That's a lot of circles you have crammed into funny places

Lagrangian points in a circle could form an equilateral triangle. 2D shapes are false.

>>16968614
>a rather complex task
yeah; no thanks

>>16968614
It could be made easier, no?

>>16968614


135
If no one can solve this problem, then it seems to me a task for a true specialist in Japanese Wasan geometry.

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a = 1/3
b = (1 – a^2)/2

c = a
d = 0

e = (2*a)/(1 + a^2)
f = (1 – a^2)/(1 + a^2)

g = (a*(7 – a^2) – 2)/(4 + (1 – a)^2)
h = (4*(1 – a^2))/(4 + (1 – a)^2)

i = (a*(7 – a^2) + 2)/(4 + (1 + a)^2)
j = (4*(1 – a^2))/(4 + (1 + a)^2)

k = (a*(5 – a^2))/(3 + a^2)
l = (4*(1 – a^2))/(3 + a^2)

q = 1/2
r = 1/150

s = ((1 + a)*(3 – a))/(4*(1 – a))
t = ((1 – a)*(3 + a))/(4*(1 + a))

zorder 1:
y = 0
x^2 + y^2 = 1^2
each edgecolor = (0, 0, 1, 1)

zorder 2:
(x – a)^2 + (y – b)^2 = b^2
edgecolor = (1, 0, 0, 1)

zorder 3:
y = (1 + x)/s
y = (1 – x)/t
each edgecolor = (0, 1, 0, 1)

zorder 4:
(x – c)^2 + (y – d)^2 = r^2
(x – e)^2 + (y – f)^2 = r^2
each edgecolor = (q, 0, q, 1)
(x – g)^2 + (y – h)^2 = r^2
(x – i)^2 + (y – j)^2 = r^2
each edgecolor = (q, q, 0, 1)
(x – k)^2 + (y – l)^2 = r^2
edgecolor = (0, q, 0, 1)

>>16968614

a) Find a specific construction that yields a maximum for gamma. (numerical)

b) Find a specific construction that yields a maximum for r*. (numerical)

c) Find a general relationship for r(R), for all constructions.

d) Find a general relationship for beta(alpha), for all constructions.

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>>16970805
>
>g = (a*(7 – a^2) – 2)/(4 + (1 – a)^2)
>h = (4*(1 – a^2))/(4 + (1 – a)^2)
>
>i = (a*(7 – a^2) + 2)/(4 + (1 + a)^2)
>j = (4*(1 – a^2))/(4 + (1 + a)^2)
>
>k = (a*(5 – a^2))/(3 + a^2)
>l = (4*(1 – a^2))/(3 + a^2)

>>16970906
You paraphrased the OP.

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>>16970927
e = (2*a)/(1 + a^2)
f = (1 – a^2)/(1 + a^2)

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>>16968614

A = (0, 0)
r1 = 1
B = (–1, 0)
C = (1, 0)

a = 1/3
b = (1 – a^2)/2
c = (1 + a^2)/2

D = (a, b)
r2 = b
E = (a, 0)
F = D/c
G = (a*(3 + b) – 1, 4*b)/(2 – a + c)
H = (a*(3 + b) + 1, 4*b)/(2 + a + c)

I = (a*(2 + b), 4*b)/(1 + c)

J = (a*(1 + b), b*(4 + b)/2)
r3 = b^2/2
K = ((3/2 + a + b)^2 – 17/4, 2*(3 – a)*b)/(2 – a + c)
L = (17/4 – (3/2 – a + b)^2, 2*(3 + a)*b)/(2 + a + c)
M = J/(1 + r3)

>>16971549
>a = 1/3
Oops, I meant a = –2/7.
>r1 = 1
>r2 = b
>r3 = b^2/2
>M = J/(1 + r3)
That's r_1, r_2, and r_3.

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>>16971549

Points A, D, and F are collinear.
So are points A, M, and J.

Cos[γ] = Dot[F, M] = (9 + 5*a^2 + 3*a^4 – a^6)/(9 + 7*a^2 – a^4 + a^6)

Solve[D[γ] = 0] has four solutions.
The real solutions are:
a^2 = 2*Sqrt[7] – 5

Thus:
γ_max = ArcCos[(7^(3/2) – 1)/18] ≈ 13.2578787°

>>16971196
>four loci
My 10th grade math teacher, Mrs. [...], was fond of loci.

>>16971549
>J = (a*(1 + b), b*(4 + b)/2)
>r3 = b^2/2
Gemini found the center and radius, of the second sought circle, for me.

