Author: P.Goubet

Aire du demi-dique de diamètre [AB]

Aire1

Aire2

 + 

Aire du triangle rectangle ABC

Aire3

Aire4

Aire5

 + 

Aire du demi-dique de diamètre [AC]

Aire du demi-dique de diamètre [BC]

Aire2

Aire3

Aire4

 + 

 + 

Aire1

Aire2

Aire3

Aire4

Aire5

+

+

 ) + 

 ( 

+ ( 

 ) 

 - ( 

Aire2

 + 

Aire3

 + 

Aire4

)

 = 

Aire1

 + 

Aire5

 + 

+

-

=

Calculer la somme des aires des deux  lunules roses ( Aire1 + Aire5 )

Les lunules d'Hippocrate

Déformer le triangle ABC

La somme des aires des deux lunules est donc égale à l'aire du triangle ABC

α = 1.1344640137963142 rad (slider from 0 to 3.141592653589793, step 0.017453292519943295)
C = (6.539999999999997, 0.040000000000000036)
B = (-0.18, 0)
O = Midpoint(B, C)
A = Rotate(C, α, O)
I = Midpoint(A, B)
h = CircleSector(I, A, B)
aireh = Area(h)
F = Midpoint(A, C)
k = CircleSector(F, C, A)
airek = Area(k)
ABC, BC, AC, AB = Polygon(B, C, A)
g = CircleSector(O, C, B)
aireg = Area(g)
lunules: aireh + ABC + airek - aireg
β = Angle(B, A, C)
f = Semicircle(A, C)
Aire2 = Point(AB)
Aire3 = Point(BC)
Aire4 = Point(AC)
Aire5 = Point(f)
Aire1 = Point(h)
T2: " \frac{ AB }{2 }  x \frac{ AB }{2 } x π = " + (AB / 2) + "x" + (AB / 2) + "x" + 3.141592653589793 + "=" + aireh
T1: "AB = " + AB
T3: "AC = " + AC
T4: "BC = " + BC
T6: " \frac{ AC }{2 }  x \frac{ AC }{2 } x π = " + (AC / 2) + "x" + (AC / 2) + "x" + 3.14 + "=" + airek
T8: " \frac{ BC }{2 }  x \frac{ BC }{2 } x π = " + (BC / 2) + "x" + (BC / 2) + "x" + 3.14 + "=" + aireg
TexteABC: "Aire " + (Name[B]) + (Name[C]) + (Name[A]) + " = " + ABC
T10: " \frac{ 1 }{2 }  x AB x AC  = " + (1 / 2) + "x" + AB + "x" + AC + "=" + ABC
T45: aireh + ""
T46: ABC + ""
T47: airek + ""
T48: aireg + ""
T53: lunules + ""