Let \(A\) and \(B\) have coordinates \((a,b)\) and \((c,d)\) respectively. Let \(\alpha\) and \(\beta\) be the angles from the positive \(x\)-axis to the lines \(OA\) and \(OB\) respectively, measured in the clockwise direction. For example, if \(A\) has coordinates \((1,1)\) and \(B\) has coordinates \((-1,-1)\), \(\alpha = 45\degree\) and \(\beta = 215\degree\).
Assume towards a contradiction, that \(a,b,c\), and \(d\) are integers and \(\angle AOB\) is \(60\degree\).
Case 1: \(a \neq 0\) and \(c \neq 0\).
Since \(0 \leq \alpha,\beta < 360\degree\), then \(\beta - \alpha = \pm 60\degree\) or \(\pm 300\degree\). Therefore, \(\tan(\beta - \alpha) = \pm \tan(60\degree) = \pm \sqrt 3\). We have \(\tan(\alpha) = \frac{b}{a}\) and \(\tan(\beta) = \frac{d}{c}\). By the angle sum formula for tan, \[\begin{align*} \tan(\beta - \alpha) &= \frac{\tan(\beta) - \tan(\alpha)}{1 + \tan(\alpha)\tan(\beta)} \\ &= \frac{\frac{d}{c} - \frac{b}{a}}{1 + \left(\frac{b}{a}\right)\left(\frac{d}{c}\right)}.\end{align*}\] Since \(a,b,c\), and \(d\) are integers with \(a \neq 0\) and \(c \neq 0\), \(\tan(\beta - \alpha)\) 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 \(\angle AOB\) is not \(60\degree\).
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 \(\angle AOB\) is not \(0\degree\), \(90\degree\), or \(180\degree\)). 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 \(B\) \(90\degree\) 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, \(\angle AOB = \angle A'OB'\). We are now back in Case 1 above but with the points \(A'\) and \(B'\), and can conclude \(\angle AOB\) cannot be \(60\degree\).
The interior angle of a regular \(n\)-gon is \(180\degree - \frac{360\degree}{n}\), and so \(\angle BCD = \angle BAE = 108\degree\). Since \(\triangle EAB\) is isosceles, \(\angle ABE = 36\degree\). Therefore, \(\angle CBF = 72\degree\). Since \[180\degree = 72\degree + 108\degree = \angle CBF + \angle BCD,\] we have that \(CD\) is parallel to \(BF\). We can similarly argue that \(BC\) and \(DF\) are parallel, and conclude that quadrilateral \(FBCD\) is a parallelogram.
Suppose \(B\), \(C\), and \(D\) have coordinates \((b_1,b_2)\), \((c_1,c_2)\), and \((d_1,d_2)\) respectively. Since quadrilateral \(FBCD\) 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 \(FBCD\) is a parallelogram, we know \(\overrightarrow{CD} + \overrightarrow{CB} = \overrightarrow{CF}\). Therefore, \(\overrightarrow{D} - \overrightarrow{C} + \overrightarrow{B} - \overrightarrow{C} = \overrightarrow{F} - \overrightarrow{C}\). This implies \(\overrightarrow{F} = \overrightarrow{D} + \overrightarrow{B} - \overrightarrow{C}\). Since \(B\), \(C\), and \(D\) are lattice points, the corresponding vectors \(\overrightarrow{B}\), \(\overrightarrow{C}\), and \(\overrightarrow{D}\) have integer coordinates. Since \(\overrightarrow{F}\) is the sum of three vectors with integer coordinates, \(\overrightarrow{F}\) also has integer coordinates and we can conclude that \(F\) is a lattice point.
We begin by showing that the angle between \(L_i\) and \(L_{i+1}\) is \(\frac{360\degree}{n}\) 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 \(\frac{360\degree}{n}\).
