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Solution by Simone Foti, Pietro Gualdi, Luca Sartori, Enrico Zonta. Part 1: $\angle BAC=45\textdegree \Rightarrow \odot(P)$ and $(CER)$ as well as $ \odot(Q)$ and $(BFS)$ are tangent to each other. Suppose $\angle BAC=45 \textdegree$. Let: - $O_1$ be the centre of $(CER)$ - $O_2$ be the centre of $(BFS)$ - $H$ be the orthocentre of $ABC$ - $\Gamma$ be the circle $(ABC)$ - $A'$ be the antipode of $A$ on $\Gamma$ - $\odot(P)$ be the circle with center $P$ and radius $PK$ - $ \odot (Q)$ be the circle with center $Q$ and radius $ QL$ - $V$ be the reflection of $S$ on $A$ [url][/url][url][/url] Lemma 1: $BS,CR$ are the altitudes from $B,C$ in triangle $\triangle ABC$, respectively. proof: $\angle RCA=45\textdegree= 90\textdegree - \angle CAB \Rightarrow CR\perp AB$ so $CR$ is the altitude from $C$, similarly we get $BS$ is the altitude from $B$. $\square$ It follows that $R,S$ are the reflections of $H$ onto lines $AB,AC$ $ \Rightarrow AR=AH=AS$, let $\omega $ be the circle with centre $A$ and radius $AH$ (i.e. circle $(RHS)$). Lemma 2: $O_1,O_2 \in BC$ proof: $\angle RO_1E=2\cdot \angle RCE=2\cdot 45\textdegree=\angle RA'S$ so clearly $\triangle RO_1E \sim RA'S \Rightarrow \angle ERO_1=\angle SRO_1= \angle SRC \Rightarrow O_1 \in RA'$. Also $\angle A'BA=90\textdegree \Rightarrow A'B\perp AB \Rightarrow RC \parallel A'B$ then $\angle O_1CR=\angle CRO_1=\angle CRA'=\angle BA'R= \angle BCR \Rightarrow O_1 \in BC$, similarly we get $O_2 \in BC$. $\square$ Lemma 3: There is a rotation of $90\textdegree$ counterclockwise centered at $A$ that takes $R,L$ to $S,K$. proof: let $X=RL \cap SK$, note that $\angle RBS=90\textdegree$ and $B$ is the midpoint of $LS$ so $RB$ is the perpendicular bisector of $LS$, hence $\angle LRB=\angle SRB$, similarly we get $\angle RSC=\angle SCR$. Now $$\angle RXS=180 \textdegree -\angle XRS- \angle XSR= 180\textdegree- (180\textdegree- 2\cdot \angle SRB)-(180\textdegree- 2\cdot \angle RSC)=$$$$=2\cdot (\angle SRB +\angle RSC-90\textdegree)= 90\textdegree=\angle RAS \Rightarrow X\in \Gamma$$Now note that $LR=RS=SK$. It is well known that centre of the spiral similarity that rakes $R,L$ to $S,K$ is centered in the midpoint of arc $RS$ of $(XSR)$ containing $A$, which is $A$ indeed. In particular this is a rotation of $90\textdegree$ counterclockwise centered in $A$ since $AR=AS$ and $\angle RAS= 90\textdegree$. $\square$ Lemma 4: let $Q'$ be the image of $Q$ under said rotation, $Q'\in RA', SC, (CER)$ proof: the rotation takes line $QR$ to a line throw $S$ perpendicular to $QR$, which is line $SC$, so $Q'\in SC$. Now $\angle RQ'C= \angle CQ'K=\angle SQ'K=\angle RQL=\angle CBA=180\textdegree \angle REC$ using the fact that $CS$ is the perpendicular bisector of $RK$ and $\triangle SQ'K\cong \triangle RQL$ by rotation, so we proved $Q' \in (CER)$. Finally note that $\angle CRQ'=90\textdegree- \angle