Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.5 AIME, harder Prove it Estonia

a) There are three numbers $a$ , $b$ , $c$ such that $a \le b \le c$ . Let $p$ , $q$ , $r$ be the pairwise sums $a+b$ , $b+c$ , $c+a$ in the order such that $p \le q \le r$ . Given that $r-q = q-p$ , is it certainly true that $c-b = b-a$ ?

b) There are four numbers $e$ , $f$ , $g$ , $h$ such that $e \le f \le g \le h$ . Let $u$ , $v$ , $w$ , $x$ , $y$ , $z$ be the pairwise sums of those numbers, in the order $u \le v \le w \le x \le y \le z$ . Given that $z-y = y-x = x-w = w-v = v-u$ , is it certainly true that $h-g = g-f = f-e$ ?

Solution

a) If $a \le b \le c$ , then $a+b \le a+c \le b+c$ , due to which $p = a+b$ , $q = a+c$ and $r = b+c$ . Equality $r-q = q-p$ can now be written as $(b+c) - (a+c) = (a+c) - (a+b)$ , simplifying to $b-a = c-b$ .

b) Let $e = 0$ , $f = 1$ , $g = 2$ and $h = 4$ . Their pairwise sums in increasing order are $u = 0+1=1$ , $v = 0+2=2$ , $w = 1+2=3$ , $x = 0+4=4$ , $y = 1+4=5$ and $z = 2+4=6$ . Thus $z-y = y-x = x-w = w-v = v-u = 1$ , but $h-g = 2 \ne 1 = g-f$ .

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