## Poll

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**7 members have voted**

April 8th, 2020 at 10:03:50 PM
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Can anyone solve this set of differential equations:

dy/dt = 1 - (y + t -12)

(dx/dt)

with dx/dt>0 before x reaches 6, and dx/dt < 0 afterwards.

x(6) = 3*sqrt(3)

y(6)=3

?

(Edit: I corrected the initial conditions' time.)

Even a numerical solution would be nice.

I believe the solution is the curved part of the path to avoid the selfish walker.

dy/dt = 1 - (y + t -12)

^{2}/18(dx/dt)

^{2}+ (dy/dt)^{2}= 1with dx/dt>0 before x reaches 6, and dx/dt < 0 afterwards.

x(6) = 3*sqrt(3)

y(6)=3

?

(Edit: I corrected the initial conditions' time.)

Even a numerical solution would be nice.

I believe the solution is the curved part of the path to avoid the selfish walker.

Last edited by: ChesterDog on Apr 9, 2020

April 9th, 2020 at 11:12:09 AM
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This solution assumes the two walkers must maintain the same constant speed. I set my unit of time to make the speed 1 foot / unit.

I couldn't solve the above differential equation, but typing y' = 1 - (y+t-12)^2/18; y(6)=3 into WolframAlpha gives a nice solution for y.

Then x = sqrt[36 - (y + t - 12)^2 ] to keep the distance to the selfish walker 6 feet.

x' = - sqrt( 1 - (y')^2 ) for after x becomes 6. Next, dy/dx = y' / x' which we set equal to the slope of the line to the destination, (12 - y) / (0 - x).

Trial and error gives t = 8.723448, x = 5.634036, and y = 5.339957.

The length of the straight line between (x, y) and (0, 12) is sqrt( (12-y)^2 + (0-x)^2 ) = 8.723448, which happens to equal the time for the sum of the initial straight path and the curved path!

So, the total distance is 17.446895 feet.

April 11th, 2020 at 5:45:47 PM
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https://fivethirtyeight.com/features/can-you-catch-the-free-t-shirt/

Answers to this puzzle are now posted. Along with new puzzles. ChesterDog appears to have the right solution (at least numerically, didn't check the formulae).

Answers to this puzzle are now posted. Along with new puzzles. ChesterDog appears to have the right solution (at least numerically, didn't check the formulae).

April 11th, 2020 at 6:59:16 PM
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There is a nice animated solution at that link. I would hate to say how many pieces of paper are in my trash right now about that problem. Not to mention a spreadsheet that attempted to do it in micro-steps. Big fail on my part with that one. I'm going to think twice about earning extra credit next time.

It's not whether you win or lose; it's whether or not you had a good bet.

April 11th, 2020 at 7:59:53 PM
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It's surprising to me that the path isn't symmetric, along the lines of what was proposed in this thread. I never would have thought outside of that box.

April 11th, 2020 at 8:35:14 PM
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Quote:rsactuaryIt's surprising to me that the path isn't symmetric, along the lines of what was proposed in this thread. I never would have thought outside of that box.

May I know what is the exact angle between the INITIAL RED STRAIGHT PATH and the horizontal BLUE line ? Not exact 60 dgeree ?

April 11th, 2020 at 9:00:51 PM
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Here's the derivation of the differential equation, but not its solution:

Starting at coordinates (0,12), the selfish walker's path is described by x=0 and y=12 - t, assuming his speed is 1 foot/unit time.

A 6-foot circle around him would be x

The other walker's trajectory is counterclockwise about this circle, with a speed of 1. His speed squared can be expressed as: x

Differentiating the path with respect to t gives: 2xx

Equations 1, 2, and 3 can be combined to eliminate x and x

18 y

This can be solved for y

y

And the initial condition is y(6) = 3.

Starting at coordinates (0,12), the selfish walker's path is described by x=0 and y=12 - t, assuming his speed is 1 foot/unit time.

A 6-foot circle around him would be x

^{2}+ (y - 12 + t)^{2}= 36 (Eqn. 1).The other walker's trajectory is counterclockwise about this circle, with a speed of 1. His speed squared can be expressed as: x

_{t}^{2}+ y_{t}^{2}= 1 (Eqn. 2).Differentiating the path with respect to t gives: 2xx

_{t}+ 2(y-12+t)(y_{t}+1) = 0. And solving for x_{t}gives: x_{t}= (12-y-t)(y_{t}+1)/x (Eqn. 3).Equations 1, 2, and 3 can be combined to eliminate x and x

_{t}to give:18 y

_{t}^{2}+ (y-12+t)^{2}y_{t}-18 + (y-12+t)^{2}= 0This can be solved for y

_{t}using the quadratic formula (using the + instead of the -) to give:y

_{t}= 1 - (y+t-12)^{2}/18.And the initial condition is y(6) = 3.

April 11th, 2020 at 9:19:27 PM
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ChesterDog, is the angle between initial red straight path and horizontal blue line EXACT 60 degree ?

April 11th, 2020 at 10:08:59 PM
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Quote:ssho88ChesterDog, is the angle between initial red straight path and horizontal blue line EXACT 60 degree ?

Yes--it's exactly 60 degrees.

A good way to prove that is this: If the selfish walker starts at B (0,12), then if he would walk for 6 feet, he would be at O (0,6). Imagine a 6-foot-radius circle centered at O (0,6).

If the other person started at A (0,0) walking 60 degrees from straight ahead, he would touch that circle at C, 6 feet from his starting point. And he would arrive at the circle at the same time the selfish walker reached (0,6).

In other words, both walkers went 6 feet to be 6 feet away from each other. Think of the circle O with an equilateral triangle AOC having the circle's radius AO as its base.

Edit: Now, I see that the above doesn't prove that the minimum total distance is achieved by starting at 60 degrees.

Last edited by: ChesterDog on Apr 11, 2020