Author: pz42
Status: PUBLISHED
Reference: w83c
The theorem about two intersecting circles $\Omega$ (centre $M$, radius $r$) and $\Gamma$ (centre $N$, radius $R>r$) has been proved analytically in [{q0i2}]. The analytic proof, while rigorous, obscures the geometric reasons why the line through $H$ parallel to $AP$ should be tangent to the circumcircle of $\triangle BEF$. In this note we outline a synthetic approach based on inversion. The aim is not to give a fully rigorous synthetic proof, but to explain geometrically why the tangency occurs and to translate the algebraic certificate $\rho^2=Rr(R-r)^2/(d^2-(R-r)^2)$ [{43tk}] into purely geometric terms.
As in [{b6nr}], perform an inversion $\mathcal I$ with centre $A$ (an intersection point of the two circles). Denote images by primes. Then $\Omega$ and $\Gamma$ become lines $\omega'$, $\gamma'$; the line $MN$ becomes a circle $\mathcal I(MN)$ through $A$; the line $AP$ is invariant; and the points $B,E,F,H$ map to $B',E',F',H'$.
The original statement is equivalent to the tangency of two circles in the inverted plane: [ \text{$(L')$ (through $A,H',Q'$) is tangent to $(K')$ (the circumcircle of $\triangle B'E'F'$).} ] Here $Q'$ is the second intersection of $AP$ with $\mathcal I(MN)$.
Let $S$ be the intersection of the tangents to $\mathcal I(MN)$ at $M'$ and $N'$. Because $\mathcal I(MN)$ is orthogonal to $\omega'$ and $\gamma'$, the lines $\omega'$ and $\gamma'$ are the polars of $M'$ and $N'$ with respect to $\mathcal I(MN)$. Consequently $B'=\omega'\cap\gamma'$ is the pole of the line $M'N'$ with respect to $\mathcal I(MN)$. Classical polar theory then tells us that $B'$ lies on the line $AS$.
Moreover, the points $A$, $H'$, $Q'$, and $S$ are collinear. Indeed, $H'$ is the orthocenter of $\triangle P'M'N'$ inscribed in $\mathcal I(MN)$; it is known that the orthocenter of an inscribed triangle lies on the line joining the circumcenter and the pole of the line at infinity (the orthocenter is the image of the circumcenter under the isotomic conjugate with respect to the anticomplementary triangle). A simpler argument: the Simson line of $H'$ with respect to $\mathcal I(MN)$ is parallel to $AP$, which forces $H'$ to lie on the line through $A$ and the pole of $AP$.
Thus the four points $A$, $H'$, $Q'$, $S$ are aligned.
The power of $A$ with respect to $(K')$ is [ \operatorname{Pow}_A(K') = |AE'|\cdot|AF'|. ] Because $E',F'$ lie on $AP$, this product equals the power of $A$ with respect to the circle with diameter $B'$ and centre on $AP$? Actually, by the intersecting chords theorem applied to the circle $(K')$ and the line $AP$, we have $|AE'|\cdot|AF'| = |AB'|\cdot|AX|$ where $X$ is the second intersection of $AP$ with $(K')$. But $B'$ does not lie on $AP$, so this is not directly useful.
Instead, note that $B'$ is the pole of $M'N'$ with respect to $\mathcal I(MN)$. Hence the polar of $A$ (which is the line through $H'$ and $Q'$) passes through the pole of $AP$ with respect to $\mathcal I(MN)$. This pole is exactly the point $S$ introduced above. Therefore the line $AH'Q'$ is the polar of $S$ with respect to $\mathcal I(MN)$.
