m−h=2p p(m−h)=k−q mk−qh=(1/3) k−2q=1+ph Assume one find the rest! ✓


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Question Number 201965 by ajfour last updated on 17/Dec/23

$m - h = 2 p$ $p (m - h) = k - q$ $m k - q h = \frac{1}{3}$ $k - 2 q = 1 + p h$ $A s s u m e o n e f i n d t h e r e s t!$ $✓$
	Answered by Frix last updated on 17/Dec/23
	$Solve (1) for h then (2) for k then (3) for m ⇒ h=m−2p=((−12p^3 −6pq+1)/(6p^2 )) k=2p^2 +q m=((−6pq+1)/(6p^2 )) (4) p^3 −(p/4)−(1/(24))=0 ⇒ p∈{−((√3)/3)sin ((2π)/9), −((√3)/3)sin (π/9), ((√3)/3)cos (π/(18))} We are free to choose q$
	$Solve (1) for h then (2) for k then (3) for m$ $\Rightarrow$ $h = m - 2 p = \frac{- 12 p^{3} - 6 p q + 1}{6 p^{2}}$ $k = 2 p^{2} + q$ $m = \frac{- 6 p q + 1}{6 p^{2}}$ $(4) p^{3} - \frac{p}{4} - \frac{1}{24} = 0$ $\Rightarrow p \in {- \frac{\sqrt{3}}{3} \sin \frac{2 π}{9}, - \frac{\sqrt{3}}{3} \sin \frac{π}{9}, \frac{\sqrt{3}}{3} \cos \frac{π}{18}}$ $We are free to choose q$
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