Equilateral triangles -- challenge problem

[lookagain]

The points are arranged in an equilateral triangular lattice style.


If this pattern continues, what will be the total number of equilateral triangles that
can be formed for Figure 6 ?

Edit: Much of the text of the original problem got deleted.

 

[James Magan]

+
M

I will take a horrendous risk and say 48.

 

[lookagain]

Big Hint:

Now that we're up into larger figures, there are equilateral triangles with

additional orientations to those in the smaller figures I showed.

 

[James Magan]

+
M

Thank you. I still only find 50. I guess that my makeshift grid is part of the problem.

 

[TchrWill]

Triangles in Triangle

If I understand you correctly:

or r
How many triangles of all sizes can be found in a triangle 'n' rows high ?
Or


A single equilateral triangle contains one triangle, n = 1.

One triangle sitting atop three others, n = 2 rows high, contains 5 triangles all together, 4 small ones and 1 large one.
......../_\
....../_\/_\

These 5 small triangles sitting atop 5 additional triangles, n = 3, contains 13 triangles all together, 9 small ones, 3 medium ones, 2 triangles high, and 1 large one, 3 triangles high.
......../_\
....../_\/_\
..../_\/_\/_\

Find the number of distinct triangles in the nth triangle "n" triangles high.

/_\......../_\................/_\................../_\
.........../_\/_\............/_\/_\............../_\/_\
............................./_\/_\/_\........../_\/_\/_\
................................................./_\/_\/_\/_\

If you draw any more, you will soon see the emerging pattern.

1.....5.....13.....26.....45.....71.....105.....148.....201.....265.....etc.

Following the totals of the first few will enable you to project the total in the 100th if you wish.
What do we now have?

Base size or height-n..................1.....2.....3.....4.....5.....6......7......8......9......10.....etc.
Triangles....................................1.....5....13...26...45...71...105..148..201...265....etc.
1st difference - d1.........................4....8....13....19...26....34...43....53.....64.........etc.
2nd difference - d2..........................4.....5.....6.....7......8......9....10....11.....12.....etc.
3rd difference – d3......................…...1.....1.....1......1......1.....1......1......1.......1......1

With the 3rd differences being constant at 1 we have a finite difference sequence with the nth term being of the form
T = an^3 + bn^2 + cn + d.

Using the data:
a + b + c + d = 1
8a + 4b + 2c + d = 5
27a + 9b + 3c + d = 13
64a + 16b + 4c + d = 26.

Solving, a = 1/6, b = 1, c = - 1/6 and d = 1.

Therefore, the general expression for the nth term is T = (n^3 + 6b^2 - n)/6.

With n = 4, we get T = (4^3 + 6(4^2) - 6 + 6)/6 = 26.

With n = 5, we get T = (5^3 + 6(5^2) - 5 + 6)/6 = 45.

With n = 6, you should get (I’ll let you calculate this.).



<----------------------------------------------------------------->

 

[John Marsh]

Total number of equilateral triangles that can be formed for Figure 6 = 50​

 

[lookagain]

big hint:

now that we're up into larger figures, there are equilateral triangles with

additional orientations to those in the smaller figures i showed.

Figure 4



.... ... . . . . . . . . . . . . . . . . . A

.... . . . . . . . . . . . . . . . . . B . . C

.. . . . . . . . . . . . . . . . . D . .. E . .. F

. . . . . . . . . . . . . .... G .. . H . .. I . . J



Please don't ignore my hint from a post earlier in this thread (in the quote box).

Look at triangle BFH and triangle CDI.

They are of different orientations from those already discussed, and including them
increases the prior count of 13 to a revised accurate total of 15 for Figure 4. \(\displaystyle \ \ \ \ \ \)<------- edit


Apply this idea to the sixth figure to get a total greater than 45.



*** The problem is still open. ***

 

[lookagain]

Agree; however, by drawing these 2 triangles, then you end up with a "star",
tips being B,C,F,I,H,D; a small equilateral triangle (smaller than ABC) is created
at these 6 tips, thus a total of 21...or do I need an eye exam?

No, Denis, as with any of the other triangles, you form the

triangles one at a time.

No six-pointed star is being made.

 

[lookagain]

Update:
---------

I'll label the 21 points in the six rows in a diagram and name the other equilateral triangles:


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A

. . . . . . . . . . . . .. . . . . . .. . . . . . . . . . . . . . . . . . . . B . . . C

. . . . . . . . . . . . . . .. . . . . . . . . .. . . . . . . . . .. . . D . . . E . . . F

. . . . . . . . . . . . .. . . .. . . ... . . . . . . . . . . . . .. G . . . H . . . I . . . J

. . . . . .. . .. . . . . . . . . . . . . . . . . . . . . . . .. K . . . L . . . M . . . N . . .O

. . . .. . . . . . . . . .. . . . . . .. .. .. .. . . . . . P . . . Q . . . R . . . S . . . T . . . U



There are the 45 equilateral triangles already mentioned in the thread.


6 with vertical sides that point to the right:

BFH
DIL
EJM
GMQ
HNR
IOS


6 with vertical sides that point to the left:

CDI
FHN
EGM
JMT
ILS
HKR


2 centered ones:

DJR
FGS


3 that slant somewhat downward to the left:

CGN
FLT
EKS


3 that slant somewhat downward to the right:

BJL
EOR
DNQ


2 others I missed: \(\displaystyle \ \ \ \ \ \) Edit

BOQ
CKT


------------------------------------------------------------


45 + 6 + 6 + 2 + 3 + 3 + 2 = 67



\(\displaystyle \boxed{\text{67 total equilateral triangles}} \ \ \ \ \ \) <------------ Edited answer