# Another Romanian Mathematical Olympiad problem

Ref: Romanian Mathematical Olympiad — Final Round, 1994

Ref: Titu Andreescu

Problem:

Let M, N, P, Q, R, S be the midpoints of the sides AB, BC, CD, DE, EF, FA of a hexagon. Prove that

$RN^{2}=MQ^{2}+PS^{2}$ if and only if MQ is perpendicular to PS.

Proof:

Let a, b, c, d, e, f be the coordinates of the vertices of the hexagon. The points M, N, P, Q, R, and S have the coordinates

$m=\frac{a+b}{2}$, $n=\frac{b+c}{2}$, $=\frac{c+d}{2}$,

$q=\frac{d+e}{2}$, $r=\frac{e+f}{2}$, $s=\frac{f+a}{2}$, respectively.

Using the properties of the real product of complex numbers, (please fill in the gaps here), we have

$RN^{2}=MQ^{2}+PS^{2}$

if and only if

$(e+f-b-c).(e+f-b-c) = (d+e-a-b).(d+e-a-b)+(f+a-c-d).(f+a-c-d)$

That is,

$(d+e-a-b).(f+a-c-d)=0$

hence, MQ is perpendicular to PS, as claimed. QED.

More later,

Nalin Pithwa

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