RQ'C=90\textdegree- \angle CBA= \angle CAA'=\angle CRA' \Rightarrow R,A',Q'$ are collinear as wanted. $\square$ Lemma 5: $A,Q',O_2$ are collinear. proof: let $W=A'B\cap SC$ and $Z$ be the point on $SC$ such that $ZO_2\parallel AA'$ and call $\angle SRC= \theta$. First note that $BA'CS$ is an isosceles trapezoid so $WO_2$ is the perpendicular bisector of $BS$, hence $O,O_2,W$ are collinear. $O$ is the centre of spiral similarity (rotation) that takes $B,C$ to $A',S$ so $OSCO_2$ is ciclic. Also $ZO_2\perp RS\Rightarrow \angle SZO_2= 90\textdegree - \angle CSR=\angle SRC=\angle SA'C=\angle O_2A'C \Rightarrow ZA'O_2C$ is ciclic. Now $\angle SA'Z= \angle SA'C + \angle CA'Z= \theta +\angle CO_2Z= \theta + \angle SO_2Z- \angle SO_2C=\theta + 135\textdegree -\angle SOC= \theta + 135\textdegree- 2\cdot \theta= 135\textdegree-\theta= 45\textdegree + (90\textdegree- \theta)= \angle SRQ'+ \angle Q'SR=\angle RQ'Z$ which implies that $ZA'$ is tangent to $Q'A'S$. Putting this together with the fact that $\angle SA'Q'=90\textdegree$ and $=A'W\perp SC$ (by rotation in $O$ that takes $B,A'$ to $C,S$) we get that $Z$ is the image of $W$ after inversion wrt $(SA'Q')$, hence $(S,Q';W,Z)=-1$. If $\bar{A}=Q'O_2 \cap AA'$ and $P_{\infty}$ is the point at infinity of line $AA'$, by projecting from $O_2$ to line $AA'$ we get $(A,A';O,P_{\infty})-1=(S,Q';W,Z)\overset{\underset{\mathrm{O_2}}{}}{=}(A',\bar{A};O,P_{\infty})=$ which means that $A \equiv \bar{A}$ by injectivity of cross ratio $\Rightarrow A,O_2,Q'$ are collinear as wanted. $\square$ * Alternatively this lemma could have been proved with complex numbers by setting $A=i, S=1,A'=-i, R=-1, B=b, C=ib$ and computing $O_2,Q'$ with the chord intersection formula $ab\cap cd=\frac{ab(c+d)-cd(a+b)}{ab-cd}$ Let $T=\omega \cap(BFS)$ and $T'= \omega \cap \odot(Q)$. Note that $\angle ASF=45 \textdegree =\angle ABS \Rightarrow \triangle ASF\sim\triangle ABS \Rightarrow AF\cdot AB=AS^2=AT^2 \Rightarrow AT,AS$ are tangent to $(BFS)$, so $AO_2$ is the internal angle bisector of $\angle SAT$, but note that $\angle O_2AQ=\angle Q'AQ=90 \textdegree $ so $AQ$ is the external angle bisector of $\angle SAT$ which is the internal angle bisector of $\angle VAT$. On the other hand note that $\triangle VQL\cong \triangle RQ'K$ for rotation ($V$ is mapped to $R$ since $AV=AS=AR$ and $\angle VAR=90\textdegree$). Recall $Q'$ is on the perpendicular bisector of $RK$ so $Q'K=Q'R \Rightarrow QL=QV \Rightarrow V\in \odot (Q)$, also note that $\angle AVQ=90\textdegree \Rightarrow AV$ is tangent to $\odot (Q)$. Now note that, since $AT', AV$ are the tangents from $A$ to $\odot(Q)$ we have that $AQ$ is the internal angle bisector of $\angle VAT'$. It follows that $\angle VAT=\angle VAT'$, but $T,T'\in \omega \Rightarrow AT=AT'$ so $T\equiv T'\Rightarrow (BFS)$ and $\odot(Q)$ are tangent to each other at $T$ as we wanted. Similarly we can prove that $\odot(P)$ and $(CER)$ are tangent as well. Part 2: $\odot(P)$ and $(CER)$ are tangent $\Rightarrow \angle BAC=45\textdegree $} First note that $\odot(P)$ and $(CER)$ are tangent $\Longleftrightarrow PO_1=PK+O_1C$ \quad (1) We will fix points $A,C$ and the circumference $\Gamma$ and let $B$ move on its arc $AA'$ non containing $C$. We wish to prove that there is a unique $B$ for which (1) is true, namely when $BAC=45\textdegree$ Note that points $R,S,A',K,O_1$ are fixed as well as $(CER)$. $\angle SPK=\angle BPK=90\textdegree- \angle SCD$ which is fixed, so $P$ moves on an arc of a circumference passing through $S,K$, call this arc $\gamma$ We define the function $f: \mathbb{R}^2\rightarrow \mathbb{R}$ such that for any point $X$ in the plane $f(X)=XO_1-XK$. Note that if we prove that $f$ is monotone as $X$ moves on $\gamma$ then there must be a unique point on $P\in \gamma$ such that $PO_1-PK=O_1C$ which is what we want. Suppose au contraire that $f$ is not monotone on $\gamma$, then there must be a point $P\in \gamma$ s.t. $f'(P)=0$, but that means that the hyperbola with foci $O_1,K$ passing through $P$ and $\gamma$ are tangent at $P$. It is well known that the tangent line to the hyperbola at $P$ is the internal angle bisector of $\angle KPO_1$, so the tangent line to $\gamma $ in $P$ must be the internal angle bisector of $\angle KPO_1$, call $l$ the reflecion of $PK$ on the tangent line at $P$ to $\gamma$ . Call angle $\angle BSA=\psi$, as $B$ varies on arc $A'A$ of $\Gamma$ we have $0 \textdegree\leq \psi\leq 90 $ Note that $\angle KPS=90\textdegree - \angle SAC=\angle CSR$ $\angle PSK=180\textdegree- \angle KSB=180\textdegree- 2\cdot \angle CSR+ \angle BSR=180\textdegree-2\cdot \angle CSR+\psi-45\textdegree$ $\angle SKP=180\textdegree- \angle KPS-\angle PSK=180\textdegree -\angle CSR-180\textdegree+2\cdot \angle CSR -\psi+45\textdegree=\angle CSR-\psi+45\textdegree $. $\angle (l,PS)=2\cdot \angle SKP+ \angle KPS= 2\cdot CSR- 2\psi+90\textdegree+\angle CSR=3\cdot \angle CSR-2\cdot \psi+90\textdegree\geq \psi-45\textdegree= \angle BSR$ since $\angle CSR\geq 45\textdegree$ and $\psi\leq 90\textdegree$. So $\angle (l,PS)\geq \angle BSR$, hence $l$ intersects line $RS$ strictly outside segment $RS$ so it cannot pass through $O_1$ and we are done. $\square$

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pietro.gualdi04 wrote: Solution by Simone Foti, Pietro Gualdi, Luca Sartori, Enrico Zonta. Part 1: $\angle BAC=45\textdegree \Rightarrow \odot(P)$ and $(CER)$ as well as $ \odot(Q)$ and $(BFS)$ are tangent to each other. Suppose $\angle BAC=45 \textdegree$. Let: - $O_1$ be the centre of $(CER)$ - $O_2$ be the centre of $(BFS)$ - $H$ be the orthocentre of $ABC$ - $\Gamma$ be the circle $(ABC)$ - $A'$ be the antipode of $A$ on $\Gamma$ - $\odot(P)$ be the circle with center $P$ and radius $PK$ - $ \odot (Q)$ be the circle with center $Q$ and radius $ QL$ - $V$ be the reflection of $S$ on $A$ [url][/url][url][/url] Lemma 1: $BS,CR$ are the altitudes from $B,C$ in triangle $\triangle ABC$, respectively. proof: $\angle RCA=45\textdegree= 90\textdegree - \angle CAB \Rightarrow CR\perp AB$ so $CR$ is the altitude from $C$, similarly we get $BS$ is the altitude from $B$. $\square$ It follows that $R,S$ are the reflections of $H$ onto lines $AB,AC$ $ \Rightarrow AR=AH=AS$, let $\omega $ be the circle with centre $A$ and radius $AH$ (i.e. circle $(RHS)$). Lemma 2: $O_1,O_2 \in BC$ proof: $\angle RO_1E=2\cdot \angle RCE=2\cdot 45\textdegree=\angle RA'S$ so clearly $\triangle RO_1E \sim RA'S \Rightarrow \angle ERO_1=\angle SRO_1= \angle SRC \Rightarrow O_1 \in RA'$. Also $\angle A'BA=90\textdegree \Rightarrow A'B\perp AB \Rightarrow RC \parallel A'B$ then $\angle O_1CR=\angle CRO_1=\angle CRA'=\angle BA'R= \angle BCR \Rightarrow O_1 \in BC$, similarly we get $O_2 \in BC$. $\square$ Lemma 3: There is a rotation of $90\textdegree$ counterclockwise centered at $A$ that takes $R,L$ to $S,K$. proof: let $X=RL \cap SK$, note that $\angle RBS=90\textdegree$ and $B$ is the midpoint of $LS$ so $RB$ is the perpendicular bisector of $LS$, hence $\angle LRB=\angle SRB$, similarly we get $\angle RSC=\angle SCR$. Now $$\angle RXS=180 \textdegree -\angle XRS- \angle XSR= 180\textdegree- (180\textdegree- 2\cdot \angle SRB)-(180\textdegree- 2\cdot \angle RSC)=$$$$=2\cdot (\angle SRB +\angle RSC-90\textdegree)= 90\textdegree=\angle RAS \Rightarrow X\in \Gamma$$Now note that $LR=RS=SK$. It is well known that centre of the spiral similarity that rakes $R,L$ to $S,K$ is centered in the midpoint of arc $RS$ of $(XSR)$ containing $A$, which is $A$ indeed. In particular this is a rotation of $90\textdegree$ counterclockwise centered in $A$ since $AR=AS$ and $\angle RAS= 90\textdegree$. $\square$ Lemma 4: let $Q'$ be the image of $Q$ under said rotation, $Q'\in RA', SC, (CER)$ proof: the rotation takes line $QR$ to a line throw $S$ perpendicular to $QR$, which is line $SC$, so $Q'\in SC$. Now $\angle RQ'C= \angle CQ'K=\angle SQ'K=\angle RQL=\angle CBA=180\textdegree \angle REC$ using the fact that $CS$ is the perpendicular bisector of $RK$ and $\triangle SQ'K\cong \triangle RQL$ by rotation, so we proved $Q' \in (CER)$. Finally note that $\angle CRQ'=90\textdegree- \angle RQ'C=90\textdegree- \angle CBA= \angle CAA'=\angle CRA' \Rightarrow R,A',Q'$ are collinear as wanted. $\square$ Lemma 5: $A,Q',O_2$ are collinear. proof: let $W=A'B\cap SC$ and $Z$ be the point on $SC$ such that $ZO_2\parallel AA'$ and call $\angle SRC= \theta$. First note that $BA'CS$ is an isosceles trapezoid so $WO_2$ is the perpendicular bisector of $BS$, hence $O,O_2,W$ are collinear. $O$ is the centre of spiral similarity (rotation) that takes $B,C$ to $A',S$ so $OSCO_2$ is ciclic. Also $ZO_2\perp RS\Rightarrow \angle SZO_2= 90\textdegree - \angle CSR=\angle SRC=\angle SA'C=\angle