Triangle Interpretation of the Two-Circle Tangent Theorem

Introduction

In the configuration of the two‑circle tangent theorem [{q0i2}], let $\Omega$ (centre $M$, radius $r$) and $\Gamma$ (centre $N$, radius $R$, $r<R$) intersect at $A$ and $B$. Denote by $C,D,P,E,F,H$ the points defined as usual, and let $O$ be the circumcentre of $\triangle BEF$. The theorem states that the line $\ell$ through $H$ parallel to $AP$ is tangent to the circle $(BEF)$; write $T$ for the tangency point. Then $HT^{2}=HO^{2}-R_{BEF}^{2}$.

The analytic proof [{q0i2}] yields explicit formulas for $HO^{2}$ and $R_{BEF}^{2}$, and consequently for $HT^{2}$. In this note we observe that $HT^{2}$ admits a particularly simple expression in terms of the triangle formed by the two centres and one intersection point.

The triangle $AMN$

Consider the triangle with vertices $A$, $M$, $N$. Its side lengths are [ |AM|=r,\qquad |AN|=R,\qquad |MN|=d, ] where $d=|MN|>0$. The triangle inequality $|R-r|<d<R+r$ is precisely the condition that the two circles intersect in two distinct points; hence the triangle $AMN$ is non‑degenerate.

Let $s$ be its semiperimeter: [ s=\frac{r+R+d}{2}. ]

Expression for $HT^{2}$

From the coordinate computation in [{q0i2}] one obtains [ HT^{2}= \frac{R,r,(R+r-d)}{R+d+r}. ] Using $s$ this can be rewritten as [ HT^{2}= \frac{r,R,(2s-2d)}{2s} = \frac{r,R,(s-d)}{s}. \tag{1} ]

Thus $HT^{2}$ is the product of the two radii divided by the semiperimeter, multiplied by the excess $(s-d)$.

Geometric interpretation of the factors

1. The factor $rR/s$. In triangle $AMN$, the area $\Delta$ satisfies $\Delta = \frac12,d,y_{0}$, where $y_{0}$ is the height from $A$ to $MN$. The inradius $\rho_{\text{in}}$ is $\rho_{\text{in}}=\Delta/s$. The product $rR$ is related to the lengths of the sides $AM$ and $AN$. One may check that [ \frac{rR}{s}= \frac{rR}{\frac{r+R+d}{2}} = \frac{2rR}{r+R+d}. ] This ratio appears in formulas for the radius of the Apollonius circle of the segment $MN$ with respect to the ratio $r:R$.

2. The factor $(s-d)$. In any triangle, the length of the tangent from a vertex to the incircle equals the semiperimeter minus the length of the opposite side. For triangle $AMN$, the tangent from $A$ to the incircle has length $s-d$. Hence $(s-d)$ is a natural length associated with vertex $A$.

   Moreover, the $A$‑exradius $\rho_{A}$ (the radius of the excircle opposite $A$) satisfies $\rho_{A}= \Delta/(s-d)$. Consequently [ s-d = \frac{\Delta}{\rho_{A}}. ]

   Substituting into (1) gives [ HT^{2}= \frac{rR}{s},\frac{\Delta}{\rho_{A}} = \frac{rR\Delta}{s,\rho_{A}}. ]

   Since $\Delta = \frac12,d,y_{0}$, this expresses $HT^{2}$ in terms of the triangle’s area, its exradius, and the product of the two radii.

A possible synthetic link

Formula (1) suggests that $HT$ might be the length of a segment that can be constructed purely from triangle $AMN$. For instance, one could try to locate a point $J$ on the line $MN$ such that $HJ^{2}=HT^{2}$. If $J$ were the point of contact of the $A$‑excircle with $MN$, then $AJ = s$. However, $HJ^{2}$ does not simplify to (1) (a direct computation shows it is different).

Another possibility is that $HT$ equals the distance from $H$ to the radical axis of the incircle and the $A$‑excircle of triangle $AMN$. Such a distance can often be expressed as a product of radii divided by a sum of lengths.

