How many triangles in the picture?

I counted over 140 then got lost..

Is there a way to calculate the number of triangles in the picture?

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- Jul 27th 2010, 02:01 AMmetlxHow many triangles?
How many triangles in the picture?

I counted over 140 then got lost..

Is there a way to calculate the number of triangles in the picture? - Jul 27th 2010, 02:49 AMFailure
140 seems like too many to me, what about

$\displaystyle \binom{7}{2}+\binom{7}{2}-1+(7-2)^2$

Well, maybe that's not exactly right: you have to check for yourself. But the basic idea that I want to convey is that you should try to use well known counting formulas to figure it out. In my attempt you can see, I hope, that $\displaystyle \scriptstyle \binom{7}{2}$, $\displaystyle \scriptstyle \binom{7}{2}-1$, and $\displaystyle \scriptstyle (7-2)^2$ are meant to count all triangles*of a certain type*. - Jul 27th 2010, 03:15 AMmetlx
apparently the right answer is 216. (6^3)

- Jul 27th 2010, 03:46 AMFailure
- Jul 27th 2010, 10:21 PMsimplependulum
The answer is $\displaystyle \binom{6}{2} \times 6 \times 2 + 6 \times 6 $

EDIT: The above formula is equivalent to $\displaystyle 6^3 ~~ ( n^3 ) $ , actually .

To derive it , i find that all the triangles found in the figure must be constructed by at least one line passing through $\displaystyle A$ and one through $\displaystyle B $ . The other line ( the third line , triangle is constructed by three lines ) can only be selected in twelve lines , try it ! However , we find that every triangle has been counted twice , so let's halve it . Therefore , the number of triangles is $\displaystyle 6 \times 6 \times 6 $ - Jul 29th 2010, 07:25 AMmetlx
- Jul 29th 2010, 03:30 PMArchie Meade
Counting by hand...

Starting at the bottom left-hand corner, using that corner as a vertex in each case,

there are 6 triangles with 2 vertices right next to each other on the right-hand side.

If we now choose any of the other 5 lines that emanate from the bottom right-hand vertex,

then there are 6 times as many triangles starting from the bottom left-hand vertex

that fit between two adjacent lines emanating from the bottom-left vertex.

If we start at the bottom right-hand vertex, we double this number while subtracting

the double-counted one at the base.

Next, we can count the triangles that are between 2 lines that have a single line between them

(emanating from the bottom left vertex).

There are 5(6) of these.

Double that and subtract the overlapping one and the two that overlap at the base

with triangles one space thick.

Next count the triangles with two lines in between.

There are 4(6) of these.

Double and subtract the overlap.

Then subtract the 2 that overlap with those that are 2 spaces thick.

Finally subtract the 2 that overlap with those that are one space thick.

Next count the triangles with 3 lines in between.

There are 3(6) of these.

Double and subtract the 1 overlap.

Subtract the 2 that overlap with those that have 2 lines between.

Subtract the 2 that overlap with those that have 1 line between.

Subtract the 2 that overlap with those that have no line between.

Next count the triangles with 4 lines between the outer edges.

There are 2(6) of these.

Double and subtract the 1 overlap.

Subtract 2 and 2 and 2 and 2 for the remaining overlaps.

Count the triangles with 5 lines between the edges.

There is only 1.

Total is $\displaystyle 6\left(2(6)+2(5)+2(4)+2(3)+2(2)\right)+1-5-4(2)-3(2)-2(2)-2$

$\displaystyle =6\left(12+10+8+6+4\right)+1-25$

$\displaystyle =6(40)-6(4)=6(36)=6^3$ - Jul 30th 2010, 12:13 PMArchie Meade
The mathematics is much clearer when we avoid having to subtract double-counted triangles.

Attachment 18388

Any triangle in the figure must have either X or Y as a vertex.

We can independently count

(1) triangles with Y as a vertex but not X

(2) triangles with X as a vertex but not Y

(3) triangles with both X and Y as a vertex.

With Y omitted and X as a vertex, we can form a triangle by selecting 2 of

the 6 blue points on the red line.

The number of ways to do this is $\displaystyle \binom{6}{2}=\frac{6(5)}{2}$

Doing the same with the other 5 lines above the red one leaving Y gives

$\displaystyle \frac{6(6)5}{2}=\frac{5}{2}6^2$ triangles with X as vertex but not Y

There are the same number of triangles containing Y but not X.

Therefore the number of triangles containing only one of those vertices is $\displaystyle (5)6^2$

With both X and Y as vertices, 6 triangles can be formed on the line with the blue dots in the 2nd diagram.

The same number of triangles can be formed on the 5 lines leaving Y immediately below that.

That gives $\displaystyle 6^2$ triangles that have both X and Y as vertices.

The total number of triangles is $\displaystyle (5)6^2+(1)6^2=(6)6^2=6^3$