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Question Number 186924 by Mingma last updated on 12/Feb/23

	Answered by mr W last updated on 12/Feb/23

	Commented by mr W last updated on 12/Feb/23
	$cos α=((s^2 +c^2 −a^2 )/(2sc)) cos γ=((4s^2 +c^2 −d^2 )/(4sc)) γ=(π/3)−α ((4s^2 +c^2 −d^2 )/(4sc))=(1/2){((s^2 +c^2 −a^2 )/(2sc))+((√(3[2(3c^2 −a^2 )s^2 −(c^2 −a^2 )^2 −s^4 ]))/(2sc))} 3s^2 −d^2 +a^2 =(√(3[2c^2 s^2 −s^4 −c^4 +2(s^2 +c^2 )a^2 −a^4 ])) 4a^4 −2(3c^2 +d^2 )a^2 +12s^4 −6(c^2 +d^2 )s^2 +3c^4 +d^4 =0 roots of this eqn. are a^2 and b^2 . a^2 +b^2 =((2(3c^2 +d^2 ))/4)=((3c^2 +d^2 )/2) that means there is following correlation between a,b,c,d: determinant (((2(a^2 +b^2 )=3c^2 +d^2 ))) a^2 b^2 =((12s^4 −6(c^2 +d^2 )s^2 +3c^4 +d^4 )/4) s^4 −(((c^2 +d^2 )s^2 )/2)+((3c^4 +d^4 −4a^2 b^2 )/(12))=0 s^2 =(1/4)[c^2 +d^2 +(√((16a^2 b^2 −(3c^2 −d^2 )^2 )/3))] ⇒s=(1/2)(√(c^2 +d^2 +(√((16a^2 b^2 −(3c^2 −d^2 )^2 )/3)))) in current case: a=5, c=8, d=10 5^2 +b^2 =((3×8^2 +10^2 )/2) ⇒b^2 =((3×8^2 +10^2 )/2)−5^2 =121 ⇒b=(√(121))=11$
	$\cos α = \frac{s^{2} + c^{2} - a^{2}}{2 s c}$ $\cos γ = \frac{4 s^{2} + c^{2} - d^{2}}{4 s c}$ $γ = \frac{π}{3} - α$ $\frac{4 s^{2} + c^{2} - d^{2}}{4 s c} = \frac{1}{2} {\frac{s^{2} + c^{2} - a^{2}}{2 s c} + \frac{\sqrt{3 [2 (3 c^{2} - a^{2}) s^{2} - {(c^{2} - a^{2})}^{2} - s^{4}]}}{2 s c}}$ $3 s^{2} - d^{2} + a^{2} = \sqrt{3 [2 c^{2} s^{2} - s^{4} - c^{4} + 2 (s^{2} + c^{2}) a^{2} - a^{4}]}$ $4 a^{4} - 2 (3 c^{2} + d^{2}) a^{2} + 12 s^{4} - 6 (c^{2} + d^{2}) s^{2} + 3 c^{4} + d^{4} = 0$ $r o o t s o f t h i s e q n . a r e a^{2} a n d b^{2} .$ $a^{2} + b^{2} = \frac{2 (3 c^{2} + d^{2})}{4} = \frac{3 c^{2} + d^{2}}{2}$ $t h a t m e a n s t h e r e i s f o l l o w i n g$ $c o r r e l a t i o n b e t w e e n a, b, c, d :$ $\begin{matrix} 2 (a^{2} + b^{2}) = 3 c^{2} + d^{2} \end{matrix}$ $a^{2} b^{2} = \frac{12 s^{4} - 6 (c^{2} + d^{2}) s^{2} + 3 c^{4} + d^{4}}{4}$ $s^{4} - \frac{(c^{2} + d^{2}) s^{2}}{2} + \frac{3 c^{4} + d^{4} - 4 a^{2} b^{2}}{12} = 0$ $s^{2} = \frac{1}{4} [c^{2} + d^{2} + \sqrt{\frac{16 a^{2} b^{2} - {(3 c^{2} - d^{2})}^{2}}{3}}]$ $\Rightarrow s = \frac{1}{2} \sqrt{c^{2} + d^{2} + \sqrt{\frac{16 a^{2} b^{2} - {(3 c^{2} - d^{2})}^{2}}{3}}}$ $i n c u r r e n t c a s e :$ $a = 5, c = 8, d = 10$ $5^{2} + b^{2} = \frac{3 \times 8^{2} + 10^{2}}{2}$ $\Rightarrow b^{2} = \frac{3 \times 8^{2} + 10^{2}}{2} - 5^{2} = 121$ $\Rightarrow b = \sqrt{121} = 11$
	Commented by Mingma last updated on 12/Feb/23
	Excellent
	Answered by nikif99 last updated on 12/Feb/23

	$\underset{―}{I n o r t h o g o n a l y A x :}$ $E q . o f c i r c l e (A, r = 5) a t p o i n t K : K_{x}^{2} + K_{y}^{2} = 5^{2} (1)$ $E q . o f (B, r = 8) a t K : {(K_{x} - B_{x})}^{2} + {(K_{y} - B_{y})}^{2} = 8^{2} \Rightarrow$ ${[K_{x} - (- \frac{a}{2})]}^{2} + {(K_{y} - \frac{a \sqrt{3}}{2})}^{2} = 64 \Rightarrow$ $a (a + K_{x} - \sqrt{3} K_{y}) = 39 (2)$ $E q . o f (K, r = 10) a t E : {(E_{x} - K_{x})}^{2} + {(E_{y} - K_{y})}^{2} = 10^{2} \Rightarrow$ ${(a + \frac{a}{2} - K_{x})}^{2} + {(\frac{a \sqrt{3}}{2} - K_{y})}^{2} = 100 \Rightarrow$ $a (3 a - 3 K_{x} - \sqrt{3} K_{y}) = 75 (3)$ $(1) (2) (3) \Rightarrow a = 8.357297$ $K_{x} = 3.101745, K_{y} = 3.921629$ $C K = \sqrt{{(K_{x} - C_{x})}^{2} + {(K_{y} - C_{y})}^{2}} \Rightarrow C K = 11$
	Commented by nikif99 last updated on 12/Feb/23

	Commented by Mingma last updated on 12/Feb/23
	Excellent
	Answered by HeferH last updated on 12/Feb/23

	Commented by HeferH last updated on 12/Feb/23

	Commented by HeferH last updated on 12/Feb/23

	$w h e r e x i s t h e h e i g h t o f t h e t r i a n g l e w i t h s i d e s$ $a, 8, 5$
	Commented by Rupesh123 last updated on 13/Feb/23
	Excellent!
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