Yoak: Pirate Treasure Map
Our band of intrepid pirates, having resolved previous squabbles over distributing booty amongst themselves and other issues have come across a treasure map fragment. The picture has been destroyed, but the following text can be read:
Stand upon the gravesite and you’ll see two great palms towering above all others on the island. Count paces to the tallest of them and turn 90 degrees clockwise and count the same number of paces and mark the spot with a flag. Return to the gravesite and count paces to the second-tallest of the trees, turn 90 degrees counter-clockwise and count off that number of paces, marking the spot with a second flag. You’ll find the treasure at the mid-point between the two flags.
Fortunately, our pirates knew which island the map referred to. Sadly, upon arriving at the island, the pirates discovered that all evidence of a gravesite had faded. The captain was preparing to order his men to dig up the entire island to find the fabled treasure when one of the more geometrically inclined pirates walked over to a particular spot and began to dig. The treasure was quickly unearthed on that very spot.
How did the pirate know where to dig?
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Sue VanHattum said,
January 25, 2010 at 5:57 pm
Nice! (Your system told me my comment was too short. Hmm…) The geometrically inclined pirate also had a very good eye for estimating distances.
Byon said,
January 26, 2010 at 12:50 am
Very cool! I like this one. SPOILER BELOW (sorry, I don’t know how to properly hide spoilers)
By emperical plotting, I found given two arbitrarily positioned tallest trees, regardless of where you choose for the gravesite, you always end up with the same treasure site (as you would expect from the geometrically inclined pirate’s actions). And, this site is at the center of a square where the line between the two trees forms one side of the square.
Looking at why, lets give the island an arbitrary grid, with the following coordinates:
The grave at (gx,gy), the tallest tree at (t1x,t1y), the second tree at (t2x,t2y).
From this, we can see that
flag1 is at ( t1x + (t1y-gy), t1y – (t1x-gx) ) and
flag2 is at ( t2x – (t2y-gy), t2y + (t2x-gx) ) and
the treasure is at ( t1x + t1y – gy + t2x – t2y + gy, t1y – t1x + gx + t2y + t2x – gx )/2 or, after removing the gravesite from the equation,
( t1x + t1y + t2x – t2y, t1y – t1x + t2y + t2x)/2
If we define a new coordinate system, with the second tree to be the origin (t2x,t2y) = (0,0) and the tall tree (t1x,t1y) = (0,1), we get the treasure at (0+1+0-0, 1-0+0+0)/2 or (1/2,1/2) in this new grid.
So start at tree 2, walk halfway to tree 1, turn 90 clockwise, walk the same distance, and there you are!
jyoak said,
February 1, 2010 at 3:58 pm
Byon, nicely done! I had looked for something of that level of simplicity in explaining the answer and hadn’t found it. It originally delayed me some in posting it until Steve found a good explanation. I still hope he posts it.
Maybe I should gamble more and just trust that one of you will solve problems elegantly when I’ve failed to. :-)
jyoak said,
February 1, 2010 at 3:59 pm
Sue, thanks! I followed the link over to your blog. It looks nice and I’ll be following it. Maybe we should set up some sort of Blog Roll here on Math Factor. I’ll shush up about that now because there is a very fine line between having an idea and volunteering. :-)
Sue VanHattum said,
February 1, 2010 at 4:15 pm
Thanks! I’m glad you liked it. (To anyone else who’d like to visit, you can google Math Mama Writes.)
jyoak said,
February 1, 2010 at 4:50 pm
Byon, one minor point though: Wouldn’t the line between the trees be the diagonal of the square? Your argument and conclusion seem to indicate that otherwise.
Stephen Morris said,
February 1, 2010 at 8:31 pm
Sue, also love your site.
As Jeff mentioned I had site of this problem before he posted it here and had a lot of fun with it. Somehow Jeff has a way of finding fantastic puzzles, even when he doesn’t have the answer yet!
I came up with three solutions. I hope this doesn’t count as a spoiler. The simplest involved similar triangles but doesn’t work so neatly for all possible locations. The second involved mirrors and the third involved rotating the whole plane.
The problem generalises. Jeff had us rotating by 90 degrees and 90 degrees. This puzzle works if you use different angles as long as they add up to 180 degrees. My mirror proof and rotating the plane proof still work.
