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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.


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.

February 5, 2013

1 comment

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

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