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Question Number 186464 by mr W last updated on 04/Feb/23

Commented by mr W last updated on 04/Feb/23

$g e n e r a l c a s e f o r Q 186302 :$ $f i n d t h e m a x i m u m a r e a o f t h e$ $t r i a n g l e .$
Commented by Frix last updated on 05/Feb/23

$Still that left vertex must be on the horizontal$ $diameter . (The area seems to be r h ?)$ $Another question : What if the right angle$ $has got a different fixed value ?$
Commented by Frix last updated on 05/Feb/23

$No . . .$ $Maybe we can try to see it the other way$ $round : Find the minimum circle for a given$ $rectangular triangle with center on the$ $intersection of a line through the vertex$ $at the right angle and a line in a right$ $angle to the first one which goes through$ $that other vertex . This seems to be a$ $special group of circles . . .$
Commented by mr W last updated on 05/Feb/23

$t h i s i s t r u e . b u t {i t}^{'} s n o t o b v i o u s t o m e .$ $o r c a n y o u s i m p l y e x p l a i n t h a t t h e$ $l e f t v e r t e x m u s t l i e o n t h e h o r i z o n t a l$ $d i a m e t e r s u c h t h a t t h e a r e a i s$ $m a x i m u m ?$
Commented by mr W last updated on 05/Feb/23

$i t i s p r o v e d a s s h o w n b e l o w t h a t f o r$ $t h e m a x i m u m t r i a n g l e t h e r i g h t$ $a n g l e v e r t e x m u s t l i e o n t h e h o r i z o n t a l$ $d i a m e t e r . t h e q u e s t i o n i s o n l y t h a t$ $t h i s r e s u l t i s n o t o b v i o u s . a l s o f o r$ $t h e q u e s t i o n f o r m i n i m u m c i r c l e$ $a s y o u s e g g u e s t e d a b o v e i t i s n o t$ $o b v i o u s w h e r e i t s c e n t e r s h o u l d l i e .$ $s o i t {d o e s n}^{'} t h e l p, b e c a u s e w e {d o n}^{'} t$ $n e e d a c a l c u l a t i o n p r o o f (w h i c h w e$ $h a v e a l r e a d y), b u t a n o b v i o u s$ $e x p l a n a t i o n .$
Commented by Frix last updated on 05/Feb/23

$Start with any triangle A B C .$ $The centers of circles through A and C lie$ $on the symmetry axe of A C .$ $Let D the center of the circle {we}^{'} re looking$ $for . The additional condition is C D ⊥ B D .$ $There are only 2 possible centers D .$ $Draw it, {you}^{'} ll see it .$
Commented by mr W last updated on 05/Feb/23

$i g o t t h e s i m p l e e x p l a n a t i o n w h y$ $f o r t h e m a x i m u m t r i a n g l e t h e v e r t e x$ $m u s t l i e o n t h e h o r i z o n t a l d i a m e t e r .$
Commented by mr W last updated on 05/Feb/23

Commented by mr W last updated on 05/Feb/23

${i t}^{'} s e a s y t o s e e$ $Δ = [B C A] = [{B C A}^{'}] = \frac{h d}{2}$ $d_{m a x} = d i a m e t e r = 2 r$ $\Rightarrow Δ_{m a x} = \frac{h \times 2 r}{2} = h r$ $w h e n v e r t e x C l i e s o n t h e h o r i z o n t a l$ $d i a m e t e r .$
Commented by ajfour last updated on 05/Feb/23

$Δ = [B C A] = [{B C A}^{'}] u n d e r s t o o d$ $b u t = \frac{h d}{2} (h o w ?)$
Commented by mr W last updated on 05/Feb/23

Commented by mr W last updated on 05/Feb/23

$[{B C A}^{'}] = [B C O] + [{B O A}^{'}]$ $= \frac{B O \times d_{1}}{2} + \frac{B O \times d_{1}}{2}$ $= \frac{B O \times 2 d_{1}}{2}$ $= \frac{h \times d}{2}$
Commented by ajfour last updated on 05/Feb/23

$i g u e s s y o u m e a n$ $= \frac{(h + s) d_{1}}{2} + \frac{(h - s) d_{1}}{2} = \frac{h (d_{1} + d_{2})}{2} = \frac{h d}{2}$ $T h a n k s s i r, i u n d e r s t a n d t h e$ $e x p l a n a t i o n p r e t t y w e l l n o w,$ $a w e s o m e r e a s o n i n g!$
Commented by mr W last updated on 06/Feb/23

$i m e a n t \frac{{h d}_{1}}{2} + \frac{{h d}_{1}}{2} = \frac{h d}{2} i n d e e d .$ $i h a d a t y p o w i t h C O \times d_{1} . i t i s B O \times d_{1} .$
Commented by mr W last updated on 05/Feb/23

	Answered by mr W last updated on 04/Feb/23

	Commented by mr W last updated on 05/Feb/23

	$a_{1} a = c_{1} c = (h - r) (h + r) = h^{2} - r^{2}$ $a_{1} = \frac{h^{2} - r^{2}}{a}$ $b^{2} = c^{2} - a^{2}$ $b^{2} = {(2 r)}^{2} - {(a - a_{1})}^{2}$ $c^{2} - a^{2} = 4 r^{2} - {(a - a_{1})}^{2}$ $c^{2} = 4 r^{2} + (2 a - a_{1}) a_{1}$ $c^{2} = 4 r^{2} + (2 a - \frac{h^{2} - r^{2}}{a}) \times \frac{h^{2} - r^{2}}{a}$ $\Rightarrow c^{2} = 2 (h^{2} + r^{2}) - \frac{{(h^{2} - r^{2})}^{2}}{a^{2}}$ $\Rightarrow b^{2} = c^{2} - a^{2} = 2 (h^{2} + r^{2}) - \frac{{(h^{2} - r^{2})}^{2}}{a^{2}} - a^{2}$ $a r e a Δ = \frac{a b}{2}$ $l e t Φ = 4 Δ^{2} = a^{2} b^{2}$ $Φ = a^{2} [2 (h^{2} + r^{2}) - \frac{{(h^{2} - r^{2})}^{2}}{a^{2}} - a^{2}]$ $Φ = 2 (h^{2} + r^{2}) a^{2} - a^{4} - {(h^{2} - r^{2})}^{2}$ $Φ = 2 (h^{2} + r^{2}) a^{2} - a^{4} - {(h^{2} + r^{2})}^{2} + 4 h^{2} r^{2}$ $\Rightarrow Φ = 4 h^{2} r^{2} - {(a^{2} - h^{2} - r^{2})}^{2}$ $w e s e e t h e m a x i m u m o f Φ i s 4 h^{2} r^{2}$ $w h e n a^{2} = h^{2} + r^{2} .$ $t h a t m e a n s t h e m a x i m u m a r e a i s$ $Δ_{m a x} = h r, w h e n t h e r i g h t a n g l e v e r t e x$ $l i e s o n t h e h o r i z o n t a l d i a m e t e r .$
	Commented by mr W last updated on 05/Feb/23

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