1989 AMC 12 Problem 23

Question

A particle moves through the first quadrant as follows. During the first minute it moves from the origin to

(1,0).

Thereafter, it continues to follow the directions indicated in the figure, going back and forth between the positive

x

and

y

axes, moving one unit of distance parallel to an axis in each minute. At which point will the particle be after exactly

1989

minutes?

(35,44)

(36,45)

(37,45)

(44,35)

(45,36)

Record the particle’s location at times

1,4,9,16,\ldots

Compare

1989

with the consecutive squares

44^2

and

45^2

At time

n^2,

the particle is at

(n,0)

for odd

n

and at

(0,n)

for even

n.

Thus at time

44^2=1936

it is at

(0,44).

It then moves

44

units right, reaching

(44,44)

at time

1980,

and moves downward for the next

9

minutes. At time

1989

it is therefore at

(44,35).

Thus the correct answer is D.

Problem 23 in other years