Question

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

Answer

Answer: 580

Concepts:arithmetic sequence distance formula vector

Difficulty rating: 2070

Small Hint:

Group the $40$ days into ten four-day cycles and sum horizontal and vertical displacements separately

Big Hint:

For cycle index $k,$ compare $(4k+1)^2$ with $(4k+3)^2$ , and similarly compare the other pair

Solution:

Index the ten cycles by $k=0,1,\ldots,9.$ The horizontal displacement is $\begin{aligned}&\frac12\sum_{k=0}^9\left((4k+1)^2-(4k+3)^2\right)\\&\quad=\sum_{k=0}^9(-8k-4)\\&\quad=-400.\end{aligned}$ Similarly, the vertical displacement is $\begin{aligned}&\frac12\sum_{k=0}^9\left((4k+2)^2-(4k+4)^2\right)\\&\quad=\sum_{k=0}^9(-8k-6)\\&\quad=-420.\end{aligned}$ Therefore the distance from the start is $\sqrt{400^2+420^2}=580.$

1993 AIME Problem 2

Answer: 580

Difficulty rating: 2070

Small Hint:

Big Hint:

Solution:

Problem 2 in Other Years