Difference between revisions of "1993 AIME Problems/Problem 2"
Mathgeek2006 (talk | contribs) m (→Solution) |
|||
Line 4: | Line 4: | ||
== 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 | 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*} | \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|. | \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*}</ | + | \end{align*}</cmath> |
The N/S displacement is | 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>. | + | 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 == |
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 , and so on. The E/W displacement is thus . Applying difference of squares, we see that The N/S displacement is 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.