排容(取捨)原理

r = 4
A: (Corner[1] + Corner[3]) / 2 + (3, 2)
B: A + (((-r)) / 2, ((-sqrt(3)) * r / 2))
C: A + (r / 2, ((-sqrt(3)) * r / 2))
c_1 = Circle(A, (3 / 4 * r))
c_2 = Circle(B, (3 * r / 4))
c_3 = Circle(C, (3 * r / 4))
D = Intersect(c_1, c_2, 2)
E = Intersect(c_2, c_3, 2)
F = Intersect(c_1, c_3, 1)
G = Intersect(c_2, c_3, 1)
H = Intersect(c_1, c_3, 2)
I = Intersect(c_1, c_2, 1)
a_0 = CircleArc(A, I, D)
a_1 = CircleArc(B, D, I)
list1: {a_0, a_1}
P_1 = Point(list1)
Q_1: P_1 + (0, 0)
lcs1 = Locus(Q_1, P_1)
a_2 = CircleArc(C, G, E)
a_3 = CircleArc(B, E, G)
list2: {a_2, a_3}
P_2 = Point(list2)
Q_2: P_2 + (0, 0)
lcs2 = Locus(Q_2, P_2)
a_4 = CircleArc(A, H, F)
a_5 = CircleArc(C, F, H)
list3: {a_4, a_5}
P_3 = Point(list3)
Q_3: P_3 + (0, 0)
lcs3 = Locus(Q_3, P_3)
a_7 = CircleArc(C, G, P_3)
a_8 = CircleArc(A, P_3, I)
a_6 = CircleArc(B, I, G)
list4: {a_6, a_7, a_8}
P_4 = Point(list4)
Q_4: P_4 + (0, 0)
lcs4 = Locus(Q_4, P_4)
a_9 = CircleArc(C, E, F)
a_a = CircleArc(A, I, F)
a_b = CircleArc(B, E, I)
list5: {a_9, a_a, a_b}
P_5 = Point(list5)
Q_5: P_5 + (0, 0)
lcs5 = Locus(Q_5, P_5)
a_c = CircleArc(A, F, D)
a_d = CircleArc(B, D, P_5)
a_e = CircleArc(C, P_5, F)
list6: {a_c, a_d, a_e}
P_6 = Point(list6)
Q_6: P_6 + (0, 0)
lcs6 = Locus(Q_6, P_6)
a_f = CircleArc(A, D, I)
a_g = CircleArc(B, I, D)
list7: {a_f, a_g}
P_7 = Point(list7)
Q_7: P_7 + (0, 0)
lcs7 = Locus(Q_7, P_7)
n = 0 (slider from 0 to 15, step 1)
text2: "" + (LaTeX[If[n ≟ 1, "A", If[n ≟ 2, "B", If[n ≟ 3, "C", If[n ≟ 4, "A∩B", If[n ≟ 5, "B∩C", If[n ≟ 6, "C∩A", If[n ≟ 7, "A∩B∩C", "A∪B∪C"]]]]]]]]) + ""
text1: "P(A∪B∪C)" + (LaTeX[If[n ≥ 9, "=P(A∪B)"]]) + "" + (LaTeX[If[n ≥ 10, "+x"]]) + "
\cr " + (LaTeX[If[n ≥ 11, "⇒x=P(C)"]]) + "" + (LaTeX[If[n ≥ 12, "-P(B∩C)"]]) + "" + (LaTeX[If[n ≥ 13, "-P(C∩A)"]]) + "" + (LaTeX[If[n ≥ 14, "+P(A∩B∩C)"]]) + "
\cr " + (LaTeX[If[n ≟ 15, "⇒P(A∪B∪C)"]]) + "
\cr " + (LaTeX[If[n ≟ 15, "=P(A)+P(B)+P(C)"]]) + "
\cr " + (LaTeX[If[n ≟ 15, "\;\;\;-P(A∪B)-P(B∩C)-P(C∩A)+P(A∩B∩C)"]]) + ""
c_a: c_1
c_b: c_2
c_c: c_3