2023 USAJMO Problems/Problem 2

Problem

(Holden Mui) In an acute triangle $ABC$ , let $M$ be the midpoint of $\overline{BC}$ . Let $P$ be the foot of the perpendicular from $C$ to $AM$ . Suppose that the circumcircle of triangle $ABP$ intersects line $BC$ at two distinct points $B$ and $Q$ . Let $N$ be the midpoint of $\overline{AQ}$ . Prove that $NB=NC$ .

Solution 1

The condition is solved only if $\triangle{NBC}$ is isosceles, which in turn only happens if $\overline{MN}$ is perpendicular to $\overline{BC}$ .

Now, draw the altitude from $A$ to $\overline{BC}$ , and call that point $X$ . Because of the Midline Theorem, the only way that this condition is met is if $\triangle{AXQ} \sim \triangle{NMQ}$ , or if $\overline{XM}=\overline{MQ}$ .

By $AA$ similarity, $\triangle{AXM} \sim \triangle{CPM}$ . Using similarity ratios, we get that $\frac{\overline{AM}}{\overline{XM}}=\frac{\overline{CM}}{\overline{PM}}$ . Rearranging, we get that $\overline{AM} \cdot \overline{MP}=\overline{XM} \cdot \overline{MC}$ . This implies that $AXPC$ is cyclic.

Now we start using Power of a Point. We get that $\overline{BX} \cdot \overline {XQ}= \overline{AM} \cdot \overline{MP}$ , and $\overline{AM} \cdot \overline{MP}=\overline{XM} \cdot \overline{MC}$ from before. This leads us to get that $\overline{BX} \cdot \overline {XQ}=\overline{XM} \cdot \overline{MC}$ .

Now we assign variables to the values of the segments. Let $\overline{BX}=a, \overline{XM}=b, \overline{MQ}=c, and \overline{QC}=d$ . The equation from above gets us that $(a+b)c=b(c+d)$ . As $a+b=c+d$ from the problem statements, this gets us that $b=c$ and $\overline{XC}=\overline{CQ}$ , and we are done.

-dragoon and rhydon516

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2023 USAJMO Problems/Problem 2

Problem

Solution 1