19. Let $S$ be a set of five different positive integers, the largest of which is $m$. It is impossible to construct a quadrilateral with non-zero area, whose side-lengths are all distinct elements of $S$.
What is the smallest possible value of $m$ ?
A 2
B 4
C 9
D 11
E 12
Solution
D
Let $W X Y Z$ be a quadrilateral, with non-zero area. We note first that the shortest route from $W$ to $Z$ is the line segment $W Z$.

Since $W X Y Z$ has non-zero area, the points $X$ and $Y$ are not both on the line segment $W Z$. Therefore the path made up of the line segments $W X, X Y$ and $Y Z$ is not the shortest route from $W$ to $Z$.

It follows that $W X+X Y+Y Z>W Z$.
Conversely, it may be seen that when the sum of the three smallest of four positive numbers is greater than the largest number of the four, then there is quadrilateral with non-zero area which has these four positive numbers as its side lengths. (You are asked to check this in Exercise 19.1)
Suppose $h, j, k, l$ and $m$ are five positive integers with $h<j<k<l<m$, which have the required property that, whichever four of them are selected, these are not the side lengths of a quadrilateral with non-zero area. Then, whichever four of these numbers are selected, the sum of the three smallest is no greater than the largest.
In Exercise 19.2 you are asked to check that, provided that $h+j+k \leq l$ and $j+k+l \leq m$, the integers $h, j, k, l$ and $m$ satisfy the condition that, whichever four of them are selected, the sum of the three smallest is no greater than the largest.

We wish to choose $m$ to be as small as possible. To achieve this we need $h, j, k$ and $l$ to be as small as possible. The smallest possible values for $h, j$ and $k$, are $h=1, j=2$ and $k=3$. This gives $h+j+k=1+2+3=6$. Because we need to have $h+j+k \leq l$, the smallest possible value of $l$ is 6 . Then $j+k+l=2+3+6=11$. Because we need to have $j+k+l \leq m$, the smallest possible value of $m$ is 11 .