Problem of the Month
Solution to Problem 6: Regular Polygons and Lattice Points

March 2025

Let $A$ and $B$ have coordinates $(a, b)$ and $(c, d)$ respectively. Let $α$ and $β$ be the angles from the positive $x$ -axis to the lines $O A$ and $O B$ respectively, measured in the clockwise direction. For example, if $A$ has coordinates $(1, 1)$ and $B$ has coordinates $(- 1, - 1)$ , $α = 45 °$ and $β = 215 °$ .

Assume towards a contradiction, that $a, b, c$ , and $d$ are integers and $∠ A O B$ is $60 °$ .

Case 1: $a \neq 0$ and $c \neq 0$ .

Since $0 \leq α, β < 360 °$ , then $β - α = \pm 60 °$ or $\pm 300 °$ . Therefore, $\tan (β - α) = \pm \tan (60 °) = \pm \sqrt{3}$ . We have $\tan (α) = \frac{b}{a}$ and $\tan (β) = \frac{d}{c}$ . By the angle sum formula for tan, $\begin{aligned} \tan (β - α) & = \frac{\tan (β) - \tan (α)}{1 + \tan (α) \tan (β)} \\ = \frac{\frac{d}{c} - \frac{b}{a}}{1 + (\frac{b}{a}) (\frac{d}{c})} . \end{aligned}$ Since $a, b, c$ , and $d$ are integers with $a \neq 0$ and $c \neq 0$ , $\tan (β - α)$ is rational. However, $\sqrt{3}$ is well known to be irrational, which gives us a contradiction. We can now conclude that if $A$ and $B$ are lattice points, then $∠ A O B$ is not $60 °$ .

Case 2: $a = 0$ or $c = 0$ .

In this case, at least one of $A$ or $B$ is on the $y$ -axis. Note that if $A$ is on the $y$ -axis, then $B$ cannot be on the $x$ - or $y$ - axis (since $∠ A O B$ is not $0 °$ , $90 °$ , or $180 °$ ). Similarly, if $B$ is on the $y$ -axis, then $A$ cannot be on the $x$ - or $y$ -axis. So actually, exactly one of $A$ or $B$ is on the $y$ -axis, and the other is not on the $x$ -axis.

Let $A^{'}$ and $B^{'}$ be the result of rotating $A$ and $B$ $90 °$ clockwise about the origin. Now exactly one of $A^{'}$ and $B^{'}$ are on the $x$ -axis, and the other is not on the $y$ -axis. Furthermore, $∠ A O B = ∠ A^{'} O B^{'}$ . We are now back in Case 1 above but with the points $A^{'}$ and $B^{'}$ , and can conclude $∠ A O B$ cannot be $60 °$ .
1. The interior angle of a regular $n$ -gon is $180 ° - \frac{360 °}{n}$ , and so $∠ B C D = ∠ B A E = 108 °$ . Since $△ E A B$ is isosceles, $∠ A B E = 36 °$ . Therefore, $∠ C B F = 72 °$ . Since $180 ° = 72 ° + 108 ° = ∠ C B F + ∠ B C D,$ we have that $C D$ is parallel to $B F$ . We can similarly argue that $B C$ and $D F$ are parallel, and conclude that quadrilateral $F B C D$ is a parallelogram.
2. Suppose $B$ , $C$ , and $D$ have coordinates $(b_{1}, b_{2})$ , $(c_{1}, c_{2})$ , and $(d_{1}, d_{2})$ respectively. Since quadrilateral $F B C D$ is a parallelogram, the difference between the $x$ -coordinates of $D$ and $C$ is equal to the difference between the $x$ -coordinates of $F$ and $B$ . The same holds for the $y$ -coordinates. Therefore, $F$ has coordinates $(b_{1} + d_{1} - c_{1}, b_{2} + d_{2} - c_{2})$ . We are assuming that $B$ , $C$ , and $D$ are lattice points, so $b_{1}, b_{2}, c_{1}, c_{2}, d_{1},$ and $d_{2}$ are all integers. Therefore, the coordinates of $F$ are integers and $F$ is a lattice point.
  
