Difference between revisions of "1993 AIME Problems/Problem 2"
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== Problem == | == Problem == | ||
During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went <math>n^{2}_{}/2</math> miles on the <math>n^{\mbox{th}}_{}</math> day of this tour, how many miles was he from his starting point at the end of the <math>40^{\mbox{th}}_{}</math> day? | During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went <math>n^{2}_{}/2</math> miles on the <math>n^{\mbox{th}}_{}</math> day of this tour, how many miles was he from his starting point at the end of the <math>40^{\mbox{th}}_{}</math> day? | ||
| − | |||
== Solution == | == Solution == | ||
| − | {{ | + | On the first day, the candidate moves <math>[4(0) + 1]^2/2\ \text{east},\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}</math>, and so on. The E/W displacement is thus <math>1^2 - 3^2 + 5^2 \ldots +37^2 - 39^2 = \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right|</math>. Applying [[difference of squares]], we see that |
| + | <cmath> | ||
| + | \begin{align*} | ||
| + | \left|\sum_{i=0}^9 \frac{(4i+1)^2 - (4i+3)^2}{2}\right| &= \left|\sum_{i=0}^9 \frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\right|\\ &= \left|\sum_{i=0}^9 -(8i+4) \right|. | ||
| + | \end{align*}</cmath> | ||
| + | The N/S displacement is | ||
| + | <cmath>\left|\sum_{i=0}^9 \frac{(4i+2)^2}{2} - \sum_{i=0}^9 \frac{(4i+4)^2}{2}\right| = \left|\sum_{i=0}^9 -(8i+6) \right|.</cmath> | ||
| + | Since <math>\sum_{i=0}^{9} i = \frac{9(10)}{2} = 45</math>, the two distances evaluate to <math>8(45) + 10\cdot 4 = 400</math> and <math>8(45) + 10\cdot 6 = 420</math>. By the [[Pythagorean Theorem]], the answer is <math>\sqrt{400^2 + 420^2} = 29 \cdot 20 = \boxed{580}</math>. | ||
== See also == | == See also == | ||
| − | {{AIME box|year= | + | {{AIME box|year=1993|num-b=1|num-a=3}} |
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:56, 13 March 2015
Problem
During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went 
miles on the 
day of this tour, how many miles was he from his starting point at the end of the 
day?
Solution
On the first day, the candidate moves ![$[4(0) + 1]^2/2\ \text{east},\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}$](http://latex.artofproblemsolving.com/b/3/2/b324e8eaf2302ed847dd08005526e4104bab8920.png)
, and so on. The E/W displacement is thus 
. Applying difference of squares, we see that
The N/S displacement is
![\[\left|\sum_{i=0}^9 \frac{(4i+2)^2}{2} - \sum_{i=0}^9 \frac{(4i+4)^2}{2}\right| = \left|\sum_{i=0}^9 -(8i+6) \right|.\]](http://latex.artofproblemsolving.com/9/6/6/9669af3dde91ab7da2f1bf716b39f186fcf118bc.png)
Since 
, the two distances evaluate to 
and 
. By the Pythagorean Theorem, the answer is 
.
See also
| 1993 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
