In triangle $ABC$ with incenter $I$ and circumcircle $\omega$, the tangent through $C$ to $\omega$ intersects $AB$ at point $D$. The angle bisector of $\angle CDB$ intersects $AI$ and $BI$ at $E$ and $F$, respectively. Let $M$ be the midpoint of $[EF]$. Prove that line $MI$ passes through the midpoint of arc $ACB$ of $w$ .
Problem
Source: OLCOMA Costa Rica National Olympiad, Final Round, 2017 Shortlist G4 day2
Tags: geometry, arc midpoint, circumcircle
BarisKoyuncu
22.09.2021 15:13
parmenides51 wrote: In triangle $ABC$ with incenter $I$ and circumcircle $w$, the tangent through $C$ to $w$ intersects $AB$ at point $D$. Let $M$ be the midpoint of $AB$. Prove that line $MI$ passes through the midpoint of arc $ACB$ of $w$. There is a typo I guess. We defined $D$ but didn't use it. Also, $MI$ doesn't pass through the midpoint of arc $ACB$ of $w$.
parmenides51
22.09.2021 21:06
This is a bit strange problem, it had defined 3 points at the start not used later, $D$,$E$,$F$ probably to make it easier. I removed all those data. I have the official solution, so indeed $MI$ passes through the arc midpoint of $ACB$. (it has an acute triangle in the solution's figure)
SerdarBozdag
22.09.2021 21:10
$I$-symmedian passes through the midpoint of arc $ACB$.
parmenides51
22.09.2021 21:15
page 5, problem 6, here is the official wording and solution
parmenides51
22.09.2021 21:43
BarisKoyuncu wrote: Also, $MI$ doesn't pass through the midpoint of arc $ACB$ of $w$. indeed, geogebra agrees with what you just said, therefore the above solved problem is wrong, unless I miss something in the translation
En el triángulo ABC con incentro I y circuncírculo w, la tangente por C a w interseca AB en el punto D. Las rectas AI yBI intersecan a la bisectriz del ∠CDB en E y F, respectivamente. M es el punto medio de AB. Pruebe que MI pasa por el punto medio del arco ACB de w
BarisKoyuncu
23.09.2021 00:22
I think I got what it is trying to say. Here is the problem. In triangle $ABC$ with incenter $I$ and circumcircle $\omega$, the tangent through $C$ to $\omega$ intersects $AB$ at point $D$. The angle bisector of $\angle CDB$ intersects $AI$ and $BI$ at $E$ and $F$, respectively. Let $M$ be the midpoint of $[EF]$. Prove that line $MI$ passes through the midpoint of arc $ACB$ of $w$.
The original wording of the question is wrong or at least it is wrong on the official pdf.
hakN
23.09.2021 09:15
Note that $F$ is the excenter of $\triangle CBD$. So by angle chase one can get $\angle EFB = \angle DFB = \angle A/2 = \angle EAB$, giving $ABEF$ is cyclic.
And so since $\triangle EIF \sim \triangle IBA$, we have that $IM$ is the $I$-symmedian of $\triangle IAB$. If we denote $K$ as the midpoint of arc $\widehat{ACB}$ of $\omega$, then it is well known (and easy to prove by angle chase) that $IK$ is the $I$-symmedian of $\triangle IAB$. So $I,M,K$ are collinear and we are done.
BarisKoyuncu
23.09.2021 10:04
Let $AI\cap \omega=G$ and $BI\cap \omega=H$. Let $P$ be the midpoint of arc $ACB$ of $\omega$.
Angle chasing gives $\measuredangle FEA=\measuredangle EAB-\measuredangle EDB=\frac 12(\measuredangle CAB-\measuredangle CDB)=\frac 12\measuredangle DCA=\frac 12\measuredangle ABC=\measuredangle FBA=\measuredangle HGA\Longrightarrow EF//GH$.
Hence, $IM$ bisects $[GH]$.
Also, we know that $GH\perp CI\Longrightarrow GH//CP\Longrightarrow \measuredangle GHB=\measuredangle CHG=\measuredangle PGH\Longrightarrow HI//PG$............$(1)$
Similarly, $GI//PH\Longrightarrow PGIH$ is a parallelogram. Thus, $IP$ bisects $[GH]$............$(2)$
From $(1)$ and $(2)$, we conclude that $I,M,P$ are collinear.