# 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
4 years, 11 months 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 - 4 years, 10 months ago