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A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

Posted September 30, 2022 12:00 AM
Pathfinder Tags: challenge question

Here's a Halloween inspired challenge question for October 2022...

The grim reaper is waiting for you on the shore of a perfectly circular lake. You are in a speed boat, with a full tank of gas, in the exact center of this lake. Although the grim reaper can fly four times as fast as the maximum speed of your boat, he cannot swim or fly over the water. As a metaphysical being, the reaper doesn’t sleep, never loses sight of you and cannot die.

But if he touches you, you die. How do you escape the grim reaper in this very logical and likely scenario?

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#1

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

09/30/2022 12:24 AM

Drive the boat around until it's out of gas, tie a rope on the boat, paddle to shore and taunt the reaper dude, when he gets on the boat in an attempt to touch me, jump off the boat grasping the rope in hand and drag the boat out to the middle of the lake with the reaper, pull the drain plug and swim to shore...

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#2

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

09/30/2022 8:28 AM

How do you escape the grim reaper in this very logical and likely scenario?

Stay put in the middle of the lake and catch fish.

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#3

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

09/30/2022 8:49 AM

I think there is something missing in this puzzle. When you get to the edge of the lake you are "home free".

The solution, as I recall, is to travel to the shore maintaining a position directly opposite from the side the grim reaper is on in a spiral path. When your distance from shore is less than 1/4 the half circumference the reaper has to travel, head directly to the shore.

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#4
In reply to #3

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

09/30/2022 10:37 AM

There must be the assumption that you are home free if you reach the shore before the GR gets to you, or an assumption that you have a friend in a fast car to pick you up and whisk you away.

I have never seen this puzzle before. If the standard solution is as you describe, that solution only works if the GR maintains a constant CW or CCW direction around the perimeter of the lake. But since the GR is free to change between CW and CWW, then if he does the boat direction must also be constantly adjusted such that a straight line from the boat to the GR will always pass through the centre point of the lake. This would result in a zig-zag path, with each zig and zag a segment of a spiral.

Of course, in the end the Grim Reaper knows that none of your efforts matter a whit. That is why the Grim Reaper gives the same greeting to everyone he meets: "Catch you later!"

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#6
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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

09/30/2022 1:35 PM

From the Grim Reaper's point of view, his best move is to circle the lake in the shortest direction to get to you. Whichever direction he chooses, you trace a path to maintain 180 degrees from his position. (If he switches direction, you do also. It makes no difference.) Since he can move 4 times as fast, you can do this out to 1/4 the radius of the lake. At this point, turn and head directly for the shore. You have to move 3/4 R, he has to move pi x R to get to you.

When you move 3/4 R, he can move 4 times as far or 3 x R, which fortunately for you is less than pi x R.

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#7
In reply to #6

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

09/30/2022 2:42 PM

If the lake is 1 mile wide, with a radius of 2640 feet, the Grim Reaper will miss you by (pi X 2640') - (3 X 2640'), which is approximately 370 feet. When you jump ashore the Grim Reaper will be coming at you at 100 miles an hour (if your boat can go 25 mph.), so he will cover the distance separating you in less than 3 seconds. Better hope your friend waiting to give you a ride has a very fast car, and better hope (as you are cheering your good fortune) that the Grim Reaper hadn't already planned for you and your friend to die in a fiery car crash when your friend sees the Grim Reaper in the rear-view mirror, panics, and loses control.

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#9
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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/01/2022 2:32 PM

I won't vote against the GA vote to this comment because I think that would be churlish. But since in Post#3 you admit to foreknowledge of the puzzle and the solution, it is my opinion that the applause, if any, should be subdued.

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#11
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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/02/2022 5:42 AM

Post #6 contains the crucial insight that (3/4)R is less than (π/4)R. Whether the guy travels in a spiral or not, as long as he stays just inside the circle with radius R/4 he can manoeuvre to a position where the reaper is at exactly the worst position when the guy makes the break for the shore.

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#16
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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/02/2022 12:09 PM

I fully understand the standard solution. It is all in Post #3. Post #6 just provides detail. My point is: Is it insight to dredge up from memory the solution to a puzzle one encountered in the past? Another poster in this thread also commented that, "As I recall the solution is... etc.". These monthly challenge questions are intended as exercises in creative problem solving. They are not intended as technical questions to be answered from memory or by google search. An analogy: If you already know how a card trick is done, keep it to yourself. Don't deny the rest of us the fun in trying to figure it out.

