eclipse12

Thanks..that was helpful. In fact, I think what I am looking for is exactly that..the isodynamic point, and it seems to me that one of the two isodynamic points does lie inside the triangle. Will confirm if that is indeed the solution I am looking for. It seems like the isodynamic points have some pretty interesting properties. Thanks for all the comments.. I managed to find an elegant solution to the problem Isodynamic points as it turns out is only a special case of the solution I was looking for. I solved the problem by finding the 'Circles of Appollonius' for the ratios that I have, and then finding the intersecting points of those two circles. I then chose the solution that lies inside the triangle. Of course, there are ratios for which the circles do not intersect.

I am aware of the 'locus of all points with distances from two points that are a common ratio'. They are called 'Appollonius Circles'. I guess it gets a little more complicated when there are more than two points involved. Thanks for your response.

D1 is unknown, only the ratio to D2 & D3 are known. Only need to solve for (x,y) so two equations with two unknowns but in the 2nd degree. Was wondering if it could be likened to an existing problem.. like if there is a formula to just plug into. The equations do get messy, so I wanted to avoid errors on my part as I have been out of touch with these things.

2 equations and 2 unknowns

Given the relative distances to the vertices of a triangle, can one determine the point. For e.g, if P is the point denoted by coordinates (x,y) and the vertices are known (x1,y1), (x2,y2) and (x3,y3), and the ratio of the distances to the vertices relative to the closest vertex (let's say P is closest to (x1,y1)) is given, then is it possible to determine P (assuming that P is within the triangle). if D1,D2 & D3 are the distances to the 3 vertices from point P, then we are only given the ratio of these distances relative to D1 i.e. (1, D2/D1, D3/D1) (and the vertices themselves are known), and P is restricted to be within the triangle.