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] 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.[ 3] [ 4]
Shape
The triangle △ABC is isosceles which has the same length of sides as AB = AC x , we can set that AB = AC = 1, BC = x
case 1) △ABC is acute triangle
The condition is
0
<
x
<
2
{\displaystyle 0<x<{\sqrt {2}}}
In this case x = 1equilateral triangle .
case 2) △ABC is right triangle
The condition is
x
=
2
{\displaystyle x={\sqrt {2}}}
In this case no value is valid.
case 3) △ABC is obtuse triangle
The condition is
2
<
x
<
2
{\displaystyle {\sqrt {2}}<x<2}
In this case the Calabi triangle is valid for the largest positive root of
2
x
3
−
2
x
2
−
3
x
+
2
=
0
{\displaystyle 2x^{3}-2x^{2}-3x+2=0}
x
=
1.55138752454832039226...
{\displaystyle x=1.55138752454832039226...}
OEIS : A046095 Example of answer
Example figure of Calabi triangle 01 Consider the case of AB = AC = 1, BC = x
0
<
x
<
2.
{\displaystyle 0<x<2.}
Let a base angle be θ and a square be □DEFG on base BC a .
Let H be the foot of the perpendicular drawn from the apex A to the base. Then
H
B
=
H
C
=
cos
θ
=
x
2
,
A
H
=
sin
θ
=
x
2
tan
θ
,
0
<
θ
<
π
2
.
{\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 HE = a / 2 EB = x - a / 2
From △DEB ∽ △AHB,
E
B
:
D
E
=
H
B
:
A
H
⇔
(
x
−
a
2
)
:
a
=
cos
θ
:
sin
θ
=
1
:
tan
θ
⇔
a
=
(
x
−
a
2
)
tan
θ
⇔
a
=
x
tan
θ
tan
θ
+
2
.
{\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}}}
△ABC is acute triangleExample figure of Calabi triangle 02 Let □IJKL be a square on side AC b .
From △ABC ∽ △IBJ,
A
B
:
I
J
=
B
C
:
B
J
⇔
1
:
b
=
x
:
B
J
⇔
B
J
=
b
x
.
{\displaystyle {\begin{aligned}&AB:IJ=BC:BJ\\&\Leftrightarrow 1:b=x:BJ\\&\Leftrightarrow BJ=bx.\end{aligned}}}
From △JKC ∽ △AHC,
J
K
:
J
C
=
A
H
:
A
C
⇔
b
:
J
C
=
x
2
tan
θ
:
1
⇔
J
C
=
2
b
x
tan
θ
.
{\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
x
=
B
C
=
B
J
+
J
C
=
b
x
+
2
b
x
tan
θ
⇔
x
=
b
x
2
tan
θ
+
2
x
tan
θ
⇔
b
=
x
2
tan
θ
x
2
tan
θ
+
2
.
{\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,
a
=
b
⇔
x
tan
θ
tan
θ
+
2
=
x
2
tan
θ
x
2
tan
θ
+
2
⇔
x
tan
θ
⋅
(
x
2
tan
θ
+
2
)
=
x
2
tan
θ
(
tan
θ
+
2
)
⇔
x
tan
θ
⋅
(
x
(
tan
θ
+
2
)
−
(
x
2
tan
θ
+
2
)
)
=
0
⇔
x
tan
θ
⋅
(
x
tan
θ
−
2
)
⋅
(
x
−
1
)
=
0
⇔
2
sin
θ
⋅
2
(
sin
θ
−
1
)
⋅
(
x
−
1
)
=
0.
{\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,
π
4
<
θ
<
π
2
,
2
sin
θ
⋅
2
(
sin
θ
−
1
)
≠
0.
{\displaystyle {\frac {\pi }{4}}<\theta <{\frac {\pi }{2}},2\sin \theta \cdot 2(\sin \theta -1)\neq 0.}
Therefore
x
=
1
{\displaystyle x=1}
△ABC is equilateral triangle.
