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BdMO Online Forum • View topic - IGO 2016 Advanced/3

For discussing Olympiad level Geometry Problems

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.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

Thamim Zahin

Posts: 98
Joined: Wed Aug 03, 2016 5:42 pm

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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joydip

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Joined: Tue May 17, 2016 11:52 am