rational
  • rational
show that if three angles in a convex polygon are equal to \(60^{\circ}\) then it must be an equilateral triangle
Geometry
schrodinger
  • schrodinger
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ikram002p
  • ikram002p
|dw:1439061574644:dw|
ikram002p
  • ikram002p
i'll try other form |dw:1439061670041:dw|
ikram002p
  • ikram002p
assume x is 60 |dw:1439061819528:dw|

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ikram002p
  • ikram002p
(if it works for quadrilateral then it works for any polygon ) geometric logic so we need to show by contradiction that we cant construct a quadrilateral with 3 60 angle
rational
  • rational
yes that looks legit
ikram002p
  • ikram002p
that would make the other angle 180 seems more like not concave nor convex
ikram002p
  • ikram002p
|dw:1439062235117:dw|
ikram002p
  • ikram002p
i feel sleepy i might write it more need in morning,gn
anonymous
  • anonymous
Sum of all the exterior angles of a polygon is 360 degrees. The exterior angle corresponding to 60 degrees is 120 degrees. Since three angles are 60 degrees, the sum of those exterior angles is 120*3 = 360 degrees. Thus there can be no other non-zero exterior angle corresponding to a vertex. So, these are the only 3 vertices, and the polygon is an equilateral triangle.
ikram002p
  • ikram002p
more neat**
rational
  • rational
Nice :) More generally : Any convex polygon with \(n\) angles equal to \(\dfrac{360}{n}\) must be a regular \(\text{n-gon}\)
ikram002p
  • ikram002p
hehehe ok that was direct
rational
  • rational
* Any convex polygon with \(n\) angles equal to \(180-\dfrac{360}{n}\) must be a regular \(\text{n-gon}\)
rational
  • rational
for example, in a square we have \(4\) angles equal to \(180-\frac{360}{4} = 90\)

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