Solution:The robot's first few movements can be described as vectors: $$\langle 10,10 \rangle, \langle 5,-5 \rangle, \langle -5/2, -5/2 \rangle, \langle -5/4, 5/4 \rangle, \langle 5/8,5/8 \rangle ...$$ The sum of the robot's horizontal movements is the following: $$10 + 5 - 5/2 - 5/4 + 5/8 + 5/16 - 5/32 - 5/64 + ...$$ By combining pairs of terms, then using the formula for the sum of an infinite series, we have: $$15 - 15/4 + 15/16 - 15/64 + ...$$ $$15(1 - 1/4 + 1/16 - 1/64 + ...)$$ $$15(\dfrac{1}{1-(-\frac14)})$$ $$15(\frac45)$$ $$12$$ The sum of the robot's vertical movements is the following, which uses the same infinite series as above: $$10 - 5 - 5/2 + 5/4 + 5/8 - 5/16 - 5/32 + 5/64 + ...$$ $$5 - 5/4 + 5/16 - 5/64 + ...$$ $$5(1 - 1/4 + 1/16 - 1/64 + ...)$$ $$5 (\frac45) $$ $$4$$ So the robot's final position is at (12 meters, 4 meters). The robot is farther from the north wall than the south, and farther from the west wall than the east, so the furthest corner is the upper left, at (0 meters, 20 meters). Using the distance formula, the distance in meters between the robot and the upper left corner is $$\sqrt{12^2 + 16^2} = 20.$$ |