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

\frac{S_G_H_I}{S_A_B_C}=\frac{\left(  p-2 \right)² }{p²-p+1}.

Feynmani kolmnurga ülesande üldistus

n = 5 (slider from 1 to 5, step 1)
p = 4 (slider from 2 to 10, step 1)
A = (-3.108922343511053, 3.0344050793650794)
B = (-5.410839921953285, -6.741270870642827)
C = (3.1690346886041247, -6.6814808036703015)
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(B, E)
i = Segment(C, F)
G = Intersect(g, i)
H = Intersect(g, h)
I = Intersect(h, i)
t2, i_1, g_1, h_1 = Polygon(G, H, I)
Punktt1 = PointIn(t1)
Punktt2 = PointIn(t2)
lug: (p - 2)^(2)
nim: p^(2) - p + 1
lug1: lug / GCD[lug, nim]
nim1: nim / GCD[lug, nim]
tekst2_{1}: "\frac{S_G_H_I}{S_A_B_C}=\frac{\left(  " + (LaTeX[p]) + "-2 \right)² }{" + (LaTeX[p]) + "²-" + (LaTeX[p]) + "+1}=\frac{" + (LaTeX[lug1]) + "}{" + (LaTeX[nim1]) + "}"
väärtus: t2 / t1
tekst2_{3}: "\frac{S_G_H_I}{S_A_B_C}=\frac{" + (LaTeX[t2]) + "}{" + (LaTeX[t1]) + "}≈" + (LaTeX[väärtus]) + "≈\frac{" + (LaTeX[lug1]) + "}{" + (LaTeX[nim1]) + "}"