Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 60313 by ajfour last updated on 19/May/19

Commented by ajfour last updated on 19/May/19

$Find side length l of largest equilateral$ $triangle circumscribing a given$ $△ ABC (sides a, b, c) .$
	Answered by ajfour last updated on 19/May/19

	$This question {wasn}^{'} t solved$ $satisfactorily before, i just now$ $find l_{m a x} = \sqrt{\frac{2}{3} (a^{2} + b^{2} + c^{2}) + \frac{8}{\sqrt{3}} (△_{ABC})} .$
	Commented by ajfour last updated on 21/May/19

	$dont you like this solution mrW Sir,$ $it is to meet your previous objection$ $to a previous coordinate geometry$ $based solution of mine which {wasn}^{'} t$ $as symmetric in a, b, c as this; if you$ $remember it at all . .$
	Commented by mr W last updated on 22/May/19

	$y o u r s o l u t i o n i s w o n d e r f u l s i r . t h e$ $r e s u l t i s i n p e f e c t f o r m . i {d o n}^{'} t$ $k n o w h o w y o u c a m e t o t a k i n g t h e p o i n t s$ $M a n d N . i w o u l d t r y t o t a k e ∠ A B R = θ$ $a s v a r i a b l e a n d g e t l = f (θ) a n d \frac{d l}{d θ} = 0 .$
	Commented by ajfour last updated on 21/May/19

	$constancy of the vertex angles at$ $P, Q, R led me into thinking this$ $Sir!$
	Commented by mr W last updated on 21/May/19

	$g r e a t t h o u g h t s i r! t h i s g i v e s t h e$ $a n s w e r t o m y q u e s t i o n h o w t o d e t e r m i n e$ $t h e m a x . c i r c u m s c r i b i n g e q u i l a t e r a l$ $t r i a n g l e g e o m e t r i c a l l y,$ $i . e . w i t h o u t u s i n g c a l c u l u s .$ $i h a v e b e e n t h i n k i n g a b o u t$ $t h i s f o r t w o d a y s . n o w i k n o w h o w .$ $i t a n s w e r s a l s o t h e q u e s t i o n t h a t$ $t h e s e a r c h e d t r i a n g l e i s a m a x i m u m,$ $n o t a m i n i m u m .$
	Commented by mr W last updated on 21/May/19

	$m y r e s u l t u s i n g g e o m e t r i c a l m e t h o d$ $i s t h e s a m e a s y o u r m e t h o d u s i n g$ $c a l c u l u s, s e e b e l o w .$
	Answered by ajfour last updated on 20/May/19

