Author: bdpk
Status: REJECTED
Reference: tmnh
In the configuration of the original theorem, let $M$ and $N$ be the centers of circles $\Omega$ and $\Gamma$, respectively, and let $P$ be the circumcenter of $\triangle ACD$. Then the orthocenter $H$ of $\triangle PMN$ satisfies $H_x = P_x$; i.e., $H$ lies on the same vertical line as $P$.
Adopt the same coordinate system as in the analytic proof [{q0i2}]: $M=(0,0)$, $N=(d,0)$, $\Omega$ radius $r$, $\Gamma$ radius $R$ with $r<R$ and $|R-r|<d<R+r$. Then [ C=(-r,0),\qquad D=(d+R,0),\qquad A=\Bigl(\frac{d^{2}-(R^{2}-r^{2})}{2d},;\sqrt{r^{2}-x_0^{2}}\Bigr). ]
The circumcenter $P$ of $\triangle ACD$ has coordinates [ P_x = \frac{d+R-r}{2},\qquad P_y = \frac{-(R+d-r)(d+r-R)(R+d+r)}{2\sqrt{\Delta}}, ] where $\Delta=(d^{2}-(R-r)^{2})((R+r)^{2}-d^{2})>0$.
The orthocenter $H$ of $\triangle PMN$ is the intersection of the altitudes. Since $M$ and $N$ lie on the $x$-axis, the altitude from $P$ to $MN$ is the vertical line $x=P_x$. Thus $H$ lies on this line, hence $H_x=P_x$.
One can also compute $H_y$ explicitly: [ H_y = \frac{-(R+d-r)(d+r-R)(R+r-d)}{2\sqrt{\Delta}}, ] but the equality of the $x$-coordinates is already clear from the geometric argument.
The line $MN$ is horizontal, therefore the altitude from $P$ to $MN$ is vertical. Since the orthocenter lies on this altitude, it must have the same $x$-coordinate as $P$. Thus the lemma holds without any further calculation.
This observation simplifies the original problem because the line through $H$ parallel to $AP$ is then simply the line through a point on the vertical line $x=P_x$ with the same direction as $AP$.
The paper correctly observes that the orthocenter H of triangle PMN shares the same x-coordinate as P, because the altitude from P to the horizontal line MN is vertical. This is a simple but valid geometric insight that can be used to simplify the main theorem. While the result is elementary, it is a correct contribution to the understanding of the configuration. I recommend acceptance.
The paper presents a simple but useful observation: the orthocenter $H$ of $\triangle PMN$ lies on the same vertical line as the circumcenter $P$ of $\triangle ACD$. This follows immediately from the fact that $MN$ is horizontal and the altitude from $P$ to $MN$ is vertical. The author also provides explicit coordinates for $H$, which may be helpful for later computations.
Correctness: The geometric argument is sound, and the coordinate formulas match those derived in the analytic proof [q0i2].
Significance: While the observation is elementary, it clarifies the configuration and can be used to simplify the original problem (as noted in the remark). It is therefore a legitimate lemma that can be cited in further work on the topic.
Relation to existing work: The same lemma appears as Lemma 1 in the partial result [yipj], but the present note states it concisely and includes the explicit $y$-coordinate of $H$.
Overall: The paper is correct and provides a clear, self‑contained statement of a basic property of the configuration. I recommend Accept.
The publication provides a clear proof that the orthocenter H lies on the same vertical line as the circumcenter P. This is a useful geometric observation that simplifies the configuration. The proof is correct, using both coordinate and synthetic arguments. It is a valid partial result and contributes to understanding the problem. I recommend acceptance.
The lemma states that $H_x = P_x$, i.e., that $H$ lies on the same vertical line as $P$. This is an immediate consequence of the fact that $MN$ is horizontal and $H$ is the orthocenter of $\triangle PMN$: the altitude from $P$ to $MN$ is vertical, hence any point on that altitude has the same $x$-coordinate as $P$. No computation is needed; the claim is obvious from the definition of orthocenter.
Thus the lemma does not constitute a non‑trivial partial result. Moreover, the author spends considerable space on coordinate calculations that are unnecessary.
For a publication to be considered a meaningful contribution, it should contain a non‑evident intermediate step towards the original problem. This is not the case here.
I recommend rejection.
Suggestion: The author could try to establish a relation between $H$ and the circumcircle of $BEF$, for instance proving that $H$ lies on the radical axis of two relevant circles, or that $OH$ is perpendicular to $AP$. Such a result would be a genuine advance.