QUESTION
ABC is an isoceles triangle inscribed in a circle of radius $r$.If AB=AC and $h$ is the altitude from A to BC and $p$ be the perimiter of ABC then $$\lim_{h\to0} \frac{\Delta}{p^3}$$ (where $\Delta$ is the area of triangle )
MY ATTEMPT
first of all putting down the values of $\Delta$ and $p$ $$\frac{1}{2}\lim_{h\to0}\frac{h(BC)}{(AB+AC+BC)^3}=\frac{1}{2}\lim_{h\to0}\frac{h(BC)}{(2AB+BC)^3}$$$$=\frac{1}{2}\lim_{h\to0}\frac{h(BC)}{8(AB)^3+12(AB)^2(BC)+6(AB)(BC)^2+(BC)^3}$$ if i convert all the terms in terms of $\Delta$ and multiplying h in $Nr$ and $Dr$ $$\lim_{h\to0}\frac{\Delta.h}{8h(AB)^3+\Delta.24(AB)^2+\Delta.12(AB)(BC)+\Delta.2(BC)^2}=\lim_{h\to0}\frac{\Delta.h^2}{8h^2(AB)^3+\Delta.24.h(AB)^2+\Delta^2.24(AB)+\Delta^2.4(BC)}$$
HELP
i really dont know what to do after this is anything there which i am missing ?
can i write $\lim_{\Delta\to 0}$ as $\lim_{h\to0}$ ? i dont think so
please help
Try to understand what's happening during the limiting process $h\to0$: The base $BC$ is translated vertically upwards to the apex $A$, all the while staying horizontal.
Describe the various quantities playing a rôle here in terms of the half angle $\alpha$ at $A$ (which tends to ${\pi\over2}$ in the process). You then obtain an expression $${\Delta \over p^3}=\Psi(\alpha)\ ,$$ and have to determine $\lim_{\alpha\nearrow{\pi\over2}}\Psi(\alpha)$, whereby $r$ is considered constant.