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

Revision as of 13:37, 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

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=(9.638133559035012,2.2984960680144826), B=(5.734005553668605,-4.81861693653532), C=(18.84453746580399,-4.836466959731463), M=(12.289271509736299,-4.827541948133391), P=(13.089191098134414,-6.97765918185881), Q=(14.950106647313914,-4.831164682073189), X=(9.628436372158689,-4.823919214193597); draw((13.41604889400778,-6.856056531808553)--(13.294446243957523,-6.529198735935188)--(12.967588448084157,-6.650801385985445)--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((10.344923836854495,-2.718588274420166),5.066624891969315),linewidth(2.)); draw(A--Q,linewidth(2.)); draw(A--X,linewidth(2.)); draw(C--P,linewidth(2.)); draw(B--P,linewidth(2.)); draw((12.294120103174464,-1.2663343070293533)--M,linewidth(2.)); dot(A,ds); label(" $A$ ",(9.703893586940504,2.462896137778213),NE*lsf); dot(B,ds); label(" $B$ ",(5.807611933540062,-4.655626882991359),NE*lsf); dot(C,ds); label(" $C$ ",(18.910297493709486,-4.6720668899677325),NE*lsf); dot(M,linewidth(4.pt)+ds); label(" $M$ ",(12.350734710136587,-4.688506896944105),NE*lsf); dot(P,linewidth(4.pt)+ds); label(" $P$ ",(13.156295051978871,-6.842147810848988),NE*lsf); dot(Q,linewidth(4.pt)+ds); label(" $Q$ ",(15.01401584030904,-4.7049469039204785),NE*lsf); dot((12.294120103174464,-1.2663343070293533),linewidth(4.pt)+ds); label(" $N$ ",(12.36717471711296,-1.1374653900475058),NE*lsf); dot(X,linewidth(4.pt)+ds); label(" $X$ ",(9.687453579964131,-4.688506896944105),NE*lsf); label("$\alpha = 90^\\circ$ (Error compiling LaTeX. Unknown error_msg)",(13.156295051978871,-6.792827789919868),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 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

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

Revision as of 13:37, 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 ==
+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=(9.638133559035012,2.2984960680144826), B=(5.734005553668605,-4.81861693653532), C=(18.84453746580399,-4.836466959731463), M=(12.289271509736299,-4.827541948133391), P=(13.089191098134414,-6.97765918185881), Q=(14.950106647313914,-4.831164682073189), X=(9.628436372158689,-4.823919214193597);
+draw((13.41604889400778,-6.856056531808553)--(13.294446243957523,-6.529198735935188)--(12.967588448084157,-6.650801385985445)--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((10.344923836854495,-2.718588274420166),5.066624891969315),linewidth(2.)); draw(A--Q,linewidth(2.)); draw(A--X,linewidth(2.)); draw(C--P,linewidth(2.)); draw(B--P,linewidth(2.)); draw((12.294120103174464,-1.2663343070293533)--M,linewidth(2.));
+dot(A,ds); label("<math>A</math>",(9.703893586940504,2.462896137778213),NE*lsf); dot(B,ds); label("<math>B</math>",(5.807611933540062,-4.655626882991359),NE*lsf); dot(C,ds); label("<math>C</math>",(18.910297493709486,-4.6720668899677325),NE*lsf); dot(M,linewidth(4.pt)+ds); label("<math>M</math>",(12.350734710136587,-4.688506896944105),NE*lsf); dot(P,linewidth(4.pt)+ds); label("<math>P</math>",(13.156295051978871,-6.842147810848988),NE*lsf); dot(Q,linewidth(4.pt)+ds); label("<math>Q</math>",(15.01401584030904,-4.7049469039204785),NE*lsf); dot((12.294120103174464,-1.2663343070293533),linewidth(4.pt)+ds); label("<math>N</math>",(12.36717471711296,-1.1374653900475058),NE*lsf); dot(X,linewidth(4.pt)+ds); label("<math>X</math>",(9.687453579964131,-4.688506896944105),NE*lsf); label("<math>\alpha = 90^\\circ</math>",(13.156295051978871,-6.792827789919868),NE*lsf,qqwuqq);
+clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 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>.