This isn’t for homework or anything like that. It’s for art. I can’t draw free hand at all, but with a compass and ruler, I can make some pretty nice looking stuff.

So, here is the setup. Circles A, B and C have radius r. Circle D has radius s = (2*sqrt(3) - 3)r/3

I want to construct a circle (E), that is tangent to circles A, B and D (and that’s not the same circle as C).

I know that the center of E will be on the line DT. I figured I could calculate what the radius of E would be, then construct that distance away from D and draw a circle there tangent to D.

I used Descartes’ Theorem to find the radius of E. With k[sub]1[/sub] = k[sub]2[/sub] = 1/r and k[sub]3[/sub] = 1/s = (2sqrt(3)+3)/r, then the radius of the fourth circle is either (9-4sqrt(3))r/33 = (2*sqrt(3)+1)s/11, or r, which is a good sanity check, since that is circle C.

But I have no idea how to construct that distance. And I’m sure there’s some simpler and probably more general way to do this that I can’t think of.

Well, that is a constructable distance, at least, so it’s doable. First, construct a line of length s. Then, construct a right angle on it, and another leg of length s. Connect the two legs, and the hypotenuse of the triangle has length sqrt(2)*s. Then construct a right angle on that hypotenuse, and put another leg of length s on it, and take the hypotenuse of this second triangle. That’s sqrt(3)s. Double that length, and then extend it out by s again, and you have (2sqrt(3)+1)*s. Then use the Euclidean construction (which is hard to explain in words, but which you can look up in any geometry book) to divide that length by 11.

Of course, doing all this would make a royal mess of your paper, and in practice, probably give you something a lot worse than if you had just used a ruler to measure it. I don’t know if there’s any simpler construction.

First note that the locus of centers of circles which are tangent to two given circles X and Y, which themselves are tangent at V, is generally one branch of a hyperbola, with foci at X and Y, and vertex V. So the “Soddy circle” tangent to three mutually-tangent circles generally has its center at the intersection of three hyperbolas.

Your problem is a little easier, because the “hyperbola” formed by circles A and B is degenerate; it’s just the line TD. So all you need to do is find the intersection of this line with (the appropriate branch of) the hyperbola h with foci A and D, and vertex V at the tangent point of A and D.

This isn’t so hard. Recall that a branch of a hyperbola (with foci X and Y) is the locus of points P such that the difference of distances XP-YP=d is constant. For the hyperbola h that we’re interested in, we know the foci (A and D) and a point on h (namely V), so we can construct the distance d. [Draw a circle centered at V with radius VD; it intersects AD again at a point W. Now AW=AV-DV=d is the difference of distances for h.]

Now you want to find the point E on TD, toward T from D, such that
AE-DE=d=AW.
Start by marking off distance d=AW, along TD from D away from T; call this point U, so that DU=d=AW. Now you want to find point E such that
AE=DE+AW=De+DU=ZU:
that is, E is equidistant from A and U. So E lies on the perpendicular bisector of AU, and also on DT, and so you’ve found it.

Omphaloskeptic, that’s an interesting approach that I wouldn’t have thought of. I’ll see what else I can apply it to.

ZenBeam, thanks!

Since I already have a circle of radius s, there’s a simpler way to get 2sqrt(3). Find two points on circle D with are distance s apart. Find the midpoint between them. The distance from that point to the center of the circle is sqrt(3)*s/2. Then quadruple that.

D’oh, of course. I saw “That’s the square root of an integer. I’ll use the standard method for constructing the square root of an arbitrary integer”, and didn’t even stop to think about easier ways to get sqrt(3) specifically.