## Thứ Bảy, 10 tháng 1, 2015

### Solution for HSGS TST 2014 Round 1 Day 1

Problem. Let $ABC$ be acute triangle inscribed $(O)$. Incircle $(I)$ touches $CA,AB$ at $E,F$. $IB,IC$ cut $(O)$ again at $M,N$. Let $S,T$ be circumcenters of triangles $IFN$ and $IEM$. $P$ lies on $ST$ such that $IP\parallel BC$. $K$ is midpoint of arc $BC$ contain $A$ of $(O)$. $IK$ cuts $(O)$ again at $L$. $J$ is midpoint of $OI$. $Q$ lies on sgement $JL$ such that $PQ=PI$. Prove that $IQ$ passes through the intersection of tangent at $B,C$ of $(O)$.

Solution (By Telv Cohl).

Lemma :

Let $I, H$ be the incenter, orthocenter of $\triangle ABC$, respectively .
Let $D, E, F$ be the tangent point of $\odot (I)$ with $BC, CA, AB$, respectively .
Let $G$ be the projection of $D$ on $EF$ .

Then $DG$ is the bisector of $\angle HGI$

Proof 1 of the lemma:

Let $A'$ be the antipode of $A$ in $\odot (ABC)$ .
Let $H^*$ be the isogonal conjugate of $H$ WRT $\triangle GBC$ .

Since $(EF, GD; GB, GC)=-1$ ,
so we get $DG$ is the bisector of $\angle BGC$ .
Since $\angle H^*BG=\angle CBH=\angle BCA', \angle GCH^*=\angle HCB=\angle A'BC$ ,
so from A useful collinearity we get $G, H^*, A'$ are collinear . ... $( \star )$

From interesting problem we get $G, I, A'$ are collinear .
so combine with $(\star )$ we get $H^* \in GI$ . ie. $DG$ is the bisector of $\angle HGI$

Proof 2 of the lemma:

Let $I'$ be the reflection of $I$ in $EF$ and $Y=BH \cap EF, Z=CH \cap EF$ .

Since $(EF, GD; GB, GC)=-1$ ,
so we get $GD$ bisect $\angle BGC$ and $\triangle BFG \sim \triangle CEG$ .
Since $\angle YBG=\angle ZCG$ ,
so we get $\triangle BFG \cap Y \sim \triangle CEG \cap Z$ .
Since $\triangle HYZ \sim \triangle IEF \sim \triangle I'FE$ ,
so from $\frac{YG}{GZ}=\frac{BG}{GC}=\frac{FG}{GE}$ we get $\triangle HYZ \cap G \sim \triangle I'FE \cap G$ .
i.e $I', G, H$ are collinear

PS. From this lemma we get the following property $( \Phi )$ :

Let $O$ be the circumcenter of $\triangle ABC$ .
Let $\triangle DEF, \triangle XYZ$ be the orthic triangle, tangential triangle of $\triangle ABC$, respectively .

Then the isogonal conjugate of $O$ WRT $\triangle DEF$ is the orthocenter of $\triangle XYZ$

Back to the main problem

Let $I_a, I_b, I_c$ be three excenters of $\triangle ABC$ .
Let $I_b', I_c'$ be the reflection of $I$ in $CA, AB$, respectively .
Let $P', Q', L'$ be the reflection of $I$ in $P, Q, L$, respectively .
Let $\triangle I_a^*I_b^*I_c^*$ be the tangential triangle of $\triangle I_aI_bI_c$ .
Let $O_b, O_c$ be the circumcenter of $\triangle II_cI_a, \triangle II_aI_b$, respectively .
Let $O'$ be the image of $I$ under the inversion WRT $\odot (I_aI_bI_c)$ and $T=\odot (II_bI_b') \cap \odot (II_cI_c')$ .

Easy to see $T, Q' \in \odot (P', P'I)$ .

Invert with center $I$ which swap $\odot (ABC)$ and $\odot (I_aI_bI_c)$ .

