A pirate has buried his treasure on an island with five...

A pirate has buried his treasure on an island with five trees located at the points $(30.0 \mathrm{~m},-20.0 \mathrm{~m}),(60.0 \mathrm{~m}, 80.0 \mathrm{~m})$ $(-10.0 \mathrm{~m},-10.0 \mathrm{~m}),(40.0 \mathrm{~m},-30.0 \mathrm{~m}),$ and $(-70.0 \mathrm{~m}$$60.0 \mathrm{~m})$, all measured relative to some origin, as shown in Figure $\mathrm{P} 3.46 .$ His ship's log instructs you to start at tree $A$ and move toward tree $B$, but to cover only one-half the distance between $A$ and $B .$ Then move toward tree $C,$ covering one-third the distance between your current location and C. Next move toward tree $D$, covering one-fourth the distance between where you are and $D .$ Finally move toward tree $E$, covering one-fifth the distance between you and $E$, stop, and dig. (a) Assume you have correctly determined the order in which the pirate labeled the trees as $A, B, C, D$, and $E$ as shown in the figure. What are the coordinates of the point where his treasure is buried? (b) What If? What if you do not really know the way the pirate labeled the trees? What would happen to the answer if you rearranged the order of the trees, for instance, to $B(30 \mathrm{~m},-20 \mathrm{~m}), A(60 \mathrm{~m},$ $80 \mathrm{~m}), E(-10 \mathrm{~m},-10 \mathrm{~m}), C(40 \mathrm{~m},-30 \mathrm{~m}),$ and $D(-70 \mathrm{~m}$ $60 \mathrm{~m}$ )? State reasoning to show that the answer does not depend on the order in which the trees are labeled.

A pirate has buried his treasure on an island with five trees located at the points $(30.0 \mathrm{~m},-20.0 \mathrm{~m}),(60.0 \mathrm{~m}, 80.0 \mathrm{~m})$ $(-10.0 \mathrm{~m},-10.0 \mathrm{~m}),(40.0 \mathrm{~m},-30.0 \mathrm{~m}),$ and $(-70.0 \mathrm{~m}$$60.0 \mathrm{~m})$, all measured relative to some origin, as shown in Figure $\mathrm{P} 3.46 .$ His ship's log instructs you to start at tree $A$ and move toward tree $B$, but to cover only one-half the distance between $A$ and $B .$ Then move toward tree $C,$ covering one-third the distance between your current location and
C. Next move toward tree $D$, covering one-fourth the distance between where you are and $D .$ Finally move toward tree $E$, covering one-fifth the distance between you and $E$, stop, and dig. (a) Assume you have correctly determined the order in which the pirate labeled the trees as $A, B, C, D$, and $E$ as shown in the figure. What are the coordinates of the point where his treasure is buried? (b) What If? What if you do not really know the way the pirate labeled the trees? What would happen to the answer if you rearranged the order of the trees, for instance, to $B(30 \mathrm{~m},-20 \mathrm{~m}), A(60 \mathrm{~m},$ $80 \mathrm{~m}), E(-10 \mathrm{~m},-10 \mathrm{~m}), C(40 \mathrm{~m},-30 \mathrm{~m}),$ and $D(-70 \mathrm{~m}$ $60 \mathrm{~m}$ )? State reasoning to show that the answer does not depend on the order in which the trees are labeled.

Key Concepts

Vector Arithmetic

This concept involves adding and subtracting position vectors and scaling them by real numbers. In many geometric problems, especially those involving movements in the plane, we use vectors to represent positions, directions, and displacements. Being able to add vectors and multiply them by scalars is essential in computing successive positions by moving fractions of the distance toward given targets.

Linear Interpolation

Linear interpolation is the process of finding a point along the straight line between two known points by using a fractional parameter. For example, moving a fixed fraction from one point to another is a form of linear interpolation. This concept is used repeatedly in the problem as the pirate moves a given fraction of the distance towards successive trees.

Affine Combinations

An affine combination of points is a weighted sum where the weights are non?negative and add up to one. The final treasure location in the problem is obtained by a series of movements that can be rewritten as an affine combination of the tree positions. This idea explains why the final answer is independent of the order in which the trees are used, as affine combinations are invariant under reordering of the terms.

Fixed Points in Iterative Processes

When an iterative process consists of taking weighted averages (or affine combinations) of fixed points, the process often converges to a single fixed point that does not depend on the order of operations. In this problem, the repeated steps of moving a fraction toward a target yield a final position that is the same regardless of the labeling order of the trees, illustrating the concept of a fixed point in an iterative affine map.

Order Independence in Affine Operations

Since the process involves repeatedly taking partial moves (i.e., affine combinations) toward fixed positions, the overall effect can be shown to be independent of the order in which these operations are executed. That is, whether you mix the fractions in one order or another, the final point is uniquely determined by the same overall weights given to each of the fixed points. This invariance is a key property of affine transformations and weighted averages.

Transcript

00:01 Okay, we're looking at question 346, 3 .46.

00:07 And we have buried treasure on an island with trees at these various points.

00:15 And they are all, the points are all relative to an origin.

00:21 We give in some instructions.

00:22 So you start at tree a, and then you move toward tree b, but you only cover tree b.

00:33 Half the distance between a and b.

00:37 You start at c, or from that point between a and b, you move towards tree c, but you only cover a third of the distance between that point and point c.

00:47 And then you move toward tree d, you only cover a quarter of the distance, and then tree e covering a fifth at the distance.

00:53 So this is kind of tricky because what you're doing is, we can draw this out, we'll draw it out really roughly here.

01:04 We've got tree a, and tree d.

01:10 We have tree c, tree e, and tree b.

01:20 Okay, so you're starting at a, right? and then you're going to move halfway between a and b.

01:28 So you're going to go to like right here.

01:32 And then based off that point, you're going to move half of the distance between or a third of the distance between that place and c.

01:42 So if you found the distance between that point and c and cut it into thirds, you would only move, you know, one third of the distance.

01:53 So it's a little bit tricky to figure this out, but you can use our vector knowledge to do it.

02:01 So what i want to do here is define vectors, vector points.

02:08 So the first one we're going to call p1.

02:13 And p1 defines the first point that you stop at.

02:20 So if you start a vector a, move halfway between a and b, you end up at p1.

02:26 And then if you start at p1 and move a third of the distance between p1 and c, you're going to end up at this point we're going to call p2 and so on until you get to the end.

02:39 Which is going to be p4, i think.

02:43 Okay, so how do we figure out where p1 is? let's start there.

02:51 So p1 is equal to wherever you start.

02:54 Where did you start? you started at vector a, vector point a.

03:00 Okay.

03:02 And then you moved from there, you moved halfway between b and a.

03:05 So what's that going to look like? b minus a over two.

03:12 Okay, so this is what defines p1.

03:16 You start at a, and then from there, you move half the distance between b and a.

03:25 So you take b minus a, b being the direction you're moving toward, right? so you're moving toward vector b, or point b, meaning you start with b and then subtract wherever you came from.

03:41 So subtract a.

03:43 But you're only going halfway, so divide 2.

03:46 Okay.

03:48 And then if we took point p2, how are we going to find p2? well, where did you start? you started at p1, right? and then add to that, it tells us we move a third of the distance between where you stopped, p1, and point c.

04:14 So this is going to be c minus p1 divided by 3...

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