Metric triangulation (theory)

In geometry, parameters such as length and angle are used to describe the parameters of geometric figures in ordinary Euclidean geometry. But as soon as we move from the plane of our Euclidean geometry to other models for describing points in space, we notice that for the description, the length parameter is described through the concepts of metrics or norms. And it is the metric parameters that give us more general connections, although more complex models are used to describe them.

In geodesy, a very important model for determining location in space is the triangulation method, which uses both metric parameters and angular parameters. But can we describe the position of a point using only metric parameters?

Let's start with a description of the task:

We have the metric parameters of the triangle (sides) a,b,c. We have distances from a point to vertices f,e,g. Describe the relationship between the metric coordinates of points with each other

Preliminary analysis

Let's consider the problem from the point of view of Geometric Point Locations (GLP). In the classical geometry problem about the intersection of two circles, we can say for sure that the GMTs located at a distance f and g from points A and B are the intersection points of the circles (A, f) and (B, g), and these are only 2 points symmetrical with respect to connecting the centers with a straight line. In the case of a triangle, another metric coordinate specifies the position of the point uniquely. But this also suggests that there is a formula connecting all 3 metric coordinates.

drawing of GMT points with metric coordinates f,g,h

Now it’s time to choose a solution strategy. The options are:

Through Heron's formula and the sum of areas
Through Van Aubel's theorem and Stewart's theorem
Through the cosine sum of angles and the cosine theorem
through the circle intersection problem

The description of the solution will be on the 3rd point. For other points, I’m ready to share a description in the comments.

Auxiliary Lemma

$\text{Formula for the cosine of the sum of angles:}\\ cos^2(\alpha) + cos^2(\beta) + cos^2(\gamma) - 2cos(\alpha)cos(\beta)cos(\gamma) = 1$

Let's start the solution with the formula we know:

$cos(\gamma) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$

Let's move the cosines to the left side and the sines to the right. Let's transform the sines:

$sin(\alpha)sin(\beta) = \sqrt{(1 - cos^2(\alpha))(1 - cos^2(\beta))}$

or:

$\sqrt{1 - cos^2(\alpha) - cos^2(\beta) + cos^2(\alpha)cos^2(\beta)}$

As we can see, there is a radical here. This means that both parts will have to be squared. Let's see what happens with cosines:

$(cos(\alpha)cos(\beta) - cos(\gamma))^2 = cos^2(\alpha)cos^2(\beta) + cos^2(\gamma) - 2cos(\alpha)cos (\beta)cos(\gamma)$

Equating both parts we get the formula above.

PS: What is important to us in this formula is that the original formula for the cosine of the sum of angles also works in this formula, despite the fact that another relation also works in this formula

Main task

Now let's apply the cosine theorem. To simplify the perception of formulas, we replace the squares of lengths with capital letters (later we will make sure that there are no lowercase letters left):

$\frac{(H + F - B)^2}{4FH} + \frac{(G + H - A)^2}{4GH} + \frac{(F + G - C)^2}{4FG} -\\ -\frac{(G + H - A)(H + F - B)(F + G - C)}{4FGH} = 1$

Now let's bring it to a common denominator:

$G(H + F -B)^2 + F(G + H - A)^2 + H(F +G - C)^2 = \\ =(G + H - A)(H + F - B) (F + G - C) + 4FGH$

And open the brackets and group:

$[(G + F)H^2 + (G + H)F^2 + (F + H) G^2+ 6GFH] + [GB^2 + FA^2 + HC^2] - \\ - 2AF(G + H) - 2BG(H + F) - 2CH(F+G) = \\ = [(F + G)H^2 + (G +H)F^2 + (H + F)G^2 + 2FGH] - A(H + F)(F + G) -\\- B(G +H)(F +G) - C(G +H)(H + F) + AB(F + G) + AC(H + F) + \\+ BC(G + H) - ABC + 4 FGH$

As you can see, monomials of the 3rd degree cancel and a polynomial of the 3rd degree remains (if A, B and C are constants). Now we need to represent the polynomial in canonical form:

$[AF^2 + (A^2 - AB - AC)F] + [BG^2 + (B^2 - AB - BC)G] + \\ + [CH^2 + (C^2 - AC - BC)H] + (A - C - B)HG + (C - A - B)FG +\\+ (B - A - C)HF + ABC = 0$

Now, to simplify things, let’s try to replace the constants:

$X = B + C - A \\ Y = A + C - B \\ Z = A + B - C$

And the reverse replacement:

$A = \frac{Y + Z}{2} \\ B = \frac{Z + X}{2} \\ C = \frac{X + Y}{2}$

Now we can see what happens (by multiplying both sides by 8):

$[4(Y + Z)F^2 - 4X(Y + Z)F] + [4(Z + X)G^2 - 4Y(Z + X)G] + \\ + [4(X + Y)H^2 - 4Z(X+ Y)H] - 8XHG -8ZFG - 8YFH +\\ + (X+Y)(Y + Z)(Z + X) = 0$

Now let’s open the brackets and group them to get the sum of squares:

$4Y(F - H)^2 + 4X(H - G)^2 + 4Z(G - F)^2 - \\ - 4X(Y + Z)F - 4Y(Z + X)G - 4Z(X + Y)H - \\ -3(Y + Z) X^2 - 3(Z + X)Y^2 - 3(X+Y)Z^2 + 2XYZ = 0$

Or transform it like this:

$4Y(F - H)^2 + 4X(H - G)^2 + 4Z(G - F)^2 + 6XYZ =\\ = 4X(Y + Z)F + 4Y(Z + X)G + 4Z( X + Y)H + 3(X+Y)(Y + Z)(Z + X)$

And so far this is the most beautiful option in representing dependency. Perhaps you can simplify it even further if you choose the following formula for the dependence of the cosines of the angles of a triangle.

PS This formula also works for points outside the triangle. Why? The picture will explain it visually, but try to figure it out with the formula):