  If you are familiar with vectors, we can rephrase the above argument in terms of vector addition. Since $F B C D$ is a parallelogram, we know $\vec{C D} + \vec{C B} = \vec{C F}$ . Therefore, $\vec{D} - \vec{C} + \vec{B} - \vec{C} = \vec{F} - \vec{C}$ . This implies $\vec{F} = \vec{D} + \vec{B} - \vec{C}$ . Since $B$ , $C$ , and $D$ are lattice points, the corresponding vectors $\vec{B}$ , $\vec{C}$ , and $\vec{D}$ have integer coordinates. Since $\vec{F}$ is the sum of three vectors with integer coordinates, $\vec{F}$ also has integer coordinates and we can conclude that $F$ is a lattice point.
1. We begin by showing that the angle between $L_{i}$ and $L_{i + 1}$ is $\frac{360 °}{n}$ for all $i$ such that $1 \leq i \leq n - 1$ . The same argument shows that the angle between $L_{n}$ and $L_{1}$ is also $\frac{360 °}{n}$ .
  
  Let $θ$ be the angle between $L_{i}$ and $L_{i + 1}$ . Then, on the original regular $n$ -gon, we can compute $θ$ by extending the end of $L_{i}$ past the beginning of $L_{i + 1}$ as in the diagram below.
  
  Since the interior angle of a regular $n$ -gon is $180 ° - \frac{360 °}{n}$ , we must have $θ = 180 ° - (180 ° - \frac{360 °}{n}) = \frac{360 °}{n} .$ Let $A$ be the center of $B_{1} B_{2} \dots B_{n}$ . Then $\frac{360 °}{n} = ∠ B_{1} A B_{2} = ∠ B_{2} A B_{3} = \dots = ∠ B_{n - 1} A B_{n} = ∠ B_{n} A B_{1} .$ Since each line segment $A B_{i}$ is the side length of the original regular $n$ -gon, the $n$ triangles $△ B_{1} A B_{2}, △ B_{2} A B_{3}, \dots, △ B_{n - 1} A B_{n}$ , and $△ B_{n} A B_{1}$ are all congruent.
  
  Therefore, the edges of the polygon $B_{1} B_{2}, B_{2} B_{3}, \dots, B_{n - 1} B_{n}$ , and $B_{n} B_{1}$ all have equal length.
  
  Consider $△ B_{1} A B_{2}$ . Since $A B_{1}$ and $A B_{2}$ have the same length, it is an isosceles triangle. Since $∠ B_{1} A B_{2} = \frac{360 °}{n}$ , we have $∠ A B_{1} B_{2} = ∠ A B_{2} B_{1} = 90 ° - \frac{180 °}{n}$ . By the congruence $△ B_{1} A B_{2}$ and $△ B_{2} A B_{3}$ , we have $∠ A B_{2} B_{3} = ∠ A B_{3} B_{2} = 90 ° - \frac{180 °}{n}$ . We can finally compute $∠ B_{1} B_{2} B_{3} = ∠ B_{1} B_{2} A + ∠ A B_{2} B_{3} = 90 ° - \frac{180 °}{n} + 90 ° - \frac{180 °}{n} = 180 ° - \frac{360 °}{n} .$ We can compute every interior angle of the $n$ -gon $B_{1} B_{2} \dots B_{n}$ in the same way to get that all interior angles are equal to $180 ° - \frac{360 °}{n}$ . Therefore, $B_{1} B_{2} \dots B_{n}$ is a regular $n$ -gon.
2. As in the previous solution, let $A$ be the center of the regular $n$ -gon $B_{1} B_{2} \dots B_{n}$ , and consider the isosceles triangle $△ B_{1} A B_{2}$ .
  
  Recall from part (a) that $∠ B_{1} A B_{2} = \frac{360 °}{n}$ . By drawing a line from the midpoint of $B_{1} B_{2}$ to $A$ , we divide $△ B_{1} A B_{2}$ into two congruent right-angled triangles. Considering one of these right-angled triangles gives $\sin (\frac{180 °}{n}) = \frac{x}{2 y} .$ Therefore, $\frac{x}{y} = 2 \sin (\frac{180 °}{n}) .$
Case 1: $n = 3$ and $n = 6$ .

Question 1 tells us that there are no equilateral lattice triangles. In fact, we can also rule out the existence of a regular lattice $n$ -gon where $n$ is a multiple of three. To see this, label the vertices $V_{1}, \dots, V_{3 n}$ in a clockwise order. Then $∠ V_{n} V_{2 n} V_{3 n} = 60 °$ . Therefore, all three of $V_{n}, V_{2 n}$ , and $V_{3 n}$ cannot be lattice points.

Case 2: $n = 5$ .

Next we will show that there are no regular lattice pentagons. We will approach this by showing that if there is a regular lattice pentagon, then we can create a smaller regular lattice pentagon. Then we can do it again, and create an even smaller regular lattice pentagon. We can continue to create smaller and smaller lattice pentagons, until we have a lattice pentagon with side length less than $1$ , so it cannot be a lattice pentagon! The only way to resolve this contradiction is to conclude that the original pentagon does not exist! Let’s execute this plan.

