Let me Count the Sides

Geometry Level 4

A regular polygon with side length 1 has area less than 123456789.

What is the maximum number of sides such polygon can have?


The answer is 39387.

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Digvijay Singh by Digvijay Singh , 21, India

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3 solutions

The area of an n n -side regular polygon A n A_n is the area of \n) isosceles triangle of equal sides r r with base 1 1 , The angle at the center of the polygon is 2 π n \frac{2\pi}{n} and the two base angles are 1 2 ( π − 2 π n ) = π 2 − π n \frac{1}{2}\left(\pi-\frac{2\pi}{n} \right) = \frac{\pi}{2} - \frac{\pi}{n} . Then, we have:

A n = n 2 ( 1 ) ( r ) sin ⁡ ( π 2 − π n ) ⇒ A n = n r 2 cos ⁡ π n \begin{aligned} A_n & = \dfrac{n}{2}(1)(r)\sin{\left(\frac{\pi}{2} - \frac{\pi}{n} \right)} \quad \Rightarrow A_n = \dfrac{nr}{2} \cos{\frac{\pi}{n}} \end{aligned}

From cosine rule, we have:

1 2 = 2 r 2 − 2 r 2 cos ⁡ 2 π n = 2 r 2 ( 1 − 2 cos ⁡ 2 π n + 1 ) = 4 r 2 sin ⁡ 2 π n \begin{aligned} 1^2 & = 2r^2 - 2r^2\cos{\frac{2\pi}{n}} = 2r^2 \left(1 - 2\cos^2{\frac{\pi}{n}} +1\right) = 4r^2 \sin^2{\frac{\pi}{n}} \end{aligned}

⇒ r 2 = 1 4 sin ⁡ 2 π n ⇒ r = 1 2 sin ⁡ π n \begin{aligned} \Rightarrow r^2 & = \frac{1}{4\sin^2{\frac{\pi}{n}}} \quad \Rightarrow r & = \frac{1}{2\sin{\frac{\pi}{n}}} \end{aligned}

⇒ A n = n r 2 cos ⁡ π n = n 4 cot ⁡ π n \begin{aligned} \Rightarrow A_n & = \dfrac{nr}{2} \cos{\frac{\pi}{n}} = \dfrac{n}{4} \cot{\frac{\pi}{n}} \end{aligned}

For the largest A n < 123456789 A_n < 123456789 , we have:

n 4 cot ⁡ π n < 123456789 ⇒ tan ⁡ π n > n 4 × 123456789 [ For small x, tan ⁡ x ≈ x ] ⇒ π n > n 4 × 123456789 n 2 < 4 π × 123456789 ⇒ n < 4 π × 123456789 = 39387.86317 n = 39387 \begin{aligned} \frac{n}{4} \cot{\frac{\pi}{n}} & < 123456789 \\ \Rightarrow \color{#3D99F6}{\tan{\frac{\pi}{n}}} & > \frac{n}{4 \times 123456789} \quad \quad \color{#3D99F6} {[\text{For small x, } \tan{x} \approx x]} \\ \Rightarrow \color{#3D99F6} {\frac{\pi}{n}} & > \frac{n}{4 \times 123456789} \\ n^2 & < 4\pi \times 123456789 \\ \Rightarrow n & < \sqrt{4\pi \times 123456789} = 39387.86317 \\ n & = \boxed{39387} \end{aligned}

We can check that A 39387 < 12345678 A_{39387} < 12345678 and A 39388 > 12345678 A_{39388} > 12345678 , confirming the answer.

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Moderator note:

Thanks. For proving the area of a regular polygon with side length 1.

Generally, the area of a regular n n -gon with side length p p is 1 4 n p 2 cot ⁡ ( π n ) \frac 14 np^2 \cot\left(\frac \pi n\right) .

I found a decent approximation by letting the radius of a similarly-sized circle be r r , calling 2 π r = n 2\pi r = n and π r 2 = 123456789 \pi r^{2} = 123456789 . Then n = 2 π 123456789 / π ≈ 39388 n = 2\pi \sqrt{123456789 / \pi} \approx 39388 , so I just used the Challenge Master's formula to test n = 39387 , n = 39388 n = 39387, n = 39388 .

Jake Lai - 5 years, 9 months ago

I forgot the 1/4 and just wrote n cot ⁡ π n n \cot \frac{\pi}{n} ...

Gian Sanjaya - 5 years, 9 months ago
Lu Chee Ket
Sep 28, 2015

A = N / [4 Tan(Pi / N)]

{Square unit area for unit side length of N-polygon.}

For N and A:

39387 for 123451377.754009 TRUE

39388 for 123457646.46933 FALSE

Answer: 39387

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from math import tan, pi

i = 40000
while i >= 39000:
    temp = (0.25*i) / tan(pi/float(i))
    if temp < 123456789:
        print i
        break
    i -= 1

# through trial and error: 40000 > i > 39000

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