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Difference between revisions of "2023 USAMO Problems/Problem 1"

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In an acute triangle <math>ABC</math>, let <math>M</math> be the midpoint of <math>\overline{BC}</math>. Let <math>P</math> be the foot of the perpendicular from <math>C</math> to <math>AM</math>. Suppose that the circumcircle of triangle <math>ABP</math> intersects line <math>BC</math> at two distinct points <math>B</math> and <math>Q</math>. Let <math>N</math> be the midpoint of <math>\overline{AQ}</math>. Prove that <math>NB=NC</math>.
 
In an acute triangle <math>ABC</math>, let <math>M</math> be the midpoint of <math>\overline{BC}</math>. Let <math>P</math> be the foot of the perpendicular from <math>C</math> to <math>AM</math>. Suppose that the circumcircle of triangle <math>ABP</math> intersects line <math>BC</math> at two distinct points <math>B</math> and <math>Q</math>. Let <math>N</math> be the midpoint of <math>\overline{AQ}</math>. Prove that <math>NB=NC</math>.
 
== Solution 1 ==  
 
== Solution 1 ==  
−
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+
Let <math>X</math> be the foot from <math>A</math> to <math>\overline{BC}</math>. By definition, <math>\angle AXM = \angle MPC = 90^{\circ}</math>. Thus, <math>\triangle AXM \sim \triangle CPM</math>, and <math>\triangle BMP \sim \triangle AMQ</math>.  
−
pen qqwuqq=rgb(0.,0.39215686274509803,0.);
 
−
pair A=(10.,3.), B=(6.,-5.), C=(18.,-5.), M=(12.,-5.), P=(12.352941176470589,-6.411764705882353), Q=(14.,-5.), X=(10.,-5.);
 
−
draw((12.691273728812305,-6.327181567796924)--(12.606690590726876,-5.988849015455207)--(12.26835803838516,-6.073432153540637)--P--cycle,linewidth(2.)+qqwuqq);
 
−
draw(A--B,linewidth(2.)); draw(B--C,linewidth(2.)); draw(C--A,linewidth(2.)); draw(A--P,linewidth(2.)); draw(circle((10.,-2.),5.),linewidth(2.)); draw(A--Q,linewidth(2.)); draw(A--X,linewidth(2.)); draw(C--P,linewidth(2.)); draw(B--P,linewidth(2.)); draw((12.,-1.)--M,linewidth(2.));
 
−
dot(A,ds); label("<math>A</math>",(10.065573740420717,3.1698164377622584),NE*lsf); dot(B,ds); label("<math>B</math>",(6.070652045162034,-4.836466959731464),NE*lsf); dot(C,ds); label("<math>C</math>",(18.071857137914453,-4.836466959731464),NE*lsf); dot(M,linewidth(4.pt)+ds); label("<math>M</math>",(12.071254591538244,-4.86934697368421),NE*lsf); dot(P,linewidth(4.pt)+ds); label("<math>P</math>",(12.416494738042081,-6.283187573652301),NE*lsf); dot(Q,linewidth(4.pt)+ds); label("<math>Q</math>",(14.060495435679398,-4.86934697368421),NE*lsf); dot((12.,-1.),linewidth(4.pt)+ds); label("<math>N</math>",(12.071254591538244,-0.8744252784255355),NE*lsf); dot(X,linewidth(4.pt)+ds); label("<math>X</math>",(10.065573740420717,-4.86934697368421),NE*lsf); label("<math>\alpha = 90^\\circ</math>",(12.449374751994828,-6.233867552723181),NE*lsf,qqwuqq);
 
−
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]
 
−
Let <math>X</math> be the foot from <math>A</math> to <math>\overline{BC}</math>. By definition, <math>\angle AXM = \angle MPC = 90^{\circ}</math>. Thus, <math>\triangle AXM \sim \triangle MPC</math>, and <math>\triangle BMP \sim \triangle AMQ</math>.  
 
  
−
From this, we have <math>\frac{MP}{MX} = \frac{MA}{MC} = \frac{MP}{MQ} = \frac{MA}{MB}</math>, as <math>MC=MB</math>. Thus, <math>M</math> is also the midpoint of <math>XQ</math>.
+
From this, we have <math>\frac{MP}{MX} = \frac{MC}{MA} = \frac{MP}{MQ}</math>, as <math>MC=MB</math>. Thus, <math>M</math> is also the midpoint of <math>XQ</math>.
  
