Geometry Question by Mithil Shah # 5

Let the side lengths and the area of the triangle be $a$ , $b$ and $c=21$ , and $A$ respectively. Then $a+b+c=48$ , $\implies a+b=27$ and $b=27-a$ . Let $s=48/2=24$ . By Heron's formula , we have:

$\begin{aligned} \sqrt{s(s-a)(s-b)(s-c)} & = A \\ \sqrt{24(24-a)(24-27+a)(24-21)} & = A \\ \sqrt{72(24-a)(a-3)} & = A \\ 6^2\cdot 2(24-a)(a-3) & = A^2 & \small \color{#3D99F6} \text{To equate perfect squares both sides} \\ \implies (24-a)(a-3) & = 2\color{#3D99F6}n^2 & \small \color{#3D99F6} \text{where }n \in \mathbb N \end{aligned}$

Equating $\begin{cases} 24-a = 2n \\ a-3 = n \end{cases}$ $\implies 24-a=2(a-3)$ $\implies a = 10$ $\implies b = 17$

The shortest side length is $a=\boxed{10}$ .

$Heron's~ formula.~~Let~integers~~a>b,~since~a+b=27,~~~b<14~~~c=21.\\ Area=\frac 1 4*\sqrt{\{a+b+c\}*\{a+b-c\}\{c+(a-b)\}*\{c-(a-b)\} }\\ Area=\frac 1 4*\sqrt{\{(a+b)^2-c^2\}\{c^2-(a-b)^2\} }\\ Area=\frac 1 4*\sqrt{\{27^2-21^2\}\{21^2-(a-b)^2\} }\\ Area=\frac 1 4*\sqrt{2*144*\{21^2-(a-b)^2\} }~~ must~be~an~integer.\\ Area=3*\sqrt{2*\{21^2-(a-b)^2\} }~~ must~be~an~integer~say=3n.\\ \implies~2*\{21^2-(a-b)^2\} =n^2 .\\ \therefore ~ \text{We can check for b=4,5. . . 13. } \\ \text{But before checking all values, let us check if (a-b)=3 or 7, factors of 21.}\\ So~~for~(a-b)=3~:-~~ 2*\{21^2-3^2\}=2*432=2^4*3^2*6,~~not~a~square~number. \\ So~~for~(a-b)=7~:-~~ 2*\{21^2-7^2\}=2*7^2(9-1)=2^4*7^2~~~a~square~number.\\ \therefore~(a-b)=7,~~so~(a,b) = (17,10). \\ So~smallest~ is ~\Large ~\color{#D61F06}{10}.$