An ant starts at the origin (0,0), facing north. On day 1, it moves 1 unit forward to (0,1) and turns right 90 degrees. On day 2, it moves 2 units forward to (2,1) and turns right 90 degrees. On day 3, it moves 3 units forward to (2,-2) and turns right 90 degrees. It keeps doing this—on day i, it moves i units forward and turns right 90 degrees. After 65 days, where is the ant?
Can someone tell me how to go about this? Thank you...
On
The motion of the ant starts off like this:
From this it is pretty clear that the ant reaches each corner of the square spiral every 4 days, starting with the NW corner on the 1st day, reaching it again on the 5th, again on the 9th etc. So, it will hit that NW corner again on the 65th day. Where is this in relation to the origin?
Note that the Westward movement of the ant, crossing the y-axis, is always symmetric about the y-axis, which is to say that the ant ends up exactly as far West of the y-axis as it was East of the y-axis at the start of that day. At the end of day 3, for example, the ant is 2 units East of the y-axis in the SE corner and at the end of day 4, it is 2 units West of the y-axis, having travelled 4 units on that day, day 4.
So, all we really need to consider is the movement of the ant on day 64 and day 65. On day 64 the ant will start off in the SE corner, 32 units away from the y-axis and end the day at the SW corner 32 units left of the y-axis. Its x coordinate is now -32.
On the 65th day, it will travel 65 units North (so it's x coordinate will not change), 32 units until it hits the x-axis and the last 33 units of travel past the x-axis until its y coordinate is now 33. So, at the end of day 65 the ant will be at (-32,33)
On
Each pair of odd numbered moves (1, 3), (5, 7), . . . results in a net y-displacement of −2. Each pair of even numbered moves (2, 4), (6, 8), . . . results in a net x-displacement of −2. After 64 days, the ant has made 16 pairs of odd numbered moves and 16 pairs of even numbered moves, so the net displacement is (−32, −32). Move 65 takes it to (-32, -32+65) => (−32, 33).
Rough Sketch
Hint - notice that the motion in the $x,y$ axis is independent - on an even day $2n$ it moves in the $x$ direction $(-1)^{n+1}\cdot 2n$ steps, and on an odd day $2n+1$ it moves in the $y$ direction $(-1)^{n}\cdot (2n+1)$ steps. You can thus solve each individual problem for $x$ and $y$. In $y$ you have - $$1-3+5-7+...=(1-3)+(5-7)+...=-2-2-...$$ And you can do something similar for $x$. Think how many terms are in the corresponding sums, and if possibly the last term in the sum might be different from $-2$ (if there's an odd number of terms).