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2023 USAJMO Problems/Problem 2

Revision as of 15:36, 16 April 2023 by Leo.euler (talk | contribs)

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 (:

Solution 2

Let $D$ be the foot of the altitude from $A$ onto $BC$. We want to show that $DM=MQ$ for obvious reasons.

Notice that $ADPC$ is cyclic and that $M$ lies on the radical axis of $(ABPQ)$ and $(ADPC)$. By Power of a Point, $(CM)(DM)=(BM)(MQ)$. As $BM=CM$, we have $DM=MQ$, as desired.

- Leo.Euler

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