# Combinatorial Geometry Minimum Number of Line Segments at Right Angles to Start and Return to Origin

Question No. 25 of the ASMO  (Math Olympiad) in Image

A Possible Solution

To Get a sum of zero, we need a minimum of 8 addends from the Counting Numbers 1 to 9 as described as follows:

Up or Down 2, 4, 6, 8,

Where positive is “up” and negative is “down”

Since

2 + 8 + (-4) + (-6) = 0

Similarly

Left or Right: 1, 3, 5, 7

1 + 7 + (-3) + (-5) = 0

Where positive is “right” and negative is “left”

What can we say?

Movement for line segments are as follows:

2 , 8 —-> both “up”

(-4) + (-6) —> both “down”

Likewise:

1 , 7 —-> both “right”

(-3) , (-5) —> both “left”

So, a Solution is

Start at (0, 0)

Move right 1 unit (1, 0)

Move up 2 units (1, 2)

Move left 3 units (-2, 2)

Move down 4 units (-2, -2)

Move left 5 units (-7, -2)

Move down 6 units (-7, -8)

Move right 7 units (0, -8)

Move up 8 units (0, 0)

There are exactly 8 segments in the shortest possible path as described above (Answer).

P/S: Here’s a Pictorial Sketch that the Solution above Works

Please pardon the horrible handwritten sketch , LOL