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Three boats are sailing on the ocean. The three boats form the corners of an equilateral triangle. Boat 1 heads southwards with a constant speed of 10 kn, boat 2 heads westwards with unknown constant speed and boat 3 heads towards northeast also with unknown constant speed. Determine the unique velocities of boat 2 and boat 3 such that the boats keep forming an equilateral triangle for all later times.

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    $\begingroup$ Hm... are the boats actually on an infinite plane? $\endgroup$
    – Weather Vane
    May 29 at 21:07
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    $\begingroup$ Yes, they are not on a sphere but on an infinite plane. $\endgroup$
    – Herbert Kociemba
    May 29 at 21:08
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    $\begingroup$ I'm voting to reopen this question because it is clearly counterintuitive to many while allowing a mildly surprising even somewhat elegant solution. I'd be interested to know what convinced the close voters that this is a "text book style" problem. $\endgroup$
    – loopy walt
    May 31 at 5:11
  • $\begingroup$ I added a second question which makes the problem even more counterintuitive. $\endgroup$
    – Herbert Kociemba
    May 31 at 12:47
  • $\begingroup$ Are you sure this is just counterintuitive and not outright wrong? Boats 1 and 2 fully determine one side of the triangle complete with position and orientation. There is no wiggle room left for the third corner other than flipping up and down which an actual physical object cannot do. $\endgroup$
    – loopy walt
    May 31 at 20:53

2 Answers 2

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As we are not given the size of the starting triangle it should work with any one or else we should find a contradiction if we work with another size. The simplest one I know has size $0$ so I will assume all three boats start at the origin. The velocity of boat $1$ is $(0,-10)$. Let the velocity of boat $2$ be $(-a,0)$ and the velocity of boat $3$ be $(b,b)$ The time will divide out so we will take it to be $1$. We need the points $(0,-10), (-a,0), (b,b)$ to form an equilateral triangle. Letting the side be $d$ we have $$a^2+100=d^2\\(a+b)^2+b^2=d^2\\b^2+(b+10)^2=d^2$$
which gives $$a=10\\b= 5(\sqrt 3-1)\\d=10\sqrt 2$$ and the speed of boat $2$ is $10$ while the speed of boat $3$ is $5\sqrt 2(\sqrt 3-1)$

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  • $\begingroup$ Did you prove that there is no contridiction with any other start conditions? $\endgroup$
    – Kris Van Bael
    May 29 at 23:02
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    $\begingroup$ Your approach is interesting. Your proof relies on the fact that indeed the speeds miraculous work for all starting conditions. For most starting conditions the three boats will not meet in one point in the past or in the future and the computations get much more complicated. The three equalities are correct, you just made a fault when solving them. You should get b=5(sqrt(3)-1) and v3= 5sqrt(2)(sqrt(3)-1) $\endgroup$
    – Herbert Kociemba
    May 30 at 1:07
  • $\begingroup$ @HerbertKociemba: Yes, I miscopied the solution and have fixed it. $\endgroup$
    – Ross Millikan
    May 30 at 4:27
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    $\begingroup$ @HerbertKociemba Not that miraculous. To see that if triangles ABC and abc equilateral then triangle A+a B+b C+c equilateral, simply subtract the barycentres That doesn't change shapes and leaves us with a clean 120° symmtery $\endgroup$
    – loopy walt
    May 30 at 16:16
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    $\begingroup$ @HerbertKociemba One triangle, ABC say, would be the boats' starting position, the other one, abc, formed by their displacements at any given time, so the absolute positions at that point in time are A+a B+b C+c. Ross's assumption of zero size starting triangle isolates the displacements, abc, and gives an easy way to ascertain that they indeed can be made to form an equilateral triangle at any time. $\endgroup$
    – loopy walt
    May 30 at 19:08
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Given coordinates of 2 vertices of an equilateral triangle,

Those of the 3rd vertex can be described as a linear expression:

$x_3= \frac{x_1+x_2}{2}+ a (y_2-y_1)$
$y_3= \frac{y_1+y_2}{2}+ a (x_1-x_2)$
Where $a = \frac{\sqrt{3}}{2}$ either positive or negative

The geometric equivalent:

Consider the edge of vertex 1 and 2 defining the triangles base. From halfway that base go perpendicular for a times the length of the base.

So, as for velocity of the third vertex:

$x’_3 = - v_2/2 + a.10$
$y’_3 = - 5 + a.v_2$
Because $x’_1=0, y’_1=-10, x’_2=-v_2, y’_2=0$

As the third boat goes straight northeast, these 2 values must be equal and positive.

Therefor:

a is positive,
$v_2=10$ And
$v_3=(10.a-5)\sqrt{2}=5\sqrt{2}(\sqrt{3}-1)$

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  • $\begingroup$ Quite counter-intuitive, actually, that it works with a rotating triangle and constant velocity vectors. $\endgroup$
    – Kris Van Bael
    May 30 at 4:58
  • $\begingroup$ So… for the mathematically challenged among us who mainly see gibberish in this (and the other) answer but also find it very counter-intuitive and mind-boggling that this could ever possibly work, the for-dummies answer is that rot-13(obng gjb tbrf ng gra xabgf, naq obng guerr tbrf ng nebhaq svir cbvag bar-frira-fvk xabgf) – correct? $\endgroup$
    – Janus Bahs Jacquet
    May 30 at 14:48
  • $\begingroup$ Yup, that is correct. $\endgroup$
    – Kris Van Bael
    May 30 at 22:05

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