O_2A'C \Rightarrow ZA'O_2C$ is ciclic. Now $\angle SA'Z= \angle SA'C + \angle CA'Z= \theta +\angle CO_2Z= \theta + \angle SO_2Z- \angle SO_2C=\theta + 135\textdegree -\angle SOC= \theta + 135\textdegree- 2\cdot \theta= 135\textdegree-\theta= 45\textdegree + (90\textdegree- \theta)= \angle SRQ'+ \angle Q'SR=\angle RQ'Z$ which implies that $ZA'$ is tangent to $Q'A'S$. Putting this together with the fact that $\angle SA'Q'=90\textdegree$ and $=A'W\perp SC$ (by rotation in $O$ that takes $B,A'$ to $C,S$) we get that $Z$ is the image of $W$ after inversion wrt $(SA'Q')$, hence $(S,Q';W,Z)=-1$. If $\bar{A}=Q'O_2 \cap AA'$ and $P_{\infty}$ is the point at infinity of line $AA'$, by projecting from $O_2$ to line $AA'$ we get $(A,A';O,P_{\infty})-1=(S,Q';W,Z)\overset{\underset{\mathrm{O_2}}{}}{=}(A',\bar{A};O,P_{\infty})=$ which means that $A \equiv \bar{A}$ by injectivity of cross ratio $\Rightarrow A,O_2,Q'$ are collinear as wanted. $\square$ * Alternatively this lemma could have been proved with complex numbers by setting $A=i, S=1,A'=-i, R=-1, B=b, C=ib$ and computing $O_2,Q'$ with the chord intersection formula $ab\cap cd=\frac{ab(c+d)-cd(a+b)}{ab-cd}$ Let $T=\omega \cap(BFS)$ and $T'= \omega \cap \odot(Q)$. Note that $\angle ASF=45 \textdegree =\angle ABS \Rightarrow \triangle ASF\sim\triangle ABS \Rightarrow AF\cdot AB=AS^2=AT^2 \Rightarrow AT,AS$ are tangent to $(BFS)$, so $AO_2$ is the internal angle bisector of $\angle SAT$, but note that $\angle O_2AQ=\angle Q'AQ=90 \textdegree $ so $AQ$ is the external angle bisector of $\angle SAT$ which is the internal angle bisector of $\angle VAT$. On the other hand note that $\triangle VQL\cong \triangle RQ'K$ for rotation ($V$ is mapped to $R$ since $AV=AS=AR$ and $\angle VAR=90\textdegree$). Recall $Q'$ is on the perpendicular bisector of $RK$ so $Q'K=Q'R \Rightarrow QL=QV \Rightarrow V\in \odot (Q)$, also note that $\angle AVQ=90\textdegree \Rightarrow AV$ is tangent to $\odot (Q)$. Now note that, since $AT', AV$ are the tangents from $A$ to $\odot(Q)$ we have that $AQ$ is the internal angle bisector of $\angle VAT'$. It follows that $\angle VAT=\angle VAT'$, but $T,T'\in \omega \Rightarrow AT=AT'$ so $T\equiv T'\Rightarrow (BFS)$ and $\odot(Q)$ are tangent to each other at $T$ as we wanted. Similarly we can prove that $\odot(P)$ and $(CER)$ are tangent as well. Part 2: $\odot(P)$ and $(CER)$ are tangent $\Rightarrow \angle BAC=45\textdegree $} First note that $\odot(P)$ and $(CER)$ are tangent $\Longleftrightarrow PO_1=PK+O_1C$ \quad (1) We will fix points $A,C$ and the circumference $\Gamma$ and let $B$ move on its arc $AA'$ non containing $C$. We wish to prove that there is a unique $B$ for which (1) is true, namely when $BAC=45\textdegree$ Note that points $R,S,A',K,O_1$ are fixed as well as $(CER)$. $\angle SPK=\angle BPK=90\textdegree- \angle SCD$ which is fixed, so $P$ moves on an arc of a circumference