HMMT November, 2015

Geometry Difficulty 4.7 AIME Prove it United States

Problem:
Let $a$ and $b$ be positive real numbers. Determine the minimum possible value of
$\sqrt{a^{2}+b^{2}}+\sqrt{(a-1)^{2}+b^{2}}+\sqrt{a^{2}+(b-1)^{2}}+\sqrt{(a-1)^{2}+(b-1)^{2}}$

Solution

Solution:
Answer: $2 \sqrt{2}$
Let $ABCD$ be a square with $A=(0,0)$ , $B=(1,0)$ , $C=(1,1)$ , $D=(0,1)$ , and $P$ be a point in the same plane as $ABCD$ . Then the desired expression is equivalent to $AP + BP + CP + DP$ . By the triangle inequality, $AP + CP \geq AC$ and $BP + DP \geq BD$ , so the minimum possible value is $AC + BD = 2 \sqrt{2}$ . This is achievable when $a = b = \frac{1}{2}$ , so we are done.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.