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An acute triangle $ABC$ is inscribed in a circumference of center $O$. Its heights are $AD$, $BE$ and $CF$. The line $EF$ intersects the circumference at two points, $P$ and $Q$.

(a) Prove that $OA$ is perpendicular to $PQ$.

(b) If $M$ is the midpoint of $BC$, prove that $AP^2 = 2 \cdot AD \cdot OM$

I can't get past the first part of the problem. It is sufficient to prove that $\triangle APQ$ is isosceles, which we can do by proving that $\widehat{AP} = \widehat{AQ}$. It is also sufficient to prove that $AO$ bissects $PQ$. Now the problem is in proving any of these equivalent statements.

I'm even more clueless on the second problem. I've tried assuming without proof that $AO$ is perpendicular to PQ and when on fiddling with the power of a point theorem, with no success.

I'd love a solution or at least some hints to this problem.

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  • $\begingroup$ geogebratube.org/student/m96341 if you download this document from geogebra , you can demonstrate (a) easily enough. ( I'm testing to see if you can download geogebra documents easily.) $\endgroup$
    – Alan
    Apr 5 '14 at 21:39
  • $\begingroup$ Click "on" Orthocenter , Circumcenter , and Circumcircle then you'll be able to create the construction very quickly. $\endgroup$
    – Alan
    Apr 5 '14 at 22:43
  • $\begingroup$ Actually, I can't demonstrate it easily enough. (I can't demonstrate it at all) I can see it is a $90^{\circ}$ angle very clearly, but I've tried to prove it in every way I could imagine, without success. $\endgroup$ Apr 6 '14 at 1:54
  • $\begingroup$ One thing I've noticed about the symmetric nature of this construction is that I can find three new points on the circumcircle and join them together to form a triangle. which I have no idea what to call. It certainly isn't new ! $\endgroup$
    – Alan
    Apr 6 '14 at 3:53
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Meant as hints if you read a line at a time, but otherwise a complete answer when you work through the details

1) consider the tangent at A, which is perpendicular to OA.
It suffices to show that the tangent is parallel to EF.
It suffices to show that angle AFE is equal to angle ACB.
This is well known, and can easily be shown because AFHE is a cyclic quad, so angle AFE is equal to angle AHE.

2) recall the Euler line. This shows us that 2OM=AH.
Since BFHD is a cyclic quad, hence AH x AD=AFxAB.
It remains to show that triangles AQF and ABQ are similar. This is true because they share a common angle, and that angle AFQ = angle ACB = angle AQB.

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  • $\begingroup$ Thank you so much, I finally arrived at the solution. $\endgroup$ Apr 8 '14 at 1:55
  • $\begingroup$ Can you give a reference for this problem? Do you have a book with Olympiad Problems? $\endgroup$
    – Alan
    Apr 8 '14 at 4:08
  • $\begingroup$ @Alan I do not know if this problem was taken from elsewhere. There are lots of books on Olympiad problems around. $\endgroup$
    – Calvin Lin
    Apr 9 '14 at 0:19
  • $\begingroup$ poti.impa.br/upload/… Problem #3. It's all in Portuguese, sorry about that, but there's a list of reference books at the end of the document (there's one from Titu Andreescu and another very famous one from Xu Jiagu, yay). $\endgroup$ Apr 16 '14 at 16:48

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