Here we go:

I've got three 3-dimensional points (coordinates given). From each of those points extends a length (distance given). I need to find what 3-dimensional point satisfies all three lengths at once.

[edit - A, B, and C are lengths that will make (x, y, z) lay exactly on the triangle]

A clarification:

I think I USED to know how to do this...

I think you might need to use the Pythagorean Theorem to find the lengths of A, B, and C. If you find values for these lines that, when raised to the power of two, add up to the same number, you should have the answer.

DISCLAIMER: I could be totally wrong.

No guarantee of right angles. Plus, I think he's given A, B, and C

I already have A, B, and C; I need to find (x, y, z) of where A, B, and C meet.

Oh. Umm this is just out of intuition but find the point on the (x, y, z)1-3 plane that's directly "under" (x, y, z) and use A, B, C plus a non-z leg (from any (x, y, z)1-3 to that point on the (x, y, z)1-3 plane) to find the z leg.

Set up a system to find the non-z leg(s) using corresponding A, B, C + pythagorean theorem

It'd be so much easier if you had an angle. Just one angle. Then you could use trig.

Actually I guess you could find some angles using slope and angle of inclination. So, that too.

Write a system of equations using the fact that the distance from, for example (x1, y1, z1) to (x, y, z) is
sqrt((x1-x)^2 + (y1-y)^2 + (z1-z)^2)
Just set the three distances equal to each other. You'll probably have two different points at least because the point could be on either side of the triangle.

Wtr.... Geometry?

Heh heh. Oh, yes.

EDIT: I might as well removem most of this due to more intelligent posts already posted.

lol by looking at it, and not being in geometery yet (next school year)

i think i know how to do it

but then again lol, i rather not say

Sorry I didn't state this before, but A, B, and C are set up so that (x, y, z) will lay exactly on the triangle, so there will only be one answer.

I've tried to set up a system, but it gets so ugly with square roots that I can't eliminate a variable and isolate another in the same equation.

Know how to use vectors?

I only know what I was taught in geometry and 10th grade algebra 2, so I know very little about vectors. That's why I asked you guys, who might give me an answer I'll understand and be able to implement into the program this is for.

Are we given A=B=C?

I think we're just given the values of A, B and C

yup.

cool.

I don't know if that will work though, as that would work only if the points laid on a 2d plane. Hmm, lemme cook up something in a sec.

3 spheres will give two possible coordinates for the point, just choose the best one.

If I understand this correctly, I think this is one of the best ways.

So you'd use:

Assuming (x,y,z) correlates with the center points and x1, y2, x3... correlate with the other points..

sqrt((x1-x)^2 + (y1-y)^2 + (z1-z)^2) = A
sqrt((x2-x)^2 + (y2-y)^2 + (z2-z)^2) = B
sqrt((x3-x)^2 + (y3-y)^2 + (z3-z)^2) = C

Square both sides and expand so you get something like:

x1^2 - 2 * x1*x + x^2 + y1^2.... = A^2, keep doing that and get the other three.

Once you've expanded that out, you whould only be left with x, y, and z as your variables. Move all your constants to the other side of the equation where the length constants are so you should then three equations like so where c's represent the constants you will get:

c1*x^2 + c2 *x + c3 * y^2 + c4 * y.... = c

Now this part is tricky, if you know how to complete the square set it up in this form:

(x1 - x) ^2 + (y2 - y) ^2 .... = c for all the equations.

Manipulate those to wheere all the x1 - x for all three are identical, i.e. if one is (3 - x)^2 while in another equation its (2 - x)^2, add 1 to the (2 - x)^2 so it becomes (3 - x)^2 and then add a 1^2 to the other side to stay consistent.

Once all the forms are identical with only differing coefficients and differing constants that are outside the parenthesis. Set up a matrix of all the coefficients and outer constants. If A is a matrix of the coefficients of the parenthesis and C is a constant of the constants on the right side of the equation, then you can just do A ^ -1 * C to get a matrix of answers that are solutions to things inside the parenthesis. So if you get (x1 - x) ^2 = 12, just solve for the x1, and continue for the others. But I think that's how you do it.

But for any three arbitrary points there is one plane that they all lie on! :p (Possibly more, but since we have a triangle here we can assume all the points are not on the same line or all on the same point.)

But the circles would have to cross through a 4th point so it could not always be a plane. But I was corrected by Detritic, I didn't know Ictus posted spheres.

I've been thinking about it a bit more. Ring said that the center point will lie on the plane and there'll be only one answer, which requires that two of spheres be tangent to each other, which means the "center" point has to be on the triangle (on the same line as two of the given points). I think. It might be possible to determine which two points, but I don't know how.

Wait, I think I do. Find the distances between the three given points and add two of given lengths (A+B, A+C, etc.) until the sum equals the length of a side.