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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/ 
Associated Topics:
College Triangles and Other Polygons

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