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The special right triangle with sides $1,\sqrt{3},2$ is well known. But consider the triangle with sides $1,\sqrt{2},2$:

enter image description here

Here $AB=2,BC=\sqrt{2},CA=1$. This seems uninteresting at first, but it turns out that this triangle can be inscribed in a regular heptagon, as shown, such that $BF \perp FE$.

enter image description here

Can you show why this is true?

I am interested in a purely geometric solution rather than the usage of coordinates.

The process should be reversible; that is, given $BF \perp FE$, we should be able to determine $BC=\sqrt{2}$ and $BA=2$.

I will add some progress that I made, I noticed that $BF$ is actually the angle bisector of $\angle IFH$. Not sure how this helps.

Perhaps there is something with heptagonal triangles. We might be able to do something with Stewart's theorem, but then this does not work for the shorter side.

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  • $\begingroup$ I don't understand why this is special. I could choose any point X on the side HI and then claim that the triangle ABX can be "inscribed" in a regular heptagon. $\endgroup$ Commented Oct 19 at 4:44
  • $\begingroup$ sory, shoul have specified that in this specific configuration, $BF \perp FE$. $\endgroup$ Commented Oct 19 at 4:45
  • $\begingroup$ You could perhaps use that $\cos(2 \pi/7), \cos(4 \pi/7), \cos(6 \pi/7)$ are the roots of the polynomial $p(x)=8x^3+4x^2-4x-1$. Then use Cardano's formula to compute exactly these cosines and the corresponding sines. Then compute all the points exactly. $\endgroup$ Commented Oct 19 at 8:49
  • $\begingroup$ I know, but I am interested in a purely geometric solution. $\endgroup$ Commented Oct 19 at 12:48
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    $\begingroup$ This is question 4/2/37 in the on-going USA Mathematical Talent Search contest. The rules state that participants "may not use 'live' help. For example, they may not ask for help on online forums...". $\endgroup$ Commented Oct 21 at 4:19

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