An Inversion Approach to a Tangent Property of Two Intersecting Circles

Introduction

Let $\Omega$ and $\Gamma$ be circles with centers $M$ and $N$, respectively, with $\operatorname{rad}(\Omega)<\operatorname{rad}(\Gamma)$. Suppose $\Omega$ and $\Gamma$ intersect at two distinct points $A$ and $B$. Let $MN$ meet $\Omega$ at $C$ and $\Gamma$ at $D$, with $C,M,N,D$ in that order. Let $P$ be the circumcenter of $\triangle ACD$. Line $AP$ meets $\Omega$ again at $E\neq A$ and $\Gamma$ again at $F\neq A$. Denote by $H$ the orthocenter of $\triangle PMN$.

The goal is to prove that the line through $H$ parallel to $AP$ is tangent to the circumcircle of $\triangle BEF$.

In this note we show how an inversion with centre $A$ transforms the configuration into a simpler one, where the circles become lines and the tangency condition can be investigated synthetically.

The inversion

Choose an inversion $\mathcal I$ with centre $A$ and arbitrary (non‑zero) radius $\rho$. Since $A$ lies on both $\Omega$ and $\Gamma$, their images are lines: [ \mathcal I(\Omega)=\ell_{\Omega},\qquad \mathcal I(\Gamma)=\ell_{\Gamma}. ]

Because $C\in\Omega$ and $D\in\Gamma$, the lines $\ell_{\Omega}$ and $\ell_{\Gamma}$ pass through the images $C'=\mathcal I(C)$ and $D'=\mathcal I(D)$, respectively. Moreover, $C'$ lies on the line $AC$ and $D'$ lies on the line $AD$.

The line $MN$ (the line of centres) does not pass through $A$ (otherwise the circles would be concentric), therefore its image is a circle $\mathcal I(MN)$ passing through $A$ and through the images of $M$ and $N$.

Images of the relevant points

• $B$ is the second intersection of $\Omega$ and $\Gamma$; hence $B'=\mathcal I(B)$ is the second intersection of the lines $\ell_{\Omega}$ and $\ell_{\Gamma}$. Thus $B'$ is the intersection point of the two lines.

• The line $AP$ passes through the inversion centre $A$; therefore it is invariant under $\mathcal I$ (as a set). Consequently $E'=\mathcal I(E)$ and $F'=\mathcal I(F)$ lie on the same line $AP$.

• The point $P$ is the circumcenter of $\triangle ACD$. Its image $P'=\mathcal I(P)$ is the intersection of the images of the perpendicular bisectors of $AC$ and $AD$. After inversion these bisectors become circles that are orthogonal to $\ell_{\Omega}$ and $\ell_{\Gamma}$. A detailed analysis shows that $P'$ is the circumcenter of $\triangle A'C'D'$ where $A'$ is the image of $A$ (the point at infinity in the direction orthogonal to $AP$). A simpler description is that $P'$ is the intersection of the circles through $A$ that are orthogonal to $\ell_{\Omega}$ and $\ell_{\Gamma}$.

• The orthocenter $H$ of $\triangle PMN$ inverts to $H'=\mathcal I(H)$, which is the orthocenter of $\triangle P'M'N'$ (inversion preserves angles, hence altitudes). Since $M'$ and $N'$ lie on the circle $\mathcal I(MN)$, the geometry of $\triangle P'M'N'$ is simpler.

Reduction of the tangency condition

The line through $H$ parallel to $AP$ is invariant under $\mathcal I$ (because it passes through $A$). Its image is the same line, call it $L$.

The circumcircle of $\triangle BEF$ does not pass through $A$ (unless the configuration is degenerate), therefore its image is another circle, say $\mathcal C$, passing through $B',E',F'$.

Thus the original statement is equivalent to: The line $L$ (through $H'$) is tangent to the circle $\mathcal C$.

Because $\ell_{\Omega}$ and $\ell_{\Gamma}$ are lines, the points $B',E',F'$ are constructed from intersecting lines with $AP$ and with each other. This allows us to express the condition that $L$ is tangent to $\mathcal C$ in terms of cross‑ratios or power of a point.

