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    caffeinemachine's Avatar
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    Mar 2012
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    Hello MHB.
    For a long time I had wanted to take part in a Polymath Project on MHB. I had tried to gather people to get us started with a polymath project here The Polymath Project
    Some members showed interest but it didn't happen.
    Recently, I wrote to Jameson asking his permission to start a Polymath thread in the Challenge subforum and he was just as excited as me.

    A Polymath Project is a discussion among mathematicians (MHBians in our case) to work on a difficult problem and finding an elegant solution to the problem. Generalizations of the original problem may be discovered and some questions related to the original problem may also be discussed.

    It is a collaboration and not a competition.

    Although we are having this in the Challenge subfroum, the spirit is very different.
    Participants are allowed, and in fact encouraged to post half-baked insights to make progress.
    Some important guidelines can be found here General polymath rules | The polymath blog. Please ignore the 'Polymath Structure' section on this page.

    Consider two congruent triangles $ABC$ and $PQR$ in the Euclidean plane.
    Say $|AB|=|PQ|$, $|BC|=|QR|$ and $|CA|=|RP|$.
    One can show, using `high school geometry' that one can superpose $ABC$ onto $PQR$ by rigidly moving (including reflections) the triangle $ABC$.
    The converse is obvious.
    That is, if one can rigidly move a triangle and superpose it on top of another, then the two triangles are congruent.
    This polymath project is devoted to proving (and possibly extending and contemplating related results) a generalized form of the above mentioned fact.

    To state the generalization, first we need to introduce some standard notations.

    Notation. The $n$-dimensional Euclidean space is denoted by $\mathbb R^n$. We use $(\mathbb R^n)^m$ as a shorthand for $\overbrace{\mathbb R^n\times \cdots \times\mathbb R^n}^{m\text{-times}}$.
    The Euclidean distance between two points $\mathbf x=(x_1,\ldots,x_n)$ and $\mathbf y=(y_1,\ldots,y_n)$ of $\mathbb R^n$ is given by $\sqrt{(x_1-y_1)^2+\cdots+(x_n-y_n)^2}$ and is denoted by $\|\mathbf x-\mathbf y\|$.

    Now. First we make precise the notion of `rigidly moving' something:

    Definition. A rigid motion in $\mathbb R^n$ is a function $L:\mathbb R^n\to\mathbb R^n$ such that $$\|L(\mathbf x-\mathbf y)\|=\|\mathbf x-\mathbf y\|$$ for all $\mathbf x,\mathbf y\in\mathbb R^n$.

    Now we generalize the notion of congruency.

    Definition. Let $\mathfrak p$ and $\mathfrak q$ be two points in $(\mathbb R^n)^m$.
    Write $\mathfrak p=(\mathbf p_1,\ldots,\mathbf p_m)$ and $\mathfrak q=(\mathbf q_1,\ldots,\mathbf q_m)$. (Note that each $\mathbf p_i$ and each $\mathbf q_i$ is a member of $\mathbb R^n$.)
    We say that $\mathfrak p$ and $\mathfrak q$ are congruent if $\|\mathbf p_i-\mathbf p_j\|=\|\mathbf q_i-\mathbf q_j\|$ for all $1\leq i,j\leq m$.

    And finally, the following will be the target of this project.

    Theorem. Let $\mathfrak p=(\mathbf p_1,\ldots,\mathbf p_m)$ and $\mathfrak q=(\mathbf q_1,\ldots,\mathbf q_m)$ be two points in $(\mathbb R^n)^m$.
    Then $\mathfrak p$ and $\mathfrak q$ are congruent if and only if there exists a rigid motion $L:\mathbb R^n\to \mathbb R^n$ such that $L(\mathbf p_i)=\mathbf q_i$ for all $1\leq i\leq m$.

    I think the problem I have posted is ideal for a Polymath on MHB. But I may be extremely wrong about since this is the first time I am doing it.

    This is not an unsolved problem.
    I solved this problem in a few hours by myself.
    I found this problem stated an used without proof (and without a reference to a proof) in a 1981 paper titled 'Tensegrity Structures' by the authors B. Roth and W. Whiteley.

    If a participant finds the problem too easy and can solve it quickly then he should refrain from posting here. In such a case please send me a Private Message saying that you have solved the problem yourself. Please provide a solution or a sketch of the solution in the PM. At the end of the project, I (or you) will post your solution here assuming it gives new insigths or a different method. At the very least your name will be mentioned in the list of members who solved it independently.
    Last edited by caffeinemachine; January 3rd, 2014 at 08:16.

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