4
$\begingroup$

enter image description here

My question is regarding this and similar problems like above. So the question is asking to find the locus of the centroid of the tetrahedron. But to my understanding, the centroid of the tetrahedron is a constant point (Because now the tetrahedron is also described according to constant coordinate points and sphere which too has constant radius). So how can a constant point like these can be represented as a equation?

$\endgroup$
4
$\begingroup$

Presumably, the intended meaning is that the sphere is allowed to vary (though the phrasing of the problem is rather poor and should indicate this more clearly). In other words, you consider all possible points $A,B,C$ that are where a sphere of radius $2k$ which passes through the origin meets the axes. Then, you consider the locus consisting of the centroids of the tetrahedron $OABC$ for all possible such $A,B,C$.

$\endgroup$
  • $\begingroup$ Ok following up your answer, i have a question. Lets say sphere is varied but it still passes through the same coordinate axes, right? making the coordinate points same. The centroid of the tetrahedron OABC only depend on those coordinate points, which is not changed, whatever variations you give to the sphere, if i am correct . So i think my question still holds. I will like some insights. Thank you. $\endgroup$ – user698033 Aug 24 '19 at 5:14
  • $\begingroup$ Why would the points $A,B,C$ be the same? Imagine taking a sphere through the origin and rotating it around the origin. As it rotates, the points where it hits the coordinate axes will change. $\endgroup$ – Eric Wofsey Aug 24 '19 at 5:16
  • 1
    $\begingroup$ Thank you. I think i get it now. I too think question is written poorly now. $\endgroup$ – user698033 Aug 24 '19 at 5:18
  • $\begingroup$ @AagatPokhrel: If you really had a constant point, say, $P=(a,b,c)$, then you could give its equation as $(x-a)^2+(y-b)^2+(z-c)^2=0$: a sphere centered at $P$ with radius $0$. $\endgroup$ – Blue Aug 24 '19 at 6:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy