Mixing
Internal angles in regular 18-gon
$begingroup$
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 angle
edited Jan 20 at 13:32
Rosie F
1,349416
asked Dec 4 '18 at 10:51
OldboyOldboy
8,59011036
$endgroup$
add a comment |
$begingroup$
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 angle
edited Jan 20 at 13:32
Rosie F
1,349416
asked Dec 4 '18 at 10:51
OldboyOldboy
8,59011036
$endgroup$
$begingroup$
You should mention what the "nice integer angle" is, to save people some time.
$endgroup$
– Blue
Dec 4 '18 at 11:47
$begingroup$
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.
$endgroup$
– Oldboy
Dec 4 '18 at 12:02
$begingroup$
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 $A_1A_7A_3angle =20^circ$.
$endgroup$
– Berci
Dec 7 '18 at 8:25
add a comment |
$begingroup$
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 angle
edited Jan 20 at 13:32
Rosie F
1,349416
asked Dec 4 '18 at 10:51
OldboyOldboy
8,59011036
$endgroup$
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 angle
euclidean-geometry polygons plane-geometry angle
edited Jan 20 at 13:32
Rosie F
1,349416
asked Dec 4 '18 at 10:51
OldboyOldboy
8,59011036
edited Jan 20 at 13:32
Rosie F
1,349416
asked Dec 4 '18 at 10:51
OldboyOldboy
8,59011036
edited Jan 20 at 13:32
Rosie F
1,349416
edited Jan 20 at 13:32
Rosie F
1,349416
edited Jan 20 at 13:32
Rosie F
1,349416
1,349416
asked Dec 4 '18 at 10:51
OldboyOldboy
8,59011036
asked Dec 4 '18 at 10:51
OldboyOldboy
8,59011036
asked Dec 4 '18 at 10:51
OldboyOldboy
8,59011036
8,59011036
$begingroup$
You should mention what the "nice integer angle" is, to save people some time.
$endgroup$
– Blue
Dec 4 '18 at 11:47
$begingroup$
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.
$endgroup$
– Oldboy
Dec 4 '18 at 12:02
$begingroup$
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 $A_1A_7A_3angle =20^circ$.
$endgroup$
– Berci
Dec 7 '18 at 8:25
add a comment |
$begingroup$
You should mention what the "nice integer angle" is, to save people some time.
$endgroup$
– Blue
Dec 4 '18 at 11:47
$begingroup$
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.
$endgroup$
– Oldboy
Dec 4 '18 at 12:02
$begingroup$
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 $A_1A_7A_3angle =20^circ$.
$endgroup$
– Berci
Dec 7 '18 at 8:25
$begingroup$
You should mention what the "nice integer angle" is, to save people some time.
$endgroup$
– Blue
Dec 4 '18 at 11:47
$begingroup$
You should mention what the "nice integer angle" is, to save people some time.
$endgroup$
– Blue
Dec 4 '18 at 11:47
$begingroup$
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.
$endgroup$
– Oldboy
Dec 4 '18 at 12:02
$begingroup$
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.
$endgroup$
– Oldboy
Dec 4 '18 at 12:02
$begingroup$
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 $A_1A_7A_3angle =20^circ$.
$endgroup$
– Berci
Dec 7 '18 at 8:25
$begingroup$
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 $A_1A_7A_3angle =20^circ$.
$endgroup$
– Berci
Dec 7 '18 at 8:25
add a comment |
1 Answer
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I don't have a direct solution. The best I have relates to a different problem, so I show the latter's solution, then show how the two problems correspond.
Lemma. Let $OAC$ be a triangle, and $X$ a point in it, and let angles $CAX=40^{circ}, XAO=10^{circ}, AOX=10^{circ}, XOC=70^{circ}$. What is $angle OCX$?
Solution of lemma.
$angle AOC=80^{circ}$ and $angle CAO=50^{circ}$ so $angle OCA=50^{circ}= angle CAO$ so $triangle ACO$ is isosceles on base $AC$, so $OA=OC$.
$angle AOX = 10^{circ} = angle XAO$, so $triangle OAX$ is isosceles on base $OA$, so $AX=OX$.
$angle OXA = 180^{circ} - angle XAO - angle AOX = 160^{circ}$.
Erect an equilateral $triangle OXE$ on base $OX$. Join $CE$.
$angle EOC = angle XOC - angle XOE = 70^{circ} - 60^{circ} = 10^{circ} = angle AOX$.
Thus $triangle$s $OAX, OCE$ are congruent (SAS, opposite sense) because $angle EOC = angle AOX, OA=OC$ and $OX=OE$.
Thus $angle OCE = angle XAO = 10^{circ}$.
$CE=AX=OX=XE$.
Thus $triangle XCE$ is isosceles on base $XC$.
$angle XEC = 360^{circ} - angle CEO - angle OEX = 360^{circ} - 160^{circ} - 60^{circ} = 140^{circ}$.
Thus $angle ECX = (180^{circ}-140^{circ})/2 = 20^{circ}$.
Thus $angle OCX = angle ECX + angle OCE = 20^{circ}+10^{circ} = 30^{circ}$, which solves the lemma.
Solution of the original problem.
Let $P_0dots P_{17}$ be a regular 18-gon. Let $P_1P_{13}$ cross $P_{5}P_{15}$ at $X$. What is $alpha=angle P_0XP_1$?
Let the centre of the 18-gon be $O$. $triangle P_0P_{1}O$ is isosceles on base $P_{0}P_1$, and $angle P_0OP_1=20^circ$, so $angle OP_1P_0=80^circ$.
$triangle P_{13}P_1O$ is isosceles on base $P_{13}P_1$, and $angle P_{13}OP_1=120^circ$, so $angle OP_1P_{13}=30^circ=angle OP_1X$ as $X$ is on $P_1P_{13}$.
$triangle P_{15}P_5O$ is isosceles on base $P_{15}P_5$, and $angle P_{15}OP_5=160^circ$, so $angle P_5P_{15}O=10^circ=angle XP_{15}O$ as $X$ is on $P_5P_{15}$.
$triangle P_{15}P_1O$ is isosceles on base $P_{15}P_1$, and $angle P_{15}OP_1=80^circ$, so $angle P_1P_{15}O=50^circ$. Thus this $triangle$ has the same angles $80^circ, 50^circ, 50^circ$ as the lemma's $triangle OAC$, so they are similar. Moreover, the angles $OP_1X=30^circ$ and $XP_{15}O=10^circ$ correspond to those in the lemma, so the $X$s correspond. Therefore, by the lemma, $OX=XP_{15}$.
$OP_0=OP_{15}$ and $angle P_{15}OP_0=60^circ$ so $triangle OP_{15}P_0$ is equilateral. Thus $triangle$s $OXP_0$ and $P_{15}XP_0$ are congruent in opposite senses (SSS) because $OX=XP_{15}$, $OP_0=P_{15}P_0$ and $P_0X$ is common. Thus $angle OP_0X=angle XP_0P_{15}$ so $angle OP_0X=30^circ=angle OP_1X$, so quadrilateral $OXP_0P_1$ is cyclic, so $alpha=angle P_0XP_1=angle P_0OP_1=20^circ$, which solves the problem.
answered Jan 19 at 12:19
Rosie FRosie F
1,349416
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add a comment |
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$begingroup$
I don't have a direct solution. The best I have relates to a different problem, so I show the latter's solution, then show how the two problems correspond.
Lemma. Let $OAC$ be a triangle, and $X$ a point in it, and let angles $CAX=40^{circ}, XAO=10^{circ}, AOX=10^{circ}, XOC=70^{circ}$. What is $angle OCX$?
Solution of lemma.
$angle AOC=80^{circ}$ and $angle CAO=50^{circ}$ so $angle OCA=50^{circ}= angle CAO$ so $triangle ACO$ is isosceles on base $AC$, so $OA=OC$.