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#12
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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/02/2022 6:05 AM

OK, I sort of remember a similar puzzle, but in my defense, I did have to figure it out again. And I'm just guessing that the goal is to escape from the lake. As stated in the puzzle the boat guy would be safe staying out there forever.

Is the solution optimal? I think it is. The boat can get as close to 1/4 R as possible and stay on the other side of the lake. From that point his best solution is to minimize his remaining distance and maximize GR's remaining distance, i.e. head directly for shore. (Remember, GR can take either way around the lake.)

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#15
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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/02/2022 11:41 AM

I agree that your described solution is optimal if the goal is simply to beat the GR to a certain point on the shore. Your described solution puts approximately 9 degrees between the boat and the GR upon landing. However, if the goal is to have the greatest distance between the GR and the boat's landing point, I suggest that minimizing the remaining distance for the boat is not the best strategy.

If, in the boat's final run to shore, the boat is angled away from the direction of the GR's approach, a small increase in the distance the boat must travel to shore results in a much larger arc, a much larger distance, that the GR must cover. Increase the boat's distance by 1/4r and this adds approximately 90 degrees to the distance the GR must travel. The boat will then travel 1r while the GR travel 4r, or approximately 228 degrees, so when the boat lands considerably more than 9 degrees will separate it from the GR. I know that the GR can change direction. So can the boat - angling towards the shore at every turn, away from the direction of approach of the GR, adding large degrees of arc and distance that the GR must travel.

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#17
In reply to #15

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/02/2022 5:41 PM

That's an interesting idea. Can you post a sketch?

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#18
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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/02/2022 7:40 PM

Here is a plot from a simulation I wrote. The outer green circle is the lake shore, the inner green circle is the R/4 circle. The boat can maneuver enough inside R/4 to be opposite the lake center '+' from GR giving a 1 1/4 lake radius head start, the optimum position. GR moves CCW over the top while the boat (red) heads for the nearest shore, and barely makes it past the shoreline as GR completes the half circle.

If I understand your idea correctly, the boat trajectory should curve downward lengthening the distance GR (and also the boat) have to travel, leaving a larger safety margin.

I will try some different trajectories for the boat and see what happens...

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#20
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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/03/2022 9:20 AM

Good point (and Answer).

Clearly you're going to set off directly for the shore then turn slightly straight away depending on which direction the Reaper takes, and I think for simplicities sake we can ignore the decision time/distance.

I can't work out what this means.

The slope never goes negative, but clearly if you turned through 180°, the Reaper would change direction.

I'm beginning to think that there is a route you can take which will totally mess him up.

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#21
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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/03/2022 10:25 AM

I don't have your and Rixter's CAD skills (I do have a nice drafting board). You both understand my idea correctly.

Yes, start for shore in a radial direction from r/4, wait a moment to see which direction the GR takes, then angle away from him. I am unsure what would be the optimum angle, but measuring directly from my drawing, if the angle is such to increase the boat's distance to shore by .25r, this will add considerably more than 1r to the distance the GR must travel to intercept the boat - in other words, although the GR is moving 4X faster than the boat, he is falling behind in the chase.

If the GR changes direction, my intuitive sense is that you should wait until he is 180 degrees opposite. This puts you and the GR in the same orientation as at the start of your run, but now you are closer to shore. At that position immediately bring the boat about and again angle away from the GR towards shore. If the GR continues to change direction and you continuously repeat this strategy, it seems possible that you will reach the shore while the GR is still on the other side of the lake (totally messed up, as you write). If I were to run a simulation this is the strategy I would investigate.

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#22
In reply to #15

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/03/2022 12:50 PM

You are correct, the safety margin improves moving out away from the direction GR is coming from.

Going straight out at 180 degrees from GR's starting point gives about 2 degrees lead. This can be increased to at least 8 degrees by turning away from GR.

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#25
In reply to #22

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/03/2022 4:33 PM

My numbers are a bit different than yours, but we agree that turning away from the GR puts the boater on shore further from the GR than the standard solution.

I have attached a photo of my drawing which shows two possible boat-runs/GR-chases. The first scenario has no directional changes of either the boat or GR. The boat path is the angled straight line in the lower left quadrant (angled more away from the GR than you show, which would account for quantitative differences in outcome). The shore landing point is labelled Boat Finish #1. At the time of boat landing, the GR, having travelled CW, is at the point GR Finish #1. By my drawing, the boat and the GR are approx. 35 degrees apart at that time.

