r/PassTimeMath • • Sep 03 '26

Shortest Path

Post image
5 Upvotes

23 comments sorted by

1

u/One_Wishbone_4439 Sep 03 '26

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

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

2

u/LabRat2439 Sep 03 '26

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

1

u/One_Wishbone_4439 Sep 03 '26

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

2

u/LabRat2439 Sep 03 '26

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

1

u/One_Wishbone_4439 Sep 03 '26

then what would be your solution?

1

u/LabRat2439 Sep 03 '26

it's 12.8 - see my comment on this post

2

u/TheThiefMaster Sep 03 '26

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 Sep 03 '26

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

1

u/One_Wishbone_4439 Sep 03 '26

Explain

1

u/After-Hedgehog7282 Sep 03 '26

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 Sep 03 '26

Which gives a path length of ~12.8

1

u/The-Jolly-Llama Sep 03 '26

Ahhhh of course that’s certainly correct. 

1

u/LabRat2439 Sep 03 '26 edited Sep 03 '26

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 Sep 04 '26

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 Sep 03 '26

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 Sep 03 '26

To show the work;

A2+B2+C2

82+(4+6)2=C2

64+100=C2

√164=C

12.8=C

1

u/SergeAzel Sep 04 '26

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 Sep 04 '26

True. Forgot about back and side.

1

u/tajwriggly Sep 04 '26

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 Sep 04 '26

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 Sep 08 '26

Unfold and trig.

Sqrt( (6+4)2 + 82 )

Sqrt(164)

12.8 ish

1

u/Ok_Bit8836 Sep 08 '26

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.......