1

The problem, asked 14 Feb at 18.18 by user @r ne, (see figure $1$ below) would appear to have been abandoned because of some adverse comment about “How to ask a good question”.

The attached figure $2$ below shows the triangle with sides $14, 15$ and $16$ in which the incenter and the centroid have been graphed. This triangle would seem to offer an answer to the above problem but it does not, the slopes in play appear as perpendicular to the eye but are not so with geometric accuracy.

If the problem had a solution, then taking as unknowns the three sides $a, b, c$ the coordinates of the incenter $I$ and centroid $G$ can be easily calculated so, using the known perpendicularity condition with the slopes $ m_ {IC} \cdot m_{IG} = -1$, we arrive at a polynomial equation in $a,b,c$. No matter of “diophantine” with the sides but of lengths of geometric segments taking any positive real value, we have the following question:

Prove or disprove that there is no triangle in the plane in which the aforementioned perpendicularity is fulfilled.

I have easily proven this and if I have made a mistake then it would not be subtle at all (instead something trivial like putting $2+3=6$ for example).

What strikes me is that a certain geometric intuition suggests that a small variation in the position of vertex $A$ of the given example (Figure $2$ below) of "almost perpendicularity" could "fix" the point, although with non-integer real sides, which does not matter in the problem at hand. In this case, it would be clear that my proof has an error that I cannot see and then the question would be devoid of any interest.

But if, as I believe, my proof is correct, then the apparent conflict between the geometric intuition in figure $2$ and the fact that such a triangle cannot exist must be explained in some way that I imagine not trivial in any respect. And I think this is due to $\pi$ which is linked to both incircle and perpendicularity.

enter image description here

cosmo5
  • 10,629
Piquito
  • 29,594
  • @cosmo5.- Yes dear friend. I tend to be careless sometimes. Regards. – Piquito Mar 14 '21 at 13:32
  • Thanks you very much for the hiperlink. I tried to use it but i could not to do. Believe me please (just in the same word "problem" I wanted to do). – Piquito Mar 14 '21 at 13:36

1 Answers1

0

An old problem answered here, asks to prove that $CI \perp IG \Rightarrow$ the following holds true.
$$6ab=(a+b)(a+b+c) \tag{A}$$

The additional conditions imposed by @rne are that $a,b,c$ be integers and $(a,b)=2$. $(16,14,15)$ does not satisfy $(A)$.

One can easily find, for example, $(a,b,c)=(14,10,11)$ satisfies all the above. Here is a drawing of this triangle in Geogebra with $\angle CIG$ measured after construction.

enter image description here

cosmo5
  • 10,629
  • Thanks for your answer. By the way, I note the absence of the remarkable MSE user Jack D'Aurizio. Your answer shows implicitely that my "proof" is wrong. If you want you can detecte my error in what follows.if you are so kind (I add another comment) – Piquito Mar 14 '21 at 13:24
  • Note in the triangle $\triangle{ABC}$ of the problem $B=(0,0), C=(a,0)$, the coordinates of $A=(X,Y)$ are determined by the system $$X^2+Y^2=c^2\(X-a)^2+Y^2=b^2$$ From this we have $$I=\left(\dfrac{(X+c)a}{p},\dfrac{aY}{p}\right)\G=\left(\dfrac{X+c}{3},\dfrac{Y}{3}\right)$$ where $p=a+b+c$ It follows $$m_{IC}=\frac{Y}{X+c-p}\m_{IG}=\frac{Y}{X+c}$$ Now the condition of perpendicularity $ m_{IC}\cdot m_{IG}=-1$ gives $$Y^2+X^2=(p-2c)X+pc-c^2$$ The system above gives $X=\dfrac{c^2+a^2-b^2}{2a}$ so we have $4ac^2=(p-2c)(c^2+a^2-b^2)+2acp$ from which $$(2c-p)((a+c)^2-b^2)=0$$ – Piquito Mar 14 '21 at 13:25
  • We get $c=a+b$ and $b=a+c$ which is not possible in a triangle. – Piquito Mar 14 '21 at 13:25
  • @Piquito G has x-coord $(X+a)/3$, this changes $m_{IG}$. – cosmo5 Mar 14 '21 at 13:36
  • @Oh my God!. Thanks you very much. Greetings. – Piquito Mar 14 '21 at 13:41
  • Welcome. This problem is also untouched. You might want to take a look at it. Regards. – cosmo5 Mar 14 '21 at 13:56
  • @cosmos.- Perfect. Is this kind of problems with no answers yet I like to study. I was one of the uncountable dreamers with FLT. Thanks you. – Piquito Mar 15 '21 at 13:50
  • 1
    @cosmos5: Hi cosmos5.The problem posted by the user learningmathematics is difficult really. The segment $\overline{ST}$ is approximately equal to $5.78$. I hope I can prove it. Regards. – Piquito Mar 15 '21 at 23:11