Patterns in Pythagorean triangles and a link to the Triple Quad Formula

Playing around with Pythagorean Triples (which are, you may know, triangles with Integer sides, or alternatively, with Rational sides, as any set of rational numbers can be converted to a set of integers with equal proportions betwixt them as were found in the Rational set)….

Playing around with them, and pondering the Altitude to the Hypoteneuse, that is, the line between the right-angle corner of the Triangle, and the Hypoteneuse, such that the line is at right angles with said Hypoteneuse….

It came to me that there is a funny pattern.

Take the Pythagorean Triangle 3 4 5 (Quadrances 9 16 25)

The altitude is 12/5, dividing the hypoteneuse into pieces of 9/5 and 16/5. The legs are 3 and 4, but this can be written 15/5 and 20/5.

But what if we cancel the ‘/5’ of each of these? Then we get a larger triangle of equal proportion…. but there is something funny about the numbers involved.

It is now a 15 20 25 pythagorean triangle…. with an atltiude of 12 dividing the Hypoteneuse into pieces of 9 and 16.

But what of the factors of these numbers? Ahh this is the odd bit.

The hypoteneuse is 5 x 5

The short leg is 3 x 5

The long leg is 4 x 5

The altitude is 3 x 4

The long-piece of hypoteneuse is 4×4

The short-piece of hypoteneuse is 3×3

Do you see the patterns? The factors are all simply the original Pythagorean triple members rearranged in pairs. Each possible pair in fact.

Call 3 the short, 4 the medium, and 5 the long. Then the pattern appears to be that in the Proportional Big Triangle,  the legs will be measured in length as short by long, and medium by long. The altitude is short by medium. The hypoteneuse is long by long, divided into pieces short by short and medium by medium.

It is almost as if you needn’t do any algebra at all to determine an Altitude of a pythagorean triangle. Let’s try 5 12 13

A proportionate triangle should have hypoteneuse 13 by 13, altitude 5 by 12, legs 5 by 13 and 12 by 13, and hypoteneuse parts of 5 by 5 and 12 by 12. Does the pattern hold?

Try a simple pythagorean test, with the hypoteneuse part, the altitude, and a leg. 12×12, 5×12, and 12×13. Square each, we have

5^2 x 12^2 + 12^2 x 12^2 =?= 12^2 x 13^2

Cancel the twelves and we ask, 5^2 + 12^2 =?= 13 ^2

Indeed, 25 + 144 = 169. The pattern worked for that case too. Interesting. What else?



OK. What about Quadrances? Basic Quadrance is the Square of Distance. It comes from Dr. Wildberger’s Rational Trigonometry. But there are an awful lot of squares on this diagram… does it link up with anything?

Indeed…. the two hypoteneuse parts, along with the hypoteneuse itself, form a collinear pair of three points. This fits into the Triple Quad Formula. Say that we name the hypoteneuse as ‘t’  and the parts as ‘p’ and ‘n’.  Qn = n*n, Qp=p*p, etc. Then.

2 * ( Q_p^2 + Q_n^2 + Q_t^2 ) = ( Q_p + Q_n + Q_t ) ^ 2

Interesting. But does it link up with anything? Yes in fact, if you look at the big proportional triangle… all of those multiplies mean that there is a relationship.

Take the short leg – 3 by 5. Let’s called it L. Now, the Quadrance will be 3×3 x 5×5… but compare this to our original Pythagorean triangle, where we had sides 3 4 5. Lets call a=3, b=4, c=5. Then Qa = 3×3, Qb=4×4. Thus, the QL of the big proprtional triangle is actually Qa * Qb.

Interesting! But where does it lead?



… it leads here. The Quadrances of the legs and altitude, when squared and summed, are twice the sum of the squares of the Quadrances of the Hypoteneuse and Hypoteneuse parts.s

That seems quite strange to me. Almost another version of the triple quad formula, but it relates the legs + altitude of a right triangle to the hypoteneuse and hypoteneuse parts.

It also is a little reminiscent of Descarte’s Circle Formula.. if one were to rearrange slightly… since

2 * ( Q_p^2 + Q_n^2 + Q_t^2 ) = ( Q_p + Q_n + Q_t ) ^ 2


( Q_p ^2 + Q_n ^2 + Q_t ^2 ) = 2 * ( Q_d ^2 + Q_m ^2 + Q_l ^2 )

then we can rewrite the first as

( Q_p^2 + Q_n^2 + Q_t^2 ) = 1/2 * ( Q_p + Q_n + Q_t ) ^ 2

then again as

( Q_p + Q_n + Q_t ) ^ 2 = 4 * ( Q_d ^2 + Q_m ^2 + Q_l ^2 )

(In this case, 925444 = 925444)

I don’t remember seeing this in Dr. Wildberger’s notes, but perhaps I have simply forgotten.



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