Let $A_1A_2A_3A_4A_5A_6$ a hexagon and $A_6 A_2 \cap A_3 A_1=B_1$ , $A_1 A_3 \cap A_2 A_4=B_2$, $ A_2 A_4 \cap A_3 A_5=B_3,$, $A_3 A_5\cap A_4 A_6=B_4$, $A_4 A_6\cap A_5 A_1=B_5$, $A_5 A_1\cap A_6 A_2=B_6$.

If $$\frac {B_2B_3}{A_2A_4}=\frac {B_4B_5}{A_4A_6}= \frac {B_6B_1}{A_6A_2}$$ show that $$\frac {B_1B_2}{A_1A_3}=\frac {B_3B_4}{A_3A_5}=\frac {B_5B_6}{A_5A_1} .$$

I tried to prove it withs areas of triangles.

  • $\begingroup$ You really ought to have a picture with this one. $\endgroup$ – Paul Sinclair Sep 22 '19 at 0:20
  • $\begingroup$ @Paul Sinclair I don't agree with you. We can draw a picture easily. $\endgroup$ – Michael Rozenberg Sep 22 '19 at 10:40
  • $\begingroup$ @MichaelRozenberg - but they likely would have gotten answers in an hour or two instead of 21 hours if a picture had been included. By leaving the picture out, they severely cut down on the number of people who were willing to attempt the problem by putting the burden of interpreting the symbolism on the reader. $\endgroup$ – Paul Sinclair Sep 22 '19 at 15:39

Hint: The figure below is not to scale:

$\hspace{1cm}$enter image description here

Note ($B_3B_4B_5B_5'$ is a parallelogram): $$\frac {B_2B_3}{A_2A_4}=\frac {B_4B_5}{A_4A_6} \Rightarrow \\ \frac {A_2A_4}{A_4A_6}=\frac {B_2B_3}{B_4B_5}=\frac {B_2B_3}{B_3B_5'}, \angle B_2B_3B_5'=\angle A_2A_4A_6 \Rightarrow \\ \Delta A_2A_4A_6\sim \Delta B_2B_3B_5' \Rightarrow \\ B_2B_5'||B_1B_6.$$


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