△ABC is right triangleExample figure of Calabi triangle 03 In this case,
x
=
2
,
tan
θ
=
1
{\displaystyle x={\sqrt {2}},\tan \theta =1}
a
=
2
3
,
b
=
1
2
.
{\displaystyle a={\frac {\sqrt {2}}{3}},b={\frac {1}{2}}.}
Then no value is valid.
△ABC is obtuse triangleExample figure of Calabi triangle 04 Let □IJKA be a square on base AC b .
From △AHC ∽ △JKC,
A
H
:
H
C
=
J
K
:
K
C
⇔
sin
θ
:
cos
θ
=
b
:
(
1
−
b
)
⇔
b
cos
θ
=
(
1
−
b
)
sin
θ
⇔
b
=
(
1
−
b
)
tan
θ
⇔
b
=
tan
θ
1
+
tan
θ
.
{\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,
a
=
b
⇔
x
tan
θ
tan
θ
+
2
=
tan
θ
1
+
tan
θ
⇔
x
tan
θ
+
2
=
1
1
+
tan
θ
⇔
x
(
tan
θ
+
1
)
=
tan
θ
+
2
⇔
(
x
−
1
)
tan
θ
=
2
−
x
.
{\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,
tan
θ
=
(
2
+
x
)
(
2
−
x
)
x
.
{\displaystyle \tan \theta ={\frac {\sqrt {(2+x)(2-x)}}{x}}.}
So, we can input the value of tanθ ,
(
x
−
1
)
tan
θ
=
2
−
x
⇔
(
x
−
1
)
(
2
+
x
)
(
2
−
x
)
x
=
2
−
x
⇔
(
2
−
x
)
⋅
(
(
x
−
1
)
2
(
2
+
x
)
−
x
2
(
2
−
x
)
)
=
0
⇔
(
2
−
x
)
⋅
(
2
x
3
−
2
x
2
−
3
x
+
2
)
=
0.
{\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,
2
<
x
<
2
{\displaystyle {\sqrt {2}}<x<2}
2
x
3
−
2
x
2
−
3
x
+
2
=
0.
{\displaystyle 2x^{3}-2x^{2}-3x+2=0.}
If x is the largest positive root of Calabi's equation:
2
x
3
−
2
x
2
−
3
x
+
2
=
0
,
2
<
x
<
2
{\displaystyle 2x^{3}-2x^{2}-3x+2=0,{\sqrt {2}}<x<2}
we can calculate the value of x by following methods.
We can set the function
f
:
R
→
R
{\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }
f
(
x
)
=
2
x
3
−
2
x
2
−
3
x
+
2
,
f
′
(
x
)
=
6
x
2
−
4
x
−
3
=
6
(
x
−
1
3
)
2
−
11
3
.
{\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
R
{\displaystyle \mathbb {R} }
f
(
2
)
=
2
−
2
<
0
,
f
(
2
)
=
4
>
0
,
f
′
(
x
)
>
0
,
∀
x
∈
[
2
,
2
]
.
{\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
2
<
x
<
2
{\displaystyle {\sqrt {2}}<x<2}
The value of x is calculated by Newton's method as follows:
x
0
=
2
,
x
n
+
1
=
x
n
−
f
(
x
n
)
f
′
(
x
n
)
=
4
x
n
3
−
2
x
n
2
−
2
6
x
n
2
−
4
x
n
−
3
.
{\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
x 0
1. 41421356237309504880168872420969807856967187537694...
x 1
1.5 8943369375323596617308283187888791370090306159374...
x 2
1.55 324943049375428807267665439782489231871295592784...
x 3
1.5513 9234383942912142613029570413117306471589987689...
x 4
1.5513875245 8074244056538641010106649611908076010328...
x 5
1.55138752454832039226 341994813293555945836732015691...
x 6
1.55138752454832039226195251026462381516359 470986821...
x 7
1.55138752454832039226195251026462381516359170380388 ...
The value of x can expressed with complex numbers by using Cardano's method :
x
=
1
3
(
1
+
−
23
+
3
i
237
4
3
+
−
23
−
3
i
237
4
3
)
.
