A4. Find the smallest positive integer $v$, such that for any two-coloring of the edges of the complete graph $K_{v}$, there are two monochromatic triangles that share exactly one vertex.
The solution to A4 is $v=9$.
On one hand, take two red complete graphs $K_{4}$, and color all the edges between them blue. The resulting complete graph $K_{8}$ has no two monochromatic triangles with exactly one common vertex.

On the other hand, consider any two-coloring of the complete graph $K_{9}$. By problem $\mathrm{B}$, there is a monochromatic triangle (let $\triangle x_{1} x_{2} x_{3}$ be red). The complete graph $K_{6}$ formed by the other six points, if all edges are red, the conclusion is evident.

Assume the conclusion does not hold. Suppose there is an edge $v_{4} v_{5}$ that is blue. By the same argument as in problem $\mathrm{B}$ (Figure 1), we can assume there is another red $\triangle x_{1} x_{2} x_{4}$.

The complete graph $K_{5}$ formed by the other five points $x_{5} 、 x_{6} 、 x_{7} 、 x_{8} 、 x_{9}$, if all edges are red, the conclusion is evident.
Suppose the edge $x_{8} x_{9}$ is blue (as in Figure 5).
If the conclusion does not hold, $x_{4} x_{8} 、 x_{4} x_{9}$ are not both blue.

Assume $x_{4} x_{8}$ is red (Figure 6). Then $x_{1} x_{8} 、 x_{2} x_{8}$ are blue, $x_{9} x_{1} 、 x_{9} x_{2}$ are red, $x_{9} x_{4} 、 x_{9} x_{3}$ are blue, $x_{3} x_{4}$ is red, $\triangle x_{2} x_{3} x_{4}$ and $\triangle x_{2} x_{1} x_{9}$ are two red triangles with exactly one common vertex.
Thus, the conclusion holds.