# Find the total number of triangles in the diagram

The title of the question says everything $\ldots$

My attempt:
We count $2(1+1+1+2+1+1+2+1+1+1+1)=26$ triangles. (On each side $13$ triangles, and then multiplied by $2$). And then we combine them and add. Are there really more than $26$ triangles?.
I got this question in a test and the options were $20,21,26$ and None of the above. I wrote None of the above. Is this right? Please tell.
Edit: Please ignore the shading, all the intersections count as vertices and it is a 2d figure.

• do intersections count as a vertex? Or should all 3 vertexes be in a point with a letter? – Novarg Oct 5 '15 at 14:45
• Oh yeah all the intersections count as points. – Aditya Agarwal Oct 5 '15 at 14:47
• In my exam, the vertices were not labelled, but I made it on geogebra and I forgot to label these intersections. – Aditya Agarwal Oct 5 '15 at 14:51
• k then my solution is incorrect – The Dark Truth Oct 5 '15 at 14:59
• Inspired by miracle173's answer, I've written a C# program that lists all triangles for any set of lines. It confirms that the answer is 21 if only labelled vertices are allowed. If the unlabelled intersections should be considered vertices as well, then the answer is 40. ideone.com/YTncBk – Dennis_E Oct 6 '15 at 10:42

I think I managed to find

40 triangles

12 singles
12 doubles
8 triple
2 fives
1 six
0 sevens
1 eight
Ignoring the 12 singles:

• That's 4 more than I found! Can you point out to me which ones I missed? – GentlePurpleRain Oct 5 '15 at 19:47
• @GentlePurpleRain ABD and AGD and mirrors. – JonTheMon Oct 5 '15 at 19:57
• Is there a way to always get correct answer in first attempt? I generally don't get correct answer in my first try and then I look at the answer only to realize in the end that my count is wrong. Then I count again and again. How do I always get correct answer in just one try? – Heisenberg May 26 '16 at 17:56
• @user109256 If you check the edits you can see even I didn't get them all the first time. Basically, split the problem up: how many triangles with 2 cells; how many triangles of 3 cells; 4 cells; etc. While doing that, I tend to look for symmetry and order (for doubles I went counter-clockwise). – JonTheMon May 26 '16 at 18:49
• @JonTheMon would you recommend leaving these questions if they appear in a test where time is restricted. Or do you think that speed of solving can be immensely increased by practicing hard. Because in test there's no time for solving a question twice. Solving a question twice means you will have to leave at least $3 -4$ questions. – Heisenberg May 27 '16 at 3:47

If the triangles must have a named(A to H) point at each corner then I think the answer is actually:

21

Count:

The triangles in one half only(left half in this case):

(B,E,D) (B,E,G) (B,G,D)

(A,D,E) (A,D,G) (A,G,E)

(A,B,E) (A,D,B)

so 16 for both halves

plus the triangles which go over both halves

(A,H,G) (A,C,B)

(D,G,H) (D,E,F)

(D,B,C)

for a total of:

16 + 5 = 21

if i missed anything please correct me

I count 40 different triangles in the picture

To check this I write them down in a sorted list of triples. Each triple is

normalized in that way that the components of the triple are alphabecticaly

ordered.

1.  A   B   C
2.  A   B   D
3.  A   B   E
4.  A   B   J
5.  A   C   D
6.  A   C   F
7.  A   C   J
8.  A   D   E
9.  A   D   F
10. A   D   G
11. A   D   H
12. A   E   G
13. A   F   H
14. A   G   I
15. A   G   H
16. A   H   I
17. B   C   D
18. B   D   E
19. B   D   G
20. B   D   J
21. B   D   K
22. B   D   L
23. B   E   G
24. B   E   K
25. B   G   K
26. C   D   F
27. C   D   H
28. C   D   J
29. C   D   K
30. C   D   L
31. C   F   H
32. C   F   L
33. C   H   L
34. D   E   F
35. D   G   H
36. D   G   I
37. D   H   I
38. D   J   K
39. D   J   L
40. D   K   L


How can I check the list?

To check if the triangle (G,H,D) is considered in my count I normalize it to

(D, G, H) and look it up in the list and find it in line 33.

To check that I did no miss a triangle that shoul be inserted in the list

between

25. B   G   K
26. C   D   F


I do the following

1. there is no triangle following (B, G, K) with vertices B an G, because the

only triple I fave to check is (B, G, L), which is not a valid triangle. (B, G,

E) is a valid triangle but it does not follow (B, G, K) alphabetically. So it

is counted elsewhere in the list and mudt not be considered here.

1. There is no triange (B,H,...) because BH is not an edge in the picture. The

same is true for (B, I, ...).

1. (B, J, ...) can be skipped because if we supply K or L for ... we get a

triangle that is deeneraed t a line and we won't count such a line.

1. similar arguments hold for (B, K, ...)

2. (B, L, ...) must not be considered because there is no letter following

3. So al triangles starting with B are exhausted and we can check trinles

starting with C, or more precise starting at least with (C, D, ...). But (C, D,

E) is no a triangle.

And the next triple (C, D, F) is a trianle and in the list.

Here is also a python program that counts and displays all triangles.

I managed to find

38 triangles in total. (Thanks to @TheDarkTruth for the last two, and to @JonTheMon for pointing out duplicates).

I may have missed some, but it seems like your answer should definitely be "None of the above."

• DGH, DEF are missing and still searching for more – The Dark Truth Oct 5 '15 at 15:17
• Row4-Col3 is the same as Row7-Col1 – JonTheMon Oct 5 '15 at 18:43
• I got 38 as well after relabelling A-L and premuting all possiblilites. – Kevan St. John Oct 5 '15 at 21:04
• Missing DHA and DGA. – Trenin Oct 6 '15 at 17:58

If every intersection counts as a vertex, then yes, the answer is indeed None of the above

I counted 29 triangles here. The whole image can be symmetrically divided by the vertical AD line, so we have to count the triangles on one side, multiply it by 2 and then add 4(EFD, GHD, ABC, AGH).

Let's call intersection of GD and BC a point X, intersection of GH and AD a point Y and intersection of BC and AD a point Z like this:

Then we have(at least, maybe more):

1. EAG
2. EGB
3. AGY
4. GBX
5. BXD
6. DXZ
7. DGY
8. DGB
9. DEA
10. DBE
11. EBX
12. ABZ
13. ABD

Then we can multiply it by 2 to get:

13*2 = 26

26 + 4 = 30

So the answer is indeed None of the above

• you forgot AGD and AEB plus if every intersection counts you have 8 triangles extra that go over both halves and not 4 – The Dark Truth Oct 5 '15 at 15:07
• @TheDarkTruth then there are even more :) – Novarg Oct 5 '15 at 15:10
• There are 6 triangles symmetric to th AD axis and the number of triangles not symmetric to AD must be even. So the number of triangles is even. – miracle173 Oct 6 '15 at 5:19

Assuming that no two points can belong to a triangle unless they are joined:

22 triangles

12 with A

5 without A but with B

3 without A,B but with C

2 without A,B,C but with D

0 without A,B,C,D

P.S. I might have missed some.