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$$\dfrac{yz(y+z-x)}{x+y+z}=p \tag1$$ $$\dfrac{xz(x+z-y)}{x+y+z}=q \tag2$$ $$\dfrac{xy(x+y-z)}{x+y+z}=r \tag3$$

This group of equations comes from the problem "if the distances of incenter of an triangle to its vertices are $\sqrt{p}$, $\sqrt{q}$ and $\sqrt{r}$. Find the length of sides of the triangle". By applying barycentric coordinates, it is easy to get the group of equations above, with $x$, $y$ and $z$ as three sides of the triangle.

I have no idea how to solve this group of equations now. Any hints? Thanks!

emacs drives me nuts
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r ne
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1 Answers1

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Rearranging, we get $$p+\frac{2xyz}{x+y+z}=yz$$ $$q+\frac{2xyz}{x+y+z}=xz$$ $$r+\frac{2xyz}{x+y+z}=xy$$ if we multiply two of these equations and divide by the third, $$\implies x=\sqrt{\frac{(q+a)(r+a)}{(p+a)}}$$ $$\implies y=\sqrt{\frac{(r+a)(p+a)}{(q+a)}}$$ $$\implies z=\sqrt{\frac{(p+a)(q+a)}{(r+a)}}$$ where $a=2xyz/(x+y+z)$ must satisfy a cubic equation $$a^3-(pq+qr+rp)a-2pqr=0$$

The cubic comes from substituting the expressions for $x,y,z$ into $a=\frac{2xyz}{x+y+z}$ and rearranging.

$$a=\frac{2\sqrt{\frac{(q+a)(r+a)}{(p+a)}}\sqrt{\frac{(r+a)(p+a)}{(q+a)}}\sqrt{\frac{(p+a)(q+a)}{(r+a)}}}{\sqrt{\frac{(q+a)(r+a)}{(p+a)}}+\sqrt{\frac{(r+a)(p+a)}{(q+a)}}+\sqrt{\frac{(p+a)(q+a)}{(r+a)}}}$$ $$=\frac{2\sqrt{(q+a)(r+a)}\sqrt{(r+a)(p+a)}\sqrt{(p+a)(q+a)}}{(q+a)(r+a)+(r+a)(p+a)+(p+a)(q+a)}$$ $$=\frac{2(p+a)(q+a)(r+a)}{(q+a)(r+a)+(r+a)(p+a)+(p+a)(q+a)}$$

So

$$3a^3+2(p+q+r)a^2+(pq+qr+rp)a=2a^3+2(p+q+r)a^2+2(pq+qr+rp)a+2pqr$$

Joshua Tilley
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  • Thanks for the reply. Can you explain how did you get the cubic equation please? – r ne Sep 13 '22 at 09:57
  • Thanks for adding the details. I got the same result for the new cubic equation. – r ne Sep 13 '22 at 11:48
  • I made a mistake the first time. If you are happy with the answer, you can accept it perhaps? :) – Joshua Tilley Sep 13 '22 at 12:14
  • I was wondering if the three equations can get values of $x,y,z$ but seems I have not got any better progress than this. – r ne Sep 13 '22 at 16:43
  • What do you mean? Is this not what the answer provides? – Joshua Tilley Sep 13 '22 at 17:22
  • The problem asks for values of $x,y,z$, if feasible. There is no provided answer as far as I know. If not feasible, then maybe get the value of $x+y+z$. – r ne Sep 13 '22 at 18:06
  • Just a nitpick, but I'd expect to see a +- sign before each of the square-root signs in the second block of equations. I understand that we're talking about sides of a triangle, but the question is just how to solve the equations. – Dawood ibn Kareem Sep 13 '22 at 20:17
  • @DawoodibnKareem This is fine, since the equation was originated from triangle sides and distances between incenter and vertices. Sure you are right to make it rigorous in formal expressions. Thanks! – r ne Sep 13 '22 at 22:22