Is the $(3,4,5)$ triangle the only rectangular triangle with this property?

Question

While solving a loosely related exercise, by luck I found out that the $(3,4,5)$ triangle has the following property:

The product of the lengths ($\sqrt{2}$ and $\sqrt{5}$) of the two shorter line segments from a corner to the center of the inscribed circle equals the length ($\sqrt{10}$) of the longest one.

Somewhat satisfied with this, for me own new found, result I now wonder if any other rectangular $(a,b,c)$ triangles have this particular property. $(a,b,c)$ does not need to be a Pythagorean triple (but it would be extra nice).

It tried some straightforward algebraic equations but failed to find answer ... Maybe finding non rectangular such triangles is easier, but ideally I ask for rectangular ones.

update

Can this property of certain pythagorean triples in relation to their inner circle be generalized for other values of $n$?

is already linked to this one but I take the freedom to explicitly mention it here in post for following reasons

• the question asked there is about generalizing answer given here
• the answers to both questions always left some exercises for reader
• myself I am not able (I continue to try) to do these exercises

Maybe someone can fully write out the missing gaps.

Answer 1

In a Pythagorean right triangle $\triangle ABC$, we know that $a^2 + b^2 = c^2$ where $a, b, c$ are positive integers. We also know that $|\triangle ABC| = rs$, where $r$ is the inradius and $s = (a+b+c)/2$ is the semiperimeter. Thus we have $$\begin{align} r &= \frac{ab}{a+b+c} \\ &= \frac{ab}{a+b+\sqrt{a^2+b^2}} \\ &= \frac{ab(a+b-\sqrt{a^2+b^2})}{(a+b+\sqrt{a^2+b^2})(a+b-\sqrt{a^2+b^2})} \\ &= \frac{ab(a+b-\sqrt{a^2+b^2})}{2ab} \\ &= \frac{1}{2}(a+b-c) \\ &= s-c. \end{align}$$ Denoting $I$ as the incenter, the respective distances from the incenter to the vertices are $$IA = \sqrt{r^2 + (s-a)^2}, \\ IB = \sqrt{r^2 + (s-b)^2}, \\ IC = \sqrt{r^2 + (s-c)^2} = r \sqrt{2}.$$ Then assuming $a < b < c$, we require $IB \cdot IC = IA$, or $$\begin{align} 0 = IB^2 \cdot IC^2 - IA^2 = \left(r^2 + (s-b)^2\right)(2r^2) - \left(r^2 + (s-a)^2\right). \end{align}$$ I leave it as an exercise to show that this condition is nontrivially satisfied if and only if $b = (a^2-1)/2$, hence $a$ must be an odd positive integer for $b$ to be an integer. Then $c$ will automatically be an integer since $$c^2 = a^2 + b^2 = \left(\frac{a^2+1}{2}\right)^2.$$ Therefore, the solution set is parametrized by the triple $$(a,b,c) = \bigl(2r+1, 2r(r+1), 2r(r+1)+1\bigr), \quad r \in \mathbb Z^+,$$ where $r$ is the inradius of such a triangle. In particular, this leads to the triples $$(3,4,5), \\ (5,12,13), \\ (7,24,25), \\ (9,40,41), \\ \ldots.$$

Answer 2

In support of the answer by @heropup , all triples where $\space B =\dfrac{A^2-1}{2}\space$ and $\space C-B=1\space$ can be generated by

\begin{align} A &=&&2k+1\\ B &= 2 k^2 + &&2 k\\ C &= 2 k^2 + &&2 k + 1 \end{align}

A similar set of all-primitive where $\space C-A=2\space$ can be generated by $\space A=4n^2-1\qquad B=4n\qquad C=4n^2+1\qquad$ et seq $(3,4,5)\quad (15,8,17)\quad (35,12,37) \quad (63,16,65)\quad (99,20,101)\space\cdots$

It would be interesting to see if the latter triples have properties similar to the former.

Answer 3

Here is an abstract way to construct a bunch of rectangular triangles satisfying the property. They don't need to have integer side lengths, though.

Take any rectangular triangle $T$. Let $x$ denote the product of the lengths of the two shorter line segments from a corner to the center of the inscribed circle, and let $y$ denote the length of the longest one. We want $x = y$.

Now let $\alpha > 0$ and consider a triangle $S$ which results from scaling $T$ by the factor $\alpha$. Then the corresponding values $x'$, $y'$ in the new triangle satisfy

$$x' = \alpha^2 x, \qquad y' = \alpha y.$$

Therefore it's possible to find $\alpha$ such that $x' = y'$, so the property holds for the triangle $S$.

This actually shows that for every rectangular triangle we can find a (unique, actually) triangle similar to it that has the desired property.