On each side of a regular pentagon a square lying outside the pentagon is constructed. (see the picture below)

$X_1,..., X_5$ are the centers of the squares.

$P$ and $Q$ are points of intersection of $X_5X_2$ with $X_1X_4$ and $X_1X_3$ respectively.

$O$ is the center of the pentagon.

Prove that the circumcircle of $OPQ$ touches $X_1X_4$ and $X_1X_3$ enter image description here


1 Answer 1


Note that, by the symmetry of the picture, $\angle OQP= \angle OPQ= \angle OPX_4$. Consider $\triangle OQP$ and the circumcircle. We have that the inscribed angle over $OP$, $\angle OQP$, is equal to the angle between $OP$ and $PX_4$, so $PX_4$ is the tangent to the circumcircle.


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