<B=60^o, circumcenter, orthocenter, concyclic wanted

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<B=60^o, circumcenter, orthocenter, concyclic wanted

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Source: 2019 Oral Moscow Geometry Olympiad grades 8-9 p3

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#1 h May 21, 2019, 7:13 PM • 1 Y

Y by Adventure10

In the acute triangle $ABC, \angle ABC = 60^o , O$ is the center of the circumscribed circle and $H$ is the orthocenter. The angle bisector $BL$ intersects the circumscribed circle at the point $W, X$ is the intersection point of segments $WH$ and $AC$ . Prove that points $O, L, X$ and $H$ lie on the same circle.

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#2 h May 21, 2019, 7:29 PM • 2 Y

Y by Adventure10, Mango247

What is L? You haven't said what point is L. Is it any random point?

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#3 h May 21, 2019, 7:32 PM • 2 Y

Y by AlastorMoody, Adventure10

$L$ is the intersection point of the angle bisector of $BL$ with the opposite site $AC$ (it felt obvious to me)

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#4 h Jul 23, 2019, 8:30 PM • 2 Y

Y by Adventure10, Mango247

Let $H'$ be reflection of $H$ over $\overline{AC}$ . Since, $\angle ABC=60^{\circ}$ $\implies$ $O$ $\in$ $\odot (AHC)$ $\implies$ $O$ is reflection of $W$ over $\overline{AC}$ $\implies$ $H'XLW$ is reflection of $HXLO$ over $\overline{AC}$ $\implies$ $HW$ $\cap$ $OH'$ $=$ $X$ $\in$ $\overline{AC}$ $\implies$ $\angle XH'W$ $=$ $90^{\circ}-\angle XBL$ $=$ $\angle BLX$ $\implies$ $H'XLW$ is cyclic $\implies$ $HXLO$ is cyclic

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#5 h Dec 29, 2021, 3:41 PM

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Let H' be reflection of H about AC.
∠AOC = 120 = ∠AWC and AO = CO , WA = WC so W is reflection of O about AC so it's enough to prove XLWH' is cyclic.
HOWH' is isosceles trapezoid sand AC is perpendicular bisector of both HH' and OW so if H,X,W are collinear then O,X,H' are collinear as well.
∠BLX = 90 - ∠HBL and ∠WH'O = 90 - ∠H'BW so ∠BLX = ∠WH'O and XLWH' is cyclic.
we're Done.

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