Two pirates decide to hide a stolen treasure on a desert island in the vicinity of a river. On one side of the river there is a birch tree located at a point they denote by B, which is directly west of a pine tree they denote by the point P. On the bank of the river nearest the trees, they drive a stake into the ground and denote this point as S. To bury the treasure, one of the pirates starts at S and walks towards B and after reaching turns right 90° and walks the same distance as SB reaching the point Q. In other words, SB = BQ. The second pirate starts at S and walks towards P , after which he turns left 90 degrees and walks the same distance as SB reaching the point R. In other words, we have SP = PR. The two pirates now advance towards each other and bury the treasure halfway between them. In other words, halfway between Q and R. Some months later the two pirates return to dig up the treasure only to discover that the stake was gone. The pirates thought they would never find the treasure until one of them said he had a plan. Is it possible to devise a way for the pirates to find the treasure without knowing the location of the stake S?

Geometric Solution

As with many problems, we can solve the problem geometrically, or we can introduce coordinates and attempt a more analytic solution. This problem can be solved in either way. We begin with a purely geometric solution by drawing the illustration below. Note that we have placed the stake S (that lies on the riverbank) in such a way that the treasure T does not lie in the river.

We assume for the moment that the position of the stake S is known and hence can find the points Q, R and the treasure point T . (When we get done, we will be surprised to see that the location of the treasure T does not depend on where we placed the stake S, but only on the points P and B !) We begin by dropping perpendiculars from Q, R, S , and T to the line segment BP . Doing this and using basic geometric arguments, we see that QQ'B and BS'S are congruent triangles as well as the two triangles SS'P and PR'R. Hence, we have the following equality of distances: BS' = QQ', S'P = RR', Q'B = SS' = R'P. And using the relations BS' = QQ', S'P = RR', the distance between the birch and pine trees can be written as

BP = BS' + S'P = QQ' + RR'

But since the midline of the right trapezoid has length

TT' = (QQ' + RR' )/2

we have TT' = BP/2. But the midline TT' of the trapezoid QQ'RR' is also the perpendicular bisector of B and P since Q'B = R'B, and so the pirates can find the treasure even if the location S of the stake is no longer known by finding the perpendicular bisector of BP, and then walking along this bisector towards the riverbank a distance of BP/2.

Analytic Geometry Solution

We can also solve this problem analytically by introducing the Cartesian plane, where we place the x-axis through the points B and P with B = (-1, 0) and P = (1, 0). See the diagram below.

If we now place the stake at an arbitrary point S = (a, b), we can use basic analytic geometry to find the slopes and equations of the lines in the figure. (Remember that slopes of perpendicular lines are negative reciprocals.) Doing this, you will discover that Q and R and are located at Q = (-1-b, 1+a), R = (1+b,1-a). Hence, to find the treasure T, we then take the average of these two points, which gets the midpoint T = (Q+R)/2 = (0,1). Hence, the treasure T is located on the perpendicular bisector of B = (-1, 0) and P = (1, 0), and the distance along the bisector is the distance from B or P to the bisector.

Complex Numbers Solution

If you are familiar with the geometry of complex numbers, there is a nice way to solve this problem without resorting to geometry or slopes of lines and so on. This is based on the fact that performing arithmetic on complex numbers acts to move points in the complex plane. In particular, multiplication of any complex number a + bi by the imaginary number i rotates the number around the origin in the complex plane in the counterclockwise direction 90 degrees; and multiplication of a + bi by -i rotates a + bi in the clockwise direction 90 degrees. Hence, we begin by placing the trees B and P at the points B = (-1,0) and P = (1,0) in the complex plane as shown in the diagram below.

We now place the stake at an arbitrary point S = a + ib. Using complex arithmetic, the point Q is simply

Q = [(a + bi)+1]i - 1 = (a + 1) i - (1 + b)

Why is this ? Well, the expression inside the brackets [ ... ] essentially moves the origin of the complex plane to the new point B = (-1, 0), after which points are multiplied by i,which rotates a+bi to Q. But then we have to subtract -1 to get back in the original coordinates of the problem. In the same way, we can find T by computing

R = [(a + bi)-1] (-i) + 1 = (1 - a)i + (1 + b)

Note that we subtract -1 from a + bi, which places the origin of the complex plane at P = (1, 0), and then if we multiply by -i, this rotates points 90 degrees in the clockwise direction, and so a + bi moves toR. Then, finally we add +1 to get R in terms of the original coordinates. So, we can now find the treasure T by taking the average of these to complex numbers, or

T = (Q + R)/2 = (1/2) [ (a + bi)i - (1 + b)] + [(1 - a)i + (1 + b)] = i

Hence, regardless of the original point S = a + bi,all remnants of S cancel and we discover that the treasure can be determined by knowing only the points Band P . In particular, as before the treasure lies on the perpendicular bisector of B and P , with the distance from the line BP being BP/2. In fact, the treasure T will always be at the vertex of a right triangle whose hypotenuse connects the trees B and P.
It is interesting to note that if 25 people placed stakes in different positions, everyone's treasure would be buried in the same place using this strategy. In other words, it wasn't a very good strategy.

This problem goes back to a very interesting book, One Two Three ... Infinity by George Gamow. Viking Press (1947). This book is still in printing and makes very interesting reading.

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Last modified on Tuesday, January 12, 1999