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Working in $R^2$, consider any continuous curve $C$ with endpoints $a$ and $b$, such that the curve does not intersect the line formed by $a$ and $b$. Then the line $ab$ and $C$ form an enclosure, $E$. Does there exist a curve $C$ and a rotation and translation of $C$, denoted $C'$, such that $C'$ strictly fits inside $E$? That is, any point in $C'$ is strictly contained in $E$, without intersecting its boundaries.

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    $\begingroup$ Would this not show that the area of $E'$ is less than the area of $E$, when in fact they are equal? $\endgroup$ – Matthew Conroy Jun 30 '16 at 18:11
  • $\begingroup$ The curve $C$ and the line $\overline{ab}$ form a closed loop in the plane, so it divides the plane into two regions. If you take $E$ to be the outside region, then it is not hard. Otherwise @MatthewConroy seems to show that it is impossible. $\endgroup$ – Servaes Jun 30 '16 at 18:18
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$C$ doesn't have to be "nice". $E$ is closed and bound in ${\Bbb R}^2$ and so there are 2 points $c,d\in E$ such that their distance is maximal, $m$ (distance is a bound and continuous function on the compact set $E\times E$ so it has a maximum).

It's pretty obvious that any such pair $c,d$ with distance $m$ must be on the boundary (if either or both are interior points then 2 points can be constructed easily that are further away: draw a line and extend it). Also, both points must be on $C$ (if either point is on $ab$, the line will be longer by moving that point into at least one of $a,b$ - the side on which the angle between $ab$ and $cd$ is obtuse or right).

Your mapping will have to move both $c$ and $d$ into $E$'s interior. But no 2 interior points have distance $m$, so it can't.

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I will assumme that a rotation and transformation means an isometry here, and that the curve $C$ must be sufficiently nice (i.e. smooth).

Suppose such a curve and a rotation and transformation exist. Let $\partial E$ denote the boundary of $E$, i.e. it is the union of $C$ and the line segment connecting $a$ and $b$. Let $C'$ denote the rotated and transformed curve and let $a'$ and $b'$ denote its endpoints corresponding to $a$ and $b$ respectively. Let $E'$ and $\partial E'$ denote the enclosed area and its boundary. Note that $\partial E$ and $\partial E'$ are closed loops.

First note that $C'$ is disjoint from $\partial E$ because $C'$ strictly fits inside $E$. By symmetry $C$ is disjoint from the line segment connecting $a'$ and $b'$, so the two boundaries $\partial E$ and $\partial E'$ can only intersect strictly in their line segments. Two line segments intersect in at most one point, whereas two closed loops cannot intersect in a single point (they would be tangent, which the two lines are clearly not).

It follows that $\partial E$ and $\partial E'$ do not intersect, and hence that $E'$ is strictly contained in $E$. Then the area of $E'$ is strictly less than that of $E$, which is impossible because $E'$ is a rotation and transformation of $E$. Hence no such curve exists.

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