−
Now, <math>NB = NC</math> iff <math>N</math> lies on the perpendicular bisector of <math>\overline{BC}</math>. As <math>N</math> lies on the perpendicular bisector of <math>\overline{XQ}</math>, which is also the perpendicular bisector of <math>\overline{BC}</math> (as <math>M</math> is also the midpoint of <math>XQ</math>), we are done.  
+
Now, <math>NB = NC</math> if <math>N</math> lies on the perpendicular bisector of <math>\overline{BC}</math>. As <math>N</math> lies on the perpendicular bisector of <math>\overline{XQ}</math>, which is also the perpendicular bisector of <math>\overline{BC}</math> (as <math>M</math> is also the midpoint of <math>XQ</math>), we are done.  
−
~ Martin2001, ApraTrip
+
~ Martin2001
 +
~ SomebodyST (minor edits)
 +
 
 +
==Solution 2 ==
 +
 
 +
We are going to use barycentric coordinates on <math>\triangle ABC</math>. Let <math>A=(1,0,0)</math>, <math>B=(0,1,0)</math>, <math>C=(0,0,1)</math>, and <math>a=BC</math>, <math>b=CA</math>, <math>c=AB</math>. We have <math>M=\left(0,\frac{1}{2},\frac{1}{2}\right)</math> and <math>P=(x:1:1)</math> so <math>\overrightarrow{CP}=\left(\frac{x}{x+2},\frac{1}{x+2},\frac{1}{x+2}-1\right)</math> and <math>\overrightarrow{AM}=\left(-1,\frac{1}{2},\frac{1}{2}\right)</math>. Since <math>\overleftrightarrow{CP}\perp\overleftrightarrow{AM}</math>, it follows that
 +
<cmath>\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*}</cmath>
 +
Solving this gives
 +
<cmath>\[
 +
x=\frac{2b^2-2c^2}{a^2-3b^2-c^2}
 +
\]</cmath>
 +
so
 +
<cmath>\[
 +
P=\left(\frac{b^2-c^2}{a^2-2b^2-2c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right).
 +
\]</cmath>
 +
The equation for <math>(ABP)</math> is
 +
<cmath>\[
 +
-a^2yz-b^2zx-c^2xy+ux+vy+wz=0.
 +
\]</cmath>
 +
Plugging in <math>A</math> and <math>B</math> gives <math>u=v=0</math>. Plugging in <math>P</math> gives
 +
<cmath>\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*}</cmath>
 +
so
 +
<cmath>\[
 +
w=\frac{2b^4-2c^4+a^4-3a^2b^2-a^2c^2}{2a^2-4b^2-4c^2}=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2}.
 +
\]</cmath>
 +
Now let <math>Q=(0,t,1-t)</math> where
 +
<cmath>\begin{align*}
 +
-a^2t(1-t)+w(1-t)&=0\\
 +
\implies t&=\frac{w}{a^2}
 +
\end{align*}</cmath>
 +
so <math>Q=\left(0,\frac{w}{a^2},1-\frac{w}{a^2}\right)</math>. It follows that <math>N=\left(\frac{1}{2},\frac{w}{2a^2},1-\frac{w}{2a^2}\right)</math>. It suffices to prove that <math>\overleftrightarrow{ON}\perp\overleftrightarrow{BC}</math>. Setting <math>\overrightarrow{O}=0</math>, we get <math>\overrightarrow{N}=\left(\frac{1}{2},\frac{w}{2a^2},1-\frac{w}{2a^2}\right)</math>. Furthermore we have <math>\overrightarrow{CB}=(0,1,-1)</math> so it suffices to prove that
 +
<cmath>\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*}</cmath>
 +
which is valid. <math>\square</math>
 +
 
 +
~KevinYang2.71
 +
 
 +
==See also==
 +
{{USAMO box|year=2023|before=First Problem|num-a=2|n=I}}

Latest revision as of 02:59, 31 August 2024

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, $\triangle AXM \sim \triangle CPM$, and $\triangle BMP \sim \triangle AMQ$.

From this, we have $\frac{MP}{MX} = \frac{MC}{MA} = \frac{MP}{MQ}$, as $MC=MB$. Thus, $M$ is also the midpoint of $XQ$.

Now, $NB = NC$ if $N$ 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 ~ SomebodyST (minor edits)

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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