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Question Number 106029 by ajfour last updated on 02/Aug/20

Commented by ajfour last updated on 02/Aug/20

$F i n d c o o r d i n a t e s o f A (x_{A}, y_{A}) i f$ $a l l f o u r q u a d r a n t s r e c e i v e e q u a l$ $a r e a p o r t i o n s o f △ A B C . (x_{A} ⩾ y_{A})$ $w i t h n e i t h e r o f A, B, C o n t h e a x e s .$
	Answered by ajfour last updated on 02/Aug/20

	$M (p, q) a n d l e t \frac{s \sqrt{3}}{2} = h$ $l e t M b e m i d p o i n t o f B C . A M = h$ $x_{A} = p - h \cos θ, y_{A} = q - h \sin θ$ $x_{B} = p - \frac{s}{2} \sin θ, y_{B} = q + \frac{s}{2} \cos θ$ $x_{C} = p + \frac{s}{2} \sin θ, y_{C} = q - \frac{s}{2} \cos θ$ $E q . o f B C :$ $y - q = - \frac{1}{\tan θ} (x - p)$ $x_{G} = p + q \tan θ$ $E q . o f A C :$ $y - q + h \sin θ = (\frac{\tan θ + \frac{1}{\sqrt{3}}}{1 - \frac{\tan θ}{\sqrt{3}}}) (x - p + h \cos θ)$ $x_{E} = (h \sin θ - q) (\frac{1 - \frac{\tan θ}{\sqrt{3}}}{\tan θ + \frac{1}{\sqrt{3}}}) + p - h \cos θ$ $y_{D} = (h \cos θ - p) (\frac{\tan θ + \frac{1}{\sqrt{3}}}{1 - \frac{\tan θ}{\sqrt{3}}}) + q - h \sin θ$ $E q . o f A B :$ $y - q + h \sin θ = (\frac{\tan θ - \frac{1}{\sqrt{3}}}{1 + \frac{\tan θ}{\sqrt{3}}}) (x - p + h \cos θ)$ $y_{F} = (h \cos θ - p) (\frac{\tan θ - \frac{1}{\sqrt{3}}}{1 + \frac{\tan θ}{\sqrt{3}}}) + q - h \sin θ$ $. . . . . v e r y t e d i o u s . . .!$
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