<s>[INST] Find $\frac{AC \times BC}{AD \times BD}$ 
$AC$ is $2004$. $CD$ bisects angle $C$. If the perimeter of $ABC$ is $6012$, find $\dfrac{AC \times BC}{AD \times BD}$.


Attempt
Let $c = AD+BD$.
We have that $\dfrac{AC}{AD} = \dfrac{BC}{BD}$. Thus, $$\dfrac{BD}{AD}+1 = \dfrac{BC}{AC}+1 \implies \dfrac{BD+AD}{AD} = \dfrac{AC+BC}{AC} = \dfrac{c}{AD} \implies AD = \dfrac{AC \cdot c}{AC+BC}.$$ Likewise it is easy to see by the angle bisector theorem that $BD = \dfrac{BC \cdot c}{AC+BC}$. Thus, we have that $\dfrac{AC \times BC}{AD \times BD} = \dfrac{AC \times BC}{\dfrac{AC \cdot BC \cdot c^2}{(AC+BC)^2}} = \dfrac{(AC+BC)^2}{c^2} = \dfrac{(2004+BC)^2}{(4008-BC)^2}$. But this answer seems to depend on $BC$, which makes no sense as the question indicates it is constant. What have I done wrong?
 [/INST] The problem is not well-posed. In fact, $BC$ can be anything between $1002$ and $3006$ (not inclusive), and the quantity in question can be any number greater than $1$ and less than $25$.
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