Internal angles in regular 18-gon
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This (seemingly simple) problem is driving me nuts.
Find angle $alpha$ shown in the following regular 18-gon.
It was easy to find the angle between pink diagonals ($60^circ$). And I was able to solve the problem with some trigonometry (getting nice integer angle). However, all my attempts to solve the problem without use of trigonometry have failed. It looked like I was close to solution all the time (so many angles are equal to $60^circ$ or $120^circ$. I felt like I had to draw just one more line and the problem would break apart. I also tried with internal symmetries and rotations but eventually I had to give up.
Is there a way to solve this kind of problem without sines and cosines?
euclidean-geometry polygons plane-geometry
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up vote
8
down vote
favorite
This (seemingly simple) problem is driving me nuts.
Find angle $alpha$ shown in the following regular 18-gon.
It was easy to find the angle between pink diagonals ($60^circ$). And I was able to solve the problem with some trigonometry (getting nice integer angle). However, all my attempts to solve the problem without use of trigonometry have failed. It looked like I was close to solution all the time (so many angles are equal to $60^circ$ or $120^circ$. I felt like I had to draw just one more line and the problem would break apart. I also tried with internal symmetries and rotations but eventually I had to give up.
Is there a way to solve this kind of problem without sines and cosines?
euclidean-geometry polygons plane-geometry
You should mention what the "nice integer angle" is, to save people some time.
– Blue
Dec 4 at 11:47
It's $20^circ$. Even more interesting is that if you extend the black line to the southeast it will pass through a vertex of the polygon.
– Oldboy
Dec 4 at 12:02
Say, the triangle with $alpha$ has vertices $A_1,A_2$ from the 18-gon. You claim the black line is $A_2A_8$. Once it's known, we can finish by observing $A_2A_8parallel A_3A_7$ and $A_1A_7A_3angle =20^circ$.
– Berci
Dec 7 at 8:25
add a comment |
up vote
8
down vote
favorite
up vote
8
down vote
favorite
This (seemingly simple) problem is driving me nuts.
Find angle $alpha$ shown in the following regular 18-gon.
It was easy to find the angle between pink diagonals ($60^circ$). And I was able to solve the problem with some trigonometry (getting nice integer angle). However, all my attempts to solve the problem without use of trigonometry have failed. It looked like I was close to solution all the time (so many angles are equal to $60^circ$ or $120^circ$. I felt like I had to draw just one more line and the problem would break apart. I also tried with internal symmetries and rotations but eventually I had to give up.
Is there a way to solve this kind of problem without sines and cosines?
euclidean-geometry polygons plane-geometry
This (seemingly simple) problem is driving me nuts.
Find angle $alpha$ shown in the following regular 18-gon.
It was easy to find the angle between pink diagonals ($60^circ$). And I was able to solve the problem with some trigonometry (getting nice integer angle). However, all my attempts to solve the problem without use of trigonometry have failed. It looked like I was close to solution all the time (so many angles are equal to $60^circ$ or $120^circ$. I felt like I had to draw just one more line and the problem would break apart. I also tried with internal symmetries and rotations but eventually I had to give up.
Is there a way to solve this kind of problem without sines and cosines?
euclidean-geometry polygons plane-geometry
euclidean-geometry polygons plane-geometry
asked Dec 4 at 10:51
Oldboy
6,1481628
6,1481628
You should mention what the "nice integer angle" is, to save people some time.
– Blue
Dec 4 at 11:47
It's $20^circ$. Even more interesting is that if you extend the black line to the southeast it will pass through a vertex of the polygon.
– Oldboy
Dec 4 at 12:02
Say, the triangle with $alpha$ has vertices $A_1,A_2$ from the 18-gon. You claim the black line is $A_2A_8$. Once it's known, we can finish by observing $A_2A_8parallel A_3A_7$ and $A_1A_7A_3angle =20^circ$.
– Berci
Dec 7 at 8:25
add a comment |
You should mention what the "nice integer angle" is, to save people some time.
– Blue
Dec 4 at 11:47
It's $20^circ$. Even more interesting is that if you extend the black line to the southeast it will pass through a vertex of the polygon.
– Oldboy
Dec 4 at 12:02
Say, the triangle with $alpha$ has vertices $A_1,A_2$ from the 18-gon. You claim the black line is $A_2A_8$. Once it's known, we can finish by observing $A_2A_8parallel A_3A_7$ and $A_1A_7A_3angle =20^circ$.
– Berci
Dec 7 at 8:25
You should mention what the "nice integer angle" is, to save people some time.
– Blue
Dec 4 at 11:47
You should mention what the "nice integer angle" is, to save people some time.
– Blue
Dec 4 at 11:47
It's $20^circ$. Even more interesting is that if you extend the black line to the southeast it will pass through a vertex of the polygon.
– Oldboy
Dec 4 at 12:02
It's $20^circ$. Even more interesting is that if you extend the black line to the southeast it will pass through a vertex of the polygon.
– Oldboy
Dec 4 at 12:02
Say, the triangle with $alpha$ has vertices $A_1,A_2$ from the 18-gon. You claim the black line is $A_2A_8$. Once it's known, we can finish by observing $A_2A_8parallel A_3A_7$ and $A_1A_7A_3angle =20^circ$.
– Berci
Dec 7 at 8:25
Say, the triangle with $alpha$ has vertices $A_1,A_2$ from the 18-gon. You claim the black line is $A_2A_8$. Once it's known, we can finish by observing $A_2A_8parallel A_3A_7$ and $A_1A_7A_3angle =20^circ$.
– Berci
Dec 7 at 8:25
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You should mention what the "nice integer angle" is, to save people some time.
– Blue
Dec 4 at 11:47
It's $20^circ$. Even more interesting is that if you extend the black line to the southeast it will pass through a vertex of the polygon.
– Oldboy
Dec 4 at 12:02
Say, the triangle with $alpha$ has vertices $A_1,A_2$ from the 18-gon. You claim the black line is $A_2A_8$. Once it's known, we can finish by observing $A_2A_8parallel A_3A_7$ and $A_1A_7A_3angle =20^circ$.
– Berci
Dec 7 at 8:25