show that $OG \perp (A' B' C') $ and that the points $O, G$ and $H$ are collinear

Question

The problem

Consider the tetrahedron $OABC$ in which $OA \perp OB \perp OC \perp OA$ . A sphere with center $X$ containing the points $A, B$ and $C$ intersect the edges $OA$, $OB$, and $OC$ a second time at points $A'$, $B'$, and $C'$, respectively. If $G$ is the centroid of triangle $ABC$ and $H$ is the orthocenter of triangle $A'B'C'$, show that $OG \perp (A' B' C') $ and that the points $O, G$ and $H$ are collinear.

drawing

my idea

Because points $A, B, C, A', B', C'$ are all on the sphere we get that $XA=XB=XC=XA'=XB'=XC'=R$, where R is the radius of the sphere

From $XA=XB=XC$ we get that the height of tetrahedron $XABC$ contains the centroid of triangle $ABC$ so $HG\perp (ABC)$

I think from here we can try expressing everything as volume...

I don't know what to do forward! I hope one of you can help me! Thank you!

Answer 1

First we prove $OG\perp A'B'$.

We have $\overrightarrow{OG}=\frac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})$,

$\overrightarrow{OG}\cdot\overrightarrow{A'B'}=\frac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\cdot(\overrightarrow{A'O}+\overrightarrow{OB'})=\frac{1}{3}(\overrightarrow{OA}\cdot\overrightarrow{A'O}+\overrightarrow{OB}\cdot\overrightarrow{OB'})=\frac{1}{3}(\overrightarrow{OB}\cdot\overrightarrow{OB'}-\overrightarrow{OA}\cdot\overrightarrow{OA'})=0$

from $OA, OB, OC$ are perpendicular to each other and $A,B,B',A'$ are on the same circle so $\triangle OAB\sim\triangle OB'A'$.

Similarly, $OG\perp B'C'$ and $OG\perp A'C'$.

Let's denote $H'$ as the intersection of $OG$ and $(A'B'C')$, and prove that $H'$ is the orthocenter of $\triangle A'B'C'$. Now that we have $OG\perp (A'B'C')$, $H'$ is the projection of $O$ onto plain $(A'B'C')$, so to prove $A'H\perp B'C'$, we just need to prove that $OA'\perp B'C'$.

$\overrightarrow{OA'}\cdot\overrightarrow{B'C'}=\overrightarrow{OA'}\cdot(\overrightarrow{B'O}+\overrightarrow{OC'})=\overrightarrow{OA'}\cdot\overrightarrow{B'O}+\overrightarrow{OA'}\cdot\overrightarrow{OC'}=0$.

Similarly we can prove $B'H\perp A'C'$, $C'H\perp A'B'$. The proof that $H'$ is the orthocenter of $\triangle A'B'C'$ is complete, thus $H$ is on $OG$.

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(Without using vectors:)

To prove $OG\perp B'C'$, denote $D$ as the midpoint of $BC$. $OB\perp OC$, so $OD=DB=DC$, $\angle BOD=\angle OBD=\angle OC'B'$, thus $OD\perp B'C'$. Since $OA\perp (OBC)$, $OA\perp B'C'$, thus $(OAD)\perp B'C'$, $OG\perp B'C'$. The rest goes similarly.

For the next part, just notice that $OA'\perp (OB'C')$ and $H'$ is the projection of $O$ onto $(A'B'C')$, we must have $AH'\perp B'C'$. The rest goes similarly.