My first geometry problem

To solve this problem, we need 3 inequalities, and $WLOG, cos A \leq cos B$ which follows that $csc A \leq csc B$ , $A \geq B$ and $B < 90^\circ$

Let $S$ be the area of the triangle

Then $S$ $= \frac{absin C}{2}$ $\leq$ $\frac {ab}{2}$ $\ldots (1)$

Also $sin (90^\circ - X)$ $\geq$ $sin (k - X)$ $\ldots (2)$ where

$k \leq$ $90^\circ$ and $X$ is acute

and $2 = \frac{cos A} {sin B} + \frac{cos B} {sin A} \leq\ \frac{cos A} {sin A} + \frac{cos B} {sin B} = cot A + cot B$ $\ldots (3)$

Which is derived from the WLOG condition and rearrangement inequality

Thus, from $(3)$ $S = \frac{absin C}{2} = \frac{absin (A+B)}{2} \geq absin A sin B$

From $(1)$ , $\frac {ab}{2} \geq absin A sin B$ which follows that

$sinA sinB \leq \frac{1}{2}$

For acute angle $B$ , $sin B sin (90 - B) \leq \frac{1}{2}$

From $(2)$ , $sin (k - B) sin B \leq sin B sin (90 - B) \leq \frac{1}{2}$

This implies $A + B \leq 90^\circ or A - B \geq 90^\circ$

Equality occurs when $A = B = 45^\circ$

Thus, the area is $(\frac{12}{\sqrt{2} + 2})^2 = 108 - 72\sqrt{2}$ as desired.

$SinA=CosB ~and~SinB=CosA~would~satisfy~the~equation.\\ Under~this~condition~for~angles~of~a~triangle~A+B=90^o.\\ So~A=B=45^o.~\implies~the~triangle~is~isosceles~right~angled.\\ So~if~a~is~one~leg,~perimeter=a+a+\sqrt2*a=12.\\ \therefore~a=\dfrac{12}{2+\sqrt2}\\ So~the~area=\frac 1 2 *a^2=\frac 1 2*\left(\dfrac{12}{2+\sqrt2} \right)^2\\ =\huge \color{#D61F06}{6.176623509}.$

Clearly A = B = 45. Triangle will then be isosceles. Apply Pythagoras Theorem to evaluate the third side in terms of the two equal sides. Equate the sum of three sides to the perimeter. Find the sides. Compute the area from this information to obtain answer as 6.206