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.

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## Comments

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TopNewestThis 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.

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