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Question Number 40857 by ajfour last updated on 28/Jul/18

Commented by ajfour last updated on 28/Jul/18

$F i n d c i r c u m s p h e r e r a d i u s R o f$ $a g e n e r a l t r i a n g u l a r p y r a m i d .$
Commented by MJS last updated on 29/Jul/18

$see my comment to qu . 40469$ $the problem is to get the coordinates of S$ $try for a triangle, obviously we can put$ $A = (\begin{matrix} 0 \\ 0 \end{matrix}) B = (\begin{matrix} c \\ 0 \end{matrix})$ $but for C we need {Heron}^{'} s formula$ $C = (\begin{matrix} x_{C} \\ h_{c} \end{matrix}) with h_{c} = \frac{2 \times area (△)}{c} =$ $= \frac{\sqrt{(a + b + c) (a + b - c) (a + c - b) (b + c - a)}}{2 c}$ $and x_{C} = \sqrt{b^{2} - h_{c}^{2}}$ $then we can find the circumcircle$ $this is possible in R^{2} but in R^{3} we need {Euler}^{'} s$ $formula and {it}^{'} s very hard, maybe impossible,$ $to handle it in any equation system$ $S = (\begin{matrix} x_{S} \\ y_{S} \\ h \end{matrix}) with h = \frac{3 \times volume (pyramid)}{area (△ A B C)}$ $then we can find the circumsphere . . .$
	Answered by ajfour last updated on 28/Jul/18

	$L e t △ A B C b e t h e b a s e a n d S t h e$ $v e r t e x . C i r c u m s p h e r e c e n t r e G .$ $O r i g i n O, t h e c e n t r e o f$ $c i r c u m c i r c l e o f b a s e (r a d i u s ρ) .$ $y a x i s t o w a r d s r i g h t, x a x i s$ $o u t w a r d s; z a x i s u p w a r d s .$ $ρ = R \cos θ . . . . . (i)$ $a = 2 ρ \cos α . . . . . . (i i)$ $b = 2 ρ \cos (α + β) . . . . (i i i)$ $c = 2 ρ \cos β . . . . (i v)$ $_____________$ $A (- ρ \sin 2 β, ρ \cos 2 β, 0)$ $B (0, - ρ, 0)$ $C (ρ \sin 2 α, ρ \cos 2 α, 0)$ $S (R \cos γ \sin ϕ, R \cos γ \cos ϕ, R \sin θ + R \sin γ)$ $_____________$ $S A = p . . . (v)$ $S B = q . . . . (v i)$ $S C = r . . . . . (v i i)$ $u n k n o w n s : α, β, γ, θ, ϕ, ρ, R$ $e q u a t i o n s : (i) \to (v i i)$ $_________________________$
	Commented by MrW3 last updated on 28/Jul/18

	$v e r y n i c e t r y s i r!$ $I t s e e m s t o b e a h a r d t a s k .$
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