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 82131 by mr W last updated on 18/Feb/20

Commented by ajfour last updated on 18/Feb/20

Commented by mr W last updated on 18/Feb/20

$F i n d t h e s i d e l e n g t h s o f t h e i n s c r i b e d$ $e q u i l a t e r a l t r i a n g l e (s) i n a g i v e n$ $t r i a n g l e w i t h s i d e l e n g t h e s a, b, c .$
Commented by ajfour last updated on 18/Feb/20

$S i r, i a m w o n d e r i n g a b o u t t h e$ $l o c u s o f t h e c e n t r o i d o f t h e$ $i n s c r i b e d e q u i l a t e r a l △ .$
Commented by mr W last updated on 18/Feb/20

$w h a t c a n b e s a i d a b o u t t h e l o c u s ?$
	Answered by mr W last updated on 19/Feb/20

	$P a r t I :$ $F i n d t h e m a x i m u m e q u i l a t e r a l s_{m a x}$ $∠ A = α, ∠ B = β, ∠ C = γ$ $i f α = β = γ, t h e n {i t}^{'} s c l e a r t h a t t h e$ $m a x i m u m e q u i l a t e r a l i s t h e t r i a n g l e$ $i t s e l f, i . e . s_{m a x} = a (= b = c) . o t h e r w i s e$ $w e w i l l h a v e a t l e a s t o n e a n g l e w h i c h$ $i s l a r g e r t h a n 60 ° .$ ${l e t}^{'} s a s s u m e β > γ > α . {i t}^{'} s c l e a r t h a t$ $β > 60 ° . w e h a v e t w o c a s e s :$ $c a s e 1 : β > γ > α, γ < 60 °$ $c a s e 2 : β > γ > α, γ ⩾ 60 °$
	Commented by mr W last updated on 19/Feb/20

	Commented by mr W last updated on 19/Feb/20

	$C a s e 1 :$ ${i t}^{'} s c l e a r t h e r e d e q u i l a t e r a l i s t h e$ $m a x i m u m .$ $\frac{s_{m a x}}{\sin γ} = \frac{a}{\sin (γ + β - 60 °)}$ $\Rightarrow s_{m a x} = \frac{a \sin γ}{\sin (β + γ - 60 °)}$
	Commented by mr W last updated on 19/Feb/20

	Commented by mr W last updated on 20/Feb/20

	$C a s e 2 :$ $b l u e e q u i l a t e r a l w i t h s_{b l u e}$ $\frac{s_{b l u e}}{\sin γ} = \frac{a}{\sin (γ + β - 60 °)}$ $\Rightarrow s_{b l u e} = \frac{a \sin γ}{\sin (γ + β - 60 °)}$ $s i m i l a r l y$ $\Rightarrow s_{r e d} = \frac{a \sin β}{\sin (β + γ - 60 °)}$ $s i n c e β > γ, s_{r e d} > s_{b l u e} .$ $\Rightarrow s_{m a x} = \frac{a \sin β}{\sin (β + γ - 60 °)}$ $t o b e e x a c t a n d f o r a l l c a s e s :$ $i f ∣ 90 ° - β ∣<∣ 90 ° - γ ∣, t h e n$ $\Rightarrow s_{m a x} = \frac{a \sin β}{\sin (β + γ - 60 °)} = \frac{a \sin β}{\sin (120 ° - α)}$ $i f ∣ 90 ° - β ∣>∣ 90 ° - γ ∣, t h e n$ $\Rightarrow s_{m a x} = \frac{a \sin γ}{\sin (β + γ - 60 °)} = \frac{a \sin γ}{\sin (120 ° - α)}$ $i . e .$ $s_{m a x} = \frac{a m a x (\sin β, \sin γ)}{\cos (30 ° - α)}$
	Answered by mr W last updated on 19/Feb/20

