mg.metric geometry – A circle with radius R, find the total area of all the circles added together

A circle with radius R is shown below in figure a. In figure b, two circles with a radius of 1/2 R are placed on top of the original circle from figure a. In figure c, four circles with a radius of 1/4 R are placed on top of the circles from figure b.

The image

Assuming this pattern continues indefinitely, find the total area of all the circles added together.

(With steps would be wonderful, I really don’t understand)

analysis – The union of 2 circles can not be the pre-image of zero

I saw at one of my class notes from Analysis that the set $Csubset mathbb{R}^2$ (where $C$ is the union of two circles of radius 1, not two balls or two discs) can not be a set of the form $f^{-1}(0)$, where $f:mathbb{R}^2 rightarrow mathbb{R}$ is a $C^1$ map with $nabla f(x)neq 0_{mathbb{R}^2}$ for every $x in f^{-1}(0)$. However, I can not see how it can be true. Could someone give me a hand?

computational geometry – Finding lowest point in circles

Given n disks in the plane, i want to compute the lowest point in their intersection area, im looking for a simple randomized incremental algorithm.

I think this problem have some similarity with 2D half-plane intersection (2D LP). In that problem we were looking for an optimal point in respect of the cost vector. But subproblems in that problem was finding the intersection between a half-plane and a convex region with can be reduced to a simpler 1D problem. That 1D problem is half-line intersection, which is easier to solve.

Here but i have trouble to define a simpler subproblem. Also in the Analysis for expected time, i don’t see how to use backward analysis here.

I also guess maybe we can solve this problem with finding the convex-hull of those circles, then we look for it’s core with half-plane intersection, but i really uncertain about this idea.

list manipulation – Connecting the closest points on circles with a line

Suppose I have the function f(c_, z_) := z^2 + c and the following collection of circles in the plane. This is a simple example with only three circles, disjoint and not nested, and lying on the real line.

Show(Table(Graphics(Line(ReIm(Nest(f(-1.755, #) &, Map(comp, CirclePoints({0, 0}, 0.1, 1000)), i)))), {i, 0, 2}))

Given one circle, I would like to connect it to the closest circle (that isn’t itself) by the shortest possible line. So in this example the lines would be the segments of the real line between the closest points of the circles. The following code does this manually.

Take comp({x_, y_}) := x + y*I and then the following gives the desired picture.

  ReIm({Nest(f(-1.755, #) &, 
     Map(comp, CirclePoints({0, 0}, 0.1, 1000)), 0), 
           Nest(f(-1.755, #) &, 
            Map(comp, CirclePoints({0, 0}, 0.1, 1000)), 0))((750)), 
           Nest(f(-1.755, #) &, 
            Map(comp, CirclePoints({0, 0}, 0.1, 1000)), 1))((750))})(
       s), {s, 0, 1, 0.001})), 
    Nest(f(-1.755, #) &, Map(comp, CirclePoints({0, 0}, 0.1, 1000)), 
     1), Nest(f(-1.755, #) &, 
     Map(comp, CirclePoints({0, 0}, 0.1, 1000)), 2), 
           Nest(f(-1.755, #) &, 
            Map(comp, CirclePoints({0, 0}, 0.1, 1000)), 0))((250)), 
           Nest(f(-1.755, #) &, 
            Map(comp, CirclePoints({0, 0}, 0.1, 1000)), 2))((750))})(
       s), {s, 0, 1, 0.001}))})))

This is messy but it works. However for more circles this will become very tedious to implement. Is there a quicker way to do this? I would still like to get the points along the connecting lines for future calculations. Also, perhaps more difficult, what if the parameter c for f(c,z) is complex so the lines are no longer in a straight line?

mg.metric geometry – A generic question on triples of circles associated with a triangle

This question is inspired by two posed by P.Terzić (both given elegant synthetic proofs by F. Petrov). The starting point is a triangle $ABC$ and a triangle centre $G_1$. There are two classical ways to use this to generate a new triangle $A_1B_1C_1$ where the new vertices are

  1. the reflections of the original ones in $G_1$;


  1. the feet of the Cevians through this point.

We now add a second centre $G_2$ to the ingredients and consider the three circles through $AG_2 A_1$, $BG_2B_1$ and $CG_2C_1$.

These three circles have the one common point $G_2$, of course, and our question is to determine under which conditions they have a second one. More specifically, our question is whether the following dichotomy holds:

For a given (distinct) pair of triangle centres, then either the above result holds or the triangle is isosceles.