passing through $S,K$, call this arc $\gamma$ We define the function $f: \mathbb{R}^2\rightarrow \mathbb{R}$ such that for any point $X$ in the plane $f(X)=XO_1-XK$. Note that if we prove that $f$ is monotone as $X$ moves on $\gamma$ then there must be a unique point on $P\in \gamma$ such that $PO_1-PK=O_1C$ which is what we want. Suppose au contraire that $f$ is not monotone on $\gamma$, then there must be a point $P\in \gamma$ s.t. $f'(P)=0$, but that means that the hyperbola with foci $O_1,K$ passing through $P$ and $\gamma$ are tangent at $P$. It is well known that the tangent line to the hyperbola at $P$ is the internal angle bisector of $\angle KPO_1$, so the tangent line to $\gamma $ in $P$ must be the internal angle bisector of $\angle KPO_1$, call $l$ the reflecion of $PK$ on the tangent line at $P$ to $\gamma$ . Call angle $\angle BSA=\psi$, as $B$ varies on arc $A'A$ of $\Gamma$ we have $0 \textdegree\leq \psi\leq 90 $ Note that $\angle KPS=90\textdegree - \angle SAC=\angle CSR$ $\angle PSK=180\textdegree- \angle KSB=180\textdegree- 2\cdot \angle CSR+ \angle BSR=180\textdegree-2\cdot \angle CSR+\psi-45\textdegree$ $\angle SKP=180\textdegree- \angle KPS-\angle PSK=180\textdegree -\angle CSR-180\textdegree+2\cdot \angle CSR -\psi+45\textdegree=\angle CSR-\psi+45\textdegree $. $\angle (l,PS)=2\cdot \angle SKP+ \angle KPS= 2\cdot CSR- 2\psi+90\textdegree+\angle CSR=3\cdot \angle CSR-2\cdot \psi+90\textdegree\geq \psi-45\textdegree= \angle BSR$ since $\angle CSR\geq 45\textdegree$ and $\psi\leq 90\textdegree$. So $\angle (l,PS)\geq \angle BSR$, hence $l$ intersects line $RS$ strictly outside segment $RS$ so it cannot pass through $O_1$ and we are done. $\square$ part 2 can be proved in an easier way by showing that $|SK-SO_1|=|SR-SO_1|<RO_1=CO_1$,which means $S$ is outside the hyperbola,that gives that there's a unique $B$,as the intersection points are on different sides of $SK$

Note that since $AO$ and $A-$altitude are isogonal, it follows that $AO\perp EF$. Now set the problem on the complex plane with $a=i, r=1, s=-1$ and the circle as the unit circle with center $o=0$. Then, $e,f$ are real and $e+ac\overline e=a+c$ but $e=\overline e$, which means $e=\frac{a+c}{ac+1}$, and $f=\frac{a+b}{ab+1}$. Also, $k=2c-r$ and $p+bs\overline p=b+s, \frac{p-k}{b-c}=\frac{(\overline p-\overline k)bc}{b-c}$. This gives $p=\frac{bc+sc+sk-bcs\overline k}{s+c}$. Let $O_1$ denote the circumcenter of $RCE$. Then, $\angle ECR = \angle ACR=\angle ECR = \frac12\angle AOR = 45^{\circ}$, which means $EO_1\perp O_1R$. So, $EO_1R$ is an isosceles right-angled triangle. This gives $o_1=\frac{e+r+(e-r)i}{2}$. We have radius of circle $REC$ is $|r-e|/\sqrt{2}$, while length $PK$ is $|\frac{2b-2c^2}{c-1}|$ when simplified, while $p=\frac{2b+1-3c}{c-1}$ when simplified. Length $PO_1=PK+|r-e|/\sqrt{2}$ gives a symmetric relation, the details are left as an exercise (not that long).