APMO 2021 P3  Geometry
 Anindya Biswas
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APMO 2021 P3  Geometry
Let $ABCD$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $AC$ and $BD$, let $L$ be the center of the circle tangent to sides $AB, BC$ and $CD$, and let $M$ be the midpoint of the arc $BC$ of $\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $BCE$ opposite $E$ lies on the line $LM$.
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 Joined: Mon Dec 21, 2020 9:26 pm
Re: APMO 2021 P3  Geometry
My Solution
Firstly note that $ L$ is the incentre of $\triangle TAC ; T= AB \cap CD $
Let $ I_a ,I_d$ be the incentres of $ \triangle BAC ,\triangle DBC$ and let $ E_a ,E_d$ be the excentres of $ \triangle ABC ,\triangle DBC$ and $ P$ be the excentre of $\triangle EBC$.
Notice $ \angle BI_aC= 90+\dfrac{\angle BAC}{2}=90+\dfrac{\angle BDC}{2}=\angle BI_dC$
So $B,I_a,I_d,C$ is cyclic then by incentre excentre lemma $ B,I_a,I_d,C,E_a,E_d$ lies on the same circle .
Also by incentre excentre lemma, $D,I_d,M,E_d$ and $ A,I_a,M,E_a$ are collinear .
Also $L=CI_a \cap BI_d$ and $ CE_d$ is the external angle bisector of $\angle ACB=\angle ECB$ which implies $C,E_d,K$ are collinear .similarly $ B,E_a,K$ are collinear .therefore by Pascal's theorem on hexagon$ I_dE_dCI_aE_aB \Longrightarrow I_dE_d \cap I_aE_a, E_dC \cap E_aB ,I_aC \cap BI_d =M,K,L$ are collinear .
Firstly note that $ L$ is the incentre of $\triangle TAC ; T= AB \cap CD $
Let $ I_a ,I_d$ be the incentres of $ \triangle BAC ,\triangle DBC$ and let $ E_a ,E_d$ be the excentres of $ \triangle ABC ,\triangle DBC$ and $ P$ be the excentre of $\triangle EBC$.
Notice $ \angle BI_aC= 90+\dfrac{\angle BAC}{2}=90+\dfrac{\angle BDC}{2}=\angle BI_dC$
So $B,I_a,I_d,C$ is cyclic then by incentre excentre lemma $ B,I_a,I_d,C,E_a,E_d$ lies on the same circle .
Also by incentre excentre lemma, $D,I_d,M,E_d$ and $ A,I_a,M,E_a$ are collinear .
Also $L=CI_a \cap BI_d$ and $ CE_d$ is the external angle bisector of $\angle ACB=\angle ECB$ which implies $C,E_d,K$ are collinear .similarly $ B,E_a,K$ are collinear .therefore by Pascal's theorem on hexagon$ I_dE_dCI_aE_aB \Longrightarrow I_dE_d \cap I_aE_a, E_dC \cap E_aB ,I_aC \cap BI_d =M,K,L$ are collinear .

 Posts: 16
 Joined: Mon Dec 21, 2020 9:26 pm
Re: APMO 2021 P3  Geometry
Here is a another cool sol by an aops user
Let $I_1$ and $I_2$ be the incircle of $\triangle ABC$ and $\triangle DBC$(Note that $L$ must be the intersection of line $BI_1$ and $CI_2$), $F$ the excircle of $EBC$, and $X = AI_1 \cap CF$, $Y = DI_2 \cap BF$.
Claim: $I_1I_2 // XY$
Proof. By incenter  excenter lemma, $AI_1$ and $DI_2$ insterect at $\Gamma$ at $M$. Moreover, since $BF$ and $CF$ are both external bisectors of $\angle DBC$ and $\angle ACB$, $MX = MI_2 = MI_1 = MB = MC = MY$ which means that $I_1 I_2 C X Y B$ is cylic with $M$ as its center. It is easy to see $I_1 I_2 // XY$ because $I_1X$ and $I_2Y$ are both diameters of $(I_1 I_2 C X Y B)$. $\blacksquare$
Now, we let $U = BI_1 \cap CF$ and $V = CI_2 \cap BF$.
Claim: $BCUV$ is cyclic
Proof. Easy angle chasing gives $\angle UBV = \angle UCV$. $\blacksquare$
Claim: $BCUV$ is cyclic
Proof. Easy angle chasing gives $\angle UBV = \angle UCV$. $\blacksquare$
Claim: $\triangle FXY$ and $\triangle LI_1I_2$ are perspective
Proof. Since, $BCXY$ and $BCUV$ are cylic. $\angle FXY = \angle FBC = \angle FUV$. So, $UV // XY // I_1I_2$. Notice that $U = FX \cap LI_1$, $V = FY \cap LI_2$, and $P_{\infty}$ (on line $XY$) $= XY \cap I_1I_2$ are collinear. So, $\triangle FXY$ and $\triangle LI_1I_2$ are perspective. $\blacksquare$
By the third claim, we get that $LF$, $XI_1$, and $YI_2$ are concurrent at $M$ (since $M$ = $XI_1 \cap YI_2$). Hence, $L, M$, and $F$ are collinear as needed
Let $I_1$ and $I_2$ be the incircle of $\triangle ABC$ and $\triangle DBC$(Note that $L$ must be the intersection of line $BI_1$ and $CI_2$), $F$ the excircle of $EBC$, and $X = AI_1 \cap CF$, $Y = DI_2 \cap BF$.
Claim: $I_1I_2 // XY$
Proof. By incenter  excenter lemma, $AI_1$ and $DI_2$ insterect at $\Gamma$ at $M$. Moreover, since $BF$ and $CF$ are both external bisectors of $\angle DBC$ and $\angle ACB$, $MX = MI_2 = MI_1 = MB = MC = MY$ which means that $I_1 I_2 C X Y B$ is cylic with $M$ as its center. It is easy to see $I_1 I_2 // XY$ because $I_1X$ and $I_2Y$ are both diameters of $(I_1 I_2 C X Y B)$. $\blacksquare$
Now, we let $U = BI_1 \cap CF$ and $V = CI_2 \cap BF$.
Claim: $BCUV$ is cyclic
Proof. Easy angle chasing gives $\angle UBV = \angle UCV$. $\blacksquare$
Claim: $BCUV$ is cyclic
Proof. Easy angle chasing gives $\angle UBV = \angle UCV$. $\blacksquare$
Claim: $\triangle FXY$ and $\triangle LI_1I_2$ are perspective
Proof. Since, $BCXY$ and $BCUV$ are cylic. $\angle FXY = \angle FBC = \angle FUV$. So, $UV // XY // I_1I_2$. Notice that $U = FX \cap LI_1$, $V = FY \cap LI_2$, and $P_{\infty}$ (on line $XY$) $= XY \cap I_1I_2$ are collinear. So, $\triangle FXY$ and $\triangle LI_1I_2$ are perspective. $\blacksquare$
By the third claim, we get that $LF$, $XI_1$, and $YI_2$ are concurrent at $M$ (since $M$ = $XI_1 \cap YI_2$). Hence, $L, M$, and $F$ are collinear as needed