Difference between revisions of "1993 AIME Problems/Problem 2"
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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 == | ||
| + | On the first day, the candidate moves <math>4(0) + 1</math> miles east, then <math>4(0) + 2</math> miles north, and so on. The E/W distance can be represented by <math>\displaystyle \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right|</math>. The N/S distance can be represented by <math>\displaystyle \left|\sum_{i=0}^9 \frac{(4i+2)^2}{2} - \sum_{i=0}^9 \frac{(4i+4)^2}{2}\right|</math>. Combining the sums and simplifying by the [[difference of squares]], we see that <math>\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|</math>. Similarily, the N/S distance turns out to be <math>\left|\sum_{i=0}^9 -(8i+6) \right|</math>. The sum of the numbers from <math>0</math> to <math>9</math> is <math>\frac{9(10)}{2} = 45</math>, so 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 = 580</math>. | ||
| − | == | + | In a similar manner, we can note that <math>1^2 - 3^2 + 5^2 \ldots +37^2 - 39^2 = -8(1 + 3 + 5 \ldots + 19)</math> and that <math>2^2 - 4^2 + 6^2 \ldots + 38^2 - 40^2 = 4(3 + 7 \ldots + 39)</math>, from which we end up with the same answer. |
| − | |||
== See also == | == See also == | ||
{{AIME box|year=1993|num-b=1|num-a=3}} | {{AIME box|year=1993|num-b=1|num-a=3}} | ||
Revision as of 15:43, 26 March 2007
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 
miles east, then 
miles north, and so on. The E/W distance can be represented by 
. The N/S distance can be represented by 
. Combining the sums and simplifying by the difference of squares, we see that 
. Similarily, the N/S distance turns out to be 
. The sum of the numbers from 
to 
is 
, so the two distances evaluate to 
and 
. By the Pythagorean Theorem, the answer is 
.
In a similar manner, we can note that 
and that 
, from which we end up with the same answer.
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 | ||
