Sunday, August 15, 2010

A Pythagorean relation for any triangle?

Physics makes extensive use of the Pythagorean law relating the squares of the sides of a right triangle. The well-known a² + b² = c² relation is of special interest for the determination of distances and lengths of vectors, but also for energy conservation laws and Lorentz transformations. There are various relations that resemble the Pythagoras law, but none of them seems to have the usefulness of Pythagoras’ original one, as well as the “beauty” originating from its sole quadratic terms. When the vertex opposite to the hypotenuse runs on the circle determined by the right triangle, we have the Pythagoras relation illustrated in Figure 1.
Some time ago, I was made aware of another interesting Pythagorean law, discovered by Nguyen Tan Tai and which I mentioned in a previous post. For any triangle with sides a, b and c, if the vertex opposite to the base (say c) runs on any fixed circle centered at the center of c, we have the relation:

a² + b² = c² + constant

In order to have a better understanding of this quadratic relation, I tried to re-derive it in my own mental representation. I will call the fixed circle with the running vertex C the “leg circle”, because it is determined by the vertex that is common to legs AC and BC of the triangle (see Figure 2).

When the leg circle is larger than the base circle, the legs AC and BC intersect the base circle and we can decompose the arbitrary triangle into two right triangles, for example as illustrated in Figure 3, the right triangles ABD and BCD.

With the Pythagoras relation we than have:

AC² + BC² = (AD + CD)² + BC²
= (AD² + CD² + 2.AD.CD) + (BD² + CD²)
= AD² + BD² + 2.CD² + 2.AD.CD
= (AD² + BD²) + 2.(AD + CD).CD
= AB² + 2.AC.CD

But the product AC.CD is the power of point C with respect to the base circle, i.e. for every point C on the leg circle, AC.CD is constant and equal to the square of the tangent ray CT (see Figure 4).

If we use another notation, AC = a, BC = b, AB = c and CT = t, the relation becomes a nice equation with only quadratic terms:

a² + b² = c² + t² + t²

One can verify that if the leg circle has same size as the base circle, we retrieve the original Pythagoras relation :

a² + b² = c²

And if the leg circle is smaller than the base circle, we have:

a² + b² + t² + t² = c²

where t is now the tangent ray to the leg circle emanating from any point of the base circle.

We thus have a Pythagoras-like relation for any triangle given its base and leg circles, as illustrated by Figures 5a, 5b and 5c.


  1. Very nice.... I like the "almost Pythagorean" idea and the fact that it is now a universal for all triangles.
    Keep up the good work

  2. Pat,

    Thanks for encouragement. I'll investigate this relation further. c² is proportional to the area of the inner circle and t² is proportional to the area of the ring, which relates to the concentric rings described in another post: Exhaustion of nested squares and Wallis product.

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