Challenge Questions

HA!

Like I'd bother pretending to be some crazy mathematician, locking myself into a small room with a whiteboard for hours on end to figure out something I'd easily figure out with a good tape and some basic geometry.

I use to do it on a coffee mill Facit but I will have to dig deep in my memory.

I think the method was based on :

Subdivide the area in 3 non overlapping triangles.

calculate the areas of the 3 triangles using the absolute coordinates of the corners. and add together.

A area triangle = 1/2 [x1y2 - x1y3 + x2y3 - x2y1 + x3y1 - x3y2]

It is not coming back yet but just have to take the sequence further ending up with - x5y4 ]

Either I'm dumber than a post (No! We're not voting on that.) or something has been omitted?

GA

As I understand it, the five posts are numbered in order from 1 to 5 following the string and the closing the loop from 5 back to 1. I get for this:

area = 1/2*{x1*(y2-y5) + x2*(y3-y1) + x3*(y4 - y2) + x4*(y5 - y3) + x5*(y1-y4)}

This is simply one half of the sum of the x value for each post multiplied by the difference in the y value of the next and previous post as we go around the string. I arrived at this solution by doing an example sketch and subtracting outlying pieces from the rectangle defined by the max - min dimensions in both x and y. I then used Mathcad to simplify.

Thanks,

Jim

There's got to be something wrong here. If you're a long way from the X origin you get a bigger number than if you're close to it.

No the formula works for any origin. (unless you exceed the capabilities of the calculator). Give it a try.

You are actually calculating triangles from the origin and will subtract the area outside of the area on the way back to point 1.

For working with bigger areas like for example a State you may need to do proper projections.

What am I misunderstanding

To make it easy for my brain, I've set the grid as unit instead of 3.

In the above picture, the area of triangles plus blue squares is;

3 + 1 + 1.5 + 2 + 4 + 1 = 12.5

Using the formula I get

1/2 [1(4-0) + 4 (3-2) + 6(0-4) + 5(0-3) + 3(2-0) ]

=1/2( 4 + 4 -24 -15 +6) = 1/2 (-5) dammit, you're correct ! Ignoring the sign, that comes out as 12.5

I don't fully grasp it yet, but nice work.

Why not this way?

....'cos working out the area can be done in the head the way I sectioned it. I wanted that so I could bounce on quickly to check the formula. The way you've sectioned it, only 2 of the triangles jump out shouting their area. Having said that, it's still easier to deduct 4 triangles from a rectangle.

It could probably be tinkered with to illustrate fyz's explanation, though I haven't been tempted to try that yet.

Not sure how you adjusted the picture, but if you don't know the program it's one of my favourite free utilities for this type of play- Geogebra. Nothing fancy, but very neat all the same. It'd take me a very long time to be able to produce anything like the wonderful stuff I've seen on your posts, I'm much more inclined to the backs of envelopes/beermats !

I just took your image into paint.. and repainted it. like this one which shows other right triangles you can use to determline line lengths and subsequent areas, if you want.

OK, how about this;

Walk the perimeter from point A, going clockwise. You can count the partial squares cut by the perimeter. Along each stretch of edge, they sum to n or n+0.5. Crossing an even nuber of partial squares will add up to n/2 full ones, and an odd number sums to ((n-1)/2 + 0.5). I'll admit that the half square in location (5,1) isn't obviously 1/2, but it must be so. Anyhow, you could walk along, adding up all those partial squares in your head. Adding on the whole white squares shouldn't be too much of a problem. If somebody lays out a grid someplace nice (maybe a tropical island), gives me a GPS and tickets, I'm willing to fly out and try this.

as you fly, you will be going right over my head, just like this did...

With one exception, the light green triangles in the picture all pair-up with dark ones.....

.

I've tried it: it works:-

I'm still trying to get my head round how.

Thanks

GA, though I would prefer write it as
A = 1/2 * ∑_i=1...n {Xi*[Y(i+1) - Y(i-1)]}
_. (and also add that Y₀=Y_n and Y_n+1=Y₁)
N.B. from a traditional integration background I would have described it as trapezoidal summation.