>>16971981
>Points A, D, and F are collinear.
|AF| = 1 and |DF| = b
thus |AD| = 1 – b

|AE|^2 + |DE|^2 = |AD|^2
a^2 + b^2 = (1 – b)^2
a^2 = 1 – 2*b
2*b = 1 – a^2

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>>16971549

α = VectorAngle[F – E, G – E]
β = VectorAngle[K – M, L – M]

If –1 < a < 3, then the following.
Cot[α] = (a^2 – a + 2)/(a + 1)
Cot[β] = (a^2 + 3)/4
β = ArcCot[([(Cot[α] + 1 – Sqrt[(Cot[α] + 7)*(Cot[α] – 1)])/2]^2 + 3)/4]

>>16973765

>α = VectorAngle[F – E, G – E]
But I used the following equation instead.
Cos[α] = Dot[F – E, G – E]/(Norm[F – E]*Norm[G – E])
And I had a hard time finding Cos[β] as a function of a.

>β = ArcCot[([(Cot[α] + 1 – Sqrt[(Cot[α] + 7)*(Cot[α] – 1)])/2]^2 + 3)/4]
I asked ChatGPT to solve the following system.
Cos[α] = [...]
Cos[β] = [...]
It solved the following system instead!
Tan[α] = [...]
Tan[β] = [...]
But the easiest system to solve is:
Cot[α] = [...]
Cot[β] = [...]

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If I had the wa-san-gaku formula
which corresponds to the OP's image,
then I would solve it for r*
and find r*_max.

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>>16973765

Cot[α] = (a^2 – a + 2)/(a + 1)
Cot[β] = (a^2 + 3)/4

in the case a = 0 : (see picture)
Cot[α] = Cot[β] must be given

but the formula yields
(a^2 – a + 2)/(a + 1) > (a^2 + 3)/4 !!!

That can't be true.

>>16975093
That's what you get, for not labeling your points!
α = angle FEG ≠ your angle GEH
(Making you nothing, but a GEH FEG ≈ gay fag.)

If a = 0, then α = ArcCot[2], β = ArcCot[3/4], and β/α = 2.
Which matches your image.

You seem to be the OP.
If you are, then isn't my β(α) equivalent to your β(α)?
Thus you owe me an apology, for trying to drag my good name around in the mud.

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>>16973765

Function: beta = ArcCot[ ( ((Cot[alpha] + 1 - Sqrt[(Cot[alpha] + 7)(Cot[alpha] - 1)]) / 2)^2 + 3 ) / 4 ]

Domain: [-ArcCot(7), pi/4]

This function exhibits a non-removable jump discontinuity at alpha = 0 due to the behavior of the nested cotangent terms.

Key Properties:

Left Boundary: Starts at alpha = -ArcCot(7), where beta = ArcCot(3).

Left Limit at Zero: As alpha approaches 0 from the negative side, beta approaches 0.

Right Limit at Zero: As alpha approaches 0 from the positive side, beta approaches pi/4.

Global Peak: Reaches an exact analytical maximum in the right quadrant at alpha = ArcCot(2), where beta = ArcCot(3/4).

Right Boundary: Ends at alpha = pi/4, where beta = pi/4.

The plot uses an exact 1:1 aspect ratio with grid spacing set at intervals of pi/16. The colored markers indicate these specific coordinate points as detailed in the legend.

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>>16975583

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>>16973765

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>>16976404

regarding the attached image:
Plot[{
ArcTan[(1 + a)/(2 – a + a^2)],
ArcTan[4/(a^2 + 3)],
ArcTan[–1/7],
ArcTan[(1 – a)/(2 + a + a^2)]
}, {a, –6, 6}]

>[f_2 is f_1 mirrored about the vertical axis]
Yes, since f_2(a) = f_1(–a).

>f_1 + f_2 = g
I can prove it without using the variable a.
Proof:

m() = measure()

α = m(angle FEG) = f_1
β = m(angle KML) = g
δ = m(angle FEH) = f_2

S = the south pole or point of circle KLM

claim:
triangle KSL ~ triangle GEH
proof:
See the image in post number 16971981.
thus:
m(angle KSL) = m(angle GEH) = α + δ

claim:
m(angle KSL) = m(angle KML)
proof:
Inscribed angles that intercept the same arc are equal in measure.
thus:
m(angle KSL) = m(angle KML) = β

in conclusion:
α + δ = β

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>>16968614
Geometric construction of the circles ...

>>16978428
You win. You're the winner.

>>16979580
The absolute winner would be the one who can explicitly calculate the value of r*. Or would it? For that is what truly demonstrates real mastery. Not only the geometric way.

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>>16978428

focus: (p, q) = D

vertex: (u, v) = G

equation(parabola): [(q – v)*(x – u) – (p – u)*(y – v)]^2 = 4*[(p – u)^2 + (q – v)^2]*[(p – u)*(x – u) + (q – v)*(y – v)]

The equation is courtesy of an unnamed Chinese AI assistant that outperformed both Grok and Gemini, though all three found the same equation in different forms.