Just Beauty (Bulgaria 2002 : Problem 4)

[sourav das]

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.

[*Mahi*]

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.

[sourav das]

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

[*Mahi*]

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.

[*Mahi*]

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

[sourav das]

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

[*Mahi*]

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!

[sourav das]

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.

$ABC$ and orthotic triangle of $P_{1}P_{2}P_{3}$ (Say $H_{1}H_{2}H_{3}$) are homothetic. So joining lines of incenter and circumcenter of $ABC$ and $H_{1}H_{2}H_{3}$ are parallel or collinear. Incenter and circumcenter of $H_{1}H_{2}H_{3}$ are orthocenter and nine point circle center of $P_{1}P_{2}P_{3}$ which are on the Euler line . But incenter of $ABC$ is circumcenter of $P_{1}P_{2}P_{3}$ which is on the Euler line of $P_{1}P_{2}P_{3}$ too. Therefor those three points are collinear.

[*Mahi*]

Hmmm... mine was same, though,

... 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.

[nayel]

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.