NY Times math problem

M Rath Inscribed thus:

Just a guess, has Pi been defined before use ?

--
Best Regards:
                        Baron.

In article Rich Grise writes: ... > Is the "agent" wading/swimming, or just walking around the perimeter > of the pond? What's each of their respective speeds? How did the rabbit > get to the middle of the pond in the first place? Was somebody using him > for Muxkie bait? ;-)

The rabbit had been caught by poachers in an helicopter, but he knew to escape.

--
dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/

In article Sylvia Else writes: > riverman wrote: ... > > I'm not convinced about the 'keep the agent on the opposite side' > > logic. Lets expand the problem to their being TWO agents, already on > > opposite sides. If the rabbit swims in any direction, off-center, then > > both agents will move together to be at the point the rabbit is > > apparently aiming for, then the rabbit moves slowly back towards > > center and the problem is reduced to the original one. So the rabbit > > can escape with TWO agents around the pool...or even an INFINITE > > number of them? (well, slightly less than infinite, in this case....) > > When the rabbit moved back towards the centre of the pool, the agents > would move back to positions oppoisite each other. > > It appears that the rabbit cannot escape if there are two agents.

Of course the rabbit can escape if the agents use a stupid strategy.

--
dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/

You mean like wandering off to get some donuts and coffee? I think it's usual to assume in these puzzles that the participants will always use the optimum strategy.

Sylvia.

Who says that the land speed of the rabbit is the same as its water speed? My assumption is that once it makes it to shore, the rabbit has no difficulty in escaping. However, the problem only requires that the rabbit makes it to shore without being captured.

Rob Johnson take out the trash before replying to view any ASCII art, display article in a monospaced font

I didn't assume that because the problem specifies each party's speed. Otherwise I might've observed that people swim a lot faster than rabbits, and the agent might have an easier time with a running dive & a swim.

However, the problem only requires

Yep. Two points.

Cheers, James Arthur

C + x. Somewhere in the neighborhood of C the agent's perception of the rabbit's position/speed/direction will be distorted.

Ed

The rabbit moves at a rate of 1, the agent at a rate of r. The edge of the lake is the unit circle centered at (0,0).

As has been discussed, the rabbit can keep the agent on the opposite side of the center of the lake until the rabbit gets to a distance of 1/r from the center of the lake. So we assume that the rabbit starts at (-1/r,0) and the agent starts at (1,0). Then the agent moves counterclockwise and the rabbit head straight down until it reaches the unit circle after time t (if it headed further left, it would meet the agent sooner, if it headed further right, it would decrease its distance to the center of the lake). The positions at time t are

rabbit: ( -1/r , -t )

agent: ( cos(rt) , sin(rt) )

To solve for the extreme case, we need to solve the equations

cos(rt) + 1/r = 0

sin(rt) + t = 0

Eliminating the sin and cos using Pythagoras' Theorem, we get

rt = sqrt(r^2 - 1)

Thus, we need to solve

r cos(sqrt(r^2 - 1)) + 1 = 0 [1]

while making sure that

sin(sqrt(r^2 - 1)) < 0 [2]

The first two solutions of [1] are

2.261826334114651437542573924426 4.603338848751700352556582029103

For the first solution, inequality [2] fails to hold, so the solution we are interested in is the second, for which [2] is satisfied.

Rob Johnson take out the trash before replying to view any ASCII art, display article in a monospaced font

Okay...with the systematic method used here...the Rabbit does get out at North and East coordinates of 10.5006 , 19.9976 where the radius point has North and East coordinates of 10 , 10. And the rabbit gets out when the agent is at a clockwise azimuth on the circle of 834 degrees !

Well...I have only plotted the first 360 degrees so far but here is the KBH Scratch Plot file:

