Triangle puzzle
 

This puzzle starts with a equilateral triangle with each side 1 unit in length.
The midpoints of each side are connected forming more triangles.

At each step, what is the count of same size triangles and the total length of all segments.
List until the number of triangles is 1024 congruent equilateral triangles.

The first two iterations:
Count 1 Length 3 Each side 1 unit
Count 4 Length 4.5 Each side 1/2 a unit.

The attached image file shows the first two iterations of the triangle.

# triangles = 1024
# unit length of a triangle = 1/63
# sides of triangles (ignoring multiplicity) = 3*1024=3072
# length of sides (ignoring multiplicity)= 3072/63=1024/21
# length of sides of original triangle = 3
# multiplicity part of sides = 1024/21 - 3 = 961/21 (multiplicity =2)
# real length of sides = 961/42+3= 1087/3= 362 + 2 / 3

:cat:

:blush: :blush: :blush:
never ever anwser a quick calculation 5 minutes before an important meeting.
Ugh, terrible error at the end

961/42+3=1087/[B]42[/B]=25+37/42

sorry

:cat:

Symmetry
 

[QUOTE=Kees]:blush: :blush: :blush:
never ever anwser a quick calculation 5 minutes before an important meeting.
Ugh, terrible error at the end

961/42+3=1087/[B]42[/B]=25+37/42

sorry

:cat:[/QUOTE]
:redface: Forgive me pointing this out but a symmetrical problem requires a symmetrical answer.
On inspection of your method and answer I strongly believe that the more likely answer is 25.5 units.
It might be something else as I have not taken a shot at it but it seems more plausible.
Reminds me of the anecdote of Hardy perusing thru the papers of Ramanujan which were submitted from India and they had not met before.
He said to the effect that the identities must be true as who else could think of such symmetry and be wrong ?
Mally :coffee:

Well, cross checking in that case: we have 1024 triangles.
If we count them from the base up there are 63+61+59+...+3+1=1024 of them
So the base (ah, getting your point :blush: ) is occupied by 32+31 triangles, so
unit length of triangle is 1/32 (which makes more sense indeed).

So total unit length (without taking multiplicity in consideration) is 1024 (triangles) * 3 (sides) * 1/32 (unit length) = 96
The sides that are on the edge are not counted double and their length totals 3 (length of sides of original triangle). The remainder 93=96-3 are counted twice because they all form the border between two triangles. So divide by 2
This gives 46.5
Adding 3 gives 99/2

:cat:

Each iteration will quadruple the number of triangles and halve the length of the segments.

So, assuming n=0 is the iteration representing a single triangle with 1 unit sides

# of triangles = 4[sup]n[/sup]
length of segments = (1/2)[sup]n[/sup]

Each iteration adds 3 segments to each of the previous iteration's triangles. If L[sub]n[/sub] represents the total length of the segments

L[sub]n[/sub] = L[sub]n-1[/sub] + 3 * 4[sup](n-1)[/sup] * (1/2)[sup]n[/sup]

For 1024 triangles, 4[sup]n[/sup] = 1024. n=5...so we're at iteration 5. The length of each segment is 1/2[sup]5[/sup] = 1/32.

Now we just need to find the total length of the sides. We can do this iteratively.

L[sub]1[/sub] = 4.5
L[sub]2[/sub] = 7.5
L[sub]3[/sub] = 13.5
L[sub]4[/sub] = 25.5
L[sub]5[/sub] = 49.5

Drew

If we don't want to iterate, we can continue with the following:

L[sub]n[/sub] = L[sub]n-1[/sub] + 3 * 4[sup]n-1[/sup] * (1/2)[sup]n[/sup]

= L[sub]n-1[/sub] + 3/4 * 2[sup]n[/sup]


L[sub]n[/sub] = 3/4 (2[sup]n[/sup] + 2[sup]n-1[/sup] + 2[sup]n-2[/sup] ... + 2[sup]0[/sup]) + C

= 3/4 * (2[sup]n+1[/sup] - 1) + C

Solve for C (L[sub]0[/sub] = 3)

3 = 3/4 + C

C = 2.25

L[sub]n[/sub] = 3/4 * (2[sup]n+1[/sup] - 1) + 2.25

We can simplify this a bit further:

L[sub]n[/sub] = 3/4 * 2[sup]n+1[/sup] + 3/2

= 3/2 * (2[sup]n[/sup] + 1)

Multiplicity and Triangles
 

[QUOTE=drew]Each iteration will quadruple the number of triangles and halve the length of the segments.