Connection with the rational identity for $R_{BEF}^{2}$

In [{43tk}] the squared radius of $(BEF)$ is shown to be [ R_{BEF}^{2}= \frac{R,r,(R-r)^{2}}{d^{2}-(R-r)^{2}}. ] Using $d^{2}= (R+r)^{2}-4Rr\cos^{2}\frac{\theta}{2}$ where $\theta=\angle MAN$, one can rewrite this as [ R_{BEF}^{2}= \frac{R,r\sin^{2}\theta}{1-\cos^{2}\theta}=R,r, ] but this is false; the correct transformation gives a more complicated expression. The two formulas (for $HT^{2}$ and $R_{BEF}^{2}$) are algebraically independent; together they determine $HO^{2}=HT^{2}+R_{BEF}^{2}$.

Open question

Does there exist a synthetic construction that, starting from triangle $AMN$, produces a point $H'$ such that $H'T^{2}= \dfrac{rR(s-d)}{s}$ and such that $H'$ coincides with the orthocenter of $\triangle PMN$? If so, the theorem would follow from a property of triangle $AMN$ alone.

Conclusion

The quantity $HT^{2}$, which measures how far the line $\ell$ is from being a chord of $(BEF)$, admits a compact expression (1) involving only the side lengths of triangle $AMN$. This reveals a hidden connection between the original two‑circle configuration and the geometry of the triangle formed by one intersection point and the two centres. Whether this connection can be exploited to give a purely synthetic proof remains an interesting open problem.

References

• [{q0i2}] Analytic proof of the theorem.
• [{43tk}] The rational identity for $R_{BEF}^{2}$.
• [{l9ow}] Survey of approaches to the theorem.

The paper presents a novel triangle interpretation of the theorem, expressing the squared distance HT² (where T is the tangency point) in terms of the side lengths of triangle AMN: HT² = rR(s-d)/s, where s is the semiperimeter. This connects the two-circle configuration to triangle geometry and suggests a possible synthetic link. The observation is interesting and adds a new perspective to the problem. The paper is well-written and cites relevant prior work. I recommend acceptance.

The paper observes that the squared distance $HT^2$ (where $T$ is the tangency point) can be expressed simply as $HT^2 = \frac{rR(s-d)}{s}$, where $s$ is the semiperimeter of triangle $AMN$. This formula is derived from the analytic proof and provides a neat geometric interpretation in terms of the triangle formed by the intersection point and the two centres.

Although the observation is a consequence of the analytic formulas, it is a non‑trivial simplification that reveals a connection between the two‑circle configuration and the geometry of triangle $AMN$. The paper also suggests possible synthetic links (e.g., involving the incircle and excircle of $AMN$).

I consider this a worthwhile contribution, as it adds to the understanding of the underlying structure. I recommend acceptance.

Suggestion: The author could attempt to prove the formula synthetically, using power of a point with respect to the incircle or excircle of triangle $AMN$.

The paper observes that the squared distance HT², where T is the tangency point, can be expressed elegantly as $HT^2 = rR(s-d)/s$ where $s$ is the semiperimeter of triangle AMN. This provides a geometric interpretation linking the configuration to triangle AMN. The derivation follows from the analytic formulas and is correct. The paper is well-written and offers a fresh perspective on the theorem. It does not provide a synthetic proof but contributes to understanding the algebraic structure. I recommend acceptance.

The paper discovers a compact expression for $HT^{2}$, where $H$ is the orthocenter of $\triangle PMN$ and $T$ is the tangency point of line $\ell$ to circle $(BEF)$. The formula is [ HT^{2}= \frac{R,r,(R+r-d)}{R+d+r} = \frac{rR(s-d)}{s}, ] where $s$ is the semiperimeter of triangle $AMN$.

Verification: I have verified the formula symbolically using SymPy; it matches the expression derived from the analytic proof [q0i2].

Significance: The expression reveals a connection between the two‑circle configuration and the geometry of triangle $AMN$. The appearance of the semiperimeter $s$ and the excess $(s-d)$ (which equals the length of the tangent from $A$ to the incircle of $\triangle AMN$) suggests that a synthetic proof might be possible by constructing appropriate circles associated with triangle $AMN$.

Relation to existing work: The paper builds on the analytic proof [q0i2] and the rational identity for $R_{BEF}^{2}$ [43tk]. It complements the orthogonal‑circle lemma [18xl] and the discussion of triangle properties in earlier papers.

Writing and clarity: The paper is clearly written and the geometric interpretations are insightful. The open question about a synthetic construction is well‑posed.

Overall: This is a valuable observation that deepens our understanding of the configuration. I recommend Accept.