Solution:

As you can see pictorially here, even if we move M to M', the intersection of the crossing line with Euclid remains the same. The first question is how to prove it. The second is: Where is the intersection point with Euclid?

Call the distance from Euclid to the pine P and the distance from Euclid to the maple M. In what follows, we will count in hundreds of meters. Thus, E is 2 units to the east of the intersection and W is 4 units to the west of the intersection.

Slope of EP is –P/2. Further, when x = 0, y = P.

So, the equation of EP is:
(1) y = (–P/2)x + P
Similarly, EM has the equation:
(2) y = –(M/2)x + M

Slope of WP is P/4, and when x = 0, y = P. So:
(3) y = (P/4)x + P
Similarly, WM has the equation:
(4) y = (M/4)x + M

L is the intersection of (1) and (4) and occurs when:
(–P/2)x + P = (M/4)x + M
x(M/4+P/2) = P – M = x(M+2P)/4

So, at L, x = 4(P–M)/(M+2P)
The y in that case is (from equation 1):
–2P(P–M)/(M+2P) + P = (2PM–2P2+MP+2P2)/(M+2P) = 3PM/(M+2P)

Checking, from equation 4:
M(P–M)/(M+2P) + M = (–M2+PM+M2+2PM)/(M+2P) = 3PM/(M+2P)

R is the intersection of (2) and (3) and occurs when:
(–M/2)x + M = (P/4)x + P

So, it follows that:
M – P = x(P/4+M/2) = x(P+2M)/4
and
x = 4(M–P)/(2M+P)

Substituting in equation (2), we get a y value:
–2M(M–P)/(2M+P) + M = (–2M2+ 2PM + 2M2+ PM)/(2M+P) = (3PM)/(2M+P)

Substituting into (3):
P(M–P)/(2M+P) + P = (MP–P2+ 2MP+P2)/(2M+P) = 3PM/(2M+P)

Now we want the line between the two points L and R:
L is (4(P–M)/(M+2P), 3PM/(M+2P))
and
R is (4(M–P)/(2M+P), 3PM/(2M+P))

prod = (M+2P)(2M+P)

Delta Y = 3PM(1/(M+2P) – 1/(2M+P))
= 3PM(2M+P–M–2P)/prod = 3PM(M–P)/prod

Delta X = 4((P–M)/(M+2P) – (M–P)/(2M+P))
= 4(P–M)(1/(M+2P) + 1/(2M+P)) = 4(P–M)(2M+P+M+2P)/prod
= 12(P–M)(M+P)

Delta Y/Delta X = PM(M–P)/4(P–M)(M+P)
= –PM/4(P+M)

Now check it on L:
3PM/(M+2P) = (4(P–M)/(M+2P))(–PM/4(P+M)) + b
= (–PM(P–M)/(M+2P)(P+M)) + b

So, b = (3PM(P+M) + (–PM(P–M)))/(M+2P)(P+M)
= (2P^2M+4PM^2)/(M+2P)(P+M)

Therefore, we can set y to 0 (meaning we are trying to find a point on Euclid) and ask for the x value at that point:

0 = (–PM/4(P+M))x + (2P^2M+4PM^2)/(M+2P)(P+M)

So, (PM/4(P+M))x = (2P^2M+4PM^2)/(M+2P)(P+M)

Therefore:

x = 4(P+M)(2P^2M+4PM^2)/PM(M+2P)(P+M)
= 4 (2PPM+4PMM)/PM(M+2P)
= 8PM(M+2P)/PM
= 8

So, just look 8 units (800 meters) east on Euclid.

Can you find a solution that requires less algebra?

The essence of this problem (like last month's) comes from the nice elementary geometry text by Edwin Moise and Floyd Downs entitled Geometry (ISBN 0-201-25335-6, Addison-Wesley, 1991).]

1 Comments

1. 1. waterbergs 02:58 PM 7/6/08

   The solution to this one is simple once one assumes that there is a solution! If this is so and the searcher can't determine which is the right pine and maple,but can still find the treasure, the site of the treasure must be independent of M and P. One can then choose M at infinity, EM and WM are then vertical. WL is then clearly twice ER, and so the treasure must be 600m east of E, i.e. 800m east of Pythagoras on Euclid - 1 minutes work.

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