2023 USAMO Problems/Problem 1

Let $X$ be the foot from $A$ to $\overline{BC}$ . By definition, $\angle AXM = \angle MPC = 90^{\circ}$ . Thus, $\triangle AXM \sim \triangle MPC$ , and $\triangle BMP \sim \triangle AMQ$ .

From this, we have $\frac{MP}{MX} = \frac{MA}{MC} = \frac{MP}{MQ} = \frac{MA}{MB}$ , as $MC=MB$ . Thus, $M$ is also the midpoint of $XQ$ .

Now, $NB = NC$ iff $N$ lies on the perpendicular bisector of $\overline{BC}$ . As $N$ lies on the perpendicular bisector of $\overline{XQ}$ , which is also the perpendicular bisector of $\overline{BC}$ (as $M$ is also the midpoint of $XQ$ ), we are done. \ApraTrip, Martin2001

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2023 USAMO Problems/Problem 1