At the risk of asking an extremely stupid question, suppose that $P\subset\mathbb{R}^2$ is a convex polygon with area $1$ that contains the origin, and let $r$ denote the farthest distance between the origin and a point in $P$, i.e. $r = \max\_{x\in P} \|x\|$. Let $S$ denote a sector of a circle with radius $r$ that is centered at the origin such that $S$ also has area $1$ (i.e. the angle of the sector must therefore be $2/r^2$ radians). My question: is there an obvious area-preserving map $f:P\to S$ such that, for any point $x\in P$, we have $\|x\|\leq \|f(X)\|$?



Map from a convex polygon that increases distance
Begin with a lemma: Let $\ T=\Delta oab$ be a triangle contained in a circle centered at the origin $o$ and let the arc $\alpha$ of the circle be such that the circle's sector $S$ based on $\alpha$ has the same area as $T$. Let the map $f\_T$ from $T$ onto $S$ be defined so that $f\_T(o)=o$, $f\_T$ maps the segment $ab$ linearly (with respect to the arc length) onto $\alpha$, and $f\_T$ maps every segment $ox$ ($x\in ab$) linearly onto the radius $of(x)$. Then $f\_T$ is area-preserving and, obviously, satisfies the condition $\| x\|\le \| f\_T(x)\|$.


That $f\_T$ is area preserving is intuitively clear and can be easily verified by the Jacobian of $f\_T$.


Now, to answer the question in the affirmative, cut $P$ into triangles $T\_1, T\_2,\ldots, T\_k$ along segments connecting $o$ with all vertices of $P$. In case $o$ is a boundary point of $P$, the triangles form a fan adjacent to, but not surrounding $o$, and if $o$ is an interior point of $P$, they form a rosette (cycle) surrounding $o$. Split your sector $S$ into subsectors $S\_1, S\_2,\ldots, S\_k$ so that the area of $S\_i$ equals the area of $T\_i$, and so that their neighbor-order is the same as the neighbor-order of the triangles $T\_1, T\_2,\ldots, T\_k$. This last condition makes sense only in case $o$ is a boundary point of $P$, otherwise the order of $S\_1, S\_2,\ldots, S\_k$ does not matter. In either case, $f$ is obtained by piecing together the functions $f\_{T\_i}$ mapping $T\_i$ to $S\_i$ as described at the beginning; in the first case $f$ will be continuous; and in the second one - only piecewise continuous. Also, in the second case $f$ remains undefined on one edge of one of the triangles. By the way, polygon $P$ need not be convex, only star-shaped with $o$ as a star center.


It remains to decide whether the construction given in this answer is *obvious* or not, as the question reads. This, I suppose, is a matter of opinion.


