Any ideas? Seems like a difficult problem. We can remove the constraint that no three points are collinear as it only makes the problem more confusing, and also WLOG the boundary of the convex hull of the points contains all the points. Some equality cases: For $3\not | k$, take $k$ points to be the vertices of a regular $k$-gon. Take $2017-k$ more points along the boundary of the $k$-gon. Now take any affine transformation on the $2017$ points and this gives $2017$ good triangles.

Elementary my friend!

Let P(n) be the proposition about n, for n is a natural number>3. P(n) is true when there exists more than n good triangles among n points. By Euler,if P(n) is true, there must exist the smallest n s.t. P(n) is true. Let n be the smallest when n=k. i.e.P(k) is true. When n=k-1, P(k-1) is true. P(k-1) is also true by removing a point inside a good triangle s.t. it is not a common point of good triangles. Assume that we need to remove that point inside a good triangleABC. Assume that there are three points D,E and F(they may not exist) s.t. ABDC,ABCE,ACBF are parallelogram. There must not exist points outside triangle DEF, if not,any point lying outside triangle DEF(e.g.G) forms a triangle with larger area than triangleABC, so make it not a good triangle. i.e. all the points lie inside DEF. If we cannot remove A,B or C, we can remove one point among the points of the other good triangles s.t. the point removed is not a common point of good triangles. After removing that point, only one good triangle is removed. It is always possible to find a point like that because the case of all vertices of good triangles are common points of good triangles if and only if D,Eor F exists. In this case, there is no point point lying outside ABDC,ABCE or ACBF in order to make triangle ABC a good triangle. But there are at most 4 good triangles in the case, where 4 points have existed. Therefore the proposition is not true for this case. Thus, if P(k) is true, P(k-1) is also true, which violates with n=k is the smallest n, s.t. P(n) is true. This implies that P(n) is not true for all n, where n is a natural number>3. In other words,among n points, there will never exist n good triangles. Sub n=2017 and we have the answer.

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Billy Can you explain why it is always possible to find a point like that because the case of all vertices of good triangles are common points of good triangles if and only if D,Eor F exists.

Well, I've been not posting this for a while hoping there'd be some interesting proofs. But it seem not, so I'll post it now, for the sake of completeness of this thread too. The problem can be written in the following form: Given $n$ distinct points on the plane, the number of triangle formed by these vertices with maximum possible area is not more than $n$. Turn out the problem was first asked, at least as far as I know, by Erdös and Purdy in 1971. However, they didn't provide the exact optimal bound $n$. For the prove of the bound $n$, I'm not sure where it first appeared, but this paper contain the prove with clear explanation.

During the test no one managed to solve problem 4.

Let $S$ be a set of all good triangles and assume $|S|\geq 2017$ and $A_{i}$ for $i={1,2,3...,2017 }$ be all given points.Then by pigeonhole principle we can find segment $A_{i}A_{j}$ which used in at least $3$ good triangles.(Obvious).Let $A_{k},A_{l},A_{m}$ be points where $X=\triangle A_{i}A_{j}A_{k};Y=\triangle A_{i}A_{j}A_{l};Z=\triangle A_{i}A_{j}A_{m}$ are good.Then $S(X)=S(Y)=S(Z)$ so if $a,b,c$ be heights from $A_{k},A_{l},A_{m}$ to$A_{i}A_{j}$ then $2S(X)=a*A_{i}A_{j}=2S(Y)=b*A_{i}A_{j}=2S(Z)=c*A_{i}A_{j}$ so $a=b=c$ so $A_{k}A_{l}A_{m}||A_{i}A_{j}$ in other words $A_{k},A_{l}$ and $A_{m}$ are colinear which is contradiction for problems condition.

$\text{NOTE:}$ We can generalise that there can not be more than $2n+1$ triangles with equal areas in triangles formed by $2n+1$ points no three of them collinear. For evens my pigeonhole may be don't works.

qweDota wrote: Then by pigeonhole principle we can find segment $A_{i}A_{j}$ which used in at least $3$ good triangles.(Obvious). This is wrong, using pigeonhole principle we can only say that we find a point $A_{i}$ which is a vertex in at least $3$ good triangles. Check it, and please, dont use Obvious if you are not sure how pigeonhole principle works.

WHO CAN THIS PROBLEM SOLUTION WITH SIMPLE PROOFS