Prove that there exists a triangle with 3 vertices of the above 16 points whose area does not exceed $10cm^2$

Question

Take 13 points inside the triangle, the area is $\text{210 cm}$$^2$, so that along with the 3 vertices of the triangle we have 16 points with no 3 points collinear. Prove that there exists a triangle with 3 vertices of the above 16 points whose area does not exceed $\text{10 cm}$$^2$ My ideas: I think with 16 points above will draw $n(n \geq21)$ triangles that do not overlap so if we assume all triangles have an area greater than $\text{10 cm}$$^2$, we will find a contradiction. And I'm trying to determine the value of $n$. I hope your help

Answer 1

There will be $27$ non-overlapping triangles in the triangulation. One way of constructing them could be:

• You start with a single triangle.

• Each additional point introduced into a (sub-)triangle can be joined to the three vertices of that (sub-)triangle, replacing the (sub-)triangle with three sub-triangles, i.e. increasing the number of non-overlapping triangles by two.

• Since you introduce $13$ points, the number of triangles in the triangulation is $1+2\times 13=27$

A less constructive approach:

• We have $n$ interior triangles, or $n+1$ if we also count the exterior triangle
• These require $\frac32(n+1)$ edges since each edge is on two triangles
• Euler's formula $v-e+f=2$ implies there must be $v= \frac32(n+1) - (n+1) +2 = (n+5)/2$ vertices
• With the number of vertices equal to $v$ gives $n=2v-5$ as the number of interior triangles in the triangulation
• Setting $v=16$ gives $n=27$ interior triangles