# INMO 2013 Question No. 1 Solution

1     Let $$(\Gamma_1)$$ and $$(\Gamma_2)$$ be two circles touching each other externally at R. Let $$(O_1)$$ and $$(O_2)$$ be the centres of $$(\Gamma_1)$$ and $$(\Gamma_2)$$, respectively. Let $$(\ell_1)$$ be a line which is tangent to $$(\Gamma_2)$$ at P and passing through $$(O_1)$$, and let $$(\ell_2)$$ be the line tangent to $$(\Gamma_1)$$ at Q and passing through $$(O_2)$$. Let $$(K=\ell_1\cap \ell_2)$$. If KP=KQ then prove that the triangle PQR is equilateral.

Discussion:

We note that $$(O_1 R O_2 )$$ is a straight line (why?)
Also $$(\Delta O_1 Q O_2 , \Delta O_1 P O_2 )$$ are right angled triangles with right angles at point Q and P respectively.
Hence $$(\Delta O_1 Q K , \Delta O_2 P K )$$ are similar (vertically opposite angles and right angles)
Thus $$( \frac{KP}{KQ} = \frac{O_2 P} {O_1 Q} = 1)$$ as KP =KQ.
Hence the radii of the two circles are equal..
This implies R is the midpoint of $$(O_1 O_2)$$ hence the midpoint of hypotenuse of  $$(\Delta O_1 Q O_2)$$
$$(O_1 R = RQ = R O_2)$$ since all are circum-radii of $$(\Delta O_1 Q O_2)$$.
Hence  $$(\Delta O_1 Q R)$$ is equilateral, similarly $$(\Delta O_2 P R)$$ is also equilateral.
Thus $$(\angle PRQ)$$ is $$(60^o)$$ also $$RQ = O_1 R = O_2 R = RP$$.
Hence triangle PQR is equilateral.