hexagon ABCDEF with <A+<C+<E=<B+<D+<F 2001 Estonia Open Senior 2.3

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hexagon ABCDEF with <A+<C+<E=<B+<D+<F 2001 Estonia Open Senior 2.3

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#1 h Sep 22, 2020, 4:00 PM • 3 Y

Y by Mango247, Mango247, Mango247

Let us call a convex hexagon $ABCDEF$ boring if $\angle A+ \angle C + \angle E = \angle B + \angle D + \angle F$ .
a) Is every cyclic hexagon boring?
b) Is every boring hexagon cyclic?

This post has been edited 2 times. Last edited by parmenides51, Sep 23, 2020, 3:15 PM

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#2 h Oct 9, 2020, 1:18 AM

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(a) Draw ALL the diagonals. Call $\theta_{XY}$ the angle inscribed in arc $\overarc{\(XY\)}$ . For example, we get $\angle A = \theta_{BC} + \theta_{CD} + \theta_{DE} + \theta_{EF}.$ If we do the same thing for $\angle B$ , $\angle C$ , $\angle D$ , $\angle E$ and $\angle F$ , we find that
$\angle A = \theta_{BC} + \theta_{CD} + \theta_{DE} + \theta_{EF}.$
$\angle B = \theta_{CD} + \theta_{DE} + \theta_{EF}+\theta_{FA} .$
$\angle C = \theta_{DE} + \theta_{EF}+\theta_{FA}+\theta_{AB} .$
$\angle D= \theta_{EF}+\theta_{FA}+\theta_{AB}+\theta_{BC} .$
$\angle E= \theta_{FA}+\theta_{AB}+\theta_{BC} + \theta_{CD} .$
$\angle F= \theta_{AB}+\theta_{BC} + \theta_{CD}+\theta_{DE} .$

Now, we can rewrite the equation $\angle A + \angle C + \angle E = \angle B + \angle D + \angle F$ as $\theta_{BC} + \theta_{CD} + \theta_{DE} + \theta_{EF}+ \theta_{DE} + \theta_{EF}+\theta_{FA}+\theta_{AB} +\theta_{FA}+\theta_{AB}+\theta_{BC} + \theta_{CD}= \theta_{CD} + \theta_{DE} + \theta_{EF}+\theta_{FA}+\theta_{EF}+\theta_{FA}+\theta_{AB}+\theta_{BC}+ \theta_{AB}+\theta_{BC} + \theta_{CD}+\theta_{DE}$ which is true after mass cancellation.

This post has been edited 1 time. Last edited by franzliszt, Oct 9, 2020, 1:21 AM

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#3 h Oct 9, 2020, 2:07 AM

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answer for (b)

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#4 h Oct 9, 2020, 1:15 PM

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Thanks for part (b). I was trying sooooo hard to show it was true, but should have found a counter example

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#5 h Oct 9, 2020, 1:39 PM • 2 Y

Y by Mango247, Mango247

easier proof for (a).
$\angle BAD+\angle BCD=\angle ABC+\angle ADC$ ;
$\angle DAF+\angle DEF=\angle ADE+\angle AFE$ .
add this, you will get it.

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#6 h Oct 9, 2020, 1:48 PM

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Lol. Looking at @franzliszt solution, I would've gotten a headache just by typing it.

This post has been edited 1 time. Last edited by eduD_looC, Oct 9, 2020, 1:48 PM

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