Now consider V 1: In addition to V 1 and 15 points, there are 15 edges * * drawn from V 1, which is the well-known pigeon hole principle.

Red, yellow, green and yellow have at least one color, and the number of edges dyed with this color is greater than or equal to 15/3=5.

Let's set this color as color A, and now consider the edges dyed with color A (≥5) from the edges drawn by V 1, and take any five of them as the research object. We set the other two ends of five sides except V 1 as V2, V3, V4, V5 and V6 respectively. According to the arrangement number, 10 edges are connected between these five points. There are two situations as follows:

(1) If one of the sides of 10 is dyed with color A, then the triangle formed by both ends of this side and V 1 is the same color triangle.

(2) If there is no side dyed with color A in the 10 side, that is, each side in the 10 side is dyed with one of the other two colors (set to colors B and C). Now the problem is transformed into a complete graph of order 5 (*** 10 edges), and each edge is colored with color B or C, which proves that there is a homochromatic triangle in this graph.

Let's assume that the edge between V2 and V3 is dyed with B, please see the picture below. * * * There are four situations:

① For case 1: If the edges (V4, V5) are not B or C, then V4 and V5 must form a homochromatic triangle with one of V2 and V3.

(2) The second case is the same as the first case.

③ For case 3: V2, V3 and V4 form the same color triangle.

④ Case 4: If one side of (V4, V5), (V5, V6) and (V4, V6) is dyed with C, obviously one triangle has the same color; If the edges (V4, V5), (V5, V6) and (v4, V6) are all colored with B, then v4, V5 and V6 form a homochromatic triangle.

To sum up, there must be triangles of the same color in the original picture.