	Commented by ajfour last updated on 20/May/19
	$cos 30° = ((BC/2)/(CM)) ⇒ ((√3)/2)=(a/2)×(1/(CM)) ⇒ PM=CM=(a/(√3)) (similarly BN=(c/(√3)) ) .....(ii) ∠BMP=180°−60°−θ BP=2PMsin (((∠BMP)/2)) BP=((2a)/(√3))sin (60°−(θ/2)) ......(i) let ∠B=β and since ∠PBR=180°, φ+30°+β+30°+(30°+(θ/2))=180° ⇒ φ=90°−β−(θ/2) BR = 2BNcos φ =((2c)/(√3))sin (β+(θ/2)) let side of equilateral △PQR be l. l= BP+BR = ((2a)/(√3))sin (60°−(θ/2))+((2c)/(√3))sin (β+(θ/2)) ⇒l(√(3 ))=a((√3)cos (θ/2)−sin (θ/2)) +2c(sin βcos (θ/2)+cos βsin (θ/2)) l(√3) =(2ccos β−a)sin (θ/2) +(2csin β+a(√3))cos (θ/2) ((d(l(√3)))/dθ)=0 ⇒ tan (θ/2)=((2ccos β−a)/(2csin β+a(√3))) l(√3)= (2ccos β−a){sin (θ/2)+((cos^2 (θ/2))/(sin (θ/2)))} l(√3) = ((2ccos β−a)/(sin (θ/2))) l_(max) (√3) = (√((2ccos β−a)^2 +(2csin β+a(√3))^2 )) =(√(4a^2 +4c^2 +4ac((√3)sin β−cos β))) = (√(4a^2 +4c^2 +8(√3)((1/2)acsin β)−4ac(((a^2 +c^2 −b^2 )/(2ac))))) ⇒l(√3) = (√(2(a^2 +b^2 +c^2 )+8(√3)△_(ABC) )) ___________________________ l_(max) =(√((2/3)(a^2 +b^2 +c^2 )+(8/(√3))△_(ABC) )) ___________________________.$
	$\cos 30 ° = \frac{BC / 2}{CM} \Rightarrow \frac{\sqrt{3}}{2} = \frac{a}{2} \times \frac{1}{CM}$ $\Rightarrow PM = CM = \frac{a}{\sqrt{3}}$ $(similarly BN = \frac{c}{\sqrt{3}}) . . . . . (ii)$ $∠ BMP = 180 ° - 60 ° - θ$ $BP = 2 PMsin (\frac{∠ BMP}{2})$ $BP = \frac{2 a}{\sqrt{3}} \sin (60 ° - \frac{θ}{2}) . . . . . . (i)$ $let ∠ B = β and since ∠ PBR = 180 °,$ $ϕ + 30 ° + β + 30 ° + (30 ° + \frac{θ}{2}) = 180 °$ $\Rightarrow ϕ = 90 ° - β - \frac{θ}{2}$ $BR = 2 BNcos ϕ$ $= \frac{2 c}{\sqrt{3}} \sin (β + \frac{θ}{2})$ $let side of equilateral △ PQR be l .$ $l = BP + BR$ $= \frac{2 a}{\sqrt{3}} \sin (60 ° - \frac{θ}{2}) + \frac{2 c}{\sqrt{3}} \sin (β + \frac{θ}{2})$ $\Rightarrow l \sqrt{3} = a (\sqrt{3} \cos \frac{θ}{2} - \sin \frac{θ}{2})$ $+ 2 c (\sin β \cos \frac{θ}{2} + \cos β \sin \frac{θ}{2})$ $l \sqrt{3} = (2 ccos β - a) \sin \frac{θ}{2}$ $+ (2 csin β + a \sqrt{3}) \cos \frac{θ}{2}$ $\frac{d (l \sqrt{3})}{d θ} = 0 \Rightarrow$ $\tan \frac{θ}{2} = \frac{2 ccos β - a}{2 csin β + a \sqrt{3}}$ $l \sqrt{3} = (2 c \cos β - a) {\sin \frac{θ}{2} + \frac{\cos^{2} \frac{θ}{2}}{\sin \frac{θ}{2}}}$ $l \sqrt{3} = \frac{2 ccos β - a}{\sin \frac{θ}{2}}$ $l_{m a x} \sqrt{3} = \sqrt{{(2 c \cos β - a)}^{2} + {(2 csin β + a \sqrt{3})}^{2}}$ $= \sqrt{{4 a}^{2} + {4 c}^{2} + 4 ac (\sqrt{3} \sin β - \cos β)}$ $= \sqrt{{4 a}^{2} + {4 c}^{2} + 8 \sqrt{3} (\frac{1}{2} acsin β) - 4 ac (\frac{a^{2} + c^{2} - b^{2}}{2 ac})}$ $\Rightarrow l \sqrt{3} = \sqrt{2 (a^{2} + b^{2} + c^{2}) + 8 \sqrt{3} △_{ABC}}$ $___________________________$ $l_{m a x} = \sqrt{\frac{2}{3} (a^{2} + b^{2} + c^{2}) + \frac{8}{\sqrt{3}} △_{ABC}}$ $___________________________.$
	Commented by ajfour last updated on 20/May/19

	$much symmetric this time!$ $say if a = b = c$ $l = \sqrt{\frac{2}{3} (3 a^{2}) + \frac{8}{\sqrt{3}} \times \frac{\sqrt{3}}{4} a^{2}} = 2 a .$ $i f c = 0 \Rightarrow a = b$ $l = \sqrt{\frac{2}{3} (2 a^{2}) + 0} = \frac{2 a}{\sqrt{3}} .$
	Answered by mr W last updated on 21/May/19

	Commented by mr W last updated on 03/Feb/24

	${P Q}_{m a x} = 2 A B, i . e . w h e n C P a n d C Q a r e$ $d i a m e t e r s o f t h e c i r c l e s . P Q / / A B .$
	Commented by mr W last updated on 21/May/19

