Geocaching - The Official Global GPS Cache Hunt Site

"Live long and prosper, Spock."  -  "I shall do neither... for I have killed my captian."

The coordinates listed for this geocache are not the actual location of the cache container.  You will need to solve a math puzzle to find the real coordinates.

Problem: You are given three points:
    A = (W 96° 45.695,  N 43° 30.251)
    B = (W 96° 45.082,  N 43° 30.006)
    C = (W 96° 45.410,  N 43° 30.719).

Draw a circle that lies on all three points. The center of that circle is the location of the cache.

The cache is camo'd well.  Please re-hide it in the same manner.  This cache has lots of real cool swag, so bring the kids.  After the First-To-Find, I'll add some hints.

There are three ways to solve this.  Method #1 is more difficult but yields an exact answer.  With Method #2, you solve it on graph paper.  Method #3 pretty much gives away the answer.

No matter which method you use,  it will be a lot easier if you use only the fractional part of each of the points. That is, the actual answer is W 96° 45.xxx,  N 43° 30.yyy.

To get credit as a find, you must sign the log book.

In addition to the explanations below, it might be useful to do a Google search on the terms:
center circle three points, or try intersection perpendicular bisector.

Method 1: (also see these links: Paul Bourke, McRae Family, Dr. Math (true math on a sphere) )

The three points A, B and C are:
A= (x_a,y_a) = (695,251)
B = (x_b,y_b) = (082,006)
C = (x_c,y_c) = (410,719).

To solve this, we need to find the equations for the perpendicular bisector for line segment AB and for line segment BC.  The intersection of these two perpendicular bisectors is the center of the circle, and hence the answer.

The equation for a line is mx + b - y = 0, where m is the slope of the line.
Then, the equation for AB is m_abx_b + b_ab - y_b = 0,
and the equation for BC is m_bcx_b + b_bc - y_b = 0.

The slope for AB is m_ab = (y_b - y_a) / (x_b - x_a), and
the slope for BC is m_bc = (y_c - y_b) / (x_c - x_b).
The slope of a perpendicular line to
AB is m₁ = - 1 / m_ab, and for
BC is m₂ = - 1 / m_bc.

The x value for the midpoint of AB is x₁ = x_b - (x_b - x_a)/2, and
the y value for the midpoint of AB is y₁ = y_b - (y_b - y_a)/2.

You now have the slope and a point for the first perpendicular bisector. With these you can solve for b₁.  Find the equation for the perpendicular bisector for BC in the same manner.

Because the center of the circle lies on both of these lines, the center is the point at which the two lines are equal to each other, e.g.,  m₁x + b₁ - y =  m₂x + b₂ - y.

Method 2: (also, see this link: a good visual of what we're doing )

You need fine-ruled or very large  graph paper, and a compass.  Place the three points on the paper.  Next, you'll need to construct the perpendicular bisectors.  For a really cool demonstration on how to do this  click here.  Or, you can follow these step-by-step instructions.  For best results, use a fine-lead mechanical pencil. Draw all lines as thin as possible.  Each .001-minute that you are off represents about 5-ft error. With care, I was able to get right on the actual coordinates.

	1)  Draw straight lines to create the line segments AB and BC. Any two pairs of the points will work.
	2)  Find the bisector of one of the lines.
	3)  Repeat for the other line.
	4)  The point where these two perpendiculars intersect is the center of the circle we desire.  The circle drawn is the only circle that will pass through all three points.

Method 3:

You really didn't think it was going to be that easy, did you? Try one, or both of the methods shown above.  After this cache's FTF, and if the answer still eludes you, email me, and I'll send you the Excel file or email the link that will solve it for you.