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

Feymani kolmnurga ülesande üldistus

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

n = 5 (slider from 1 to 5, step 1)
p = 6 (slider from 2 to 10, step 1)
A = (0.58, 1.2)
B = (-2.3, -2.08)
C = (5.02, -2.16)
t1, c, a, b = Polygon(A, B, C)
d = Circle(A, c / p)
e = Circle(B, a / p)
f = Circle(C, b / p)
F = Intersect(d, c)
D = Intersect(e, a)
E = Intersect(f, b)
g = Segment(D, A)
h = Segment(F, C)
i = Segment(E, B)
G = Intersect(g, i)
H = Intersect(h, i)
I = Intersect(g, h)
t2, h_1, i_1, g_1 = Polygon(I, G, H)
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{(" + (LaTeX[p]) + "-2)^{2}}{" + (LaTeX[p]) + "^2-" + (LaTeX[p]) + "+1}=\frac{" + (LaTeX[lug1]) + "}{" + (LaTeX[nim1]) + "}"
väärtus: t2 / t1
tekst5: "\frac{S_{GHI}}{S_{ABC}}≈\frac{" + (LaTeX[t1]) + "}{" + (LaTeX[t2]) + "}≈" + (LaTeX[väärtus]) + "≈\frac{" + (LaTeX[lug1]) + "}{" + (LaTeX[nim1]) + "}"