663&137@F+wtN0E0 41&759@S+wtN20E20 0 352&448@P+bkN10E10 353&447@L+rdN9.9564E9.9992 365&446@L+rdN9.5661E9.9588 378&443@L+rdN9.1452E9.8458 390&437@L+rdN8.7477E9.6669 402&430@L+rdN8.3826E9.4288 412&421@L+rdN8.0572E9.1387 421&410@L+rdN7.7774E8.8044 428&399@L+rdN7.5477E8.4338 433&386@L+rdN7.3716E8.0349 437&373@L+rdN7.2515E7.6158 439&360@L+rdN7.1886E7.1844 439&346@L+rdN7.1834E6.7484 437&333@L+rdN7.2357E6.3155 434&320@L+rdN7.3441E5.8932 429&307@L+rdN7.5067E5.4886 422&295@L+rdN7.7209E5.1089 414&285@L+rdN7.9834E4.7607 405&275@L+rdN8.2899E4.4506 394&267@L+rdN8.6357E4.1851 382&260@L+rdN9.0151E3.9703 369&255@L+rdN9.4215E3.8124 356&252@L+rdN9.8469E3.7173 343&251@L+rdN10.2819E3.6911 329&253@L+rdN10.7150E3.7398 316&257@L+rdN11.1311E3.8687 305&263@L+rdN11.5107E4.0819 295&273@L+rdN11.8285E4.3792 289&284@L+rdN12.00538E4.7515 285&297@L+rdN12.1525E5.1743 286&311@L+rdN12.1075E5.6063 292&323@L+rdN11.9231E5.9996 301&333@L+rdN11.6270E6.318 312&340@L+rdN11.2554E6.5445 325&344@L+rdN10.8406E6.6769 339&346@L+rdN10.4072E6.7212 325&347@L+rdN10.8414E6.7579 312&350@L+rdN11.2621E6.8715 352&448@P+bkN10E10 41&453@P+blN19.9985E10.1745 352&448@+blN10E10 45&502@P+blN19.8481E11.7365 352&448@+blN10E10 59&554@P+blN19.3969E13.4202 352&448@+blN10E10 82&603@P+blN18.6603E15.0000 352&448@+blN10E10 113&647@P+blN17.6604E16.4279 352&448@+blN10E10 152&686@P+blN16.4279E17.6604 352&448@+blN10E10 196&717@P+blN15.0000E18.6603 352&448@+blN10E10 245&740@P+blN13.4202E19.3969 352&448@+blN10E10 297&754@P+blN11.7365E19.8481 352&448@+blN10E10 352&759@P+blN10.0000E20.0000 352&448@+blN10E10 406&754@P+blN8.2635E19.8481 352&448@+blN10E10 458&740@P+blN6.5798E19.3969 352&448@+blN10E10 507&717@P+blN5.0000E18.6603 352&448@+blN10E10 551&686@P+blN3.5721E17.6604 352&448@+blN10E10 590&647@P+blN2.3396E16.4279 352&448@+blN10E10 621&603@P+blN1.3397E15.0000 352&448@+blN10E10 644&554@P+blN0.6031E13.4202 352&448@+blN10E10 658&502@P+blN0.1519E11.7365 352&448@+blN10E10 663&448@P+blN0.0000E10.0000 352&448@+blN10E10 658&393@P+blN0.1519E8.2635 352&448@+blN10E10 644&341@P+blN0.6031E6.5798 352&448@+blN10E10 621&292@P+blN1.3397E5.0000 352&448@+blN10E10 590&248@P+blN2.3396E3.5721 352&448@+blN10E10 551&209@P+blN3.5721E2.3396 352&448@+blN10E10 507&178@P+blN5.0000E1.3397 352&448@+blN10E10 458&155@P+blN6.5798E0.6031 352&448@+blN10E10 406&141@P+blN8.2635E0.1519 352&448@+blN10E10 352&137@P+blN10.0000E0.0000 352&448@+blN10E10 297&141@P+blN11.7365E0.1519 352&448@+blN10E10 245&155@P+blN13.4202E0.6031 352&448@+blN10E10 196&178@P+blN15.0000E1.3397 352&448@+blN10E10 152&209@P+blN16.4279E2.3396 352&448@+blN10E10 113&248@P+blN17.6604E3.5721 352&448@+blN10E10 82&292@P+blN18.6603E5.0000 352&448@+blN10E10 59&341@P+blN19.3969E6.5798 352&448@+blN10E10 45&393@P+blN19.8481E8.2635 352&448@+blN10E10 51&448@Agent->+blN19.65E10 352&463@Rabbit Begins Here+rdN10E10.5

Copy to a text file, save the text file as "Name.plt", and the file will run in KBH Scratch Plot...

Copy to a text file, save the text file as "Name.plt" with quotation marks, and the file will run in KBH Scratch Plot...

I didn't look up the official solution, but from my thinking:

You don't need any calculus of variaton approach to proof, that the rabbit can escape. (and you can easily show one examplary non optimal trajectory). The basic four operators are enough ( + - / * )

Getting the optimal trajectory (the shortest time for the rabbit to escape) might be a little more tricky. I didn't think too much about this one. (Typical engineering thinking: Why finding the optimal trajectory if one can escape and if one can see the agent run in vain ;-) )

You need the additional assumption, that the time for the rabbit to escape can be infinite if you want to find highest possible speed for the agent to escape.

Looking at the plot...it seems that the rabbit path should continue to spiral westward rather than turn back northward and eastward at the sharp point.

But I don't yet see why the coordinate output of the KBH Code would be correct at 340 degrees but incorrect at 350 degrees. Why I get this worked out it will probably require a new subject posting...

