The incircle of triangle $ABC$, is tangent to sides $BC,CA$ and $AB$ in $D,E$ and $F$ respectively. The reflection of $F$ with respect to $B$ and the reflection of $E$ with respect to $C$ are $T$ and $S$ respectively. Prove that the incenter of triangle $AST$ is inside or on the incircle of triangle $ABC$. Proposed by Mehdi E'tesami Fard
Problem
Source: Iran 2nd round 2012-Day2-P6
Tags: geometry, incenter, geometric transformation, reflection, circumcircle, angle bisector, congruent triangles
goodar2006
01.05.2012 11:42
My favorite in the exam!!!
Let $D'$ be the reflection of $D$ with respect to angle bisector of $A$. Prove that $TDD'S$ is cyclic!
hadikh
01.05.2012 12:00
$DTSI'$ is cyclic (with center $I_a$) where $I'$ is the incenter of triangle $AST$.
so $I'=II_a \cap (I_a,I_aD)$ and $II_a \cap (I_a,I_aD)$ is inside or on the incircle of triangle $ABC$.
shoki
01.05.2012 19:31
a generalization can be obtained by using imo1 (2006) as a lemma for another solution : consider an arbitrary point like $D'$ on $(I)$ and an arbitrary line passing thru it like $d$. $d$ intersects AB and AC in $X$ and $Y$. now choose $W$ on $AB$ such that $WX$=$XD'$ and $W$ is not on the same side of $A$ wrt $d$. choose $Z$ similarly on $AC$ such that $ZY$=$YD'$.now the incenter of $AWZ$ is inside or on $(I)$.
mohamad4
04.05.2012 14:15
here is my solution at the exam: let $I'$ be the incenter of the triangle $ATS$ and $P$ be the intersection point of incircle of the triangle $ABC$ and $AI_a$ that's clear that $I,I',I_a$ lies on the interior angle bisector of $A$ what is not really hard is obtaining the following equality: $TF'=DD'=E'S=b-c$ so $I_aT=I_aD=I_aS$ and from the congruent triangles $F'TI_a$ and $E'SI_a$ we get $\widehat{TI_aS}=\widehat{F'I_aE'}$ that means that $ATI_aS$ is concyclic . because of the equality$I_aT=I_aS$ we get $I_a$ lies at the middle of the Arc$TS$ from the circumcircle of the triangle $ATS$ in addition, from the well known lemma we obtain that $I_aI'=I_aT=I_aS$ and from above equalities we see: $I_aI'=I_aT=I_aS=I_aD$ in order to complete the solution it remains to show that$I_aI\geI_aI'\geI_aP$ in the triangle $IDI_a$ we have following properties: $ID+DI_a \ge II_a $ $ \Rightarrow$ $ IP+I_aD \ge II_a$ $\Rightarrow$ $I_aD \ge I_aP $ $\Rightarrow$ $I_aI' \ge I_aP$ on the other hand we have: $\widehat{IDI_a}=\widehat{IDC}+\widehat{CDI_a}=90+\widehat{CDI_a}>90$ $\Rightarrow \widehat{IDI_a}>90 \Rightarrow \widehat{IDI_a}>\widehat{DII_a} \Rightarrow I_aD<II_a \Rightarrow I_aI'<II_a$ so we conclude that $I'$ lies between $I$ and $P$ (Q.E.D)
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Nevergiveupbtw
09.05.2012 18:56
We can have a simplify solution. Let a,b,c,p be length of BC,CA,AB and p=(a+b+c)/2 We have AT+AS=p. We need (2p-TS)/(2cos(A/2))-AI<=r congruent 2BC-TS<=2rcos(A/2)=EF. But we have EF+TS>=2BC because B, C are midpoints of EF and ES.
Profhereinafter
23.06.2012 09:36
Quote: because of the equality we get lies at the middle of the Arc from the circumcircle of the triangle in addition, from the well known lemma we obtain that Please say this lemma!! I don't know.
Mahi
23.06.2012 19:59
Profhereinafter wrote: Please say this lemma!! I don't know.
If $M$ is the midpoint of arc $BC$(not containing $A$) of circumcircle of $ABC$ and $I$ is the incenter of $\triangle ABC$ then $MI=MB=MC$
Let $I'$ be the incenter of $ATS$. Then $I,I',I_a$ collinear (bisector of $\angle A$).
Now, $\angle TDS = 90^\circ +\frac {\angle A}2 = \angle TI'S$.
So $I'DTS$ concyclic.
Again, center of circle $I'TDS$ = (perpendicular bisector of $DT$) $\cap$ (perpendicular bisector of $DS$)
But if A- excircle of $ABC$ touches $AB,AC,BC$ at $X,Y,Z$ then $BXZ$ and $BTD$ are homothetic, similarly $BYZ$ and $BSD$ are homothetic.
So center of circle $I'TDS$ = $I_a$.
So $I_aD =I_aI'$.
But the smallest distance from a point to a circle is the through the line connecting the center of the circle and that point.
So $I'$ is inside the incircle of $\triangle ABC$
hofamo
13.12.2012 18:50
Ia is the incenter of ATS.its easy to know IaDTS is cyclic.we know the circumcenter of IaDTS is on bisector of BAC angle.we know incircle of ABC and circumcircle of DTSIa have one meet point(itcan be touch point)and because I is on bisector of A its easy to show Ia is in (I).(I is the incenter of ABC)
Mahdi_Mashayekhi
17.02.2022 18:16
Let $I'$ be incenter of $ATS$. Let $I_a$ be excenter of $ABC$. Let $r$ be Radius of incircle of ABC. Claim1 : $TDI'S$ is cyclic and $I_a$ is it's circumcenter. Proof : $\angle TI'S = \angle 90 + \angle A/2 = \angle 180 - \angle B/2 - \angle C/2 = \angle TDS$ so $TDI'S$ is cyclic and Note that $I_a$ lies on angle bisectors of $\angle TBC$ and $ \angle SCB$ witch are perpendicular bisectors of $TD$ and $DS$. Claim2 : $I'$ lies between $I$ and $I_a$. Proof : $\angle TI'S = \angle 90 + \angle A/2 > \angle 90$ so $I'$ and $I_a$ are on different sides of $TS$ and we know $I$ and $I'$ are on same side of $TS$. Claim3 : $II_a - I'I_a \le r$. Proof : $I'I_a = DI_a$ and $r = ID$ so we need to prove $ID + DI_a \le II_a$ witch is true because $IDI_a$ is a triangle and equality holds if $D$ is exactly $I'$. we're Done.