First, let's consider the different medal combinations that can be awarded to the 3 winners:
(1) If there are NO TIES then the three medals awarded are: GOLD, SILVER, BRONZE.
(2) What if there is a 2-WAY tie? 
--If there is a 2-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER.
--If there is a 2-WAY tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER.
--There cannot be a 2-WAY tie for THIRD (because exactly three medals are awarded in total).
(3) What if there is a 3-WAY tie?
--If there is a 3-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD.
--There are no other possible 3-WAY ties.
Thus, there are 4 possible medal combinations:
(1) G, S, B      (2) G, G, S      (3) G, S, S      (4) G, G, G
 

COMBINATION 1: Gold, Silver, Bronze

Number of possibilities = $4\times 3\times 2$ =24 different Winner's triangles that will contain 1 GOLD, 1 SILVER, and 1 BRONZE medalist. 
COMBINATION 2: Gold, Gold, Silver.

Selection =${}^4{C}_3$ and arrangement =$\frac{4!}{3!(4-3)!}$= $4\times 3$=4 ways
COMBINATION 3: Gold, Silver, Silver.
Using the same reasoning as for Combination 2, we see that there are 24 possible Winner's triangles, but only 12 unique Winner's triangles that contain 1 GOLD medalist and 2 SILVER medalists.
COMBINATION 4: Gold, Gold, Gold
Finally , then we have the following:
Selection =${}^4{C}_3$ and arrangement $=\frac{3!}{2!}=$ 4 $\times$1=4 ways
Thus, there are 24 + 12 + 12 + 4 = 52 unique Winner's triangles. 
The correct answer is B.