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.^[1] It is an isosceles triangle which is obtuse with an irrational but algebraic ratio between the lengths of its sides and its base.^[2]

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.

Shape

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 ${\displaystyle 0<x<{\sqrt {2}}}$.

In this case x = 1 is valid for equilateral triangle.

case 2) △ABC is right triangle

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

In this case no value is valid.

case 3) △ABC is obtuse triangle

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

In this case the Calabi triangle is valid for the largest positive root of ${\displaystyle 2x^{3}-2x^{2}-3x+2=0}$ at ${\displaystyle x=1.55138752454832039226...}$ (OEIS: A046095).

Example of answer

Example figure of Calabi triangle 01

Consider the case of AB = AC = 1, BC = x. Then

${\displaystyle 0<x<2.}$

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

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

Then HB = ⁠x/2⁠ and HE = ⁠a/2⁠, so EB = ⁠x - a/2⁠.

From △DEB ∽ △AHB,

${\displaystyle {\begin{aligned}&EB:DE=HB:AH\\&\Leftrightarrow {\bigg (}{\frac {x-a}{2}}{\bigg )}:a=\cos \theta$ :\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{aligned}}}

case 1) △ABC is acute triangle

Example figure of Calabi triangle 02

Let □IJKL be a square on side AC with its side length as b. From △ABC ∽ △IBJ,

${\displaystyle {\begin{aligned}&AB:IJ=BC:BJ\\&\Leftrightarrow 1:b=x:BJ\\&\Leftrightarrow BJ=bx.\end{aligned}}}$

From △JKC ∽ △AHC,

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

Then

${\displaystyle {\begin{aligned}&x=BC=BJ+JC=bx+{\frac {2b}{x\tan \theta }}\\&\Leftrightarrow x=b{\frac {x^{2}\tan \theta +2}{x\tan \theta }}\\&\Leftrightarrow b={\frac {x^{2}\tan \theta }{x^{2}\tan \theta +2}}.\end{aligned}}}$

Therefore, if two squares are congruent,

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

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

Therefore ${\displaystyle x=1}$, it means that △ABC is equilateral triangle.

case 2) △ABC is right triangle

Example figure of Calabi triangle 03

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

Then no value is valid.

case 3) △ABC is obtuse triangle

Example figure of Calabi triangle 04

Let □IJKA be a square on base AC with its side length as b.

From △AHC ∽ △JKC,

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

Therefore, if two squares are congruent,

${\displaystyle {\begin{aligned}&a=b\\&\Leftrightarrow {\frac {x\tan \theta }{\tan \theta +2}}={\frac {\tan \theta }{1+\tan \theta }}\\&\Leftrightarrow {\frac {x}{\tan \theta +2}}={\frac {1}{1+\tan \theta }}\\&\Leftrightarrow x(\tan \theta +1)=\tan \theta +2\\&\Leftrightarrow (x-1)\tan \theta =2-x.\end{aligned}}}$

In this case,

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

So, we can input the value of tanθ,

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

In this case, ${\displaystyle {\sqrt {2}}<x<2}$, we can get the following equation:

${\displaystyle 2x^{3}-2x^{2}-3x+2=0.}$

Root of Calabi's equation

If x is the largest positive root of Calabi's equation:

${\displaystyle 2x^{3}-2x^{2}-3x+2=0,{\sqrt {2}}<x<2}$

we can calculate the value of x by following methods.

Newton's method

We can set the function ${\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }$ as follows:

${\displaystyle {\begin{aligned}f(x)&=2x^{3}-2x^{2}-3x+2,\\f'(x)&=6x^{2}-4x-3=6{\bigg (}x-{\frac {1}{3}}{\bigg )}^{2}-{\frac {11}{3}}.\end{aligned}}}$

The function f is continuous and differentiable on ${\displaystyle \mathbb {R} }$ and

${\displaystyle {\begin{aligned}f({\sqrt {2}})&={\sqrt {2}}-2<0,\\f(2)&=4>0,\\f'(x)&>0,\forall x\in [{\sqrt {2}},2].\end{aligned}}}$

Then f is monotonically increasing function and by Intermediate value theorem, the Calabi's equation f(x) = 0 has unique solution in open interval ${\displaystyle {\sqrt {2}}<x<2}$.

