3d Circumcircle problem

[shy author]

There are 3 lines in 3d space given by their parametric equations in
[x,y,z] coordinates:

(a) [75/13*r, 4, 180/13*r + 52/5]
(b) [33/13*s, 56/5*s + 1388/165, 396/65*s + 52/5]
(c) [-42/13*t, 56/5*t + 308/15, -504/65*t + 52/5]

Find out if these lines intersect or not. If they do intersect, do the
points determine a triangle? If there is a triangle, what are the
lengths of its sides? Is it a Heronian triangle or not?

Extra credit:

Find the center of the circumcircle and its radius for this triangle, if
the triangle exists.

[Alexander Bogomolny]

Two lines intersect if they have a common point. For lines (a) and (b) this would mean that, for some r and s, you have three equations:

75/13*r = 33/13*s
4 = 56/5*s + 1388/165
180/13*r + 52/5 = 396/65*s + 52/5.

Incidently, these system is solvable (the first and the third equations are equivalent, s is found from the second and r from either first or third.) Thus you find a point (x, y, z) - the intersection of (a) and (b).

You have to try doing the same with the other two pairs of lines, (a)&(c) and (b)&(c). If these give you two additional points, then three points form a triangle - in 2d as well as in 3d.

There is a standard formula for the distance between two points

dist((x1, y1, z1), (x2, y2, z2)) =
sqrt((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2)

that will give you the side lengths.

For the extra credit check

http://www.cut-the-knot.org/triangle/MetricRelationsInTriangle.shtml

especially, #29.

A. Bogomolny
http://www.cut-the-knot.org

[shy author]

This is a partial solution.

For lines (a) & (b) the intersection parameters r,s = [-13/75,-13/33] and
let the point be called A at [-1,4,8]

For lines (a) & (c) the intersection parameters r,t = [62/75, -31/21] and
let the point be called B at [62/13,4,284/13]

For lines (b) & (c) the intersection parameters s,t = [20/33, -10/21] and
let the point be called C at [20/13,76/5,916/65]

Thus any pair of lines do intersect in a point and there are 3 points.

Do they determine a triangle?

We find the distance, using the distance formula d=sqrt((x0-x1)^2+(y0-y1)
^2+(z0-z1)^2) and according

distance (side) a = distance B to C = 14
distance (side) b = distance A to C = 13
distance (side) c = distance A to B = 15

Thus the sides are 14,13,15 which is a triangle.

The area of this triangle is A = sqrt(s(s-a)(s-b)(s-c)) = 84

Thus this triangle is Heronian

The next partial solution will show how to calculate the circumcenter and
radius for this 3d Heronian triangle