MathDB

2

Part of 2019 Brazil Team Selection Test

Problems(2)

Prove that 3 tangents are concurrent

Source: 2019 Brazil IMO TST 3.2

6/13/2023
Let ABCABC be a triangle, and A1A_1, B1B_1, C1C_1 points on the sides BCBC, CACA, ABAB, respectively, such that the triangle A1B1C1A_1B_1C_1 is equilateral. Let I1I_1 and ω1\omega_1 be the incenter and the incircle of AB1C1AB_1C_1. Define I2I_2, ω2\omega_2 and I3I_3, ω3\omega_3 similarly, with respect to the triangles BA1C1BA_1C_1 and CA1B1CA_1B_1, respectively. Let l1≠BCl_1 \neq BC be the external tangent line to ω2\omega_2 and ω3\omega_3. Define l2l_2 and l3l_3 similarly, with respect to the pairs ω1\omega_1, ω3\omega_3 and ω1\omega_1, ω2\omega_2.
Knowing that A1I2=A1I3A_1I_2 = A_1I_3, show that the lines l1l_1, l2l_2, l3l_3 are concurrent.
geometryconcurrencyEquilateral Triangletangentexternal tangent
Always possible to choose some friends

Source: 2019 Brazil Ibero TST P2

6/14/2023
We say that a distribution of students lined upen in collumns is <spanclass=′latex−italic′>bacana</span><span class='latex-italic'>bacana</span> when there are no two friends in the same column. We know that all contestants in a math olympiad can be arranged in a <spanclass=′latex−italic′>bacana</span><span class='latex-italic'>bacana</span> configuration with nn columns, and that this is impossible with n−1n-1 columns. Show that we can choose competitors M1,M2,⋯ ,MnM_1, M_2, \cdots, M_n in such a way that MiM_i is on the ii-th column, for each i=1,2,3,…,ni = 1, 2, 3, \ldots, n and MiM_i is a friend of Mi+1M_{i+1} for each i=1,2,…,n−1i = 1, 2, \ldots, n - 1.
combinatorics