I dont see anything hard though?

pretty sure that there are only $10$ possible cases for making a triangle, then i could possible then be able to do this

thdnder wrote: I dont see anything hard though? Okay bro 3 points? The following solution was awarded $5$ points which resulted in me not qualifying IMO. Throughout the solution, we take the indices modulo $5$. Answer: $4$. Construction: Obvious from attachment #2. Bound: Assume the contrary and let $A_1A_2A_3A_4A_5$ (in that order) be a pentagon which contains at least $5$ big triangles. Let $S$ denote the total area of the pentagon. 1. There exist an index $i$ such that the triangles $A_iA_{i+1}A_{i+2}$ and $A_{i+1}A_{i+2}A_{i+3}$ have areas smaller than $\frac{S}{4}$. We say that a triangle is a side triangle if it consists of $3$ consecutive vertices of the pentagon. I claim that a side triangle cannot be big. Suppose the contrary and by symmetry, we may assume that $A_5A_1A_2$ is big. Then, the area of the quadrilateral $A_2A_3A_4A_5$ is smaller than $\frac{S}{2}$, which implies that all big triangles must have a vertex in $A_1$. We can divide the pentagon into three triangles $A_1A_2A_3$, $A_1A_3A_4$ and $A_1A_4A_5$ upon which it is evident that at most one of these triangles are big. It folows that there are at most $4$ big triangles, a contradiction. Since there are exactly $5$ triangles $(A_1A_3A_4, A_2A_4A_5, A_3A_5A_1, A_4A_1A_2, A_5A_2A_3)$ that are not side triangles, each one of them must be big. Therefore, by PHP, for each $i$, at least one of the triangles $A_iA_{i+1}A_{i+2}$ and $A_iA_{i+4}A_{i+3}$ have area smaller than $\frac{S}{4}$ since $A_iA_{i+1}A_{i+2}$ is a big triangle. We may conclude that at least $3$ of the side triangles have area smaller than $\frac{S}{4}$, which implies our claim. $\square$ 2. Conclusion. So, the idea here is to fix the index $i$ and the vertices $A_i, A_{i+1}, A_{i+2}$ and $A_{i+3}$ satisfying 1. and show that we can't pick the final vertex so that it satisfies our conditions. For the sake of convenience, let $A \equiv A_i$, $B \equiv A_{i+1}$, $C \equiv A_{i+2}$ and $D \equiv A_{i+3}$, and let the final vertex be $T$. Let $AB \cap CD = Y$. 2.1 $Y$ has to be in the opposite side of $B$, $C$ with respect to the line $AD$. Suppose the contrary. Then, it is evident that $\frac{S}{2} < S_{ADC} < S_{BDC}$ which contradicts 1. (View attachment #3.) $\square$ 2.2 The point $T$ doesn't exist. View attachment #1 while reading this part. Suppose henceforth that our pentagon doesn't satisfy 2.1. Note that since the points $ABCDT$ have to be in that order, $T$ has to be in the opposide side of the points $B$ and $C$ about $AD$. Let $l_1$ be the line parallel to $CD$ passing through $A$ and let $l_2$ be the line parallel to $AB$ passing through $D$. And let $X = l_1 \cap l_2$. I claim that $T$ must be inside the triangle $ADX$. Indeed, if $T$ is on the other side of the points $B$ and $C$ with respect to the line $AX$, we get $d(T, CD) > d(A, CD)$ which implies that $S_{TCD} > S_{ACD} > \frac{S}{2}$, a contradiction. By symmetry $T$ must be on the same side as the points $B$ and $C$ with respect to the line $DX$. This implies our claim. Now, let $B'$ be the point symmetric to $B$ about $A$ and let $l_3$ be the line passing through $B'$ parallel to $BC$. Define $C'$ and $l_4$ similarly. Let $E$ and $F$ be the intersection points of the line $B'C'$ and the lines $AX$ and $BX$ respectively. We prove that $T$ lies in $XEF$. Observe that it suffices to show that $T$ lies on the opposite side of the points $B$ and $C$ with respect to the lines $l_3$ and $l_4$. Indeed, if $T$ is on the same side as the points $B$ and $C$ with respect to the line $l_3$, we get $d(T, BC) \leq 2d(A, BC)$ which implies that $S_{TBC} \leq 2S_{ABC} \leq \frac{S}{2}$ which contradicts the fact that $TBC$ is a big triangle. By symmetry, $T$ lies on the opposide side of $B$ and $C$ with respect to $l_4$. Now, we are in the home stretch. Observe that the value of $d(T, AD)$ shifts linearly when $T$ shifts across the line $B'C'$. Since the points $E$ and $F$ are located between the points $B'$ and $C'$, this implies that $\min (d(E', AD); d(F', AD)) \geq \min(d(B', AD); d(C', AD))$. This implies that $S_{TAD} \geq \min(S_{B'AD}, S_{C'AD}) = \min (S_{BAD}, S_{CAD}) > \frac{S}{2}$, a contradiction by 1. $\blacksquare$

