INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

February 5, 2013

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.


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.

One comment on “INMO 2013 Question No. 1 Solution”

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.