What Path? - April 22-26, 1996
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I will be in San Diego this week for the big national math teachers' conference, and while I'm there, I'll try to do some boating or something.If I head out of one spot, and follow a path so that my lines of sight back to the dock and to a bridge up the shore are always perpendicular, what sort of path will I travel?
Annie says: Solutions
Didn't fool many of you this week - 19 correct answers, and 5 incorrect. With all of you along, I surely wasn't going to get lost, was I?
David Love provided a nice explanation, as did several other students, but the real whiz-bang solution came from Brian Gordon, who proved to us that the path is going to be a semicircle.
Several folks neglected to consider the path a semicircle, assuming that we could drive the boat on land - pay careful attention to the setting in which the problem is posed. They got credit, but I won't let them drive if I'm ever in town!
Following are highlights. The names of all the people who submitted correct solutions and most of the solutions are also available, and we welcome two new schools this week.
David Love Grade 11 Akiba Hebrew Academy For this problem you are asked to start at one point with your boat, and continue riding so that the lines of sight back to the dock and to a bridge up the shore are always perpendicular. The path you will follow is simply a semicircle starting at the point point, and going around to the other point. You know this based on a law of geometry that says in a circle, any angle formed using two points on the circle that are opposite each other (or form the diameter), if you draw lines from each of them to the same place anywhere on the circle it will form 90 degree angles, and thus perpendicular lines. Because a boat can not travel on land, it can only make a half circle. Its path is as follows. this is the bridge | | | | | | | | | (this is the path traveled) | | | | | | | | | | | (this is the dock) and you start the button going up and to the right, then halfway through heading back leftish.
Mark Winesett and Rita Beckner Grade 9 Franklin County High School, Rocky Mount, Virginia A: bridge B: dock BF is tangent, AF is radius (of c3). Since a tangent and radius are always perpendicular, they form right angles at the point of intersection. Therefore, the path must follow the semicircle between points A and B for the sight lines to be perpendicular to each other.
Amy Forster Grade 7, age 11 Wilkins/Forster family Crooked Tree Point Cygnet, Tasmania, Australia The path of the boat is a semicircle. Let DB = a line joining the bridge and the Dock. If you plot the points where the line of sight of the dock and bridge are at right angles, you get a series of right angled triangles, each with the same length hypotenuse, line DB. I remembered that joining any point on the circumference of a circle to each end of the diameter will form a right angle at the circumference. The possible points where the boat traces a semicircular path from the dock to the bridge, and the line DB, become the diameter of that circle.
Brian Gordon Dartmouth '92 Your path must be the arc of a semicircle from the dock to the bridge. Here's an explanation that I hope nobody else comes up with. (I like to do my geometry analytically.) Let the dock be (0,0). Let the bridge be at (a,0) along the shore, indicated by the x-axis. Let (x,y) be the coordinates of your boat at any time. Then the slopes from the boat to the dock and bridge must be negative reciprocals. Slope from boat to dock: y/x Slope from boat to bridge: y/(x-a) So we have the equation y/x = (a-x)/y Cross-multiplying yields y^2 = ax - x^2 which can turn into (x - .5a)^2 + y^2 = .25a^2 This is the equation of a circle. Considering that y>0 so that you don't ram the boat on land, you get a semicircle. --bri
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