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First divide the unit square into four squares with sides $s$ of length $\frac{1}{2}$. Now given $9$ points and $4$ squares, by the pigeon hole principle at least one square $S$ which contains three of those points. Those three points form a triangle whose area $A$ is bounded by half the area of $S$ (see here for a proof). Since the sides $s$ of $S$ have ...


If all six points lie on the convex hull, they form a hexagon with interior angle sum $(6-2)\pi=4\pi$, so at least one interior angle is at least $\frac23\pi$, so at least one of the other two angles in the corresponding triangle is at most $\frac\pi6$. Else, one point lies in a triangle formed by three others. At least one angle of that triangle is at most ...


Divide the track length $x$ into $\left\lceil\frac x\epsilon\right\rceil$ segments. Let $S$ denote the set of these segments, and note where the runners are in $S^{10}$ once per time unit. Since this set has a finite number of elements, the runners will at some point have to return to an element they'd visited before. Now shift the movement such that the ...


For each degree two vertex, if there exist a Hamiltonian cycle, will contain the two edges adjacent to the vertex. Since there are $k+1$ vertices of degree two of which none are adjacent, that means there will be $2k+2$ edges you must choose, but a Hamiltonian cycle in a graph of $2k+1$ vertices cannot have that many edges.


Yes. If there are seven people, then they may all have one day of the week for themselves. If there are more than seven (i.e. at least eight) then someone will have to share. In my opinion, you shouldn't really need a formula to come to this answer, but whatever floats your boat.

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