Question Number 15765 by ajfour last updated on 13/Jun/17
Commented by ajfour last updated on 13/Jun/17
$${Q}.\mathrm{15736}\:\:\:\left({alternate}\:{solution}\right) \\ $$$${Area}\:\Delta{APF}\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({Area}\:\Delta{ABP}\right) \\ $$$$ \\ $$$${Maximum}\:{area}\:{of}\:\Delta{APF} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\left({maximum}\:{Area}\:{of}\:\Delta{ABP}\right) \\ $$$$\:\:\:\:\:\:{Area}\:{of}\:{inscribed}\:\Delta{ABP}\:{is} \\ $$$${maximum}\:{when}\:{it}\:{is}\:{equilateral}. \\ $$$$\:\:\:\:\:\:\:\:\:{x}={r}\mathrm{cos}\:\mathrm{30}°\:=\frac{{r}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{y}={x}\mathrm{tan}\:\mathrm{60}°\:=\:\frac{\mathrm{3}{r}}{\mathrm{2}} \\ $$$${Max}.{Area}\:{of}\:\Delta{APF}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{r}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}{r}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}\sqrt{\mathrm{3}}{r}^{\mathrm{2}} }{\mathrm{8}}\:. \\ $$
Commented by mrW1 last updated on 13/Jun/17
$$\mathrm{very}\:\mathrm{clever}! \\ $$