Mongolian Mathematical Olympiad

Answer: 4.

First, we construct a convex pentagon with 4 large triangle. Let $A{B}_1{B}_2{C}_1{C}_2$ be the convex pentagon such that $A{B}_1=B_2{C}_1=C_{2}A=1$ and $B_1{B}_2=C_1{C}_2=ϵ>0$. When $ϵ$ is sufficiently small each triangle $A{B}_i{C}_j$ is big for each $i$ and $j$.

Now we show that the number of large triangles is at most 4. We define a triangle as "boundary" if its sides are formed by 2 sides of the given pentagon and 1 diagonal of the pentagon. Similarly, we define a triangle as "center" if its sides are formed by 1 side of the given pentagon and 2 diagonals of the pentagon. Furthermore, we define a triangle as "small" if it is not a large triangle.

Assume that there exists a boundary triangle, which we will call ABC. Then the four triangles formed from the quadrilateral ACDE must all be small. Additionally, at most one of the triangles BCD, BDE or BEA can be large. Therefore, the number of small triangles is at least 6.

Now we assume that all boundary triangles are small. Let $P$ be the intersection point of diagonals AD and CE. Without loss of generality, we can assume that $d(B,AD)\le d(C,AD)$, where $d(X,YZ)$ denotes the distance from point $X$ to line $YZ$ in the Euclidean plane. Then we can see that $S_{\{BPD\}}\le S_{\{CPD\}}\le S_{\{CED\}}$. Also $S_{\{ABP\}}\le \max \{S_{\{ABC\}},S_{\{ABE\}}\}$. Since $S_{\{ABD\}}=S_{\{ABP\}}+S_{\{BPD\}}$, we have either $S_{\{ABD\}}\le S_{\{ABC\}}+S_{\{CED\}}=S-S_{\{ACE\}}<S/2$ or $S_{\{ABD\}}\le S_{\{ABE\}}+S_{\{CED\}}=S-S_{\{BCE\}}<S/2$. In both cases, we have a contradiction. Therefore, there must exist small and center triangle. The completes the solution.