In the second scenario the GR decides to reverse direction from CW to CCW at the point labelled on the upper right quadrant. The boater sees this and continues his bearing until the GR is 180 degrees opposite (at the dotted line), then he comes about and sets an angled course to the point on the shore labelled Boat Finish #2 in the lower right quadrant. When the boat reaches that point, the GR is at the point labelled GR Finish #2 in the lower left quadrant. At the finish, the GR and the boater are still separated by about 35 degrees in this scenario.

I drew a third scenario in which the GR reverses direction again, back to a CW direction (with an accompanying boat reversal of direction), but I didn't include it because the drawing was becoming too cluttered. However, in that scenario, almost 70 degrees separate the GR from the boat when the boat lands, so it appears that the more times the GR changes direction (with the boater also responding with a directional change) the more distance is put between the boater and the GR at the finish.

Now, if we could just apply this to the real world, the Grim Reaper wouldn't have a chance.

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#27
In reply to #25

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/04/2022 7:11 AM

Once the boat is outside the inner ¼R circle there is never any reason for the Reaper to change direction, unless the boat re-enters the circle.

So the optimum angle to turn at is 90°

If R is 1 then you miss the reaper by 0.5893 instead of 0.1416 by going straight.

Turning more than 90° you re-enter the circle.

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#28
In reply to #27

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/04/2022 11:07 AM

Very good. In my first sketch (not posted) I did draw the boat's path at a tangent (90 degrees) to the r/4 circle. Your rigorous analysis shows that my first grab turned out to be the optimum path. In my posted drawing I pulled back that angle to 80 degrees. From your left chart the distance between the boat and the GR when the boat lands is only incrementally increased by altering the boat path from 80 to 90 degrees - from 33.6 degrees on the circumference of the lake to 33.7 degrees. I got out my protractor again, placed it on my drawing and squinted with my good eye on the markings. 34 degrees is what I read. Not bad for old technology. (as I wrote, I have a nice drafting table, fitted with a vintage Vemco drafting machine).

GA for your analysis. We can add this boat run wrinkle to the standard solution to the puzzle.

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#31
In reply to #28

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/10/2022 6:34 AM

You should really grab a free or very cheap 2D CAD package. I'm a fan of TurboCAD, but, they don't seem to do a free version any more (They used to do a brilliant Learning Edition for windows 95.) ; they do still do DoubleCAD which is free, and has a user interface which is similar to Autocad:-

https://www.turbocad.com/content/doublecad-xt-v5

Another very simple thing to try would be one of the free geometry programs like Geogebra

Try it here:-

https://www.geogebra.org/geometry?lang=en

Or download one of the programs here:-

https://www.geogebra.org/download

probably just go for "geometry" or "geogebra classic 6"

Avoid 3D calculator: I think you have to pay for it.

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#32
In reply to #31

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/10/2022 9:50 AM

I agree. Knowing a bit of CAD would be very useful (and not just for CR4 puzzles). I will look at the links you provided. Thanks.

I would never completely abandon my drafting table though. Twenty years ago I got it from a local high school. I fitted it with a new pad and trimmed it with cherry wood edging and pencil tray. Because of CAD the school was getting rid of all their drafting tables. Now none of the technical students will learn how to draw on paper. I think that's a shame. It is a pleasure to use precision drawing instruments - to create a clean, nicely rendered drawing. My father was the drafting supervisor at General Electric. I suppose I feel a connection to him by keeping the old ways alive. Also, unlike a computer-generated image, people respond positively to something drawn by hand. Not long ago I took a drawing to a machine shop to have a part made. The owner looked at the drawing appreciatively. "Hand drawn," he said. "Very nice. We don't see hand drawings often anymore."

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#29
In reply to #27

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/09/2022 5:23 AM

Correction to the above: I kind of mis-placed Theta.

When the boat turns at a right angle the angle at the centre of the circle will be 75.52°, and

the reaper will miss by 0.5867

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#30
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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/09/2022 9:36 AM

Since the optimum path determined by your post #27 calculations corresponds so closely to the angle between the GR and the boat measured directly from my drawing, I just assumed your original calculations were correct. Converting your corrected figure to degrees, it shows that your amendment has enabled the GR to close the gap between himself and the boater by 0.14 degree. Miniscule, but I understand your wish to post the correction. This optimum boat path puts 33.62 degrees between the GR and the boater when the boat lands. This is more than 4 times the distance separating them in the standard solution.

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#33
In reply to #3

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/25/2022 2:31 PM

Hear ye, hear ye!

Here be the logic that sets you free from the grim reaper.

