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Given any non-isosceles triangle $\triangle ABC$, and denoting $AB$ its longest side, the following construction enter image description here

determines the points $DFGE$ (see this post for details).

My conjecture is that if $\triangle ABC$ is an integer triangle, then the pentagon $DFCGE$ is a Robbins pentagon, and viceversa.

enter image description here

I am not sure of this claim. Therefore, I ask your advice both to assess if this is true and, otherwise,

to find the supplementary conditions needed.

Thanks for your help!

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    $\begingroup$ In a $3,4,5$ triangle where $AD=AF=2$ we get $DF=\sqrt{8/5}$, so the conjecture is false. There needs to be, at least, a condition that will force $DF$ and $EG$ to be rational. $\endgroup$ – nickgard Aug 3 '18 at 10:09
  • $\begingroup$ @nickgard Thanks!!! Please, can you sketch me how you got $DF=\sqrt{8/5}$? $\endgroup$ – user559615 Aug 3 '18 at 10:13
  • $\begingroup$ OK, I see it. Thanks again! $\endgroup$ – user559615 Aug 3 '18 at 10:20
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No, let $AD=x=2$ and $EB=z=1$ and $DE = y=2$. Then

$$\begin{eqnarray}c=x+y+z &=& 5\\ b=x+y &=& 4\\ a=y+z &=& 3\end{eqnarray}$$

Now, $$ \cos \alpha = {b^2+c^2-a^2\over 2bc} = {4\over 5}\in\mathbb{Q}$$ so $$DF^2 = 2x^2-2x^2\cos \alpha = x^2{2\over 5}\implies DF\notin \mathbb{Q}$$

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  • $\begingroup$ True, but if $AD=EB$ then $\triangle ABC$ is isosceles, isn't it? $\endgroup$ – user559615 Aug 3 '18 at 9:51
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    $\begingroup$ But I'm almost sure if you play a little with $x,y,z$ you will get negative result. $\endgroup$ – Aqua Aug 3 '18 at 9:52
  • $\begingroup$ Yes. I also believe it. But then, which other conditions may be needed to make the conjecture true? $\endgroup$ – user559615 Aug 3 '18 at 9:52
  • $\begingroup$ I edited also the title of the post, this way there's less confusion. $\endgroup$ – user559615 Aug 3 '18 at 10:07
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    $\begingroup$ I think that additional question is better to put as another question. $\endgroup$ – Aqua Aug 3 '18 at 10:23

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