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This geometrical problem was proposed in a Mathematics Contest for high school students of my country. It is truly hard to find its solution.


Let $ABC$ be an acute triangle inscribed in the circle with its center $O$. The line which is perpendicular to $AO$ at $O$ intersects $AB$ and $AC$ at $E$ and $F$ respectively.

Let $D$ be the intersection point of $BF$ and $CE$. The circumscribed circle of triangle $BDC$ intersects $AB$ and $AC$ at $M$ and $N$ respectively and the circumscribed circle of triangle $DEF$ intersects $AB$ and $AC$ at $P$ and $Q$ respectively.

Let $S$ be the intersection point of $BC$ and $EF$, and $K$ be the intersection point of $PN$ and $MQ$.

Prove that $AK\perp SD$.

I am happy if someone could propose some fresh ideas to attack this problem.

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Let $(WXYZ)$ denote the circumcircle of any cyclic quadrilateral $WXYZ$.

$EFCB$ is cyclic as $\angle OEA=\frac{\pi}{2}-\angle BAO=\angle ACB.$ Therefore, $PQ\|BC $ and $EF\|MN$. Also, it follows that $MNPQ$ is cyclic.

Now, $\angle DBM=\angle DCN$ as $EFCB$ is cyclic. Therefore, $DM=DN$. Similarly, $\angle DEP=\angle DFQ$ as $EFCB$ is cyclic. Consequently, $DP=DQ$. Hence, $D$ is the centre of $(MNPQ)$.

Let $MN$ intersect $PQ$ at $T$. By construction, $S$ is the radical center of $(PQEF), (MNCB)$ and $(EFCB)$. So, $SD$ is the radical axis of $(PQEF)$ and $(MNCB)$. By construction, $T$ is the radical center of $(PQEF), (MNCB)$ and $(MNPQ)$. So, $TD$ is the radical axis of $(PQEF)$ and $(MNCB)$, whence, $S,\ T, D$ are collinear. Thus, $AK\perp SD \iff AK \perp TD.$

Applying Brokard's theorem on $MNPQ$, we have, $AK \perp TD$, as $D$ is the centre of $(MNPQ)$ and $A, K$ and $T$ are the three diagonal points of the complete quadrangle $MNPQ$.

$\blacksquare$

Note that for any $E,\ F$ on $AB$ and $AC$ respectively such that $AEF\sim ACB$, the problem statement is true. The circumcentre of $ABC$ is irrelevant.

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  • $\begingroup$ Thank you for your solution. Could you introduce me the references for this problem? I am grateful to you if you could explain why you could solve the problem. $\endgroup$ – Blind Dec 30 '18 at 12:37
  • $\begingroup$ Which contest did this problem appear in? $\endgroup$ – Anubhab Ghosal Dec 30 '18 at 12:53
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    $\begingroup$ Read about radical axes at maa.org/sites/default/files/pdf/ebooks/pdf/EGMO_chapter2.pdf if you are not familiar with them. Read about Brokard's theorem from the link in the answer. $\endgroup$ – Anubhab Ghosal Dec 30 '18 at 12:56
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    $\begingroup$ As for why I could solve the problem, I have trained in olympiad geometry. If you want to know my thoughts while solving the problem, they are as follows. First, I noticed that EFCB is cyclic. Next the radical axis SD occured automatically. With a diagram on geogebra, I noticed that D is the centre of MNPQ. proving that was not too difficult. Now the problem concerned entirely of the complete quadrangle MNPQ except for the line SD. I linked this line to MNPQ by constructing T. The problem was then simply the statement of Brokard's Theorem. $\endgroup$ – Anubhab Ghosal Dec 30 '18 at 12:57
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    $\begingroup$ OK. I will wait one week to accept the solution if there are not another interesting solutions. $\endgroup$ – Blind Dec 30 '18 at 13:59

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