	$P a r t I I$ $F i n d t h e m i n i m u m e q u i l a t e r a l s_{m i n} .$ ${l e t}^{'} s s a y f o r t h e g i v e n t r i a n g l e A B C$ $w i t h s i d e l e n g t h e s a, b c a n d i n n e r$ $a n g l e s α, β, γ t h e m i n i m u m i n s c r i b e d$ $e q u i l a t e r a l t r i a n g l e i s t h a t o n e w i t h$ $t h e s i d e l e n g t h s .$ $n o w w e i m a g e t h a t w e h a v e a n$ $e q u i l a t e r a l t r i a n g l e w i t h s i d e l e n g t h s .$ $w e c a n c o n s t r u c t i n f i n i t e c i r c u m -$ $t r i a n g l e s w i t h t h e i n n e r a n g l e s$ $α, β, γ . a l l o f t h e s e c i r c u m t r i a n g l e s$ $a r e s i m i l a r t o e a c h o t h e r a n d s i m i l a r$ $t o t h e g i v e n t r i a n g l e A B C . i t i s c l e a r$ $t h a t t h e l a r g e s t o n e o f t h e s e c i r c u m -$ $t r i a n g l e s h a s e x a c t l y t h e s a m e s i z e a s$ $t h e g i v e n t r i a n g l e . t h i s c o n s i d e r a t i o n$ $g i v e s u s f o l l o w i n g w a y t o s o l v e o u r$ $p r o b l e m :$ $f o r a n e q u i l a t e r a l t r i a n g l e w i t h s i d e$ $l e n g t h s = 1 w e c o n s t r u c t c i r c u m t r i a n g l e s$ $w i t h i n n e r a n g l e s α, β, γ a n d f i n d$ $t h e l a r g e s t o n e . i f t h e l a r g e s t o n e$ $h a s t h e s i d e l e n g t h e s a^{'}, b^{'} a n d c^{'}, t h e n$ $w e h a v e$ $\frac{s_{m i n}}{1} = \frac{a}{a^{'}} = \frac{b}{b^{'}} = \frac{c}{c^{'}} .$ $t h r o u g h t h i s w a y w e g e t t h e s i d e$ $l e n g t h s_{m i n} o f t h e m i n i m u m i n s c r i b e d$ $e q u i l a t e r a l f o r t h e g i v e n t r i a n g l e A B C .$
	Commented by mr W last updated on 19/Feb/20

	Commented by mr W last updated on 19/Feb/20

	$n o w w e j u s t n e e d t o f i n d t h e l a r g e s t$ $c i r c u m t r i a n g l e f o r a n e q u i l a t e r a l$ $t r i a n g l e w i t h s i d e l e n g t h s = 1 . t h e$ $c i r c u m t r i a n g l e m u s t h a v e t h e s a m e$ $a n g l e s a s t h e g i v e n t r i a n g l e A B C,$ $n a m e l y α, β, γ .$
	Commented by mr W last updated on 19/Feb/20

	Commented by mr W last updated on 19/Feb/20

	Commented by mr W last updated on 20/Feb/20

	$r_{A} = \frac{s}{2 \sin α} = \frac{1}{2 \sin α}$ $r_{B} = \frac{s}{2 \sin β} = \frac{1}{2 \sin β}$ $r_{C} = \frac{s}{2 \sin γ} = \frac{1}{2 \sin γ}$ $w h e n i t s v e r t e x e s l i e o n t h e t h r e e$ $c i r c l e s w i t h r a d i i r_{A}, r_{B}, r_{C} a n d c e n t e r s$ $a t A^{'}, B^{'}, C^{'}, t h e n t h e c i r c u m t r i a n g l e$ $a l w a y s h a s t h e i n n e r a n g l e s α, β, γ .$ $t h e m a x i m u m c i r c u m t r i a n g l e i s$ $A^{″} B^{″} C^{″} . w e k n o w t h i s i s t h e c a s e$ $w h e n A^{″} D, B^{″} D a n d C^{″} D a r e t h e$ $d i a m e t e r s o f t h e c i r c l e s .$ $B^{'} C^{' 2} = r_{B}^{2} + r_{C}^{2} - 2 r_{B} r_{C} \cos (60 ° + 90 ° - β + 90 ° - γ)$ $B^{'} C^{' 2} = r_{B}^{2} + r_{C}^{2} - 2 r_{B} r_{C} \cos (60 ° + α °)$ $B^{'} C^{' 2} = \frac{1}{4} [\frac{1}{\sin^{2} β} + \frac{1}{\sin^{2} γ} - \frac{2 \cos (60 ° + α)}{\sin β \sin γ}]$ $B^{'} C^{'} = \frac{1}{2} \sqrt{\frac{1}{\sin^{2} β} + \frac{1}{\sin^{2} γ} - \frac{2 \cos (60 ° + α)}{\sin β \sin γ}}$ $a^{'} = B^{″} C^{″} = 2 B^{'} C^{'}$ $\Rightarrow a^{'} = \sqrt{\frac{1}{\sin^{2} β} + \frac{1}{\sin^{2} γ} - \frac{2 \cos (60 ° + α)}{\sin β \sin γ}}$ $s i m i l a r l y$ $\Rightarrow b^{'} = \sqrt{\frac{1}{\sin^{2} γ} + \frac{1}{\sin^{2} α} - \frac{2 \cos (60 ° + β)}{\sin γ \sin α}}$ $\Rightarrow c^{'} = \sqrt{\frac{1}{\sin^{2} α} + \frac{1}{\sin^{2} β} - \frac{2 \cos (60 ° + γ)}{\sin α \sin β}}$ $a s d e s c r i b e d a b o v e,$ $s_{m i n} = \frac{a}{a^{'}}$ $s_{m i n} = \frac{a}{\sqrt{\frac{1}{\sin^{2} β} + \frac{1}{\sin^{2} γ} - \frac{2 \cos (60 ° + α)}{\sin β \sin γ}}}$ $o r$ $s_{m i n} = \frac{b}{\sqrt{\frac{1}{\sin^{2} γ} + \frac{1}{\sin^{2} α} - \frac{2 \cos (60 ° + β)}{\sin γ \sin a}}}$ $o r$ $s_{m i n} = \frac{c}{\sqrt{\frac{1}{\sin^{2} α} + \frac{1}{\sin^{2} β} - \frac{2 \cos (60 ° + γ)}{\sin α \sin β}}}$ $\sin α = \frac{a}{2 R}, \sin β = \frac{b}{2 R}, \sin γ = \frac{c}{2 R}$ $\cos (60 ° + γ) = \frac{1}{2} (\cos γ - \sqrt{3} \sin γ)$ $\cos (60 ° + γ) = \frac{1}{2} (\frac{a^{2} + b^{2} - c^{2}}{2 a b} - \frac{\sqrt{3} c}{2 R})$ $s_{m i n} = \frac{c}{\sqrt{\frac{4 R^{2}}{a^{2}} + \frac{4 R^{2}}{b^{2}} - \frac{4 R^{2}}{a b} (\frac{a^{2} + b^{2} - c^{2}}{2 a b} - \frac{\sqrt{3} c}{2 R})}}$ $s_{m i n} = \frac{c}{\sqrt{4 R^{2} (\frac{a^{2} + b^{2} + c^{2}}{2 a^{2} b^{2}}) + \frac{2 R \sqrt{3} c}{a b}}}$ $s_{m i n} = \frac{1}{\sqrt{\frac{2 R^{2}}{a^{2} b^{2} c^{2}} (a^{2} + b^{2} + c^{2}) + \frac{2 R \sqrt{3}}{a b c}}}$ $w i t h R = \frac{a b c}{4 Δ}$ $s_{m i n} = \frac{1}{\sqrt{\frac{1}{8 Δ^{2}} (a^{2} + b^{2} + c^{2}) + \frac{\sqrt{3}}{2 Δ}}}$ $\Rightarrow s_{m i n} = \frac{2 \sqrt{2} Δ}{\sqrt{a^{2} + b^{2} + c^{2} + 4 \sqrt{3} Δ}}$ $w i t h Δ = a r e a o f Δ A B C .$
	Commented by mr W last updated on 19/Feb/20

