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Question Number 164174 by mr W last updated on 15/Jan/22

Commented by mr W last updated on 15/Jan/22

$f i n d t h e p o s i t i o n o f p o i n t P i n a$ $g i v e n t r i a n g l e s u c h t h a t t h e s u m$ $o f i t s d i s t a n c e s t o t h e v e r t i c e s i s$ $m i n i m u m . f i n d t h e c o r r e s p o n d i n g$ $d i s t a n c e s t o t h e v e r t i c e s .$
	Answered by mr W last updated on 15/Jan/22

	Commented by mr W last updated on 15/Jan/22

	Commented by mr W last updated on 15/Jan/22

	$t h i s p o i n t P i s t h e s o - c a l l e d$ $F e r m a t p o i n t, s e e d i a g r a m a b o v e .$ $s a y i t s d i s t a n c e s t o t h e v e r t i c e s a r e$ $x, y, z r e s p e c t i v e l y .$ $w e h a v e$ $x^{2} + y^{2} - 2 x y \cos 120 ° = c^{2}$ $x^{2} + y^{2} + x y = c^{2}$ $s i m i l a r l y$ $y^{2} + z^{2} + y z = a^{2}$ $z^{2} + x^{2} + z x = b^{2}$ $s a y t h e a r e a o f t r i a n g l e i s Δ, t h e n$ $Δ = \frac{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}{4}$ $o n t h e o t h e r s i d e,$ $\frac{x y \sin 120 °}{2} + \frac{y z \sin 120 °}{2} + \frac{z x \sin 120 °}{2} = Δ$ $\frac{\sqrt{3}}{4} (x y + y z + z x) = Δ$ $x y + y z + z x = \frac{4 Δ}{\sqrt{3}}$
	Commented by mr W last updated on 16/Jan/22

	${B A}^{'} = \frac{2}{3} \times \frac{\sqrt{3} a}{2} = \frac{\sqrt{3} a}{3}$ ${B C}^{'} = \frac{\sqrt{3} c}{3}$ $∠ A^{'} {B C}^{'} = ∠ B + 60 °$ $C^{'} A^{' 2} = {(\frac{\sqrt{3} a}{3})}^{2} + {(\frac{\sqrt{3} c}{3})}^{2} - 2 \times \frac{\sqrt{3} a}{3} \times \frac{\sqrt{3} c}{3} \cos (∠ B + 60 °)$ $C^{'} A^{' 2} = \frac{a^{2} + c^{2} - a c (\cos ∠ B - \sqrt{3} \sin ∠ B)}{3}$ $C^{'} A^{' 2} = \frac{1}{3} [a^{2} + c^{2} - a c (\frac{a^{2} + c^{2} - b^{2}}{2 a c} - \frac{2 \sqrt{3} Δ}{a c})]$ $C^{'} A^{' 2} = \frac{a^{2} + b^{2} + c^{2} + 4 \sqrt{3} Δ}{6}$ $C^{'} A^{'} = A^{'} B^{'} = B^{'} C^{'} = \sqrt{\frac{a^{2} + b^{2} + c^{2} + 4 \sqrt{3} Δ}{6}} = d, s a y$ $(t h i s i s t h e s i d e l e n g t h o f N a p o l e o n$ $t r i a n g l e A^{'} B^{'} C^{'})$ $\frac{y d}{2} = \frac{\sqrt{3} a}{3} \times \frac{\sqrt{3} c}{3} \times \sin (∠ B + 60 °) = a r e a [{P A}^{'} {B C}^{'}]$ $y = \frac{a c}{3 d} \times (\sin ∠ B + \sqrt{3} \cos ∠ B)$ $y = \frac{a c}{3 d} \times (\frac{2 Δ}{a c} + \sqrt{3} \times \frac{a^{2} + c^{2} - b^{2}}{2 a c})$ $y = \frac{4 Δ + \sqrt{3} (a^{2} - b^{2} + c^{2})}{6 d}$ $y = \frac{4 \sqrt{6} Δ + 3 \sqrt{2} (a^{2} - b^{2} + c^{2})}{6 \sqrt{a^{2} + b^{2} + c^{2} + 4 \sqrt{3} Δ}}$ $s i m i l a r l y$ $x = \frac{4 \sqrt{6} Δ + 3 \sqrt{2} (- a^{2} + b^{2} + c^{2})}{6 \sqrt{a^{2} + b^{2} + c^{2} + 4 \sqrt{3} Δ}}$ $z = \frac{4 \sqrt{6} Δ + 3 \sqrt{2} (a^{2} + b^{2} - c^{2})}{6 \sqrt{a^{2} + b^{2} + c^{2} + 4 \sqrt{3} Δ}}$ $x + y + z = \sqrt{\frac{a^{2} + b^{2} + c^{2} + 4 \sqrt{3} Δ}{2}}$ $◼$
	Commented by behi834171 last updated on 15/Jan/22

	$v e r y n i c e s o l u t i o n, d e a r m a s t e r . t h a n k s a l o t .$ $a = b = c \Rightarrow Δ = a^{2} . \frac{\sqrt{3}}{4}$ $4 \sqrt{3} Δ + b^{2} + 3 c^{2} - a^{2} = 4 \sqrt{3} \times a^{2} . \frac{\sqrt{3}}{4} + 3 a^{2} = 6 a^{2}$ $4 \sqrt{3} Δ + a^{2} + b^{2} + c^{2} = 4 \sqrt{3} \times a^{2} . \frac{\sqrt{3}}{4} + 3 a^{2} = 6 a^{2}$ $\Rightarrow y = x = z = \sqrt{\frac{4 a^{2}}{3} - a^{2}} = a . \frac{\sqrt{3}}{3} . ◼$ $[a l s o s e e q \neq 14157]$
	Commented by mr W last updated on 15/Jan/22

	$t h a n k s a l o t s i r!$ $i^{'} v e f o r g o t t h a t o l d p o s t .$ $n o w i h a v e f o u n d a b e t t e r f o r m u l a i n$ $s y m m e t r i c f o r m f o r x, y, z, s e e a b o v e,$ $p l e a s e r e v i e w!$
	Commented by Tawa11 last updated on 15/Jan/22

	$Great sir .$
	Commented by Tawa11 last updated on 15/Jan/22

	$Thanks for your time .$
	Commented by behi834171 last updated on 15/Jan/22

	$t h i s i s m u c h m o r e b e a u t i f u l .$ $d e a r m a s t e r! y o u r a p p r o a c h t o q u e s t i o n s$ $i s u n i q u e .$ $t h i s m e t h o d s a r e e s p e c i a l l y y o u r s .$ $i l e a r n m u c h f r o m y o u r p o s t s .$ $g o d b l e s s y o u f o r t h i s l o v e l y f u r o m .$
	Commented by mr W last updated on 15/Jan/22

	$t h a n k s f o r r e v i e w i n g m y p o s t s s i r!$
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