20060517, 17:24  #1 
Sep 2002
1010010110_{2} Posts 
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. 
20060517, 23:24  #2 
Sep 2002
296_{16} Posts 
The attached image file shows the first two iterations of the triangle.

20060518, 07:19  #3 
Dec 2005
C4_{16} Posts 
# 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 
20060518, 10:45  #4 
Dec 2005
196_{10} Posts 
never ever anwser a quick calculation 5 minutes before an important meeting. Ugh, terrible error at the end 961/42+3=1087/42=25+37/42 sorry 
20060519, 16:24  #5  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Symmetry
Quote:
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 

20060520, 06:51  #6 
Dec 2005
C4_{16} Posts 
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 ) 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=963 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 
20060520, 10:06  #7 
Jun 2005
2·191 Posts 
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^{n} length of segments = (1/2)^{n} Each iteration adds 3 segments to each of the previous iteration's triangles. If L_{n} represents the total length of the segments L_{n} = L_{n1} + 3 * 4^{(n1)} * (1/2)^{n} For 1024 triangles, 4^{n} = 1024. n=5...so we're at iteration 5. The length of each segment is 1/2^{5} = 1/32. Now we just need to find the total length of the sides. We can do this iteratively. L_{1} = 4.5 L_{2} = 7.5 L_{3} = 13.5 L_{4} = 25.5 L_{5} = 49.5 Drew Last fiddled with by drew on 20060520 at 10:07 
20060520, 10:27  #8 
Jun 2005
2·191 Posts 
If we don't want to iterate, we can continue with the following:
L_{n} = L_{n1} + 3 * 4^{n1} * (1/2)^{n} = L_{n1} + 3/4 * 2^{n} L_{n} = 3/4 (2^{n} + 2^{n1} + 2^{n2} ... + 2^{0}) + C = 3/4 * (2^{n+1}  1) + C Solve for C (L_{0} = 3) 3 = 3/4 + C C = 2.25 L_{n} = 3/4 * (2^{n+1}  1) + 2.25 We can simplify this a bit further: L_{n} = 3/4 * 2^{n+1} + 3/2 = 3/2 * (2^{n} + 1) Last fiddled with by drew on 20060520 at 10:28 
20060521, 16:05  #9  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Multiplicity and Triangles
Quote:
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. 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 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 

20060522, 02:30  #10 
May 2004
New York City
4235_{10} Posts 
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. 
20060522, 17:01  #11  
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
Quote:
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 

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