We have a convex polygon $P$ in the plane and two points $S,T$ in the boundary of $P$, dividing the perimeter in a proportion $1:2$. Three distinct points in the boundary, denoted by $A,B,C$ start to move simultaneously along the boundary, in the same direction and with the same speed. Prove that there will be a moment in which one of the segments $AB, BC, CA$ will have a length smaller or equal than $ST$.
Problem
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Tags: geometry
bobthesmartypants
27.11.2016 03:14
By Pigeonhole one of the arcs that $AB, BC, CA$ dictate must have size at most 1/3rd the perimeter of $P$. WLOG let this arc be $AB$.
If the length of arc that $AB$ creates is exactly 1/3rd the perimeter of $P$, then we're done, since at some point during the movement $AB=ST$.
Otherwise, consider the lines parallel to $ST$ that intersect $P$ at two points $A', B'$ such that $A'B'$ determines an arc equal in length to the arc $AB$ determines. Note that exactly two of these lines exist, as a continual movement of the line starting from $ST$ and going one direction until the line does not intersect $P$ anymore creates a continuous decrease in the arc length it dictates; the same is true for the line going the opposite direction.
Let the two lines explained above intersect the interior of $P$ at segments $\ell_1, \ell_2$. If the length of $\ell_1, \ell_2$ is less than or equal to the length of $ST$, then we are done as at some point $A,B$ will rotate onto the intersection of one of these lines with $P$. Thus, for the sake of contradiction, assume both the segments $\ell_1$ and $\ell_2$ must be strictly greater than the length of $ST$. Thus, if we take the convex hull of $\ell_1$ and $\ell_2$, we see that at least one of $S, T$ must be strictly inside it; but this gives a contradiction of the convexity of $P$, so we're done. $\Box$