While specifying triangles using rational-valued distances between vertices ensures the quadrances are at least squares, valid solutions to the spreads of a triangle do not depend on this property. Thus a triangle with quadrances 18, 32 and 50 still has spreads of 3/5, 4/5 and 1 even though it would formally have side lengths of 3√2, 4√2 and 5√2, and therefore would be non-constructible.
EDIT 19/10/2015: Actually, such a triangle is constructible!
A slightly weaker irrationality occurs when one quadrance is a square, but not the other two. For instance a triangle with unit quadrance AC, having an obtuse spread of 5/7 (and therefore lying 1 – 5/7 or 2/7 outside the referent semi-circle, as shown in the figure below) has non constructible vertex B at (-2/7,√(10/49)) and quadrance AB of 14/49.
The third quadrance BC (from the larger right angled triangle) is then (9/7)² + 10/49 = 91/49.
Multiplying by 49 to clear denominators on these three values 1 : 14/49 : 91/49 gives quadrances of 49, 14 and 91. Then using the formula for the quadrea (16 times square of signed area) of this similar triangle will give the same rational spreads:
49² Æ (A,B,C) = Æ (a,b,c)
Æ (a,b,c) = 49² [ (QAB + QAC + QBC)² – 2 (QAB² + QAC² + QBC²) ]
= (Qab + Qac + Qbc)² – 2 (Qab² + Qac² + Qbc²)
≡ (14 + 49 + 91)² – 2 (14² + 49² + 91²)
SA = Æ (a,b,c)/(4 QabQac) ≡ 1960/2744 = (5 x 392)/(7 x 392) = 5/7
SB = Æ (a,b,c)/(4 QabQbc) ≡ 1960/5096 = (35 x 56)/(91 x 56) = 35/91
SC = Æ (a,b,c)/(4 QacQbc) ≡ 1960/17836 = (10 x 196)/(91 x 196) = 10/91
Triangle abc (not shown) would therefore have rational quadrances (only one a square) and spreads, with vertices dilated by √49 (x7) at (say) (0,0), (7,0) and (-2,√10) and a non-square qudarea of 1960. It is apparent therefore that no rescaling of the basic triangle ABC can make all of its points rational.
What will make a triangle completely ‘rational’, geometrically as well as metrically?
Provided all quadrances of a triangle are square AND the quadrance of one vertex to an opposite side is also a square, the resulting triangle will contain three rational points and thus be constructible.
This is shown considering two right triangles, with common height, meeting to form a third triangle. Without loss of generality, the base of this triangle can be a unit, as shown in the figure below (the semi-circle, given for reference, shows the triangle is itself non-right):
For vertex D to be rational only requires that y² exists (given the prior assumption that x and 1 – x are rational lengths)
QAD = x² + y², QCD = (dist(A,C) – x)² + y² = QAC – 2dist(A,C)x + x² + y²
QCD – QAD = QAC – 2dist(A,C)x
⇒ (QCD – QAD – QAC)² = 4QACx²
⇒ x² = (QCD – QAD – QAC)²/4QAC
⇒ y² = QAD – x² = [4QACQAD – (QCD – QAD – QAC)²]/4QAC
The denominator is a square, so we require the numerator [4QACQAD – (QCD – QAD – QAC)²] be square also.
We note this numerator is the non-symmetric expression for the quadrea of the triangle Æ (A,B,C), 16 times the square of its area, which we have already seen above. That is:
4QABQAC – (QAB – QBC – QAC)² ≡ (QAB + QAB + QAC)² – 2 (QAB² + QAB² + QAC²)
Thus Q(y) is square ⇒ dist(D,E) is rational ⇒ Æ (A,B,C) is square.
The square property of the quadrea assures only that y is a square therefore, but constraints still exist on choosing (square) quadrances QAD and QCD to give all three vertices rational coordinates.