I missed the start of this thread, so I hope I'm not repeating something.
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.
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 > firstname.lastname@example.org