In a square, there are K points ( $K>2$ ). Into what smallest number of triangles should the square be divided so that no more than one point is located in each triangle?
The solution assumes that points are not allowed to lie on the boundary of the square or on the boundaries of the triangles. If we assume that they can, the solution remains the same, except for the case $K=1$.

We need the following lemma:

Lemma. Suppose there are $k>0$ points located in triangle ABC. Then the triangle can be divided into $k$ triangles such that each contains exactly one point.

Proof of the lemma. We will prove this lemma by induction on $k$. The base case $k=1$ is obvious: the only triangle in the partition will be the original triangle. Now, suppose for all $k1$ the square can be divided into $K+1$ triangles. To do this, note that there will be at least one vertex of the square for which not all points lie on a line passing through it. Let this vertex be A of the square ABCD. Then there exists a point P on side BC such that none of the given points lie on segments AP and DP, and not all given points lie in one of the triangles ABP, APD, and DCP. Applying the lemma to each of these triangles, we get that there exists a partition into $K+1$ triangles that satisfies the problem's condition.

Now we will prove that a smaller number of triangles is not always sufficient. Let $O$ be the center of the given square ABCD. Place $K$ points on the interval BO. Suppose there exists a partition of the square into $K$ triangles that satisfies the problem's condition. Then each triangle contains exactly one marked point, and each marked point lies in exactly one triangle (i.e., a point cannot lie on the boundary of two triangles). In this partition, consider all triangles with vertex $D$. If there are several such triangles, then at least one of them either does not contain the line $BD$, or one of its sides lies on the line $BD$. In neither of these cases can there be marked points in this triangle, which contradicts what has been proven. Therefore, if such a triangle exists, it can only be one. But then two of its sides lie on segments $AD$ and $CD$, meaning this triangle is contained in triangle $ACD$, and thus no marked point can lie in it. We have reached a contradiction. Thus, we have proven that a partition of the square into $K$ triangles that satisfies the problem's condition is impossible.

## Answer

3, if $k=1$, $k+1$ triangles, if $k>1$.