For discussing Olympiad level Geometry Problems
Thamim Zahin
Posts: 98
Joined: Wed Aug 03, 2016 5:42 pm

3. Let $P$ be the intersection point of sides $AD$ and $BC$ of a convex qualrilateral $ABCD$. Suppose that $I_1$ and $I_2$ are the incenters of triangles $PAB$ and $PDC$, respectively. Let $O$ be the circumcenter of $PAB$, and $H$ the orthocenter of $PDC$. Show that the circumcircles of triangles $AI_1B$ and $DHC$ are tangent together if and only if the circumcircles of triangles $AOB$ and $DI_2C$ are tangent together.
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joydip
Posts: 47
Joined: Tue May 17, 2016 11:52 am

A generalization of the problem :

Let $P$ be the intersection point of sides $AD$ and $BC$ of a convex qualrilateral $ABCD$. Suppose that $I_1$ and $I_2$ are the incenters of $\triangle PAB$ & $\triangle PDC$, respectively. Let $O$ be the circumcenter of $\triangle PAB$, and $H$ the orthocenter of $\triangle PDC$. Show that the circumcircles of $\triangle AI_1B$ and $\triangle DHC$ intersects at an angle $\theta$ if and only if the circumcircles of $\triangle AOB$ and $\triangle DI_2C$ intersects at an angle $\theta$.

Proof:
Lemma :
Let $\alpha$ & $\beta$ be two circles meeting at point $A , B$. If $C \in \alpha$ & $D \in \beta$, then the angle between $\alpha$ & $\beta$ is equal to the angle between $(ADC)$ & $(BDC)$.

proof :Let an inversion with center $D$ and an arbitary radius take $A,B,C$ to $A',B',C'$ respectively .Then the angle between $A'C'$ & $B'C'$ is equal to the angle between $A'B'$ & $(A'B'C')$ by alternate segment theorem ,completing the proof .

Let $(AI_1B),(DI_2C),(DHC)$ meet $BC$ at $M,N,S$ respectively .Let $(AI_2B) \cap (DHC)={ U,V }$.
Let $(BVC)$ & $(AVD)$ meet again at $X$.
$\angle DXC= \angle VAD +\angle VBC=\angle VAD + \angle MAV=\angle MAD=\angle DNC$ So, $X \in (DI_2C)$

$\angle AXB= \angle VXB + \angle AXV= \angle VCB + \angle VDA=\angle SDV + \angle VDA= \angle SDA=2\angle APB=\angle AOB$. So, $X \in (AOB)$
Let $(BUC)$ & $(AUD)$ meet again at $Y$.Then Similerly we get $(DI_2C) \cap (AOB) ={X,Y}$
So, by our lemma angle between $(AUV)$ & $(DUV)$ , angle between $(AUD)$ & $(AVD)$,and angle between $(AXY)$ & $(DXY)$ are same, completing the main proof.

(the original problem can be solved more easily, only by considering $X$ & $Y$)
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