Feymani kolmnurga ülesande üldistus

Kui kolmnurga ABC küljed jagada p osaks ning
seejärel ühendada kolmnurga tipud vastupäeva vastaskülje
sellise punktiga, mis jaotab vastaskülje suhtes
1/(p-1), siis tekkiva kolmnurga GHI ja kolmnurga ABC
pindalade suhe on

\frac{S_{GHI}}{S_{ABC}}=\frac{\left(p-2   \right)^2 }{p^2-p+1}.

n = 5 (slider from 1 to 5, step 1)
p = 4 (slider from 2 to 10, step 1)
A = (0.58, 3.06)
B = (-2.8, -3.8)
C = (5.7, -3.98)
t1, c, a, b = Polygon(A, B, C)
d = Circle(A, c / p)
e = Circle(B, a / p)
f = Circle(C, b / p)
D = Intersect(e, a, 1)
E = Intersect(f, b, 1)
F = Intersect(d, c, 1)
g = Segment(A, D)
h = Segment(C, F)
i = Segment(B, E)
G = Intersect(g, h)
H = Intersect(g, i)
I = Intersect(h, i)
t2, i_1, g_1, h_1 = Polygon(G, H, I)
Lug: (p - 2)^(2)
Nim: p^(2) - p + 1
Lug1: Lug / GCD[Lug, Nim]
Nim1: Nim / GCD[Lug, Nim]
tekst4: "\frac{S_{GHI}}{S_{ABC}}=\frac{\left(" + (LaTeX[p]) + "-2   \right)^2 }{" + (LaTeX[p]) + "^2-" + (LaTeX[p]) + "+1}=\frac{" + (LaTeX[Lug1]) + "}{" + (LaTeX[Nim1]) + "}"
väärtus: t2 / t1
tekst5: "\frac{S_{GHI}}{S_{ABC}}=\frac{" + (LaTeX[t2]) + "}{" + (LaTeX[t1]) + "}≈" + (LaTeX[väärtus]) + "≈\frac{" + (LaTeX[Lug1]) + "}{" + (LaTeX[Nim1]) + "}"