The Orthocenter Of Triangle Joining The Circumcenters

Geometry Level 4

Let $O$ be the circumcenter of an acute $\triangle ABC.$ Let $O_A, O_B, O_C$ be the circumcenters of $\triangle BCO, \triangle CAO, \triangle ABO$ respectively, and let $S$ be the circumcenter of $\triangle O_AO_BO_C.$ Let $H$ be the orthocenter of $\triangle ABC.$ Find $\angle OSH \pmod{ \pi}$ in degrees.

Details and assumptions

The notation $\angle OSH \pmod{\pi}$ means you have to enter the remainder when $\angle OSH$ is divided by $180$ in degrees. For example, if you think $\angle OSH = 275^{\circ},$ you should enter $95.$ In particular, if you think $\angle OSH = 180^{\circ},$ (that is, $O,S,H$ are collinear), enter $0.$
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The answer is 0.

2 solutions

Cody Johnson
Apr 14, 2014

I noticed that everything was exact on your diagram except for the location of $S$ , it should be a bit lower. Since it seems like pretty general conditions, $0$ is the best guess.

Phil Peters
Apr 12, 2014

Not a solution: The triangle is the kosnita triangle. Try and prove stuff about perspectivity, and homothecy (but obviously not with the reference triangle, think about other triangles which are similar, they might have a homothecy with the kosnita triangle). Also think about the point O in relation to the kosnita triangle (it might not be the circumcentre of the kosnita triangle, but it could be something else).

The Orthocenter Of Triangle Joining The Circumcenters

The answer is 0.

2 solutions

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