TITLE: If I stretch a convex polygon, does the original fit into the streched version?
QUESTION [32 upvotes]: 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.

REPLY [2 votes]: Yes, this is always possible.
Sandwich $P$ between two lines of some variable slope $p$, touching $P$ at $A$ and $B$ respectively, where $AB$ has slope $q$.  Without loss of generality, assume $q > 0$ when $p = 0$.  We can vary this configuration continuously such that $p$ increases from $0$ while $q$ increases toward $\infty$, so $pq$ will increase continuously from $0$ toward $\infty$, while $p < q$ at all times.  (The one exception is if $AB$ becomes vertical while $p = 0$, but then we’re already done: we can fit $P$ into $P^\alpha$ without any rotation.)  Stop when $pq = \alpha$.
We claim that rotating $P$ clockwise about $A$ by $\theta = \tan^{-1} \frac{pq - 1}{p + q}$, then translating it by $\vec{AA^\alpha}$, sends it to a subset of $P^\alpha$.
To see this, it suffices to show that the linear transformation about origin $A$ composed of this rotation by $\theta$ followed by a stretching in the $x$ direction by ratio $\frac{1}{\alpha}$:
$$\begin{bmatrix}
  \frac1\alpha \cos \theta & \frac1\alpha \sin \theta \\
  -\sin \theta & \cos \theta
\end{bmatrix} = \frac{1}{\sqrt{(1 + p^2)(1 + q^2)}}\begin{bmatrix}
  \frac{p + q}{pq} & \frac{pq - 1}{pq} \\
  1 - pq & p + q
\end{bmatrix}$$
sends $P$ to a subset of itself.  You can verify that this transformation has eigenvectors
$$v_1 = \begin{bmatrix}1 \\ p\end{bmatrix}, v_2 = \begin{bmatrix}1 \\ q\end{bmatrix},$$
with eigenvalues
$$\lambda_1 = \frac1p \sqrt{\frac{1 + p^2}{1 + q^2}}, \lambda_2 = \frac1q \sqrt{\frac{1 + q^2}{1 + p^2}},$$
where $\lambda_1 < \lambda_2 < 1$ because
$$\begin{gather*}
\lambda_2 - \lambda_1 = \frac{(q - p)(pq - 1)}{pq\sqrt{(1 + p^2)(1 + q^2)}} > 0, \\
1 - \lambda_2^2 = \frac{(pq)^2 - 1}{(1 + p^2)q^2} > 0.
\end{gather*}$$
So we can reinterpret this transformation as the composition of a stretching along each line of slope $p$ about its intersection with $AB$ by ratio $\frac{\lambda_1}{\lambda_2}$, followed by a dilation about $A$ by ratio $\lambda_2$; both these component transformations send $P$ to a subset of itself because it’s convex.  This proves the claim.