$angle AOX = 10^{circ} = angle XAO$, so $triangle OAX$ is isosceles on base $OA$, so $AX=OX$.
$angle OXA = 180^{circ} - angle XAO - angle AOX = 160^{circ}$.
Erect an equilateral $triangle OXE$ on base $OX$. Join $CE$.
$angle EOC = angle XOC - angle XOE = 70^{circ} - 60^{circ} = 10^{circ} = angle AOX$.
Thus $triangle$s $OAX, OCE$ are congruent (SAS, opposite sense) because $angle EOC = angle AOX, OA=OC$ and $OX=OE$.
Thus $angle OCE = angle XAO = 10^{circ}$.
$CE=AX=OX=XE$.
Thus $triangle XCE$ is isosceles on base $XC$.
$angle XEC = 360^{circ} - angle CEO - angle OEX = 360^{circ} - 160^{circ} - 60^{circ} = 140^{circ}$.
Thus $angle ECX = (180^{circ}-140^{circ})/2 = 20^{circ}$.
Thus $angle OCX = angle ECX + angle OCE = 20^{circ}+10^{circ} = 30^{circ}$, which solves the lemma.
Solution of the original problem.
Let $P_0dots P_{17}$ be a regular 18-gon. Let $P_1P_{13}$ cross $P_{5}P_{15}$ at $X$. What is $alpha=angle P_0XP_1$?
Let the centre of the 18-gon be $O$. $triangle P_0P_{1}O$ is isosceles on base $P_{0}P_1$, and $angle P_0OP_1=20^circ$, so $angle OP_1P_0=80^circ$.
$triangle P_{13}P_1O$ is isosceles on base $P_{13}P_1$, and $angle P_{13}OP_1=120^circ$, so $angle OP_1P_{13}=30^circ=angle OP_1X$ as $X$ is on $P_1P_{13}$.
$triangle P_{15}P_5O$ is isosceles on base $P_{15}P_5$, and $angle P_{15}OP_5=160^circ$, so $angle P_5P_{15}O=10^circ=angle XP_{15}O$ as $X$ is on $P_5P_{15}$.
$triangle P_{15}P_1O$ is isosceles on base $P_{15}P_1$, and $angle P_{15}OP_1=80^circ$, so $angle P_1P_{15}O=50^circ$. Thus this $triangle$ has the same angles $80^circ, 50^circ, 50^circ$ as the lemma's $triangle OAC$, so they are similar. Moreover, the angles $OP_1X=30^circ$ and $XP_{15}O=10^circ$ correspond to those in the lemma, so the $X$s correspond. Therefore, by the lemma, $OX=XP_{15}$.
$OP_0=OP_{15}$ and $angle P_{15}OP_0=60^circ$ so $triangle OP_{15}P_0$ is equilateral. Thus $triangle$s $OXP_0$ and $P_{15}XP_0$ are congruent in opposite senses (SSS) because $OX=XP_{15}$, $OP_0=P_{15}P_0$ and $P_0X$ is common. Thus $angle OP_0X=angle XP_0P_{15}$ so $angle OP_0X=30^circ=angle OP_1X$, so quadrilateral $OXP_0P_1$ is cyclic, so $alpha=angle P_0XP_1=angle P_0OP_1=20^circ$, which solves the problem.
answered Jan 19 at 12:19
Rosie FRosie F
1,349416
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add a comment |
$begingroup$
I don't have a direct solution. The best I have relates to a different problem, so I show the latter's solution, then show how the two problems correspond.
Lemma. Let $OAC$ be a triangle, and $X$ a point in it, and let angles $CAX=40^{circ}, XAO=10^{circ}, AOX=10^{circ}, XOC=70^{circ}$. What is $angle OCX$?
Solution of lemma.
$angle AOC=80^{circ}$ and $angle CAO=50^{circ}$ so $angle OCA=50^{circ}= angle CAO$ so $triangle ACO$ is isosceles on base $AC$, so $OA=OC$.