	$t h i s i s t h e g e o m e t r i c a l s o l u t i o n :$ $c o n s t r u c t p o i n t M a n d p o i n t N s u c h$ $t h a t ∠ B M C = 120 ° a n d ∠ A N B = 120 ° .$ $c o n s t r u c t c i r c l e w i t h p o i n t M a s$ $c e n t e r a n d t h r o u g h B a n d C .$ $c o n s t r u c t c i r c l e w i t h p o i n t N a s$ $c e n t e r a n d t h r o u g h A a n d B .$ $b o t h c i r c l e s i n t e r s e c t a t p o i n t D .$ $c o n s t r u c t d i a m e t e r D P a n d D R .$ $P R i s t h e s i d e l e n g t h o f t h e m a x i m u m$ $e q u i l a t e r a l t r i a n g l e P Q R w h i c h w e$ $a r e s e a r c h i n g .$
	Commented by mr W last updated on 21/May/19

	$M B = M C = \frac{a}{2} \times \frac{2}{\sqrt{3}} = \frac{a}{\sqrt{3}}$ $N B = N A = \frac{c}{2} \times \frac{2}{\sqrt{3}} = \frac{c}{\sqrt{3}}$ $∠ M B N = ∠ B + 60 °$ ${M N}^{2} = {B N}^{2} + {B M}^{2} - 2 \times B M \times B N \times \cos (B + 60 °)$ $= \frac{a^{2}}{3} + \frac{c^{2}}{3} - \frac{2 a c}{3} (\frac{1}{2} \cos B - \frac{\sqrt{3}}{2} \sin B)$ $= \frac{a^{2} + c^{2}}{3} - \frac{a c \cos B}{3} + \frac{1}{\sqrt{3}} a c \sin B$ $= \frac{a^{2} + c^{2}}{3} - \frac{a c (a^{2} + c^{2} - b^{2})}{6 a c} + \frac{1}{\sqrt{3}} a c \sin B$ $= \frac{a^{2} + b^{2} + c^{2}}{6} + \frac{1}{\sqrt{3}} a c \sin B$ $l = P R = 2 M N = 2 \sqrt{\frac{a^{2} + b^{2} + c^{2}}{6} + \frac{1}{\sqrt{3}} a c \sin B}$ $= \sqrt{\frac{2 (a^{2} + b^{2} + c^{2})}{3} + \frac{4 a c \sin B}{\sqrt{3}}}$ $s i n c e Δ = \frac{a c \sin B}{2} = a r e a o f Δ A B C$ $\Rightarrow l = \sqrt{\frac{2 (a^{2} + b^{2} + c^{2})}{3} + \frac{8 Δ}{\sqrt{3}}}$
	Commented by ajfour last updated on 22/May/19

	$Why this way we get the maximum$ $size △ PQR, i fail to follow the reason,$ $Sir; please enlighten . .$
	Commented by mr W last updated on 22/May/19

	$w e c a n s e e t h a t R m u s t b e o n t h e c i r c l e$ $w i t h c e n t e r N a n d P o n t h e c i r c l e w i t h$ $c e n t e r M . n o w t h e q u e s t i o n i s t o f i n d$ $t h e p o s i t i o n s o f P a n d R s u c h t h a t P R$ $r e a c h e s t h e m a x i m u m .$ $w e c a n s e e ∠ D P B a n d ∠ D R B a r e b o t h$ $c o n s t a n t, t h e r e f o r e ∠ P D R i s a l s o$ $c o n s t a n t . i n t h i s c a s e t h e P R r e a c h e s$ $i t s m s x i m u m i f b o t h D P a n d D R$ $r e a c h t h e i r m a x i m u m, i . e . w h e n$ $D P a n d D R a r e t h e d i a m e t e r s o f t h e$ $c i r c l e s . i c a n r e m e m b e r o n c e i p u t$ $t h i s q u e s t i o n i n a n e a r l i e r p o s t, a n d$ $y o u h a v e s o l v e d i t u s i n g c a l c u l u s$ $m e t h o d . u n f o r t u n a t e l y i c a n n o t$ $f i n d t h i s o l d p o s t, b u t m a y b e y o u c a n$ $a l s o r e m e m b e r i t, s e e i m a g e b e l o w .$
	Commented by mr W last updated on 22/May/19

	Commented by ajfour last updated on 22/May/19

	$Thanks Sir, for the geometric way!$
	Commented by mr W last updated on 22/May/19

	$t h i s i s a v e r y i n t e r e s t i n g q u e s t i o n .$ $i s p e n t a l o n g t i m e t o f i n d a w a y t o$ $u n d e r s t a n d a n d s o l v e i t g e o m e t r i c a l l y .$ $i^{'} d n e v e r c o m e t o t h i s s o l u t i o n w i t h o u t$ $y o u r i m a g e . t h e r e f o r e t h a n k s a l o t!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum