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Zeus93

#1 h May 19, 2011, 1:56 PM • 4 Y

Y by doxuanlong15052000, baladin, Adventure10, and 1 other user

$A, B \in$ circle $(S)$ , C is inside $(S)$ . Circle $(S')$ is tangent to $AC, BC$ and $(S)$ at $P, Q, R$ respectively. Prove that the circumcircle of triangle $APR$ is go through the incenter of triangle $ABC$

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#2 h May 21, 2011, 6:16 AM • 8 Y

Y by buratinogigle, Mengsay, esi, doxuanlong15052000, baladin, Adventure10, and 2 other users

$CB,CA$ cut $(S)$ again at $D,E$ and $I,J$ are the incenters of $\triangle ABC,\triangle ABD.$ By Sawayama's lemma, we have that $PQ$ passes through $J.$ Since $B,I,J$ are collinear on the angle bisector of $\angle CBA,$ we obtain

$\angle CPJ =90^{\circ}-\frac{_1}{^2}\angle ACB=\angle AIJ$ $\Longrightarrow$ $P,A,I,J$ are concyclic $(\star)$

Let $RA,RE$ cut $(S')$ again at $A',E'.$ Since $R$ is the exsimilicenter of $(S) \sim (S'),$ then $AE \parallel A'E'$ $\Longrightarrow$ Arcs $PA',PE'$ of $(S')$ are congruent $\Longrightarrow$ $RP$ bisects $\angle ARE.$ On the other hand, angle chase gives

$\angle PJA=\angle CPQ-\angle CAJ=90^{\circ}-\frac{_1}{^2}\angle ACB-\angle CAJ \Longrightarrow$

$2\angle PJA=180^{\circ}-\angle ACB-2(\angle EAB-\frac{_1}{^2}\angle DAB)$

$2\angle PJA=\angle EAB+\angle DBA-2\angle EAB+\angle DAB=\angle DBA+\angle DAE=\angle ARE$

Since $RP$ bisects $\angle ARE,$ then $2\angle PJA=2\angle PRA$ $\Longrightarrow$ $P,J,R,A$ are concyclic. Together with $(\star),$ we conclude that $P,A,R,I$ are concyclic, as desired. In addition, this result yields that $RI$ bisects $\angle ARB,$ which is another celebrated property.

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jayme

#3 h May 21, 2011, 6:44 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
this nice problem is used as a lemma to prove the Protassov's result.
For a synthetic proof see
http://perso.orange.fr/jl.ayme vol. 2 Un remarquable résultat de Vladimir Protassov p. 2
Sincerely
Jean-Louis

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#4 h Mar 10, 2014, 2:55 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

By Sawayama's lemma, the incenter $I_D$ of $\triangle ABD$ lies on $(ARP)$ and on $PQ \implies \angle ARI_D = \angle CPQ = \angle PRQ$ . Therefore, $2 \cdot \angle ARI_D \pm \angle PRI_D = \angle ARQ = \angle ARD + \angle DRQ = \angle ABD + \tfrac{1}{2} \angle BAD.$ But $\pm \angle PRI_D = \pm \angle PAI_D = \tfrac{1}{2} \angle BAD - \angle CAB$ , i.e. $\angle ARI_D = \tfrac{1}{2} \angle ABD + \tfrac{1}{2} \angle CAB \implies \angle ARI_D = \angle AII_D$ and $I \in (APRI_D)$ .

Another interesting result is that, if $T = AD \cap BE$ , then the incenter of $ABT$ lies on $IX$ , the radical axis of the two circles, for $AI_D I_E B$ is cyclic.

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mathuz

#5 h Mar 10, 2014, 6:35 AM • 1 Y

Y by Adventure10

curvilinear circles!
$(S')$ is curvilinear circle of the triangles $ABD$ , $ABE$ .

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jayme

#6 h Jun 13, 2015, 6:12 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,

You can see : http://www.artofproblemsolving.com/community/c6h407366 p. 80

Sincerely
Jean-Louis

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