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Revision as of 21:55, 18 October 2022

Rectangles $BCC_1B_2,$ $CAA_1C_2,$ and $ABB_1A_2$ are erected outside an acute triangle $ABC.$ Suppose that\[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]Prove that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent.

Solution

2021 USAMO 1.png

Let $D$ be the second point of intersection of the circles $AB_1B$ and $AA_1C.$ Then \[\angle ADB = 180^\circ – \angle AB_1B,\angle ADC = 180^\circ – \angle AA_1C \implies\] \[\angle BDC = 360^\circ – \angle ADB – \angle ADC =\] \[= 360^\circ – (180^\circ – \angle AB_1B) – (180^\circ – \angle AA_1C) =\] \[=\angle AB_1B + \angle AA_1C \implies \angle BDC + \angle BC_1C = 180^\circ \implies\] $BDCC_1B_2$ is cyclic with diameters $BC_1$ and $CB_2 \implies \angle CDB_2 = 90^\circ.$ Similarly, $\angle CDA_1 = 90^\circ \implies$ points $A_1, D,$ and $B_2$ are collinear.

Similarly, triples of points $A_2, D, C_1$ and $C_2, D, B_1$ are collinear.

(After USAMO 2021 Solution Notes – Evan Chen)

vladimir.shelomovskii@gmail.com, vvsss

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