Triangle Division: CR4 Challenge (04/29/08)

1. 100

2. Yes

1. 64

2. n squared plus 2 time the sum (from i = 1 to n-1) of n time 2 to i

1. 64

2. n squared plus (2 time the sum (from i = 1 to n-1) of n time 2 to i) - n

I am iterating to a solution. Sorry about the lack of an easy to read equation.

1. 60

2. n squared plus (2 time the sum (from i = 1 to n-1) of n time 2 to i) - n

I changed my mind again.

I can not find a triangle that doesn't include point A, Point B, or both A and B.

These are those I find that have Points A and/or B included.

1. 64 Triangles

2. n squared plus (2 time the sum (from i = 1 to n-1) of n time 2 to i)

That's all well and good but you forgot one. The outside one counts too, so there is 65 total. Look real carefully at the question's wording.

Look real carefully at the last figure in the second column.

64

n^3

I had noticed post #13 and it did not show the whole shape. Is one of these doubling on a configuration or omitting one? I just do not have time to pour over it to chack them.

I did perform a detail check and came up with only 64, so Post #13 must have made an error.

It is not obviousl how the ordering method guarantees non-duplication*, which would make it prone to errors. Most (but apparently not all) were avoided...

*e.g. the lowest three wide by two high counting from A is the same as two wide by three high counting from B

80

n²(n+1)

1. 8

2. 2n

My answer is the same as yours. even before I saw it.

1. 8

2. 2n

My above answer referred to pure triangles without lines through it. The answer would be 8, and 2n, however, including the outer one. So, that is wrong.

However, counting all possibilities of lines circumscribing any formed triangle regardless of lines within it (regardless of being crossed-out) then I agree with the 64 in total and n^3.

It is easier to start with one divider line or two sections. The total number is 8 and the formula is also n^3.

On three division emanating per side we have 27 triangles and n^3 still applies.

Now another question: Assuming we add such divider lines also from the top apex of the triangle, how many triangles in total will be formed? What is the formula then?

Regards

Your question got me to looking at another possibility: How many triangles would there be if we included those formed by 3 vertices, but one line is not drawn? For example, CDE, CDG, and CDI. We have 18 vertices to work with. I won't say for sure, but this might be why some answered with 100.

This could be a valid consideration because the Challenge Question asks how many triangles are in the figure, not how many are formed by drawn lines.

Now I need to get out my statistics book and review permutations and combinations.

In order for this extension to be properly answerable, you would need to define whether there are any common intersections within the area of the triangle. If there are none, then the answer to your extension is _3n+1C₃ - 3._n+1C₃ = [(3n+1).3n.(3n-1) - 3.(n+1).n.(n-1)]/6 = 4.n³.

I really admire your know-how of math and your ability to development formulae, 'Physicist?'

I was away for a few days.

Your point is well taken of needing to define the lines coming from point C (top point of the triangle). To make it simpler, the lines from C should go through the existing intersections of the lines from points A and B.

Assuming the lines from A and B have equally distributed angles, or go to equally distributed points on the opposite sides of the triangle, then the number of lines from C is not 'n', but the number is 2* (n-2) +1. At n= 4 the number of lines from C is 5, to go through every intersection of lines from A and B.

Doing this, the added number of triangles is defined compared to the number of triangles added when the lines from C are at random angles, which is undefined as it changes with the location of the lines coming from C.

Now, having the definition in place, a solution is specific.

To be honest, I am not able to find it.

Regarding the comment by Doug I am not sure what the difference is: "… because the Challenge Question asks how many triangles are in the figure, not how many are formed by drawn lines."

Can you specify what you meant by that?

Regards

(1) 110

(2) AB+AC+CB = 2n(1+2+3+n)+2n²-2

Regards JD

Can't argue with the pictures provided by Jim. Looks like he has all of the triangles covered.

While I agree with Jim that the answer for n=4 is 64 I think the formula for the number of triangles = n³. Looking at the number of triangles for n=1 to n=5 gives 1, 8, 27, 64, 125.

n² + 2 (sum of 1 to n-1) x n

or

Simply

n³

The answer will be same.

Tehseen

65 triangles

pls include the big one..

I wasn't going to do this, but I wanted to check for myself:

I hope this is easier to see than the one that Jim did. I went in a sequence from bottom to top, beginning with 1 wide by 1 tall, then 1 wide by 2 tall, etc. Also, had to watch for duplicates. And as carboncopy says, don't forget the original triangle ABC.

The figure in the second row from the bottom and seventh column is the same as the fifth row from the bottom second column. And the full triangle is missing.

1. 64

2. n^3

You're right, Edgar! Thanks for catching that. I didn't shade in the original triangle as it should have been understood to be in the total count.

