Dot Triangles - "By Logic"
Here is my quick triangle count - only a half hour to get all the errors out of it - which may still have an error:

A (6) - ABG, ACG, ACI, AGI, AHJ, ANQ
B (3) - ABG, BDK, BGO
C (8) - ACG, ACI, CEH, CGI, CGM, CHL, CON, CNT
D (6) - BDK, DEJ, DGH, DMK, DOS, DQS
E (7) - CEH, DEJ, EGL, EGN, EIL, EJK, ELN
F (2) - FHI, FST
G (13) - ABG, ACG, AGI, BGO, CGI, CGM, DGH, EGL, EGN, GHK, GIO, GLN, GNR
H (10) - AHJ, CEH, CHL, DGH, FHI, GHK, HJO, HIK, HLR, HMN
I (10) - ACI, AGI, CGI, EIL, FHI, GIO, HIK, IJN, ILU, IOP
J (12) - AHJ, DEJ, EJK, HJO, IJN, JKR, JKS, JLM, JQR, JQU, JRS, JRU
K (12) - BDK, DKM, EJK, GHK, HIK, JKR, JKS, KLS, KMT, KNO, KPT, KRS
L (11) - CHL, EGL, EIL, ELN, GLN, HLR, ILU, JLM, KLS, LNU, LOR
M (5) - CGM, DKM, HNM, JML, KMT
N (15) - ANQ, CON, CNT, ELN, EGN, GLN, GNR, HMN, IJN, KNO, LNU, NOT, NQS, NSO, NST
O (11) - BGO, CON, DOS, GIO, HJO, IOP, KNO, LOR, NOS, NOT, OTS
P (2) - IOP, KPT
Q (5) - AQN, DQS, JQR, JQU, NQS
R (8) - GNR, HLR, JKR, JQR, JRS, JRU, KRS, LOR
S (11) - DOS, DQS, FST, JRS, JKS, KLS, KRS, NQS, NST, NOS, OST
T (7) - CNT, FST, KMT, KPT, NOT, NST, OST
U (4) - ILU, JQU, JRU, LNU

(Edit: My horrible inaccurate list has been updated to reflect the comments below. If I had seen the BGO triangle, I may never have finished the puzzle as it is the kind of triangle I believe a "mean" constructor will certainly choose since it is the largest one without using any grid-lines possible. Of course, CGM and DOS also work in this way so its good I missed those too. The first part of my old logical path to eliminate KPT/FHI has the extra wrinkle of needing to test BGO but this will also fail)

So, my count above is 141 for 3x47 triangles. Considering my first draft of this list had 106 triangles (seemingly short of 36 which was my triangle count last year), a counting puzzle here is also difficult. Ok, so the logical path needs to focus on B, F, and P (assuming you notice those are the critical points and I only noticed B and P during the test). Either IOP and FST, or KPT and FHI (which also forces ABG). If you draw in the second, it forces CON. This forces U-J (but either QR), but then E cannot go anywhere. Great.

So IOP and FST. Now, it is either BDK or ABG. Let's say BDK. M must be in JLM or HMN, so either of these takes a vertex from NLU, so U connects to J (and one of QR) and M is HMN. Now, that forces EGL, but now C can't be ACG so it fails. So ABG! Great.

IOP, FST, ABG. Now CH (but either E or L). QJ (R or U), which finally forces DKM. R must now be to JQ, leaving LNU, and CEH.

Um, you testsolvers totally did not catch how unfair this puzzle is unless you guess and guess well. Does anyone have a good "logical" way through this? I really hit the bad side of the variance curve here, and I hope it doesn't cost me the title. Over 22% of my total time for 20 points.