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.

How many equations and how many unknowns?

2 equations and 2 unknowns

 

2 equations and 2 unknowns, but 2nd degree equations

2 equations and 2 unknowns

 

Even for 2nd order, I think it should be solvable though it may be ugly. But you have 3 equations, one each D1, D2 and D3. If you don't know D1 you have 3 unknowns, if you know it you have 2: x and y.

Even for 2nd order, I think it should be solvable though it may be ugly. But you have 3 equations, one each D1, D2 and D3. If you don't know D1 you have 3 unknowns, if you know it you have 2: x and y.

 

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.

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.

 

You have to solve for D1 as well, so there are 3.

Whilst more complicated - this question is similar to that asked by Vastor in the thread coordinate geometry. He was asking about the locus of all points with distances from two points that are always in a common ratio. You are asking the same thing but with two sets of common ratio. I must admit I was very surprised when the locus of all points with a common ratio came out as a circle (with the proviso that when the ratio is 1:1 the circle is of infinite ratio - ie a straight line). Your solution would be when the circles defined by the loci intercept - so you could have 1 (circles touching) 2 or no results - no idea whether one or more must be in the triangle. I tried to battle through the algebra for the general solution but got lost - Schrodingers Hat polished it off, you can see some working in this thread in the sandpit

Whilst more complicated - this question is similar to that asked by Vastor in the thread coordinate geometry. He was asking about the locus of all points with distances from two points that are always in a common ratio. You are asking the same thing but with two sets of common ratio. I must admit I was very surprised when the locus of all points with a common ratio came out as a circle (with the proviso that when the ratio is 1:1 the circle is of infinite ratio - ie a straight line). Your solution would be when the circles defined by the loci intercept - so you could have 1 (circles touching) 2 or no results - no idea whether one or more must be in the triangle. I tried to battle through the algebra for the general solution but got lost - Schrodingers Hat polished it off, you can see some working in this thread in the sandpit

 

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.

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.

 

Not at all - in fact, thank you for the name Appollonius circles. I have just flicked to the Wikipedia page on them and will read more later - have you read the section on isodynamic points? Still not the same as your problem but ...

Not at all - in fact, thank you for the name Appollonius circles. I have just flicked to the Wikipedia page on them and will read more later - have you read the section on isodynamic points? Still not the same as your problem but ...

 

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.

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.

 

 

Glad that worked! With reference to your last sentence "Of course, there are ratios for which the circles do not intersect." - there must also be ratios at which the circles merely touch (ie just one point of intersection); one set of ratios per triangle perhaps - that's gotta define an interestng point