circles and triangles problem

Question

Hello everyone,

I have an interesting problem here I've been trying to solve for the last little while:

I have 3 circles on a 2D xy plane, each with the same known radius. I know the coordinates of each of the three centers (they are arbitrary and can be anywhere).

What is the largest triangle that can be drawn such that each vertice of the triangle sits on a separate circle, what are the coordinates of those verticies?

I've been looking at this problem for hours and asked a bunch of people but so far only one person has been able to suggest a plausible solution (though i have no way of proving it).

The solution that we have come up with involves first creating a triangle about the three circle centers. Next we look at each circle individually and calculate the equation of a line that passes through the circle's center and is perpendicular to the opposite edge. We then calculate two intersection points of the circle. This is then done for the next two circles with a result of 6 points. We iterate over the 8 possible 3 point triangles that these 6 points create (the restriction is that each point of the big triangle must be on a separate circle) and find the maximum size.

The results look reasonable (at least when drawn out on paper) and it passes the special case of when the centers of the circles all fall on a straight line (gives a known largest triangle). Unfortunate i have no way of proving this is correct or not.

I'm wondering if anyone has encountered a problem similar to this and if so, how did you solve it?

Note: I understand that this is mostly a math question and not programming, however it is going to be implemented in code and it must be optimized to run very fast and efficient. In fact, I already have the above solution in code and tested to be working, if you would like to take a look, please let me know, i chose not to post it because its all in vector form and pretty much impossible to figure out exactly what is going on (because it's been condensed to be more efficient).

Lastly, yes this is for school work, though it is NOT a homework question/assignment/project. It's part of my graduate thesis (abet a very very small part, but still technically is part of it).

Thanks for your help.

Edit: Heres a new algorithm that i came up with a little while ago.

Starting at a circle's centre, draw a line to the other two centres. Calculate the line that bisects the angle created and calculate the intersections between the circle and the line that passes through the centre of your circle. You will get 2 results. Repeat this for the other two circles to get a total of 6 points. Iterate over these 6 points and get 8 possible solutions. Find the maximum of the 8 solutions.

This algorithm will deal with the collinear case if you draw your lines in one "direction" about the three points.

From the few random trials i have attempted using CAD software to figure out the geometries for me, this method seems to outperform all other methods previously stated However, it has already been proven to not be an optimal solution by one of Victor's counter examples.

I'll code this up tomorrow, for some reason I've lost remote access to my university computer and most things are on it.

Answer 1

Not optimal, works well when all three are not colinear:

I don't have a proof (and therefore don't know if it's guaranteed to be biggest). Maybe I'll work on one. But:

We have three circles with radius R with positions (from center) P0, P1, and P2. We wish to find the vertices of a triangle such that the area of the triangle is maximum, and the vertices lie on any point of the circles edges.

Find the center of all the circles and call that C. Then C = (P0 + P1 + P2) / 3. Then we find the point on each circle farthest from C.

Find vectors V0, V1, and V2, where V_i = P_i - C. Then find points Q0, Q1, and Q2, where Q_i = norm(V_i) * R + P_i. Where norm indicates normalization of a vector, norm(V) = V / |V|.

Q0, Q1, and Q2 are the vertices of the triangle. I assume this is optimal because this is the farthest the vertices could be from each other. (I think.)

Answer 2

My first thought is that you should be able to find an analytic solution.

Then the equations of the circles are:

(x1-h1)^2 + (y1-k1)^2 = r^2
(x2-h2)^2 + (y2-k2)^2 = r^2
(x3-h3)^2 + (y3-k3)^2 = r^2

The vertices of your triangle are (x1, y1), (x2, y2), and (x3, y3). The side lengths of your triangle are

A = sqrt((x1-x2)^2 + (y1-y2)^2)
B = sqrt((x1-x3)^2 + (y1-y3)^2)
C = sqrt((x2-x3)^2 + (y2-y3)^2)

So the area of the triangle is (using Heron's formula)

S = (A+B+C)/2
area = sqrt(S(S-A)(S-B)(S-C))

So area is a function of 6 variables.

At this point I realize this is not a fruitful line of reasoning. This is more like something I'd drop into a simulated annealing system.

So my second thought is to choose the point on circle with centre A as follows: Construct line BC joining the centres of the other two circles, then construct the line AD that is perpendicular to BC and passes through A. One vertex of the triangle is the intersection of AD and circle with centre A. Likewise for the other vertices. I can't prove this but I think it gives different results than the simple "furthest from the centre of all the circles" method, and for some reason it feels better to me. I know, not very mathematical, but then I'm a programmer.

Answer 3

Let's assume the center of the circles to be C0,C1 and C2; and the radius R.

Since the area of a triangle is .5*base*height, let's first find the maximum base that can be constructed with the circles. Base = Max {(|C0-C1|+2R),(|C1-C2|+2R,(|C2-C0|+2R}

Once the base length is determined between 2 circles, then we can find the farthest perpendicular point from the base line to the third circle. (product of the their slopes is -1)

For special cases such as circles aligned in a single line, we need to perform additional checks at the time of determining the base line.

Answer 4

It appears that finding the largest Apollonius circle for the three circles and then inscribing an equilateral triangle in that circle would be a solution. Proof left as an exercise ;).

EDIT

This method has issues for collinear circles like other solutions here, too and doesn't work.

Answer 5

I believe this is a convex optimization problem (no it's not, see below), and hence can be solved efficiently using well known methods.

You essentially want to solve the problem:

maximize:  area(v1,v2,v3) ~ |cross((v2-v1), (v3-v1))|
such that: v1 in C1, v2 in C2, v3 in C3   (i.e., v_i-c_i)^2 - r_i^2 <= 0)

Each of the constraints are convex, and the area function is convex as well. Now, I don't know if there is a more efficient formulation, but you can at least use an interior point method with derivatives since the derivative of the area with respect to each vertex position can be worked out analytically (I have it written down somewhere...).

Edit: grad(area(v1,v2,v3))(v_i) = rot90(vec(vj,vk)), where vec(a,b) is making a 2D vector starting at a and ending at b, and rot90 means a positive orientation rotation by 90 degrees, assuming (vi,vj,vk) was positively oriented.

Edit 2: The problem is not convex, as should be obvious considering the collinear case; two degenerate solutions is a sure sign of non-convexity. However, the configuration starting at the circle centers should be in the globally optimal local maximum.

Answer 6

Some initial thoughts.

1. Definition Call the sought-after triangle, the maximal triangle. Note that this might not be unique: if the circles all have the same centre, then there are infinitely many maximal triangles obtained by rotation around the center, and if the centres are colinear, then there will be two maximal triangles, each a mirror image of the other.
2. Definition Call the triangle (possibly, degenerately, either a point or a line) whose vertices are the centres of the circles the interior triangle.
3. Observation The solution can be expressed as three angles, indicating where on the circumference of each circle the triangle is to be found.
4. Observation Given two exterior vertices, we can determine a third vertex that gives the maximal area: draw the altitude of the triangle between the two exterior vertices and the centre of the other circle. This line intersects the circumference in two places; the further away point is the maximising choice of third vertex. (Fixed incorrect algorithm, Federico's argument can be adapted to show correctness of this observation)
5. Consequence The problem is reduced to from a problem in three angles to one in two.
6. Conjecture Imagine the diagram is a pinboard, with three pins at the three centres of the circles. Imagine also a closed loop of string of length equal to the perimiter of the interior triangle, plus the radius of a circle, and we place this loop around the pins. Take an imaginary pen and imaginarily draw the looping figure where the loop is always tight. I conjecture that the points of the maximal triangle will all lie on this looping figure, and that in the case where the interior triangle is not degenerate, the vertices of the maximal triangle will be the three points where the looping figure intersects one of the circle circumferences. Many counterexamples

More to follow when I can spare time to think about it.

Answer 7

Let A, B, C be the vertexes of your triangle, and suppose they are placed as in your solution. Notice that the key property of your construction is that each of the vertexes lies on a tangent to its circle which is parallel to the opposite side of the triangle. Obviously, the circle itself lies entirely on one side of the tangent, and in the optimal solution each tangent leaves its circle on the same side as the other vertexes.

Consider AB as the "base" of the triangle, and let C float in its circle. If you move C to another position C' within the circle, you will obtain another triangle ABC' with the same base but a smaller height, hence also with a smaller area:

For the same reason, you can easily see that any position of the vertexes that doesn't follow your construction cannot be optimal. Suppose, for instance, that each one of the vertexes A', B', C' does not lie on a tangent parallel to the side connecting the other two.

