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Let there be an angle $x\widehat{O}y $ and A a random point inside it(excluding the rays Ox and Oy). Is it true that the product $d(A,Ox)*d(A,Oy)$ is constant regarless of A? If so, provide the proof (or at least a hint of it)

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    No, that's part of a circle, and there will be in general only $1$ other point on that arc of a circle with product of distances the same.2011-12-02

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For simplicity consider the angle made by the origin, the positive $x$-axis, and the positive $y$-axis. If $c$ is positive, the locus of points $A(x,y)$ in the first quadrant such that the product of the distances to the $x$-axis and to the $y$ axis is $c$ is the half of the hyperbola with equation $xy=c$. Almost all of the points in the first quadrant do not lie on that hyperbola. So certainly the product of the distance to the $x$-axis and to the $y$-axis is not independent of the position of $A$.

A similar remark applies to two rays that do not form a right angle. The locus of points $A$ inside the ray such that the product of the distances from $A$ to the two arms of the ray is constant is half a hyperbola, with the arms of the ray as asymptotes.

Added: In a comment, the OP asked whether one has constant product if the distance $AO$ is constant. The answer is no. Distance $AO$ constant means that $A$ lies on an arc of a certain circle with centre $O$. Let $A$ be a point on that arc, and let $c$ be the product of the distances for $A$. Then the points with constant product $c$ lie on a hyperbola, and that hyperbola will meet the arc of the circle in at most two points. So there are at most two points on the arc that have equal product of distances. If you move towards one arm of the ray, and away from the angle bisector of the ray, your distance from $O$ must increase.

(There are easier ways to see that on the arc of the circle, the product of the distances cannot be constant. Just let the point on the arc approach an arm of the ray. One of the distances remains bounded, and the other approaches $0$, so the product approaches $0$. The comment about the hyperbola just provides more geometric detail.)