Author Topic: Kong Lake  (Read 355 times)

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Online Weehawk

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Kong Lake
« on: July 11, 2017, 04:48:06 PM »


Mario may be in for it now. Kong has him trapped in a perfectly circular pond. Fortunately for Mario, Kong can't swim, but he can run 10 miles per hour on dry land. If Mario can make it to he shore ahead of Kong, he can outrun Kong from there. Mario is able to determine the optimal strategy, even if you aren't.

For the first person who can correctly post the minimum speed Mario is going to have to row, along with an brief explanation of why that is the correct answer, I have a DK Forum t-shirt:



If you have already earned a shirt, you may PM me your answer just for fun and I will credit you in the thread.

Please recuse yourself if you are familiar with this general problem.

Offline Barra

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Re: Kong Lake
« Reply #1 on: July 11, 2017, 05:05:58 PM »
6.4 mph?

I just assumed an arc distance of 10 at a 90 degree angle to work out the radius of 6.36 and hence he has to row slightly faster than that to get to shore before Kong makes his way around.
Could be way off seems like an age since I've done any of this lol
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Online Weehawk

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Re: Kong Lake
« Reply #2 on: July 11, 2017, 05:10:32 PM »
Could be way off

Right about that, at least.

Offline Barra

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Re: Kong Lake
« Reply #3 on: July 11, 2017, 05:11:52 PM »
Lol well excuse me then

Edit: guess if he rowed downwards that'd decrease his minimum required speed by half
« Last Edit: July 11, 2017, 05:56:53 PM by Barra »
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Offline krehztim

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Re: Kong Lake
« Reply #4 on: July 11, 2017, 05:54:05 PM »
How many digits do we need to go?  Needs to be faster than 10/pi.  Mario should go down, however far it is for Mario to go down, Kong has to go pi times longer.  So, if Mario only has to go 1/pi distance, he only needs to go 10/pi speed (mph).  3.183099, for example, rounded.

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Re: Kong Lake
« Reply #5 on: July 11, 2017, 05:59:19 PM »
How many digits do we need to go?  Needs to be faster than 10/pi.  Mario should go down, however far it is for Mario to go down, Kong has to go pi times longer.  So, if Mario only has to go 1/pi distance, he only needs to go 10/pi speed (mph).  3.183099, for example, rounded.

10/pi (~3.18) would indeed be fast enough for Mario to escape, in fact it is how fast he would have to row if he simply started out in the opposite direction of Kong's original position and never altered course.

It is not the minimum necessary, however. That is, there is a better strategy that would allow him to escape without having to row that fast.

Offline krehztim

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Re: Kong Lake
« Reply #6 on: July 11, 2017, 06:07:15 PM »
Now that I think about it, he could go in a nearly infinite spiral, but eventually reach the shore.  Seems a nearly impossible answer.
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Offline Adam_Mon

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Re: Kong Lake
« Reply #7 on: July 12, 2017, 01:37:26 AM »
From a visual point of view I guess if Mario traveled south slowly while switching directions left and right over the center line/point of the circle it might cause Kong to travel back over ground he's already covered along the top outside edge (assuming Kong is taking the path of least distance to Mario).

As for the math I have no idea, the only information I can see is the circle (360*) and the speed (10mph) but how to work with those I have no clue  <confused>



Offline CaptainJivePants

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Re: Kong Lake
« Reply #8 on: July 12, 2017, 05:35:56 AM »
Mario knows how to use a paddle like a hammer. Kong's in for it when Mario gets to the shore. Mario can row any speed he damned well pleases.
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Re: Kong Lake
« Reply #9 on: July 12, 2017, 05:57:28 AM »
Mario knows how to use a paddle like a hammer. Kong's in for it when Mario gets to the shore. Mario can row any speed he damned well pleases.
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Kong isn't fazed by hammers (or oars), but rest assured if there are any barrels floating in that pond Mario will make short work of them.

And I think he's pretty safe from fireballs anyway.

Offline Boognish

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Re: Kong Lake
« Reply #10 on: July 12, 2017, 06:40:40 AM »
0 mph for the boat

.mario is a excellent fast swimmer.
 he leaves the boat in the lake and swims under water, fakes out kong. and makes a clean escape


« Last Edit: July 12, 2017, 06:50:59 AM by Boognish »
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Online Weehawk

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Re: Kong Lake
« Reply #11 on: July 12, 2017, 06:59:28 AM »
0 mph for the boat

.mario is a excellent fast swimmer.
 he leaves the boat in the lake and swims under water, fakes out kong. and makes a clean escape





Unfortunately, the water is crystal clear, making it easy for Kong to see Mario, and for Mario to see the piranha.

