Three Dimensional Proof: Radical Axis Theorem

The Radical Axis Theorem states the following:

Given three distinct circles Γ1,Γ2,Γ3\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 Γ1\Gamma_1 and Γ2\Gamma_2 has radical axis 3\ell_3, and Γ2\Gamma_2 and Γ3\Gamma_3 has radical axis 1\ell_1. Let P=13P=\ell_1\cap \ell_3. Then by the definition of radical axis, pow(P,Γ1)=pow(P,Γ2)\text{pow}(P, \Gamma_1)=\text{pow}(P, \Gamma_2) because P3P\in\ell_3 and pow(P,Γ2)=pow(P,Γ3)\text{pow}(P, \Gamma_2)=\text{pow}(P, \Gamma_3) because P1P\in \ell_1. Naturally, this means pow(P,Γ1)=pow(P,Γ3)\text{pow}(P, \Gamma_1)=\text{pow}(P, \Gamma_3) so P2P\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 Γi\Gamma_i be OiO_i for 1i31\le i\le 3. In addition, let the intersection of circles Γ1\Gamma_1 and Γ2\Gamma_2 that is closer to O3O_3 be P3P_3, and define P1,P2P_1, P_2 similarly.

Note that by definition, O1P2=O1P3O_1P_2=O_1P_3, and cyclically similar expressions. Thus, if we imagine rotating O1P2O3\triangle O_1P_2O_3 about line O1O3O_1O_3 and O1P3O2\triangle O_1P_3O_2 about line O1O2O_1O_2, then at some point, the image of P2P_2 and the image of P3P_3 about these two rotations will coincide at a point PP above the plane containing O1O2O3\triangle O_1O_2O_3. In other words, PO1O2O3PO_1O_2O_3 is a tetrahedron with base O1O2O3\triangle O_1O_2O_3. However, note that because we performed a rotation, P2O3=PO3P_2O_3=PO_3 and P3O2=PO2P_3O_2=PO_2. Since P1O3=P2O3P_1O_3=P_2O_3 and P1O2=P3O2P_1O_2=P_3O_2, O2P1O3O2PO3\triangle O_2P_1O_3\cong \triangle O_2PO_3, so we can rotate O2P1O3\triangle O_2P_1O_3 about line O2O3O_2O_3 so that the image of P1P_1 coincides with PP.

But here's the punchline: if we track the projections of P1,P2,P3P_1, P_2, P_3 onto the plane containing O1O2O3\triangle O_1O_2O_3 as we rotate the triangles, it is clear to see that the locus is just a perpendicular from PiP_i to line Oi+1Oi1O_{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 PP onto the plane containing O1O2O3\triangle O_1O_2O_3 lies on the radical axis of each pair of circles, and the proof is complete.

Note by Daniel Liu
2 years, 12 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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.

Calvin Lin Staff - 2 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...