Hexagonal Tilings with Triangular Tiles

Relevant wiki: General Term Pattern Recognition

If you focus on one of the six equilateral triangles the hexagon splits into, as above, it's easy to see that the number of triangular tiles is given by $1 + 3 + 5 + 7 + .... + (2n-1) = n^2$ where $n$ is the number of colored "layers". Thus the number of triangles in the hexagon must be $6n^2$ for some $n \in \mathbb{N}$ , and the only answer that fits the bill is $6(10)^2 = \boxed{600}$

Relevant wiki: Sum of n, n², or n³

If we denote $E_{n}$ to be the number of equilateral triangles needed to form regular hexagon with side length $n$ (we take side of small equilateral as a unit of measure), we can write recurrence relation: $E_{n} = E_{n-1} + 6\left ( 2n - 1 \right ).$ Now, since $E_{0} = 0$ , we have: $\begin{aligned}E_{n} &= \sum_{k = 1}^{n}6(2k - 1) \\ &=12\sum_{k = 1}^{n}k - 6\sum_{k = 1}^{n}1 \\ &=12 \cdot \dfrac{n(n+1)}{2} - 6n \\ &= 6n^{2}\end{aligned}$ The only number of this form given is $\boxed{600}.$

Relevant wiki: Method of Differences

Working out the next in the sequence gives a total of 96 triangles used. The sequence of tiles required to form each hexagon is therefore as follows: 6, 24, 54, 96, ... If we work out the nth term for this quadratic sequence:

Sequence: 6, 24, 54, 96, ...

First difference: 18, 30, 42, ...

Second difference: 12, 12, ... ( $\frac{12}{2}$ = 6: the coefficient of n²)

(Since the second difference is constant, the sequence is quadratic.)

n²: 1, 4, 9, 25, ...

6n²: 6, 24, 54, 96, ... (the coefficient of n² is 6)

which is equal to the sequence so the nth term of the sequence is 6n². Since 600 is the only option divisible by 6 which gives a square number (600 / 6 =100 = 10²) (which is necessary for 6n² to be a whole number), then 600 must be the answer.

Every Hexagon can be divided into 6 big triangles. Each triangle is the same and is further divided into a number of small triangles. This further division into small equilateral triangles follows a pattern of 1, 4, 9, 16 . . . So, we know, that the * total number of little triangles is a multiple of 6 and a perfect square. * All the answers are divisible by 6 but only 1 answer leaves a perfect square i.e. 100.

It could be a Quadratic Function... Let S(n) be the number of equilateral triangle in a hexagon with n triangles for each side...

S(n)=an^2 + bn + c

6=a+b+c 24=4a+2b+c 54=9a+3b+c

Solve using systems: We will get a=6, b=0, and c=0

So: S(n)=6n^2

Let n=10 S(10)=6(10^2)=600

So 600

600 is the only answer divisible by 6

First its 6, then 18 triangles are added with total of 24, then 30 are added with a total of 54 and so on... 6.1 + 6.3 + 6.5 +.... We can see that this is a series, just keep on adding successive numbers until you get one of the four numbers!

1st Hexagon = 6 Equilateral triangles = 6x $1^2$ 2nd Hexagon = 24 Equilateral triangles = 6x $2^2$ 3rd Hexagon = 54 Equilateral triangles = 6x $3^2$

It can be conjectured that the number of hexagons = 6 $n^2$ . 600 is the only number given which is 6x(a square).

The sequence is as 6, 24 , 54 ...

Which can be broken down as 6, 6 + 6×3 , 6 +6×3 +6×5, .....

So for the the nth sequence that would up to requisite no. of triangle

N th = 6 + 6×3 +6×5 + ..... 6× n

     = 6( 1 +3 +5 ...+ n)


     =6{ (n/2)×(2×1 + (n-1)×2)}


     = 6n^2

And the only option that gives the perfect square on dividing by 6 is 600

Notice that the first phase has 6 1 triangles, the 2nd phase has 6 3 triangles, and added up together give us 6 4 triangles. And this is true because is easy to see that when we look up the following phase there is always 12 more triangles (6 2) this means that the number of triangles at the n th phase is 6(2n-1), because the phases are always growing in pairs (so we know that the next phase is just the previous one plus 6*2). But we are not supposed to calculate the number of triangles at one phase, we were asked for calculating the sum of the phases. Since the triangles of a phase n is defined by an odd number, and the sum of n phases must be defined by the sum of every single odd number from 1 to 2n-1. This can also be expressed as n². It's a very common example when teaching proof by induction (since it's true for n=1, we suppose that the statement is true for n=k, and then prove that it holds for n=k+1. If that's true, it will hold for all the positive integer values due to the fact, 2 is 1+1, and 3 is 2+1, etc.). So we can express the sum of n phases as 6n² and try to substitute the values purposed below. If it's a valid number, n will be an integer. The only number which matches is 600.

Tiles increase as follows: 6(1) 6(1+3) 6(1+3+5) In general six times an odd number summation. Since: 1+3+5+...+kth_odd = k^2 All options are divisible by 6, but only √100 gives an integer, only 600 can be a number of tiles.

