MathLabs

Problem 2

Let ABCABC be an acute-angled triangle. The internal bisector of angle AA meets the circumcircle of ABCABC again at A1A_1; points B1B_1 and C1C_1 are defined similarly. Let A0A_0 be the point where the line AA1AA_1 meets the external bisectors of angles BB and CC; points B0B_0 and C0C_0 are defined similarly. Prove that the area of triangle A0B0C0A_0B_0C_0 is twice the area of the hexagon AC1BA1CB1AC_1BA_1CB_1, and that this area is at least four times the area of triangle ABCABC.
Step 6 of 6: Sum the three inequalities
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In plain words

The orthocenter partitions the acute triangle; the three reflected-cap comparisons show that the hexagon has at least twice the area of ABCABC.

[ABC]≤[BCA1]+[CAB1]+[ABC1]=[AC1BA1CB1]−[ABC][ABC]\le[BCA_1]+[CAB_1]+[ABC_1]=[AC_1BA_1CB_1]-[ABC]
Detailed analysis

Apply Step 5 cyclically. Since ABCABC is acute, HH is inside it and [ABC]=[BCH]+[CAH]+[ABH][ABC]=[BCH]+[CAH]+[ABH]. Thus [ABC]≤[BCA1]+[CAB1]+[ABC1][ABC]\le[BCA_1]+[CAB_1]+[ABC_1]. The three terms on the right are exactly the three outward caps of the hexagon, so their sum is [AC1BA1CB1]−[ABC][AC_1BA_1CB_1]-[ABC]. Therefore 2[ABC]≤[AC1BA1CB1]=12[A0B0C0]2[ABC]\le[AC_1BA_1CB_1]=\tfrac12[A_0B_0C_0], and the desired [A0B0C0]≥4[ABC][A_0B_0C_0]\ge4[ABC] follows.