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{1}{7}

3) Pöörame kolmnurki BJH, CKI ja ALG 180⁰ 
    ümber punktide J, K ja L. 

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

Feynmani kolmnurga ülesande üldistus

n = 4 (slider from 1 to 4, step 1)
α = 3.141592653589793 rad (slider from 0 to 180°, step °)
Q_4: n > 2
p = 4 (slider from 2 to 10, step 1)
A = (0.6418202191043687, 5.07634482553771)
B = (-0.020675757501811054, -0.5019052019818842)
C = (6.375421803473796, -0.45897837271359143)
poly1_1, c, a, b = Polygon(A, B, C)
d_1 = Circle(A, (1 / p * c))
e_1 = Circle(B, (1 / p * a))
f_1 = Circle(C, (1 / p * b))
F = Intersect(d_1, c, 1)
D = Intersect(e_1, a, 1)
E = Intersect(f_1, b, 1)
g = Segment(B, E)
h = Segment(A, D)
i = Segment(F, C)
G = Intersect(h, i)
H = Intersect(g, h)
I = Intersect(g, i)
poly2_1, i_2, g_2, h_2 = Polygon(G, H, I)
text1_1: "Area of ABC: " + poly1_1 + ""
text2_1: "Area of GHI: " + poly2_1 + ""
i_3 = Segment(G, H)
p_1 = Segment(H, I)
q = Segment(I, G)
Q_1: n ≥ 7
tulemus1: (p - 2)^(2)
tulemus2: p^(2) - p + 1
tulemus3: Area[poly2_1] / Area[poly1_1]
tulemus4: tulemus1 / GCD[tulemus1, tulemus2]
tulemus5: tulemus2 / GCD[tulemus1, tulemus2]
tekst4: "\frac{S_{GHI}}{S_{ABC}}= 
\frac{\left(" + (LaTeX[p]) + "-2\right)^2 }{" + (LaTeX[p]) + "^2-" + (LaTeX[p]) + "+1}=
\frac{" + (LaTeX[tulemus4]) + "}{" + (LaTeX[tulemus5]) + "}"
tekst5: "\frac{S_{GHI}}{S_{ABC}}= 
\frac{" + (LaTeX[poly2_1]) + "}{" + (LaTeX[poly1_1]) + "}≈" + (LaTeX[tulemus3]) + "≈
\frac{" + (LaTeX[tulemus4]) + "}{" + (LaTeX[tulemus5]) + "}"
Export_1 = (-1.097714664616557, 8.309833197630729)
Export_2 = (23.358099288871763, -2.1069109884157684)