Friday Puzzler -- Buried Treasure

Re: Friday Puzzler -- Buried Treasure

Larry Barrett

Pencil and paper test cases says the answer is yes. Can't prove it so far.

Or, you could do it like the Oak Island boys and dig 100 foot holes all over the island and maybe find the treasure if it is really there!

It's Complex

Alex Y

The proof, that is. Your test cases have undoubtedly led to the right answer.

Answer

Alex Y

Larry got this one, working with examples.

It doesn't matter where the wooden flag pole was, following the instructions will lead you to the same place for digging up the treasure. So just pick a starting point (midway between the other columns is a nice candidate, but anywhere will do), drive in your own "flagpole", and get ready for the riches!

Proof (for math geeks only ;-) )

Alex Y

Two essentially identical proofs that the flagpole doesn't matter:

1) Random points on a plane

Let the flagpole be at (a,b), the stone column at (c,d), and the brick column at (e,f)

To get from the flagpole to the stone column, you walk along (add) the vector V[c-a, d-b] (I'm making up this notation as I go; it's been too long for me to remember, but hopefully it will be clear.)

A vector 90 degrees to the right, and three times as long as the walk you just took would be V[3(d-b),-3(c-a). Adding this to the position of the stone column gets you (c+3d-3b,d-3c+3a) as the position for the first stake.

Following the same process with the brick column gets you (e-3f+3b,f+3e-3a) as the position of the second stake. Adding these two and dividing by 2 to get the midpoint ends up cancelling the a's and b's, so the flagpole doesn't matter.

This method can be made somewhat easier if you translate, rotate, and rescale your coordinate system to ;up some key points on the origin, and on an axis at unit distance, but it can be harder to follow conceptually, IMO

2) Complex plane

Ironically, a much less complex proof is to use complex numbers, and the complex plane where the complex number a+bi is plotted as (a,b) from the previous proof

Let the positions of the wooden, stone, and brick markers be given by the complex numbers W, S, and B

Walk from W to S on the vector [S-W] Rotate that vector 90 degrees clockwise by multiplying by -i. , then multiply magnitude by 3 and add to the position of the stone column to get

S-3iS+3iW

Going to the Brick column, and multiplying by i to turn left, you get the position of the second stake to be B+3iB-3iW

And the W's cancel out when you go to find the midpoint.

Now, non-math geeks, aren't you sorry you ignored the subject line? ;-)