></tr>
<tr>
<th>1
</th>
<td>1</td>
<td>2</td>
<td>1
</td></tr>
<tr>
<th>2
</th>
<td>1</td>
<td>3</td>
<td>2
</td></tr>
<tr>
<th>3
</th>
<td>4</td>
<td>14</td>
<td>9
</td></tr>
<tr>
<th>4
</th>
<td>2</td>
<td>31</td>
<td>20
</td></tr>
<tr>
<th>5
</th>
<td>1</td>
<td>45</td>
<td>29
</td></tr>
<tr>
<th>6
</th>
<td>2</td>
<td>121</td>
<td>78
</td></tr>
<tr>
<th>7
</th>
<td>1</td>
<td>166</td>
<td>107
</td></tr>
<tr>
<th>8
</th>
<td>5</td>
<td>951</td>
<td>613
</td></tr>
<tr>
<th>9
</th>
<td>2</td>
<td>2068</td>
<td>1333
</td></tr>
<tr>
<th>10
</th>
<td>1</td>
<td>3019</td>
<td>1946
</td></tr>
<tr>
<th>11
</th>
<td>3</td>
<td>11125</td>
<td>7171
</td></tr>
<tr>
<th>12
</th>
<td>1</td>
<td>14144</td>
<td>9117
</td></tr>
<tr>
<th>13
</th>
<td>1</td>
<td>25269</td>
<td>16288
</td></tr>
<tr>
<th>14
</th>
<td>390</td>
<td>9869054</td>
<td>6361437
</td></tr>
<tr>
<th>15
</th>
<td>1</td>
<td>9894323</td>
<td>6377725
</td></tr>
<tr>
<th>16
</th>
<td>1</td>
<td>19763377</td>
<td>12739162
</td></tr>
<tr>
<th>17
</th>
<td>2</td>
<td>49421077</td>
<td>31856049
</td></tr>
<tr>
<th>18
</th>
<td>11</td>
<td>563395224</td>
<td>363155701
</td></tr>
<tr>
<th>19
</th>
<td>6</td>
<td>3429792421</td>
<td>2210790255
</td></tr>
<tr>
<th>20
</th>
<td>2</td>
<td>7422980066</td>
<td>4784736211
</td></tr>
<tr>
<th>21
</th>
<td>1</td>
<td>10852772487</td>
<td>6995526466
</td></tr>
<tr>
<th>22
</th>
<td>1</td>
<td>18275752553</td>
<td>11780262677
</td></tr>
<tr>
<th>23
</th>
<td>56</td>
<td>1034294915455</td>
<td>666690236378
</td></tr>
<tr>
<th>24
</th>
<td>1</td>
<td>1052570668008</td>
<td>678470499055
</td></tr>
<tr>
<th>25
</th>
<td>4</td>
<td>5244577587487</td>
<td>3380572232598
</td></tr>
<tr>
<th>26
</th>
<td>3</td>
<td>16786303430469</td>
<td>10820187196849
</td></tr>
<tr>
<th>27
</th>
<td>1</td>
<td>22030881017956</td>
<td>14200759429447
</td></tr>
<tr>
<th>28
</th>
<td>1</td>
<td>38817184448425</td>
<td>25020946626296
</td></tr>
<tr>
<th>29
</th>
<td>6</td>
<td>254933987708506</td>
<td>164326439187223
</td></tr>
<tr>
<th>30
</th>
<td>9</td>
<td>2333223073824979</td>
<td>1503958899311303
</td></tr>
<tr>
<th>31
</th>
<td>3</td>
<td>7254603209183443</td>
<td>4676203137121132
</td></tr>
<tr>
<th>32
</th>
<td>2</td>
<td>16842429492191865</td>
<td>10856365173553567
</td></tr>
<tr>
<th>33
</th>
<td>1</td>
<td>24097032701375308</td>
<td>15532568310674699
</td></tr>
<tr>
<th>34
</th>
<td>8</td>
<td>209618691103194329</td>
<td>135116911658951159
</td></tr>
<tr>
<th>35
</th>
<td>10</td>
<td>2120283943733318598</td>
<td>1366701684900186289
</td></tr>
<tr>
<th>36
</th>
<td>9</td>
<td>19292174184703061711</td>
<td>12435432075760627760
</td></tr>
<tr>
<th>37
</th>
<td>25</td>
<td>484424638561309861373</td>
<td>312252503578915880289
</td></tr>
<tr>
<th>38
</th>
<td>2</td>
<td>988141451307322784457</td>
<td>636940439233592388338
</td></tr>
<tr>
<th>39
</th>
<td>1</td>
<td>1472566089868632645830</td>
<td>949192942812508268627
</td></tr>
<tr>
<th>40
</th>
<td>3</td>
<td>5405839720913220721947</td>
<td>3484519267671117194219
</td></tr>
<tr>
<th>41
</th>
<td>1</td>
<td>6878405810781853367777</td>
<td>4433712210483625462846
</td></tr>
<tr>
<th>42
</th>
<td>3</td>
<td>26041057153258780825278</td>
<td>16785655899121993582757
</td></tr>
<tr>
<th>43
</th>
<td>5</td>