Offline danman123456

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Re: Kong Lake
« Reply #12 on: July 12, 2017, 07:46:59 AM »
This feels like Calc or Trig to me I swear ive seen a problem like this before.

Simple math is :
Circumference = 2PieR
Radius = 10 miles (arbitrary number)
Circ = 62.83 miles
half = 31.415 miles (Distance Kong must travel)

Kong can run to opposite side of circle in 3.1415 hours (10 MPH)
Mario needs to row 10 miles in this time frame 10/3.1415 = 3.18319274232
Hence he needs to row at least 3.19 (I'm rounding up)
I would imagine you row to a certain point in a straight line and then change direction keeping Kong as far away as possible as you move closer to the edge obviously cuts down the speed you require but I just don't have a clue on the math that would be involved. There has to be a 10/x number that applies on the minimum speed you have to row to "escape". 2mph maybe as 3mph is definitely fast enough and 1mph seems way too slow :D
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Online Weehawk

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Re: Kong Lake
« Reply #13 on: July 12, 2017, 08:24:08 AM »
This feels like Calc or Trig to me I swear ive seen a problem like this before.

Simple math is :
Circumference = 2PieR
Radius = 10 miles (arbitrary number)
Circ = 62.83 miles
half = 31.415 miles (Distance Kong must travel)

Kong can run to opposite side of circle in 3.1415 hours (10 MPH)
Mario needs to row 10 miles in this time frame 10/3.1415 = 3.18319274232
Hence he needs to row at least 3.19 (I'm rounding up)
I would imagine you row to a certain point in a straight line and then change direction keeping Kong as far away as possible as you move closer to the edge obviously cuts down the speed you require but I just don't have a clue on the math that would be involved. There has to be a 10/x number that applies on the minimum speed you have to row to "escape". 2mph maybe as 3mph is definitely fast enough and 1mph seems way too slow :D

See reply #5 in this thread.

Offline up2ng

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Re: Kong Lake
« Reply #14 on: July 12, 2017, 09:04:54 AM »
The path is not going to be an infinite spiral but that's sort of on the right track.  The problem with that is that as Mario meanders close to the shore Kong will be able to just stroll around on the shore and Mario will never be able to safely disembark.

I believe the correct path will be an arc such that at every moment in time Mario is travelling in a direction that is directly opposite from Kong's position (instantaneously -- feels like calculus to me!).  So, if Kong starts out deciding to run in a clockwise direction from 12 o clock to 1 o clock, then by the time he reaches 1 o clock Mario has curved his downward path, starting out by rowing towards 6 o clock but gradually curving his path so that he is now pointed in the same direction as a vector from the center point to 7 o clock would point.  This makes it so that at every moment in time, Kong is indifferent about whether or not to continue running clockwise or counterclockwise, and if he switches then Mario just switches the direction of his arc.  By the time Mario reaches the coast, he must be travelling fast enough so that this strategy still yields an arc that is outward as compared to an orbital path.  (My hunch is that Mario could use this strategy but travel so slowly that Kong could simply run around in circles and Mario would never make outward progress beyond a certain radius).

As for calculating the actual number . . . that just hurts my brain!   <confused>

Instant Edit:  I think the above explanation is incorrect but I left it there to show the idea of what we are trying to achieve (Kong's indifference).  We don't travel in the opposite "direction" from Kong (the opposite of his vector from the circle's center point to his position).  Instead, we constantly travel in a direction towards the POSITION that is the opposite of Kong's POSITION.  So, if Kong has travelled to the 1 o clock position, Mario has now curved his arc so that he is now pointed towards 7 o clock!  Note that this direction of travel is NOT the opposite of the direction of the vector from the circle's center to Kong's position since Mario is no longer at the circle's center, nor is he positioned anywhere along the line from the circle's center to 7 o clock (he is on a curve towards that position -- the shape of the curve is NOT an arc of a circle . . . it's curvature is instantaneously becoming more -- "curvey?" as Mario travels farther and farther from the circle's center.

Calculating the number still hurts my brain!   <confused>
 
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