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

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

@@ Line 1: / Line 1: @@
 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 ==
-[asy]import graph; size(28.013771887739892cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.278031073276777,xmax=26.735740814463117,ymin=-9.456108920092317,ymax=4.780937121446827;
+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=(9.,3.), B=(6.,-5.), C=(19.,-5.), M=(12.5,-5.), P=(13.544262295081968,-7.386885245901639), Q=(16.,-5.), X=(9.,-5.);
-draw((13.863767779500606,-7.247101596468485)--(13.723984130067452,-6.927596112049847)--(13.404478645648814,-7.067379761483001)--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((11.,-2.3125),5.67650035232977),linewidth(2.)); draw(A--Q,linewidth(2.)); draw(A--X,linewidth(2.)); draw(C--P,linewidth(2.)); draw(B--P,linewidth(2.)); draw((12.5,-1.)--M,linewidth(2.));
-dot(A,ds); label("<math>A</math>",(9.062733314861951,3.1698164377622584),NE*lsf); dot(B,ds); label("<math>B</math>",(6.070652045162033,-4.836466959731464),NE*lsf); dot(C,ds); label("<math>C</math>",(19.058257556496844,-4.836466959731464),NE*lsf); dot(M,linewidth(4.pt)+ds); label("<math>M</math>",(12.564454800829438,-4.86934697368421),NE*lsf); dot(P,linewidth(4.pt)+ds); label("<math>P</math>",(13.616615247317322,-7.253147985258316),NE*lsf); dot(Q,linewidth(4.pt)+ds); label("<math>Q</math>",(16.066176286796924,-4.86934697368421),NE*lsf); dot((12.5,-1.),linewidth(4.pt)+ds); label("<math>N</math>",(12.564454800829438,-0.8744252784255355),NE*lsf); dot(X,linewidth(4.pt)+ds); label("<math>X</math>",(9.062733314861951,-4.86934697368421),NE*lsf); label("<math>\alpha = 90^\\circ</math>",(13.600175240340947,-7.187387957352823),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}}

2023 USAMO (Problems • Resources)
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Difference between revisions of "2023 USAMO Problems/Problem 1"

Latest revision as of 02:59, 31 August 2024

Solution 1

Solution 2

See also