Raja Oktovin wrote: Let $ ABC$ be an acute-angled triangle such that there exist points $ D,E,F$ on side $ BC,CA,AB$, respectively such that the inradii of triangle $ AEF,BDF,CDE$ are all equal to $ r_0$. If the inradii of triangle $ DEF$ and $ ABC$ are $ r$ and $ R$, respectively, prove that \[ r + r_0 = R.\] Soewono, Bandung Indonesia IMO 2010 TST : : it is too early to choose team IMO 2010 .

cool enough uh? fyi, regional selection test was in april 2009, national olympiad was in august 2009, about 30 contestants go to 1st stage the 1st stage of tst is in october-november 2009, about 12 contestants go to 2nd stage the 2nd stage of tst will be in about march 2010 (together with APMO 2010), the 2nd stage is to choose 6 IMO 2010 contestants the 3rd stage is the last preparation to IMO.

Let $ O_1,O_2,O_3$ denote the centers of the incircles of $ \triangle AFE,$ $ \triangle BDF,$ $ \triangle CED.$ $ (O_1),(O_2),(O_3)$ touch $ EF,FD,DE,$ at $M,N,L.$ $ (O_2),(O_3)$ touch $ BC,$ at $P,Q.$ $ (O_3),(O_1)$ touch $ CA$ at $R,S$ and $ (O_1),(O_2)$ touch $ AB$ at $T,U.$ Since $ O_1O_2UT,$ $ O_2O_3QP$ and $O_3O_1SR$ are rectangles and $ O_1,O_2,O_3$ lie on the internal bisectors of $ A,B,C,$ it follows that $ \triangle ABC$ and $ \triangle O_1O_2O_3$ are centrally similar through their common incenter $ I.$ The similarity coefficient equals the ratio between their perimeters/inradii $ \frac {r - r_0}{r} = \frac {O_1O_2 + O_2O_3 + O_3O_1}{a + b + c} = \frac {PQ + RS + TU}{a + b + c}$ $\frac {r - r_0}{r} = \frac {DN + DL + EL + EM + FM + FN}{a + b + c} = \frac {DE + EF + FD}{a + b + c} \ (1)$ On the other hand, we have: $r(a + b + c) = r_0(a + b + c + DE + EF + FD) + \varrho(DE + EF + FD)$ $\Longrightarrow \ r - r_0 = \frac {(DE + EF + FD)( r_0 + \varrho)}{a + b + c} \ (2)$ Substituting $ (r - r_0)$ from $ (1)$ into $ (2)$ yields $ r = r_0 + \varrho.$