Count Triangles: N1=6 N2= 6+18=24 N3=6+18+30=54

All have common factor 6: N1=6•1 N2=6•4 N3=6•9

General Form: Ni=6•i^2 Or Ni/6=Ni^2

The only answer when divided by 6 that results in a perfect square is 600. 600/6=100=10^2; i=10.

Since the question of the sum of consecutive odd integers being a perfect square was raised elsewhere .....

We start with a hexagon with an outer layer of 6 triangles. Adding 12 gives one with an outer layer of 18 triangles, then 30, etc. The total number of triangles in each hexagon is equal to its predecessor plus the number of triangles on the outer layer. This means our sums will be 6, (6 + 18), (6 + 18 + 30), etc. Eventually, such a pattern gives a total of 600, which is the correct answer.

In an Excel spreadsheet, place the series of odd number 3, 5, 7... in column A starting at cell A2. Place the number 6 in cell B1. In cell B2, place the formula =B1+(A2*6). Drag this formula down column B for as may cells as you have filled in column A. This will compute the number of triangles required for successively larger hexagons.

2×3=6 4×6=24 6×9=54 ..... .. ..... 20×30=600 it follows the same pattern that's all I know

Triangle must be made of perfect square amount of equilateral triangles. The first triangle has 1 the next has 4 then the next triangle has 9 and so on. You need 6 of those triangles to make a regular triangle. Dividing all of the solutions by 6 we see 600 / 6 = 100 which is the only perfect square and can therefore make a subtriangle.

We can note that the number of triangles to make one side of the hexagon is a perfect square (c.f. table below).

In general, if we have $N$ interactions so we have $N^2$ triangles by side. Then, with we have a number $n$ , just divide $n$ by $6$ and if the result is a perfect square so we can make a regular hexagon. We can see that $600/6 = 100$ and $100$ is a perfect square soon we can build a regular hexagon.

"Algorithm" in c++ for(int i = 1; i < 800; i+=2){ x+=i*6; cout<<x<<' '; }

The number of tiles can be treated as a sequence: 6, 24, 54. When I divided each by it's greatest common factor, I got another amazing sequence: 1, 4, 9 - the first three integers squared. This makes the sequence: $\sum_{n=1}^{\infty} 6n^{2}$

Number of tiles, T=6+3.6+5.6+7.6+9.6+.............+(2n-1).6 =6[1+3+5+7+.........+(2n-1)]=6*S(suppose the series of n odd numbers=S) Now we need to calculate S= [1+3+5+7+.......+(2n-1)]

S=1+3+5+7+.....(2n-3)+(2n-1)...............(1) S=(2n-1)+(2n-3)+.....+3+1...............(2) By adding (1) and (2)- 2S=n.2n S=n^2 So we got number of tiles T=6*n^2 here we should get a natural number for n. From the options only 600 gives us a natural number for n. Because 600/6=100 is a square of 10. So we can get n=10. So the answer is 600.

Using Progressions, we can solve the question.

Sequence of number of triangles in each layer: 6, 18, 30, ... Starting term, a = 6 Common difference, d = 18 - 6 = 12

To get the total number of triangles, we sum up the sequence of number of triangles in each layer using the formula Sn = n/2 (2a + (n - 1) d). Therefore, Sn = n/2 (2(6) + (n - 1) 12) = 6n^2

Then set 6n^2 equal to the options that are given. If n is an integer, that is the answer. 6n^2 = 600 n^2 = 100 n = 10

6(n)^2 = number of triangles in reg hexagon; 1=6, 2=24,3=54 etc... 10=600

6,24,54,96,150,216,294,384,486,600...

If you take a look at the example you will notice that the number of triangles needed to make a regular hexagon of any size is 6 x n, where n is a square number. As shown in the example, 6 x 1 = 6, 6 x 4 = 24 and 6 x 9 = 54 (1, 4 and 9 are square numbers). The only option where 6 is multiplied by a square number is 600, which is 6 x 100.

Just get the numbers from 1 to 10 squared and multiple them by the base 6.

The pattern is 6 × n squared To check the numbers you divide them by six and look if the square root results in an integer.

The amount of tiles is 6 times a triangular number, i.e. a number of the form $1+3+5+ ... +(2n-1)=n^2$ . Only 600 has this property: $600=6*10^2$

Sum of triangles is = 6 (1) + 6 (3) + 6 (5) ...........+ 6 (1+2(n-1))

So it is concluded that the sum of the triangles divided by 6 must be the sum of the following arithmetic sequence: 1,3,5,.......(2n-1)

Which is equivalent to (n/2)(2a+(n-1)d) where d=2 and a=1, so the sum is n^2 , in other words it must be a square number.

The only square number in the above options is 600/6=100

Every time the outer layer will consist of the number of triangles= 6 odd. So for first its 6 1, for second total triangles will be 6+6x3=24, for third it will be 24+6x5=54 and so on we can write the function as f(n)=f(n-1) + 6x(2xn-1) where f(0) =0.