Unit circle

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Source: May Olympiad (Olimpiada de Mayo) 1999

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#1 h Feb 26, 2018, 7:37 PM • 2 Y

Y by Adventure10, Mango247

In a unit circle where $O$ is your circumcenter, let $A$ and $B$ points in the circle with $\angle BOA = 90$ . In the arc $AB$ (minor arc) we have the points $P$ and $Q$ such that $PQ$ is parallel to $AB$ . Let $X$ and $Y$ be the points of intersections of the line $PQ$ with $OA$ and $OB$ respectively. Find the value of $PX^2 + PY^2$

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#2 h Feb 26, 2018, 11:20 PM • 3 Y

Y by mathisreal, Adventure10, Mango247

Let the distance $OA$ be $R$ . Let the point $X'$ be the point different from $X$ on the line $OA$ such that $PX'=PX$ . It is easy to see by symmetry that triangle $OXY$ is right angled isosceles at O, and hence: $\angle OX'P = 180 - \angle XX'P = 180 - \angle PXX' = 135$ . Now, we can observe that
$\angle X'PY = \angle YOX' = 90$ . Hence quadrilateral $OX'PY$ is cyclic with diameter $X'Y$ . As triangle $X'PY$ is right angled, $PX^2+PY^2=PX'^2+PY^2=X'Y^2$ . If we can find the diameter of the circumcircle of $OX'PY$ we will be done. The circumcircle of $OX'PY$ is the circumcircle of triangle $X'PY$ . Let the radius of this circle be $x$ . Then by the sine rule: $2x=\frac{OP}{\sin\left(135\right)}=\sqrt{2}R$ . Then
$PX^2+PY^2=X'Y^2=4x^2=2R^2$ .

This post has been edited 1 time. Last edited by Kaskade, Feb 26, 2018, 11:20 PM

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#3 h Sep 23, 2022, 11:05 AM • 3 Y

Y by Mango247, Mango247, Mango247

another solution given here

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