IMO PROBLEM 1965

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MATHPRITOM
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IMO PROBLEM 1965

Unread post by MATHPRITOM » Fri May 06, 2011 9:42 pm

Given a triangle OAB such that ∠AOB = α < 90◦, let M be an
arbitrary point of the triangle different from O. Denote by P and Q the
feet of the perpendiculars from M to OA and OB, respectively. Let H be
the orthocenter of the triangle OPQ. Find the locus of points H when:
(a) M belongs to the segment AB;
(b) M belongs to the interior of
$\triangle OAB$.

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Moon
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Re: IMO PROBLEM 1965

Unread post by Moon » Thu May 26, 2011 7:48 pm

Copying from the IMO Compendium pdf is a real art! (I am just joking, there is nothing wrong!)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

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Tahmid Hasan
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Re: IMO PROBLEM 1965

Unread post by Tahmid Hasan » Thu May 26, 2011 9:41 pm

hehehe,i used isogonal conjugate to solve this one(my first IMO problem with this technique)
বড় ভালবাসি তোমায়,মা

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Tahmid Hasan
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Re: IMO PROBLEM 1965

Unread post by Tahmid Hasan » Sat May 28, 2011 12:28 pm

here's my solution:
(a)observe that the locus of the circumcentre of $\triangle OPQ$ is a straight line.since the isogonal conjugate of cirumcentre is the orthocentre then the locus of orthocentrre is also a straiht line.
consider the two case when $M$ conincides with $B,C$.then you will find that the orthocentre is the foot of perpendiculars from $AB,AC$.since only 1 straight line can be drawn form those two point(suppose $Q$ and $T$),
$QT$ is indeed the locus.
(b)the only trick here is to draw a line through $M$ parallel to $BC$.then the results follows from (a)
বড় ভালবাসি তোমায়,মা

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