AK x BN independent of point M, equilateral (2018 Izumrud Olympiad 9 p1)

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AK x BN independent of point M, equilateral (2018 Izumrud Olympiad 9 p1)

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#1 h May 4, 2021, 8:40 PM

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An equilateral triangle $ABC$ is inscribed in a circle. On an arc $AB$ not containing a point $C$ , a point $M$ is chosen that is different from fixed points $A$ and $B$ . Let lines $AC$ and $BM$ intersect at point $K$ , let lines $BC$ and $AM$ intersect at point $N$ . Prove that the product of the lengths of the segments $AK$ and $BN$ does not depend from the choice of point $M$ .

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#2 h Oct 30, 2021, 3:21 PM

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Given the circle, midpoint $O(0,0)$ and radius $r=2$ .
Points $A(2,0),B(-1,\sqrt{3}),C(-1,-\sqrt{3})$ .

Choose $M(2\cos \alpha,2\sin \alpha)$ .

The line $AM$ cuts $BC$ in the point $N(-1,\frac{3\sin \alpha}{1-\cos \alpha})$ .
Then $BN=\frac{3\sin \alpha}{1-\cos \alpha}-\sqrt{3}$ .

The line $BM$ cuts $AC$ in the point $K(\frac{5\cos \alpha+\sqrt{3}\sin \alpha +1}{\cos \alpha -\sqrt{3}\sin \alpha +2},\frac{x_{K}-2}{\sqrt{3}})$ .
Then $AK=\frac{2(x_{K}-2)}{\sqrt{3}}$ .

The product $BN \cdot AK =12=3r^{2}$ .

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