Draw any three circles of different radii in a plane such that no circle is inside another - each must have interior points not in the interior or on the circumference of either of the other circles, to rule out the inside-but-tangent case. It is easiest to visualize this with non-overlapping circles, but overlapping will not "break" the result.

For each of the three pairs, draw the two common external tangents. For each two tangents, mark the point at which they intersect.

Prove that these three points are colinear.

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TopNewestOne way you could approach this is with vectors (which is rather whack based; there's definitely a synthetic solution for this, perhaps inversion based)

Let the three circles have radius r

1, r2, r_3, and be centered at A,B,C respectively.WLOG let r

1 < r2 < r_3Define the intersection of the tangents for the circles centered at A,B as D.

Observe that D = A - (B-A)(r

1)/(r2 - r1) = (Ar2 - Br1) / (r2-r_1)We can also define E and F, the other two intersections similarly, giving us:

D = (Ar

2 - Br1) / (r2 - r1)E = (Br

3 - Cr2)/ (r3 - r2)F = (Cr

1 - Ar3)/ (r1 - r3)D-E = k { A(r

3 - r2) + B (r1 - r3) + C(r2 - r1) } for some constant k based on r1, r2, r_3after expanding, you can remove the (r

2 - r1)(r3-r2) at the bottom, and all the top terms are multiples of r2; k = (r2)/((r2 - r1)(r3 - r2)). We do not care about the magnitude of D-E, so we just look at the direction.D-F = l { A(r

2 - r3) + B (r3 - r1) + C(r1 - r2) } for some constant lThus D-E and D-F are parallel, the three points are collinear.

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It struck me later that there is a much simpler continuation after you obtain the ratios of AD to AB, BE to BC, and CF to AC. Just use Menelaus' theorem, and the rest is trivial.

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I'm going to have to learn how to read that notation...

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