Another point to note is that the way I read the challenge, run around the stakes means that the string will not touch all the stakes if the polygon defined by the stakes themselves is re-entrant.
If on the other hand we fix the string to each of the stakes, I agree that it fails if the polygon is self-intersecting: depending on the topology, areas can be multiply counted or even counted negatively (it's really calculating area*number_of_directional_enclosures - good for magnetic field calculations and the like, but not so good for lawn-cutting).

I think we need a more complete definition of the problem. But maybe something will come along in the discusson to make this clearer. In the figure below I've made up a example with the dotted lines being the 3 x 3 feet squares and the solid lines being the string.

By "corners of your grid" I am assuming that this is any of the corners of the 3 x 3 squares. Clearly, some of the area inside the string could be non-lawn. Depending on the shape of the lawn, it might not be possible to have the entire lawn inside the string.

Perhaps the challenge wants us to assign (x,y) coordinates to the stake locations and then use those coordinates to get an area.

Good luck,

Jim

And what if one of those stakes is inside the area made by the other four?

And what if my lawn is a circle?

And what if I choose to mow it anyway?

A circle can be done as long as the segments are short enough to approximate the circle. your points need to be increased.

It does not matter as long as you are not crossing any lines.

To understand the principle take a figure with any 5 points. (or n points for that matter)

Take the origin outside.

Draw connecting lines from the origin to each point.

Calculate the area of each triangle.

The areas of the triangles in the negative direction (outside of the figure) will be subtracted from the positive ones.

The missing piece of information is whether or not we are required to turn the corner at each stake. If we aren't, the simplest shape we have is something like this:



Of course, I'm only thinking of the simplest shapes right now. I'll have to cogitate on this issue more later, because it's late and I want to gaze at other parts of CR4 before I partake of slumber tonight.

This assumes a flat lawn. I live in the foothills of the Blue Ridge mountains, and my (~ 1 acre) lawn is nowhere near flat. It's a curving slope with a saddle shape at one end and a domed area at the other.

I need a general, non-Euclidian solution.

And you need to know this why?

I too have an acer in the hills by Lake Tahoe and I need to know this so I can figure out just how bad my Gardener is ripping me off when he does the weed clean up each year. Though I do not mind paying since I drink beer and watch him sweat. I guess it all works out in the end.

Humm maybe I should buy goats..........................

Maybe I should say it is projected onto a flat surface and then the area is calculated from the projection.

The projection is needed because the pieces of land must be compiled into an area plan, which would be impossible to do if the plans are distorted.

About 25 years ago I had the privilege of experiencing the American way of surveying and plans and was actually amazed. Really primitive compared to SA.

I have a Tall Fescue lawn, and I need to re-seed it this Fall. I'd like to know the exact amount of grass seed I'll need.

Yes, I'm kidding.

I just wanted to make the point that even a 'general solution' is not truly general unless it works for non-Euclidian spaces, too. My lawn is non-Euclidian.

give an exact mathematician equiation of the surface of the lawn and it will be possible!

Okay I have 1 post in 5 out of the six areas of my property that get mowed. What do I do now?

It takes me 8 hours with a 46 inch cut mower to do all of my yard sections. The post and grid concept doesn't factor in for the hills trees, junk, stuff, and other things I have to mow around.

My mower does not make perfect angular corners either. So do I have to subtract the radius for each corner from the whole as well?

Its just another impractical application of nearly useful mathematics.

I still believe something is missing. The general equation for area enclosed by a polygon (the surveyor's area equation) has been stated by many, but what about the question, "What is the area of the lawn inside the string?" I still don't see how to get that.

You know the positions of the posts, so you can count their Cartesian co-ordinates (in units of one yard) from an arbitrary point. You know the equations, so (assuming that the string is sensibly routed) you can calculate the enclosed area; this will initially be in square yards - multiply by 9 to convert to square feet if you wish.

That's obviously too easy - so can you say where the problem lies?

OK. Let me give the coordinates (in yards) of my five stakes.

