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# 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.

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

1. the solution to first question isn't complete...you will also hav to contradict the case where tangents are opposite