"Holography" in Euclidean Geometry

Let's make one last summary of all below.

Since I learned on my own to address problems in Euclidean geometry from a purely synthetic approach, I've been fascinated by it. If only I had learned this on my teenage years...

Yet this time the passion for a visual approach seems to have mislead me down a windy rabbit hole just come out at what seems trivial had I chosen a more analytical approach of counting degrees of freedom.

We need 6 numbers to specify the coordinates of three points in the plane. If it's a right triangle, only five coordinates are independent. 

But we don't have to stick to cartesian coordinates. 

If we fix one point on the origin, we need four number for the two remaining points plus two more for defining the translation that sat this triangle on the origin.

We can repeat that thought process and further constrain a second point to be on the vertical axis. We can bring it there by a rotation, after using a translation to fix the first point. This leaves then three coordinates for one point "and a half", plus a number for specifying the rotation we had to apply, and two more numbers for the translation. For a right triangle, the third point is constrained as well, in this case, to be on the horizontal axis. Again it amounts to 6 and 5 dof, respectively.

If we fix the length of the "vertical" side to 1, we reduce the previous count of coordinates by one, which we use to specify a dilation. Fir right triangles then we have one horizontal coordinate, one scaling factor, one rotation, and the two numbers for a translation.

And this is nothing more what all the rambling below amounts to.

So, yeah, we have reduced a right triangle to a single point on a line; the whole 2d figure "emerges" from that single point. But the number of degrees of freedom have not changed, we simply hid them in the form of transformations.

I think one take home lesson is to always explore a problem from multiple perspectives, e.g, synthetic & analytic , very soon at the start. Some paths offer much less resistance for making solid steps forward.

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The following statement in planar geometry is true: Given an arbitrary quadrilateral, build squares on each side. Then the line-segments joining the centers of opposite squares are perpendicular and of equal length. (T1)

This can be quickly deduced from the following statement: (T2) As a first step towards a purely geometric explanation of [this] result, consider [the setup where] squares have been constructed on two sides of an arbitrary triangle[. Then] the line-segments from their centres to the midpoint m of the remaining side are perpendicular and of equal length. (From Needham's Visual Complex Analysis 2023)

Both claims have a purely geometric proof as well as a straightforward algebraic, yet coordinate-free proof using complex numbers. 

[Btw, I hate the names "complex" or worse "imaginary" numbers. They are neither more nor less complex than other numbers, nor more imaginary; in fact, they are more real than the reals. See Needham. Traditions exist only for us to break out free of them. They are but a reflection of our flaws and ignorance. This isn't more true than in this case. See Needham historical review. Calling them, say, Geometric numbers would be much more useful, both, for thinking about them and for pedagogical reasons] 

Be it as it may, these results lead me to think on how to encode a triangle on a straight line, which in turn led to the digression below on coming up with a meaningful construction.

Alas, it's all but a moot point.

I see now where the mistake is, namely allowing for construction procedures that "create" new, additional points.

Doing so even two points define 12 isosceles right triangles -all congruent. It's obviously true that all isosceles right triangles of a given size can be described by just one number, namely the length of a leg. And this is what two points define, a  segment of a given length.

Analogously, all right triangles can be described by just two numbers that give the size of the two legs. And once more, three co-linear points provide enough information. In fact they provide three lengths!

So if I were to make this into anything useful, it would hinge on providing a meaningful encoding and construction procedures.

Can we device a procedure for encoding all three sides into the three lengths defined by three co-linear points?

Also curious about: Usually we define a point as origin and choose to have one point of any triangle located there. This is tantamount to taken modulo by any translation. We still have degree of freedom of talking about dilations (scaling) and rotations. 

However, in the present case, we are defining an origin but as well a privileged line. This is equivalent of taken modulo by translations *and* rotations. There is then only one degree of freedom left: dilations.

Thus, given a fixed construction procedure based on three co-linear points, how do those points need to be changed if we scale the triangle by a factor $\gamma$? Answer: The ratio of the segments formed by two free points against the fixed third one is constant. Let's explain: Fix points C as the origin. Then the lengths AC and BC keep a constant ratio when we scale all side by the same factor $\gamma$. It is straightforward to recognize when drawn (see figure). 

We can then mod by dilations by fixing $\|AC\|\,=\,1$. This is obviously the same as fixing the length of one leg to 1. Then we can parametrize all right triangles by a single number, the length of the other leg, or one single point in this fixed line, B. 