(0,0)

(3,1)

(5,0)

(2,4)

(2,2)

Now, what is the area? I cannot determine it.

Try drawing it out as I did in #22 (starting at point A - ignore the other letters, just work clockwise). I still can't see how the formula was derived

There's probably some way by counting the number of grid-lines crossed as you 'walk' around the perimeter, though I can't see a 'neat' way of expressing it. Alternatively, there may be a way by counting the number of intersections enclosed by the pentagon + corners of 3' squares that are just excluded by the pentagon. Haven't had time to think that through.

When I posted my #141, I was just responding to Randall's solution in #130. I hadn't looked carefully enough into the earlier posts which have by-and-large anticipated what I tried to put across. For instance: jim35848 (#8), Hendrik (#14 & #64), SlideRuler (#84), and some more of yours. Your #44 provides a fairly complete account of how the formula works. Since I get easily intimidated by algebraic expressions, I did a bit of rough sketching and numerical checking using 'easy' coordinates to satisfy myself.

My pentagon with the sequence (-1, -2), (1, -1), (3, 1), (2, 3), (-2, 2) had an area of 30/2 = 15, whereas a crossed pentagon with the same points in altered sequence (-1, -2), (1, -1), (2, 3), (3, 1), (-2, 2) had an area of 15/2 = 7.5 because a negative term reduces the overall value.

It will take far too much effort to represent all that pictorially in CR4, but using a rough sketch or points on square-ruled paper it is easy to just keep your fingers in sequence on the three relevant points for each term [ x_i (y_i+1 - y_i-1) ] as indicated in your #21, and write down the relevant rectangle area by inspection (if the integers are small enough and the minus signs not too confusing). It's also quite straightforward on a calculator.

I also verified that the symmetrical crossed quadrilateral (1, 1), (4, 2), (2, 4), (5, 5) has zero area.

All this may be trivial stuff to many folks, but I find it interesting and of practical value.

=TeeSquare=

Assuming flat ground and that the string joins the posts in order and the last to the first (and it's all grass in the centre):

Hendrik gave the formula for five stakes as:

Area = 1/2 [x1y2-x1y5+x2y3-x2y1+x3y4-x3y2+x4y5-x4y3+x5y1-x5y4],

Using your positions, this gives:

1/2*(0*2-0*1+3*0-3*0+5*4-5*1+2*2-2*0+2*0-2*4)

= 5.5 square yards (49.5 sq-ft)

(I checked for careless mistakes by drawing it on paper - same answer).

Am I missing something subtle, or are you complaining about the necessary assumptions (given the cryptic wording), of how the stakes are joined, plane ground, and everything enclosed being grass?

No, you're missing something the size of a battleship. Perhaps you have to have really screwed up surveying a plot to see the problem.... Or perhaps you live in a state that doesn't use metes and bounds.

I think that you've managed to be 'superior' without even trying to explain the problem you perceive.

I'm assuming of course that you gave absolute stake positions in the standard order - with the sequence of defined stakes corresponding to string connections. (Note that the calculations ask for the "area inside the string" rather than the area the stakes were initially intended to measure).
I also assumed that you gave absolute positions, as required by use of the term "coordinates" in an engineering forum.

But I see that you referred to the method of metes and bounds. That would usually mean using relative positions. Obviously you can't have done that, because the relative positions you give would have the string crossing itself, which metes and bounds methods wouldn't allow. Plus, Hendrik already explained that if the string crosses itself the method breaks down.

If you believe there is a different and significant difficulty I humbly suggest it's more than time to get your act together and explain it.

Snippy today?

What standard order? Reread the challenge. I didn't cross any strings.

Did I ever tell you about the time I surveyed a right-of-way and got a negative area? I went down to the tax office and told them they owed me. Threw me out, they did.

(This is a standard surveying calculation and has a standard caveat; only dumbasses like me forget that caveat - but only once!)

The question did not say that 'you' only knew the position of the stakes and not how you ran the string. It asked if you could calculate the area after 'you' had run the string. So of course that information was available to you. Whether you communicated accurately to me is irrelevant - according to the challenge 'you' had knowledge of the position of the string, so 'you' could calculate the area. To say that you (TVP45) couldn't was simply false.

