<s>[INST] Beautifully looking little geometry/trigonometry problem Given triangle ABC, a,b,c as its sides, p is a half perimeter, such that $\dfrac{p-a}{11}=\dfrac{p-b}{12}=\dfrac{p-c}{13}$. 
We need to find $(\tan\dfrac{A}{2})^2$
(A)$\dfrac{143}{432}$  B)$\dfrac{13}{33}$ C)$\dfrac{11}{39}$ D)$\dfrac{12}{37}$
I've tried to substitute the area of the triangle as $S=(1/2)(ab\sin C)=(1/2)(ac\sin B)=(1/2)(bc\sin A)$ and I know that $(\tan A/2)^2=(1-\cos A)/(1+\cos A)$ And some other formulas, but how is $\tan A/2$  derived from the given? 
Sorry, the problem's given was confused, p is half-perimeter.
 [/INST] Let:  $$x=\dfrac{p-a}{11}=\dfrac{p-b}{12}=\dfrac{p-c}{13}$$
Thus, $11x+12x+13x=3p-(a+b+c)$, and so $36x=p$.
Denoting $r$ as inradius and $\Delta$ as area, we that:
$$r=\frac{\Delta}{s}=\frac{\sqrt{p(p-a)(p-b)(p-c)}}{p}=\sqrt{\frac{11\cdot 12\cdot 13\cdot x^3}{36x}}$$
And:
$$\tan^2\frac{A}2=\frac{r^2}{(p-a)^2}=\frac{11\cdot 12\cdot 13\cdot x^2}{36\cdot (11x)^2}=  { ...}$$
As to why $\tan\frac{A}2=\frac{r}{p-a}$, look at this diagram, and ask if you have any further doubts :)

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