$angle AOX = 10^{circ} = angle XAO$, so $triangle OAX$ is isosceles on base $OA$, so $AX=OX$.
$angle OXA = 180^{circ} - angle XAO - angle AOX = 160^{circ}$.
Erect an equilateral $triangle OXE$ on base $OX$. Join $CE$.
$angle EOC = angle XOC - angle XOE = 70^{circ} - 60^{circ} = 10^{circ} = angle AOX$.
Thus $triangle$s $OAX, OCE$ are congruent (SAS, opposite sense) because $angle EOC = angle AOX, OA=OC$ and $OX=OE$.
Thus $angle OCE = angle XAO = 10^{circ}$.
$CE=AX=OX=XE$.
Thus $triangle XCE$ is isosceles on base $XC$.
$angle XEC = 360^{circ} - angle CEO - angle OEX = 360^{circ} - 160^{circ} - 60^{circ} = 140^{circ}$.
Thus $angle ECX = (180^{circ}-140^{circ})/2 = 20^{circ}$.
Thus $angle OCX = angle ECX + angle OCE = 20^{circ}+10^{circ} = 30^{circ}$, which solves the lemma.
Solution of the original problem.
Let $P_0dots P_{17}$ be a regular 18-gon. Let $P_1P_{13}$ cross $P_{5}P_{15}$ at $X$. What is $alpha=angle P_0XP_1$?
Let the centre of the 18-gon be $O$. $triangle P_0P_{1}O$ is isosceles on base $P_{0}P_1$, and $angle P_0OP_1=20^circ$, so $angle OP_1P_0=80^circ$.
$triangle P_{13}P_1O$ is isosceles on base $P_{13}P_1$, and $angle P_{13}OP_1=120^circ$, so $angle OP_1P_{13}=30^circ=angle OP_1X$ as $X$ is on $P_1P_{13}$.
$triangle P_{15}P_5O$ is isosceles on base $P_{15}P_5$, and $angle P_{15}OP_5=160^circ$, so $angle P_5P_{15}O=10^circ=angle XP_{15}O$ as $X$ is on $P_5P_{15}$.
$triangle P_{15}P_1O$ is isosceles on base $P_{15}P_1$, and $angle P_{15}OP_1=80^circ$, so $angle P_1P_{15}O=50^circ$. Thus this $triangle$ has the same angles $80^circ, 50^circ, 50^circ$ as the lemma's $triangle OAC$, so they are similar. Moreover, the angles $OP_1X=30^circ$ and $XP_{15}O=10^circ$ correspond to those in the lemma, so the $X$s correspond. Therefore, by the lemma, $OX=XP_{15}$.
$OP_0=OP_{15}$ and $angle P_{15}OP_0=60^circ$ so $triangle OP_{15}P_0$ is equilateral. Thus $triangle$s $OXP_0$ and $P_{15}XP_0$ are congruent in opposite senses (SSS) because $OX=XP_{15}$, $OP_0=P_{15}P_0$ and $P_0X$ is common. Thus $angle OP_0X=angle XP_0P_{15}$ so $angle OP_0X=30^circ=angle OP_1X$, so quadrilateral $OXP_0P_1$ is cyclic, so $alpha=angle P_0XP_1=angle P_0OP_1=20^circ$, which solves the problem.
answered Jan 19 at 12:19
Rosie FRosie F
1,349416
$endgroup$
add a comment |
$begingroup$
I don't have a direct solution. The best I have relates to a different problem, so I show the latter's solution, then show how the two problems correspond.
Lemma. Let $OAC$ be a triangle, and $X$ a point in it, and let angles $CAX=40^{circ}, XAO=10^{circ}, AOX=10^{circ}, XOC=70^{circ}$. What is $angle OCX$?
Solution of lemma.
$angle AOC=80^{circ}$ and $angle CAO=50^{circ}$ so $angle OCA=50^{circ}= angle CAO$ so $triangle ACO$ is isosceles on base $AC$, so $OA=OC$.
