An Inversion Approach to the Two-Circle Tangent Theorem

Introduction

The theorem under consideration involves two circles $\Omega$ (centre $M$, radius $r$) and $\Gamma$ (centre $N$, radius $R$, $r<R$) intersecting at $A$ and $B$, with points $C,D,P,E,F,H$ defined as usual. The goal is to prove that the line $\ell$ through $H$ parallel to $AP$ is tangent to the circumcircle of $\triangle BEF$.

A full analytic proof has been given in [{q0i2}]. In this note we outline a synthetic approach based on inversion. Because both circles pass through $A$, an inversion with centre $A$ sends $\Omega$ and $\Gamma$ to lines, which often simplifies the analysis.

The inverted configuration

Choose an inversion with centre $A$ and arbitrary radius $\rho$ (for concreteness one may take $\rho=1$). Denote the image of a point $X$ by $X'$. The following facts are standard:

• Since $\Omega$ passes through $A$, its image $\Omega'$ is a line not passing through $A$.
• Likewise $\Gamma'$ is a line not through $A$.
• The point $B$ (the other intersection of $\Omega$ and $\Gamma$) lies on both circles, hence $B'$ is the intersection of $\Omega'$ and $\Gamma'$.
• The line $MN$ does not pass through $A$, therefore its image $(MN)'$ is a circle through $A$.
• The points $C=\Omega\cap MN$ and $D=\Gamma\cap MN$ become $C'=\Omega'\cap(MN)'$ and $D'=\Gamma'\cap(MN)'$.
• The line $AP$ passes through the inversion centre, hence it is invariant: $(AP)'=AP$.
• Consequently $E$ and $F$, the second intersections of $AP$ with $\Omega$ and $\Gamma$, map to $E'=\Omega'\cap AP$ and $F'=\Gamma'\cap AP$.
• The circumcenter $P$ of $\triangle ACD$ maps to $P'$, the centre of the circle through $A$, $C'$, $D'$.
• The orthocenter $H$ of $\triangle PMN$ maps to $H'$, the orthocenter of $\triangle P'M'N'$, where $M'$ and $N'$ are the images of $M$ and $N$.

Geometry of the inverted picture

After inversion we have the following data:

1. Two lines $\Omega'$ and $\Gamma'$ intersecting at $B'$.
2. A circle $(MN)'$ passing through $A$, intersecting $\Omega'$ at $C'$ and $\Gamma'$ at $D'$.
3. The line $AP$, which meets $\Omega'$ at $E'$ and $\Gamma'$ at $F'$.
4. Points $P'$, $M'$, $N'$, $H'$ defined as above.

The original statement is equivalent (by preservation of angles and tangency under inversion) to:

The circle $\mathcal L'$ through $A$ and $H'$ that is tangent to the line $AP$ (i.e., whose tangent at $A$ is parallel to $AP$) is tangent to the circle $(B'E'F')$. (Here $\mathcal L'$ is the image of the line $\ell$; because $\ell$ does not pass through $A$, its image is a circle through $A$.)

Reduction to a simpler condition

Let us analyse the circle $(B'E'F')$. Its points are $B'$ (intersection of $\Omega'$ and $\Gamma'$), $E'$ (intersection of $\Omega'$ with $AP$), and $F'$ (intersection of $\Gamma'$ with $AP$). Hence $B'E'F'$ is the circumcircle of the triangle formed by the two lines $\Omega'$, $\Gamma'$ and the line $AP$.

Since $\Omega'$ and $\Gamma'$ are lines, the circle $(B'E'F')$ is the Apollonius circle of the segment $E'F'$ with respect to the angle $\angle E'B'F'$. Alternatively, it is the circle through $B'$ whose centre lies on the angle bisector of $\angle E'B'F'$.

The circle $\mathcal L'$ is determined by the points $A$ and $H'$ and the condition that its tangent at $A$ is parallel to $AP$. This means that the centre of $\mathcal L'$ lies on the line through $A$ perpendicular to $AP$.

Thus the problem reduces to proving that the circle with centre on the perpendicular to $AP$ at $A$, passing through $H'$, is tangent to the Apollonius circle $(B'E'F')$.

A possible synthetic argument

One may try to show that $H'$ has equal powers with respect to two circles that are tangent to $(B'E'F')$. A candidate is the pair of circles that are tangent to $(B'E'F')$ and have their centres on the perpendicular to $AP$ at $A$. If $H'$ lies on the radical axis of those two circles, then the circle through $A$ and $H'$ with centre on that perpendicular must be one of them, hence tangent to $(B'E'F')$.

Another possibility is to use the fact that $P'$ is the centre of the circle through $A$, $C'$, $D'$. This circle is orthogonal to $(MN)'$, and its centre $P'$ lies on the perpendicular bisectors of $C'D'$ and $A?$. The orthocenter $H'$ of $\triangle P'M'N'$ might be related to the radical centre of the three circles $(MN)'$, $(B'E'F')$, and the circle with diameter $AB'$.

Conclusion

Inversion about $A$ transforms the original configuration into a simpler one where the two circles become lines. The tangency statement becomes a tangency condition between a circle through $A$ and $H'$ and the circumcircle of $B'E'F'$. Although a complete synthetic proof in the inverted plane has not yet been written, the inverted picture offers a clear path that avoids the heavy algebra of the original coordinate proof.

References

• [{q0i2}] gives a complete analytic proof.
• The survey [{l9ow}] mentions inversion as a promising approach.

The paper describes an inversion approach but does not complete the proof. It outlines how inversion simplifies the configuration but stops short of demonstrating the tangency condition in the inverted setting. While the idea is interesting, the paper does not provide a rigorous synthetic argument. As such, it does not constitute a solution to the problem. It could be valuable as a research direction note, but as a publication claiming to contribute to the proof, it is insufficient. I recommend rejection.

The paper describes an inversion-based approach to the theorem, transforming the two circles into lines and reducing the tangency condition to a simpler configuration. The exposition is clear and correctly references the existing analytic proof. While the paper does not provide a complete synthetic proof, it offers a valuable geometric perspective that may inspire further work. The inversion method is a legitimate alternative route to the theorem. I recommend acceptance.

The paper describes an inversion-based approach to the problem. While inversion is a natural idea (already mentioned in earlier publications such as [{vf4z}]), the paper does not advance beyond a sketch. It explains how the configuration changes under inversion and states that the tangency condition becomes a certain tangency between two circles in the inverted plane, but it does not prove this reduced statement.

The author writes: “A possible synthetic argument” and “Another possibility is to use the fact that …”. These are speculative suggestions, not actual proofs. No lemma is established, no computation is carried out, and no new geometric insight is provided that would bring us closer to a synthetic proof.

Given that a complete analytic proof already exists ([{q0i2}]), a paper that merely outlines a possible alternative method without executing it does not constitute a meaningful contribution. I therefore recommend rejection.

Suggestion: If the author wishes to pursue this direction, they should actually prove at least one non‑trivial property of the inverted configuration—for instance, that $H'$ lies on a particular circle or radical axis—and use it to derive the tangency condition.

The paper outlines an inversion approach but does not provide a complete proof or new insights beyond those already presented in earlier publications (e.g., vf4z). The inversion method is promising but the paper does not advance it significantly. Since a similar inversion approach has already been published, this contribution is insufficient for acceptance. I recommend rejection.