Let \(\theta\) be the angle between \(L_i\) and \(L_{i+1}\). Then, on the original regular \(n\)-gon, we can compute \(\theta\) 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 \(180\degree - \frac{360\degree}{n}\), we must have \[\theta = 180\degree - (180\degree - \frac{360\degree}{n}) = \frac{360\degree}{n}.\] Let \(A\) be the center of \(B_1B_2\cdots B_n\). Then \[\frac{360\degree}{n} = \angle B_1AB_2 = \angle B_2AB_3 = \cdots = \angle B_{n-1}AB_n = \angle B_nAB_1.\] Since each line segment \(AB_i\) is the side length of the original regular \(n\)-gon, the \(n\) triangles \(\triangle B_1AB_2, \triangle B_2AB_3,\ldots,\triangle B_{n-1}AB_n\), and \(\triangle B_nAB_1\) are all congruent.
Therefore, the edges of the polygon \(B_1B_2, B_2B_3,\ldots,B_{n-1}B_n\), and \(B_nB_1\) all have equal length.
Consider \(\triangle B_1AB_2\). Since \(AB_1\) and \(AB_2\) have the same length, it is an isosceles triangle. Since \(\angle B_1AB_2 = \frac{360\degree}{n}\), we have \(\angle AB_1B_2 = \angle AB_2B_1 = 90\degree - \frac{180\degree}{n}\). By the congruence \(\triangle B_1AB_2\) and \(\triangle B_2AB_3\), we have \(\angle AB_2B_3 = \angle AB_3B_2 = 90\degree - \frac{180\degree}{n}\). We can finally compute \[\angle B_1B_2B_3 = \angle B_1B_2A + \angle AB_2B_3 = 90\degree - \frac{180\degree}{n} + 90\degree - \frac{180\degree}{n} = 180\degree - \frac{360\degree}{n}.\] We can compute every interior angle of the \(n\)-gon \(B_1B_2\cdots B_n\) in the same way to get that all interior angles are equal to \(180\degree - \frac{360\degree}{n}\). Therefore, \(B_1B_2\cdots B_n\) is a regular \(n\)-gon.
As in the previous solution, let \(A\) be the center of the regular \(n\)-gon \(B_1B_2 \cdots B_n\), and consider the isosceles triangle \(\triangle B_1AB_2\).
Recall from part (a) that \(\angle B_1AB_2 = \frac{360\degree}{n}\). By drawing a line from the midpoint of \(B_1B_2\) to \(A\), we divide \(\triangle B_1AB_2\) into two congruent right-angled triangles. Considering one of these right-angled triangles gives \[\sin\left(\frac{180\degree}{n}\right) = \frac{x}{2y}.\] Therefore, \[\frac{x}{y} = 2\sin\left(\frac{180\degree}{n}\right).\]
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 \(V_1,\ldots,V_{3n}\) in a clockwise order. Then \(\angle V_nV_{2n}V_{3n} = 60\degree\). Therefore, all three of \(V_n, V_{2n}\), and \(V_{3n}\) 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 \(A_1B_1C_1D_1E_1\). 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 \(E_1B_1\) and \(A_1C_1\) intersect at \(A_2\),
the lines \(A_1C_1\) and \(B_1D_1\) intersect at \(B_2\),
the lines \(B_1D_1\) and \(C_1E_1\) intersect at \(C_2\),
the lines \(C_1E_1\) and \(D_1A_1\) intersect at \(D_2\), and
the lines \(D_1A_1\) and \(E_1B_1\) intersect at \(E_2\).
Then \(A_2B_2C_2D_2E_2\) is a regular pentagon (see if you can prove this!). Suppose the side length of pentagon \(A_1B_1C_1D_1E_1\) is \(l_1\), and the side length of \(A_2B_2C_2D_2E_2\) 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 \(2\), \(\angle E_1A_1E_2 = \angle B_1A_1A_2 = 36\degree\). Since the interior angle of a regular pentagon is \(108\degree\), we have \(\angle E_2A_1A_2 = 108\degree - 36\degree - 36\degree = 36\degree\).
Let \(G\) be a point on \(E_2A_2\) so that \(GA_1\) is perpendicular to \(E_2A_2\). Let \(F\) be the point on \(A_1B_1\) so that \(FA_2\) is perpendicular to \(A_1B_1\). Then \(G\) and \(F\) are the midpoints of \(E_2A_2\) and \(A_1B_1\) respectively (this fact needs proof, but I will leave it up to you!).
