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Question Number 36270 by ajfour last updated on 30/May/18

Commented by ajfour last updated on 30/May/18

$F i n d m a x i m u m a r e a o f a t r i a n g l e$ $w h e n i t s v e r t i c e s l i e o n t h e$ $c i r u m f e r e n c e o f t h r e e c i r c l e s .$ $(A s s h o w n i n f i g . t h e c i r c l e s$ $t o u c h e a c h o t h e r) .$
Commented by tanmay.chaudhury50@gmail.com last updated on 31/May/18

$i h a v e d e l e t e d t h e p o s t . . . y e s i w a s w r o n g$
Commented by MJS last updated on 31/May/18

$anyway thank you for all the input, {it}^{'} s$ $good to have many people with many$ $different points of view$
Commented by ajfour last updated on 31/May/18

Commented by ajfour last updated on 31/May/18

$m r W S i r, i m a g a i n w i s h i n g$ $y o u r c o o p e r a t i o n . . .$
Commented by ajfour last updated on 31/May/18

$l e t D E b e x - a x i s w i t h D a s o r i g i n .$ $l e t ∠ F D E = α$ $A \equiv (x_{1}, y_{1}), B \equiv (x_{2}, y_{2}), C \equiv (x_{3}, y_{3})$ $x_{1}^{2} + y_{1}^{2} = p^{2}$ ${(x_{2} - p - q)}^{2} + y_{2}^{2} = q^{2}$ ${[x_{3} - (p + r) \cos α]}^{2} + [y_{3} - (p + r) \sin α] = r^{2}$ $△ = \frac{1}{2} \| \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} \| w i t h$ $\frac{\partial △}{\partial x_{1}} = 0, \frac{\partial △}{\partial x_{2}} = 0, \frac{\partial △}{\partial x_{3}} = 0$ $\Rightarrow \| \begin{matrix} 1 & - \frac{x_{1}}{y_{1}} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} \| = 0$ $\Rightarrow (y_{2} - y_{3}) + \frac{x_{1}}{y_{1}} (x_{2} - x_{3}) + (x_{2} y_{3} - x_{3} y_{2}) = 0$ $s i m i l a r l y t w o m o r e e q u a t i o n s$ $f o r \frac{\partial △}{\partial x_{2}} = 0, \frac{\partial △}{\partial x_{3}} = 0 .$ $u s i n g t h e s e s i x e q u a t i o n s w e f i n d$ $(x_{i}, y_{i}) a n d t h u s △_{m a x} .$
Commented by MrW3 last updated on 11/Jun/18
Great!
Commented by MrW3 last updated on 12/Jun/18

$t h e t r i a n g l e w i t h t h e m a x . a r e a s h o u l d$ $b e t h a t o n e w h i c h f u l f i l l s :$ $t a n g e n t a t A i s p a r a l l e l t o B C$ $t a n g e n t a t B i s p a r a l l e l t o C A$ $t a n g e n t a t C i s p a r a l l e l t o A B$
	Answered by MJS last updated on 02/Jun/18

	$in A the tangent must be parallel to the line$ $M_{q} M_{r}, same goes for B & M_{p} M_{r} and C & M_{p} M_{q}$ $I came to this conclusion by calculating it for$ $p = q = r and then reducing r . I tried to find the$ $formulas but {couldn}^{'} t handle them . with given$ $values for p, q, r {it}^{'} s easy to calculate and see$ $that changing one of the verticles A, B, C$ $(use △ θ) by any small value will decrease the$ $area .$ $I^{'} ll try to show . . .$
	Commented by MrW3 last updated on 12/Jun/18

	$I t h i n k y o u r c o n c l u s i o n i s n o t c o r r e c t, s i r .$ $T h e t a n g e n t a t A s h o u l d b e p a r a l l e l t o t h e$ $l i n e B C, n o t t o t h e l i n e M_{q} M_{r}, e t c .$
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