Remarks. 1. The results of Petrov display two situations where the first situation is true.

  1. we are using the term “circle” in the generalised sense which includes lines. These occur in the specal case of an isosceles triangle (for an equilateral one we have three lines through the centre–the second point of intersection is at infinity).

  2. we refer to the online Encyclopedia of Triangle Centers for the concepts and notations we are using.

  3. since there are now over $40,000$ registered centres there are potentially that number squared such results. Hence a synthetic proof can’t be expected and one is forced to apply Tate’s maxim–think geometrically, prove algebraically.

mg.metric geometry – Three circles intersecting at one point

Proposition. Let $triangle ABC$ be an arbitrary triangle with nine-point center $N$ and circumcenter $O$. Let $A’,B’,C’$ be a reflection points of the points $A,B,C$ with respect to the point $N$. Consider the three circles $k_1,k_2,k_3$ defined by the points $AOA’$ , $BOB’$ and $COC’$ , respectively. I claim that $k_1$,$k_2$ and $k_3$ meet at a common point $𝑃$.

geometry – 45-45-90 Triangles and the Relation to Circles.

So, The other day I was playing around with Geogabra, as one does, when I made a discovery I couldn’t find elsewhere. I found that if you have 45-45-90 special right triangles within a circle, you can do some math to find the area of the circle if your only given the area of the triangle, and vice-versa. This is already somewhat easy if you know basic algebra, but there is a way to do it by just multipling and dividing one time. Let me explain.

If you have an image like this, where the triangle within it is a 45-45-90 triangle, and the congrent sides are equivelent to the radius of the circle. (The areas are already filled in for easyness sake):
Image for Example.

You can find the area of the circle if you’re given the area of the triangle by multipling the area of the triangle by 2(pi).

The example given in the image, the area of the triangle is 50, so 50*pi = about 314.16, which is about the area of the circle. Dividing the area of the cricle by 2(pi) would give you 50, the area of the triangle.

You can do something similar with a 45-45-90 triangle, but the hypotenuse is equal to the diameter of the circle: Image 2 for Example

to solve this, simple multiply the area of the triangle, in this case 100, by pi to get the circle’s area, or divide the area of the circle by pi to get the triangle’s area.

This works for any triangle/circle combonation that follow these rules:

  1. The triangle is 45-45-90
  2. Either the hypotenuse is equal to the circle’s diameter, or the other two sides are both equal to the circle’s radius.

I hope you enjoyed this math jargin. My question is: is this an actal thing that already exists and people know about, or is it something that I discovered? I dought that, but hey, it’s worth a shot. I’m asking because I’m currently taking geometry, and we are long past our special right triangle unit, but I have never heard of this before. Thanks for coming to my Ted Talk.

equation solving – Mathematica crashing on Solve (finding points on 4 circles on a sphere). How to reformulate?

I define a small circle on a unit sphere by the direction of the plane’s normal and its distance to the sphere center (i.e. origin) like this (parametrized by the angle t):

sphereCircleRadiusFromOfs(ofs_) := Sqrt(1 - ofs^2);
pointOnSphereCircle(dir_, ofs_) := dir*ofs + Normalize@Cross(Cross({0, 1, 0}, dir), dir) * sphereCircleRadiusFromOfs(ofs);
sphereCircle(dir_, ofs_, t_) := dir + RotationMatrix(t, dir).(pointOnSphereCircle(dir, ofs) - dir);

Now, given 4 such circles with plane normals towards the 4 vertices of a regular tetrahedron and a distance of 1/Sqrt(2), I want to find solutions for the 4 angles such that the sums of the 4 points on the 4 circle are {0,0,0}.

I attempt to this by:

ofs = 1/Sqrt(2);
sol = Solve(
  sphereCircle(Normalize@{+1, +1, +1}, ofs, tPPP) + 
  sphereCircle(Normalize@{+1, -1, -1}, ofs, tPNN) + 
  sphereCircle(Normalize@{-1, +1, -1}, ofs, tNPN) + 
  sphereCircle(Normalize@{-1, -1, +1}, ofs, tNNP) == {0, 0, 0}, {tPPP, tPNN, tNPN, tNNP})

Unfortunately, Mathematica keeps computing forever, consuming more and more memory and will eventually crash. Is there a way to reformulate the problem such that Mathematica is more successful in solving it ?