Wikipedia:Reference desk/Archives/Mathematics/2021 May 20

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May 20

biased circle

Given three points $z_{1},z_{2},z_{3}$ and three constants $h_{1},h_{2},h_{3}$, find $z_{0}$ such that

${\frac {|z_{1}-z_{0}|}{h_{1}}}={\frac {|z_{2}-z_{0}|}{h_{2}}}={\frac {|z_{3}-z_{0}|}{h_{3}}}$

Each equation defines a circle (add: or line), and I could find the intersection of the three circles; but there may be a more direct way (add: perhaps involving a Möbius transformation?). —Tamfang (talk) 03:18, 20 May 2021 (UTC)

While Möbius transformations preserve circlehood, they do not preserve centres.  --Lambiam 22:50, 20 May 2021 (UTC)

Let $(x_{i},y_{i})=z_{i}.$ Each circle equation is a quadratic equation in the $(x,y)$ coordinates, but by taking their pairwise differences, you get three linear equations. These can be solved like any system of linear equations. After the smoke cleared up, I ended up with this:

Define

$d_{i}=h_{i}^{2}-(x_{i}^{2}+y_{i}^{2}),$

$a=\textstyle {\sum (x_{i}-x_{j})^{2}},$

$b=\textstyle {\sum (x_{i}-x_{j})^{2}(y_{i}-y_{j})^{2}},$

$c=\textstyle {\sum (y_{i}-y_{j})^{2}},$

$s=\textstyle {\sum (x_{i}-x_{j})^{2}(d_{i}-d_{j})^{2}},$

$t=\textstyle {\sum (y_{i}-y_{j})^{2}(d_{i}-d_{j})^{2}},$

where the summations are over pairs of indices $1\leq i<j\leq 3.$ Then

$z_{0}=\left({\frac {bt-cs}{2(ac-b^{2})}},{\frac {bs-at}{2(ac-b^{2})}}\right).$

 --Lambiam 07:10, 20 May 2021 (UTC)

I haven't gone through this in detail, but I think there's something off about it. For one thing, you're finding the intersection of two circles so there should be 2 points in common. As a concrete example, take z₁=(0,0), z₂=(1,0), z₃=(0,1), h₁=1, h₂=2, h₃=√2. Then one solution is z₀=(-1,0) with |z₁-z₀|/h₁ = |z₂-z₀|/h₂ = |z₃-z₀|/h₃ = 1. Another solution is z₀=(.2,.4) with |z₁-z₀|/h₁ = |z₂-z₀|/h₂ = |z₃-z₀|/h₃ = √.2 --RDBury (talk) 12:20, 20 May 2021 (UTC)

You're right; I have solved a different and more specific problem:

${\frac {|z_{1}-z_{0}|}{h_{1}}}={\frac {|z_{2}-z_{0}|}{h_{2}}}={\frac {|z_{3}-z_{0}|}{h_{3}}}=1$

under the assumption that this system has a solution. The first of your two solutions for the counterexample satisfies the extra condition and is the one found by the formulas I gave. The second will be found after dividing each $h_{i}$ by ${\sqrt {5}}.$  --Lambiam 12:49, 20 May 2021 (UTC)

That would explain it. I tried rewriting the system as |z_i-z₀| = k h_i where k is some fixed but unknown value. With your method you can find z₀ if you know k, but I don't see an easy way of finding k in the general case. I suppose you could use the Cayley–Menger determinant to get quadratic equation in k², but I don't know if that's an improvement over the OP's method. --RDBury (talk) 14:19, 20 May 2021 (UTC)

Another, more geometric approach: By this system, the points $z_{1},z_{2},z_{3}$ for a triangle with side lengths $\lambda h_{1},\lambda h_{2},\lambda h_{3}$ for an unknown $\lambda$. Hence (essentially by the side-side-side congruence theorem), the solution space is four-dimensional and is given by rotating, translating, and scaling an ur-triangle (which can be found in many ways, e.g. assuming $z_{1}$ to be at the origin and $z_{2}$ to be on the positive x-axis, then finding $z_{3}$ using stuff like Heron's formula). Duckmather (talk) 04:34, 21 May 2021 (UTC)

not quite that

Going over my algebra again, I think that what I want is instead

$|z_{n}-z_{0}|=e^{ks_{n}}$

where $k$ is also unknown; which looks not nearly so easy. Oh well, I'll probably have to do it iteratively. —Tamfang (talk) 16:48, 20 May 2021 (UTC)

Based on those equations alone, I think I know how to parametrize the solution space: it's $z_{n}=z_{0}+e^{ks_{n}+i\theta _{n}}$, where the free parameters are $z_{0},k,\theta _{n}$. Duckmather (talk) 04:43, 21 May 2021 (UTC)

It's easy to solve with s all equal; from there, maybe Newton's method will get somewhere. —Tamfang (talk) 22:14, 21 May 2021 (UTC)

Er, make that $|z_{n}-z_{0}|={\frac {e^{ks_{n}}}{k}}$. —Tamfang (talk) 21:58, 21 May 2021 (UTC)

Too bad there's no obvious way to extract k as a function of z₀. —Tamfang (talk) 22:40, 21 May 2021 (UTC)

But see Lambert W function – it depends on what one considers obvious. The quite specific form of the rhs suggests that you have an equally quite specific application in mind.  --Lambiam 22:59, 21 May 2021 (UTC)

relative distance between {n/2} gram inner and outer points?

For a {n/2} gram, is there an easy formula for the ratio of the distances between a) the distance from the center to the inner points and b) the distance from the center to the outer points. It is clear that the ratio goes to 1 as n goes to infinity, but I don't know if there is a (relatively) clean formula.Naraht (talk) 12:29, 20 May 2021 (UTC)

By the law of sines, $a/b=\sin(\pi /2-2\pi /n)/\sin(\pi /2+\pi /n)=\cos(2\pi /n)/\cos(\pi /n)=2\cos(\pi /n)-1/\cos(\pi /n)$. --116.86.4.41 (talk) 14:36, 20 May 2021 (UTC)

The triangle

What is a {n/2} gram? Which centre is being discussed - things like triangles have more than one. -- SGBailey (talk) 10:10, 22 May 2021 (UTC)

See Regular polygon § Regular star polygons. --116.86.4.41 (talk) 11:21, 22 May 2021 (UTC)

Thanx. I have a shower with the jets in the shape of an N / 2 gram and was wondering how many points to the gram were needed for the inner points to be at least 90% of the way out to the outer points. This gives me way to calculate!Naraht (talk) 12:07, 24 May 2021 (UTC)