Difference between revisions of "1986 AHSME Problems/Problem 19"
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| + | We imagine this problem on a coordinate plane and let Alice's starting position be the origin. We see that she will travel along two edges and then go halfway along a third. Therefore, her new <math>x</math>-coordinate will be <math>1 + 2 + \frac{1}{2} = \frac{7}{2}</math> because she travels along a distance of <math>2 \cdot \frac{1}{2} = 1</math> km because of the side relationships of an equilateral triangle, then <math>2</math> km because the line is parallel to the <math>x</math>-axis, and the remaining distance is <math>\frac{1}{2}</math> km because she went halfway along and because of the logic for the first part of her route. For her <math>y</math>-coordinate, we can use similar logic to find that the coordinate is <math>\sqrt{3} + 0 - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}</math>. Therefore, her distance is <cmath>\sqrt{\left(\frac{7}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{49}{4} + \frac{3}{4}} = \sqrt{\frac{52}{4}} = \sqrt{13},</cmath> giving an answer of <math>\boxed{A}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 19:45, 9 October 2017
Problem
A park is in the shape of a regular hexagon 
km on a side. Starting at a corner,
Alice walks along the perimeter of the park for a distance of 
km.
How many kilometers is she from her starting point?
Solution
We imagine this problem on a coordinate plane and let Alice's starting position be the origin. We see that she will travel along two edges and then go halfway along a third. Therefore, her new 
-coordinate will be 
because she travels along a distance of 
km because of the side relationships of an equilateral triangle, then km because the line is parallel to the -axis, and the remaining distance is 
km because she went halfway along and because of the logic for the first part of her route. For her 
-coordinate, we can use similar logic to find that the coordinate is 
. Therefore, her distance is ![\[\sqrt{\left(\frac{7}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{49}{4} + \frac{3}{4}} = \sqrt{\frac{52}{4}} = \sqrt{13},\]](http://latex.artofproblemsolving.com/7/a/d/7ad22f1a96150a6c4b502c0e7667ccee8f75e297.png)
giving an answer of 
.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