The treasure will be in a different location but that location is neatly tied to the angles we choose.
I will post these if no-one else posts them first, promise. It will probably be a separate post as diagrams really help.
Stephen Morris said,
February 5, 2010 at 6:36 pm
So this is how I see it. If you happen to be a brilliant school teacher (Hi Sue) you can get your students to do this with tracing paper (do they still have tracing paper in schools?)
Put a pin in the tallest tree and rotate the island through 90 degrees clockwise. This maps one of the flags to the gravesite.
Put the island back just how you found it.
Now put a pin in the other tree and rotate the island through 90 degrees clockwise. This maps the gravesite to the other flag.
If we do the first rotation and then the second we will have mapped one flag onto the other and rotated the whole island through 180 degrees. The halfway point between the flags is the centre of rotation. This is the same whatever our choice of gravesite/flags.
Hopefully you see that this works for other angles as long as they add up to 180 degrees.
Sean McCloskey said,
February 22, 2010 at 2:01 pm
I have a slightly different version of Byon’s solution, which was easier for me, though it might be more complicated to some.
Assume a coordinate plane centered at the large tree, with the x-axis passing through both trees. Your coordinates from any arbitrary point on the island are (a,b). The function f(a,b) –> (-b,a) maps to your destination after following the first set of instructions.
Now go back to your starting point (a,b) and assume a coordinate plane centered on the smaller tree. Your coordinates with respect to this new origin are (a-x, b), where x is the distance between the two trees going in the direction from the larger tree to the smaller tree. The function g(a,b) –> (b,-a) describes your destination after following the second set of instructions, so you arrive at (b, -a+x). But this is relative to the small tree. Relative to the large tree your second destination is (b+x, -a+x).
All that’s left to do is find the midpoint between (-b,a) and (b +x ,-a+x). So the midpoint is (x/2 , x/2).
Ron said,
February 23, 2010 at 11:12 am
Is this the same (or simlilar ) to aproblem in George Gamow’s book One two three…..Infinity ? He used (as I recall–don’ have the book any more) complex numbers to solve it>
jyoak said,
February 23, 2010 at 2:30 pm
Ron,
I’ve never seen that book, but it sure does look interesting. Thanks for the pointer.
I got the inspiration for this puzzle, if I remember correctly, from the IBM site Ponder This which can be found here: http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/pages/index.html . I’m not sure how literally I took it nor am I even 100% sure this is where I got it, but regardless it is an interesting source of puzzles.
Stephen Morris said,
February 23, 2010 at 3:23 pm
Just checked, yes you can do it with complex numbers.
Byon said,
February 23, 2010 at 10:07 pm
Thanks, Jeff. Yes, I guess you could also consider a different square where the trees mark opposite corners, and the treasure is one of the remaining corners. But I saw it as a square sqrt(2) larger than that, where the trees marked adjacent corners, and the treasure was in the center of that square. Both should work. Thanks again for the fun problem!
Ron said,
February 24, 2010 at 11:17 am
Yep ! I found the Gamow book I told you about on line w/ ability to go in and read some pages– and the problem and solution using imaginary numbers is around pages 35-37
One could also use vector algebra could we not?
I enjoy the MAA site very much. I am a 76 year old retired high school math teacher who tries to keep my brain from going dead!!
Sue VanHattum said,
December 31, 2010 at 9:50 am
My first comment makes me think I had a solution when I wrote it. But I now have no idea what my solution was. Oh well, that allows me the pleasure of solving it again. But my efforts this time don’t give me any elegant answers…
I’m writing to point out that all of our ideas seem to allow for two different possible locations for the treasure. Perhaps one of those locations was in the water, so the pirate knew to dig at the other.
Jeremy said,
January 24, 2012 at 5:19 pm
Can someone explain why the treasure is always the midpoint between the two flags?
jyoak said,
January 24, 2012 at 5:42 pm
Jeremy, that’s part of the statement of the problem. It is the last sentence in the second paragraph.
The Pirate’s Treasure « Where the Arts Meet the Sciences said,
May 10, 2012 at 5:44 pm
[…] on-line versions of it (physicsforums.com, mathpages.com, Bradley University, geometer.org, the mathfactor podcast, and University of Georgia), I couldn’t remember how I had solved it! It apparently appears […]