Quadrilateral 1-2-3-4

Diagram for side lengths 1-2-3-4 clockwise; $90^\circ$ between $m=1, n=2;$ $a=\sqrt{5}, b=3, c=4, \text{Area}[a\text{-}b\text{-}c]=\frac{1}{4}\sqrt{4a^2b^2-\big(a^2+b^2-c^2\big)^2}$

The diagram to the right is drawn with the case of quadrilateral 1-2-3-4 (clockwise) in mind: $m=1, n=2, a=\sqrt{1^2+2^2}=\sqrt{5}, b=3, c=4.$ . But we can also use it to explain other cases. Note that, with $m$ and $n$ fixed, switching $b$ and $c$ doesn't change the area of quadrilateral $m$ - $n$ - $b$ - $c$ .

Case 1: $m=2, n=4$
$\quad$ Since $a=\sqrt{2^2+4^2}=2\sqrt{5}>1+3=4,$ we cannot make a quadrilateral in this case.

Case 2: $m=3, n=4$
$\quad$ Since $a=\sqrt{3^2+4^2}=5>1+2=3,$ we cannot make a quadrilateral in this case either.

Case 3: $m=1, n=2, b=3, c=4 \implies \text{Area} = 1+\sqrt{11}\approx 4.32$
$\quad$ This is the case in the diagram. With $m=1$ and $n=2,$ we have $a=\sqrt{1^2+2^2}=\sqrt{5}.$ The remaining two sides are $b=3$ and $c=4.$ By Heron's formula , the area of triangle $a$ - $b$ - $c$ is $\frac{1}{4}\sqrt{4a^2b^2-\big(a^2+b^2-c^2\big)^2}=\frac14\sqrt{4\times 5\times9-(-2)^2}=\frac14\sqrt{176}=\frac14\sqrt{16\times 11}=\sqrt{11}.$ Since the area of right triangle $m$ - $n$ - $a$ is $\frac12\times 1 \times 2=1,$ the area of quadrilateral 1-2-3-4 is $1+\sqrt{11}.$

Case 4: $m=1, n=3, b=2, c=4 \implies \text{Area} = \frac32+\sqrt{\frac{39}2}\approx 4.62$

Case 5: $m=1, n=4, b=2, c=3 \implies \text{Area} = 2+2\sqrt{2}\approx 4.83$

Case 6: $m=2, n=3, b=1, c=4 \implies \text{Area} = 3+\sqrt{3}\approx 4.73$

So, we can conclude as follows:

• There are 4 cases where we can make quadrilaterals. Since we can switch $b$ and $c$ in each case, the number of different shapes is $w=4\times 2=8.$
• Since switching $b$ and $c$ doesn't change the area, the number of distinct areas that these 8 shapes have is $x=4.$
• Since the area of the quadrilateral is the smallest when the right angle is between side lengths 1 and 2, $y=1$ and $z=2$ $($ or $y=2$ and $z=1).$

Therefore, $w+x+y+z=8+4+1+2=15.$ $_\square$