# If I stretch a convex polygon, does the original fit into the streched version?

Suppose you have a convex polygon $$P=\mathrm{conv}(\{(x_1,y_1),\dots, (x_k,y_k)\})$$ and you stretch it in one dimension, that is, we choose $$\alpha>1$$ and get a new polygon $$P^\alpha=\mathrm{conv}(\{(\alpha x_1,y_1),\dots, (\alpha x_k,y_k)\})$$.

Is it true that you can translate and rotate $$P$$ to make it fit into $$P^\alpha$$?

This seems true to me, as you are somehow making it bigger and "keeping the shape", but I have no additional insight in how one could prove such a thing.

• I feel a possible counterexample in slightly offset rhombuses - like take a rhombus and rotate it so the diagonals is slightly off of parallel to the axes - but I don't have the tools here to confirm. – Dan Uznanski Oct 4 at 11:20
• You need to prove this only for a triangle, since you can triangulate polygons. – lightxbulb Oct 4 at 12:01
• @lightxbulb But you will not necessarily be able to put the pieces compatibly together. – Michal Adamaszek Oct 4 at 12:49
• @DanUznanski: I had the same thought, but my sketching convinced me that you are OK for those shapes. – Ross Millikan Oct 4 at 15:02
• The special (limiting) case of ellipsoids can be addressed via linear algebra: Let the original ellipsoid be $E_1=\{Mx:\|x\|=1\}$, and the stretched ellipsoid be $E_2=\{Nx:\|x\|=1\}$ with $N=AM$ where $A=\operatorname{diag}(\alpha,1,\dots,1)$. Then $E_1$ can fit inside $E_2$ if and only if the singular values of the corresponding matrices satisfy $\sigma_1(M)\le\sigma_1(N), \dots, \sigma_n(M)\le\sigma_n(N)$. But this is true, since $\sigma_i(N)\in[\sigma_{\min}(A)\sigma_i(M),\sigma_{\max}(A)\sigma_i(M)]=[\sigma_i(M),\alpha\sigma_i(M)]$. So ellipsoids cannot provide a counterexample. – Rahul Oct 4 at 16:35

I think this hipothesis not true.

With a simple transformation we can make the stretch direction as parallel to X-axis. So, I draw two consecutive edges $$A,B,C$$ of a convex polygon, let's say they are at "north", and do a X-scale to get $$D,E,F$$:

As you can see, the original $$ABC$$ must be translated to $$UEV$$, by an horizontal distance $$BE=d=\alpha·B_x-B_x=(\alpha-1)B_x$$. It must be also rotated so $$U,V$$ be inner to $$DEF$$

The problem arises when you want to apply the same translation to the rest of vertices.

$$BE$$ must be the greatest distance of all possible-applied translations, otherwise other vertices will fall outside of the stretched polygon. This itself does not seem a great issue, just choose the biggest one. But it may be that that chosen distance is "$$d_{max}>>d$$; in this case translating $$B$$ with $$d_{max}$$ will make it to fall at the very right of $$E=\alpha·B$$

To get things worst, each translation has an associated rotation. The likelihood of any $$translation, rotation$$ pair fitted for one vertex, to be valid for the entire polygon is very low, zero in general.

It's also obvious that there are cases where such a pair can be obtained. For example the triangle in the image above.

• Sorry, I can't make sense of your argument. You have to apply a single translation/rotation pair to the entire polygon, not a different translation and rotation to each vertex. Of course, if you have a nonzero rotation, every vertex will be translated by a different amount -- but it will always be (translation of some chosen reference point) + (rotation of vertex relative to reference point). // Anyway, you cannot simply assert "The likelihood of any translation, rotation pair fitted for one vertex, to be valid for the entire polygon is very low, zero in general" without any justification. – Rahul Oct 4 at 17:46
• Triangles aren't enough in general here; you will always end with all three edges longer and at least one of the angles larger. Just fit the matching angle and the rest takes care of itself. – Dan Uznanski Oct 4 at 18:59