We denote the circumcircle of a triangle $XYZ$ by $(XYZ)$.
Let $ABC$ be an acute angled triangle and $P$ be a point inside the triangle such that $\angle BPC=180^{\circ}−\angle A$.$BP,CP$ intersect $CA,AB$ at $E,F$.Circle $(AEF)$ intersects $(ABC)$ again at $G$.The circle with diameter $PG$ intersects $(ABC)$ again at $K$. $D$ is the projection of $P$ on $BC$ and $M$ is the midpoint of $BC$.Prove that the circle $(KPG)$ and circle $(KDM)$ are tangent to each other.
Tangent Circles
"Questions we can't answer are far better than answers we can't question"
- Raiyan Jamil
- Posts:138
- Joined:Fri Mar 29, 2013 3:49 pm
Re: Tangent Circles
$Hints:$
A smile is the best way to get through a tough situation, even if it's a fake smile.
- asif e elahi
- Posts:185
- Joined:Mon Aug 05, 2013 12:36 pm
- Location:Sylhet,Bangladesh
Re: Tangent Circles
Which two circles?Raiyan Jamil wrote:$Hints:$
Apply a negative inversion with center $P$ that sends $\bigodot ABC$ to itself.We denote the circumcircle of a triangle $XYZ$ by $(XYZ)$.
Let $ABC$ be an acute angled triangle and $P$ be a point inside the triangle such that $\angle BPC=180^{\circ}-\angle A$.$BP,CP$ intersect $CA,AB$ at $E,F$.Circle $(AEF)$ intersects $(ABC)$ again at $G$.The circle with diameter $PG$ intersects $(ABC)$ again at $K$. $D$ is the projection of $P$ on $BC$ and $M$ is the midpoint of $BC$.Prove that the circle $(KPG)$ and circle $(KDM)$ are tangent to each other.
Re: Tangent Circles
This problem is an application of IMO 2015/3.
"Questions we can't answer are far better than answers we can't question"
- Raiyan Jamil
- Posts:138
- Joined:Fri Mar 29, 2013 3:49 pm
Re: Tangent Circles
$KPG ,KDM$asif e elahi wrote: Which two circles?
A smile is the best way to get through a tough situation, even if it's a fake smile.