The Longest Side

Let $a$ , $b$ , $c$ be the lengths of the three sides of the triangle. Denote the semiperimeter of the triangle by $s$ and its circumradius by $R$ . Let $a+b=S$ and $ab=P$ . Then we have

$a+b+c=20\Leftrightarrow c=20-a-b\Leftrightarrow c=20-S \ \ \ \ \ (1)$ By Heron’s formula for the area of the triangle,

$\begin{aligned} \left[ ABC \right]=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)} & \Rightarrow \sqrt{10\left( 10-a \right)\left( 10-b \right)\left( a+b-10 \right)}=18 \\ & \Leftrightarrow 10\left( 100-10\left( a+b \right)+ab \right)\left( a+b-10 \right)={{18}^{2}} \\ & \Leftrightarrow 10\left( 100-10S+P \right)\left( S-10 \right)=324 \ \ \ \ \ (2) \\ \end{aligned}$ Again, for the triangle's area we have

$\left[ ABC \right]=\dfrac{abc}{4R}\overset{\left( 1 \right)}{\mathop{\Rightarrow }}\,18=\dfrac{P\left( 20-S \right)}{4\times 4}\Leftrightarrow P=\dfrac{288}{20-S} \ \ \ \ \ (3)$ Substituting, $\begin{aligned} \left( 2 \right) & \overset{\left( 3 \right)}{\mathop{\Leftrightarrow }}\,10\left( 100-10S+\dfrac{288}{20-S} \right)\left( S-10 \right)=324 \\ & \Leftrightarrow 25{{S}^{3}}-1000{{S}^{2}}+13301S-58820=0 \\ & \Leftrightarrow \left( 5S-68 \right)\left( 5{{S}^{2}}-132S+865 \right)=0 \\ & \Leftrightarrow S=\dfrac{68}{5}\text{, or }S=\dfrac{66\pm \sqrt{31}}{5} \\ \end{aligned}$

For $S=\dfrac{68}{5}$ , $\left( 3 \right)\Leftrightarrow P=45$ .

Now, we can find the values of $a$ and $b$ , solving the quadratic equation ${{x}^{2}}-Sx+P=0\Leftrightarrow {{x}^{2}}-\dfrac{68}{5}x+45=0$ which gives $a,b=\dfrac{34\pm \sqrt{31}}{5}$ Using $(1)$ , we find $c=20-\dfrac{68}{5}\Leftrightarrow c=\dfrac{32}{5}$ If we take $S=\dfrac{66+\sqrt{31}}{5}$ , or $S=\dfrac{66-\sqrt{31}}{5}$ , repeating the same process we get the same three values for the side-lengths of the triangle (perhaps in different order):

$\dfrac{34+\sqrt{31}}{5}, \ \ \ \ \ \dfrac{34-\sqrt{31}}{5}, \ \ \ \ \ \dfrac{32}{5}$ Thus, the length of the longest possible side is $\dfrac{34+\sqrt{31}}{5}$ For the answer, $p=34$ , $q=31$ , $r=5$ , thus, $p+q+r=\boxed{70}$ .

From the perimeter equation $P = a + b + c$ , we know that:

$a + b + c = 20$

From the circumradius equation $2R = \cfrac{abc}{2T}$ , we know that $2 \cdot 4 = \cfrac{abc}{2 \cdot 18}$ , which rearranges to:

$abc = 288$

By Heron's formula , $T = \sqrt{s(s - a)(s - b)(s - c)}$ . Substituting $s = \frac{1}{2}P = \frac{1}{2} \cdot 20 = 10$ and $T = 18$ gives $18 = \sqrt{10(10 - a)(10 - b)(10 - c)}$ , which rearranges to $\cfrac{162}{5} = 1000 - 100(a + b + c) + 10(ab + ac + bc) - abc = 1000 - 100(20) + 10(ab + ac + bc) - 288$ , so that $ab + ac + bc = \cfrac{3301}{25}$ .

Now $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)$ , and substituting the above values gives $20^2 = a^2 + b^2 + c^2 + 2 \cdot \cfrac{3301}{25}$ which rearranges to:

$a^2 + b^2 + c^2 = \cfrac{3398}{25}$

From $a + b + c = 20$ we know that $c = 20 - a - b$ , and substituting this into $a^2 + b^2 + c^2 = \cfrac{3398}{25}$ and solving for $b$ gives $b = \frac{1}{10} (\sqrt{-75 a^2 + 1000 a - 3204} - 5a + 100)$ , and substituting $c$ and then $b$ into $abc = 288$ and then rearranging gives:

$25a^3 - 500a^2 + 3301a - 7200 = (5a - 32)(5a^2 - 68a + 225) = 0$

The three solutions for $a$ , which correspond to the three possible sides of the triangle, are $a = \cfrac{32}{5}$ and $a = \cfrac{34 \pm \sqrt{31}}{5}$ , the largest of which is $a = \cfrac{34 + \sqrt{31}}{5}$ .

Therefore, $p = 34$ , $q = 31$ , $r = 5$ , and $p + q + r = \boxed{70}$ .