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# Three Dimensional Proof: Radical Axis Theorem

The Radical Axis Theorem states the following:

Given three distinct circles $$\Gamma_1, \Gamma_2, \Gamma_3$$, the radical axes of the three circles taken pairwise concur at a point, called the radical center.

The vanilla method of proving this statement is not hard:

Suppose $$\Gamma_1$$ and $$\Gamma_2$$ has radical axis $$\ell_3$$, and $$\Gamma_2$$ and $$\Gamma_3$$ has radical axis $$\ell_1$$. Let $$P=\ell_1\cap \ell_3$$. Then by the definition of radical axis, $$\text{pow}(P, \Gamma_1)=\text{pow}(P, \Gamma_2)$$ because $$P\in\ell_3$$ and $$\text{pow}(P, \Gamma_2)=\text{pow}(P, \Gamma_3)$$ because $$P\in \ell_1$$. Naturally, this means $$\text{pow}(P, \Gamma_1)=\text{pow}(P, \Gamma_3)$$ so $$P\in\ell_2$$, and the proof is finished.

Amazingly, there is also a very nice alternate way to prove this statement by extending the diagram to three dimensions. This proof will only work if the three circles pairwise intersect, but I believe it's possible to extend the proof to any three circles using complex points. But let's focus on real situations for now.

Let the center of circle $$\Gamma_i$$ be $$O_i$$ for $$1\le i\le 3$$. In addition, let the intersection of circles $$\Gamma_1$$ and $$\Gamma_2$$ that is closer to $$O_3$$ be $$P_3$$, and define $$P_1, P_2$$ similarly.

Note that by definition, $$O_1P_2=O_1P_3$$, and cyclically similar expressions. Thus, if we imagine rotating $$\triangle O_1P_2O_3$$ about line $$O_1O_3$$ and $$\triangle O_1P_3O_2$$ about line $$O_1O_2$$, then at some point, the image of $$P_2$$ and the image of $$P_3$$ about these two rotations will coincide at a point $$P$$ above the plane containing $$\triangle O_1O_2O_3$$. In other words, $$PO_1O_2O_3$$ is a tetrahedron with base $$\triangle O_1O_2O_3$$. However, note that because we performed a rotation, $$P_2O_3=PO_3$$ and $$P_3O_2=PO_2$$. Since $$P_1O_3=P_2O_3$$ and $$P_1O_2=P_3O_2$$, $$\triangle O_2P_1O_3\cong \triangle O_2PO_3$$, so we can rotate $$\triangle O_2P_1O_3$$ about line $$O_2O_3$$ so that the image of $$P_1$$ coincides with $$P$$.

But here's the punchline: if we track the projections of $$P_1, P_2, P_3$$ onto the plane containing $$\triangle O_1O_2O_3$$ as we rotate the triangles, it is clear to see that the locus is just a perpendicular from $$P_i$$ to line $$O_{i+1}O_{i-1}$$ (where indices are taken mod 3)--but this is just another definition of the radical axis! Thus, the projection of $$P$$ onto the plane containing $$\triangle O_1O_2O_3$$ lies on the radical axis of each pair of circles, and the proof is complete.

Note by Daniel Liu
3 months, 1 week ago

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This is one of my favourite theorems. It is really simple / obvious, but has surprisingly many applications.

It comes in handy for Olympiad problems in which numerous circles are involved, and we want to show that 3 lines are concurrent. Then, I go start hunting for the corresponding circles. Staff · 2 months, 3 weeks ago