concyclic wanted twice, orthocenter and circumcenter related

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concyclic wanted twice, orthocenter and circumcenter related

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Source: 2017 France JBO TST 2.4

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#1 h Jul 20, 2020, 5:08 AM • 1 Y

Y by Mango247

Let $ABC$ be an acute triangle. The altitudes $[AA_1], [BB_1]$ and $[CC_1]$ intersect at point $H$ . Let $A_2$ be the symmetric of $A$ wrt $B_1C_1$ and $O$ be the center of the circle circumscribed to the triangle $ABC$ .
a) Prove that points $O, A_2, B_1$ and $C$ are concyclic.
b) Prove that $O, H, A_1$ and $A_2$ are concyclic.

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#2 h Jul 21, 2020, 9:10 AM

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a) Let $M$ be the midpoint of side $AB$ . Since $\angle AA_1B=90^\circ$ we have $AM=A_1M$ . Because $\angle MAA_1=90^\circ-\angle ABC=\angle OAC$ there is $\triangle AMA_1\sim\triangle AOC\implies \frac{AC}{AO}=\frac{AA_1}{AM}=\frac{2AA_1}{AB}.$ Denote $X$ as the intersection of lines $B_1C_1$ and $AO.$ Quadrilateral $BCB_1C_1$ is cyclic ( $\angle BC_1C=90^\circ=\angle BB_1C$ ), hence $\triangle AXB_1\sim\triangle AA_1B\implies \frac{2AA_1}{AB}=\frac{2AX}{AB_1}=\frac{AA_2}{AB_1}.$ Finally $AB_1\cdot AC=AO\cdot AA_2$ QED.
b) $\angle HA_1C=90^\circ=\angle HB_1C\implies$ quadrilateral $HA_1CB_1$ is cyclic $\implies AH\cdot AA_1=AB_1\cdot AC=AO\cdot AA_2$ QED

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