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= (xa,ya) = (695,251)
B = (xb,yb) = (082,006)
C = (xc,yc) = (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
mab
xb +
bab
- yb =
0,
and the equation for BC is
mbc
xb +
bbc
- yb =
0.
The slope for AB is
mab
= (yb - ya) /
(xb - xa),
and
the slope for BC is
mbc
= (yc - yb) /
(xc -
xb).
The slope of a perpendicular line to
AB is
m1
= - 1 /
mab
, and for
BC is
m2
= - 1 /
mbc
.
The x value for the midpoint of AB
is
x1
= xb - (xb -
xa)/2, and
the y value for the midpoint of AB is
y1
= yb - (yb -
ya)/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.
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.
