Max area of triangle -PHP How do i prove that the maximum area that can be obtained among 3 random points in a square is half the area of the square?-
I need it to for the following question
" Show that among any 9 points inside a triangle of area 1 there are
three points which form a triangle of area at most 1/4." Hence there would be at least 3 points in a square of side 1/2. But then how do i proceed?
 A: Actually you don't need to prove this to solve your problem, although it is a good question. For your problem, just take the triangle formed by the three midpoints of the sides of the bigger triangle. This triangle divides the bigger in 4 triangles with area $1/4$. Then, given 9 points in the bigger triangle, by the pigeonhole principle, at least 3 of them are in one triangle of area $1/4$.
For the different question, here is the proof. Given 3 points in the square, if the points are not at the sides, we can have a different triangle with its vertex at the sides of the square with bigger area than the previous one, e.g.

If we now have a triangle with the vertives at the sides, if one of them is not a vertex of the square, we can get a new triangle with bigger area by moving the vertex such that the new side is parallel to the previous one:

Now, let $a$ and $b$ be the sizes of the segments in the image:

The area $A$ of the triangle is
$$A=1-\frac{a}{2}-\frac{b}{2}-\frac{(1-a)(1-b)}{2}=\frac{1-ab}{2}.$$
So, for the maximum area, we must minimize $a$ or $b$ to $0$, which yields $A=1/2$.
A: Let's call the vertices of the triangle $A,B,C$ and the ones of the square $a,b,c,d$.
If the three points are all in a half of the square (by a half I mean one of the triangles $abc, bcd, cda, dab$) you have the claim, since the area of $ABC$ is less than 1/2.
If the points are not so "nice", then:
1) consider the vertex of the triangle that has the shortest distance from all vertices of the square (let's say that that the minimum distance is from $A$ to $a$);
2) translate the triangle $ABC$ by the vector $Aa$. Now $A$ coincides with $a$ and it's easy to see that $B$ and $C$ are always inside the square;
3) draw the line containing $B$ and parallel to $AC$ and call $D$ the intersection between this line and one of the side of the square adjacent to $A$. The triangle $ACD$ has the same area as $ABC$.
4) draw the line containing $C$ and parallel to $AD$ and call $E$ the intersection between this line and one of the side of the square adjacent to $A$. The triangle $ADE$ has the same area as $ACD$.
Now, the triangle $ADE$ is equivalent to $ABC$ and is contained in one "half" of the square. You can reach this even at the end of point 2) or 3).
So the area of $ABC$ is always less or equal to half the area of $abcd$.
