A question on orthocentre of a triangle.

Question

If origin is the orthocentre of a triangle formed by $A(\cos X,\sin X)$, $B(\cos Y, \sin Y)$, $C(\cos Z, \sin Z)$, then what is the angle $BAC$?

I tried using the formula for orthocentre which involves $\tan$ of an angle and stuff but it's becoming difficult.

Please help.

Answer 1

(Expanded from the comments.)  The problem implies that all $3$ vertices are at the same distance from the orthocenter, meaning that the orthocenter coincides with the circumcenter, which in turn means that the triangle must be equilateral.

Answer 2

The three points are all one unit away from the origin so they are all on the circumference of the unit circle.

The origin is the orthocenter so any line through $O$ and one of the vertices is perpendicular to the line through the other two vertices (e.g. line through $OA$ is perpendicular to $BC$).

$|OB| = |OC|$ so $OBC$ is isosceles. The altitude of the isosceles triangle $OBC$ is perpendicular to segment $BC$ and goes through $O$. As $OA$ is another line through $O$ that is perpendicular to $BC$, this means that $OA$ produced coincides with the altitude of $OBC$ and therefore it also bisects $BC$

This means that $B$ and $C$ are symmetric in the line $OA$. A similar argument shows that $A$ and $C$ are symmetric in the altitude $OB$ and that $A$ and $B$ are symmetric in the altitude $OC$. So the triangle is equilateral and the required angle is $2 \pi/3$.

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This argument tries to avoid any explicit facts about the circumcenter.

Answer 3

It is a classical theorem that the reflection $H'$ of the orthocenter $H=O$ with respect to any side belongs to the circumscribed circle (http://www.cut-the-knot.org/Curriculum/Geometry/AltitudeAndCircumcircle.shtml#Explanation). Then, $HH'$ is the perpendicular bissector of the corresponding side, say $AB$. Thus the angle $AOH'$ is $\pi/3$, thus angle $AOB$ is $2 \pi/3$. Doing the same for the other angles, we end up with triangle $ABC$ equilateral.