Please if someone can provide a solution !

Notice the bound $2n-5$ cannot be lowered, since a planar triangulation using $n$ vertices has precisely $2n-5$ (bounded) facets (triangles).

Let $S=\{S_1,S_2,\cdots, S_n\}.$ A triangle, henceforth, will be referred to as the union of its perimeter and interior. The crucial lemma is as follows: Lemma: Given a segment $\overline{AB}$ and a set of points $X=\{X_1,X_2,\cdots, X_m\}$ such that all of the $X_i$ lie on the same half-plane WRT $AB,$ the intersection $\bigcap_{i=1}^{m}\triangle ABX_i$ is nonempty. Proof: We will actually show that the intersection is a nondegenerate triangle with $AB$ as a side. The case $m=3$ is trivial. Assume the claim is true for $m=k,$ and that the desired intersection is $\triangle ABP.$ Then from the convexity of $ABPX_{k+1},$ we have exactly one of $AP\cap BX_{k+1},BP\cap AX_{k+1}$ nonempty. WLOG let the latter case be true. Then if $BP\cap AX_{k+1}=P',$ $\triangle ABP\cap \triangle ABX_{k+1}=\triangle ABP'$ which is clearly nondegenerate. Hence by induction, we may conclude. Back to the main problem: Let the convex hull of $S$ have $k$ vertices. The constructions for $k=3,4$ are trivial. If $k>4,$ we can have a construction with $n$ points by taking a point in $\triangle S_iS_{i+1}S_{i+2}\cap \triangle S_{i+1}S_{i+2}S_{i+3}$ for all $1\leq i\leq n,$ where indices are taken $\mod n.$ Since $n<2n+5,$ this suffices. Now the problem is trivial. For each point $X$ within the convex hull of $S,$ the line $XS_1$ partition the rest of the $S_i$ into two half planes, so by Lemma, at most $1$ new point is necessary for each half plane, for a total of $2$ new points for each point in the convex hull of $S,$ so we may conclude.

Generic_Username wrote: ...Let the convex hull of $S$ have $k$ vertices. The constructions for $k=3,4$ are trivial. If $k>4,$ we can have a construction with $n$ points by taking a point in $\triangle S_iS_{i+1}S_{i+2}\cap \triangle S_{i+1}S_{i+2}S_{i+3}$ for all $1\leq i\leq n,$ where indices are taken $\mod n.$ Since $n<2n+5,$ this suffices. Now the problem is trivial. For each point $X$ within the convex hull of $S,$ the line $XS_1$ partition the rest of the $S_i$ into two half planes, so by Lemma, at most $1$ new point is necessary for each half plane, for a total of $2$ new points for each point in the convex hull of $S,$ so we may conclude. But apart from that triangles, there are many others, such as: $S_i S_j S_k$ (where i,j,k are not consequtive), $X_iX_jS_k$, $X_i X_jX_k$, where $X_i,X_j, X_k$ are inside the convex hull. You should also take care of them!?

orl wrote: $ S$ be a set of $ n$ points in the plane. No three points of $ S$ are collinear. Prove that there exists a set $ P$ containing $ 2n - 5$ points satisfying the following condition: In the interior of every triangle whose three vertices are elements of $ S$ lies a point that is an element of $ P.$ Let $\vec v$ be a vector with $\vec v\nparallel \vec{S_iS_j}$ for $S_i,S_j\in S$. $\{\vec s\pm\varepsilon\vec v\mid \vec s\in S\}$ is a set of $2n$ points satisfying the condition for sufficent small $\varepsilon>0$. At least 5 points are not inside the convex hull and therefore not needed.

pi_quadrat_sechstel wrote: Let $\vec v$ be a vector with $\vec v\nparallel \vec{S_iS_j}$ for $S_i,S_j\in S$. $\{\vec s\pm\varepsilon\vec v\mid \vec s\in S\}$ is a set of $2n$ points satisfying the condition for sufficent small $\varepsilon>0$. At least 5 points are not inside the convex hull and therefore not needed. Let $s_1,s_2\in S$. Why are you sure that there is a point of $S$ in the interior of the triangle determined by the points $\vec{s_1}+\varepsilon\vec{v}, \vec{s_1}-\varepsilon\vec{v}, \vec{s_2}+\varepsilon\vec{v}$?

dgrozev wrote: pi_quadrat_sechstel wrote: Let $\vec v$ be a vector with $\vec v\nparallel \vec{S_iS_j}$ for $S_i,S_j\in S$. $\{\vec s\pm\varepsilon\vec v\mid \vec s\in S\}$ is a set of $2n$ points satisfying the condition for sufficent small $\varepsilon>0$. At least 5 points are not inside the convex hull and therefore not needed. Let $s_1,s_2\in S$. Why are you sure that there is a point of $S$ in the interior of the triangle determined by the points $\vec{s_1}+\varepsilon\vec{v}, \vec{s_1}-\varepsilon\vec{v}, \vec{s_2}+\varepsilon\vec{v}$? Of course, there is not necessarily a point of $S$ in a triangle with vertices in $P$. The condition is that in the interior of any triangle whose vertices are in $S$ is a point of $P$. You switched $S$ and $P$.

My bad, I obviously forgot the statement! Nice solution!

Let the convex hull of $S$ form an $m$-gon. Can the given condition be modified such that "$2n+5$" is changed to "$2n-m-2$" and still hold true? If anyone can construct a counterexample (perhaps I missed something silly), please let me know. thx.