2021 USAMO Problems/Problem 1

Problem

Rectangles $BCC_{1}B_{2}$ , $CAA_{1}C_{2}$ , and $ABB_{1}A_{2}$ are erected outside an acute triangle $ABC$ . Suppose that $\angle BC_{1}C + \angle CA_{1}A + \angle AB_{1}B = 180^{\circ}.$ Prove that lines $B_{1}C_{2}$ , $C_{1}A_{2}$ , and $A_{1}B_{2}$ are concurrent.

Solution

Let $D$ be the second point of intersection of the circles $AB_1B$ and $AA_1C.$ Then: $\begin{align*} \angle ADB &= 180^\circ – \angle AB_1B,&\angle ADC &= 180^\circ – \angle AA_1C\\ \angle BDC &= 360^\circ – \angle ADB – \angle ADC\\ &= 360^\circ – (180^\circ – \angle AB_1B) – (180^\circ – \angle AA_1C)\\ &= \angle AB_1B + \angle AA_1C\\ \angle BDC + \angle BC_1C &= 180^\circ \end{align*}$ Therefore, $BDCC_1B_2$ is cyclic with diameters $BC_1$ and $CB_2$ , and thus $\angle CDB_2 = 90^\circ.$ Similarly, $\angle CDA_1 = 90^\circ$ , meaning points $A_1$ , $D$ , and $B_2$ are collinear.

Similarly, the 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

Video Solution

https://youtube.com/watch?v=6e_IGnpQGEg

2021 USAMO (Problems • Resources)
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All USAMO Problems and Solutions

2021 USAJMO (Problems • Resources)
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All USAJMO Problems and Solutions

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2021 USAMO Problems/Problem 1

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Video Solution

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