QUESTION: Let $P=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ and $Q=(−\frac{1}{\sqrt{2}},−\frac{1}{\sqrt{2}})$ be two vertices of a regular polygon having $12$ sides such that $PQ$ is a diameter of the circle circumscribing the polygon. Which of the following points is not a vertex of this polygon?
(A) $(\frac{\sqrt{3}-1}{2\sqrt{2}},\frac{\sqrt{3}+1}{2\sqrt{2}})$
(B) $(\frac{\sqrt{3}+1}{2\sqrt{2}},\frac{\sqrt{3}-1}{2\sqrt{2}})$
(C) $(\frac{\sqrt{3}+1}{2\sqrt{2}},\frac{1-\sqrt{3}}{2\sqrt{2}})$
(D) $({-\frac{1}2},\frac{\sqrt{3}}{2})$
MY APPROACH: If $P$ and $Q$ are the end points of the diameter, it is quite clear that the equation of the circle must be $$x^2+y^2=1$$ Therefore, all the vertices must lie on this circle. Now, checking from the options, we find that every point given in the options satisfies the above equation. Now I am stuck.
How else should I tackle the sum? Thank you in advance 😊.
Best Answer
D)
But how did I get that answer? Well, firstly have you drawn a picture of the diagram, with angles labelled? If so, then you’ll notice that the arg (angle from the x axis) of any point on the polygon must be 15 mod 30. The last option doesn’t satisfy this property (the angle it forms is 120 degrees).
However, another approach is to notice — again from a well drawn diagram — that the figure is symmetric in the x axis, the y axis, and the x=y line. Since the first three options are just reflections of each other in these lines, the answer must be D. Meta analysis is useful in multiple choice!