1989 AHSME Problems/Problem 23

Problem

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?

$[asy] import graph; Label f; f.p=fontsize(6); xaxis(0,3.5,Ticks(f, 1.0)); yaxis(0,4.5,Ticks(f, 1.0)); draw((0,0)--(1,0)--(1,1)--(0,1)--(0,2)--(2,2)--(2,0)--(3,0)--(3,3)--(0,3)--(0,4)--(1.5,4),blue+linewidth(2)); arrow((2,4),dir(180),blue); [/asy]$

$\text{(A)}\ (35,44)\qquad\text{(B)}\ (36,45)\qquad\text{(C)}\ (37,45)\qquad\text{(D)}\ (44,35)\qquad\text{(E)}\ (45,36)$

Solution

Squares of size $1\times1,\ 2\times2,\ 3\times3,\ ...$ are successively enclosed between the path and the axes.

$[asy] import graph; axialshade((0,0)--(0,3)--(3,3)--(3,0)--cycle,yellow,(-6,-6),white,(3,3)); axialshade((0,0)--(0,2)--(2,2)--(2,0)--cycle,orange,(-3,-3),white,(2,2)); axialshade((0,0)--(0,1)--(1,1)--(1,0)--cycle,red,(-2,-2),white,(1,1)); Label f; f.p=fontsize(6); xaxis(0,3.5,Ticks(f, 1.0)); yaxis(0,4.5,Ticks(f, 1.0)); draw((0,0)--(1,0)--(1,1)--(0,1),red+linewidth(2)); draw((0,1)--(0,2)--(2,2)--(2,0),orange+linewidth(2)); draw((2,0)--(3,0)--(3,3)--(0,3),yellow+linewidth(2)); draw((0,3)--(0,4)--(1.5,4),green+linewidth(2)); arrow((2,4),dir(180),green); dot((0,1),red);dot((2,0),orange);dot((0,3),yellow); [/asy]$

It takes $1+1+1$ minutes to enclose the first square, $1+2+2$ minutes to enclose the second, $1+3+3$ minutes to enclose the third, and so on. After odd squares, the particle is on the Y axis; after even squares, the particle is on the X axis.

First we find the highest integer $n$ such that $\sum_{k=1}^n1+2k\le1989$ . The sum is equal to $\sum_{k=1}^n1+2\sum_{k=1}^nk=n+n(n+1)=n(n+2)=(n+1)^2-1$ so the highest value is $n=43$ for which the sum is $1935$ .

After $1935$ minutes the 43rd square is enclosed and the particle is at the point $(0,43)$ . During the 1936th minute it moves up to $(0,44)$ . At the end of the 1980th minute it has moved right to $(44,44)$ . After this it moves downward, and at the end of the 1989th minute it is at $(44,35)$ . The answer is $\boxed{\textbf{(D)}}$ .

Solution 2

Note that it takes $n^2+n$ minutes to travel to any point in the format $(n,n)$ <-- aka the top-right corners of the "square-like" shapes. (try this yourself if you dont believe it)

Additionally, if $n$ is divisible by 2, the next step would be to go down parallel to the y-axis from $(n,n)$ . Else, the next step would be to go left parallel to the x-axis from $(n,n)$ . We now find a value for $n$ such that $n^2+n = n(n+1) \le 1989$ .

Note that $40^2 = 1600$ . After testing values over 40, we notice that $44(44+1) = 1980$ . Therefore it takes 1980 seconds to move to $(44,44)$ . Since $n$ is divisible by 2, we just need to move down $9$ units from $(44, 44)$ , yielding the final answer of $\boxed{\textbf{(D)}}$ .

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1989 AHSME Problems/Problem 23

Contents

Problem

Solution

Solution 2

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