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


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