A problem from multiple angles(pun intended)

Geometry Level 2

Which of the following represents the interior angle of a regular polygon, having n n sides ( n > 3 ) (n>3) , whose respective ratio of the number of sides ( n ) (n) to the number of diagonals ( d ) (d) is P : 1 P:1 , i.e., n d = P 1 \dfrac{n}{d}=\dfrac{P}{1}

3 + 2 P 1 + 3 P × 18 0 ∘ \dfrac{3+2P}{1+3P} \times 180^{\circ} 2 + P 2 + 3 P × 18 0 ∘ \dfrac{2+P}{2+3P} \times 180^{\circ} 1 + 3 P 1 + P × 18 0 ∘ \dfrac{1+3P}{1+P} \times 180^{\circ} 3 + P 3 + 2 P × 18 0 ∘ \dfrac{3+P}{3+2P} \times 180^{\circ} 1 + P 2 + 3 P × 18 0 ∘ \dfrac{1+P}{2+3P} \times 180^{\circ}

Vinayak Srivastava by Vinayak Srivastava , 15, India

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1 solution

If a convex polygon has n n sides, then it has

d = n ( n − 3 ) 2 d=\dfrac {n(n-3)}{2} diagonals.

So, according to the given condition of the problem,

n d = P ⟹ n = 3 P + 2 P \dfrac nd=P\implies n=\dfrac {3P+2}{P}

Now, each internal angle of the polygon is

( 1 − 2 n ) π \left (1-\dfrac 2n\right )π radians.

Substituting for n n we get the value of each internal angle as

( 1 − 2 P 3 P + 2 ) π \left (1-\dfrac {2P}{3P+2}\right ) π

= P + 2 3 P + 2 π =\boxed {\dfrac {P+2}{3P+2}π} .

1 Helpful 2 Interesting 1 Brilliant 0 Confused

Nice solution, Sir!

Vinayak Srivastava - 10 months, 2 weeks ago

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