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2023 USAMO Problems/Problem 1

Revision as of 18:17, 28 April 2023 by Lpieleanu (talk | contribs) (→‎Solution 1)

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

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. ~ Martin2001

Solution 2

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

See also

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