We had some users saying that that there are no closed loops, but there are if you search carefully!

We received two answers: one from Joseph from Crewe and, which was very visual and one which made full use of algebra, which was submitted by an anonymous user. We particularly liked the way that the solvers appealed to both algebraic and geometrical considerations to solve the problem. Very well done, whoever you were!

FIRST SOLUTION :

Each vector will be expressed in the form (x y), where the x-component represents the units moved to the right (negative means left), and the y-component represents the units moved upwards (negative means down).
Each vector will also be assigned to a letter from A-G:
A: Black
B: Dark blue
C: Green
D: Light blue
E: Pink
F: Red
G: Yellow

Grid 1: The direction of each vector is as follows:

A: (-6 -3) B: (2 3) C: (3 1) D: (-3 0) E: (4 0) F: (-4 1) G: (-2 4)

By considering only the y-direction at first, the only vector with a negative y-value is A. To make the y-component 0, B must be added to A. D and E have a y-component of 0, so adding D and/or E to AB (or even DE by itself) will still keep the y-component of 0. Hence there are only 5 permutations with a y-component of 0: AB, ABD, ABE, ABDE and DE. To make a closed loop, the x-component must also be 0; analysing the x-components of the 4 permutations, only ABE gives an x-component of 0 (-6 + 2 + 4), and so it is the only closed loop.

Grid 2: The direction of each vector is as follows:

A: (-2 3) B: (2 3) C: (2 0) D: (-2 -2) E: (0 -2) F: (2 -3) G: (-2 0)

By visualisation alone, it is easy to see that AF (black and red) and CG (green and yellow) create a closed loop. A combination of ACFG also creates a closed loop. However, these may not be the only closed loops present. Again by considering only the y-direction at first, D and E cannot be used to make a closed loop, since any permutations with D and/or E give y-components of -2 or -4. Adding A and/or B, which are the only vectors with positive y-components, cannot make a y-component of 0. By eliminating D and E, F is the only vector with a negative y-component, which is balanced by addition of A or B. Hence there are only 9 permutations with a y-component of 0: AF, AFC, AFG, ACFG, BF, BFC, BFG, BCFG and CG. To make a closed loop, the x-component must be 0, and so AF, ACFG and CG are the only 3 closed loops.

Grid 3: The direction of each vector is as follows:

A: (0 1) B: (-2 0) C: (3 0) D: (1 -5) E: (-3 3) F: (4 2) G: (-1 -6)

Once again, looking at the y-components, there are only two vectors with negative y-components, D and G. The y-component of D can be made 0 by adding E and F, possibly B and/or C, which have a y-component of 0. The y-component of G can be made 0 by adding E, F and A, possibly B and/or C. There are only 9 permutations with a y-component of 0: AEFG, ABEFG, ACEFG, ABCEFG, DEF, CDEF, BDEF, BCDEF and BC. Only AEFG and BDEF have an x-component of 0, so they are the only two closed loops.

SECOND SOLUTION:

For Grid one: Only one arrow points downwards (the black one). so the black arrow must be part of the closed loop, if such a loop exists. The x-offset of the black arrow is greater than the x offset of all of the other arrows, thus there must be at least two right facing arrows in the closed loop. Thus two or three of pink, green and dark blue must be present. Both dark blue and green togther would be too high and just pink and green would be too low: thus the loop contains pink and dark blue. These form a closed loop.

For Grid two: It is obvious that Black + Red + Green + Yellow make a parallelogram closed loop, and it is obvious that this is the only one.

For Grid 3: The coordinates of the vectors are
Red (4, 2)
Yellow (-1, -6)
Purple (-2, 0)
Green (3,0)
Black (0,1)
Light Blue (1, -5)
Pink (-3, 3)


Looking at the y coordinates of these, I can see that any closed loop must contain Red, Pink, Black and Yellow or Red, Pink and Light Blue.

Red + Pink + Black + Yellow = (0, 0)

So this is a closed loop.

Red + Pink + Light Blue = (2,0)

This sorts out the Y-coordinate. I now see that I need to include Purple to make another closed loop.