Consider a regular pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$ . Create a smaller regular pentagon by drawing all five diagonals of the pentagon, and taking the five intersections to be the vertices of our smaller pentagon (see the diagram below).

More precisely,
- the lines $E_{1} B_{1}$ and $A_{1} C_{1}$ intersect at $A_{2}$ ,
- the lines $A_{1} C_{1}$ and $B_{1} D_{1}$ intersect at $B_{2}$ ,
- the lines $B_{1} D_{1}$ and $C_{1} E_{1}$ intersect at $C_{2}$ ,
- the lines $C_{1} E_{1}$ and $D_{1} A_{1}$ intersect at $D_{2}$ , and
- the lines $D_{1} A_{1}$ and $E_{1} B_{1}$ intersect at $E_{2}$ .
Then $A_{2} B_{2} C_{2} D_{2} E_{2}$ is a regular pentagon (see if you can prove this!). Suppose the side length of pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$ is $l_{1}$ , and the side length of $A_{2} B_{2} C_{2} D_{2} E_{2}$ is $l_{2}$ . The goal now is to write down an expression for $l_{2}$ in terms of $l_{1}$ . By the solution to Question $2$ , $∠ E_{1} A_{1} E_{2} = ∠ B_{1} A_{1} A_{2} = 36 °$ . Since the interior angle of a regular pentagon is $108 °$ , we have $∠ E_{2} A_{1} A_{2} = 108 ° - 36 ° - 36 ° = 36 °$ .

Let $G$ be a point on $E_{2} A_{2}$ so that $G A_{1}$ is perpendicular to $E_{2} A_{2}$ . Let $F$ be the point on $A_{1} B_{1}$ so that $F A_{2}$ is perpendicular to $A_{1} B_{1}$ . Then $G$ and $F$ are the midpoints of $E_{2} A_{2}$ and $A_{1} B_{1}$ respectively (this fact needs proof, but I will leave it up to you!).

There are now two right-angled triangles, $△ A_{1} F A_{2}$ and $△ A_{1} G A_{2}$ . Consider the former triangle. Note that the length of $A_{1} F$ is $\frac{1}{2} l_{1}$ and $∠ A_{2} A_{1} F = 36 °$ . Let $x$ be the length of $A_{1} A_{2}$ . Then $\frac{l_{1}}{2 x} = \cos (36 °) .$ Now consider right-angled triangle $△ A_{1} G A_{2}$ . We have $∠ G A_{1} A_{2} = 18 °$ , and the length of $G A_{2}$ is $\frac{1}{2} l_{2}$ . Therefore, $\sin (18 °) = \frac{l_{2}}{2 x} .$ Solving for $x$ in both of these equations and rearranging gives $l_{2} = l_{1} \frac{\sin (18 °)}{\cos (36 °)} .$ At the end of this solution, there is an extra section which shows how to deduce that $\sin (18 °) = \frac{\sqrt{5} - 1}{4}$ and $\cos (36 °) = \frac{\sqrt{5} + 1}{4}$ . Using these exact values we have $l_{2} = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} l_{1} = \frac{3 - \sqrt{5}}{2} l_{1} .$ Note that $2 < \sqrt{5} < 3$ and so $0 < \frac{3 - \sqrt{5}}{2} < 1$ .

Great, we can now repeat this process of creating smaller and smaller pentagons, and we can compute the side length of each one as follows.

