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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.

3-4-5 trangle with inscribed circle and the three line segments

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.

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3 Answers 3

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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 $$(3,4,5), \\ (5,12,13), \\ (7,24,25), \\ (9,40,41), \\ \ldots.$$

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    $\begingroup$ I wonder if all pythagorean triples where the longest sides differ by 2 share a similar property. $\endgroup$ Commented Jul 8, 2022 at 22:23
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    $\begingroup$ @FirstName LastName See my answer below for $(3,4,5)\quad (15,8,17)\quad (35,12,37) \quad (63,16,65)\quad (99,20,101)\space\cdots$ $\endgroup$
    – poetasis
    Commented Jul 8, 2022 at 23:16
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    $\begingroup$ @FirstName LastName Let $\quad B=\dfrac{A^2-1}{2}=\dfrac{(2k+1)^2-1}{2} =2 k^2 + 2 k$ $C-B=(2 k^2 + 2 k +1)-(2 k^2 + 2 k)=1$ so, in all cases, we have $\space (A,B,B+1).$ $\endgroup$
    – poetasis
    Commented Jul 9, 2022 at 0:18
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    $\begingroup$ @FirstNameLastName The solution set I provided uniquely characterizes all Pythagorean triples satisfying your property. This means that if a triple has this form, then it will satisfy your property, and if a right triangle with integer sides satisfies your property, it will have the form $(2r + 1, 2r(r+1), 2r(r+1)+1)$ for some positive integer $r$. $\endgroup$
    – heropup
    Commented Jul 9, 2022 at 7:02
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    $\begingroup$ @FirstNameLastName Please ask that as a separate question, because the answer is detailed enough that it should not be relegated to a comment. $\endgroup$
    – heropup
    Commented Jul 10, 2022 at 0:10
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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.

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  • $\begingroup$ The property in question for the 3-4-5 case resulted from relevant coordinates being integer and therefore lengths being square roots of integers and thereby being able to spot the multiplicative property easily. Maybe similar properties can be spotted thanks to similar conditions and some luck. $\endgroup$ Commented Jul 8, 2022 at 23:26
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    $\begingroup$ Triplets of the form $(4n^2 - 1, 4n, 4n^2 + 1)$ do not in general satisfy the property that the product of the two shorter distances from the incenter to a vertex equals the longest distance from the incenter to a vertex. For instance, $(15, 8, 17)$ with $n = 2$ has an inradius of $3$, and distances to vertices $\sqrt{18}, \sqrt{34}, \sqrt{153}$, but $18 \cdot 34 > 153.$ $\endgroup$
    – heropup
    Commented Jul 9, 2022 at 6:58
  • $\begingroup$ @heropup : But $18*34$ happens to be $4*153$ and same multiplication factor $4$ for case $n=3$ and $r=5$, $50*74=4*925$. Perhaps a coincidence? Also : multiplication factor $4=2*2$ (as we had $1=1*1$). $\endgroup$ Commented Jul 9, 2022 at 14:17
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    $\begingroup$ @FirstNameLastName The triple $(4n^2 - 1, 4n, 4n^2 + 1)$ will satisfy $IA \cdot IC = 2 IB$. Again, I leave the proof of this as an exercise for the reader. Note that in the case $(3,4,5)$, this does not contradict your original finding since that was $IB \cdot IC = IA$ (the product of the shorter distances equals the longest). This is because $4n^2 - 1 < 4n $ when $n = 1$. $\endgroup$
    – heropup
    Commented Jul 9, 2022 at 15:59
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    $\begingroup$ @FirstNameLastName Using \begin{align*} &A=(2n-1)^2+&&2(2n-1)k\\ &B= &&2(2n-1)k+2k^2\\ &C=(2n-1)^2+&&2(2n-1)k+2k^2 \end{align*} perhaps there are interesting attributes when $\space n=k.$ \begin{array}{c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 \\ \hline \end{array} $\endgroup$
    – poetasis
    Commented Jul 9, 2022 at 16:53
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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.

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  • $\begingroup$ Your method also works for non rectangular triangles, i.e. for any triangle. I up-voted your answer. In fact one may now wonder if any non rectangular triangles exist having integer side lengths and satisfying the property. Of course @heropup's classification of such rectangular ones certainly is very neat. $\endgroup$ Commented Jul 9, 2022 at 11:30
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    $\begingroup$ @FirstNameLastName Your question about the existence of general integer-sided triangles (not necessarily Pythagorean) with $a < b < c$ satisfying $IB \cdot IC = IA$ is answered in the affirmative; e.g., $(7,15,20), (8,26,30)$. In general, if $$c = \frac{(a^2-1)b + a\sqrt{(a^2+1-2b)(a^2+1+2b)}}{a^2+1}$$ is an integer, then the desired property holds. I leave it to others to parametrize this family of solutions as it leads to a Diophantine condition that I have not examined in further detail. I recommend that you ask this as a separate question if you would like others to provide more analysis. $\endgroup$
    – heropup
    Commented Jul 9, 2022 at 19:28
  • $\begingroup$ @heropup : nice result! $\endgroup$ Commented Jul 9, 2022 at 20:33

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