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

Revision as of 12:41, 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=4.780937121446827; 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(" $A$ ",(9.062733314861951,3.1698164377622584),NE*lsf); dot(B,ds); label(" $B$ ",(6.070652045162033,-4.836466959731464),NE*lsf); dot(C,ds); label(" $C$ ",(19.058257556496844,-4.836466959731464),NE*lsf); dot(M,linewidth(4.pt)+ds); label(" $M$ ",(12.564454800829438,-4.86934697368421),NE*lsf); dot(P,linewidth(4.pt)+ds); label(" $P$ ",(13.616615247317322,-7.253147985258316),NE*lsf); dot(Q,linewidth(4.pt)+ds); label(" $Q$ ",(16.066176286796924,-4.86934697368421),NE*lsf); dot((12.5,-1.),linewidth(4.pt)+ds); label(" $N$ ",(12.564454800829438,-0.8744252784255355),NE*lsf); dot(X,linewidth(4.pt)+ds); label(" $X$ ",(9.062733314861951,-4.86934697368421),NE*lsf); label("$\alpha = 90^\\circ$ (Error compiling LaTeX. Unknown error_msg)",(13.600175240340947,-7.187387957352823),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 12:41, 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=4.780937121446827;
+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>.