# 2021 USAMO Problems/Problem 1

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

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)