In triangle $ABC$, point $I$ is the center of the inscribed circle points, points $I_A$ and $I_C$ are the centers of the excircles, tangent to sides $BC$ and $AB$, respectively. Point $O$ is the center of the circumscribed circle of triangle $II_AI_C$. Prove that $OI \perp AC$
Problem
Source: 2012 Oral Moscow Geometry Olympiad grades 8-9 p4
Tags: geometry, incenter, excenters, excenter, perpendicular, circumcircle
MathMonster0923
10.09.2019 12:46
Extend rays $\vec{AI}$ and $\vec{CI}$ to intersect $(ABC)$ at $D$ and $F$ respectively, and let $E$ be the intersection of lines $OI$ and $AC$.
By the Incenter-Excenter lemma, $D$ is the midpoint of $II_A$ and $F$ is the midpoint of $II_C$.
So $OF \perp FI$, and $OD \perp DI$, and $OFID$ is cyclic.
We need to show $\angle IEA = 90^{\circ}$.
$\angle AIE = \angle OID = 90^{\circ} - \angle IOD = 90^{\circ} - \angle IFD = 90^{\circ} - \angle CFD = 90^{\circ} - \angle CAD = 90^{\circ} - \frac{A}{2}$
$\angle IEA = 180^{\circ} - \angle EAI - \angle AIE = 180^{\circ} - \frac{A}{2} - (90^{\circ} - \frac{A}{2}) = 90^{\circ}$
And we are done.
khanhnx
10.09.2019 15:31
Just note that: $AC$ is anti - parallel in $\triangle II_AI_C$ then: $IO$ $\perp$ $AC$
dikhendzab
18.06.2020 16:27
Let $OI$ and $AC$ intersect in point $D$. From angle chase: $\angle IBI_A=90^{\circ}+\frac{\beta}{2}-\frac{\beta}{2}=90^{\circ}$. So $\angle ABI_A=\angle IBI_A+\angle ABI=90^{\circ}+\frac{\beta}{2}$. From triangle $ABI_A$: $\angle AI_AB=180^{\circ}-\angle BAI_A-\angle ABI_A=180^{\circ}-\frac{\alpha}{2}-90^{\circ}-\frac{\beta}{2}=90^{\circ}-\frac{\alpha}{2}-\frac{\beta}{2}$. Since $O$ is circumcenter of triangle $II_AI_C$ we have that $\angle I_COI=2\angle BI_AI=180^{\circ}-\alpha-\beta=\gamma$ Now $\angle OI_CI=\angle OII_C=\angle CID=90^{\circ}-\frac{\gamma}{2}$. Finally: $\angle IDC=180^{\circ}-\angle ICD-\angle CID=180^{\circ}-\frac{\gamma}{2}-90^{\circ}+\frac{\gamma}{2}=90^{\circ} \implies OI \perp AC$
Bexultan
16.05.2023 15:24
The circumcenter of triangle $AIC$ is the midpoint of arc $AC$ no containing the point $B$. This point lies on $BI$. In other words, we want $BI$ and $BO$ be isogonal conjugates with respect to $\angle AIC$. This is same as saying that they are isogonal wrt $\angle I_AII_C$. But this is true, since $IB$ is the altitude and $IO$ is the diameter of triangle $II_AI_C$
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