# 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!

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 [ (Q_{AB} + Q_{AC} + Q_{BC})² – 2 (Q_{AB}² + Q_{AC}² + Q_{BC}²) ]

= (Q_{ab} + Q_{ac} + Q_{bc})² – 2 (Q_{ab}² + Q_{ac}² + Q_{bc}²)

≡ (14 + 49 + 91)² – 2 (14² + 49² + 91²)

= 1960

S_{A} = Æ (a,b,c)/(4 Q_{ab}Q_{ac}) ≡ 1960/2744 = (5 x 392)/(7 x 392) = 5/7

S_{B} = Æ (a,b,c)/(4 Q_{ab}Q_{bc}) ≡ 1960/5096 = (35 x 56)/(91 x 56) = 35/91

S_{C} = Æ (a,b,c)/(4 Q_{ac}Q_{bc}) ≡ 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 *y*² exists (given the prior assumption that *x* and 1 – *x* are rational lengths)

Q_{AD} = *x*² + *y*², Q_{CD} = (dist(A,C) – *x*)² + *y*² = Q_{AC} – 2dist(A,C)*x + x²* + *y*²

Thus

Q_{CD} – Q_{AD} = Q_{AC} – 2dist(A,C)*x*

⇒ (Q_{CD} – Q_{AD} – Q_{AC})² = 4Q_{AC}*x*²

⇒ *x*² = (Q_{CD} – Q_{AD} – Q_{AC})²/4Q_{AC}

#### ⇒ *y*² = Q_{AD} – x² = **[4Q**_{AC}Q_{AD} – (Q_{CD} – Q_{AD} – Q_{AC})²]/4Q_{AC}

_{AC}Q

_{AD}– (Q

_{CD}– Q

_{AD}– Q

_{AC})²]/4Q

_{AC}

The denominator *is* a square, so we require the numerator **[4Q _{AC}Q_{AD} – (Q_{CD} – Q_{AD} – Q_{AC})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:

**4Q**_{AB}Q_{AC} – (Q_{AB} – Q_{BC} – Q_{AC})² ≡ (Q_{AB} + Q_{AB} + Q_{AC})² – 2 (Q_{AB}² + Q_{AB}² + Q_{AC}²)

_{AB}Q

_{AC}– (Q

_{AB}– Q

_{BC}– Q

_{AC})² ≡ (Q

_{AB}+ Q

_{AB}+ Q

_{AC})² – 2 (Q

_{AB}² + Q

_{AB}² + Q

_{AC}²)

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 Q_{AD} and Q_{CD} to give all three vertices rational coordinates.

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