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In $\mathbb{R}^2$ consider a square (call it $S$) and three triangles (one acute $T_2$ and two obtuse $T_1$ and $T_3$) such that each triangle shares one different side with the square and the triangles and the square are disposed exactly as in the following picture.

enter image description here

Define $P:=S\cup T_1\cup T_2\cup T_3$. Call $x_i$ the vertex of $T_i$ opposed to the side of $T_i$ shared with the square $S$. Choose any point $p$ inside the square $S$ such that all three segments $\overline{px_i}$ are entirely contained in $P$.

Now move all the vertexes of $P$ in a continuous way in such a manner that all the following lengths are not increased:

  • the lengths of all the sides of the square and of the three triangles

  • the lengths of the two diagonals of the square and the lengths of the segments from each $x_i$ to the vertexes of the square which are contained in $P$.

In the following picture I've drawn in blue all the segments whose lengths are not increased moving the vertexes of $P$:

enter image description here

Call $P':=S'\cup T_1'\cup T_2'\cup T_3'$ the polygon obtained in such a manner and $x_i'$ the vertexes of the new triangles.

For any $p'\in S'$ define $d(x_i',p')$ in the following way:

  • If $\overline{x_i'p'}\subset P'$, then $d(p',x_i'):=|p'x_i'|$

  • If $\overline{x_i'p'}\not\subset P'$, then $d(p',x_i'):=$ minimum of $|v'p'|+|v'x_i'|$ for $v'$ which varies between the vertexes of $S'\cap T_i$

I'm wondering if it's always possible to find $p'\in S'$ such that $d(p',x_i')\le |px_i|$ for $i=1,2,3$.

So my question is of course how to prove the existence of $p'$. I'm trying to consider all possible cases in which the vertexes of $P$ could move (given the bonds of the lengths), but it's quite complicated. Do you thing there's a better way to proceed?

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  • $\begingroup$ Should you add the condition that all three segments $\overline{p'x_i}$ are entirely contained in $P$? Otherwise I believe the answer is trivially Yes. $\endgroup$ – Joseph O'Rourke Apr 7 '17 at 15:59
  • $\begingroup$ And must the triangles remain acute/obtuse as in the original? $\endgroup$ – Joseph O'Rourke Apr 7 '17 at 16:00
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    $\begingroup$ It depends on the allowed motions. I would try to represent any motion as the composition of two or more simple motions (e.g. moving only one point slightly), and show that you can find the point with such a representation. Even if you can't decompose every allowed motion this way, maybe the simpler ones are dense in the allowed ones, and thus the result is within epsilon of being true. Gerhard "Then Take Epsilon To Zero" Paseman, 2017.04.07. $\endgroup$ – Gerhard Paseman Apr 7 '17 at 16:04
  • $\begingroup$ @JosephO'Rourke about the first question: I've edited what I mean. About your second question: the triangles do not have to remain acute/obtuse $\endgroup$ – user23116 Apr 7 '17 at 16:13
  • $\begingroup$ @GerhardPaseman the problem with your suggestion is that each simple motion depends on the lengths of the original polygon $P$ and not on the lengths of the previous simple motion. So each step can not be considered independently $\endgroup$ – user23116 Apr 7 '17 at 17:15
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Let me prove a bit more general statement.

Let $P=[v_1\dots v_n]$ and $P'=[v_1'\dots v_n']$ be two solid polygons such that if $[vw]$ is a side of $P$ or a diagonal which lies in $P$ completely then $|v-w|\ge |v'-w'|$. Then there is a short map $P\to P'$ which sends any vertex $v$ of $P$ to the corresponding vertex $v'$ of $P'$.

Note that if there is one diagonal $[vw]$ such that $|v-w|=|v'-w'|$ then we can cut $P$ in two polygons and reduce the question to a simpler case.

First note that we can assume that $|v-w|=|v'-w'|$ for any side of $[vw]$ of $P$. If this is note the case choose a vertex $x'$ so that the triangle $[x'v'w']$ lies in $P'$. Put joints in these vertexes and increase $|v'-w'|$ up to $|v-w|$ so that the rest of diagonals stay the same or increase. The new poygon $P''$ admits a short map to $P'$ so we can make each side to be the same as in $P$ one by one and reduce the question to this case.

Note that the convex angles get smaller in $P'$. Therefore one concave angle, say $a$, should go up, in particular it stays concave. Choose 4 vertexes: $a$, its neighbors $x$ and $y$ and a vertex $b$ visible from $a$. Put joints in these vertexes. We can move the polygon straighting the angle at $a$, keeping the distances $|a-x|$, $|a-y|$, $|b-x|$ and $|b-y|$ and nondecreasing the rest of diagonals. Repeating if necessary, we will get one more diagonal fixed.

P.S. Doing these operations, you can end in a generalized solid polygon --- a flat disc with polygonal boundary, when it is mapped to the plane you can get overlaps.

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  • $\begingroup$ @user23116 it was totally wrong, now it should be correct --- please check it. (Funny problem) $\endgroup$ – Anton Petrunin Apr 7 '17 at 23:14
  • $\begingroup$ I just noticed one problem: if $[bx]$ crosses $[ay]$, then the motion deformation is decreasing the distance $|a-b|$ for a while --- I am not sure how to fix it, but it seems not to be a problem for you 7-gon. $\endgroup$ – Anton Petrunin Apr 8 '17 at 15:55
  • $\begingroup$ I don't follow when you say "$P''$ admits a short map to $P'$": it doesn't seem trivial to me and surely it doesn't follow from Kirszbraun's theorem since we can not take into account all diagonals of the polygons. Am I missing something on why there should be such a short map? $\endgroup$ – user23116 Apr 9 '17 at 9:54
  • $\begingroup$ @user23116 You need to construct a map from triangle $[xyz]$ to $[xyz']$ such that $|y-z|=|y-z'|$ and $\measuredangle[y\,_x^z]\ge\measuredangle[y\,_x^{z'}]$. Let $\ell$ be the bisector of $[y\,_z^{z'}]$, reflect the part of $[xyz]$ lying outside of $\ell$ in $\ell$ and apply the closest-point projection to $[xyz']$. $\endgroup$ – Anton Petrunin Apr 9 '17 at 13:58
  • $\begingroup$ Thank you, I understand there's a short map between the two triangles (as it is guaranteed by Kirszbraun's theorem for the euclidean plane) but I still can't understand how does this map extend to a short map $P''\rightarrow P'$. Also, the map you defined with the nearest point projection won't just collapse the part of $[xyz]$ lying outside of $[yz′]$ on $[yz′]$? $\endgroup$ – user23116 Apr 9 '17 at 22:27
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This was a counterexample under one interpretation of the question, where distance was Euclidean straight-line distance. Now that the OP has defined a new distance measure in cases where the line-of-sight segments are not inside $P'$, I would have to rethink this...

MovingTriangles

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  • $\begingroup$ It is no longer a counterexample with your new definition of distance. But $p$ could be at the original $x_2$. The point was that there is no point $p'$ in $P'$ that can see (by line-of-sight) each of $x_1,x_2,x_3$. $\endgroup$ – Joseph O'Rourke Apr 7 '17 at 16:25

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