Let $k$ be a positive integer, and suppose $A_{k} B_{k} C_{k} D_{k} E_{k}$ is a regular pentagon with side length $l_{k}$ . Create the pentagon $A_{k + 1} B_{k + 1} C_{k + 1} D_{k + 1} E_{k + 1}$ as above by declaring that
- the lines $E_{k} B_{k}$ and $A_{k} C_{k}$ intersect at $A_{k + 1}$ ,
- the lines $A_{k} C_{k}$ and $B_{k} D_{k}$ intersect at $B_{k + 1}$ ,
- the lines $B_{k} D_{k}$ and $C_{k} E_{k}$ intersect at $C_{k + 1}$ ,
- the lines $C_{k} E_{k}$ and $D_{k} A_{k}$ intersect at $D_{k + 1}$ , and
- the lines $D_{k} A_{k}$ and $E_{k} B_{k}$ intersect at $E_{k + 1}$ .
Then $A_{k + 1} B_{k + 1} C_{k + 1} D_{k + 1} E_{k + 1}$ is a regular pentagon with side length $l_{k + 1}$ where $l_{k + 1} = \frac{3 - \sqrt{5}}{2} l_{k} .$ Repeatedly applying the equation $l_{k + 1} = \frac{3 - \sqrt{5}}{2} l_{k}$ we have that for any positive integer $k$ , $l_{k} = {(\frac{3 - \sqrt{5}}{2})}^{k - 1} l_{1} .$ We will now prove that there is some $k$ large enough so that $l_{k} < 1$ . Choose a positive integer $k$ so that $k > \frac{- \log (l_{1})}{\log (3 - \sqrt{5}) - \log (2)} + 1.$ Here we don’t care what the base is for the logs, so we will be lazy and not write anything down as the base. After rearranging the inequality a little, and applying some log laws we have $k - 1 > \frac{\log (\frac{1}{l_{1}})}{\log (\frac{3 - \sqrt{5}}{2})} .$ Since $\frac{3 - \sqrt{5}}{2} < 1$ , $\log (\frac{3 - \sqrt{5}}{2}) < 0$ . Therefore, $\begin{aligned} (k - 1) \log (\frac{3 - \sqrt{5}}{2}) & < \log (\frac{1}{l_{1}}) \\ \Rightarrow {(\frac{3 - \sqrt{5}}{2})}^{k - 1} & < \frac{1}{l_{1}} \\ \Rightarrow l_{1} {(\frac{3 - \sqrt{5}}{2})}^{k - 1} & < 1 \\ \Rightarrow l_{k} & < 1. \end{aligned}$ Great! Let’s put everything together. Suppose $A_{1} B_{1} C_{1} D_{1} E_{1}$ is a regular lattice pentagon with length $l_{1}$ . Then by Question 2(b), the regular pentagon $A_{k} B_{k} C_{k} D_{k} E_{k}$ is a lattice pentagon for all positive integers $k$ . However, when $k > \frac{- \log (l_{1})}{\log (3 - \sqrt{5}) - \log (2)} + 1$ we have shown that the side length $l_{k}$ of $A_{k} B_{k} C_{k} D_{k} E_{k}$ satisfies $l_{k} < 1$ . Since the smallest distance between two lattice points in the plane is $1$ , $A_{k} B_{k} C_{k} D_{k} E_{k}$ cannot be a lattice pentagon, which is a contradiction! Therefore, a regular pentagon cannot be a lattice pentagon.

Case 3: $n \geq 7$ .

The general strategy for this case will be the same as in the case $n = 5$ . The difference here will be our construction of successive smaller regular polygons.

Let $P_{1}$ be a regular $n$ -gon with side length $l_{1}$ . Let $P_{k}$ be the regular $n$ -gon obtained from $P_{1}$ by applying $k$ times the process from Question 3. Let $P_{k}$ have side length $l_{k}$ . Then from the solution to 3(b) above we have $l_{k} = l_{1} {(2 \sin (\frac{180 °}{n}))}^{k} .$ Between $0 °$ and $90 °$ , the sine function is strictly increasing. Therefore, for $n \geq 7$ we have $0 < 2 \sin (\frac{180 °}{n}) < 2 \sin (\frac{180 °}{6}) = 1.$ As in the $n = 5$ case, we want to show that if we choose $k$ to be big enough, $l_{k} < 1$ . To that end, let $a = 2 \sin (\frac{180 °}{n})$ and choose a positive integer $k$ so that $k > \frac{- \log (l_{1})}{\log (a)} .$ Then again, since $0 < a < 1$ , $\log (a) < 0$ and we have $\begin{aligned} k \log (a) & < \log (\frac{1}{l_{1}}) \\ \Rightarrow a^{k} & < \frac{1}{l_{1}} \\ \Rightarrow l_{1} a^{k} & < 1 \\ \Rightarrow l_{k} & < 1. \end{aligned}$ It remains to show that if $P_{k}$ is a lattice polygon, then so is $P_{k + 1}$ . Let $P_{k}$ be the lattice polygon $A_{1} A_{2} \dots A_{n}$ , where $A_{i}$ has coordinates $(a_{i}, b_{i})$ . Now, translate the polygon $P_{k + 1}$ so that its center is at the origin $O$ , with coordinates $(0, 0)$ . Let $P_{k + 1}$ be the polygon $B_{1} B_{2} \dots B_{n}$ . Then for each $i < n$ , $B_{i}$ has coordinates $(a_{i + 1} - a_{i}, b_{i + 1} - b_{i})$ , and $B_{n}$ has coordinates $(a_{1} - a_{n}, b_{1} - b_{n})$ . Since each of the $a_{i}$ and $b_{i}$ are integers, we have that each of the $B_{i}$ is a lattice point and $P_{k + 1}$ is a lattice polygon.

