AoPS - Problem

Source: Saudi Arabia IMO TST Day IV Problem 4

Tags: geometry, circumcircle, incenter, geometric transformation, reflection, perpendicular bisector, angle bisector

TheMaskedMagician

22.07.2014 22:17

Points $A_1,~ B_1,~ C_1$ lie on the sides $BC,~ AC$ and $AB$ of a triangle $ABC$, respectively, such that $AB_1 -AC_1 = CA_1 -CB_1 = BC_1 -BA_1$. Let $I_A,~ I_B,~ I_C$ be the incenters of triangles $AB_1C_1,~ A_1BC_1$ and $A_1B_1C$ respectively. Prove that the circumcenter of triangle $I_AI_BI_C$, is the incenter of triangle $ABC$.

I liked this problem the most of the geos on the test This is much shorter, I think: Let $I$ be the incenter of $\triangle ABC$ and reflect $A_1$ over $BI_b$ to $A_2$, $B_1$ over $CI_c$ to $B_2$, $C_1$ over $AI_a$ to $C_2$. Now our condition is $A_1B_2=B_1C_2=C_1A_2$. Since the perpendicular bisector of each segment is an angle bisector of $ABC$, we see $(A_1B_1C_1A_2B_2C_2)$ has center $I$. It suffices to show $I_a,I_b,I_c$ lie on the same circle. Now $A_2I_b=I_bA_1$ and $\angle A_2C_1I_b=\angle I_bC_1A_1$, so $I_b$ lies on the circle, and similarly for the others, as desired. The weird length conditions and angle bisectors scream "reflections!"