Hard to Imagine an Equiangular 18-gon

Geometry Level 3

In a recent Problem of the Week , we learned that an equiangular hexagon (6-sided figure whose interior angles are all the same) exists if and only if the side lengths, represented by a i , a_i, satisfy S = a 1 − a 4 = − ( a 2 − a 5 ) = a 3 − a 6 = − ( a 4 − a 1 ) = a 5 − a 2 = − ( a 6 − a 3 ) . S = a_1 - a_4 = - (a_2 - a_5 ) = a_3 - a_6 = - (a_4 - a_1) = a_5 - a_2 = - (a_6 - a_3 ).

True or False?

An equiangular 18-gon exists if and only if the side lengths, represented by a i , a_i, satisfy S = a 1 − a 10 = − ( a 2 − a 11 ) = a 3 − a 12 = ⋯ = − ( a 18 − a 9 ) . S = a_1 - a_{10} = - (a_2 - a_{11} ) = a_3 - a_{12} = \cdots = - (a_{18} - a_9 ).

True False

Calvin Lin by Calvin Lin

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2 solutions

Jon Haussmann
Oct 8, 2017

This is a follow-up to Calvin's post. Borrowing the same notation, we have a 1 ω + a 2 ω 2 + ⋯ + a 18 ω 18 = 0 , a_1 \omega + a_2 \omega^2 + \dots +a_{18} \omega^{18} = 0, where ω \omega is a primitive 18th root of unity. The minimal polynomial of ω \omega is x 6 − x 3 + 1 = 0 x^6 - x^3 + 1 = 0 , the 18th cyclotomic polynomial. Using this result, the equation above reduces to ( − a 6 − a 9 + a 15 + a 18 ) + ( a 1 − a 7 − a 10 + a 16 ) ω + ( a 2 − a 8 − a 11 + a 17 ) ω 2 + ( a 3 + a 6 − a 9 − a 12 − a 15 + a 18 ) ω 3 + ( a 4 + a 7 − a 13 − a 16 ) ω 4 + ( a 5 + a 8 − a 14 − a 17 ) ω 5 = 0. \begin{aligned} &(-a_6 - a_9 + a_{15} + a_{18}) + (a_1 - a_7 - a_{10} + a_{16}) \omega + (a_2 - a_8 - a_{11} + a_{17}) \omega^2 \\ &\quad + (a_3 + a_6 - a_9 - a_{12} - a_{15} + a_{18}) \omega^3 + (a_4 + a_7 - a_{13} - a_{16}) \omega^4 + (a_5 + a_8 - a_{14} - a_{17}) \omega^5 = 0. \end{aligned} Hence, the a i a_i must satisfy − a 6 − a 9 + a 15 + a 18 = 0 , a 1 − a 7 − a 10 + a 16 = 0 , a 2 − a 8 − a 11 + a 17 = 0 , a 3 + a 6 − a 9 − a 12 − a 15 + a 18 = 0 , a 4 + a 7 − a 13 − a 16 = 0 , a 5 + a 8 − a 14 − a 17 = 0. \begin{aligned} -a_6 - a_9 + a_{15} + a_{18} &= 0, \\ a_1 - a_7 - a_{10} + a_{16} &= 0, \\ a_2 - a_8 - a_{11} + a_{17} &= 0, \\ a_3 + a_6 - a_9 - a_{12} - a_{15} + a_{18} &= 0, \\ a_4 + a_7 - a_{13} - a_{16} &= 0, \\ a_5 + a_8 - a_{14} - a_{17} &= 0. \end{aligned}

One solution is a 1 = a 4 = a 7 = 2 a_1 = a_4 = a_7 = 2 , a 13 = 3 a_{13} = 3 , and the rest of the a i a_i are equal to 1.

This example shows that the condition in the problem does not necessarily hold.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Indeed! This solution provides some insight into the underlying theory, and how it goes wrong / right (depending on your point of view).

For those who are interested in grappling with this further, think about how the triangle case (where equiangular implies equilaterial) relates to the hexagon case, particularly in terms of how the 3rd and 6th roots of unity are related. (For example, rewrite my solution using the 6th root of unity, and see how much neater it becomes.)

In particular, conclude that when n n is prime, the equiangular n − n- gon is regular, and the equiangular 2 n − 2n- gon has the simple description as stated. However, for composite n n , the 2 n − 2n- gon starts getting really interesting.

If you want to take this further, think about how the 6th and 18th root of unity interact to give the weird 18-gon that we see above.

Calvin Lin Staff - 3 years, 8 months ago

FYI I'm not sure how you came up with the counterexample, but all that you needed was the observation that x 6 − x 3 + 1 = 0 x^ 6 -x^3 + 1 = 0 .

In addition, there is an easy/obvious counterexample, which is symmetric of order 3. Though that might not have misled people into thinking that there must always be symmetry, especially since Jon's example also has a line of symmetry (connecting midpoints of sides a 4 a_4 and a 13 a_{13} .

So, as an exercise, find a counterexample where there is no symmetry.

Hint: We just need to find ∑ a i ω i = 0 \sum a_i \omega^i = 0 where the coefficients are "lopsided".

Calvin Lin Staff - 3 years, 8 months ago
Calvin Lin Staff
Oct 6, 2017

[This is not a complete solution.]

Turns out that this problem is harder than I anticipated (in part because I already knew how to think of the underlying theory).

First, let's review why the condition occurs for a hexagon. Let's represent the sides of the hexagon using complex numbers as vectors.
The first side, which (WLOG) goes right will be represented by a 1 a_1 .
The second side, which (WLOG) goes 6 0 ∘ 60 ^\circ north of east, will be represented by − a 2 ω 2 - a_2 \omega^2 , where ω \omega is the cube root of unity.
The third side, which goes 6 0 ∘ 60^\circ north of west, will be represented by a 3 ω a_3 \omega .
The fourth side, which goes left, will be represented by − a 4 -a_4 .
The fifth side, which goes 6 0 ∘ 60^\circ south of west, will be represented by a 5 ω 2 a_5 \omega^2 .
The sixth side, which goes 6 0 ∘ 60 ^ \circ south of east, will be represented by a 6 ω a_6 \omega .


Then, in order for this to form a polygon, the path must be closed, meaning that we're back to the same point. Thus,

a 1 − a 2 ω 2 + a 3 ω − a 4 + a 5 ω 2 − a 6 ω = 0 ( a 1 − a 4 ) + ( a 3 − a 6 ) ω + ( a 5 − a 2 ) ω 2 = 0 a_1 - a_2 \omega^2 + a_3 \omega - a_4 + a_5 \omega^2 - a_6 \omega = 0 \\ (a_1 - a_4) + (a_3 - a_6) \omega + (a_5 - a_2) \omega^2 =0

Recall that for real numbers a , b , c a, b, c , a + b ω + c ω 2 = 0 a + b \omega + c \omega^2 = 0 if and only if a = b = c a = b= c . Hence, we can conclude that

a 1 − a 5 = − ( a 2 − a 5 ) = a 3 − a 6 a_1 - a_5 = - (a_2 - a_5) = a_3 - a_6

Now, apply this to the 18-gon. What is the conclusion? Why do we arrive at a different result? At what point does the logic diverge?

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