Will the Triangle People Meet?
Date: 11/09/2007 at 00:23:24 From: Haim Subject: Puzzle: are the triangle people going to meet Triangle ABC has equal sides of 10 meters each. At each corner stand person A, B, and C respectively. At the same moment all of them start moving at 1 meter / second velocity: person-A is moving to person- B, person-B is moving to person-C, and person-C is moving to person- A. Questions: Will they meet? How long it will take them to meet? I think I need to build an integral equation
Date: 11/09/2007 at 19:57:04 From: Doctor Vogler Subject: Re: Puzzle: are the triangle people going to meet Hi Haim, Thanks for writing to Dr. Math. I found this done most easily using polar coordinates. But since I don't know all of the formulas for polar coordinates, I switched back and forth between polar and rectangular a lot. So let's make the origin/pole be the center of the equilateral triangle. Then A is at angle s (or theta) and distance from the pole r. So we may assume that s starts at 0, and r starts at 10m. We also have the rectangular coordinates x = r cos s y = r sin s r^2 = x^2 + y^2 y/x = tan s Furthermore, rotating the whole system by 120 degrees only changes the places of A, B, and C, which means that B is at angle s + 120 degrees and distance r, and C is at angle s + 240 degrees (or s - 120 degrees) and distance r. Next, taking derivatives of the above formulas, we get the relations r r' = x x' + y y' and r^2 s' = x y' - y x' and therefore (by inverting these two equations) x' = x (r'/r) - y s' y' = y (r'/r) + x s' Now, the velocity requirement is (x')^2 + (y')^2 = (1 m/s)^2. Substituting our formulas for x' and y' in polar coordinates, we get (r')^2 + (r s')^2 = (1 m/s)^2. Next, the direction of movement (x', y') of A is the vector from B to A, which is the difference B - A, where B = (-(x + y*sqrt(3))/2, (x*sqrt(3)-y)/2) A = (x, y) and therefore y'/x' = (-x + y*sqrt(3))/(x*sqrt(3) + y) which, converted to polar coordinates, is r r' + (r^2 s')sqrt(3) = 0 or s' = -r'/(r*sqrt(3)). Substituting this into our previous differential equation, we get (r')^2 + (r'/sqrt(3))^2 = (1 m/s)^2 (4/3)(r')^2 = (1 m/s)^2 (r')^2 = (3/4)(1 m/s)^2 and therefore r' = + or -(1/2)*sqrt(3) m/s. I think it's reasonable to assume that the radius is decreasing at the beginning, and therefore, r' is a constant -(1/2)sqrt(3) meters per second, which means that, since r = 10m at time = t = 0, we have r = 10m - (1/2)sqrt(3)t m/s, and so r = 0 at t = 20/sqrt(3) seconds. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
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