Angles of convex polygons
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For a convex polygon, show that the sum of any two interior angles is greater than the difference between any two interior angles. (the polygon has more than 3 sides)
If I pick 4 dots A,B,C,D and say the theorem applies(angleA+ angleB>angleC-angleD), then it is same as proving that angleA+angleB+angleD>angleC, so I found that this theorem is the same meaning by saying that sum of any three interior angles is greater than any one angle. However, since this is a convex polygon, one Angle is bigger than zero and smaller than one hundred eighty. Then I want to prove that the sum of three angles in a convex polygon is bigger than 180(or equal). Is there any proof that works for every convex polygon?
angle convex-geometry
asked Dec 28 '18 at 5:12
mathheromathhero
235
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add a comment |
$begingroup$
For a convex polygon, show that the sum of any two interior angles is greater than the difference between any two interior angles. (the polygon has more than 3 sides)
If I pick 4 dots A,B,C,D and say the theorem applies(angleA+ angleB>angleC-angleD), then it is same as proving that angleA+angleB+angleD>angleC, so I found that this theorem is the same meaning by saying that sum of any three interior angles is greater than any one angle. However, since this is a convex polygon, one Angle is bigger than zero and smaller than one hundred eighty. Then I want to prove that the sum of three angles in a convex polygon is bigger than 180(or equal). Is there any proof that works for every convex polygon?
angle convex-geometry
asked Dec 28 '18 at 5:12
mathheromathhero
235
$endgroup$
add a comment |
$begingroup$
For a convex polygon, show that the sum of any two interior angles is greater than the difference between any two interior angles. (the polygon has more than 3 sides)
If I pick 4 dots A,B,C,D and say the theorem applies(angleA+ angleB>angleC-angleD), then it is same as proving that angleA+angleB+angleD>angleC, so I found that this theorem is the same meaning by saying that sum of any three interior angles is greater than any one angle. However, since this is a convex polygon, one Angle is bigger than zero and smaller than one hundred eighty. Then I want to prove that the sum of three angles in a convex polygon is bigger than 180(or equal). Is there any proof that works for every convex polygon?
angle convex-geometry
asked Dec 28 '18 at 5:12
mathheromathhero
235
$endgroup$
For a convex polygon, show that the sum of any two interior angles is greater than the difference between any two interior angles. (the polygon has more than 3 sides)
If I pick 4 dots A,B,C,D and say the theorem applies(angleA+ angleB>angleC-angleD), then it is same as proving that angleA+angleB+angleD>angleC, so I found that this theorem is the same meaning by saying that sum of any three interior angles is greater than any one angle. However, since this is a convex polygon, one Angle is bigger than zero and smaller than one hundred eighty. Then I want to prove that the sum of three angles in a convex polygon is bigger than 180(or equal). Is there any proof that works for every convex polygon?
angle convex-geometry
angle convex-geometry
asked Dec 28 '18 at 5:12
mathheromathhero
235
asked Dec 28 '18 at 5:12
mathheromathhero
235
edited Dec 28 '18 at 6:03
mathhero
asked Dec 28 '18 at 5:12
mathheromathhero
235
asked Dec 28 '18 at 5:12
mathheromathhero
235
asked Dec 28 '18 at 5:12
mathheromathhero
235
235
add a comment |
add a comment |
1 Answer
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It's not true. Try in triangle $alpha=120^{circ}$ and $beta=gamma=30^{circ}.$
Thus, $$beta+gamma>alpha-beta$$ is wrong.
answered Dec 28 '18 at 5:30
Michael RozenbergMichael Rozenberg
108k1895200
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1 Answer
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$begingroup$
It's not true. Try in triangle $alpha=120^{circ}$ and $beta=gamma=30^{circ}.$
Thus, $$beta+gamma>alpha-beta$$ is wrong.
answered Dec 28 '18 at 5:30
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
It's not true. Try in triangle $alpha=120^{circ}$ and $beta=gamma=30^{circ}.$
Thus, $$beta+gamma>alpha-beta$$ is wrong.
answered Dec 28 '18 at 5:30
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
It's not true. Try in triangle $alpha=120^{circ}$ and $beta=gamma=30^{circ}.$
Thus, $$beta+gamma>alpha-beta$$ is wrong.
answered Dec 28 '18 at 5:30
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
It's not true. Try in triangle $alpha=120^{circ}$ and $beta=gamma=30^{circ}.$
Thus, $$beta+gamma>alpha-beta$$ is wrong.
answered Dec 28 '18 at 5:30
Michael RozenbergMichael Rozenberg
108k1895200
edited Dec 28 '18 at 5:37
answered Dec 28 '18 at 5:30
Michael RozenbergMichael Rozenberg
108k1895200
answered Dec 28 '18 at 5:30
Michael RozenbergMichael Rozenberg
108k1895200
answered Dec 28 '18 at 5:30
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
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