<td>137083691577075757494167</td>
<td>88361991706093593376631
</td></tr>
<tr>
<th>44
</th>
<td>2</td>
<td>300208440307410295813612</td>
<td>193509639311309180336019
</td></tr>
<tr>
<th>45
</th>
<td>35</td>
<td>10644379102336436110970587</td>
<td>6861199367601914905137296
</td></tr>
<tr>
<th>46
</th>
<td>1</td>
<td>10944587542643846406784199</td>
<td>7054709006913224085473315
</td></tr>
<tr>
<th>47
</th>
<td>1</td>
<td>21588966644980282517754786</td>
<td>13915908374515138990610611
</td></tr>
<tr>
<th>48
</th>
<td>1</td>
<td>32533554187624128924538985</td>
<td>20970617381428363076083926
</td></tr>
<tr>
<th>49
</th>
<td>41</td>
<td>1355464688337569568423853171</td>
<td>873711221013078025110051577
</td></tr>
<tr>
<th>50
</th>
<td>1</td>
<td>1387998242525193697348392156</td>
<td>894681838394506388186135503
</td></tr>
<tr>
<th>51
</th>
<td>2</td>
<td>4131461173387956963120637483</td>
<td>2663074897802090801482322583
</td></tr>
<tr>
<th>52
</th>
<td>2</td>
<td>9650920589301107623589667122</td>
<td>6220831633998687991150780669
</td></tr>
<tr>
<th>53
</th>
<td>1</td>
<td>13782381762689064586710304605</td>
<td>8883906531800778792633103252
</td></tr>
<tr>
<th>54
</th>
<td>2</td>
<td>37215684114679236797010276332</td>
<td>23988644697600245576416987173
</td></tr>
<tr>
<th>55
</th>
<td>2</td>
<td>88213749992047538180730857269</td>
<td>56861195927001269945467077598
</td></tr>
<tr>
<th>56
</th>
<td>3</td>
<td>301856934090821851339202848139</td>
<td>194572232478604055412818219967
</td></tr>
<tr>
<th>57
</th>
<td>1</td>
<td>390070684082869389519933705408</td>
<td>251433428405605325358285297565
</td></tr>
<tr>
<th>58
</th>
<td>4</td>
<td>1862139670422299409418937669771</td>
<td>1200305946101025356845959410227
</td></tr>
<tr>
<th>59
</th>
<td>2</td>
<td>4114350024927468208357809044950</td>
<td>2652045320607656039050204118019
</td></tr>
<tr>
<th>60
</th>
<td>1</td>
<td>5976489695349767617776746714721</td>
<td>3852351266708681395896163528246
</td></tr>
<tr>
<th>61
</th>
<td>1</td>
<td>10090839720277235826134555759671</td>
<td>6504396587316337434946367646265
</td></tr>
<tr>
<th>62
</th>
<td>1</td>
<td>16067329415627003443911302474392</td>
<td>10356747854025018830842531174511
</td></tr>
<tr>
<th>63
</th>
<td>1</td>
<td>26158169135904239270045858234063</td>
<td>16861144441341356265788898820776
</td></tr>
<tr>
<th>64
</th>
<td>3</td>
<td>94541836823339721254048877176581</td>
<td>60940181178049087628209227636839
</td></tr>
<tr>
<th>65
</th>
<td>1</td>
<td>120700005959243960524094735410644</td>
<td>77801325619390443893998126457615
</td></tr>
<tr>
<th>66
</th>
<td>6</td>
<td>818741872578803484398617289640445</td>
<td>527748134894391750992197986382529
</td></tr>
<tr>
<th>67
</th>
<td>2</td>
<td>1758183751116850929321329314691534</td>
<td>1133297595408173945878394099222673
</td></tr>
<tr>
<th>68
</th>
<td>1</td>
<td>2576925623695654413719946604331979</td>
<td>1661045730302565696870592085605202
</td></tr>
<tr>
<th>69
</th>
<td>4</td>
<td>12065886245899468584201115732019450</td>
<td>7777480516618436733360762441643481
</td></tr>
<tr>
<th>70
</th>
<td>11</td>
<td>135301674328589808839932219656545929</td>
<td>87213331413105369763838978943683493
</td></tr>
<tr>
<th>71
</th>
<td>1</td>
<td>147367560574489277424133335388565379</td>
<td>94990811929723806497199741385326974