I was looking at your collection and I noticed the one on the top row 2nd from the right does not have the mirror image shown anywhere. I did not check all but also the whole shape option is not shown.

As of now, a second "good answer" rating? (The systematic approach to filling in the squares makes it easy to see what's going on).

I don't criticize the approach, but similar explanations have been given. N=1, n=2 etc

Tell me what am I missing?

Post #26, in reply to post #19, where Edgar pointed out a duplicate image I missed.

HAHA! I like the name you gave it:

http://cr4.globalspec.com/PostImages/200804/well_you_bloody_asked_A0ABB9D6-BD58-494B-593C4FA7E3BE3EA9.JPG

Actually, I had 2 duplicates. I'll tell you later how I found out.

I have confirmed through another method that there are 64 triangles, including the original. This method is probably close to what we will see in the official answer, because it requires only one diagram to draw.

1. 77

2. n³ +n² -n +1

No! More than that.

OK: good answer to guest #1

1. 100

2. [n(n+1)/2]²

Phyz, I agree with you formulation with 1 exception.

1. Triangles including A: 4(4+3+2+1) = 40
... Triangles including B but not A: 4(3+2+1) = 24

Triangles including a,b,c=1
-> 65

2. N(N(N+1)/2+(N-1)N/2) =1+N³

I altered your premise and formula slightly to reflect the inclusion of the original triangle based on the figures above that show 64 triangles without the inclusion of the original. Hope your not offended and if I am incorrect I apologize.

If I remember my geometry even slightly correct the inner number of triangles will be divisible by eight because each of the 2 line segments is divided into 4 line segments for a total of 8, however this will not include the original triangle in a formula if 1 is not added to the end of the sequence.

But then again I could be totally bonkers and not remember anything.

No the fourth 1 in 4(4+3+2+1) is the big one. Fyz is right again.

I disagree with the "... +1"

If youconsider his equation as follows: 4*(4+3+2+1) + 4*(3+2+1)

the first term implies 4 triangles for each "section(s)" from vertex A

there are 4x single section sets, 3x 2-section sets, 2x 3-section sets and a single 4-section set - the largest of which is the ABC triangle.

the second term are those triangles from vertex B less the overlap of those that included vertex A and were thus already accounted for.

therefore n^3 is sufficient to account for all triangles

Ah I see the wisdom of it and stand corrected.

Another misunderstood equation on my part.

I didn't think phiz could be wrong as he usually is right on.

That what i get for not getting enough sleep last night.

However hard I might try, I'm certainly not always right (and I didn't get too much sleep last night either)

Fyz

How do I classify thee? Let me count the ways
I classify thee by the breadth and depth and height...

Choose any pair of lines passing through A. This has n lines crossing it, so each pair form part of n triangles. There are n pairs of lines passing through A that are adjacent, n-1 pairs spaced by 1, etc, so a total of n+(n-1)+... = n(n+1)/2 pairs of lines.
Thus the total number of triangles that include vertex A is n.n.(n+1)/2

Now we do the equivalent for vertex B, but we do not want to duplicate triangles that have a vertex at A. That just means we can't count triangles that include the side AB. So the number of lines passing through B that we can use is n instead of n+1. The number of triangles using each pair is unchanged. So the total number of triangles that include vertex B but not vertex A is n.(n-1).n/2.
^{_____________________________________________________
At least I tried}

While the answer has been given, it seems some people are still having trouble systematically counting all the triangles. Every triangle will have point A or point B, or both contained in it. These triangles can be divided into two groups, those that contain both points, and those that contain only 1 of them, these later can be divided into two basically similar groups based on which of A or B they contain (and each of these groups will have the same number of triangles, so we can count just ne set and double it).

For the triangles that contain both A & B, we can choose any of the 4 lines (excluding the baseline A-B) from each of the points and make a triangle (note that one of these will be the full triangle), thus we get 4*4 = 16 triangles in this group (or in general n*n)

For the triangles that only contain 1 of the points, first choose the point (2 ways), then we choose 2 of the 4 non-baseline lines from that point (6 ways, or n*(n-1)/2 ) and one of the 4 lines from the other point (4 or n) giving 2*6*4 = 48 (or 2*n*n*(n-1)/2 = n*n*(n-1) )

adding these together gives us 64 triangles or n*n + n*n*(n-1) = n*n*n.

1. 64

2. n³

Would it be possible for those answer's that ≠64

Could you show your working? I'm curious how you actually found 8, or 100 triangles, or is there another "4th" dimension that I just cannot see?

For my 2-cents, I like the answer/approach that BobD gave at post #11. I still suspect there is a much more intuitive way of 'seeing' the general solution (although a number of people have made excellent posts to illustrate).

