For various reasons, I am currently trying to solve the following problem:

For three circles, which are positioned at 0,0, 1,0, and 0.5, sqrt(0.75), find some x and y such that these circles intersect at this point, where each has an integer radius.

I've done the intersection of two circles with fixed radius, but I'm having problems with this one. The only thing I've managed to achieve is going in circles substituting things. Any suggestions? I tried subtracting some equations from others, but all I've got now is a giant, single, equation.

What's the difference between point A (0, 0) and point C (0.5, sqrt(3)/2).

Can you see any symmetry between AC and BC?

[EDIT:] That won't work. Yet.

Yes, AC and BC both have the length of 1. That's how it was intended. As does AB, of course.

Having messed around a bit with the problem, I'm starting to be convinced that there is no solution to this problem if you restrict yourself to integer radii. Here's why I think so:

First, consider only two circles (call them C1 and C2) with integer radii centred at (0,0) and (0,1). If these are to cut each other, there are three possibilities:

- The radii of C1 and C2 are equal. In that case the intersection lies on the bisector of the line joining the centres of C1 and C2.

- The radius of C1 compared to that of C2 is smaller by 1. In that case the intersection lies along the x-axis, somewhere on the left.

- The radius of C1 compared to that of C2 is larger by 1. In that case the intersection lies along the x-axis, somewhere on the right.

If the radii of C1 and C2 differ by more than 1, one of the circles is completely inside the other and hence don't intersect.

Suppose we have C1 and C2 intersecting. Having exhausted these three possibilities, I don't see how it is possible to add the third point and letting the circle centred on that point (C3) intersect in the same point where C1 and C2 intersect.

For example, if C3 intersects C1, it will (by the logic above) be either on the line joining the centres of C1 and C3 OR on the bisector of that line. Neither of these lines will pass through the possible points where C1 and C2 can intersect.

I hope I'm missing something, but that's the best I could do in the small time frame I have here.

C3 already exists. That's the 0.5, sqrt(0.75) centered circle- which is an intersection between C1 and C2. I'm trying to find a point on which to place a fourth circle which intersects all three existing circles- of unbounded integer radius which is volatile- i.e., effectively, I'm talking about the intersection of *all* circles centered at those points with an integer radius.

I'm assuming you mean that the fourth circle intersects the centres of the 3 other circles (otherwise you could pretty much draw ANY circle).

This will in many cases be impossible EVEN IF you manage to get an intersection of three circles.

Imagine C1 at (0,0) with r=3, C2 at (0,1) with r=2, C3 at (0,2) with r=1.

They intersect (well not truly intersect but that's not the problem here) at (0,3), but if you then draw a circle at (0,3) with some radius, the centres of C1,2,3 will never all be on the arc of C4.

So what you are stating is no requirement for the solution of your problem, though it is a nice bonus. This bonus can only be achieved if C1,2,3 all have the same radius.

However, it is possible to have a C4 intersecting the centres of C1,2,3 at the same time. I just can't get all of the equations on here out of my head. (That's a shame for a mathematician).

Now back to the matter at hand.
You not only want them to intersect (easily done), you also want them all to have r∈ℤ. There is only one solution, due to the location of the centre of C3 (0.5;sqrt(3/4)).

I found this solution while searching for formulas in the situation of r1=r2. Then the arcs of C1 intersect at (0.5;sqrt(n^2-0.5^2)). If you discard the requirement that n should be an integer and take n=0,5 then you'll find them intersecting at the x-axis (which is just right because since the distance between C1 and C2 equals 1 and take the radii of 0.5, they'll meet in the middle at the x axis because C1 and C2 lie on the x-axis).

Now, n must be an integer, so it's 0,1,2,... (integers are a selection of Z not N so 0 and negative numbers can join in here, but negative numbers aren't an option because a radius is nonnegative).

If we take n=1, then we get the intersection at (0.5,sqrt(3/4)), which is the centre of C3.

So take r3=0 and you have solved the problem. Additionally, Dots can be named circles with r=0 so C3 is still a perfectly fine circle. The other intersection is at (0.5;-sqrt(3/4)) so C3 should have r3=2sqrt(3/4) to join in that intersection but then r3 is no integer)

Taking n>1, we'll get the y-coordinate of the intersection of C1 and C2 to be at sqrt(n^2-1/4). This will be 15/4, 35/4, 63/4, 99/4, 143/4,... for n=2,3,4,5,6...
These will never give an integer as coordinate.

That means that the radius of C3 will be (y-coordinate minus sqrt(3/4)) or (minus y-coordinate + 2sqrt(3/4)) The latter for the intersection in the negative area. Also these will never give an integer.

That means that the only solution for r1=r2 will be (r1,r2,r3)=(1,1,0)

It gets a lot more complicated for having r1 != r2, but here you can still work with symmetry because your figure is symmetrical.

If r1 > r2, the intersections will move to the right, but also the y-coordinate will change (it will increase first - positive region- , as r1 increases, but then it'll decrease - the y-coordinate will be zero if r1=r2+1 leaving only 1 intersection).

If r1 < r2 then you'll have exactly the same, but with the intersections moving to the left.


TL:DR - there is only one solution, that is (r1,r2,r3)=(1,1,0)

No, I meant intersects their circumference. The reason you can't draw ANY circle is because the radius can only be integral.

In that case drawing a circle that intersects the circumference of the three others is not related three circles having to intersect each other's circumference at a given point.

Anyway, what do you think of the integer solution?

Okay, I'm completely lost now what kind of problem we're dealing with. A picture would be greatly appreciated.

Here are the three circles.

The left one (red dot at the left side) is at (0,0)
The right one is at (1,0)
The middle one is at (0.5;sqrt(3/4))

We have to find three radii (they may all be different) that will let the three circles intersect at a certain point (The black blob is where the three intersect in my image).
That's easily done (just make the picture).

The added restriction is to let all radii be of an integer value. Then there is only one solution: r1=r2=1, r3=0. They will all intersect at (0.5;sqrt(3/4))

Okay, that's the problem I solved in my first post as well and I came to pretty much the same conclusion, except I didn't consider a radius r3=0 as a legitimate solution. I just considered the kind of points where C1 and C2 can possibly intersect and the kind of points where C1 and C3 can intersect and couldn't find any overlap there. Apparently this didn't satisfy DeadMG, so I'm not quite sure where he thinks the problem lies.

Surely the green circle is incorrect as it has to have an integer radii, therefore >/= 1, therefore hitting the center of the blue circle.

Yeah, it's just an artist's impression.

It's just a sketch. That's the good thing about mathematical problems: you make a sketch of the part that isn't clear, so you can make it easier to understand.

It was just the part vleessju needed. The fact that the sketch doesn't give a solution to the problem is irrelevant.

One picture speaks louder than a thousand words. This is true in mathematics and science as well.