Given your reservations (and contributions elsewhere), I would have expected you to clarify the proviso that the stakes needed to be listed clockwise in the order that the string connected them (assuming that you had listed the positions in standard Cartesian order); that would have been positive. What you actually wrote was merely irritating. In addition you made reference to metes and bounds; as you should know, this system uses differential measures* taken in standard order, and taking your numbers to represent differentials would indeed have given a self-intersecting boundary.

*Professionally used these would usually include a "check" measure from the final point to the first - which is why I assumed absolute measures in the first instance.

You mean like this ;

? That looks to be 6 square yards. Are you saying you can't get the formula to work ?

Yep, that one's 5.5 square yards. No problem. And, the next one is...

huh?

I think you have plotted point C one square too high and the area is really 5.5 square yards.

Nah, it's TVP's fault - he listed the point as (5,1) when he intended to write (5,0) .

Cheers, jim. My booby-wooby.

Do you mean 3 square feet or squares having sides equal to 3 feet (9 square feet)?

I still have some concern we are not considering that part of the area enclosed by the string is outside the lawn. Unless the lawn is a rectangle, I think there are ways to enclose non-lawn areas inside the string. However, I don't think that was what was intended by the challenge question.

One point to be made is the area inside the string is some multiple of half a square yard. Assuming all of the stakes are at unique corners, the smallest area is 1.5 square yards and the possible areas go up in half square yard increments. Enclosed areas of 1.5, 2, and 2.5 square yards are shown below.

I have interpreted "take a long string and run it around the five stakes" to mean that the string does not cross itself and we end up with some configuration as shown in the following figure.

My next question is, "Given a string of length L, is there a general equation to calculate the maximum area inside the string?"

Thanks,

Jim

Obviously a pentagon, except you have to approach it from discrete coordinates.

Rectangle, or ellipse, or parallelogram, or...I think a sufficient (though not always necessary) condition that everything inside the string is lawn is that the lawn perimeter is convex.

A point of perhaps marginal interest: if we are talking about actual area of lawn enclosed, your comment about 0.5-square yard multiples only breaks down if the string crosses itself.

I'm afraid that the 1.5 square yards minimum requires an unstated constraint for the positions of the stakes.
We can get 1-square yard with each stake needed to define the shape of non-zero enclosed area - as in the example below:
1.N.4
2.5.3

If we allow some stakes to be redundant (so far as defining non-zero area is concerned), we can enclose areas of 0.5 square yards (or even zero)

I omitted to address the "next question". I'm reasonably confident that with an unlimited number of stakes and discrete locations the answer will be no*, but such negatives are notoriously difficult to prove.
*Other than an iterative method.

If we restrict ourselves to a triangle I think we could manage a discrete (but quantised) equation. If that is correct, there is probably an increasingly complex hierarchy (depending on the length of the string) that involves stacked max(min(...)) type statements.

My next question is, "Given a string of length L, is there a general equation to calculate the maximum area inside the string?"

This sort of problem is solved by a branch of calculus called "Calculus of Variations". By these means it can be proven that the circle encloses the greatest area for a given perimeter.

I tried to find a simple reference for this but was unsuccessful.

the biggest area in a given lenght of line is still a circle!

There only problem that previous solutions had with re-entrant polygons was if the "run around the five stakes" means that the string is tightened around the outside of the stakes, rather than running from stake to stake. (That of course depends on how you read the challenge).
I've had a look at solving both that interpretation, and the problem of the string intersecting itself; but I can only find iterative solutions, which I wouldn't describe as "general"

I might have stumbled upon a simple method that could be used with a particular class of shapes. Irregular polygons, with 5 vertices and a generally convex shape, can be divided into 3 triangles.

To understand the formula for n points first have a look at 3 points (triangle)

Start with any oblique triangle ABC

(0,-1) (2,2) (4,0)

Change the Y value of point 1 to zero. Note that the Area did not change

Now change the X value of point 2 to 4. No change in Area

The area of the triangle is now 1/2 * base * height

What is important in a triangle is the width and the height.

