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:
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.
geometry contest-math polygons angle
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up vote
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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:
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.
geometry contest-math polygons angle
When you tag a question contest-math you should tell us what contest it's from.
â Ethan Bolker
Aug 25 at 12:45
add a comment |Â
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:
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.
geometry contest-math polygons angle
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:
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.
geometry contest-math polygons angle
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.
â Ethan Bolker
Aug 25 at 12:45
add a comment |Â
When you tag a question contest-math you should tell us what contest it's from.
â Ethan Bolker
Aug 25 at 12:45
When you tag a question contest-math you should tell us what contest it's from.
â Ethan Bolker
Aug 25 at 12:45
When you tag a question contest-math you should tell us what contest it's from.
â Ethan Bolker
Aug 25 at 12:45
add a comment |Â
2 Answers
2
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, $$angle APB=angle BAP=50^o.$$
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 $BA=BC$.
â 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 $(APC).$
â mengdie1982
Aug 25 at 13:24
add a comment |Â
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:
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 $angle APB = angle AP'B$.
This is clever! I presume you found the answer first then constructed around that?
â user
Aug 25 at 13:09
add a comment |Â
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, $$angle APB=angle BAP=50^o.$$
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 $BA=BC$.
â 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 $(APC).$
â mengdie1982
Aug 25 at 13:24
add a comment |Â
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, $$angle APB=angle BAP=50^o.$$
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 $BA=BC$.
â 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 $(APC).$
â mengdie1982
Aug 25 at 13:24
add a comment |Â
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, $$angle APB=angle BAP=50^o.$$
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, $$angle APB=angle BAP=50^o.$$
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 $BA=BC$.
â 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 $(APC).$
â mengdie1982
Aug 25 at 13:24
add a comment |Â
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 $BA=BC$.
â 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 $(APC).$
â 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 $BA=BC$.
â 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 $BA=BC$.
â 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 $(APC).$
â 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 $(APC).$
â mengdie1982
Aug 25 at 13:24
add a comment |Â
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:
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 $angle APB = angle AP'B$.
This is clever! I presume you found the answer first then constructed around that?
â user
Aug 25 at 13:09
add a comment |Â
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:
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 $angle APB = angle AP'B$.
This is clever! I presume you found the answer first then constructed around that?
â user
Aug 25 at 13:09
add a comment |Â
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:
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 $angle APB = angle AP'B$.
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:
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 $angle APB = angle AP'B$.
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?
â user
Aug 25 at 13:09
add a comment |Â
This is clever! I presume you found the answer first then constructed around that?
â user
Aug 25 at 13:09
This is clever! I presume you found the answer first then constructed around that?
â user
Aug 25 at 13:09
This is clever! I presume you found the answer first then constructed around that?
â user
Aug 25 at 13:09
add a comment |Â
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When you tag a question contest-math you should tell us what contest it's from.
â Ethan Bolker
Aug 25 at 12:45