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Topic: Re: Triangle Coordinates Calculation
Replies: 2   Last Post: Jan 6, 2004 9:35 PM

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Peter Ash

Posts: 13
Registered: 12/6/04
Re: Triangle Coordinates Calculation
Posted: Dec 14, 1998 7:07 AM
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I missed the start of this thread, so I hope I'm not repeating

I just wanted to mention that the problem is essentially that of
converting from bipolar coordinates to Cartesian coordinates. In a
bipolar coordiate system, the coordinates of a point are the distances
from the point to two fixed points (poles). Of course, there is
generally a one-to-two relationship between points and coordinate pairs,
as a point and its reflection in the line through the poles share the
same coordinates.

--Peter Ash

Mary Krimmel wrote:
> -> > vertex. What I know is the following:
> -> >
> -> > 1) x,y values of 2 of the vertices(as mentioned above)
> -> > 2) the lengths of all 3 sides of the triangle
> -> >
> -> > I expect that there will be 2 solutions because of the quadratic
> -> > relationship. The simpler the solution the better, as this problem
> -> > will plug into a computer program. Please help!
> -> >
> -> > Thanks,
> -> > Mark Dumas
> -> Given point (a,b) at distance r from your unknown point, (x,y),
> -> you must have (x-a)^2 + (y-b)^2 = r^2.
> -> With two points and two distances, you have two equations.
> -> Since the equations represent circles, there will usually
> -> be two solutions ( there may be 0 or 1 solutions).
> There are two different sets of two equations, since if l1 and l2 are
> the sides and P1 (x1, y1) and P2 (x2, y2) are the two known vertices, we
> could have l1 being the radius of the circle centered at P1 and l2 being
> the radius of the circle centered at P2, or vice versa. If l1 = l2 then
> the triangles are isosceles, one vertex on each side of the known base.
> So in general you have four solutions. You have no real solution if the
> triangle inequality is violated, i.e., if one side is given as longer
> than the sum of the other two. I can't think of any possibility for
> just one solution.
> Geometrically it is so simple - just draw the circles and see the
> intersections. But although it is straightforward to solve the pair of
> simultaneous equations it is tedious. Maybe there is a simpler way to
> solve them than I can see.
> You also have no solution if it happens that the distance between the
> points you have does not equal one of the three lengths of the sides! Of
> course you take l1 and l2 to be the two other two sides.
> Mary Krimmel

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