Search and Rescue: Part II
Posted January 29, 2001 as a Math Forum EPoW

Introduction: In this problem, students are asked to place a helicopter rescue base between two campgrounds. The location of the base should correspond to certain constraints, such as the population of each campground, and the fact that a minimal distance is favorable. The location is determined by dragging a dot across the map, and the heading and distance information is automatically computed and displayed.

Where’s the Math: This problem encourages students to investigate the concept of locus. By dragging the dot across the map, students can discover that, even if the distance between the base and each of the two campgrounds is required to be the same, there are an infinite number of locations for the base. However, all these locations lie on a certain line: the perpendicular bisector of the line connecting the two campgrounds. This result can be obtained directly from certain theorems in geometry dealing with equidistance and perpendicular bisectors; conversely, students can discover these theorems on their own by experimenting with the applet. The problem also deals with the concepts of minimum distance and proportion, which result from students trying to make the base twice as close to one campground, while still making the resulting distances as small as possible. Standards: Measurement, geometry Role of Components: Geometer’s Sketchpad is used to show the movement of the base and the connecting lines. Number Entry displays the heading values and distances.

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Sample submitted solutions:

From:  Will, age 15
School:  McLean High School, McLean, Virginia

1. How many places could you put the new rescue base in order to guarantee equal response time to either campground?
An infinite number of places, you can position the base anywhere as long as there is an equal distance to each camp.

2. Explain to the engineers where they should build the rescue base and why, using precise information from your map. Remember that the engineers need very precise instructions.
The Base should be at a heading of 135 degrees from the Moose camp and 287.6 from the trout camp. This way it is exactly 2.7 Miles from each camp, allowing equal rescue time.

3. The engineers have come back and reported that there will likely be twice as many campers in the Moose campground. This means that there will be on average twice as many rescue missions required to the Moose campground. Based on this new information, where would you now position the rescue base?
Beacuse it needs to twice as close to the Moose Camp you'd put the new base at a heading of 147 degress from the Moose camp so it is 1.8 miles away from it. And 288.1 degrees from the Trout camp so it's 3.6 miles away. This way the number of rescues will be equally expressed with the number of people.

1. How many places could you put the new rescue base in order to guarantee equal response time to either campground?
An infinite number of places but only two with minimized travel distance.

2. Explain to the engineers where they should build the rescue base and why, using precise information from your map. Remember that the engineers need very precise instructions.
2 places minimize the distance.
A) North of the mountains 132.4 to moose, and 290.2 to trout. This means that it is 2.7 miles to each camp.
B)South of the mountain 112.3 to moose, and 310.3 to trout.
This means 2.6 miles to each park. This position is the closest base that is equal distance between the two camps. The distance could have been shorter if the helicopters would have been allowed to fly over the mountains.

3. The engineers have come back and reported that there will likely be twice as many campers in the Moose campground. This means that there will be on average twice as many rescue missions required to the Moose campground. Based on this new information, where would you now position the rescue base?
In this problem I thought that "average twice as many rescue missions required to the Moose campground" was very important. I took this to mean that, I should form a ratio using the distances between the two camps. The ratio would have to consist of a number that was half of the other number. This base would also still have to be as close to the two camps as possible. I chose 103.8-310 as my position in this problem. This position puts the base 3.6 miles to trout, and 1.8 miles to moose.
I have changed my mind however and decided to keep the camps in the same place. Having the distance to trout longer was not acceptable. Responce time to either camp needs to be minimized. In this position all emergency call responce times will be the same.

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