Four Dogs Squared: November 2021 challenge question

Mathematically, the dogs would never meet, assuming point size dogs. (Realistically, it would depend on the size of the dogs.) The dogs converge on the center in a square formation, spinning faster and faster!

Total distance would be infinite...

Correction...

Software bug...my apologies...

Octave simulation, .001 second step size

Time = 120 seconds = 2 minutes

Speed = 30 mph = 1/2 mile/minute

Distance = Time x Speed = 1 mile

Nice logic!

I like it. Each of the four dogs always on the corners of a rotating shrinking square.

I've been playing around with this for a little bit. If you have ND dogs arranged in a circle spaced 1 mile apart. The radius of the circle will be

r=.5/sin(pi/ND).

ND=3, r=.5774, r²=.3333, T=80 sec

ND=4, r=.7071, r²=.5, T=120 sec

ND=5, r=.8507, r²=.7236, T=174 sec

ND=6, r=1.0, r²=1.0, T=240 sec

The time to reach the center (and the distance) is proportional to the square of the ratio r.

Your clever scheme works for 4 dogs because the target dog is always running at a 90 degree angle to the pursuing dog, so the closing speed is the same as the pursuing dog's speed. It doesn't work for 3 dogs where the dogs are running toward each other or for more than 4 dogs where the target dog is running away from the pursuing dog.

As I said in #9 "If in fact the dogs did meet - how long were the dogs?" because the question raises Zeno's dichotomy because unless the dogs have a length and are just theoretical points then ∞ becomes the answer to the two parts of the question - how far and how long?
The dog's length introduces a limit to the equation/calculus. But only as long as the dog remains in one piece. BUT 4 greyhound kind of dogs (remember 30mph) crashing into each other at exactly the same moment while trying to reach the centre could lead to bits and pieces.

Zeno's dichotomy is normally called a paradox, and, the explanation is that the "calculator" has deliberately designed the steps to avoid reaching the goal. But the sum of the infinite series (normally used in Zeno) ½ + ¼ + ¹/₈ +¹/₁₆^..... is still just 1.

In the same way, even if the dogs are infinitesimally small, the distance they travel is still just 1 mile (covered in 2 minutes).

Unless they are over a micro-metre long and they meet outside the centre ... creating a limit.
And no Zeno never gets to 1, dichtomy or paradox, because it was never meant to. I'm sure Frege and Russells' "paradox" would answer the question in a similar manner.
Until we make Pi less than infinite - the question has no answer and even if "the dogs are infinitesimally small, the distance they travel is still just 1 mile" and maybe only a picometre long it is still 1 mile minus half a picometre, possible a quarter of a picometre - dunno and can't be bothered trying to calc it, but a mile less any distance isn't a mile, I have dinner and a very presentable Malbec waiting.

I'll concede to it being less than a mile by some distance related to their body size, but, you've flipped from ∞ to less than a mile in two consecutive posts: are you sure you didn't start that Malbec sooner?

As Yogi Berra said, "It's deja vu all over again." In post #53 in the Sep. 21 Challenge Question you assured us, on the authority of your Prof. of Mathematics niece, that the bird flying between the two trains would fly an infinite distance. Quite an accomplishment for the bird, flying at a fixed speed for a fixed time - but then maybe you were in no condition to think about flying birds (or in the present case, running dogs) because in post #57 of the same thread you wrote that, at that time also, you were enjoying "a perfectly good Malbec". Cheers!

I was lucky. Because I was working on it a long time I posted mine before I saw your post #3: it was only when seeing that my result was the same as yours that I was sure I was right.

Correctomundo! First answer to get all three parts correct.

For further info, the curve each dog follows is a logarithmic spiral.

SE's post 4 shows a nice rendering known as "Huddy's Doodle". It consists entirely of straight line segments, but observers don't always believe that.

If you repeat the pattern in a grid of squares,"spinning" inward oppositely in adjacent squares, you get an effect like a mat of ginkgo leaves.

The three other Archimedean tessellations with an even number of polygons around each vertex give a similar fern-like effect. (4444, 3636, 333333, 3464)

Y'all do realise that if "...each dog will run in a straight line after the dog to their left..." is fact, the question has no solution? Unless the straight line has a heap of jinks in it ... but then it wouldn't be "A" straight line it would be many?
Can the dog maintain 30mph with all these right hand "jinks" especially as the "jinks" will become quite abrupt the closer to the intersection?
If in fact the dogs did meet - how long were the dogs?
To quote the US FHWA of the DOT (gotta luv tha Yanks)
"Spirals are curves used to transition between a circular curve with a specific radius and degree of curvature and a straight tangent (whose radius is infinity). The term spiral is interchangeable with easement or transition curve. The radius and sharpness of a spiral curve increase uniformly along its length. The length and degree of curvature of a spiral curve are based on the anticipated speed of traffic and the sharpness of the circular curve that the spiral must meet."
You would do Mr. Euler (or Cornu) proud with a solution...

The direction can be a straight line, the path is the trajectory....

Then the direction should have been 45 degrees from the corner a distance of 0.7071 miles - to arrive at that point the dog, given the set parameter of "after the dog on their left", therefore, required a curved trajectory of 1 mile in 2 minutes.

So the quaestio debet read:

Four dogs are placed in opposite corners of a 1 mile wide square.
These dogs are fast. 30 mph fast. And they are stamina freaks.

At their trainer's command, each dog will run in a trajectory after the dog to their left.

At what part of the square do the dogs meet and where is it relative to the starting point? How long does it take?
Bonus: How far did each dog go?

I don't think a dog would understand a trajectory command...

You've obviously not seen a dog chase a boomerang

At some instant after the dogs meet,they will form a circle and become one long dog,perhaps a Dachshund.

You're doggone right!

Answer #2 was correct. The dogs converge asymptotically. The distance and time are infinite. How? Coincidentally as the dogs get closer together they also get closer to following the same path at the same speed, not allowing them to ever touch, assuming as answer #2 did that the dogs are points. This is perhaps harder to visualize with a converging square/circle but it is nevertheless true. Like fractals, if you keep zooming into the center dot, it will always look the same.

Your answer does not "square" with Randall's answer and simple explanation in Post #4, in which the dogs are visualized as always being on the corners of a rotating, shrinking square. At the start of their run, the square is one mile per side. Since the dogs run at 30 miles per hour, the sides of the rotating, shrinking square will be reduced to zero in 2 minutes. By this explanation the square sides do not tend asymptotically towards zero. They become zero.

Se post #18