Since $O_b, O_c$ is the image of $I_b', I_c'$, respectively ,
so from $(\Phi)$ we get the image $T'=BO_b \cap CO_c$ of $T$  is the orthocenter of $\triangle I_a^*I_b^*I_c^*$ . ... $( \star)$
Since the center $P'$ of $\odot (ITQ')$ lie on the line passing through $I$ and parallel to $I_b^*I_c^*$ ,
so combine with $(\star)$ we get the image of $\odot (ITQ')$ is $I_a^*-$ altitude of $\triangle I_a^*I_b^*I_c^*$ .

Since $O'$ is the image of $O$ under the inversion ,
so the image of $Q$ is the intersection of $I_a^*T'$ and $\odot (IO'K)$ ,
hence from $\odot (IO'K) \perp \odot (I_aI_bI_c)$ we get $I_A^*$ is the image of $Q$ .

Since $I_a^*I_b, I_a^*I_c$ are the tangent of $\odot (I_aI_bI_c)$ ,
so $\odot (IBQ'), \odot (ICQ')$ are internal tangent to $\odot (ABC)$ at $B, C$, respectively ,
hence from radical center theorem we get the tangent of $\odot (O)$ at $B, C$ and $IQ' \equiv IQ$ are concurrent .

Q.E.D

Solution (By Tran Quang Hung).

Lemma. Let $ABC$ be a triangle. Incircle $(I)$ touches $BC,CA,AB$ at $D,E,F$. $K$ is projection of $D$ on $EF$. Then $KD$ is bsector of $\angle IKH$.

Proof.  Let $L$ be reflection of $I$ through $EF$. We will prove that $L,H,K$ are collinear then $KD$ is bisector of $\angle IKH$, indeed.

$BM,CN$ are altitude of $ABC$. Easily seen $L$ is orthocenter of $AEF$. $EQ,FP$ are altitudes of $AEF$. We have $HM.HB=HN.HC$ and $LE.LQ=LF.LP$, this means $H$ and $L$ lie on radical axis of circle diameter $BE,CF$.

Let $S,T$ be projection of $B,C$ on $EF$. We have $\frac{KS}{KT}=\frac{DB}{DC}=\frac{BF}{CE}=\frac{KF}{KE}$, deduce $KF.KT=KE.KS$. Thus $K$ lie on radical axis of circle diameter $BE,CF$. Thus $L,H,K$ are collinear.

Corollary 1. Let be triangle $ABC$ inscribed circle $(O)$. $BE,CF$ are altitudes. $K,L$ are reflection of $O$ through $BE,CF$. $EK$ cuts $FL$ at $S$. Then the line passing though $S$ and perpendicular to $EF$ is concurrent with tangent at $B,C$ of $(O)$.

Using the inversion center $H$ and power $\overline{HB}.\overline{HE}=\overline{HC}.\overline{HF}$ we get Corollary 2.

Corollary 2. Let be triangle $ABC$ inscribed circle $(O)$. $AD,BE,CF$ are altitudes concurrent at $H$. Euler circle is $(N)$. $(AEF)$ cuts $(O)$ again at $G$. $Q$ is inversion of $G$ through $(N)$. $K,L$ are reflection of $H$ through $DF,DE$. $(HKB)$ intersects $(HLC)$ again at $R$. Then circles $(HQR)$ and $(HBC)$ are orthogonal.

Using the dilation center $H$ ratio $\frac{1}{2}$ we get the original problem on triangle $I_aI_bI_c$.

Let $ABC$ be acute triangle inscribed $(O)$. Incircle $(I)$ touches $CA,AB$ at $E,F$. $IB,IC$ cut $(O)$ again at $M,N$. Let $S,T$ be circumcenters of triangles $IFN$ and $IEM$. $P$ lies on $ST$ such that $IP\parallel BC$. $K$ is midpoint of arc $BC$ contain $A$ of $(O)$. $IK$ cuts $(O)$ again at $L$. $J$ is midpoint of $OI$. $Q$ lies on sgement $JL$ such that $PQ=PI$. Prove that $IQ$ passes through the intersection of tangent at $B,C$ of $(O)$.

Reference.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=619179