	$e x a m p l e 1 : a = b = c = 4$ $\sin α = \sin β = \sin γ = \frac{\sqrt{3}}{2}$ $s_{m i n} = \frac{4}{\sqrt{2 {(\frac{2}{\sqrt{3}})}^{2} + {(\frac{2}{\sqrt{3}})}^{2}}} = 2$ $e x a m p l e 2 : a = 3, b = 4, c = 5$ $\sin α = \frac{3}{5}, \sin β = \frac{4}{5}, \sin γ = 1$ $s_{m i n} = \frac{3}{\sqrt{\frac{25}{16} + 1 - 2 (\frac{1}{2} \times \frac{4}{5} - \frac{\sqrt{3}}{2} \times \frac{3}{5}) \times \frac{5}{4} \times 1}} = \frac{12}{\sqrt{25 + 12 \sqrt{3}}}$ $s_{m i n} = \frac{4}{\sqrt{\frac{25}{9} + 1 - 2 (\frac{1}{2} \times \frac{3}{5} - \frac{\sqrt{3}}{2} \times \frac{4}{5}) \times \frac{5}{3} \times 1}} = \frac{12}{\sqrt{25 + 12 \sqrt{3}}}$ $s_{m i n} = \frac{5}{\sqrt{\frac{25}{16} + \frac{25}{9} - 2 (\frac{1}{2} \times 0 - \frac{\sqrt{3}}{2} \times 1) \times \frac{5}{3} \times \frac{5}{4}}} = \frac{12}{\sqrt{25 + 12 \sqrt{3}}}$
	Commented by mr W last updated on 19/Feb/20

	$o r$ $s_{m i n} = \frac{2 \sqrt{2} Δ}{\sqrt{a^{2} + b^{2} + c^{2} + 4 \sqrt{3} Δ}} = \frac{12 \sqrt{2}}{\sqrt{50 + 24 \sqrt{3}}} = \frac{12}{\sqrt{25 + 12 \sqrt{3}}}$
	Commented by ajfour last updated on 22/Feb/20

	$Sir, would you please let me know$ $You can't use 'macro parameter character #' in math mode$ $the concept used here for largest$ $cirumtriangle construction . .$ $rest of your solution, i have$ $followed entirely, very clear$ $and clever Sir . .$
	Commented by mr W last updated on 23/Feb/20

	$t h a n k s f o r r e v i e w i n g s i r!$ $s e e Q 60313 .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum