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

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!

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 $$OG$$.

(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.

• Thank you so much for your idea! May I ask if there is any way to transform this idea into one without vectors? I didn't learn about them yet! Thanks again! Commented May 2 at 8:56
• I added a section that doesn't use vectors. Still, I think using vectors is more convenient and intuitive. Commented May 2 at 9:16
• I have one more question...How did you know that A,B,A',B' are on the same circle and how did it help you arrive at the fact those triangles are similar? Commented May 2 at 14:43
• Oh, $A,B,C,A',B',C'$ are all on the sphere, and the cross section of a sphere is a circle. Then, $\angle AA'B'+\angle ABB'=\pi$, so $\angle OA'B'=\angle ABB'$, the two triangles are similar. Commented May 2 at 15:56
• $\triangle OAB\sim\triangle OA'B' \implies \frac{OA’}{OA}=\frac{OB’}{OB}$; since $\vec{OB}\cdot\vec{OB’}=OB \;OB’$ and $\vec{OA}\cdot\vec{OA’}=OA \;OA’$ I cannot see how you get that $OB \;OB’ - OA \;OA’=0$ Commented May 3 at 1:42