All the above is voided...the agent coordinates should have been based on a circle radius of 10 instead of a circle radius of 1 .

The KBH code below is edited to reset the value of "flg" to zero at the end of the loop.

And according to this systematic method...the rabbit gets out at North and East coordinates of 8.1332 , 0.1612 with the agent at a clockwise azimuth of

535 degrees on the circle edge.

The KBH Scratch Plot file below is correct until the sharp point...but is edited below to stop before the sharp point.

KBH Code:

{The rabbit at the center of a 10 radius circular pond and the chasing agent is at the edge of the pond.}

Var ads, nag, eag, rba, rbn, rbe, nd, ed, dgr, dst: double; flg: integer;

Procedure D; Begin If (flg = 1) Then Begin {Procedure E} rba:= rba + Pi; Exit; End; rba:= rba + (Pi * 2); {The structure of procedures D, E, & F are from the KBH Survey Progam for the HP35S} End;

Procedure F; Begin If (flg = 1) Then Exit; rba:= rba + Pi; End;

begin {KBH Code} flg:= 0; rbn:= 10; rbe:= 10; dgr:= 1;

While (dgr < 361) Do {Change value for desired output range} Begin ads:= (dgr * Pi) / 180; nag:= 10 + (Cos(ads) * 10); eag:= 10 + (Sin(ads) * 10); nd:= rbn - nag; If (nd = 0) Then nd:= 0.000000000001; ed:= rbe - eag; If (ed >= 0) Then flg:= 1; rba:= ArcTan(ed / nd); If (rba < 0.000000000001) Then D Else F; rbn:= rbn + (Cos(rba) * 0.043633231); rbe:= rbe + (Sin(rba) * 0.043633231); dst:= Sqrt(Sqr(10 - rbn) + Sqr(10 - rbe)); rba:= (rba / Pi) * 180; WriteLn(rbn:15:4, rbe:15:4, dgr:15:4, dst:15:4); dgr:= dgr + 1; flg:= 0; End;

ReadLn; end.

And the above KBH code runs in Delphi Console Mode...

Now the rabbit gets out at North and East coordinates of 8.1332 , 0.1612 while the agent is at an azimuth of 535 degrees on the edge of the circular pond and on his second trip around the pond. Of course this is a systematic method only...

And here is the KBH Scratch Plot file for the rabbit path:

