Question

# ABC is an isosceles triangle inscribed in a circle of radius r. If $$AB=AC$$ and h is the altitude from A to BC. If the triangle ABC has perimeter $$P$$ and area $$\Delta$$ then $$\displaystyle \lim_{h\rightarrow 0}512\:r\:\frac{\Delta }{P^{3}}$$ is equal to

Solution

## We have $$BC=2BD, AD=h$$ and $$OD=h-r$$.$$\therefore \displaystyle BC=2\sqrt{r^{2}-(h-r)^{2}}=2\sqrt{2hr-h^{2}}$$$$\Rightarrow \displaystyle AB=\sqrt{2hr-h^{2}+h^{2}}=\sqrt{2hr}$$so that $$P=2AB+BC$$$$=\displaystyle 2[\sqrt{2hr-h^{2}}+\sqrt{2hr}]$$Also the area of $$\Delta ABC$$ is$$\displaystyle \Delta =BD\cdot AD=h\sqrt{2hr-h^{2}}$$$$\therefore \displaystyle \frac{\Delta }{P^{3}}=\frac{h\sqrt{2hr-h^{2}}}{8(\sqrt{2hr-h^{2}}+\sqrt{2hr})^{3}}$$$$=\displaystyle \frac{h\sqrt{2hr-h}}{8(\sqrt{2hr-h}+\sqrt{2r})^{3}}$$$$\Rightarrow \displaystyle \lim_{h\rightarrow 0}512r\frac{\Delta }{P^{3}}=512\:r\:\frac{\sqrt{2r}}{8(2\sqrt{2r})^{3}}=4$$Mathematics

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