Great, now we can put everything together. Suppose $P_{1}$ is a regular lattice $n$ -gon with side length $l_{1}$ . Then for each positive integer $k$ , we can create another regular lattice $n$ -gon with side length $l_{k} = l_{1} a^{k}$ , where $a = 2 \sin (\frac{180 °}{n})$ . If $k > \frac{- \log (l_{1})}{\log (a)}$ , then $l_{k} < 1$ , contradicting the fact that $P_{k}$ is a lattice polygon.

Therefore, we can conclude that for $n \geq 7$ , there is no regular lattice $n$ -gon.

Through the cases we have ruled out the existence of a regular lattice $n$ -gon for all $n \geq 3$ except for $n = 4$ . Of course, there are plenty of lattice squares!

There are a couple of things worth discussing about the solutions above.

With a little bit of love and care, the solution to Question 1 can be massaged to show that if $A$ , $B$ , and $C$ are distinct lattice points, and if $θ = ∠ A B C$ , then $\tan (θ)$ is a rational number. There is a theorem called Niven’s Theorem which says the following:

Niven’s Theorem: Let $θ = (180 °) (\frac{a}{b})$ , where $a$ and $b$ are integers. If $\tan (θ)$ is rational, then $\frac{a}{b}$ is an integer or $\frac{a}{b} = \frac{2 k + 1}{4}$ where $k$ is an integer.

This theorem can be used to rule out the existence of regular lattice $n$ -gons for all $n$ except for $n = 8$ . The case of regular lattice octagons can then be dealt with separately.
When dealing with the case $n = 5$ and the case $n \geq 7$ in the solution to Question 4, we obtained an infinite sequence of positive numbers $l_{1}, l_{2}, l_{3}, \dots$ with the property that $l_{i} > l_{i + 1}$ for all positive integers $i$ . We needed to show that there is some $k$ large enough so that $l_{k} < 1$ . The fact that the sequence $l_{1}, l_{2}, l_{3}, \dots$ is decreasing does not guarantee that the sequence eventually becomes smaller than $1$ . To see this, consider the sequence $1 + \frac{1}{2}, 1 + \frac{1}{3}, 1 + \frac{1}{4}, \dots$ . This is a sequence of positive numbers that is decreasing, but is never less than $1$ . This is why we had to go through so much trouble to find an explicit $k$ and prove that $l_{k} < 1$ .

A computation of $\sin (18 °)$ and $\cos (36 °)$

Here we will prove that $\sin (18 °) = \frac{1}{4} (\sqrt{5} - 1)$ and $\cos (36 °) = \frac{1}{4} (\sqrt{5} + 1)$ .

Consider the regular pentagon $A B C D E$ with side length $1$ as shown in the diagram below.

Extend the line $D C$ in both directions and let $S$ and $T$ be points on the extended line so that $E S$ and $B T$ are perpendicular to $D C$ . From our solution to Question $2$ (a), we know $S T$ is parallel to $E B$ , and therefore, the length of $E B$ is equal to the length of $S T$ .

In our solution to Question $2$ (a) we showed that $∠ A E B = ∠ A B E = 36 °$ . Therefore the length of $E B$ is $2 \cos (36 °)$ .

Since $∠ E D C = ∠ B C D = 108 °$ , we have $∠ E D S = ∠ B C T = 72 °$ and so $∠ S E D = ∠ C B T = 18 °$ . Then the length of $S T$ is the sum of the lengths of $S D, D C$ , and $C T$ . The length of $D C$ is $1$ , and the lengths of $S D$ and $C T$ are both $\sin (18 °)$ . Since the length of $E B$ is equal to the length of $S T$ we have $2 \cos (36 °) = 2 \sin (18 °) + 1.$ By the double angle formula for cosine, we have $2 (1 - 2 (\sin (18 °))^{2}) = 2 \sin (18 °) + 1.$ If we let $x = \sin (18 °)$ we have that $x$ satisfies $4 x^{2} + 2 x - 1 = 0.$ The quadratic formula then gives us $x = \frac{- 1 \pm \sqrt{5}}{4} .$ Since $\sin (18 °) > 0$ we must have $\sin (18 °) = \frac{1}{4} (\sqrt{5} - 1)$ . Using the equation $2 \cos (36 °) = 2 \sin (18 °) + 1$ gives us $\cos (36 °) = \frac{\sqrt{5} - 1}{4} + \frac{1}{2} = \frac{\sqrt{5} + 1}{4} .$

Problem of the Month Solution to Problem 6: Regular Polygons and Lattice Points

Problem of the Month
Solution to Problem 6: Regular Polygons and Lattice Points