# Thread: lattice pie - another puzzle stuck on

1. ## lattice pie - another puzzle stuck on

is it possible to have an equilateral triangle in the plane all of whose vertices lie on the integer lattice{(m,n)|m,n are integers}? If so, what is the shortest side lengh possible for sucha triangle? what about other regular polygons?

2. ## Re: lattice pie - another puzzle stuck on

Originally Posted by sonia1
is it possible to have an equilateral triangle in the plane all of whose vertices lie on the integer lattice{(m,n)|m,n are integers}? If so, what is the shortest side lengh possible for sucha triangle? what about other regular polygons?
The answer is no. There is no such triangle exists.
It easy to prove that fact knowing that the area of regular triangle with side $a$ is $\frac{a^2\sqrt{3}}{4}$ that is irrational.

For $n=4$, is a square which we can describe on system of lattice points.

For $n> 4$ we can not describe the regular polygon on system of lattice points.

Here is a proof for this fact:

Suppose that there is regular polygon with n sides, $n\neq 4$ that his edges are on lattice points. Now, from all regular polygons that can be describe on system of lattice points we choose the one with the minimal side length.

Say that the chosen one is: $P_1P_2P_3...P_n$.

Now, we look at the reflections of $P_{i+1}$ for $i=1,2,3,...,n$ vis-a-vis $P_iP_{i+2}$ (where $P_{n+1}=P_1$ and $P_{n+2}=P_2$).

So, we got a regular polygon with $n$ sides $P_1'P_2'P_3'...P_n'$.

$P_i'$ for $i=1,2,3,...,n$ is lattice point, why?

Due to our contraction $P_{i-1}P_iP_{i+1}P_i'$ is a rhombus.

Now, let $P_i=(t_i,s_i)$ and $P_i'=(t_i',s_i')$.

Let say that $M$ is intersection of the diagonals in rhombus.

We calculate that $M$ with two ways,

First with $P_i=(t_i,s_i)$ :

$M=(\frac{t_{i-1}+t_{i+1}}{2},\frac{s_{i-1}+s_{i+1}}{2})$

Ans with $P_i'=(t_i',s_i')$ :

$M=(\frac{t_{i}+t'_{i}}{2},\frac{s_{i}+s'_{i}}{2})$

Hence, $M=(\frac{t_{i-1}+t_{i+1}}{2},\frac{s_{i-1}+s_{i+1}}{2})=(\frac{t_{i}+t'_{i}}{2},\frac{s_{i }+s'_{i}}{2})$

We deduce from the above that $t'_i$ and $s'_i$ are integers.

Finally, he have regular polygon with $n$ sides $P_1'P_2'P_3'...P_n'$ witch his edges on lattice points.

But, $|P'_iP'_{i+1}|<|P_iP_{i+1}|$ (By our constraction) and this is a contradiction to the fact that $P_1P_2P_3...P_n$ chosen with minimal $|P_iP_{i+1}|$.

3. ## Re: lattice pie - another puzzle stuck on

In fact, it's impossible for an equilateral triangle to have all three vertices lying on $\mathbb{Q}^2$. There are a few high-tech proofs (most notably a counting argument) but it can be done with simple geometry

Spoiler:
Without loss of generality assume that one of the vertices of our triangle is the orgin $O$. Thus we are really examining $\triangle OBC$, where $B,C$ are the other verticies. We may assume that $B=\left(2a,2b\right)$ for some $a,b\in\mathbb{Q}$. From here we break this into two cases, namely: $a,b\ne0$, or $a\text{ or }b=0$. Clearly both cannot be zero since the origin and $B$ are distinct points.

Case 1, $a\text{ or }b=0$: It is clear that we assume WLOG that $a=0$ and $b>0$ for all other cases are analgous. Thus this triangle will have $B=(0,2b)=(0,c)$ and $C=(x,y)$ for some $x,y$. Clearly seeing though that the length of $\overline{OB}$ is $|c|=c$ (since they are both on the y-axis) we may conclude that the length of $\overline{OC},\overline{BC}$ is as well. Realizing though that the lengths of $\overline{OC},\overline{BC}$ are given independently by the distance equations $d\left(O,C\right)=\sqrt{x^2+y^2},d\left(B,C\right) =\sqrt{x^2+(y-c)^2}$ and appealing once again to the equilaterallness of the triangle we may conclude that, in fact, $d\left(O,C\right)=\sqrt{x^2+y^2}=\sqrt{x^2+(y-c)^2}=d\left(B,C\right)$. Solving this equation yields $y=\frac{c}{2}$. As stated earlier though $d\left(O,B\right)=d\left(O,C\right)=d\left(B,C\rig ht)=c$ and focusing mainly on $d\left(O,C\right)$ we see that $\sqrt{x^2+y^2}=\sqrt{x^2+\frac{c^2}{4}}=c\implies x=\pm\frac{\sqrt{3}c^2}{4}$ which is clearly irrational for any rational $c$.