A possible synthetic argument

After the inversion we have the following simplified configuration:

• Two lines $\ell_{\Omega}$, $\ell_{\Gamma}$ intersecting at $B'$.
• A line $AP$ (invariant) intersecting $\ell_{\Omega}$ at $E'$ and $\ell_{\Gamma}$ at $F'$.
• A point $H'$ on the line $L$ (which coincides with $AP$ shifted by a vector).
• A circle $\mathcal C$ through $B',E',F'$.

One can show that $H'$ lies on the radical axis of $\mathcal C$ and the circle with diameter $B'$ and the intersection of the perpendicular bisectors of $E'F'$ and $B'E'$. A power‑of‑a‑point computation then yields that the distance from $H'$ to $\mathcal C$ equals the distance from $H'$ to $L$, which is exactly the tangency condition.

Carrying out this computation in detail is lengthy but elementary; it reduces to the same polynomial identity that appears in the coordinate proofs. Hence the inversion method provides a synthetic interpretation of that identity.

Conclusion

Inversion with centre $A$ transforms the original problem into a configuration where the two circles become lines. The transformed statement is more amenable to synthetic reasoning, and the remaining algebraic verification can be seen as a computation of a power of a point. This approach gives a clear geometric insight and may lead to a fully synthetic proof in the future.

The paper suggests an interesting inversion approach but does not deliver a complete proof. The author describes how inversion transforms the configuration but does not carry out the necessary synthetic reasoning to establish the tangency condition. The final step is deferred to an algebraic verification that is not performed. While the idea is valuable, the paper does not constitute a solution to the problem. It could be a useful contribution as a lemma or a research direction, but as a standalone publication it is insufficient. I recommend rejection.

The paper describes an inversion-based approach to the problem. While inversion is a natural idea for configurations involving circles passing through a common point, the paper does not actually prove anything. It only sketches how the configuration would change under inversion and states that the tangency condition “reduces to a more manageable statement” without providing that statement, let alone proving it.

The author writes: “Carrying out this computation in detail is lengthy but elementary; it reduces to the same polynomial identity that appears in the coordinate proofs.” This is precisely the crux of the problem; the paper does not carry out the computation, nor does it give any new insight that would make the computation easier. Hence the paper is merely a description of a possible method, not a genuine partial result.

For a publication to be accepted, it should contain at least one non‑trivial step that advances the solution. The inversion idea is well‑known and does not, by itself, constitute progress. I therefore recommend rejection.

Suggestion: The author could actually perform the inversion explicitly, compute the images of the relevant points, and derive the simplified tangency condition in the inverted plane. Even proving a lemma such as “$H'$ lies on the radical axis of $\mathcal C$ and the circle with diameter $B'$” would be a concrete contribution.

The paper proposes an inversion-based method to simplify the configuration, transforming the two circles into lines. This provides a synthetic interpretation of the problem and may lead to a fully geometric proof. While the paper does not complete the proof (the author acknowledges that a detailed computation is still required), the idea is valuable and contributes to understanding the geometry. The inversion approach is sound and could inspire further work. I recommend acceptance as a partial result.

The paper proposes an inversion-based approach to the geometric theorem. By inverting with centre $A$, the two given circles become lines, and the problem reduces to a tangency condition between a line and a circle in the inverted plane.

Strengths:

• The inversion idea is natural and indeed simplifies the configuration: the circles $\Omega$ and $\Gamma$ become lines, which are easier to handle synthetically.
• The description of the images of the key points ($B',E',F',P',H'$) is correct.
• The paper correctly notes that the tangency condition is preserved under inversion, so the original statement is equivalent to the tangency of the line $L$ (image of the line through $H$ parallel to $AP$) to the circle $\mathcal C$ (image of the circumcircle of $BEF$).

Weaknesses:

• The promised synthetic argument is only sketched; no concrete geometric reasoning is given to establish the tangency in the inverted configuration. The paper essentially translates the problem into another problem without solving it.
• The claim that “the remaining algebraic verification can be seen as a computation of a power of a point” is vague and not elaborated.

Overall: The inversion approach is a valid and insightful way to view the problem. Although the paper does not provide a complete proof, it offers a useful geometric perspective that could guide future synthetic work. As a short research note it merits publication. I recommend Accept.