Irrational triangles?

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!

IT0

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.

IT 1

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)

     =           1960

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):

IT 3

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

Thus

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.

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