.....# 43 wasn't one of them, though I'm not feeling brave enough to question it. Yet. I have to draw the line somewhere !

I'm afraid it's one of those "kick-yourself-for-not-seeing" methods that I find is a pig to explain. So I would really welcome a more transparent exposition.

Me too. Imagine standing on point 'B'. Each of the 'n' sectors you can look along is crossed by 'n' lines drawn from 'A'. That gives a nice n X n, but hopping over to A gives some repetition, and you'd also miss some triangles. That's not the answer, and I can't offer anything good, but my hunch is that somewhere along the line a factorial is inevitable !*

*Pondering this, is far worse punishment than pulling the trigger as I lament my word-play.

Perhaps count sectors that are larger than a single width (such as EAB, EAG, EAI in #41)?
(Factorials are not strictly inevitable when you are only at the cubic level as in this case)

Many things can be explained by cubics.

That is not cubic-cool.
^{More like the strongest possible temptation for an electric-chair practical joke? (I doubt it would be foiled)}

OK, now I present the different approach I mentioned in post #35:

1. Label all the intersections. They are now vertices that can serve as vertices of triangles.
2. List all the triangles by their vertices.
3. Count the triangles.

I labeled the outside vertices first, then the inside vertices, top to bottom. I listed the triangles in alphabetic order: ABC, ABD, ABE, ABF, etc. Working through this led me to discover a second duplication, and a missing image from my first illustration. This is a corrected version of that illustration:

The first five rows show triangles with vertex A, the bottom 3 show those with vertex B.

The catch with this method is not to skip a letter when working alphabetically.

I'm not going to type in the whole list. If you want to reproduce it from what I've given here, go ahead and have a blast!!

Label each of the lines as shown below. Any combination of three lines might define a triangle. However, any combination of (0,1,3,5,7) or (0,2,4,6,8) do not define a triangle since they meet at point. For each triplet of lines there are 6 possible number sequences but only one triangle. For example 123, 132, 213, 231, 312, and 321 are all the same triangle (in this case outer). The table below lists all possible combinations of the nine lines with the lowest numerical value of the six options shown. If we eliminate all the combinations that are lines meeting at a single point (in bold), we have 64 possible triangles.

In general, if we divide each lateral side of the outer triangle by m, we have 2m+1 lines with (2m+1)(2m)(2m-1)/6 possible triangles. The number of combinations that meet at a single points are 2 m!/(m-3)!/6. Somehow this leaves m³ triangles but that part of the math eludes me.

Consider an extension of this problem in which there can be an arbitrary number of lines dividing each side of the outer triangle. In the original problem, two sides were divided into four sections and one side was not sub-divided. Consider the figure below in which one side is divided by two, one is divided by three, and one is divided by four.

There are a total of nine lines which I have numbers as above. There is a total number of combinations of 9*8*7 or 504 possible triplet combinations of these nine lines. However, anyone triangle has six different combinations that are associated with it. (123, 132, 213, 231, 312, 321 all outline the same triangle.) This leaves 84 possible triangles. I always use the triplet with the lowest numerical value.

In the above figure, 123, 124, 134, and 234 meet in the lower left corner, 456, 457, 458, 467, 468, 478, 567, 568, 578, and 678 meet in the lower right corner and 159 meet at the top corner. Since they meet in a corner, they have a single point in common and do not define triangles. In the above figure, there are no locations on the interior of the outer triangle where three lines intersect. If there were three lines intersecting at a single point inside the outer triangle, those combination of lines would also not define triangles. Following is a table of all the line triplets that might define triangles in the figure above. Those in bold do not define triangles.

Arranged in the same pattern in the figure below are the triangles defined by the lines.

In general, the number of sub-triangle when the sides are divided into m, n, and of segments is:

(m+n+0)*(m+n+o-1)*(m+n+o-2)/6 minus (the combination of (m+1) taken 3 at at time) minus (the combination of (n+1) taken 3 at at time) minus (o+1) taken 3 at a time minus the number of line triplets intersecting inside the outer triangle.

In our case this is:

m=2, n=3, m=4

(m+n+0)*(m+n+o-1)*(m+n+o-2)/6 =9*8*7/6 = 84

(the combination of (m+1) taken 3 at at time) = 1

(the combination of (n+1) taken 3 at at time) = 4

(the combination of (o+1) taken 3 at at time) = 10

the number of line triplets intersecting inside the outer triangle = 0

84 - 1 -4 -10 -0 = 69

Agreed - it really does become quite easy once you use the more general result:
₉C₃ - ₅C₃ - ₄C₃ - ₃C₃ = 84 - 10 - 4 - 1 = 73