# Quadrilateral inside a square

There is a square with a side length of 1. Inside this square there is a quadrilateral. Each vertex of the quadrilateral is on a side of the square. The area of the quadrilateral is bigger than half (the area of the square is 1).

You need to prove that there is a line segment inside the quadrilateral that is parallel to one of the square's sides and is longer than 1/2. • Each point of the quadrilateral is on a side of the square. Wouldn't this imply that the quadrilateral is the square? Or do you mean vertex? – CodeNewbie Jul 12 '15 at 11:53

## 2 Answers

This can be shown relatively trivially using basic integration (or equivalently, breaking the problem into parts).

Suppose that you have two functions, $f_1(x)$ and $f_2(x)$ - each is a piecewise linear function, with $f_1(0)=f_2(0)$ and $f_1(1)=f_2(1)$, as well as $\max f_1(x) = 1$ and $\min f_2(x)=0$.

Now, the area of the quadrilateral is $$\int_0^1 (f_1(x)-f_2(x))dx\leq 1\times \max (f_1(x)-f_2(x))$$ Because the integral must be greater than $\frac12$, the maximal value of $f_1(x)-f_2(x)$ must also be greater than $\frac12$.

• Interesting. Since the orientation of the $x$ axis is not specified, this proof shows that there is an internal line segment longer than 0.5 in any orientation, not just parallel to one of the sides of the square. – Lawrence Jul 12 '15 at 14:38
• One too might note that the only critical points of $f_1(x)-f_2(x)$ are at the vertices of the quadrilateral - so such a line may originate from one of the corners. (And, this still holds if we generalize as @Lawrence suggests) – Milo Brandt Jul 12 '15 at 16:25

Since each vertex must lie on a side of the square, here's one simple example to disprove your claim. Without having a line parallel to any side, we can have a quadrilateral whose area is more than half of the square (although it would have a side that needs to have a length greater than 0.5). The square in black is the square of unit side and unit area, while the quadrilateral in red has an area that is quite visibly more than half of the square and also has each vertex on the square. Hence, what you want proven has been easily disproved with this simple example.

• But here you CAN draw vertical line segment inside, which will be longer than 1/2. So this is not a counterexample. – Somnium Jul 12 '15 at 13:37
• For future reference, this answer and frodoskywalker's answer were both posted before the question received a clarifying edit. – Lopsy Jul 12 '15 at 13:42