rational one year ago show that if three angles in a convex polygon are equal to $$60^{\circ}$$ then it must be an equilateral triangle

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1. ikram002p

|dw:1439061574644:dw|

2. ikram002p

i'll try other form |dw:1439061670041:dw|

3. ikram002p

assume x is 60 |dw:1439061819528:dw|

4. 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

5. rational

yes that looks legit

6. ikram002p

that would make the other angle 180 seems more like not concave nor convex

7. ikram002p

|dw:1439062235117:dw|

8. ikram002p

i feel sleepy i might write it more need in morning,gn

9. 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.

10. ikram002p

more neat**

11. rational

Nice :) More generally : Any convex polygon with $$n$$ angles equal to $$\dfrac{360}{n}$$ must be a regular $$\text{n-gon}$$

12. ikram002p

hehehe ok that was direct

13. rational

* Any convex polygon with $$n$$ angles equal to $$180-\dfrac{360}{n}$$ must be a regular $$\text{n-gon}$$

14. rational

for example, in a square we have $$4$$ angles equal to $$180-\frac{360}{4} = 90$$

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