So, assuming n=0 is the iteration representing a single triangle with 1 unit sides

# of triangles = 4[sup]n[/sup]
length of segments = (1/2)[sup]n[/sup]

Each iteration adds 3 segments to each of the previous iteration's triangles. If L[sub]n[/sub] represents the total length of the segments

L[sub]n[/sub] = L[sub]n-1[/sub] + 3 * 4[sup](n-1)[/sup] * (1/2)[sup]n[/sup]

For 1024 triangles, 4[sup]n[/sup] = 1024. n=5...so we're at iteration 5. The length of each segment is 1/2[sup]5[/sup] = 1/32.

Now we just need to find the total length of the sides. We can do this iteratively.

L[sub]1[/sub] = 4.5
L[sub]2[/sub] = 7.5
L[sub]3[/sub] = 13.5
L[sub]4[/sub] = 25.5
L[sub]5[/sub] = 49.5

Drew[/QUOTE]

:bow:
Hats off to you Drew! this is really beautiful mathematics- to condense complicated thought into a minimum of symbols- the sheer economy of words- is mind boggling!
In this problem the difference in the lengths of the various inputs follows a simple law - the G.P. 1.5, 3, 6, 12, 24 corresponding to lengths [3,] 4.5,7.5,
13.5, 25.5 ,49.5 The next will be 49.5 + 48 and one can go on and on by just doubling the difference each time.

:love: Thanks to you Kees to help me convince myself that Drews iteration stands the test.
But your multiplicity (Never heard the term before, we called it duplication )
opened a can of worms for me and these last few days I spent staring at an equilateral triangle of 4^3 or 64 triangles trying to find some pattern with the previous ones.

But in all fairness to you your last post hit the nail on the head as I wanted to prove to myself the correctness of Drew's solution.

Of course the sheer symmetry of his answers no doubt convinced me to the hilt:redface:

So we can put paid to this problem.

A word for Desousa 123. I have about 150 books on math recreations on my shelves but besides the 'circle dot' problem we discussed when I first joined, this is by far the second best I have come across. I am also curious to know as compiler what your solution is to this problem.

If you have a site for these problems I would love to have a go.
Thank you one and all.

Mally :coffee:

This is simpler as a double check:
The #triangles sequence is 1,4,16,64,256,1024,4096,...
Cut off at 1024. This gives 5 steps.
The lengths are 3, 3/2 * 3, 3/2 * 5, 3/2 *17, 3/2 * 33, 3/2 * 65, ...
i.e. 3, 4.5, 7.5, 25.5, 49.5, 97.5, ...
Cut off at five steps gives: 49.5.

:cool:

[QUOTE=davar55]This is simpler as a double check:
The #triangles sequence is 1,4,16,64,256,1024,4096,...
Cut off at 1024. This gives 5 steps.
The lengths are 3, 3/2 * 3, 3/2 * 5, 3/2 *17, 3/2 * 33, 3/2 * 65, ...
i.e. 3, 4.5, 7.5, 25.5, 49.5, 97.5, ...
Cut off at five steps gives: 49.5.

:cool:[/QUOTE]
:exclaim:
If Im not mistaken you are looking at the problem in retrospect.

If you had to start just at what is given i.e. 3 how would you arrive at 3/2 and the multipliers 3 , 5 , 17 , 33 and 65? ....
They dont seem to be related in any way to me. Can you kindly clarify?
Mally :coffee: