Polygon with minimum internal angle 120^\circ

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Polygon with minimum internal angle 120^\circ

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#1 h Aug 9, 2007, 4:27 PM • 2 Y

Y by Adventure10, Mango247

How many sides can have the polygon with minimum internal angle of $120^\circ$ by adding every $5^\circ$ ?
Sorry for my poor English.

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#2 h Apr 28, 2025, 2:12 PM

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For any convex polygon with $N$ sides, the total internal angle sum (in degrees) computes to $180^{\circ}(N-2)$ (i). The polygon in question has internal angles whose sum is represented by the arithmetic series: $\sum_{k=1}^{N} 120^{\circ} + 5^{\circ}(k-1)$ (ii). Equating (i) with (ii) yields a quadratic in $N$ :

$180^{\circ}(N-2) = \sum_{k=1}^{N} 120^{\circ} + 5^{\circ}(k-1)$ ;

or $180^{\circ}(N-2) = \sum_{k=1}^{N} 115^{\circ} + 5^{\circ}k$ ;

or $36(N-2) = \sum_{k=1}^{N} 23 + k$ ;

or $36(N-2) = 23N +\frac{N(N+1)}{2}$ ;

or $N^2-25N+144=0$ ;

or $(N-9)(N-16)=0$ ;

or $N=9, 16$ .

In order for the polygon in question to be convex we require all of its interior angles $< 180^{\circ}$ . At $N=9$ the maximum angle is $120^{\circ}+5^{\circ}(9) = 165^{\circ}$ (admissible), and $N=16$ yields $120^{\circ}+5^{\circ}(16) = 200^{\circ}$ (not admissible). Hence, our desired polygon has $\boxed{9}$ . sides.

This post has been edited 1 time. Last edited by Mathzeus1024, Apr 29, 2025, 9:57 AM

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