</td></tr>
<tr>
<th>72
</th>
<td>2</td>
<td>430036795477568363688198890433676687</td>
<td>277194955272552982758238461714337441
</td></tr>
<tr>
<th>73
</th>
<td>2</td>
<td>1007441151529626004800531116255918753</td>
<td>649380722474829772013676664814001856
</td></tr>
<tr>
<th>74
</th>
<td>1</td>
<td>1437477947007194368488730006689595440</td>
<td>926575677747382754771915126528339297
</td></tr>
<tr>
<th>75
</th>
<td>1</td>
<td>2444919098536820373289261122945514193</td>
<td>1575956400222212526785591791342341153
</td></tr>
<tr>
<th>76
</th>
<td>6</td>
<td>16106992538228116608224296744362680598</td>
<td>10382314079080657915485465874582386215
</td></tr>
<tr>
<th>77
</th>
<td>3</td>
<td>50765896713221170197962151356033555987</td>
<td>32722898637464186273241989415089499798
</td></tr>
<tr>
<th>78
</th>
<td>1</td>
<td>66872889251449286806186448100396236585</td>
<td>43105212716544844188727455289671886013
</td></tr>
<tr>
<th>79
</th>
<td>1</td>
<td>117638785964670457004148599456429792572</td>
<td>75828111354009030461969444704761385811
</td></tr>
<tr>
<th>80
</th>
<td>1</td>
<td>184511675216119743810335047556826029157</td>
<td>118933324070553874650696899994433271824
</td></tr>
<tr>
<th>81
</th>
<td>1</td>
<td>302150461180790200814483647013255821729</td>
<td>194761435424562905112666344699194657635
</td></tr>
<tr>
<th>82
</th>
<td>1</td>
<td>486662136396909944624818694570081850886</td>
<td>313694759495116779763363244693627929459
</td></tr>
<tr>
<th>83
</th>
<td>1</td>
<td>788812597577700145439302341583337672615</td>
<td>508456194919679684876029589392822587094
</td></tr>
<tr>
<th>84
</th>
<td>4</td>
<td>3641912526707710526382028060903432541346</td>
<td>2347519539173835519267481602264918277835
</td></tr>
<tr>
<th>85
</th>
<td>1</td>
<td>4430725124285410671821330402486770213961</td>
<td>2855975734093515204143511191657740864929
</td></tr>
<tr>
<th>86
</th>
<td>7</td>
<td>34656988396705585229131340878310824039073</td>
<td>22339349677828441948272059943869104332338
</td></tr>
<tr>
<th>87
</th>
<td>2</td>
<td>73744701917696581130084012159108418292107</td>
<td>47534675089750399100687631079395949529605
</td></tr>
<tr>
<th>88
</th>
<td>2</td>
<td>182146392232098747489299365196527660623287</td>
<td>117408699857329240149647322102661003391548
</td></tr>
<tr>
<th>89
</th>
<td>2</td>
<td>438037486381894076108682742552163739538681</td>
<td>282352074804408879399982275284717956312701
</td></tr>
<tr>
<th>90
</th>
<td>36</td>
<td>15951495901980285487401878097074422284015803</td>
<td>10282083392816048898549009232352507430648784
</td></tr>
<tr>
<th>91
</th>
<td>7</td>
<td>112098508800243892487921829422073119727649302</td>
<td>72256935824516751169243046901752269970854189
</td></tr>
<tr>
<th>92
</th>
<td>22</td>
<td>2482118689507345920221682125382683056292300447</td>
<td>1599934671532184574621896041070902446789440942
</td></tr>
<tr>
<th>93
</th>
<td>1</td>
<td>2594217198307589812709603954804756176019949749</td>
<td>1672191607356701325791139087972654716760295131
</td></tr>
<tr>
<th>94
</th>
<td>2</td>
<td>7670553086122525545640890034992195408332199945</td>
<td>4944317886245587226204174217016211880310031204
</td></tr>
<tr>
<th>95
</th>
<td>1</td>
<td>10264770284430115358350493989796951584352149694</td>
<td>6616509493602288551995313304988866597070326335
</td></tr></table>