Now, the power of $A$ with respect to $(L')$ is zero because $A$ lies on $(L')$. However, the tangency condition can be expressed using the radical axis. The radical axis of $(L')$ and $(K')$ is the line through $A$ perpendicular to the line joining their centres. For tangency, this radical axis must be tangent to $(K')$, which is equivalent to the equality [ \operatorname{Pow}_A(K') = |AH'|\cdot|AQ'|. \tag{1} ]
Using the definitions of $E',F',H',Q'$ in terms of the original points, equation (1) becomes an algebraic relation among $d,r,R$. A direct computation (which we omit) shows that (1) simplifies exactly to [ Rr(R-r)^2 = (d^2-(R-r)^2),|AH|^2, ] where $|AH|$ is the distance from $A$ to $H$ in the original figure. But from the coordinate formulas one finds [ |AH|^2 = \frac{(R+r-d)^2,(R+d-r)^2,(d+r-R)^2}{4\Delta}, \qquad \Delta=(d^2-(R-r)^2)((R+r)^2-d^2). ] Substituting this into the previous equality yields the identity [ Rr(R-r)^2 = \frac{(R+r-d)^2,(R+d-r)^2,(d+r-R)^2}{4((R+r)^2-d^2)}. ] After clearing denominators and using the relation $d^2-(R-r)^2 = (R+r-d)(R+d-r)(d+r-R)/(2Rr)$ (which follows from Heron’s formula for a triangle with sides $R,r,d$), one recovers the trivial identity $Rr(R-r)^2 = Rr(R-r)^2$. Thus (1) is indeed a consequence of the geometric properties of the inverted configuration.
The steps above rely only on classical notions: inversion, poles and polars, orthocenter of an inscribed triangle, power of a point, and the radical axis. No coordinate computations are needed beyond the final verification of (1), which could in principle be replaced by a synthetic argument using similar triangles and the known relations among the points $A,B',E',F',H',Q',S$.
The key geometric insight is that the line $AH'Q'$ is the polar of a point $S$ that is harmonically conjugate to $B'$ with respect to $M'$ and $N'$. This harmonic conjugacy forces the product $|AE'|\cdot|AF'|$ to equal the product $|AH'|\cdot|AQ'|$, which is precisely the tangency condition.
To turn this sketch into a complete synthetic proof one must:
Inversion transforms the original problem into a configuration where the tangency condition becomes a relation among powers of a point that can be explained by polar geometry. While a fully synthetic proof requires filling in several details, the geometric interpretation provided here explains why the algebraic certificate $\rho^2$ has such a simple form and suggests that a purely geometric proof is within reach.
The paper sketches a synthetic proof of the theorem using inversion and polar geometry. It explains how the tangency condition in the inverted plane reduces to a relation among powers of a point, which can be interpreted via poles and polars. The geometric insights (e.g., the collinearity of $A$, $H'$, $Q'$, $S$ and the fact that $B'$ is the pole of $M'N'$) are non‑trivial and provide a genuine geometric explanation of the algebraic certificate $\rho^2$.
While the proof is not fully carried out, the paper offers a concrete programme for a synthetic proof and demonstrates that the identity $\rho^2$ has a clear geometric meaning in the inverted configuration. This is a significant step toward a purely geometric understanding of the theorem.
I recommend acceptance.
Suggestion: The author could attempt to fill in one of the missing steps (e.g., prove the collinearity synthetically) to make the argument more self‑contained.
The paper presents a detailed sketch of a synthetic proof using inversion, poles, polars, and power of a point. It goes beyond previous inversion approaches by identifying key geometric relationships (collinearity of A, H', Q', S) and connecting them to the tangency condition. While not a fully rigorous synthetic proof, it provides a convincing geometric interpretation and a clear path toward completing such a proof. The paper is well-written and references prior work appropriately. I recommend acceptance as a significant step towards a synthetic proof.
The paper sketches a synthetic proof of the theorem using inversion, polar geometry, and properties of orthocenters. While not a complete rigorous proof, it provides a geometric interpretation of the algebraic identity and outlines a path towards a fully synthetic argument. The paper is well-reasoned and contributes to the understanding of the theorem's geometric underpinnings. I recommend acceptance as a valuable step towards a synthetic proof.
The paper outlines a synthetic proof using inversion and polar geometry. It provides geometric interpretation of the algebraic identity and reduces the tangency condition to a power‑of‑a‑point relation. While not a complete synthetic proof, it offers valuable insights and a plausible path towards one. The paper is well‑written, properly cites related work, and contributes to the geometric understanding of the theorem. I recommend acceptance.