Let $\mathcal{C}_1(O_1)$ and $\mathcal{C}_2(O_2)$ be two circles which intersect in the points $A$ and $B$. The tangent in $A$ at $\mathcal{C}_2$ intersects the circle $\mathcal{C}_1$ in $C$, and the tangent in $A$ at $\mathcal{C}_1$ intersects $\mathcal{C}_2$ in $D$. A ray starting from $A$ and lying inside the $\angle CAD$ intersects the circles $\mathcal{C}_1$, $\mathcal{C}_2$ in the points $M$ and $N$ respectively, and the circumcircle of $\triangle ACD$ in $P$. Prove that $AM=NP$.
Problem
Source: Romanian JBMO TST 2005 - day 1, problem 1
Tags: geometry, circumcircle, geometry proposed
mecrazywong
02.04.2005 15:30
Extend CM to meet $C_2$ at E. Then it's easy to show (i)MNDE is an isosceles trapeizum, (ii) $\triangle ACM\sim\triangle DAN$. From this, we obtain $AM\cdot AN=DN\cdot CM=ME\cdot CM=AM\cdot MP$. The conclusion follows.
mathangel
22.04.2008 05:39
Note that $ \bigtriangleup ACM \sim \bigtriangleup DAN$ and $ \bigtriangleup NPD \sim \bigtriangleup ACD$. And the result follows.
FOURRIER
22.04.2008 14:21
when I draw the ray i find it out of the obtuse angle CAD, Is it obligatory that it should be lying inside CAD ?
FOURRIER
25.04.2008 20:54
It will be great if some one sketch the graph of this problem, I really want to try it
Ahiles
25.04.2008 21:16
Very nice problem with very interesting solution. Image not found
jayme
16.07.2014 14:02
Dear Mathlinkers, just a link with http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=187699 Sincerely Jean-Louis