Difference between revisions of "2023 USAJMO Problems/Problem 2"

Revision as of 18:58, 6 October 2023

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

Solution 3

We are going to use barycentric coordinates on $\triangle ABC$ . Let $A=(1,0,0)$ , $B=(0,1,0)$ , $C=(0,0,1)$ , and $a=BC$ , $b=CA$ , $c=AB$ . We have $M=\left(0,\frac{1}{2},\frac{1}{2}\right)$ and $P=(x:1:1)$ so $\overrightarrow{CP}=\left(\frac{x}{x+2},\frac{1}{x+2},\frac{1}{x+2}-1\right)$ and $\overrightarrow{AM}=\left(-1,\frac{1}{2},\frac{1}{2}\right)$ . Since $\overleftrightarrow{CP}\perp\overleftrightarrow{AM}$ , it follows that $\begin{align*} a^2\left(\frac{1}{2}\cdot\frac{1}{x+2}+\frac{1}{2}\left(\frac{1}{x+2}-1\right)\right)+b^2\left(\frac{1}{2}\cdot\frac{x}{x+2}-\left(\frac{1}{x+2}-1\right)\right)\\ +c^2\left(\frac{1}{2}\cdot\frac{x}{x+2}-\frac{1}{x+2}\right)=0. \end{align*}$ Solving this gives $x=\frac{2b^2-2c^2}{a^2-3b^2-c^2}$ so $P=\left(\frac{b^2-c^2}{a^2-2b^2-2c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right).$ The equation for $(ABP)$ is $-a^2yz-b^2zx-c^2xy+ux+vy+wz=0.$ Plugging in $A$ and $B$ gives $u=v=0$ . Plugging in $P$ gives $\begin{align*} -a^2\left(\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right)^2-b^2\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\\ -c^2\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}+w\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}=0 \end{align*}$ so $w=\frac{2b^4-2c^4+a^4-3a^2b^2-a^2c^2}{2a^2-4b^2-4c^2}=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2}.$ Now let $Q=(0,t,1-t)$ where $\begin{align*} -a^2t(1-t)+w(1-t)&=0\\ \implies t&=\frac{w}{a^2} \end{align*}$ so $Q=\left(0,\frac{w}{a^2},1-\frac{w}{a^2}\right)$ . It follows that $N=\left(\frac{1}{2},\frac{w}{2a^2},1-\frac{w}{2a^2}\right)$ . It suffices to prove that $\overleftrightarrow{ON}\perp\overleftrightarrow{BC}$ . Setting $\overrightarrow{O}=0$ , we get $\overrightarrow{N}=\left(\frac{1}{2},\frac{w}{2a^2},1-\frac{w}{2a^2}\right)$ . Furthermore we have $\overrightarrow{CB}=(0,1,-1)$ so it suffices to prove that $\begin{align*} a^2\left(-\frac{w}{2a^2}+\frac{1}{2}-\frac{u}{2a^2}\right)+b^2\left(-\frac{1}{2}\right)+c^2\left(\frac{1}{2}\right)=0\\ \implies w=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2} \end{align*}$ which is valid. $\square$

~KevinYang2.71

@@ Line 65: / Line 65: @@
 ~KevinYang2.71
+==See Also==
+{{USAJMO newbox|year=2023|num-b=1|num-a=3}}
+{{MAA Notice}}

2023 USAJMO (Problems • Resources)
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