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.