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PostPosted: Fri, 3 Jul 2009 18:01:34 UTC 
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
Disclaimer: This is an olympiad-training question, not my invention.

Let ABCD be a convex quadrilateral. Let P be a point on the plane, and let E,F,G,H be the points of intersections of angle bisectors of angles APB, APC, BPC and CPD with sides AB,AC,BC,CD respectively. Find all points P such that EFGH is a parallelogram.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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