wasnt Q5 in Iran TST 2006 the one with the radius of the exscribed A - circle qual to the circumradius and proving O was the orthocentre of the triangle MNL? or do you have two sets of Iran TST's or something?

Yes we have two sets of TST's. Now both are finished and you can see them in National Olympiads section.

Remark: $AP,BQ,CR$ concur in the circumcenter of the triangle $ABC$.

Quadrilateral $OQDR$ being cyclic (see upper remark) we have $QR=OD\sin D=OD \sin 2A$ and the analogous relations. But $OD\geq OD'$ with $D'$ being the midpoint of $BC$ and $DE=R\sin 2C$ so it will be enough to prove: \[ \sum 2R\cos A\sin 2A\geq \sum R\sin 2A\Leftrightarrow \sum \sin 2A(2\cos A - 1)\geq 0 (1) \] If we take $A\leq B \leq C$ we distinguish 2 cases: when all angles are greater than $\frac{\pi}4$ and when one is less than $\frac{\pi}4$ and the other ones are greater. In the second case we have $\frac{\pi}2-2A<2B-\frac{\pi}2$ so $\sin 2A\geq \sin 2B\geq \sin 2C$. The previous inequalities are also clear for the first case. All we have to do now is apply Abel's inequality in $(1)$ (and take into account that $\cos A\geq \frac 12$, $\cos A+\cos B\geq 1$ and $\sum \cos A\geq \frac 32$). Equality for an equilateral triangle.

Ikap, I don't quite think that \[ \cos A+\cos B+\cos C\geq\frac{3}{2}. \]

ikap wrote: Quadrilateral $OQDR$ being cyclic (see upper remark) we have $QR=OD\sin D=OD \sin 2A$ and the analogous relations. But $OD\geq OD'$ with $D'$ being the midpoint of $BC$ and $DE=R\sin 2C$ so it will be enough to prove: \[ \sum 2R\cos A\sin 2A\geq \sum R\sin 2A\Leftrightarrow \sum \sin 2A(2\cos A - 1)\geq 0 (1) \] If we take $A\leq B \leq C$ we distinguish 2 cases: when all angles are greater than $\frac{\pi}4$ and when one is less than $\frac{\pi}4$ and the other ones are greater. In the second case we have $\frac{\pi}2-2A<2B-\frac{\pi}2$ so $\sin 2A\geq \sin 2B\geq \sin 2C$. The previous inequalities are also clear for the first case. All we have to do now is apply Abel's inequality in $(1)$ (and take into account that $\cos A\geq \frac 12$, $\cos A+\cos B\geq 1$ and $\sum \cos A\geq \frac 32$). Equality for an equilateral triangle. though your proof is wrong but the (1) is right ,and this one is equivalent to the Modalva TST 2006 day2 problem2

Can any one give the proof why $AP,BQ,CR $ are concurrent at point $O$.I got it but by complex geometry,it was very complicated.

Babai wrote: Can any one give the proof why $AP,BQ,CR $ are concurrent at point $O$. $O,H$ are isogonal conjugates with respect to $\triangle{ABC}$. Also, we can just project $PQ$ onto $DE$, etc. (giving lengths of $2R\sin{A}(\cos{A}+\cos{B}\cos{C}(\cos{2B}+\cos{2C}))$, etc.). Since $DE=2R\sin{C}\cos{C}$, etc. it eventually simplifies to \[\sum\sin{4A}+\sin{2A}\ge 0\Longleftrightarrow\sum\sin(\pi-2B+\pi-2C)\ge\sum\sin(2\pi-4A),\]which, if WLOG $0\le A\le B\le C\le\pi/2$, follows from Karamata on the concave function $f(x)=\sin{x}$ (for $x\in(0,\pi)$) and \[(2\pi-4A,2\pi-4B,2\pi-4C)\succ(\pi-2A+\pi-2B,\pi-2A+\pi-2C,\pi-2B+\pi-2C).\] BTW, note that by simple trig calculations ($H$ is the incenter of $\triangle{DEF}$) $\triangle{PQR}$ is the extouch triangle of $\triangle{DEF}$, so this is the same as USA TSTST 2011.7 (but I do think this problem disguised the fact better, since I didn't realize the equivalence until I was already done). Now, I'm probably missing something obvious, but does anyone have a better reason for this?

who can give a nice resolution to the problem?

Even though this is not the cutest, I actually kinda enjoyed it. It makes more sense to consider the problem as in/ex reference frame rather than ortho. Indeed, reformulate the problem as follows: Quote: Let $B_e$ be the Bevan point of $\triangle ABC$. Let $D,E,F$ be feet from $B_e$ to $\overline{AB},\overline{BC},\overline{AC}$, respectively. Prove that $DE+EF+FD\geq \tfrac{1}{2}(AB+BC+CA)$. Let $I_AI_BI_C$ be the excentral triangle. Using properties of Bevan point, we have $B_e$ the circumcenter of $(I_AI_BI_C)$ and $B_e$ the concurrency of $\overline{I_AD},\overline{I_BE},\overline{I_CF}$. Also, $AEFB_e,BDFB_e,CDEB_e$ are cyclic. Hence, using Ptolemy and Law of Sines, we obtain that \begin{align*} DE^2&=DE\cdot B_eC\cdot \sin{\angle ACB}\\&=\frac{AB}{2R}(CE\cdot B_eD+CD\cdot B_eE)\\&=\frac{c}{2R}((s-a)(2R-r_A)+(s-b)(2R-r_B)), \end{align*}where $r_A$ is the radius of $A$-excircle and so on, $R$ the circumradius, $s$ the semiperimeter, as usual. Here, we also used the fact that the radius of $(I_AI_BI_C)$ is twice the radius of $(ABC)$. Observe that \begin{align*} \frac{r_A}{2R}=\frac{r_A}{r}\cdot \frac{r}{2R}=\frac{s}{(s-a)}\cdot \frac{r}{2R}. \end{align*}Thus, furthermore, \begin{align*} DE^2&=\frac{c}{2R}((s-a)(2R-r_A)+(s-b)(2R-r_B))\\&=c((s-a)(1-\frac{r_A}{2R})+(s-b)(1-\frac{r_B}{2R})) \\&=c((s-a)(1-\frac{s}{(s-a)}\cdot \frac{r}{2R})+(s-b)(1-\frac{s}{(s-b)}\cdot \frac{r}{2R}))\\&=c\left(c-\frac{sr}{R}\right). \end{align*}Now, we use $\frac{sr}{R}=\frac{S}{R}=c\cdot \sin{\alpha}\cdot \sin{\beta}$, where $S$ denotes the area of $\triangle ABC$, $\alpha$ denotes $\angle BAC$ and so on. Therefore, the given inequality is now equivalent to \begin{align*} \sum_{cyc}c\sqrt{1-\sin{\alpha}\cdot \sin{\beta}}\geq \frac{a+b+c}{2}. \end{align*}Now, we do algebraic manipulations as such:\begin{align*} \sqrt{1-\sin{\alpha}\cdot \sin{\beta}}=\sqrt{\frac{\sin^2{\alpha}+\sin^2{\beta}+\cos^2{\alpha}+\cos^2{\beta}-2\sin{\alpha}\sin{\beta}}{2}}\geq \sqrt{\frac{\cos^2{\alpha}+\cos^2{\beta}}{2}}\geq \frac{\cos{\alpha}+\cos{\beta}}{2}, \end{align*}where we used the trivial inequality and QM-AM. Now just note that $\sum_{cyc}c(\cos{\alpha}+\cos{\beta})=a+b+c$ as all projections simply cover the whole perimeter. $\blacksquare$