$angle AOX = 10^{circ} = angle XAO$, so $triangle OAX$ is isosceles on base $OA$, so $AX=OX$.
$angle OXA = 180^{circ} - angle XAO - angle AOX = 160^{circ}$.
Erect an equilateral $triangle OXE$ on base $OX$. Join $CE$.
$angle EOC = angle XOC - angle XOE = 70^{circ} - 60^{circ} = 10^{circ} = angle AOX$.
Thus $triangle$s $OAX, OCE$ are congruent (SAS, opposite sense) because $angle EOC = angle AOX, OA=OC$ and $OX=OE$.
Thus $angle OCE = angle XAO = 10^{circ}$.
$CE=AX=OX=XE$.
Thus $triangle XCE$ is isosceles on base $XC$.
$angle XEC = 360^{circ} - angle CEO - angle OEX = 360^{circ} - 160^{circ} - 60^{circ} = 140^{circ}$.
Thus $angle ECX = (180^{circ}-140^{circ})/2 = 20^{circ}$.
Thus $angle OCX = angle ECX + angle OCE = 20^{circ}+10^{circ} = 30^{circ}$, which solves the lemma.
Solution of the original problem.
Let $P_0dots P_{17}$ be a regular 18-gon. Let $P_1P_{13}$ cross $P_{5}P_{15}$ at $X$. What is $alpha=angle P_0XP_1$?
Let the centre of the 18-gon be $O$. $triangle P_0P_{1}O$ is isosceles on base $P_{0}P_1$, and $angle P_0OP_1=20^circ$, so $angle OP_1P_0=80^circ$.
$triangle P_{13}P_1O$ is isosceles on base $P_{13}P_1$, and $angle P_{13}OP_1=120^circ$, so $angle OP_1P_{13}=30^circ=angle OP_1X$ as $X$ is on $P_1P_{13}$.
$triangle P_{15}P_5O$ is isosceles on base $P_{15}P_5$, and $angle P_{15}OP_5=160^circ$, so $angle P_5P_{15}O=10^circ=angle XP_{15}O$ as $X$ is on $P_5P_{15}$.
$triangle P_{15}P_1O$ is isosceles on base $P_{15}P_1$, and $angle P_{15}OP_1=80^circ$, so $angle P_1P_{15}O=50^circ$. Thus this $triangle$ has the same angles $80^circ, 50^circ, 50^circ$ as the lemma's $triangle OAC$, so they are similar. Moreover, the angles $OP_1X=30^circ$ and $XP_{15}O=10^circ$ correspond to those in the lemma, so the $X$s correspond. Therefore, by the lemma, $OX=XP_{15}$.
$OP_0=OP_{15}$ and $angle P_{15}OP_0=60^circ$ so $triangle OP_{15}P_0$ is equilateral. Thus $triangle$s $OXP_0$ and $P_{15}XP_0$ are congruent in opposite senses (SSS) because $OX=XP_{15}$, $OP_0=P_{15}P_0$ and $P_0X$ is common. Thus $angle OP_0X=angle XP_0P_{15}$ so $angle OP_0X=30^circ=angle OP_1X$, so quadrilateral $OXP_0P_1$ is cyclic, so $alpha=angle P_0XP_1=angle P_0OP_1=20^circ$, which solves the problem.
answered Jan 19 at 12:19
Rosie FRosie F
1,349416
$endgroup$
I don't have a direct solution. The best I have relates to a different problem, so I show the latter's solution, then show how the two problems correspond.
Lemma. Let $OAC$ be a triangle, and $X$ a point in it, and let angles $CAX=40^{circ}, XAO=10^{circ}, AOX=10^{circ}, XOC=70^{circ}$. What is $angle OCX$?
Solution of lemma.
$angle AOC=80^{circ}$ and $angle CAO=50^{circ}$ so $angle OCA=50^{circ}= angle CAO$ so $triangle ACO$ is isosceles on base $AC$, so $OA=OC$.