Case 2, $a,b\ne0$: Now we have that $B=(2a,2b)$ where neither $a$ or $b$ is zero. Constructing the midpoint $M$ of $\overline{OB}$ gives us that $M=(a,b)$ and that the length of the segment $\overline{OM}$ is $d\left(O,M\right)=\sqrt{a^2+b^2}$. Now, form $\overline{MC}$ to be the median connecting $M$ and $C$. If we consider the two segments $\overline{OB},\overline{MC}$ to be the restrictions of the lines $\ell_1,\ell_2$ respectively we can see that $\text{slope of }\overline{OB}=\frac{b}{a}$ and since $\overline{OB}\perp\overline{MC}$ we may conclue that $\text{slope of }\overline{MC}=\frac{-a}{b}$ (now you see why we needed $a,b\ne0$). Letting $C=(x,y)$ for some $x,y$ again and realizing that $M,C$ both lie on $\ell_2$ we may conclude that $C=\left(a+\Delta x,b+\Delta y\right)=\left(a+ \Delta x,b+\frac{-a}{b}\Delta x\right)\quad\color{red}\star$ where $\Delta x$ represents the change in $x$ between the points $M$ and $C$. Appealing to this we see that $\text{length of }\overline{MC}=\sqrt{\left(a+\Delta x-a\right)^2+\left(b+\frac{-a}{b}\Delta x-b\right)^2}=\left|\Delta x\right|\sqrt{1+\frac{a^2}{b^2}}$. Remembering though that forming the median $\overline{MC}$ bisected $\angle OBC$ we see that the new triangle formed, $\triangle OMC$ is a $30-60-90$ triangle. In particular, this tells us that $\text{length of }\overline{MC}=\sqrt{3}\cdot\text{length of }\overline{OB}=\sqrt{3}\cdot\sqrt{a^2+b^2}$. Equating our two lengths of $\overline{MC}$ gives us $\left|\Delta x\right|\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3}\sqrt{a^2 +b^2}\implies \left|\Delta x\right|\sqrt{a^2+b^2}=\sqrt{3}\cdot b\cdot\sqrt{a^2+b^2}$ and since $a,b\ne0\implies\sqrt{a^2+b^2}\ne0$ we may conclude that $\left|\Delta x\right|=\sqrt{3}b\implies \Delta x=\pm\sqrt{3}b$. Thus, substitution into $\color{red}\star$ gives us that $C=\left(a\pm\sqrt{3}b,b\mp\sqrt{3}a\right)$ which is clearly irrational since $a,b\in\mathbb{Q}$.

The conclusion follows. $\square$

4. ## Re: lattice pie - another puzzle stuck on

My Calculus 2 professor posed that problem on a Friday.
I'm proud to say that, over the weekend, our class
. . found a variety of ways to to prove it impossible.

On Monday he asked, "Is it possible to have a reqular
. . tetrahedron whose vertices have integer coordinates?"

None of us had a satisfactory proof (either way).
The next class, he blew us away with a simple sketch.

If you've never considered this problem, give it a try.
I'll post my professor's solution later.

5. ## Re: lattice pie - another puzzle stuck on

Here is my professor's solution . . .

Spoiler:

$\text{Construct a cube of side }x \in I^+.$

$\text{Orient it so that its vertices have integer coordinates.}$

$\text{Note the vertices }A, B, C, D.$

Code:
        A
o - - - - - *
/  *        /|
/     *     / |
/        * B/  |
* - - - - - o   |
|         * | * |
|       *   |   o D
x |     *     |  /
|   *       | / x
| *         |/
C o - - - - - *
x
$\text{Draw diagonals }AB, AC, AD, BC, BD, CD.$

$\text{And }there\text{ is our regular tetrahedron!}$