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Interesting tangency
Mr.ARS   3
N Feb 15, 2024 by Mr.ARS
Let I - incenter of ABC. Tangent at point B of circumcenter ABC intersects AI and CI at P and Q respectively. If R - orthocenter of PIQ, prove that the circumscribed circles of AIC and PQR touch each other.
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Mr.ARS
Feb 14, 2024
Mr.ARS
Feb 15, 2024
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Mr.ARS
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Mr.ARS
#1 h  Feb 14, 2024, 11:01 AM
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Let I - incenter of ABC. Tangent at point B of circumcenter ABC intersects AI and CI at P and Q respectively. If R - orthocenter of PIQ, prove that the circumscribed circles of AIC and PQR touch each other.
https://bytepix.ru/ib/k4ihdlbulp.png
This post has been edited 3 times. Last edited by Mr.ARS, Feb 14, 2024, 11:02 AM
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MathLuis
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MathLuis
#2 h  Feb 14, 2024, 5:21 PM • 5 Y
Y by v4913, Elainedelia, OronSH, crazyeyemoody907, Mr.ARS
Solved with a hint from v4913
Let $O$ the center of $(PQR)$ and let $M$ the midpoint of minor arc $BC$ in $(ABC)$, let $(AIC)$ hit $AB,BC$ at $E,F$ respectivily, let $PQ \cap AC=T$, let $I_B$ the B-excenter of $\triangle ABC$, let $TI_B \cap (AIC)=U$, and denote $(\cdot)'$ the inverse of $(\cdot)$ in $\sqrt{ac}$ inversion.
Claim 1: $\triangle EMF \sim \triangle POQ$
Proof: Since both are isosceles we only need to prove by angle chasing that: $$\angle POQ=2\angle PRQ=2(180-\angle AIC)=2\angle AI_BC=180-\angle ABC=\angle AMC=\angle EMF$$Which is true hence the claim is done. (It is easy to see $EAFC$ is an Isosceles Trapezoid btw)
Claim 2: $QBCU, PBAU$ are cyclic
Proof: By PoP we get $TB^2=TA \cdot TC=TU \cdot TI_B$ hence by angle chase
$$\angle QBU=\angle BI_BU=\angle ICU \implies QBCU \; \text{cyclic}$$Same symetric process gives $PBAU$ cyclic, thus claim finished.
Claim 3: $Q,E,U$ and $P,F,U$ are colinear triples.
Proof: By $\sqrt{ac}$ inversion we get that $A,U',Q'$ and $C,U',P'$ are colinear, note that $U'I_B \perp AC$ follows by symetry on the perpendicular bisector of $BC$ since $(BIT'U')$ is cyclic and $BT' \parallel AC$ is also known. Since $P', Q'$ both lie in $BT'$ we have by double reim's that $BU'FQ'$ and $BU'EP'$ are cyclic and inverting back gives the result.
Finishing: Now by Claim 3 we can in fact get that $U$ is center of homothety of triangles $\triangle EMF$ and $\triangle QOP$, thus $M,O,U$ are colinear, now since $\angle QUP=2\angle EMF=2\angle QOP$ we get that $UQPR$ is cyclic, plus since $M,O,U$ are colinear we can safely say that $(PQR), (AIC)$ are tangent at $U$ as desired, thus we are done :cool:.
This post has been edited 3 times. Last edited by MathLuis, Feb 14, 2024, 5:25 PM
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OronSH
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OronSH
#3 h  Feb 14, 2024, 9:12 PM
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solved with geogebra

We solve another problem: In $\triangle ABC$ with incenter $I,$ excenter opposite $A$ as $I_A$ we let $D,E$ be points on $(ABI_A),(ACI_A)$ respectively with $AD,AE\parallel BC.$ Then let $J$ be the point on $(BICI_A)$ with $IJ\parallel BC.$
Claim. $BJD,CJE$ collinear.
Proof. $\measuredangle BJI=\measuredangle BI_AI=\measuredangle BI_AA=\measuredangle BDA$ and $IJ\parallel AD$ gives $BJD$ collinear, similarly for $CJE.$
Claim. $\measuredangle EI_AD=-\measuredangle EJD.$
Proof. $\measuredangle EI_AD=\measuredangle EI_AA+\measuredangle AI_AD=\measuredangle I_AEA+\measuredangle EAI_A+\measuredangle ABJ=\measuredangle I_ACA+\measuredangle (BC,AI)+\measuredangle ABC+\measuredangle ACI=\measuredangle I_ACI+\measuredangle BAI=90+\measuredangle BAI=\measuredangle BIC=\measuredangle BJC=-\measuredangle EJD.$
Thus $J$ lies on the reflection $\omega$ of $(EI_AD)$ over line $ADE.$ Thus the homothety at $J$ taking $BC$ to $DE$ takes $(BICI_A)$ to $\omega,$ so the two circles are tangent.

To finish, invert the diagram about $A$ and rename points.
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Mr.ARS
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Mr.ARS
#4 h  Feb 15, 2024, 8:25 PM
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MathLuis wrote:
Solved with a hint from v4913
Let $O$ the center of $(PQR)$ and let $M$ the midpoint of minor arc $BC$ in $(ABC)$, let $(AIC)$ hit $AB,BC$ at $E,F$ respectivily, let $PQ \cap AC=T$, let $I_B$ the B-excenter of $\triangle ABC$, let $TI_B \cap (AIC)=U$, and denote $(\cdot)'$ the inverse of $(\cdot)$ in $\sqrt{ac}$ inversion.
Claim 1: $\triangle EMF \sim \triangle POQ$
Proof: Since both are isosceles we only need to prove by angle chasing that: $$\angle POQ=2\angle PRQ=2(180-\angle AIC)=2\angle AI_BC=180-\angle ABC=\angle AMC=\angle EMF$$Which is true hence the claim is done. (It is easy to see $EAFC$ is an Isosceles Trapezoid btw)
Claim 2: $QBCU, PBAU$ are cyclic
Proof: By PoP we get $TB^2=TA \cdot TC=TU \cdot TI_B$ hence by angle chase
$$\angle QBU=\angle BI_BU=\angle ICU \implies QBCU \; \text{cyclic}$$Same symetric process gives $PBAU$ cyclic, thus claim finished.
Claim 3: $Q,E,U$ and $P,F,U$ are colinear triples.
Proof: By $\sqrt{ac}$ inversion we get that $A,U',Q'$ and $C,U',P'$ are colinear, note that $U'I_B \perp AC$ follows by symetry on the perpendicular bisector of $BC$ since $(BIT'U')$ is cyclic and $BT' \parallel AC$ is also known. Since $P', Q'$ both lie in $BT'$ we have by double reim's that $BU'FQ'$ and $BU'EP'$ are cyclic and inverting back gives the result.
Finishing: Now by Claim 3 we can in fact get that $U$ is center of homothety of triangles $\triangle EMF$ and $\triangle QOP$, thus $M,O,U$ are colinear, now since $\angle QUP=2\angle EMF=2\angle QOP$ we get that $UQPR$ is cyclic, plus since $M,O,U$ are colinear we can safely say that $(PQR), (AIC)$ are tangent at $U$ as desired, thus we are done :cool:.

It is wonderful! But in claim 3 you can use Pascal for UEACII_B
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