Then, constructing the tangent to the circle that contains (say) C', which is parallel to A'B' and leaves the circle on the same side as A'B', and moving C' to the point of tangency C, it is always possible to construct a triangle A'B'C which has the same base, but a greater height, hence also a greater area:

Since any triangle that does not follow your construction cannot be optimal, I do believe that your construction is optimal. In the case when the centers of the circles are aligned I'm a bit confused, but I guess that it is possible to prove optimality along the same lines.

Answer 8

I've created an HTML5 canvas app that may be useful for people to play with. It's pretty basic (and the code is not beautiful), but it lets you move three circles of equal radius, and then calculates a maximal triangle using gradient/steepest descent. You can also save bitmaps of the diagram. The diagram also shows the triangle whose vertices are the circle centers, and one of the altitudes. Edit1: the "altitude" is really just a line segment through one of the circle centers and perpendicular to the opposite edge of the triangle joining the centers. It's there because some of the suggested constructions use it. Edit2: the steepest descent method sometimes gets stuck in a local maximum. You can get out of that maximum by moving a circle until the black triangle flips and then bringing the circle back to its original position. Working on how to find the global maximum.

This won't work in IE because it doesn't support canvas, but most other "modern" browsers should work.

I did this partially because I found some of the arguments on this page questionable, and partially because I've never programmed a steepest descent and wanted to see how that worked. Anyways, I hope this helps, and I hope to weigh in with some more comments later.

Edit: I've looked at the geometry a little more and have written up my findings in a separate answer.

Answer 9

I've taken the liberty of submitting a second answer, because my original answer referred to an online app that people could play with to get insight. The answer here is more a geometric argument.

The following diagram illuminates, I hope, what is going on. Much of this was inspired by @Federico Ramponi's observation that the largest triangle is characterized by the tangent at each vertex being parallel to the opposite side.

The picture was produced using a trial version of the excellent desktop program Geometry Expressions. The diagram shows the three circles centered at points A,E, and C. They have equal radii, but the picture doesn't really depend on the radii being equal, so the solution generalizes to circles of different radii. The lines MN, NO, and OM are tangent to the circles, and touch the circles at the points I,H, and G respectively. The latter points form the inner triangle IHG which is the triangle whose size we want to maximize.

There is also an exterior triangle MNO which is homethetic to the interior triangle, meaning that its sides are parallel to that of IHG.

@Federico observed that IHG has maximal area because moving any of its vertices along the corresponding circle will result an a triangle that has the same base but less height, therefore less area. To put it in slightly more technical terms, if the triangle is parameterized by angles t1,t2,t3 on the three circles (as pointed out by @Charles Stewart, and as used in my steepest descent canvas app), then the gradient of the area w.r.t to (t1,t2,t3) is (0,0,0), and the area is extremal (maximal in the diagram).

So how is this diagram computed? I'll admit in advance that I don't quite have the full story, but here's a start. Given the three circles, select a point M. Draw tangents to the circles centered at E and C, and designate the tangent points as G and I. Draw a tangent OHN to the circle centered at A that is parallel to GI. These are fairly straightforward operations both algebraically and geometrically.

But we aren't finished. So far we only have the condition that OHN is parallel to GI. We have no guarantee that MGO is parallel to IH or that MIN is parallel to GH. So we have to go back and refine M. In an interactive geometry program it's no big deal to set this up and then move M until the latter parallel conditions are met (by eyeballs, anyways). Geometry Expressions created the diagram, but I used a bit of a cheat to get it to do so, because its constraint solver was apparently not powerful enough to do the job. The algebraic expressions for G, I, and H are reasonably straightforward, so it should be possible to solve for M based on the fact that MIHG is a parallelogram, either explicitly or numerically.

I should point out that in general if you follow the construction starting from M, you have two choices of tangent for each circle, and therefore eight possible solutions. As in the other attempted answers to the question, unless you have a good heuristic to help you choose in advance which of the tangents to compute, you should probably compute all eight possible triangles and find the one with maximum area. The other seven will be extremal in the sense of being minimal area or saddle points.

That's it. This answer is not quite complete in that it leaves the final computation of M somewhat open ended. But it's reduced to either a 2D search space or the solution of an ornery but not humongous equation.

Finally, I have to disagree with @Federico's conclusion that this confirms that the solution proposed by the OP is optimal. It's true that if you draw perpendiculars from the circle centers to the opposite edge of the inner triangle, those perpendiculars intersect the circle to give you the triangle vertex. E.g. H lies on the line through A perpendicular to GI), but this is not the same as in the original proposed solution (which was to take the line through A and perpendicular to EC - in general EC is not parallel to GI).