MHB Can Two Points in a Cube Be Less Than √(3)/2 Apart?

lfdahl
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There are $9$ points in the interior of a cube of side $1$.

a) Show that at least two of them are less than $\frac{1}{2}\sqrt{3}$ apart.

b) Can $\frac{1}{2}\sqrt{3}$ be replaced by a smaller number?
 
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Hint:

Divide the unit cube into $8$ cubes with side length $\frac{1}{2}$
 
Suggested solution:
(a). The cube may be partitioned into eight equal cubes, each with side of $\frac{1}{2}$, so that
at least one of these cubes contains more than one point. The greatest distance
between any two points in the interior of a cube is less than the length of its
longest diagonal, which in the case of the smaller cubes is precisely $\frac{1}{2}\sqrt{3}$.

(b). Placing a point at the center of the unit cube, and the remaining ones arbitrarily
close to the cube's vertices yields, among the distances between any two points,
one arbitrarily close to $\frac{1}{2}\sqrt{3}$. Hence, this number cannot be replaced by a smaller
one in the result of (a).
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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