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Can we always find two parallel lines tangent to a strictly convex plane body with smooth differentiable border such that two curve arcs formed by body’s border between two points at which body touches these parallel lines are equal in length? The statement is true for ellipse, but is it true in general?

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Yes, it is true in general. Consider two arbitrary parallel lines $L1$ and $L2$ touching arbitrary strictly convex plane body $B$ with smooth border at points $P1$ and $P2$. Assume that arc $A1$, which passes body’s border from $P1$ to $P2$ clockwise, is not equal in length to arc $A2$, which passes body’s border from $P1$ to $P2$ anticlockwise. Let’s fix positions of $P1$ and $P2$ at body’s border as well as arcs $A1$ and $A2$. Let’s start rotating body $B$ clockwise $180$ degrees, while keeping lines $L1$ and $L2$ parallel, preserving their initial direction, and keeping them always touching body $B$ at some points $P1*$ and $P2*$ (we may have to change distance between $L1$ and $L2$ when doing so). In the process of rotation, point $P2*$ passes exactly arc $A1$ of body’s border from $P2$ to $P1$ and point $P1*$ passes arc $A2$ from $P1$ to $P2$. $A1*$, arc of body’s border between $P1*$ and $P2*$ going clockwise (when $P1*$, $P2*$ are sliding along body’s border), changes its length during rotation, when we start rotation, it coincides with arc $A1$ and at the end of rotation coincides with arc $A2$. Because length of $A1*$ changes continuously during rotation and $A1$ is not equal to $A2$ in length, we can assume that length of $A1*$ will be equal to $0,5*(length(A1) + length(A2))$ at some moment during rotation and thus equal to length of $A2*$

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