$angle AOX = 10^{circ} = angle XAO$, so $triangle OAX$ is isosceles on base $OA$, so $AX=OX$.
$angle OXA = 180^{circ} - angle XAO - angle AOX = 160^{circ}$.
Erect an equilateral $triangle OXE$ on base $OX$. Join $CE$.
$angle EOC = angle XOC - angle XOE = 70^{circ} - 60^{circ} = 10^{circ} = angle AOX$.
Thus $triangle$s $OAX, OCE$ are congruent (SAS, opposite sense) because $angle EOC = angle AOX, OA=OC$ and $OX=OE$.
Thus $angle OCE = angle XAO = 10^{circ}$.
$CE=AX=OX=XE$.
Thus $triangle XCE$ is isosceles on base $XC$.
$angle XEC = 360^{circ} - angle CEO - angle OEX = 360^{circ} - 160^{circ} - 60^{circ} = 140^{circ}$.
Thus $angle ECX = (180^{circ}-140^{circ})/2 = 20^{circ}$.
Thus $angle OCX = angle ECX + angle OCE = 20^{circ}+10^{circ} = 30^{circ}$, which solves the lemma.
Solution of the original problem.
Let $P_0dots P_{17}$ be a regular 18-gon. Let $P_1P_{13}$ cross $P_{5}P_{15}$ at $X$. What is $alpha=angle P_0XP_1$?
Let the centre of the 18-gon be $O$. $triangle P_0P_{1}O$ is isosceles on base $P_{0}P_1$, and $angle P_0OP_1=20^circ$, so $angle OP_1P_0=80^circ$.
$triangle P_{13}P_1O$ is isosceles on base $P_{13}P_1$, and $angle P_{13}OP_1=120^circ$, so $angle OP_1P_{13}=30^circ=angle OP_1X$ as $X$ is on $P_1P_{13}$.
$triangle P_{15}P_5O$ is isosceles on base $P_{15}P_5$, and $angle P_{15}OP_5=160^circ$, so $angle P_5P_{15}O=10^circ=angle XP_{15}O$ as $X$ is on $P_5P_{15}$.
$triangle P_{15}P_1O$ is isosceles on base $P_{15}P_1$, and $angle P_{15}OP_1=80^circ$, so $angle P_1P_{15}O=50^circ$. Thus this $triangle$ has the same angles $80^circ, 50^circ, 50^circ$ as the lemma's $triangle OAC$, so they are similar. Moreover, the angles $OP_1X=30^circ$ and $XP_{15}O=10^circ$ correspond to those in the lemma, so the $X$s correspond. Therefore, by the lemma, $OX=XP_{15}$.
$OP_0=OP_{15}$ and $angle P_{15}OP_0=60^circ$ so $triangle OP_{15}P_0$ is equilateral. Thus $triangle$s $OXP_0$ and $P_{15}XP_0$ are congruent in opposite senses (SSS) because $OX=XP_{15}$, $OP_0=P_{15}P_0$ and $P_0X$ is common. Thus $angle OP_0X=angle XP_0P_{15}$ so $angle OP_0X=30^circ=angle OP_1X$, so quadrilateral $OXP_0P_1$ is cyclic, so $alpha=angle P_0XP_1=angle P_0OP_1=20^circ$, which solves the problem.
answered Jan 19 at 12:19
Rosie FRosie F
1,349416
edited Jan 19 at 12:52
answered Jan 19 at 12:19
Rosie FRosie F
1,349416
answered Jan 19 at 12:19
Rosie FRosie F
1,349416
answered Jan 19 at 12:19
Rosie FRosie F
1,349416
1,349416
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$begingroup$
You should mention what the "nice integer angle" is, to save people some time.
$endgroup$
– Blue
Dec 4 '18 at 11:47
$begingroup$
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.
$endgroup$
– Oldboy
Dec 4 '18 at 12:02
$begingroup$
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 $A_1A_7A_3angle =20^circ$.
$endgroup$
– Berci
Dec 7 '18 at 8:25