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Question Number 47657 by behi83417@gmail.com last updated on 12/Nov/18

Commented by mr W last updated on 13/Nov/18

$d e a r f a t h e r f r o m B e h i :$ $i t t o o k m e s o m e t i m e, b u t I c o u l d f i n a l l y$ $You can't use 'macro parameter character #' in math mode$ $t h i s i n t e r e s t i n g q u e s t i o n w a s o n c e$ $t r e a t e d . I g a v e t h e r e a m e t h o d h o w$ $t o d r a w t h e r e q u e s t e d e q u i l a t e r a l$ $t r i a n g l e a n d t h e f o r m u l a t o c a l c u l a t e$ $i t s s i d e l e n g t h w h i c h i s$ $l = \sqrt{\frac{a^{2} + b^{2} + c^{2} \pm \sqrt{6 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}) - 3 (a^{4} + b^{4} + c^{4})}}{2}}$ $w i t h a, b, c = d i s t a n c e s f r o m a p o i n t$ $t o t h e v e r t e x e s o f t h e t r i a n g l e .$ $i n o u r c a s e w e h a v e$ $a = 5$ $b = 7$ $c = 11$ $\Rightarrow l \approx 11.9342 o r \approx 7.2509$
Commented by MJS last updated on 13/Nov/18

$great!$ $but how to decide if the given point is inside$ $our outside the triangle (s) ?$
Commented by behi83417@gmail.com last updated on 13/Nov/18

$i h a v e n o w o r d s t o s a y m a s t e r .$ $i j u s t s a y t h a n k s i n a d v a n c e s i r .$ $. . . . . . a n d m e m o r i e s a t t a c k!$
Commented by mr W last updated on 11/Feb/20

$t o s e e i f t h e p o i n t p i s i n s i d e o r o u t s i d e$ $t h e t r i a n g l e, w e c a l c u l a t e t h e a n g l e s$ $b e t w e e n a a n d b, b a n d c a s w e l l a s$ $c a n d a . i f a l l o f t h e s e a n g l e s a r e$ $o b t u s e, t h e n p o i n t p l i e s i n s i d e .$ $i f a t l e a s t o n e o f t h e s e a n g l e s i s a c u t e,$ $t h e n p o i n t p l i e s o u t s i d e .$ $i n o u r e x a m p l e :$ $w i t h l = 11.9342$ $\cos α = \frac{b^{2} + c^{2} - l^{2}}{2 b c} \Rightarrow α = 79.7 ° < 90 °$ $\cos β = \frac{c^{2} + a^{2} - l^{2}}{2 c a} \Rightarrow β = 88.1 ° < 90 °$ $\cos γ = \frac{a^{2} + b^{2} - l^{2}}{2 a b} \Rightarrow γ = 167.8 °$ $\Rightarrow p i s o u t s i d e .$ $w i t h l = 7.2509$ $\cos α = \frac{b^{2} + c^{2} - l^{2}}{2 b c} \Rightarrow α = 40.3 ° < 90 °$ $\cos β = \frac{c^{2} + a^{2} - l^{2}}{2 c a} \Rightarrow β = 31.9 ° < 90 °$ $\cos γ = \frac{a^{2} + b^{2} - l^{2}}{2 a b} \Rightarrow γ = 72.2 ° < 90 °$ $\Rightarrow p i s o u t s i d e .$ $[c o n c l u s i o n n o t c o r r e c t, n e e d s t o b e$ $a m e n d e d . . .]$
Commented by MJS last updated on 13/Nov/18

$thank you$
Commented by MJS last updated on 14/Nov/18

$l = \sqrt{\frac{a^{2} + b^{2} + c^{2} \pm \sqrt{6 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}) - 3 (a^{4} + b^{4} + c^{4})}}{2}} =$ $= \sqrt{\frac{a^{2} + b^{2} + c^{2} \pm \sqrt{3 δ}}{2}}; δ = (a + b + c) (a + b - c) (a - b + c) (- a + b + c)$
Commented by mr W last updated on 14/Nov/18

$t h a n k s f o r t h i s e a s y t o r e m e m b e r$ $f o r m!$
	Answered by MJS last updated on 13/Nov/18

	$A = (\begin{matrix} 0 \\ 0 \end{matrix}) B = (\begin{matrix} s \\ 0 \end{matrix}) C = (\begin{matrix} \frac{1}{2} s \\ \frac{\sqrt{3}}{2} s \end{matrix})$ $P = (\begin{matrix} p \\ q \end{matrix})$ $∣ A P ∣^{2} = 25 \Rightarrow p^{2} + q^{2} - 25 = 0$ $∣ B P ∣^{2} = 49 \Rightarrow p^{2} + q^{2} + s^{2} - 2 p s - 49 = 0$ $∣ C P ∣^{2} = 121 \Rightarrow p^{2} + q^{2} + s^{2} - p s - \sqrt{3} q s - 121 = 0$ $Solution 1$ $p \approx 4.96 q \approx - . 62 s \approx 11.93$ $A = (\begin{matrix} 0 \\ 0 \end{matrix}) B = (\begin{matrix} 11.93 \\ 0 \end{matrix}) C = (\begin{matrix} 5.97 \\ 10.34 \end{matrix}) P = (\begin{matrix} 4.96 \\ - . 62 \end{matrix})$ $Solution 2$ $p \approx 1.97 q \approx - 4.60 s \approx 7.25$ $A = (\begin{matrix} 0 \\ 0 \end{matrix}) B = (\begin{matrix} 7.25 \\ 0 \end{matrix}) C = (\begin{matrix} 3.63 \\ 6.28 \end{matrix}) P = (\begin{matrix} 1.97 \\ - 4.60 \end{matrix})$ $\Rightarrow no solution within the triangle$ $these are symmetrically the same :$ $Solution 3$ $p \approx - 4.96 q \approx . 62 s \approx - 11.93$ $Solution 4$ $p \approx - 1.97 q \approx 4.60 s \approx - 7.25$
	Commented by behi83417@gmail.com last updated on 13/Nov/18

	$t h a n k y o u s o m u c h d e a r M J S .$ $t h i s i s n i c e a n d s i m p l e .$ $s = 11.93, 7.25 i s t h e c o r r e c t a n s w e r .$
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