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

Revision as of 15:27, 13 April 2023

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

[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=4.7809371214468275; 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(" $A$ ",(10.065573740420717,3.1698164377622584),NE*lsf); dot(B,ds); label(" $B$ ",(6.070652045162034,-4.836466959731464),NE*lsf); dot(C,ds); label(" $C$ ",(18.071857137914453,-4.836466959731464),NE*lsf); dot(M,linewidth(4.pt)+ds); label(" $M$ ",(12.071254591538244,-4.86934697368421),NE*lsf); dot(P,linewidth(4.pt)+ds); label(" $P$ ",(12.416494738042081,-6.283187573652301),NE*lsf); dot(Q,linewidth(4.pt)+ds); label(" $Q$ ",(14.060495435679398,-4.86934697368421),NE*lsf); dot((12.,-1.),linewidth(4.pt)+ds); label(" $N$ ",(12.071254591538244,-0.8744252784255355),NE*lsf); dot(X,linewidth(4.pt)+ds); label(" $X$ ",(10.065573740420717,-4.86934697368421),NE*lsf); label("$\alpha = 90^\\circ$ (Error compiling LaTeX. Unknown error_msg)",(12.449374751994828,-6.233867552723181),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy] 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, ApraTrip

Page

Toolbox

Search

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

Revision as of 15:27, 13 April 2023

Solution 1

@@ 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=4.7809371214468275;
+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>.