Iveela wrote: thdnder wrote: I dont see anything hard though? Okay bro 3 points? The following solution was awarded $5$ points which resulted in me not qualifying IMO. Throughout the solution, we take the indices modulo $5$. Answer: $4$. Construction: Obvious from attachment #2. Bound: Assume the contrary and let $A_1A_2A_3A_4A_5$ (in that order) be a pentagon which contains at least $5$ big triangles. Let $S$ denote the total area of the pentagon. 1. There exist an index $i$ such that the triangles $A_iA_{i+1}A_{i+2}$ and $A_{i+1}A_{i+2}A_{i+3}$ have areas smaller than $\frac{S}{4}$. We say that a triangle is a side triangle if it consists of $3$ consecutive vertices of the pentagon. I claim that a side triangle cannot be big. Suppose the contrary and by symmetry, we may assume that $A_5A_1A_2$ is big. Then, the area of the quadrilateral $A_2A_3A_4A_5$ is smaller than $\frac{S}{2}$, which implies that all big triangles must have a vertex in $A_1$. We can divide the pentagon into three triangles $A_1A_2A_3$, $A_1A_3A_4$ and $A_1A_4A_5$ upon which it is evident that at most one of these triangles are big. It folows that there are at most $4$ big triangles, a contradiction. Since there are exactly $5$ triangles $(A_1A_3A_4, A_2A_4A_5, A_3A_5A_1, A_4A_1A_2, A_5A_2A_3)$ that are not side triangles, each one of them must be big. Therefore, by PHP, for each $i$, at least one of the triangles $A_iA_{i+1}A_{i+2}$ and $A_iA_{i+4}A_{i+3}$ have area smaller than $\frac{S}{4}$ since $A_iA_{i+1}A_{i+2}$ is a big triangle. We may conclude that at least $3$ of the side triangles have area smaller than $\frac{S}{4}$, which implies our claim. $\square$ 2. Conclusion. So, the idea here is to fix the index $i$ and the vertices $A_i, A_{i+1}, A_{i+2}$ and $A_{i+3}$ satisfying 1. and show that we can't pick the final vertex so that it satisfies our conditions. For the sake of convenience, let $A \equiv A_i$, $B \equiv A_{i+1}$, $C \equiv A_{i+2}$ and $D \equiv A_{i+3}$, and let the final vertex be $T$. Let $AB \cap CD = Y$. 2.1 $Y$ has to be in the opposite side of $B$, $C$ with respect to the line $AD$. Suppose the contrary. Then, it is evident that $\frac{S}{2} < S_{ADC} < S_{BDC}$ which contradicts 1. (View attachment #3.) $\square$ 2.2 The point $T$ doesn't exist. View attachment #1 while reading this part. Suppose henceforth that our pentagon doesn't satisfy 2.1. Note that since the points $ABCDT$ have to be in that order, $T$ has to be in the opposide side of the points $B$ and $C$ about $AD$. Let $l_1$ be the line parallel to $CD$ passing through $A$ and let $l_2$ be the line parallel to $AB$ passing through $D$. And let $X = l_1 \cap l_2$. ($1$ point deduction for not stating that $X \equiv P_{\inf}$ when $AB \parallel CD$) I claim that $T$ must be inside the triangle $ADX$. Indeed, if $T$ is on the other side of the points $B$ and $C$ with respect to the line $AX$, we get $d(T, CD) > d(A, CD)$ which implies that $S_{TCD} > S_{ACD} > \frac{S}{2}$, a contradiction. By symmetry $T$ must be on the same side as the points $B$ and $C$ with respect to the line $DX$. This implies our claim. Now, let $B'$ be the point symmetric to $B$ about $A$ and let $l_3$ be the line passing through $B'$ parallel to $BC$. Define $C'$ and $l_4$ similarly. Let $E$ and $F$ be the intersection points of the line $B'C'$ and the lines $AX$ and $BX$ respectively. We prove that $T$ lies in $XEF$. Observe that it suffices to show that $T$ lies on the opposite side of the points $B$ and $C$ with respect to the lines $l_3$ and $l_4$. Indeed, if $T$ is on the same side as the points $B$ and $C$ with respect to the line $l_3$, we get $d(T, BC) \leq 2d(A, BC)$ which implies that $S_{TBC} \leq 2S_{ABC} \leq \frac{S}{2}$ which contradicts the fact that $TBC$ is a big triangle. By symmetry, $T$ lies on the opposide side of $B$ and $C$ with respect to $l_4$. (Here, i simply wrote $d(T, BC) \leq 2d(A, BC) \leq \frac{S}{2}$ by accident which lead to an another $1$ point deduction. Bye bye IMO.) Now, we are in the home stretch. Observe that the value of $d(T, AD)$ shifts linearly when $T$ shifts across the line $B'C'$. Since the points $E$ and $F$ are located between the points $B'$ and $C'$, this implies that $\min (d(E', AD); d(F', AD)) \geq \min(d(B', AD); d(C', AD))$. This implies that $S_{TAD} \geq \min(S_{B'AD}, S_{C'AD}) = \min (S_{BAD}, S_{CAD}) > \frac{S}{2}$, a contradiction by 1. $\blacksquare$

I've heard that problem selection commitee actually found way easier solution than your solution, which led to give 3 points on deriving 5 big triangles must be $A_iA_{i+2}A_{i+4}$ for $1 \le i \le 5$. And they actually tried that problem before putting it, and some guy in problem selection commitee found the following solution. So unfortunate for you. Let $ABCDE$ be the our pentagon. It's easy to conclude that 5 big triangles must be $ACD, BDE, CEA, DAB, EBC$. Let $P = AD \cap CE$. WLOG, we may assume that $d(B, AD) \le d(C, AD)$. In this case, we have $[BPD] \le [CPD] \le [CED]$. (As usual, $[P_1P_2\dots P_n]$ is the area of $P_1P_2\dots P_n$) And we have $[ABP] \le \max([ABC], [ABE])$ and $[ABD] = [ABP] + [BPD]$ implies that $[ABD] \le [ABC] + [CED] = [ABCDE] - [ACE] < \frac{[ABCDE]}{2}$, if $\max([ABC], [ABE]) = [ABC]$ and $[ABD] \le [ABE] + [CED] = [ABCDE] - [BCE] < \frac{[ABCDE]}{2}$, otherwise. Thus, in either case, we get a contradiction. $\blacksquare$

how was this the only C lmao