{\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] [ a]
The value of x can also be expressed without complex numbers by using Viète's method :
x
=
1
3
(
1
+
22
cos
(
1
3
cos
−
1
(
−
23
11
22
)
)
)
=
1.55138752454832039226195251026462381516359170380389
⋯
.
{\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}}}
The value of x has continued fraction representation by Lagrange 's method as follows:
1
+
1
1
+
1
1
+
1
4
+
1
2
+
1
1
+
1
2
+
1
1
+
1
5
+
1
2
+
1
1
+
1
3
+
1
1
+
1
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] [ b]
base angle and apex angle
The Calabi triangle is obtuse with base angle θ and apex angle ψ as follows:
θ
=
cos
−
1
(
x
/
2
)
=
39.13202614232587442003651601935656349795831966723206
⋯
∘
,
ψ
=
180
−
2
θ
=
101.73594771534825115992696796128687300408336066553587
⋯
∘
.
{\displaystyle {\begin{aligned}\theta &=\cos ^{-1}(x/2)\\&=39.13202614232587442003651601935656349795831966723206\cdots ^{\circ },\\\psi &=180-2\theta \\&=101.73594771534825115992696796128687300408336066553587\cdots ^{\circ }.\\\end{aligned}}}
See also
Notes
^ If we set the polar form of complex number, we can calculate the value of x as follows:
α
=
r
e
i
φ
=
r
(
cos
φ
+
i
sin
φ
)
=
−
23
+
3
i
237
4
,
r
=
1
4
(
−
23
)
2
+
9
⋅
237
=
1
4
2
⋅
11
3
=
(
11
2
)
3
,
cos
φ
=
−
23
4
1
r
=
−
23
⋅
2
2
4
⋅
11
11
=
−
23
11
22
,
α
3
=
r
3
e
i
φ
3
=
r
3
(
cos
(
φ
3
)
+
i
sin
(
φ
3
)
)
,
α
3
+
α
¯
3
=
2
r
3
cos
(
φ
3
)
=
22
cos
(
1
3
cos
−
1
(
−
23
11
22
)
)
,
x
=
1
3
(
1
+
α
3
+
α
¯
3
)
=
1
3
(
1
+
22
cos
(
1
3
cos
−
1
(
−
23
11
22
)
)
)
.
{\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}}}
Viète's method .
^ If a continued fraction [a 0 , a 1 , a 2 , ...] are found, with numerators h 1 , h 2 , ... and denominators k 1 , k 2 , ... then the relevant recursive relation is that of Gaussian brackets :
h n a n h n − 1h n − 2k n a n k n − 1k n − 2h n / k n = a n h n − 1h n − 2/ a n k n − 1k n − 2 [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] x is as follows:
The value of numerators hn and denominators kn 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
The rational approxmation of x is h 95 / k 95 and an error bounds ε is as follows:
x
≈
h
95
k
95
=
10264770284430115358350493989796951584352149694
6616509493602288551995313304988866597070326335
=
1.5513875245483203922619525102646238151635917038038871995280071201179267425542569572957604536135484903
⋯
,
ε
=
1
k
95
(
k
95
+
k
94
)
=
1.82761...
×
10
−
91
.
{\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}}}
Citations
^ 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 .^ a b c Weisstein, Eric W. "Calabi's Triangle" . MathWorld ^ Conway, J.H. ; Guy, R.K. (1996). "Calabi's Triangle" . The Book of Numbers . New York: Springer-Verlag. p. 206.^ 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. ^ 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. ^ (sequence A046096 OEIS )
References
External links
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66e4afeff8b2655b441e6111d842ff62bb27fcc3)
![{\displaystyle x={1 \over 3}{\Bigg (}1+{\sqrt[{3}]{-23+3i{\sqrt {237}} \over 4}}+{\sqrt[{3}]{-23-3i{\sqrt {237}} \over 4}}{\Bigg )}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/823bc02ed3a453a10f81d062a8caaeaf450096b5)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e29491473aa1576936006ec8df2184968a3a5724)