The reaper is 4x times as fast as the boat, but he must stay on the land. If the boat circles the lake at radius/4 (or less), the boat will eventually outpace the GR in your respective circles. Once he is 180° behind, immediately drive to the shore and make your get away.

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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

09/30/2022 12:37 PM

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Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/01/2022 12:02 PM

So, you manoeuvre your boat in an outward spiral or zig-zag path, depending on the Grim Reaper's directional constancy, until you are r/4 distance from the centre of the lake. At that distance, and no further, you can keep the centre point of the lake on an imaginary straight line between you and the GR.

From there, if you make a radial run to shore you will reach the shore approximately 9 degrees ahead of the GR on the circumference of the lake (Rixter posts #3 and #6). It is unspecified in the puzzle, but if this means you have (for now) eluded the GR, then this strategy is a win. But if to escape the GR you still have to dash to your car, you want to maximize the time to do this. Therefore if there is a more optimal path for your final boat run to the shore you would be wise to choose it.

My math skills to calculate optimal paths have long gone into the memory hole, but from a circle drawing of this puzzle, a better strategy would be as follows:

Once you have manoeuvred and reached a point on the circumference of the imaginary circle r/4 from the centre of the lake, start your final boat run along a radial path to shore and simultaneously check to see in what direction the GR has decided to take, then immediately turn away from the GR and follow a line tangent to the described imaginary circle. The distance to shore along this tangent is approximately 1r (measured from my drawing). By the time you reach shore the GR will have travelled 4r, or approximately 228 degrees. The point where you land your boat will be approximately 62 degrees beyond him (again measured from my drawing), allowing you considerably more time for a dash to your car than if you had followed a radial path. (During your boat run to shore the GR is unlikely to change direction because he is gaining on you, but he is gaining more slowly than if you had chosen a radial path to shore with a 3/4 r distance.)

Note to other posters: If you are familiar with the monthly challenge question and know the solution, kindly withhold the solution to allow others the fun of grappling with the puzzle.

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#10

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/02/2022 3:32 AM

Not exactly as specified, but since this is inspired by Halloween some of your readers may like to find out what Halloween refers to.

See the posts here: https://mythandmegalithica.blogspot.com/

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#13

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/02/2022 11:00 AM

From memory it starts with the boat spiral;ing away from the centre until it has covered 75% of the distance to shore at which point it can turn and run for the shore.

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#14

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/02/2022 11:09 AM

In order for me to beat the Grim Reaper, I have to get to the shore from the middle of the lake in a less amount of time than it takes for the Grim Reaper to go half way around the circumference. That is, my time, T1, must be less than the Grim Reapers time, T2.

Time = distance / velocity.

My distance is r and the Grim Reapers distance is pi(r).

My velocity is v and the Grim Reapers velocity is 4 time mine or 4(v).

This sets up the equation T1 < T2

or r/v < pi(r)/(4v)

r/v cancels from either side so we get the following...

1<(pi/4)

This is never true. So my response is "THE GRIM REAPER GETS YOU EVERY TIME."

Some of us fish in the middle of the lake longer than others, but eventually the Grim Reaper will touch all us. So enjoy that time in the time in the lake. Use up that entire tank of gas. Using all that gas to outsmart the Grim Reaper is futile. So use most of that gas for other purposes.

If there is a solution, please clue me in. :) Or sell it. You could get very rich.

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#26
In reply to #14

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/03/2022 8:31 PM

Interesting. Your 'proof' that there is no solution (when there is), reminds me of an old 'proof' that 2=1 (when it isn't):

a=b

therefore a2=ab

therefore a2-b2=ab-b2

therefore (a-b)(a+b)=b(a-b)

therefore a+b=b

therefore b+b=b (since a=b)

therefore 2b=b

therefore 2=1

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#19

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/02/2022 11:05 PM

This puzzle appeared a few decades ago in a Martin Gardner Mathematical Games column. An additional stipulation was that once on shore, you could outrun the GR. In other words, home free if you can get to a point on the edge of the lake before the GR can get there. The radial run suffices, but further maneuvering can increase your margin of safety.

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#23

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/03/2022 2:56 PM

The only way to be safe from the grim reaper is to trap him someplace.....like in a tree for instance....

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#24

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

10/03/2022 3:35 PM

For extra credit, what is the Grim Reaper's name?

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Jack, of course!

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#34

Re: A Spooky Challenge Question with the Grim Reaper (Oct. 2022)

11/01/2022 4:05 PM

Or… carry full size snickers candy bars,,,

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