BdMO Online Forum

Problem 4 Let $I$ be the incenter of non-equilateral triangle $ABC$,
and let $T_{1}, T_{2}, T_{3}$ be the tangency points of the incircle with sides $BC,
CA, AB$ respectively. Prove that the orthocenter of triangle $T_{1}T_{2}T_{3}$
lies on line $OI$, where $O$ is the circumcenter of triangle $ABC$.
Note: Official solution is toooo big. So try a unique way to solve it and discover why Geometry consider to be beauty of Mathematics.

Complex!
The actual solution is just too long to be LATEXed, so I'm posting only the scheme.

1.Let $T_{1}, T_{2}, T_{3}$ be complex numbers $a,b,c$ respectively.
2.It will lead to $I=0$ and $A= \frac {2ab} {a+b} ,B= \frac {2bc} {b+c},C= \frac {2ca} {c+a}$
3.They will lead to $O=\frac {abc(a+b+c)} {(a+b)(b+c)(c+a)}$
4.Again from $\triangle T_{1}T_{2}T_{3}$ the orthocenter is $a+b+c$
5. So at last we have $0,a+b+c,\frac {abc(a+b+c)} {(a+b)(b+c)(c+a)}$ collinear ,which proves the claim.

Waiting to see your "beautiful" solution.Please post it soon.

I'm expecting rather beautiful solution Mahi. Your solution is also complex like the official solution. Don't you find out the short proof??

Where complex surely solves a problem, I think it's more professional to complete it that way. But now you tell about it, I think I'd try and find out where the beauty is. By the way it is not "Complex" at all, just the calculations are too lengthy, so I didn't latexed it. In fact the answer revolves before you all the time.

BTW, welcome back.

And there is also another thing, there is a hell lot of cases in this kind of problem. So I think it is wiser to take the shortest path and do away with the problem in one calculation

Don't you try homothety? Seriously I'm expecting the solution with homothety. One more day and i will post my solution.

It was done actually... but it's nothing that beautiful,is it? Or perhaps I was expecting too much from it? Don't know.

But you are always welcome to post the solutions... though such hurry won't do any good in these Eid days!

I thought the problem would be solved in Euclidean way .I was so amused to solve the problem that it was more than beauty for me. I don't think it is a very good idea to start with complex and not with Euclidean (My personal comment, don't take it seriously ). I am posting my solution now as i think i won't be even able to open my laptop for a few days.

Hmmm... mine was same, though,

sourav das wrote:... I don't think it is a very good idea to start with complex and not with Euclidean (My personal comment, don't take it seriously ) ...

Why do you think that? I never said I do or I have to solve every problem with complex, but if it is "Doable" with complex, it can be checked within a few (when I say few, I really do mean FEW, say five) minutes. And if there is a sure solve, why one would take a leap towards the unknown? And this isn't cowardliness, because I think you've got to do anything that's necessary to solve a problem.

Thanks for your personal comment, though.

Let $X,Y,Z$ denote the feet of the altitudes of triangle $T_1T_2T_3$ respective to $T_1,T_2,T_3$.

$\triangle XYZ\sim\triangle ABC$, because they have parallel pairs of sides. Let $K$ be the centre of similitude.

Let $I'$ be the incentre of $\triangle XYZ$. $II'$ is the Euler line of $\triangle T_1T_2T_3$, hence the circumcentre $O'$ of its pedal triangle $XYZ$ lies on (is the midpoint of) $II'$. Hence $K, I, O'$ are collinear. But $K,O,O'$ are collinear. Hence $IO$ is the Euler line of $\triangle T_1T_2T_3$, and we are done.