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  1. #1
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    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?
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: lattice pie - another puzzle stuck on

    Quote Originally Posted by sonia1 View Post
    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}|.
    Attached Thumbnails Attached Thumbnails lattice pie - another puzzle stuck on-mhf1.bmp  
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  3. #3
    MHF Contributor Drexel28's Avatar
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    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

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  4. #4
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    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.

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  5. #5
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    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!}


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