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

${\displaystyle {\begin{aligned}x_{0}&={\sqrt {2}},\\x_{n+1}&=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}={\frac {4x_{n}^{3}-2x_{n}^{2}-2}{6x_{n}^{2}-4x_{n}-3}}.\end{aligned}}}$

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

NO	itaration value
x0	1.41421356237309504880168872420969807856967187537694...
x1	1.58943369375323596617308283187888791370090306159374...
x2	1.55324943049375428807267665439782489231871295592784...
x3	1.55139234383942912142613029570413117306471589987689...
x4	1.55138752458074244056538641010106649611908076010328...
x5	1.55138752454832039226341994813293555945836732015691...
x6	1.55138752454832039226195251026462381516359470986821...
x7	1.55138752454832039226195251026462381516359170380388...

Cardano's method

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

${\displaystyle x={1 \over 3}{\Bigg (}1+{\sqrt[{3}]{-23+3i{\sqrt {237}} \over 4}}+{\sqrt[{3}]{-23-3i{\sqrt {237}} \over 4}}{\Bigg )}.}$^{[3][5][lower-alpha 1]}

Viète's method

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

${\displaystyle {\begin{aligned}x&={1 \over 3}{\bigg (}1+{\sqrt {22}}\cos \!{\bigg (}{1 \over 3}\cos ^{-1}\!\!{\bigg (}\!-{23 \over 11{\sqrt {22}}}{\bigg )}{\bigg )}{\bigg )}\\&=1.55138752454832039226195251026462381516359170380389\cdots .\end{aligned}}}$^[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, ...] =

${\displaystyle 1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{4+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{5+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{3+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{390+\cdots }}}}}}}}}}}}}}}}}}}}}}}}}}}}}$.^{[3][6][7][lower-alpha 2]}

base angle and apex angle

The Calabi triangle is obtuse with base angle θ and apex angle ψ as follows:

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

See also

• Biggest little polygon
• Cubic equation

Footnotes

Notes

1. If we set the polar form of complex number, we can calculate the value of x as follows:

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

   Then this Cardano's method is equivalent as Viète's method.
2. If a continued fraction [a₀, a₁, a₂, ...] are found, with numerators h₁, h₂, ... and denominators k₁, k₂, ... then the relevant recursive relation is that of Gaussian brackets:

   h_n = a_nh_{n − 1} + h_{n − 2},

   k_n = a_nk_{n − 1} + k_{n − 2}.

   The successive convergents are given by the formula

   ⁠h_n/k_n⁠ = ⁠a_nh_{n − 1} + h_{n − 2}/a_nk_{n − 1} + k_{n − 2}⁠.

   If the continued fraction is
   

   [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, ...],^[8]

   we can calculate the rational approxmation of x is as follows: The rational approxmation of x is ⁠h₉₅/k₉₅⁠ and an error bounds ε is as follows:

   ${\displaystyle {\begin{aligned}x&\approx {\frac {h_{95}}{k_{95}}}\\&={\frac {10264770284430115358350493989796951584352149694}{6616509493602288551995313304988866597070326335}}\\&=1.5513875245483203922619525102646238151635917038038871995280071201179267425542569572957604536135484903\cdots ,\\\varepsilon &={\frac {1}{k_{95}(k_{95}+k_{94})}}\\&=1.82761...\times 10^{-91}.\end{aligned}}}$

   The value of numerators h_n and denominators k_n of continued fraction

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

Citations

1. Calabi, Eugenio (3 Nov 1997). "Outline of Proof Regarding Squares Wedged in Triangle". Archived from the original on 12 December 2012. Retrieved 3 May 2018.
2. Stewart 2004, p. 15.
3. Weisstein, Eric W. "Calabi's Triangle". MathWorld.
4. Conway, J.H.; Guy, R.K. (1996). "Calabi's Triangle". The Book of Numbers. New York: Springer-Verlag. p. 206.
5. Stewart 2004, pp. 7–10.
6. Joseph-Louis, Lagrange (1769), "Sur la résolution des équations numériques", Mémoires de l'Académie royale des Sciences et Belles-lettres de Berlin, 23 - Œuvres II, p.539-578.
7. Joseph-Louis, Lagrange (1770), "Additions au mémoire sur la résolution des équations numériques", Mémoires de l'Académie royale des Sciences et Belles-lettres de Berlin, 24 - Œuvres II, p.581-652.
8. (sequence A046096 in the OEIS)

References

• Stewart, Ian (2004), Galois Theory (3rd ed.), Chapman and Hall/CRC, ISBN 978-1-58488-393-7 - Galois Theory Errata.

External links

• Weisstein, Eric W. "Calabi's Triangle". MathWorld.