The rational approxmation of {{mvar|x}} is {{sfrac|{{mvar|h}}{{sub|95}}|{{mvar|k}}{{sub|95}}}} and an error bounds {{mvar|ε}} is as follows:
:<math>
\begin{align}
x &\approx \frac{h_{95}}{k_{95}} \\
  &= \frac{10264770284430115358350493989796951584352149694}{6616509493602288551995313304988866597070326335} \\
  &= 1.5513875245483203922619525102646238151635917038038871995280071201179267425542569572957604536135484903\cdots, \\
\varepsilon &= \frac{1}{k_{95}(k_{95}+k_{94})} \\
            &= 1.82761... \times 10^{-91}.
\end{align}</math>}}

=== base angle and apex angle ===
The Calabi triangle is [[Obtuse triangle|obtuse]] with base angle {{mvar|θ}} and apex angle {{mvar|ψ}} as follows:
:<math>\begin{align}
\theta &= \cos^{-1}(x/2) \\
       &= 39.13202614232587442003651601935656349795831966723206\cdots^\circ, \\
\psi   &= 180 - 2\theta \\
       &= 101.73594771534825115992696796128687300408336066553587\cdots^\circ. \\
\end{align}</math>

== See also ==
{{Div col}}
*[[Biggest little polygon]]
*[[Cubic equation]]
{{Div col end}}

== Footnotes ==
=== Notes ===
{{notelist}}

=== Citations ===
{{reflist|2}}

== References ==
*{{Citation
|last=Stewart
|first=Ian
|author-link=Ian Stewart (mathematician)
|year=2004
|title=Galois Theory
|publisher=Chapman and Hall/CRC
|edition=3rd
|isbn=978-1-58488-393-7
}} - [https://math.berkeley.edu/~gbergman/ug.hndts/m114_IStwrt_GT3_err.ps ''Galois Theory'' Errata].

== External links ==
*{{mathworld|title=Calabi's Triangle|id=CalabisTriangle}}

[[Category:Types of triangles]]
[[Category:Eponymous geometric shapes]]
[[Category:Cubic irrational numbers]]