A Spiral for This Year

Calculus Level 4

Chong begins at his house and runs 2015 miles north, 2014 miles east, 2013 miles south, 2012 miles west, 2011 miles north... 2 miles east, and 1 mile south. What is the straight line distance from where he began to where he ended in miles?

If the answer can be represented by the form a b a\sqrt{b} , where a a and b b are positive integers and b b is squarefree, find a b ab .

Problem credit to my best friend William Asai


The answer is 2016.

Trevor Arashiro by Trevor Arashiro , 22, USA

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2 solutions

Jake Lai
May 10, 2015

It's actually quite simple!

First, let's consider the x x direction, ie east/west. If Chong travels 2014 miles east and 2012 west, it is clear his net displacement will be 2 miles east. This is the same with 2010 miles east and 2008 west, and 2006 miles east and 2004 miles west, and... you get the idea. There are a total of 2014 + 2 4 \frac{2014+2}{4} "net displacements" in the x x direction, and hence a total of 1008 miles displaced!

Likewise for the y y direction, ie north/south; there are a total of 2015 + 1 4 \frac{2015+1}{4} "net displacements" in the y y direction, and thus a total of 1008 miles displaced as well.

Hence, by Pythagoras's theorem, we get that the total displacement is

100 8 2 + 100 8 2 = 1008 2 \sqrt{1008^{2}+1008^{2}} = 1008\sqrt{2}

and so a b = 1008 × 2 = 2016 ab = 1008 \times 2 = \boxed{2016} .

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Gj ~~~~~~~~~~

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lol

Trevor Arashiro - 6 years, 1 month ago

Excellent solution, I also divided them into components, except I wrote 4 separate arithmetic series (which later became 2). After that, finding the [final] net displacement in component form was a breeze, then the Pythagorean theorem was used to find the distance.

Chris M. - 6 years ago
Aravind Vishnu
May 11, 2015

When you complete one round (you come nearest to the starting point after start.),you are 2 2 miles away from the starting point. Take this point as the new starting point and continue with 2011 to north, 2010 to east and so on. You will have 2016 4 = 504 such points each at a distance of 2 2 miles from the previous starting point. So the total distance is 504 × 2 2 = 1008 2 . Here a = 1008 and b = 2. So, ab = 2016. \text{When you complete one round (you come nearest to the starting point after start.),you are}\\ 2\sqrt{2}\text{miles away from the starting point. Take this point as the new starting point}\\ \text{and continue with 2011 to north, 2010 to east and so on.}\\ \text{You will have} \frac{2016}{4}=504 \text{such points each at a distance of}{2}\sqrt{2} \text{miles from the previous starting point.}\\ \text{So the total distance is} 504\times2\sqrt{2}=1008\sqrt{2}.\text{Here a}=1008\text{and b}=2.\text{So, ab}=2016.

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