The inscribed circle $\omega$ of the non-isosceles acute-angled triangle $ABC$ touches the side $BC$ at the point $D$. Suppose that $I$ and $O$ are the centres of inscribed circle and circumcircle of triangle $ABC$ respectively. The circumcircle of triangle $ADI$ intersects $AO$ at the points $A$ and $E$. Prove that $AE$ is equal to the radius $r$ of $\omega$.
Problem
Source: All-Russian Olympiad 2012 Grade 10 Day 1, UZMO 2012
Tags: geometry, inradius, circumcircle, geometric transformation, reflection, perpendicular bisector, geometry unsolved
RSM
29.05.2012 15:35
$ AO $ is the reflection of $ ID $ on the perpendicular bisector of $ AI $, since $ ID $ and $ AO $ are anti-parallel wrt $ \angle BAC $. Now since $ AEDI $ is cyclic. So $ AEDI $ is an isosceles trapezium. So $ AE=r $.
hatchguy
31.05.2012 23:16
We approach indirectly. Take a point $E'$ in $AO$ such that $A$ lies between $E'$ and $O$ and $AE' = r$. If we show that $AI \parallel E'D$ we will be done since this implies $AE'DI$ being cyclic and therefore $E' = E$. Denote by $L$ the intersection of $AO$ and $ID$. Let $N \equiv AI \cap BC$. It is enough to show that $LA = LI$. The last is equivalent to $\angle IAO = \angle LIA = \angle DIN$. So we only need to show that $\angle IAO + \angle INB = 90$ which is just simple angle chase. EDIT: Weird, RSM's post didnt appear when I wrote my solution, and it was posted some hours ago...
WakeUp
01.06.2012 14:56
hatchguy wrote: EDIT: Weird, RSM's post didnt appear when I wrote my solution, and it was posted some hours ago... What happened was that I posted this problem (it's a problem from the ARO 2012), but later a fellow Geo moderator brought to my attention that it had already been posted recently. So I was about to delete my thread and replace it on the resources page with this one, however you posted your solution before I could So I moved your post here and deleted the old thread.
anantmudgal09
28.02.2017 23:10
Let $K$ be the intersection of the $A$ altitude with $\odot(ADI)$. Since $AK, AO$ are symmetric in $AI$, we get $\overarc{IE}=\overarc{IK}$. From $AK \parallel DI$ we get $\overarc{AI}=\overarc{DK} \Longrightarrow \overarc{AE}=\overarc{ID}$ so $AE=r$ as required.
jayme
18.05.2018 15:04
Dear Mathlinkers, also at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=481639 Sincerely Jean-Louis
Synthetic_Potato
20.05.2018 09:10
WoLoG, assume that $AB>AC$. Let $ID\cap AO = P$. See that $\angle PIA = \angle IAC + \angle ACD + \angle CDI - 180^\circ = \angle C + \frac{\angle A}2 - 90^\circ = \frac{\angle A}2 - (90^\circ - \angle C) = \angle IAB - \angle OAB = \angle PAI$. Hence, $\angle PAI = \angle PIA$. Now the conclusion is obvious.
mathetillica
07.01.2019 10:46
Since $AEDI$ is cyclic so $\angle AID=360-\angle B-\frac{\angle A}{2}-90=270-\angle B- \frac{\angle A}{2}=180-\angle AED$ which implies $\angle AED=\frac{\angle A}{2}-(90-\angle B)=\angle OAI$.Thus $AI||ED$ which gives $AE=ID=r$ as $AEDI$ now becomes isosceles trapezium.
IAmTheHazard
03.09.2023 17:06
... Let $H$ be the orthocenter. Then since $H$ and $O$ are isogonal conjugates, $$\measuredangle AID=\measuredangle IAH=\measuredangle OAI=\measuredangle EAI,$$hence $AIDE$ is an isosceles trapezoid, so $AE=ID$. $\blacksquare$