A treasure map locates the treasure using four landmarks, two of which have vanished.

geometry/treasure.map New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

A treasure map locates the treasure using four landmarks, two of which
have vanished.  Can the treasure still be found exactly?

I have a treasure map which says:

"Desert Road and Mountain Road intersect at right angles in downtown
Dry Gulch.  On Desert Road, within 10 miles east of this intersection
grows the old hanging tree (H).  Within 10 miles west of the
intersection on Desert Road is the old well (W).  On Mountain Road,
within 10 miles north, there is the stagecoach stop (S) in one place
and Barry's Farmhouse (F) in another spot.  Draw a first line from H
to F and a second line from W to S.  These two lines intersect at
Magic Point A.  Draw a third line from H to S and a fourth from W to
F; they cross at Magic Point B.  The treasure is buried at the point
where line AB crosses Desert Road."

I went searching for the treasure.  I found the hanging tree 3 miles
east of town and the well 6 miles west of town.  But there was no
longer any sign of the stagecoach stop or of Barry's farmhouse.  Yet,
algebraically (using coordinates) I was able to find the *exact*
location of the treasure.

-- Edmund Taylor <etaylor@interaccess.com>

Show solution

This problem can easily be solved by means of projective geometry.
Let O be the intersection of Desert Road & Mountain Road, let Y be the
point where AB crosses Mountain Road, and let X be the treasure spot
where AB crosses Desert Road.

There is a so-called "cross-ratio" associated to every quadruple of
points C, D, E and G on a line.  It is defined as

[CDEG] = ((CD)(EG))/((CG)(ED)).

Here the distances CD etc. are oriented distances, so that, for
example, CD=-DC.  Given any two lines L and M, and a point P not on L
or M, we can project L to M by sending any point Q on L to the
intersection of QP with M.  (This is called `projection from P'.)  It
is a theorem that the cross-ratio is invariant under such a
projection.

Now the points W, O, H, and X are collinear, BYAX are collinear, and
projection from F sends WOHX to BYAX.  Consequently [WOHX]=[BYAX].
But also, projection from S sends HOWX to BYAX.  Thus [HOWX]=[BYAX].
It follows that [HOWX]=[WOHX], so ((HO)(WX))/((HX)(WO)) =
((WO)(HX))/((WX)(HO)).  Hence (HO)(WX) = (+ or -)(HX)(WO), i.e.,
-3(9+HX) = (+ or -)6(HX).  (We let CD>0 iff D is east of C.)  Solving,
we find that HX is -3 or 9.  Since HX=-3 would mean that X=O, we must
take HX=9.  Therefore the treasure is 9 miles to the east of the old
hanging tree, or 12 miles east of the crossroads O.

-- David Moews <dmoews@xraysgi.ims.uconn.edu>

[ Editor: two more solutions from the 1994 thread, "Geometry Fun".  In
posting the problem, Edmund Taylor added: "For reference, I recall the
answer is something like 15 miles east of town on Desert Rd." ]

Well, since the puzzle is solvable without any knowledge of where the
farm and the coach stop are, I expect we can choose any places for them
and get the same answer.  Choosing two simple distances, I make the
coach 3 miles north and the house 6 miles north.

        For A:

        H-F:  y=2x+6
        W-S:  y=-1/2x+3

        X:  5/2x=-3  x=-6/5  y=18/5

        For B:

        H-S:  y=x+3
        W-F:  y=-x+6

        X:  2x+3  x=3/2  y=9/2

        Treasure Hunt:

        M:  m=(9/2-18/5)/(3/2+6/5)=1/3

        B:  9/2=1/3*3/2+b  b=4

        X:  y=1/3x+4  1/3x=-4  x=-12 or 12 miles E

        Done.

The map gives no information beyond the fact that the sites aren't on
the same spot and both are north, so if using different numbers gave us
a different answer, the problem would be unsolvable.

-- Scott "Fool" Purdy <spurdy@pomona.claremont.edu>

[ Editor: Purdy's coordinates put east at negative x (H at x = -3, W
at x = 6). ]

This is a problem in projective geometry.  I am not very knowledgable
in this area, so I had to do some research.  Here is what I have
figured out.

If w,x,y,z are four points on a line, then the cross-ratio is
(w,x;y,z) =  (|wy| |xz|) / (|wz| |xy|), where |ab| denotes the directed
distance from a to b.  (That is, |ba| = -|ab|.)

Now for the theorem:  Let W, H, F, and S be any four points, no three
of which are collinear.  Let P = WH intersect FS, A = FH intersect WS,
B = FW intersect HS, and T = WH intersect AB.  Then the pairs of points
W,H and P,T separate each other harmonically, i.e. (W,H;P,T) = -1.

Now we plug in the given data: |WP| = 6, |PH| = 3, and |WH| = 9.
Let x = |HT|.  Then |WT| = x + 9, so

-1 =  (W,H;P,T) = (|WP| |HT|) / (|WT| |HP|) = 6x / ((x+9)(-3))

==>  3(x+9) = 6x   ==>  x = 9.

Therefore, the treasure (T) is 9 miles east of the hanging tree (H).

My source was the VNR Concise Encyclopedia of Mathematics.  This is
probably not the best reference, but it is all that I have available at
the moment.  A better reference would probably be "Geometry, a
Comprehensive Course" by Daniel Pedoe.

-- Dave Radcliffe <radcliff@alpha2.csd.uwm.edu>