1993 AIME Problems/Problem 2

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 $n^{2}_{}/2$ miles on the $n^{\mbox{th}}_{}$ day of this tour, how many miles was he from his starting point at the end of the $40^{\mbox{th}}_{}$ day?

Solution

On the first day, the candidate moves $4(0) + 1$ miles east, then $4(0) + 2$ miles north, and so on. The E/W distance can be represented by $\displaystyle \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right|$ . The N/S distance can be represented by $\displaystyle \left|\sum_{i=0}^9 \frac{(4i+2)^2}{2} - \sum_{i=0}^9 \frac{(4i+4)^2}{2}\right|$ . Combining the sums and simplifying by the difference of squares, we see that $\left|\sum_{i=0}^9 \frac{(4i+1)^2 - (4i+3)^2}{2}\right| = \left|\sum_{i=0}^9 \frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\right| = \left|\sum_{i=0}^9 -(8i+4) \right|$ . Similarily, the N/S distance turns out to be $\left|\sum_{i=0}^9 -(8i+6) \right|$ . The sum of the numbers from $0$ to $9$ is $\frac{9(10)}{2} = 45$ , so the two distances evaluate to $8(45) + 10\cdot 4 = 400$ and $8(45) + 10\cdot 6 = 420$ . By the Pythagorean Theorem, the answer is $\sqrt{400^2 + 420^2} = 29 \cdot 20 = 580$ .

In a similar manner, we can note that $1^2 - 3^2 + 5^2 \ldots +37^2 - 39^2 = -8(1 + 3 + 5 \ldots + 19)$ and that $2^2 - 4^2 + 6^2 \ldots + 38^2 - 40^2 = 4(3 + 7 \ldots + 39)$ , from which we end up with the same answer.

1993 AIME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1993 AIME Problems/Problem 2

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