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Let $ABC$ be an acute triangle. Circle $\omega_1$, with diameter $AC$, intersects side $BC$ at $F$ (other than $C$). Circle $\omega_2$, with diameter $BC$, intersects side $AC$ at $E$ (other than $C$). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$. Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$. Prove that lines $AB$, $ML$, $NK$ are concurrent

My Progress:

Claim : $K,M,L,N$ is cyclic

Proof : Let $NM\cap KL=H$ . Note that $H$ will be the orthocenter of $ABC$ .

By POP, $NH\cdot HM= CH\cdot CF=KH\cdot HL$.

Claim: $C$ is the centre of $(KMLN)$

Proof: Since $CA$ is the diametre , we have CA as the perpendicular bisector of $LN$ .

Similarly $CB$ is the perpendicular bisector of $KM$ .

enter image description here

Now , I just want to show AB is the Polar of $H$ wrt $(KLMN)$ . Then by Brocard's theorem, I know that $NK\cap LM \in AB $.

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  • $\begingroup$ To clarify the problem, based on my understanding the points shown in the picture as $L,N,K,M$ are the points $K,L,X,Y$ in the original problem. If possible please edit the problem to match the picture ( I didn't edit as I might be wrong). $\endgroup$ – cr001 Aug 20 '20 at 5:20
  • $\begingroup$ Also the points $A,C$ are actually $C,A$ in the original problem (swapped) $\endgroup$ – cr001 Aug 20 '20 at 5:23
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    $\begingroup$ @cr001 done thanks! I think it's okay , right now ? $\endgroup$ – Sunaina Pati Aug 20 '20 at 5:27
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    $\begingroup$ The question looks good now. In your proof of $C$ is the center, I think you mistyped $KM$ and $NL$ to be $MN$ and $KL$ but I think that's easily understandable so it's fine. $\endgroup$ – cr001 Aug 20 '20 at 5:30
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It suffices to show that the polar of $H$ passes through $A$ as well as $B$. By symmetry it suffices to show the polar of $H$ passes through $A$ or equivalently, the polar of $A$ passes through $H$.

You know the polar of $A$ is perpendicular to $AC$

Observe that $$AC.AE=AK.AM= AC^2-r^2$$ where $r$ is the radius of the circle $KLMN$.

Rewriting this as $$AC^2-r^2= AC.(AC-EC)$$ $$ \implies AC.EC=r^2$$

Thus the polar of $A$ wrt $KLMN$ is the line perpendicular to $AC$ and passes through $E$. In other words it is the line $BE$ and hence passes through $H$.

Note: There is probably some disparity between the labeling in the question and that in the diagram. My answer follows the labeling of the diagram.

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    $\begingroup$ Great Proof ! I always forget about La hire's theorem $\endgroup$ – Sunaina Pati Aug 20 '20 at 7:45
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    $\begingroup$ Thanks. It felt nice doing a problem like this after years. :) $\endgroup$ – Soumik Aug 20 '20 at 8:13
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    $\begingroup$ Great application of La Hire! By the way, one could also have used $AN^2=AE\cdot AC=AK\cdot AM\implies AN$ is tangent to $(KLMN)$ - where the first equation follows from well-known relations in right triangles. From there, it is straightforward to infer that $EN$ is the polar of $A$ wrt $(KLMN)$. $\endgroup$ – Dr. Mathva Aug 20 '20 at 18:48

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