Let $ABC$ be an acute-angled triangle with $AB \ne AC$, and let $I$ and $O$ be its incenter and circumcenter, respectively. Let the incircle touch $BC, CA$ and $AB$ at $D, E$ and $F$, respectively. Assume that the line through $I$ parallel to $EF$, the line through $D$ parallel to$ AO$, and the altitude from $A$ are concurrent. Prove that the concurrency point is the orthocenter of the triangle $ABC$. Petar Nizic-Nikolac, Croatia
Problem
Source: 2019 RMM Shortlist G3
Tags: incenter, geometry, Circumcenter, orthocenter
19.06.2020 02:03
Let the concurrency point be $H$. Then since $H$ is on the $A-$altitude and $AI$ is the angle bisector of $HAO$ and $IH$ is perpendicular to $AI$, the reflection of $H$ in $I$ lies on $AO$. By reflecting $DIH$ in $I$, we see that since $DH$ is parallel to $AO$, and the image of $H$ is on $AO$, the image of $D$ is on $AO$ as well, i.e., if $DD'$ is a diameter of the incircle, then $D'$ is on $AO$. By the parallelogram $AHDD'$, we have $AH = 2r$. Now note that if the $A-$excircle is tangent to $BC$ at $D''$, then $A, D', O, D''$ are collinear, and since the $D, D''$ are reflections in the perpendicular bisector of $BC$, $O$ is the midpoint of $D'D''$. Thus $\mathrm{dist}(O, BC) = \frac12 \,\mathrm{dist}(D', BC) = r$. Thus $AH = 2\,\mathrm{dist}(O, BC)$, and this means $H$ is the orthocenter of $ABC$.
19.06.2020 07:31
Seems like the author resubmitted this for EMC 2019 P3.
20.09.2023 20:41
26.05.2024 15:49
Nice problem with a nice solution Let the $A$-altitude and the line through $I$ parallel to $EF$ meet at $H'$, and $J$ be the reflection of $H'$ across $AI$, which of course lies on $AO$. Also let $W = DI \cap AO$. As $H'D \parallel AO$ and $H'I=IJ$, we must have $DI = IW$, meaning $W$ is actually the $D$-antipode in the incircle, so if $D'$ is the $A$-extouch point then $D \in AO$. Now we can easily finish: noting that $OD$ and $AOD'$ are reflections across the perpendicular bisector of $BC$, we have that $OD$ passes through the point $H_A \neq A$ on $(ABC)$ such that $AH_A \perp BC$, so $\angle H'DB = \angle OD'D = \angle ODD' = \angle H_ADB$ implies that $H'$ is indeed the orthocenter.