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Suppose you're given a circle with center $O$, I'm curious, how can one construct with ruler and compass three circles inside the larger circle such that each is tangent to the larger circle as well as to the other two?

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Isn't this a sub-problem of an Apollonius problem: ? – mbaitoff Feb 24 '11 at 6:08
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If the radius of the circle is 1, the radius of three circles that will be internally tangent is $2\sqrt{3}-3\approx 0.464$ by Soddy's formula, given in the Soddy's circles section of this. You can construct this length, then mark it off on a diameter of the circle to find one of the centers, and finish constructing the equilateral triangle.

Added: For the construction, make a right angle with 1 on one side, swing 2 as a hypotenuse, and you have $\sqrt{3}$. Then you can mark a line to get $2\sqrt{3}-3$. Draw a diameter of the circle you are given and mark off $2\sqrt{3}-3$ from the circumfrence to find the center of one circle. Construct an equilateral triangle of $2(2\sqrt{3}-3)$ bisected by the diameter and you have the three centers.

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Thanks Ross Millikan, I think I should have been more explicit. Do you know of a ruler and compass construction? – user7440 Feb 24 '11 at 5:05
Just make another line and mark off two segments of $\sqrt{3}$ end to end and back up 1 three times. This is using the modern compass that allows you to transfer a length. I have read the the Greeks used a compass that collapsed when you pick it up, but there is a proof that you can still do everything you want with one of those. – Ross Millikan Feb 24 '11 at 6:23
Ok, so once I get a length of $2\sqrt{3}-3$, I'm thinking of inscribing an equilateral triangle in the given circle. I could then bisect each of the angles of that triangle, and then cut of points from $2\sqrt{3}-3$ from the circumference along each of those bisectors. Like you said, these will be the three centers, correct? Thank you for your time. – user7440 Feb 24 '11 at 6:34
No, you can't inscribe the triangle in the unit circle as it is too small. It is sized so when you draw a circle of $2\sqrt{3}-3$ around each vertex they are tangent to the outer circle. As they need to be tangent to each other, the side of the triangle (correction to above) is $2(2\sqrt{3}-3)$. You really just need to construct an equilateral triangle of this side with its center matching the circle center. – Ross Millikan Feb 24 '11 at 14:40

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