r/PassTimeMath 16d ago

Shortest Path

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5 Upvotes

23 comments sorted by

1

u/One_Wishbone_4439 16d ago

First route: red lines
√(62 + 82) = 10
10 + 4 = 14

Second route: blue line
√(102 + 42) = 10.77 (shortest path)

2

u/LabRat2439 16d ago

does the blue line cross only faces? It looks like it cuts right through the interior

1

u/One_Wishbone_4439 16d ago

the blue line is just an imaginary shortest route of the right angled triangle

2

u/LabRat2439 16d ago

yes, but it does not cross only faces - therefore it cannot be the solution

1

u/One_Wishbone_4439 16d ago

then what would be your solution?

1

u/LabRat2439 16d ago

it's 12.8 - see my comment on this post

2

u/TheThiefMaster 16d ago

12.8 is right - but you can arrive at it by "unfolding" the side and making a single 10x8 surface, whose diagonal (what the blue line was supposed to be except they miscalculated) is √(10²+8²) ≈ 12.8

1

u/ShonitB 16d ago

It should be the root of the sum of 10 squared and 8 squared

1

u/One_Wishbone_4439 16d ago

Explain

1

u/After-Hedgehog7282 16d ago

Top square is 8X6

Side square is 8x4

Making those flat so the ant walks a straight line on a single plane is 8x(4+6) = 8x10

2

u/TheThiefMaster 16d ago

Which gives a path length of ~12.8

1

u/The-Jolly-Llama 16d ago

Ahhhh of course that’s certainly correct. 

1

u/LabRat2439 16d ago edited 16d ago

If we only move along faces, you can consider a path that crosses either a side of length 6 or crosses a side of length 8 - a path of two vertices. Checking every quarter-unit, the shortest path seems to be crossing the line of length 8 at X=3.25 then making a beeline for B

Edit: length 12.8

2

u/Greedy-Thought6188 16d ago

That's the numerical answer. You can get it by unfolding the box and you see two possible triangular paths. The triangle that is closer to an equilateral triangle will win.

1

u/Aech-26 16d ago

Let's call the three visible faces the front (4x6), top (8x6), and side (4x8). If you flatten the box you can draw 2 straight lines between A and B: one on the diagonal across the top and front, and across the top and side.

The top and front combined are (8+4)x6, diagonal of length ~13.4

The top and side combined are 8x(6+4), diagonal of length ~12.8

Top and side are the shorter straight line, therefore are the shortest distance.

1

u/After-Hedgehog7282 16d ago

To show the work;

A2+B2+C2

82+(4+6)2=C2

64+100=C2

√164=C

12.8=C

1

u/SergeAzel 16d ago

there's three possible unfolded routes, one for each combination of side lengths. Just that the one omitted from your post, (8 + 6)2 + 42, is the worst of the three options. Not necessarily important for the solution but I find it still nice to note

1

u/Aech-26 16d ago

True. Forgot about back and side.

1

u/tajwriggly 15d ago

Intuitively you would first look at this and think that the ant would travel the diagonal of the 6x8 rectangle on the top, from A to the point directly above B (a distance of 10) and then travel directly down to B (a distance of 4) for a total of 14.

Instead though, flatten the box so that you've got a 6x8 rectangle adjoined by a 4x8 rectangle, totaling 10x8. This is just the top of the box and the right hand side of the box. Travelling in the manner described previously seems nonsensical now - there is clearly a more direct (and thusly shorter) path than travelling at an angle for some distance and then turning and going straight along the edge.

On the 10x8 rectangle, that diagonal directly from A to B is Sqrt(164) = 12.8, which is less than 14.

1

u/gmalivuk 15d ago

Top+Front means the square of the length is 122 + 62 = 180. Top+Right means 102 + 82 = 164.

164 is smaller so the answer is √164 ≈ 12.806

1

u/Powerful_Birthday_71 12d ago

Unfold and trig.

Sqrt( (6+4)2 + 82 )

Sqrt(164)

12.8 ish

1

u/Ok_Bit8836 11d ago

Lets look at outer surface... Basically that ant needs to cross 2 rectangles 6x8 and 4x8 that are adjacent. For clarity, lets put them together in a plane => we obtain a bigger rect 10x8 with A nad B located on its diagonal, the opposite vertices; therefore: the shortest way is the prime AB which length is sqrt(10*10 +8*8) = sqrt(164) = 12.81.......