222. An extension of probl. 2 from IMO 2009.

by Virgil Nicula, Feb 16, 2011, 1:08 PM

IMO 2009, problem 2. Let $ ABC$ be a triangle with circumcircle $(O)$ . For $ P\in (AC)$ and $ Q\in (AB)$ consider the midpoints $K$ , $L$ , $M$ of $[BP]$ , $[CQ]$

and $ [PQ]$ respectively. Let $w$ be the circumcircle of $\triangle KLM$ . Suppose that $ PQ$ is tangent to $w$ . Prove that $ OP = OQ$ (proposed by Sergei Berlov - Russia).



Extension. Let $ ABC$ be a triangle with the circumcircle $ C(O,R)$ . For $ P\in (AC)$ and $ Q\in (AB)$ denote $M\in (PQ)$ for which $ MQ = m\cdot MP$ ,

$m>0$ . Let $\begin{array}{c}
 K\in PB\ ;\ MK\parallel AB\\\\
L\in QC\ ;\ ML\parallel AC\end{array}$ . Suppose that $ PQ$ is tangent to the circumcircle of $ \triangle KLM$ . Prove that $ \boxed {\ OQ^2 - m\cdot OP^2 = R^2(1 - m)\ }$ .


Proof. Exists $m> 0$ so $\begin{array}{ccc} \frac {MK}{QB} = \frac {PM}{PQ} = \frac {1}{m + 1} & \implies & MK = \frac {1}{m + 1}\cdot QB \\
 \\
\frac {ML}{PC} = \frac {QM}{QP} = \frac {m}{m + 1} & \implies & ML = \frac {m}{m + 1}\cdot PC\end{array}$ and $ \left\|\begin{array}{c} \widehat {KLM}\equiv\widehat {QMK}\equiv\widehat {PQA} \\
 \\
\widehat {LKM}\equiv\widehat {PML}\equiv\widehat {QPA}\end{array}\right\|\ \implies$

$ \triangle KLM\ \sim\ \triangle PQA\ \implies\ \frac {MK}{AP} = \frac {ML}{AQ}\ \implies$ $ \frac {1}{m + 1}\cdot\frac {QB}{PA} = \frac {m}{m + 1}\cdot\frac {PC}{QA}$ $ \implies$

$ QB\cdot QA = m\cdot PA\cdot PC$ $ \implies$ $ OQ^2 - R^2 = m\left(OP^2 - R^2\right)$ $ \implies$ $ OQ^2 - m\cdot OP^2 = R^2(1 - m)$ .
This post has been edited 18 times. Last edited by Virgil Nicula, Nov 22, 2015, 3:09 PM

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