Maximin diameter of a transformed convex figure You are given a compact convex figure in the plane, $C$.
Your goal is to transform $C$ to a different figure $C'$ with the smallest possible diameter, as long as:


*

*The trasformation from $C$ to $C'$ is affine and invertible (e.g. scaling in one or more directions);

*$C'$ contains a disc with diameter $1$.


For example:


*

*If $C$ is an ellipse, you can transform it to a disc with diameter $1$.

*If $C$ is a parallelogram, you can transform it to a square with side-length $1$ and diameter $\sqrt{2}$.

*If $C$ is a triangle, you can transform it to an equilateral triangle with diameter $\sqrt{3}$.


An ellipse is obviously the best case; what is the worst case?
Cross-post from math.se
 A: Second version after incomplete first version, using John's theorem and some remarks thereon by Keith Ball.  
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The worst case is the triangle (as intuition predicts). A few images would be really helpful, but at the moment I don't have time for this, so let me just indicate the main arguments.
To see why the triangle is the worst case, we consider a compact convex $C$, with boundary $\delta C$. 
By John's theorem we know that $C$ contains a maximal-area (inscribed) ellipse $E_C$. W.l.o.g. the center of $E_C$ is the origin $(0,0)$ (otherwise translate). Our first affine transformation (say $G$) is to downscale $C$ in the longitudinal direction of the semi-major axis of $E_C$, with factor $\frac{a}{b}$ where $a,b$ are the lengths of the semi-minor and semi-major axis of $E_C$ respectively. 
We then have that $G(E_C)$ is a circle, say $M$. $M$ is the maximal-area inscribed ellipse of our transformed compactum $G(C)$. Our second affine transformation $H$ is a global scaling with a factor $\frac{1}{r}$ where $r$ is the radius of $M$.
Then $H\circ G (C)=C'$ has $H(M)$ as maximal-area inscribed ellipse, and $H(M)=M'$ is the unit circle with radius $1$.
Now we apply (Keith Ball's comments on) John's theorem, to find $m$ real numbers $c_i>0$ (for $1\leq i\leq m$) and corresponding intersection points $u_i$ of $M'$ with $\delta C'$, the boundary of $C'$, such that:
a) $\Sigma c_i\cdot u_i=0$
b) $\Sigma c_i\cdot \langle x,u_i\rangle\cdot u_i=x$ for all $x\in \mathbb{R}^2$ 
John's theorem actually states that the unit circle (which is $M'$) is the maximal-area inscribed ellipse of a convex $K$ if and only if $K$ has intersection points $u_i$ with the unit circle admitting real numbers $c_i>0$ such that criteria a) and b) hold.
In Keith Ball's words: Condition a) shows that the contact points do not all lie "on one side" of the unit circle, and condition b) that they do not all lie "close to a proper subspace". In other words, they are sufficiently spread out over the unit circle. 
$C'$ is contained entirely within the polygon $P$ delineated by the $m$ tangent lines $l_i$ to the unit circle $M'$ in the points $u_i$ respectively (picture!). The trivial reason for this is that $C$ is convex. 
The polygon $P$ also has the unit circle as maximal-area inscribed ellipse, once again by John's theorem since a) and b) also hold for $P$.
In Keith Ball's paper it is proved that any convex $K$ for which the unit circle is the maximal-area inscribed ellipse, is contained in the disc around the origin with radius $2$ ($2\cdot B_2$, globally scaled twice the unit disc).
Therefore $P$ is contained in $2\cdot B_2$. From this it follows that the triangle is the worst case, using conditions a) and b). This is elegantly proved in the final remark on page 11,12 of Keith Ball's paper, stating for dimension $n$: 
If $C$ is a convex body with the Euclidean unit ball $B_n$ as maximal-volume ellipsoid, then diam$(C)\leq\sqrt{2n(n+1)}$.
