Let P = ( 1 <msqrt> 2 </msqrt> </mfrac> , 1 <msqrt>

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Let $P=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ and $Q=\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$ 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) $\left(\frac{\sqrt{3}-1}{2\sqrt{2}},\frac{\sqrt{3}+1}{2\sqrt{2}}\right)$
(B) $\left(\frac{\sqrt{3}+1}{2\sqrt{2}},\frac{\sqrt{3}-1}{2\sqrt{2}}\right)$
(C) $\left(\frac{\sqrt{3}+1}{2\sqrt{2}},\frac{1-\sqrt{3}}{2\sqrt{2}}\right)$
(D) $\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
If $P$ and $Q$ are the end points of the diameter, it is quite clear that the equation of the circle must be

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?
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Karla Hull
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!

Chant6j
MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document A)
B)
C)