In $\Delta ABC$, $\frac{AP}{AQ}=\frac{BP}{BQ}= \frac{CP}{CQ}=2$ find circumcentre given coordinates of $P,Q$.

Question

Let $A,B,C$ be the vertices of $\Delta ABC$. Let $P(1,1,2)$ and $Q(4,7,8)$ be two points such that $\frac{AP}{AQ}=\frac{BP}{BQ}= \frac{CP}{CQ}=2$. Find the coordinates of circumcentre of $\Delta ABC$.

It seems to be some property of triangles which I unfortunately not knowing/able to figure out.

My try was using vectors, assigning the position vectors of $A(\overrightarrow {a}),B(\overrightarrow {b}), C(\overrightarrow {c})$ and these would be the solution to the equation $\frac{\overrightarrow{x}-\overrightarrow{p}}{\overrightarrow{x}-\overrightarrow{q}} = 2$. But squaring and simplifying I just get a relation with $|\overrightarrow{x}|$ and dot product, but I don't see it lead to anything useful.

Answer 1

Relation $\frac{AP}{AQ}=\frac{BP}{BQ}=\frac{CP}{CQ}=2$ and the fact that circumcenter M has equal distance from vertices of triangle deduce that M can be on line PQ as an option. In fact M can be the intersection of PQ and the plain the triangle locates on such that $\frac{PM}{QM}=2$. For this particular case we can write:

$x_M=\frac{4+1}3=\frac{5}3$

$y_M=\frac{7+1}3=\frac{8}3$

$z_M=\frac{8+2}3=\frac{10}3$

Answer 2

The given conditions on $A,B,C$ show that they can be randomly chosen on the ellipsoid with equation $$ (x-1)^2+(y-1)^2+(z-2)^2=4(x-4)^2+4(y-7)^2+4(z-8)^2\ . $$ (It is a quadric in the space, it is bounded since for points $X$ going to "infinity" we have $XP/XQ\to 1$, it passes through the point on the segment $PQ$ at $1/3$ distance (from this length) from $Q$, and through the symmetric point of $P$ w.r.t. $Q$.)

In particular our random choice can use a "great variety" of planes $ABC$. Also pairs of such triangle planes that not in intersect.

We can also make a choice of three "very near points" $A,B,C$ (distance less $\epsilon=10^{-10}$) to a given point $E$ on this ellipsoid, so that the circumcenter of $\Delta ABC$ (chosen to be with angles $<90^\circ$) should be interior and thus also "near" $E$.

We can also fix a normal direction, say one corresponding to the line $PQ$, and intersect the ellipsoid with a "moving plane" normal to the fixed normal direction. We obtain an ellipse (a circle for the specific choice), and the circumcenters of triangles with vertices on such an ellipse (or circle) are not determined.

From all this discussion, we cannot "find" the exact position of the circumcenter.

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The geometric locus of the circumcenter is an other problem.