The area of the triangle is less than $30 cm^2$.

Question

Given a rectangle with dimensions $10$cm and $6$cm, show that for every $3$ points in the interior of the rectangle, the area of the triangle is less than $30$ cm$^2$.

I draw the diagonals but now I am stuck.

Answer 1

The area of the rectangle is $60 cm^2$. Consider a triangle with the base as one of the sides of the rectangle.The the 3rd vertex be located on the opposite side. Clearly the area of this triangle is $30cm^2$. But since all vertices of the triangle need to be inside the rectangle, the distance from the base to the opposite vertex or the height will be reduced, resulting in an area lesser than $30cm^2$. You can extend this argument to the triangle have base as one diagonal and and the opposite vertex as a vertex of the rectangle.The area of this triangle is also $30cm^2$. But the base has to be smaller than the diagonal, again resulting in a smaller area.

Answer 2

Let $x_{min}$ be the smallest x coordinate among the three points. Similarly for $x_{max}$, $y_{min}$, and $y_{max}$. Clearly $x_{max}-x_{min} < 10$ and $y_{max}-y_{min} < 6$, so all you have to do is show that the area of the triangle is equal to $(x_{max}-x_{min})(y_{max}-y_{min})/2$

Answer 3

Let's have the rectangle $ABCD$ and $P$, $Q$, and $R$ three points inside. Let's use $A=(0,0)$, $B=(a,0)$, $C=(a,b)$, and $D=(0,b)$. The $x$ coordinates of the three interior points are in the interval$(0,a)$ and the $y$ coordinates are in the $(0,b)$. From $P,\ Q,\ R$, choose the point with minimum $x$ and draw a parallel to $y$ axis. Repeat the procedure for the point with maximum $x$ coordinate. Now do the same for the $y$ coordinate. You get a rectangle $A'B'C'D'$ that is in the interior of $ABCD$ (so the area is smaller). At least one of $P,\ Q,\ R$ is in the corner of this rectangle. Without loss of generality, assume $P=A'$, $Q$ is on $B'C'$, and $R$ is on $C'D'$. The lengths of the rectangle sides are $a'$ and $b'$. The coordinates of $Q$ with respect to $P$ are $(a', y)$, and $R$ is at $(x,b')$. We have $0\le x\le a'$ and $0\le y\le b'$. Now let's write the area of the $PQR$ triangle by subtracting from the area of $A'B'C'D'$ the triangles that are formed with the sides: $$\begin{align}A_{PQR}&=A_{A'B'C'D'}-A_{QA'B'}-A_{RA'D'}-A_{QRC'}\\&=a'b'-\frac12a'y-\frac12b'x-\frac12(a'-x)(b'-y)\\&=\frac12 a'b'-\frac12xy\le\frac12a'b'\lt \frac12 ab\end{align}$$