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A point $P$ is marked within regular hexagon $ABCDEF$ so that we have $\angle BAP=\angle DCP = 50^\circ $ If $\angle APB$ has measure $x$ degrees, find $x$.

Here's a diagram:

enter image description here

Referring to the diagram, I've tried to do some angle chasing:

$\angle PAF=\angle PGE=\angle CPH=\angle PCB=70, \angle PGF=110, \angle GPA=HPC=60$, and all the interior angles are equal to $120$.

However, it seems like this isn't sufficient to solve the problem and I think I might need a construction of some sort - but I'm stuck here.

Assuming the solution needs more than just angle chasing, I'd appreciate constructions/observations and explanations of why they'd solve the problem.

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  • $\begingroup$ When you tag a question contest-math you should tell us what contest it's from. $\endgroup$ – Ethan Bolker Aug 25 '18 at 12:45
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Solution

As the figure shows, we may readily obtain that $$\angle OAP=\angle OCP=10^o,$$ which implies that $A,O,P,C$ are concyclic. But we know the center of the circle $(AOC)$ is $B$, hence $$BA=BP.$$ As a result, $$\angle APB=\angle BAP=50^o.$$

enter image description here

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  • $\begingroup$ Very nice, thank you. (I'd upvote if I could). Looking back, I guess you could also see that $B$ is the centre of $(APC)$ because the external angle of $ABC$ is twice $\angle APC$ and $BA=BC$. $\endgroup$ – user574848 Aug 25 '18 at 13:21
  • $\begingroup$ Notice that the center of a circle is unique. Now that $(AOC)$ and $(APC)$ are the same circle, and $B$ is the center of $(AOC)$, so it is also the center of $(APC).$ $\endgroup$ – mengdie1982 Aug 25 '18 at 13:24
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Suppose we construct an alternative point $P'$ in a slightly different way. Draw the line through $C$ of the hexagon such that $\angle DCG = 50$, same as in your diagram, and ignore all the rest. Then draw point $P'$ on that line such that $BC=BP'$. That gives this picture:

enter image description here

Using the fact that $AB=BC=BP'$ you now have isosceles triangles $BCP'$ and $BP'A$, and you should have no trouble finding all the angles in those two triangles.

Happily it turns out that $\angle BAP' = 50$, so that the point $P'$ is actually the same as point $P$, and $\angle APB = \angle AP'B$.

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  • $\begingroup$ This is clever! I presume you found the answer first then constructed around that? $\endgroup$ – user574848 Aug 25 '18 at 13:09

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