# Problem of three circles

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.

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.

• 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. – Blind Dec 30 '18 at 12:37
• Which contest did this problem appear in? – Anubhab Ghosal Dec 30 '18 at 12:53