If the trees are plotted on a Cartesian plan, the following rules are true:

- When tree A > B, then the treasure is at a point (x, y)
- When tree A < B, then the treasure is at a point (x, -y)

Start with the gallows at point (3, -4) and call it G

_{1}. The resulting vector G

_{1}B can be demonstrated with δy/δx, which translates to {(4 - 0)/(3 - 6)} = -4/3, which would give it a positive slope of 4/3. The directions state that a 90

^{o}turn to the right must be made, resulting in a vector equal in length to the original, but with a negative slope of 3/4, which would make the stake (S

_{1}) at point (10, -3). The placement of the gallows was chosen for being halfway between the distance of the trees, so following the directions gives the solution mirror symmetry, placing the S

_{2}at the inverse of S

_{1}along the x-axis, (-10, -3). Following the directions one step further puts the point "treasure" at (3, -3) just one unit ahead of the gallows on the y-axis.

Simple ways to check to see if the answer is true is to:

- Pick more random points and follow the directions
- Put the gallows on the opposite side of the y-axis (3, 4) and follow the directions

By choosing the 2

^{nd}option, the first vector, GB, would be an extension of the second vector in the first step thereby having the same negative slope of 3/4, and a 90

^{o}turn to the right would place it at (2, -3). Again, the left turn directions for point A gives the destination of the resulting stake at a mirror point (4, -3), leaving the treasure at the point (3, -3).

What's fascinating about the solution to the problem is not that the treasure always falls at the same point no matter where the gallows are, it that the treasure ALWAYS falls along a point that is (x = (0.5 * the distance to B), y = negative (0.5 * the distance to B)) when B > A no matter what the distance between the two. I will leave this to the reader to experiment with.

When the points are reversed, that is B = (0, 0) and A = (6, 0), the location of treasure is also reversed or (x = (0.5 * the distance to B), y = (0.5 * the distance to B)) when B < A. This is not that spectacular in itself, but visualize the problem. If one were to show up on an island with only the clues "One tree is A, and the other is B", how would one decide which tree was which when they got there? Whose to say my tree A is the same as your tree A? This gives the original puzzle two possible solutions, which means if you find a parchment with this riddle on it, it's best to bring an extra friend with their own shovel.

I'm going to take a moment to plug this months Philadelphia Math Counts Meetup. This month we have been watching Michael Starbird's videos on "How to model the Continuum." Starbird uses fairly simple math as a way a to demonstrate how build good problem solving skills, which in turn helps lead to better reasoning skills and better proof building skills. Even if you don't live in Philadelphia, I suggest you go to the link and watch the videos in the description. They are well done and are aimed for people who aren't good at math.

Banned complain !! Complaining only causes life and mind become more severe. Enjoy the rhythm of the problems faced. No matter ga life, not a problem not learn, so enjoy it :)

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