Gallivan's Triangle

The first right triangle has side lengths 512, 512, and $512 \sqrt{2}$ (by Pythagorean Theorem). Folding the paper once creates a right angle opposite a side that was previously length 512, so the leg lengths are $512/ \sqrt{2}$ . A pattern emerges where, for each subsequent fold, the dimensions of the triangle are divided by $\sqrt{2}$ . With 13 folds, the length of the hypotenuse is $512sqrt(2) / \sqrt{2} ^{13} = 512 / 2^{6} = 2^{9} / 2^{6} = 2^{3} = 8$ .

For a right isosceles triangle, when folded in half by bring the equal sides together give rises to a new right isosceles triangle with its longest side equal to $1/sqrt{2}$ times the longest side of the previous triangle. Hence the reduction of side length by folding the triangle each time follows a geometric sequence.

Lets assume the longest side of the very first triangle is $a_{0}$ . $a_{k}$ be the longest side after folding the triangle $k$ times which can be expressed as $a_{k}$ = $a_{0}$ $r^k$

where $r =$ $1/sqrt{2}$

Hence the longest side length after 13th fold is

$a_{13}$ = $a_{0}$ $(1/sqrt{2})^{13}$ in this case $a_{0}$ = $sqrt{2}$ x $(512)$

Hence $a_{13} = 8$

I started with the original largest triangle. let the measure of it's leg be L and r = \sin 45^\circ

Let L_i\ be the leg of the triangle after the i\^th fold

Starting the first fold which bring the the leg as the hypotenuse of the new triangle. Thus, L 1 = L 2nd fold takes L 1 as the new hypotenuse thus L 2 = r \times L 1

continue folding..

i^th fold takes L {i-1} as the new hypotenuse and thus L i = r \times L_{i-1} for 2 \leq i \leq 13

thus L_{13} = r^{12} \times L = \frac {1} { sqrt{2}^{12}} \times 512 = \frac {1}{64} \times 512 = \frac {512}{64} = 8 units

In the initial triangle, the length of the longest side is $(2^{0.5})(512)$ by Pythagoras' Theorem. Once a fold is made, half the length of the longest side becomes a side of the triangle, and the longest side therefore becomes 512. Once repeated, we may observe that now that 256 is the new side and therefore the longest side is $(2^{0.5})(256)$ . It can be therefore observed that after each fold, the new length for the longest side becomes $(2^{0.5})$ times the previous one. Using the formula for the general term for a Geometric Progression with first term $(512)(2^{0.5})$ and (2^-0.5) as the common ratio, for the 14th term, we can compute the length of the longest side to be 8.

Calculation example of the next term of the sequence; 2nd, Length = Sqrt(512^2*2)/2=362.038672

Sequence; 512, 362.038672, 256, 181.019336, 128 ...

The sequence itself can be separated into two distinct geometric sequences as given below:

Sequence 1... 512, 256, 128 and so on. (Odd terms of Sequence) Sequence 2... 362.038672, 181.019336, 90.509668 and so on. (Even terms of Sequence)

Determining which sequence is of relevance to the current situation, the piece of paper was folded 13 times thus their are 13th terms in the pattern. Since the last term is odd then sequence 1 is of relevance, which in turn the 13th term then becomes the 7th term of Sequence 1 (12/2+1=7).

Using sequence 1, we can see that the common multiplier is 0.5 (this is the same for the other sequence). Thus from this observation we can use the geometric progression formula [T(n)=ar^(n-1), where a is the initial value and r is the ratio].

T(n)=512(0.5^(n-1))

So the 7th term of the sequence is;

T(7)=512*(0.5^(7-1))=8

Thus 8 is the length of the isosceles triangle after being folded 13 times.

The very first triangle has the equal sides of 512 length. The first folding yields a triangle having a hypotenuse of 512 and the equal sides of 362.039 or 512/(2^.5). The second folding yields another triangle having a hypotenuse of 362.039 and the equal sides of 256. It is clear that the longest side of each right isosceles triangle is the hypotenuse. And the equal side of the triangle is the hypotenuse of the next triangle. So in the 2nd triangle the length of hypotenuse, 2^.5=1.41 512/1.41=362.038 So in the 13th triangle " " " " ,2^(12/2)= 2^6=64 512/64=8 (Sorry, my English is a bit weak and my keyboard doesn't support mathematical signs)

The original triangle has 512 \sqrt{2} as its longest side. When you fold it once, the longest side is now 512 \sqrt{2} \times \sqrt{2}/2, fold it twice and the longest side is now 512 \sqrt{2} \times (\sqrt{2}/2)^2 fold it thrice and the longest side is now 512 \sqrt{2} \times (\sqrt{2}/2)^3 ... fold it thirteen times and the longest side is now 512 \sqrt{2} \times (\sqrt{2}/2)^13, which is equal to 8.

The second fold will result in the length of the shorter side being 512/2 Therefore , the thirteenth fold will result in the shorter side being of length 512/(2^13/2) , which is equal to 2^2.5 .

