3
$\begingroup$

Triangle $ABC$ has centroid $ G$ and orthcenter $H$. Line (through $A$) is perpendicular to $GA$, line (through $B$) is perpendicular to $GB$, line (through $C$) is perpendicular to $GC$ cut at three points which form a new triangle $A_1B_1C_1$. This new triangle has centroid $G_1$. Show that three point $G,H,G_1$ are collinear.

I have tried to so this problem with lots of theorems. But I can't find the way to solve. Or using any lemma? Help me to find and draw any auxiliary geometry element.

$\endgroup$
  • $\begingroup$ There is no reason why $G=H$, let alone $G,H,G_1$ concur. Do you mean $G.H,G_1$ collinear instead? $\endgroup$ – user10354138 Dec 7 '18 at 14:38
  • $\begingroup$ Oh. I'm sorry. It is "colinear". $\endgroup$ – Trong Tuan Dec 7 '18 at 15:02
  • $\begingroup$ Recall the Euler line... $\endgroup$ – user10354138 Dec 7 '18 at 15:24
2
$\begingroup$

Taking @user10354138's comment, here's how we attack the problem: We will show that the midpoint $O$ of $GG_1$ is the circumcenter of $\triangle ABC$. In particular, we can actually show that $G$ is the midpoint of $HG_1$. enter image description here

In the picture above, $A_2,B_2,C_2$ are midpoints of $GA_1,GB_1,GC_1$ respectively. Then $C_2$ is on the perpendicular bisector of $AB$. So it is sufficient to show that $C_2O\perp AB$, or $C_1G_1\perp AB$.

Now, if we consider a triangle $XYZ$ with sides (parallel to) the medians of $\triangle ABC$, then

  1. The sides of $\triangle XYZ$ and $\triangle A_1B_1C_1$ are pairwise perpendicular.
  2. The medians of $\triangle XYZ$ are parallel to the sides of $\triangle ABC$.

(The existence/construction of $\triangle XYZ$ and the proofs of the above statements are classical and left to you.)

From (1), it follows that $\triangle XYZ$ and $\triangle A_1B_1C_1$ are similar and their medians are pairwise perpendicular. This and (2) yield that $C_1G_1\perp AB$ and so on, which is what we are looking for.

$\endgroup$
  • $\begingroup$ "From (1), it follows that △XYZ and △A1B1C1 are similar and their medians are pairwise perpendicular. This and (2) yield that $C_1G_1$⊥AB". Can you make it more clear? I still hav not understood. Thank you $\endgroup$ – Trong Tuan Dec 9 '18 at 13:44

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.