# Chaitanya's Random Pages

## August 31, 2012

### A Pythagorean triangle from the incentre of another Pythagorean triangle

Filed under: mathematics — ckrao @ 1:00 pm
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A Pythagorean triple is a set of three positive integers $a, b, c$ satisfying $a^2 + b^2 = c^2$. Two examples are $(3,4,5)$ and $(5,12,13)$. These form the side lengths of a right-angled triangle, called a Pythagorean triangle.

Some time ago I found that a (3,4,5) triangle can be found inside a (7,24,25) right-angled triangle as follows.

In this post we will generalise the above figure. That is, in a right-angled triangle with integer side lengths consider a smaller right-angled triangle formed by one of its vertices, the incentre, and the incircle’s point of tangency with a side as shown above. For what side lengths of the orignal triangle does this inner triangle also have integer side lengths? Below is a second example.

To find the general solution we use the fact that primitive Pythagorean triples (those without a common factor) have the following form:

$\displaystyle (m^2-n^2, 2mn, m^2 + n^2), \quad(1)$

where $m$ and $n$ are positive integers with no common factor, $m>n$ and one of them is even. Any other (non-primitive) triple is formed by multiplying each of the three elements by the same positive integer.

Here is an easy way to derive (1) that I found here . We have $b^2 = c^2 - a^2 = (c-a)(c+a)$ from which $b/(c-a) = (c+a)/b = m/n$ where $m$ and $n$ are coprime integers. Hence $n/m = (c-a)/b$ and we have the two equations

$\displaystyle \frac{c}{b} + \frac{a}{b}= \frac{m}{n}$

$\displaystyle \frac{c}{b} - \frac{a}{b}= \frac{n}{m}.$

Adding and subtracting the two equations gives $c/b = (m/n + n/m)/2 = (m^2 + n^2)/2mn$ and $a/b = (m/n - n/m)/2 = (m^2 - n^2)/2mn$. We can then equate numerator and denominator leading to the desired result provided the right sides are reduced. This will be the case if $m$ and $n$ have no common factor and one of them is even (this ensures the numerator $m^2 \pm n^2$ is odd).

Since a line from a vertex of a triangle to its incentre is an angle bisector, we have angles of $\theta$ and $2\theta$ in our inner and outer triangles respectively. Since the inner triangle has side lengths forming a Pythagorean triple we have

$\displaystyle \tan \theta = \frac{m^2 - n^2}{2mn}$ or $\displaystyle \tan \theta = \frac{2mn}{m^2 - n^2}$.

Only one of these is less than 1 since they are reciprocals of each other.

Using the identity $\tan 2\theta = 2 \tan \theta /(1 - \tan^2 \theta)$,

either $\displaystyle \tan 2\theta = \frac{(m^2 - n^2)/(2mn)}{1 - [(m^2 - n^2)/(2mn)]^2} = \frac{4mn(m^2-n^2)}{(2mn)^2 - (m^2 - n^2)^2}$

or $\displaystyle \tan 2\theta = \frac{(2mn)/(m^2 - n^2)}{1 - [(2mn)/(m^2 - n^2)]^2} = \frac{4mn(m^2-n^2)}{(m^2 - n^2)^2 - (2mn)^2}$.

These are equivalent up to sign and we choose the case for which $\tan 2\theta > 0$ (i.e. $2\theta < \pi/2$) depending on whether $m^2 - n^2$ is smaller or larger than $2mn$.

Next we show that the numerator $4mn(m^2-n^2) = 4mn(m+n)(m-n)$ is coprime with the denominator $(m^2 - n^2)^2 - (2mn)^2 = (m^2 - n^2 + 2mn)(m^2 - n^2 - 2mn)$ (which is odd when $m$ and $n$ have different parity).  Suppose an odd prime $p$ divides both $(m+n)$ in the numerator and $(m^2 - n^2)^2 - (2mn)^2 = (m+n)^2(m-n)^2 - (2mn)^2$ in the denominator. Then $p$ divides $(2mn)^2$ and so $p$ divides $m$ or $n$. But $p$ divides $(m+n)$ which implies $p$ is a factor of both $m$ and $n$: contradiction. A similar contradiction can be used to show any odd prime factor of $(m-n)$ cannot divide the denominator.

Next if an odd prime $p$ divides both $m$ in the numerator and $(m^2 - n^2)^2 \pm (2mn)^2$ in the denominator, then $m$ divides $n^4$ implying $p$ divides both $m$ and $n$: contradiction. A similar contradiction can be used to show any odd prime factor of $(m-n)$ cannot divide the denominator. We conclude that the numerator and denominator have no common factors.

From these values of $\tan 2\theta$, the numerator and denominator (possibly times a common multiple) correspond directly to side lengths for the outer triangle and we find the hypotenuse to be $\left([4mn(m^2-n^2)]^2 + [(m^2 - n^2)^2 - (2mn)^2]^2\right)^{1/2} = (m^2 +n^2)^2$. Hence the outer triangle’s side lengths correspond to the following Pythagorean triples:

$\displaystyle \left(k(4mn(m^2-n^2)), k((m^2 - n^2)^2 - (2mn)^2), k(m^2 + n^2)^2\right)$

Hence we find the interesting fact that the hypotenuse of the outer triangle will always be a multiple of a perfect square, as is seen in the two above examples. With this as the side lengths of the outer triangle, the inradius will be $(a + b - c)/2 = k(4mn(m^2-n^2) - 8(mn)^2)/2 = 2kmn(m^2 - n^2 - 2mn)$ (the formula $(a+b-c)/2$ is valid for any right angled triangle with hypotenuse $c$). This inradius corresponds to one of the side lengths of the inner triangle. The three side lengths of the inner triangle will have the form

$\displaystyle \left(k(m^2-n^2)(m^2 - n^2 - 2mn), 2kmn(m^2 - n^2 - 2mn), k(m^2 + n^2)(m^2 - n^2 - 2mn)\right)$

The first diagram above corresponds to $m=2,n=1,k=1$ while the second corresponds to $m=3,n=2,k=1$. The next simplest type would be $m=4, n=1,k=1$ leading to a $(56,105,119)$ inner triangle and an $(161, 240, 289)$ outer triangle.

1. That’s a neat sequence. Thanks for writing it up.

Comment by Joseph Nebus — September 1, 2012 @ 4:20 am

2. I have found three formulas that generate all primitive Pythagorean Triples and also give you the incircle radius and diameter as well. s and n are any positive integers.

a = s2 + 2sn ( s squared + 2sn )
b = 2n2 + 2sn ( 2n squared + 2sn
c = s2 +2n2+ 2sn ( s squared +2n squared + 2sn)

the incirle radius is sn and the incircle diameter 2sn.

Comment by Terry Furler — April 29, 2017 @ 9:05 am

• Nice, these become the formulas in this post with the substitution $s = m-n$. Note that we require $s$ to be odd for the triple to be primitive, otherwise all of $a,b$ and $c$ will be even.

Comment by ckrao — April 29, 2017 @ 1:06 pm

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