AN AMO LONGLISTED PROBLEM OF 1970

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MATHPRITOM
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AN AMO LONGLISTED PROBLEM OF 1970

Unread post by MATHPRITOM » Fri Apr 29, 2011 8:55 pm

Let,ABCD be a quadrangle.On the sides of AB,BC,CA & AD 4 squares are drawn.Let,P,Q,,R,S are the centers of the squares respectively.Prove that,PR=QS & PR is perpendicular to QS.
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*Mahi*
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Re: AN AMO LONGLISTED PROBLEM OF 1970

Unread post by *Mahi* » Thu May 05, 2011 2:11 pm

It becomes very easy if you use complex numbers ... just get the values of $P,Q,R,S$ and then it directly follows that $PR=QS$ & $PR \perp QS$.
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