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Difference between revisions of "1989 AHSME Problems/Problem 23"

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After <math>1935</math> minutes the 43rd square is enclosed and the particle is at the point <math>(0,43)</math>. During the 1936th minute it moves up to <math>(0,44)</math>. At the end of the 1980th minute it has moved right to <math>(44,44)</math>. After this it moves downward, and at the end of the 1989th minute it is at <math>(44,35)</math>. The answer is <math>\boxed{\textbf{(D)}}</math>.
 
After <math>1935</math> minutes the 43rd square is enclosed and the particle is at the point <math>(0,43)</math>. During the 1936th minute it moves up to <math>(0,44)</math>. At the end of the 1980th minute it has moved right to <math>(44,44)</math>. After this it moves downward, and at the end of the 1989th minute it is at <math>(44,35)</math>. The answer is <math>\boxed{\textbf{(D)}}</math>.
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==Solution 2==
 +
 +
Note that it takes <math>n^2+n</math> minutes to travel to any point in the format <math>(n,n)</math> <-- aka the top-right corners of the "square-like" shapes. (try this yourself if you dont believe it)
 +
 +
Additionally, if <math>n</math> is divisible by 2, the next step would be to go down parallel to the y-axis from <math>(n,n)</math>. Else, the next step would be to go left parallel to the x-axis from <math>(n,n)</math>. We now find a value for <math>n</math> such that <math>n^2+n = n(n+1) \le 1989</math>.
 +
 +
Note that <math>40^2 = 1600</math>. After testing values over 40, we notice that <math>44(44+1) = 1980</math>. Therefore it takes 1980 seconds to move to <math>(44,44)</math>. Since <math>n</math> is divisible by 2, we just need to move down <math>9</math> units from <math>(44, 44)</math>, yielding the final answer of <math>\boxed{\textbf{(D)}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 22:09, 30 October 2025

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 $\boxed{\textbf{(D)}}$.

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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