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In the Cartesian plane let $O(0,0)$ , $P(\cos A,\sin A)$, $Q( - \sin A , \cos A)$ be the vertices of a triangle. Two circles are drawn with $OP$, $OQ$ as diameter intersecting at $R$. Then find $OR$.

I know that $Q$ is the reflection of $P$ in $y$-axis. However I have not been able to come up with anything substantial. Please help me solve this.

Arbuja
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Yami Kanashi
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  • $Q$ is the rotation of $P$ 90 degrees anticlockwise, not a reflection in the $y$-axis. – LtSten Apr 19 '17 at 18:31
  • @LtSten ouh so i guess that's the mistake I'm making – Yami Kanashi Apr 19 '17 at 18:32
  • @amd so? – Yami Kanashi Apr 19 '17 at 18:33
  • I might ask you the same thing. Draw it and see. I was mistakenly taking $Q$ as a reflection per your question, but even rotated the three points are colinear. The fact that the two circles are the same size is important, too. Draw it and see if that gives you any ideas. – amd Apr 19 '17 at 18:38
  • Use https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference to format equations. – Arbuja Apr 19 '17 at 18:40

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It's easy with pure geometry: the right triangle $OPQ$ is isosceles, and $OR$ is an altitude of this right triangle, hence a perpendicular bisector of the hypotenuse $PQ$. As a result, $OR^2=\dfrac12$, and the polar angle of $R$ is $A+\dfrac\pi4$.

Bernard
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