# USATST 2013/2 Prove that the intersection of $XL$ and $KY$ lies on $BC$.

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 $$KM$$ .

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

• 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). – cr001 Aug 20 '20 at 5:20
• Also the points $A,C$ are actually $C,A$ in the original problem (swapped) – cr001 Aug 20 '20 at 5:23
• @cr001 done thanks! I think it's okay , right now ? – Sunaina Pati Aug 20 '20 at 5:27
• 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. – cr001 Aug 20 '20 at 5:30

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.

• Great Proof ! I always forget about La hire's theorem – Sunaina Pati Aug 20 '20 at 7:45
• Thanks. It felt nice doing a problem like this after years. :) – Soumik Aug 20 '20 at 8:13
• 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 $(KLMN)$. – Dr. Mathva Aug 20 '20 at 18:48