Since this is the shorter side , the longer side will be sqrt((2^2.5)^2 + (2^2.5)^2) = sqrt ( 2 ^5 + 2 ^ 5 ) = sqrt 64 = 8

when we fold one time the length of isosceles side is divided by √2. so, when i fold 13 times, it will be divided by (√2)^13. but it asked the longest side so to get it from the right isosceles-triangle, i need to multiple the √2. so in total 512 will be divide by the 2^6. 2^9/2^6=2^3=8

After the first fold the hypothenuse of the right triangle is 512. After another two folds the hypothenuse is 256. so after two folds the longest side is halfed. That means in the first fold it is 512. in the third it is 256. for the fold number n+1 it is 512/2^n-1/2. so in the fold number 13 it is 512/2⁶. Since 512=2⁹: 2⁹/2⁶=2³=8

Let $u_n$ be the length of the longest side of the triangle after $n$ folds. Knowing that one the two equal sides at step $n$ becomes the hypotenuse at step $n+1$ , we have : $u_{n+1} = u_n \times \frac{1}{\sqrt{2} }$ and $u_1 = 512$ .

That is, $u_n$ is a geometric progression with ratio $\frac{1}{\sqrt{2} }$ . Therefore, we have $u_n = 512 \times (\frac{1}{\sqrt{2}})^{n-1}$ for $n \ge 1$ .

We compute then $u_{13} = 512 \times \frac{1}{2^6}$ = 8.

when the triangle is folded for the first time its longest side is of length 512. for every next fold it gets reduced by \sqrt{2}. thus next triangle has longest side equal to 2^{8.5}. thus repeating it 11 more times we get the answer as 8 i.e 2^{3}

after 2nd fold side of isosceles triangle is 256. similarly after 12th fold side is 8. and in 13th fold greatest side will be the smallest of 12th fold which is 8.

Note that the first time the triangle is folded the original cathetus of it becomes the hypotenuse of the new one. On the second fold, the catheti of the new triangle are exactly half of the original catheti, or $256$ (since $\frac{512}{2}$ ). On the third fold however, we again have the new cathetus be the hypotenuse of the new triangle. This pattern goes on such as that every even number of folds the original cathetus is divided by $2$ , whereas every odd number of folds the corresponding cathetus becomes the hypotenuse of the new triangle.

Since we have $13$ folds, which is an odd number, we can see that it is the hypotenuse of the final triangle we are looking for. Since there are $6$ even numbers from $1$ to $13$ , it is evident that the hypotenuse equals $\frac{512}{2^{6}}=8$ .

Another important point is that in any right isosceles triangle the hypotenuse is unarguably the longest side in it. Therefore, the longest side of our requested triangle is $8$ .

The relationship of the sides of a right isosceles triangle is 1:1:sqrt(2), where sqrt(2) is the hypotenuse. So when you fold it along the hypotenuse, the new sides have length half of sqrt(2) times the old sides. sqrt(2)/2 = 1/sqrt(2), so if you fold it 13 times, you divide the length by sqrt(2) 13 times. 12 times is like dividing by 2 six times, so you get 512/64 = 8, and then you divide that by sqrt(2) one last time.

So the equal sides are 8/sqrt(2) = 4sqrt(2). But the longest side would be sqrt(2) times this, which is 8.

The consecutive triangle sizes are halved with each fold. Since the ratio of area to side length is (x^2)/x, then the triangle side lengths are multiplied by a ratio of 1/(sqrt2) or (sqrt2)/2. This process is repeated twelve more times, so 512 * [(sqrt2)/2)]^12 = 512 * 1/(2^6) = 512/64 = 8 .

Draw out the triangle and show the working step for 3 times.

Get their relationship as the factor of sqrt 2

512 / (sqrt 2)^13 = 5.6569 is the length of equal side of the isosceles triangle after 13th fold

Use Theorem Pythagoras to solve the longest length as

l^2 = (5.6569)^2 + (5.6569)^2

l = 8

fold 1) for the first fold, hypotenuse = 512 2 x (length of equal sides)^2 = hyp ^2 = 512^2 => equal side = 512 / sqrt 2

fold 2) Here length of equal sides in fold 1 = hyp. 2 x ( length of equal side)^2 = (512 / sqrt 2)^2 => 512 / 2 = 256

Similarly for fold 3) , length of equal side = 256 / sqrt 2 therefore for nth fold , length of equal sides = 512 / (sqrt 2)^n

length of equal sides in 12 th fold = Longest side( hypotenuse) in the 13th fold

so length of equal sides in 12th fold = 512 / (sqrt2)^12 = 512 / 64 = 8 units length of longest side = 8 units

First, the size of initial triangle is 512 = 2^9 By pythagoras: The first folding gives to us a size of 2^(17/2). The second folding -> 2^(16/2) So, the thirteenth folding ->: 2^(5/2.) By phytagoras again, we have the largest size is 2^3 = 8.