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.