There are now two right-angled triangles, \(\triangle A_1FA_2\) and \(\triangle A_1GA_2\). Consider the former triangle. Note that the length of \(A_1F\) is \(\frac12 l_1\) and \(\angle A_2A_1F = 36\degree\). Let \(x\) be the length of \(A_1A_2\). Then \[\frac{l_1}{2x} = \cos(36\degree).\] Now consider right-angled triangle \(\triangle A_1GA_2\). We have \(\angle GA_1A_2 = 18\degree\), and the length of \(GA_2\) is \(\frac 12 l_2\). Therefore, \[\sin(18\degree) = \frac{l_2}{2x}.\] Solving for \(x\) in both of these equations and rearranging gives \[l_2 = l_1 \frac{\sin(18\degree)}{\cos(36\degree)}.\] At the end of this solution, there is an extra section which shows how to deduce that \(\sin(18\degree) = \frac{\sqrt 5 - 1}{4}\) and \(\cos(36\degree) = \frac{\sqrt 5 + 1}{4}\). 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 \(A_kB_kC_kD_kE_k\) 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 \(E_kB_k\) and \(A_kC_k\) intersect at \(A_{k+1}\),
the lines \(A_kC_k\) and \(B_kD_k\) intersect at \(B_{k+1}\),
the lines \(B_kD_k\) and \(C_kE_k\) intersect at \(C_{k+1}\),
the lines \(C_kE_k\) and \(D_kA_k\) intersect at \(D_{k+1}\), and
the lines \(D_kA_k\) and \(E_kB_k\) 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 \(k\), \[l_k = \left(\frac{3 - \sqrt 5}{2}\right)^{k-1}l_1.\] 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 \[k - 1> \frac{\log\left(\frac{1}{l_1}\right)}{\log\left(\frac{3 - \sqrt 5}{2}\right) }.\] Since \(\frac{3-\sqrt 5}{2} < 1\), \(\log\left(\frac{3-\sqrt 5}{2}\right) < 0\). Therefore, \[\begin{align*} (k-1)\log\left(\frac{3 - \sqrt 5}{2}\right) &< \log\left(\frac{1}{l_1}\right) \\ \Rightarrow \quad \quad \left(\frac{3 - \sqrt 5}{2}\right)^{k-1} &< \frac{1}{l_1} \\ \Rightarrow \quad \quad l_1\left(\frac{3 - \sqrt 5}{2}\right)^{k-1} &< 1 \\ \Rightarrow \quad \quad l_k&< 1.\end{align*}\] Great! Let’s put everything together. Suppose \(A_1B_1C_1D_1E_1\) is a regular lattice pentagon with length \(l_1\). Then by Question 2(b), the regular pentagon \(A_kB_kC_kD_kE_k\) 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 \(A_kB_kC_kD_kE_k\) satisfies \(l_k < 1\). Since the smallest distance between two lattice points in the plane is \(1\), \(A_kB_kC_kD_kE_k\) 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 \[l_k = l_1\left( 2\sin\left(\frac{180\degree}{n}\right)\right)^k.\] Between \(0\degree\) and \(90\degree\), the sine function is strictly increasing. Therefore, for \(n \geq 7\) we have \[0 < 2\sin\left(\frac{180\degree}{n}\right) <2\sin\left(\frac{180\degree}{6}\right) =1.\] 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 \(a = 2\sin\left(\frac{180\degree}{n}\right)\) 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 \[\begin{align*} k\log(a) &< \log\left(\frac{1}{l_1}\right) \\ \Rightarrow \quad \quad a^k &< \frac{1}{l_1} \\ \Rightarrow \quad \quad l_1a^k &<1 \\ \Rightarrow \quad \quad l_k &<1.\end{align*}\] 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 \(A_1A_2\cdots A_n\), 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 \(B_1B_2\cdots B_n\). 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 \(l_k = l_1a^k\), where \(a = 2\sin\left(\frac{180\degree}{n}\right)\). 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!
