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>. | 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>. | ||
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| + | == See also == | ||
| + | {{AHSME box|year=1989|num-b=22|num-a=24}} | ||
| + | |||
| + | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 08:03, 22 October 2014
Problem
A particle moves through the first quadrant as follows. During the first minute it moves from the origin to 
. 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]](http://latex.artofproblemsolving.com/b/5/9/b59fc45d3e82134723e6aeede797eafa9e65567d.png)
Solution
Squares of size 
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]](http://latex.artofproblemsolving.com/f/0/0/f00c48f86de6b301f3b223ff2c0285b404447a0f.png)
It takes 
minutes to enclose the first square, 
minutes to enclose the second, 
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 
such that 
. The sum is equal to
![\[\sum_{k=1}^n1+2\sum_{k=1}^nk=n+n(n+1)=n(n+2)=(n+1)^2-1\]](http://latex.artofproblemsolving.com/7/4/a/74a094c00816602bdcbb729e5eef40c9db981ba6.png)
so the highest value is 
for which the sum is 
.
After minutes the 43rd square is enclosed and the particle is at the point 
. During the 1936th minute it moves up to 
. At the end of the 1980th minute it has moved right to 
. After this it moves downward, and at the end of the 1989th minute it is at 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