The formula for a any triangle is

Area = 1/2( x1(y2 -y3) + x2(y3 -y1) + x3(y1 - y2))

In a right triangle and the origin at one of the points only the term containing the base and the height will have a value.

In an oblique triangle or an outside origin extra terms will cancel out.

This principle will hold true for triangles in a n sided polygon.

This formula can also be derived from the law of sines, law of cosines etc.

"... Change the Y value of point 1 to zero. Note that the Area did not change ..."

I think that statement and a later one are not true unless the vertex shift is parallel to the opposite side. The formula is however correct, and I tried to give a geometrical reasoning in #141.

=TeeSquare=

In reply to #40:

The problem does not appear to state in what order the points have to be connected.

For example:

If you connect P1 to P2 to P3 to P4 to P5 to P1 the area is 5.5

If you connect them P1 to P4 to P5 to P2 to P3 to P1 the area is 6

At the risk of irritating a few people for whom this is too obvious I'd like to just pull together a few posts from this thread.

First the area of a triangle from the origin to two points can be found like this:-

The area of the green triangle is clearly the area of the whole rectangle (x₁*y₂) minus the area of the yellow triangles

x₁*y₂-½[x₁*y₁+x₂*y₂+(x₁-x₂)*(y₂-y₁)]

Which all simplifies (surprisingly to me) to

½(x1*y2 - x2*y1)

It doesn't matter what the orientation or position of the original line is it always works out.

This is apparently the cross product of the two vectors and SlideRuler referred to it in his post #55. But I can't see why it's true. Also because the points are arbitrary you can't guarantee whether you'll get a positive or negative number.

So now if you want to calculate the area of a pentagon like this:-

You just have to add up triangles eaO, abO, bcO, and then subtract triangles cdO and edO (where O is the origin' sorry left it off the diagram). It turns out that as long as you do the calculation consistently you don't need to worry about which triangles to add, and which to subtract: so the answer is:-

½[(x_ay_b-x_by_a) + (x_by_c-x_cy_b) + (x_cy_d-x_dy_c) + (x_dy_e-x_ey_d) + (x_ey_a-x_ay_e)]

And, this works for all polygons and shapes (but clearly you need to either make sure you start with a "positive" triangle or just take the absolute value of the result).

This formula can be re-arranged easily to form which Hendrik gave early on but seems far more intuitive in the form above. Also Hendrik did point out the intuitive idea of adding and subtracting triangles in post #19.

Good challenge question. I had no idea about the cross product thing, but, I still have the curious geometric puzzle:-

Why is the green triangle's area equal to half the blue rectangle minus the red rectangle?

There must be a good dissection which makes it obvious, but I can't see it at the moment.

It might help to project to the axes and then you have trapezoids. The area of a trapezoid is 1/2 the sum of the bases x the height. Then subtract the trapezoids that aren't there.

Just another way to see it.

Don't know if you solved this geometric puzzle to your satisfaction or not, but I offer this interpretation.

Fig 1 The blue (which is also under the yellow triangle) should have double the area of the triangle.

FIG 2. Move the three blue areas as shown. The bulk of the triangle is sandwiched in blue, and therefore using 2X the area for this portion. We can therefore discard it and concentrate on the 2 blue and 1 yellow shapes left.

Fig 3. Splitting the yellow in two as shown, the yellow triangle can be skewed without affecting the area to the position shown where it is clearly 1/2 of its blue companion. This is a special case however, since in the case chosen the base of the yellow triangle is equal to the side of companion rectangle. In general this will not be the case and the base of the triangle and the side of the rectangle will form a rational fraction. Nevertheless it is true for the figure chosen (where each length is 1/2 the unit square) and the more general case is shown in figs 4a and 4b

Fig 4a The triangle has been skewed and it is seen that the base is twice the dimension of the companion rectangle.

Fig 4b compacts the base by 2x and stretches the altitude by 2X leaving the area of the triangle unchanged and it is now seen to have an area of 1/2 the rectangle.

Q.E.D. (I hope)

Elegant!