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- Mar 10, 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 http://mathhelpboards.com/questions-comments-feedback-25/polymath-project-5054.html

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

Although we are having this in the Challenge subfroum, the spirit is very different.

Some important guidelines can be found here General polymath rules | The polymath blog.

__________________

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.

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:

Now we generalize the notion of congruency.

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

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

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

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 http://mathhelpboards.com/questions-comments-feedback-25/polymath-project-5054.html

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.*__________________

**THE PROBLEM.**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.

*The $n$-dimensional Euclidean space is denoted by $\mathbb R^n$. We use $(\mathbb R^n)^m$ as a shorthand for $\overbrace{\mathbb R^n\times \cdots \times\mathbb R^n}^{m\text{-times}}$.***Notation.**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:

__A__**Definition.****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.

__Let $\mathfrak p$ and $\mathfrak q$ be two points in $(\mathbb R^n)^m$.__**Definition.**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.

__Let $\mathfrak p=(\mathbf p_1,\ldots,\mathbf p_m)$ and $\mathfrak q=(\mathbf q_1,\ldots,\mathbf q_m)$ be two points in $(\mathbb R^n)^m$.__**Theorem.**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 $1\leq i\leq m$.

_________________________

**IMPORTANT NOTE.**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

*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.***refrain**
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