Angle chasing with a marked point within a regular hexagon

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














up vote
2
down vote

favorite












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.







share|cite|improve this question






















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












up vote
2
down vote

favorite









up vote
2
down vote

favorite











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.







share|cite|improve this question














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.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 25 at 12:02









John Glenn

1,884324




1,884324










asked Aug 25 at 9:48









user

527




527











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
















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















When you tag a question contest-math you should tell us what contest it's from.
– Ethan Bolker
Aug 25 at 12:45




When you tag a question contest-math you should tell us what contest it's from.
– Ethan Bolker
Aug 25 at 12:45










2 Answers
2






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oldest

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up vote
0
down vote



accepted










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






share|cite|improve this answer




















  • 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 $BA=BC$.
    – user
    Aug 25 at 13:21











  • 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 $(APC).$
    – mengdie1982
    Aug 25 at 13:24


















up vote
2
down vote













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 $angle APB = angle AP'B$.






share|cite|improve this answer






















  • This is clever! I presume you found the answer first then constructed around that?
    – user
    Aug 25 at 13:09










Your Answer




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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










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






share|cite|improve this answer




















  • 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 $BA=BC$.
    – user
    Aug 25 at 13:21











  • 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 $(APC).$
    – mengdie1982
    Aug 25 at 13:24















up vote
0
down vote



accepted










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






share|cite|improve this answer




















  • 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 $BA=BC$.
    – user
    Aug 25 at 13:21











  • 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 $(APC).$
    – mengdie1982
    Aug 25 at 13:24













up vote
0
down vote



accepted







up vote
0
down vote



accepted






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






share|cite|improve this answer












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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 25 at 13:17









mengdie1982

3,628216




3,628216











  • 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 $BA=BC$.
    – user
    Aug 25 at 13:21











  • 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 $(APC).$
    – mengdie1982
    Aug 25 at 13:24

















  • 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 $BA=BC$.
    – user
    Aug 25 at 13:21











  • 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 $(APC).$
    – mengdie1982
    Aug 25 at 13:24
















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 $BA=BC$.
– user
Aug 25 at 13:21





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 $BA=BC$.
– user
Aug 25 at 13:21













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 $(APC).$
– mengdie1982
Aug 25 at 13:24





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 $(APC).$
– mengdie1982
Aug 25 at 13:24











up vote
2
down vote













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 $angle APB = angle AP'B$.






share|cite|improve this answer






















  • This is clever! I presume you found the answer first then constructed around that?
    – user
    Aug 25 at 13:09














up vote
2
down vote













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 $angle APB = angle AP'B$.






share|cite|improve this answer






















  • This is clever! I presume you found the answer first then constructed around that?
    – user
    Aug 25 at 13:09












up vote
2
down vote










up vote
2
down vote









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 $angle APB = angle AP'B$.






share|cite|improve this answer














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 $angle APB = angle AP'B$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 25 at 12:47

























answered Aug 25 at 12:41









Jaap Scherphuis

3,648616




3,648616











  • This is clever! I presume you found the answer first then constructed around that?
    – user
    Aug 25 at 13:09
















  • This is clever! I presume you found the answer first then constructed around that?
    – user
    Aug 25 at 13:09















This is clever! I presume you found the answer first then constructed around that?
– user
Aug 25 at 13:09




This is clever! I presume you found the answer first then constructed around that?
– user
Aug 25 at 13:09

















 

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