Proving that the orthocenter lies on $OD$?

Question

While trying to solve this question using GeoGebra, I realized the following curious thing:

If $I$ is the incenter of $\triangle ABC$, $ID \perp BC$ with $D$ on $BC$, $AD \perp IO$ with $O$ on $AD$, then if $H$ is the orthocenter of $\triangle ABC$, $O$, $I$, $H$ are collinear.

Edit: Sorry people, its wrong and I came up with it by mistake, don't waste your time on it.

i am sorry i had assumed $O$ is the circumcenter. is that true? — abel, Jan 03 '15 at 20:17
I drew it in Geogebra and it does not look perpendicular at all. — Calvin Lin, Jan 03 '15 at 20:39
@CalvinLin I am not sure what you mean. What does not look perpendicular? O is not the circumcenter. — Sawarnik, Jan 03 '15 at 20:41
Sorry, I meant that OIH does not look like a straight line. Can you add an image of your diagram? — Calvin Lin, Jan 03 '15 at 20:43
@CalvinLin I am really sorry, Calvin. But could you look into the other question? :) — Sawarnik, Jan 03 '15 at 20:50
@Sawarnik The OP doesn't bother to respond to request for a diagram, so I can't really be bothered to do work for him either. — Calvin Lin, Jan 03 '15 at 20:52
I have just posted an answer showing that in a scalene triangle the incenter never belongs to the $OH$ line. — Jack D'Aurizio, Jan 03 '15 at 20:54
@CalvinLin Ok, I will add a diagram. It seems to be a nice question, though :) — Sawarnik, Jan 03 '15 at 20:56
@Sawarnik It is an olympiad problem. I've seen it in the past. What I meant is that if the OP has no interest, then I can't be bothered. — Calvin Lin, Jan 03 '15 at 20:58

score 1 · Accepted Answer · answered Jan 03 '15 at 20:42

1

My belief is that the statement is false.

Assume it is true, that OIH is a straight line. Then, AOD and IOH are perpendicular, meaning that AD is perpendicular to IH.

Similarly, this implies that BE and CF (where E, F are the foot of the perpendiculars from I to AC, and AB respectively) are perpendicular to IH.

But this is clearly not possible.

Hence, the statement is false.

The following image shows the points A, D, I, H as defined (without the point O as yet). When we add the point O such that $IO \perp AD$, it does not look like $OIH$ is a straight line.

enter image description here

answered Jan 03 '15 at 20:42

Calvin Lin

68,864

Well, yes, its true. By mistake, I came to the conclusion. – Sawarnik Jan 03 '15 at 20:47
Sorry for your time. – Sawarnik Jan 03 '15 at 20:48

score 1 · Answer 2 · answered Jan 03 '15 at 20:52

By using trilinear coordinates (today is the day) we have that a point belongs to the Euler line $OH$ iff its trilinear coordinates $[x:y:z]$ satisfy $$ \sum_{cyc} \sin(2A)\sin(B-C)x = 0. \tag{1}$$ For the incenter $I=[1:1:1]$, the last identity is equivalent to: $$ \left(\sin(A)+\sin(B)+\sin(C)\right)\prod_{cyc}\sin\frac{A-B}{2},\tag{2} $$ so, if we assume that $ABC$ is a scalene triangle, $I$ belongs to the Euler line iff $$ \cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} = 0, \tag{3} $$ no way.

Proving that the orthocenter lies on $OD$?

2 Answers2