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Claim: Any right triangle can be uniquely determined by three co-linear points. And vice versa, given a right triangle we can determine three co-linear points that completely describe the triangle.

Hence all the information that describes a right triangle can be "squashed into a line". 

This squashing reminds me of the holographic principle in theoretical physics. I couldn't resist such a catchy title. 

Modulo any similarity transformation, this line can be fixed at hoc for all triangles. 

Wait... this is bit of a silly thing. If we fix the location at the origin of one point for all triangles these are then determined solely by 2 points, or four numbers. For right triangles we need only one point (2 numbers) plus the distance along a line through the origin perpendicular to the one joining the point with the origin, i.e., only three numbers.

That "fixing" and "modulo" hand-waving does a lot work...

However, what isn't evident is that three (3) co-linear points in the plane determine uniquely a right triangle. The proof is given in the Geogebra construction in Fig.1 below.

Well, about unique... not quite.

Given three co-linear points A, B and C, let's choose A as the midpoint of the hypotenuse and B that of one side of the right triangle, with the segment BC determining half the length of that side. This determines points G&I of the green right triangle, and the segment AG determines its third point H. 

If we choose C as the midpoint of the hypotenuse, we obtain the orange right triangle.

These are right triangles because the they are inscribed in a circle with center the midpoint of the hypotenuse and the latter being a diameter of that circle.

Furthermore, if for each choice of midpoint of the hypotenuse we agree that either of the other points can be the midpoint of one leg, with their segment just determining the length of said leg, then three colinear points in the plane may determine up to 6 right triangles uniquely. 

Some pending questions: are all those triangles similar? what can we say precisly about the  transformations relating them all? how does properties of triangles map to relationships between those three points?

We need to show how a right triangle determines three co-linear points. But this is trivial: we take an endpoint of the leg through B and map it onto the line AB through a rotation centered in B. This gives us 2 points -they are equivalent, because we "say" the leg passes through B. We could have done this using the other leg's midpoint (that between H&I). 

Thus, a right triangle can be encoded into four possible set of three points partitioned into two perpendicular lines that cross at the hypotenuse midpoint (see Aubel's theorem).

Lema: All right triangles obtained from three co-linear points are congruent between themselves.

The following diagram proves it for a couple such right triangles. Given that all are obtained following analogous steps, this prove should apply to any such pair.

Assuming previous Lema, this means this set of congruent triangles ontained from three co-linear points determines a subset of the set of all isometries of the plane. That is, we exclude scaling/dilation transformations. At most rotations, translations and reflections.

Clearly I rushed into a conclusion the other day -too lazy to check and went with a hunch.

The following diagram does show two of those right triangles that are congruent. By Thales, HI = 2 AB = JL. Finally, the circle with center in L and radius LK' contains the vertex I. Thus both green and orange triangles are congruent.

But not all right triangles we can obtain from those three points are neither congruent nor even similar to the rest! 

There are 3 possible choices for the hypotenuse midpoint. For each of those, the segment defined by the other two points determines the length of the "main" leg of the right triangle. However, the length of the "secondary" leg is given by the segment spanned between the hypotenuse midpoint and the point we choose as midpoint of the main leg. The two choices we have for the latter have in general different lengths. This already gives us 6 different right triangles.

I think now that for each of those six triangles we have two ways of building them that are related by a general similarity transformation. In this diagram it's a rotation with reflection; in other cases it's just a reflection. 

But I need to work out more cases. I'm still not seeing things right.

When choosing the first or last point as the one lying on the hypotenuse midpoint, we can build 2 pairs of  different right triangles for each of the two choices of hypotenuse, with each pair of triangles being one the reflection of the other along the defining line. The  diagram shows just one triangle of each such pair. Not shown are the 2 isosceles right triangles defined by each of the two hypotenuse midpoints.

Choosing the middle point as the midpoint of the hypotenuse gives two possible right triangles for each choice of the main leg midpoint. Dashed-line triangles are the reflected constructions of the solid-line ones. Again, the two isosceles solutions having point B on their hypotenuse are not shown.

If the triangle isn't a right triangle, then we don't have any "compression". Initially I caught carried away by this compression idea and thought that all triangles could be thus encoded into three co-linear points. But, this is not the case because in order to determine a leg we would need say an angle. This then makes it three co-linear points *and* a number. However, as mentioned above, there is then no compression in terms of the amount of data needed for it's description -the amount of numbers.