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 (:

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 (Less technical bary)

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

Solution 4 (Less bashy bary)

We employ barycentric coordinates. Set $AMC$ as the reference triangle with $A = (1, 0, 0)$ , $M = (0, 1, 0)$ , and $C = (0, 0, 1)$ . We immediately have, $P = (S_B : S_A : 0); B = (0, 2, -1)$ Since it passes through $A$ , for some $v, w$ , the equation of circle $(ABP)$ is, $(ABP): -a^2 yz - b^2 zx - c^2 xy + (vy + wz)(x + y + z) = 0$ Plugging in $P$ , $- c^2 S_{AB} + (v S_A)(c^2) = 0$ $\iff v = S_B$ Plugging in $B$ , $2a^2 + (2v - w) = 0$ $\iff 2a^2 + 2 S_B = 3 a^2 - b^2 + c^2 = w$ In conclusion the circle has formula, $(ABP): -a^2 yz - b^2 zx - c^2 xy + ((S_B)y + (3 a^2 - b^2 + c^2) z)(x + y + z) = 0$ $Q$ is the second intersection of circle $(ABP)$ with $\overline{CM}$ . We let $Q = (0, 1 - t, t)$ for some $t \neq -1$ . Plugging this in, $-a^2 (1-t)t + ((S_B)(1-t) + (3 a^2 - b^2 + c^2) t) = 0$ We claim that $t = -\frac{S_B}{a^2}$ is the other solution. $\left(1+ \frac{S_B}{a^2} \right) S_B + \left((S_B)\left(1+ \frac{S_B}{a^2}\right) - (3 a^2 - b^2 + c^2) \frac{S_B}{a^2}\right) = 0$ $\iff \left(\frac{3a^2 - b^2 + c^2}{2a^2} \right) + \left(\left(\frac{3a^2 - b^2 + c^2}{2a^2}\right) - (3 a^2 - b^2 + c^2) \frac{1}{a^2}\right) = 0$ Factoring out the $\frac{3a^2 - b^2 + c^2}{2a^2}$ , this is clearly true.

We also check that, these are not the same value. $-\frac{S_B}{a^2} = -1$ $\iff a^2 - b^2 + c^2 = 2a^2$ $\iff c^2 = a^2 + b^2$ The triangle is acute, so this is impossible.

Since we had a quadratic in $t$ with at most two solutions, the second intersection $Q$ is indeed, $Q = \left( 0, 1 + \frac{S_B}{a^2}, -\frac{S_B}{a^2} \right)$ Therefore, $N = \frac{A + Q}{2} = \left( \frac{1}{2}, \frac{1}{2} - \frac{S_B}{2a^2}, \frac{S_B}{2a^2} \right)$

2023 USAJMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

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