663&137@F+wtN0E0 41&759@S+wtN20E20 0 352&448@P+bkN10E10 353&447@L+rdN9.9564E9.9992 365&446@L+rdN9.5661E9.9588 378&443@L+rdN9.1452E9.8458 390&437@L+rdN8.7477E9.6669 402&430@L+rdN8.3826E9.4288 412&421@L+rdN8.0572E9.1387 421&410@L+rdN7.7774E8.8044 428&399@L+rdN7.5477E8.4338 433&386@L+rdN7.3716E8.0349 437&373@L+rdN7.2515E7.6158 439&360@L+rdN7.1886E7.1844 439&346@L+rdN7.1834E6.7484 437&333@L+rdN7.2357E6.3155 434&320@L+rdN7.3441E5.8932 429&307@L+rdN7.5067E5.4886 422&295@L+rdN7.7209E5.1089 414&285@L+rdN7.9834E4.7607 405&275@L+rdN8.2899E4.4506 394&267@L+rdN8.6357E4.1851 382&260@L+rdN9.0151E3.9703 369&255@L+rdN9.4215E3.8124 356&252@L+rdN9.8469E3.7173 343&251@L+rdN10.2819E3.6911 329&253@L+rdN10.7150E3.7398 316&257@L+rdN11.1311E3.8687 305&263@L+rdN11.5107E4.0819 295&273@L+rdN11.8285E4.3792 289&284@L+rdN12.00538E4.7515 285&297@L+rdN12.1525E5.1743 286&311@L+rdN12.1075E5.6063 292&323@L+rdN11.9231E5.9996 301&333@L+rdN11.6270E6.318 312&340@L+rdN11.2554E6.5445 325&344@L+rdN10.8406E6.6769 339&346@L+rdN10.4072E6.7212 352&344@L+rdN9.9729E6.6856 365&341@L+rdN9.5502E6.5793 378&336@L+rdN9.1482E6.4106 390&329@L+rdN8.7736E6.1875 400&321@L+rdN8.4315E5.9171 410&311@L+rdN8.1258E5.6061 418&300@L+rdN7.8594E5.2609 425&288@L+rdN7.6345E4.8873 431&276@L+rdN7.4526E4.4909 435&263@L+rdN7.3148E4.0771 438&250@L+rdN7.2216E3.6511 439&237@L+rdN7.1730E3.2176 440&223@L+rdN7.1687E2.7815 438&209@L+rdN7.2079E2.3471 436&196@L+rdN7.2895E1.9186 432&183@L+rdN7.4120E1.5 427&171@L+rdN7.5736E1.0949 421&158@L+rdN7.7720E.7064 414&147@L+rdN8.0048E0.3376 410&142@L+rdN8.1332E0.1612 352&448@P+bkN10E10 41&453@P+blN19.9985E10.1745 352&448@+blN10E10 45&502@P+blN19.8481E11.7365 352&448@+blN10E10 59&554@P+blN19.3969E13.4202 352&448@+blN10E10 82&603@P+blN18.6603E15.0000 352&448@+blN10E10 113&647@P+blN17.6604E16.4279 352&448@+blN10E10 152&686@P+blN16.4279E17.6604 352&448@+blN10E10 196&717@P+blN15.0000E18.6603 352&448@+blN10E10 245&740@P+blN13.4202E19.3969 352&448@+blN10E10 297&754@P+blN11.7365E19.8481 352&448@+blN10E10 352&759@P+blN10.0000E20.0000 352&448@+blN10E10 406&754@P+blN8.2635E19.8481 352&448@+blN10E10 458&740@P+blN6.5798E19.3969 352&448@+blN10E10 507&717@P+blN5.0000E18.6603 352&448@+blN10E10 551&686@P+blN3.5721E17.6604 352&448@+blN10E10 590&647@P+blN2.3396E16.4279 352&448@+blN10E10 621&603@P+blN1.3397E15.0000 352&448@+blN10E10 644&554@P+blN0.6031E13.4202 352&448@+blN10E10 658&502@P+blN0.1519E11.7365 352&448@+blN10E10 663&448@P+blN0.0000E10.0000 352&448@+blN10E10 658&393@P+blN0.1519E8.2635 352&448@+blN10E10 644&341@P+blN0.6031E6.5798 352&448@+blN10E10 621&292@P+blN1.3397E5.0000 352&448@+blN10E10 590&248@P+blN2.3396E3.5721 352&448@+blN10E10 551&209@P+blN3.5721E2.3396 352&448@+blN10E10 507&178@P+blN5.0000E1.3397 352&448@+blN10E10 458&155@P+blN6.5798E0.6031 352&448@+blN10E10 406&141@P+blN8.2635E0.1519 352&448@+blN10E10 352&137@P+blN10.0000E0.0000 352&448@+blN10E10 297&141@P+blN11.7365E0.1519 352&448@+blN10E10 245&155@P+blN13.4202E0.6031 352&448@+blN10E10 196&178@P+blN15.0000E1.3397 352&448@+blN10E10 152&209@P+blN16.4279E2.3396 352&448@+blN10E10 113&248@P+blN17.6604E3.5721 352&448@+blN10E10 82&292@P+blN18.6603E5.0000 352&448@+blN10E10 59&341@P+blN19.3969E6.5798 352&448@+blN10E10 45&393@P+blN19.8481E8.2635 352&448@+blN10E10 51&448@Agent->+blN19.65E10 352&463@Rabbit Begins Here+rdN10E10.5 352&448@P+bkN10E10 661&475@P+blN0.0381E10.8716 599&475@L+blN2.0381E10.8716 597&475@Agent Stops Here on 2nd Loop+blN2.1E10.88

Copy to a text file, save the text file as "Name.plt" with quotation marks, and the file will run in KBH Scratch Plot...

And the rabbit path is a small-cap handwritten-style "e"...

ent

ar

ic

s,

I've totally lost track of what you are trying to do. Can you recap, and possibly post the graph on the web and give us the link?

Dave

GIGO.

RL

e

I was thinking about generalizations...

Setting maximal accelerations seems difficult. Generalizing to more dimensions is pointless. Other shapes seem ugly. But here are two that may be interesting and feasible:

1) The agent has different maximal counterclockwise and clockwise speeds (perhaps he must run backwards when going clockwise).

2) There are n agents.

Dave wrote:

I've totally lost track of what you are try But you replied to the final version and thanks for that...

Now...with each step of the agent around the circle...the rabbit moves on a line from the agent to the rabbit and away from the agent. So just say that the rabbit turns away from each new position of the agent. And the rabbit moves one-fourth the distance that the agent moves...each time that the rabbit turns away.

And this is a systematic method. The agent keeps moving around the circle and the rabbit keeps turning away...

The graphics that I have is a coordinate plot and not an animation movement. But you can find Scratch Plot with google, download it in ten seconds, install it, and make a file for it with the above data. And the rabbit path is really nice looking...and I think it has a point to make.

Now with the rabbit path you can scale a tangent line off the path at any point...knowing that the path runs away from the opposite side...and see which agent point on the circle produced the rabbit path at that rabbit path point.