domingo, 29 de septiembre de 2019

1995 International Mathematical Olympiad, problem 1


Proof. Notice that because of the Radical Axis theorem, the  quadrilateral $BCMN$ is cyclic. It follows that $\angle{BCM}=\angle{BNM}$. Observe that $\angle{MAC}+\angle{BCM}=90^\circ$ ($\triangle{ACM}$ is a right triangle). It turns out that $AMND$ is cyclic. Indeed, $\angle{MAD}+\angle{MNB}+\angle{BND}=180^\circ$.
Now, by the Radical Axis theorem, $AM$, $ND$ and $XY$ must concur.



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