AE is equal to inradius of triangle ABC

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AE is equal to inradius of triangle ABC

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Source: All-Russian Olympiad 2012 Grade 10 Day 1, UZMO 2012

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#1 h May 29, 2012, 12:17 PM • 1 Y

Y by Adventure10

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 $\omega$ .

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RSM

#2 h May 29, 2012, 12:35 PM • 3 Y

Y by MahsaMorady, Adventure10, Mango247

$AO$ 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 $AE=r$ .

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#3 h May 31, 2012, 8:16 PM • 2 Y

Y by Adventure10, Fardad

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...

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WakeUp

#4 h Jun 1, 2012, 11:56 AM • 2 Y

Y by Adventure10, Mango247

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.

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#5 h Feb 28, 2017, 8:10 PM • 2 Y

Y by Adventure10, Mango247

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.

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jayme

#6 h May 18, 2018, 12:04 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
also at

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=481639

Sincerely
Jean-Louis

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Synthetic_Potato

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Synthetic_Potato

#8 h May 20, 2018, 6:10 AM • 2 Y

Y by Adventure10, Mango247

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.

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#10 h Jan 7, 2019, 7:46 AM • 1 Y

Y by Adventure10

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.

This post has been edited 6 times. Last edited by mathetillica, Jan 7, 2019, 7:51 AM

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#11 h Sep 3, 2023, 2:06 PM • 1 Y

Y by centslordm

...

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 $AE=ID$ . $\blacksquare$

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