# July 2008 Puzzle Solution

A great-grandfather makes his family work for their land and fortune

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).]

Dennis Shasha is at the Courant Institute of Mathematical Sciences, New York University. His most recent puzzle book, Puzzles for Programmers and Pros, was published in May 2007 by John Wiley and Sons/Wrox.

View
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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