 # Migrating Angle Defects

NOTE: This is a draft for a geometry tutorial essay; it would be better if it were a little longer and I could find something else of interest to say — should you read it and have any thoughts, please let me know. Diagram for Essay

In Theorem 24 Lobachevski remarks offhandedly that Line JH* must be perpendicular to DE just as it is to FG because the quadrilaterals DFHJ and JHGE ” fit one another if we so place one on the other that the line EF remains in the same position.” My initial reaction was disbelief, because with that construction the angle defect, whatever that is, isn’t evenly distributed, and if it’s not evenly distributed then how can the triangle FHJ not equal π? But that is impossible in the Imaginary Geometry according to Theorem 20. Can the angle defect be said to “reside” somewhere in particular, or does it shift as the area is divided?

In quadrilateral DFHJ, DF=FH, and FJ is in common between triangles FDJ and FHJ. Are these two triangles equal? Looking at it one way, they must not be, because angle FDJ < angle FHJ, while two of the sides are equal. Also, according to Theorem 8 equal sides lie opposite equal angles and inversely, but the angles FHJ and FDJ are not equal, while two of the sides are. So DJ and JH must be different. Does JH then equal DF and EG, while DJ is larger (because the greater side lies opposite the greater angle)? That seems likely, since DE looks about as much like a Euclidean parallel as one can arrive at in a world with angle defects, lying equidistant from FG at least at FD and GE; as a straight line it shouldn’t be expected to sag, and thus JH should equal DF and EG. But DJH is right, so it looks like whatever the angle defect is, is all hanging out over at D, and is presumably not at F or J, so how could JFH and HJF not each equal half a right angle? But they must be less since there’s a right angle at H and the sum of all the angles must equal <π.

Theorem 22 demonstrates that the angle defect of the larger figure equals the sum of the defects of those figures inside it, and in theorem 23 Lobachevski says that if the angle sum of AA’B’ is π=a then the angle sum of triangle AA”B’ must be π=2a, because AA’=A’A”, A’B’ are in common, and A’B’ is perpendicular to AC; therefore they must be congruent triangles. Since the angles at A’ and A” are all right, there must be a deficiency up at B’, just as there is at D. Triangles FHJ and JHG (imagining a line JG) would be a case similar to A’A”B’ and AA’B”, because they both share two sides and the included angle, so if the angles of triangle FHJ = π-a then the angle sum of FJG must be π-2a. The -2a part of things has to be hiding out somewhere other than H, since each side of H is 1/2π. If HJ =FD as I suspect then JFH=FHJ, and each is 1/4π-1/2a. So then would DJF and DFJ each be 1/4π while their compliments are less? That might make the angle sum of triangle FHJ similar to that of FDJ, but it would also make the angles DFH and HJD less than right, unless they exceed 1/4π on the side of FDJ; but that shouldn’t happen either.

As far as I can tell, then the angle defect in a right angled figure must show up in whichever angle the geometer is not examining or folding or constructing at the moment, especially in a case like the angles like J, which would be acute if it were in the original construction, but is right in the event of being folded. That is very unsatisfactory.; more so even than if it were to be simply impossible to construct multiple right angles in the same figure that are quite that. Or than if a straight line is not quite π., though it must be since a geometry where the line of parallelism is ½ π is still the reference point used in describing other possibilities.

I’ll have to stay, then, with the unsatisfactory answer that either the angle defect in a rectilinear figure of Imaginary Geometry is to be found in whichever angle the geometer is not currently looking at, or the geometer is required to be a bit less precise about the rightness of his right angles than might be desired, or it’s not possible to construct a figure such as DEFG that is pivot-able along JH.

* I found the diagrams for Theorems 23 and 24 easier to look at if I combined them, but as a result the lettering on 24 is different from that in the text.