A Pre-RMO question! -15

Note that $12$ can't be written as sum of two squares of integers $\because 12=2^2\times3;3≡3(\mod 4)$ see Sum of two squares theorem

Therefore, $12$ can't be the hypotenuse of the triangle. $\therefore \exists a∈Z^+,b∈Z^+|12^2+b^2=a^2$ $\rightarrow 144+b^2=a^2$ $\rightarrow 144=(a-b)(a+b)$ $\because a∈Z^+;b∈Z^+,a>b \therefore (a-b)∈Z^+,(a+b)∈Z^+;a-b<a+b$ Pair wise factors of $144=(a+b)(a-b)=144\times1;72\times2;48\times3;36\times4;24\times6;18\times8;16\times9$ $(a+b)(a-b)=144\times1;72\times2;48\times3;36\times4;24\times6;18\times8;16\times9$

Suppose it is given $a+b=k_1;a-b=k_2\rightarrow 2a=k_1+k_2 \rightarrow a=\frac{k_1+k_2}{2}$ as $a∈Z \therefore (k_1+k_2)≡0(\mod 2)$ $\rightarrow (a+b)(a-b)=\cancel{144\times1};\blue{72\times2};\cancel{48\times3};\blue{36\times4};\blue{24\times6};\blue{18\times8};\cancel{16\times9};$ Now $perimeter=12+a+b\therefore$ for largest perimeter we need largest $a+b$ . $\therefore a+b=72;a-b=2$ $perimeter=12+72=\boxed{84}$

To maximize the perimeter 12 should not be hypotenuse.

Let the hypotenuse be h and other side be b.

Now,from pythagoras theorem,

12^2+b^2=h^2

(h+b)(h-b)=144

since we want to maximize perimeter so,we should maximize (h+b)

(h+b)(h-b)=144*1

h+b = 144 and h-b =1

But,in this case h and b ain't integer.

so,our remaining option is (h+b)(h-b) = 72*2

h+b=72

h-b=2

So,h=37 and b=35

Therefore maximum perimeter is 84

Cheers😊