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

(→‎Solution 1)
(→‎Solution 1)
Line 3: Line 3:
  
 
import graph; size(28.013771887739892cm);  
 
import graph; size(28.013771887739892cm);  
−
real labelscalefactor = 0.5; /* changes label-to-point distance */
+
real labelscalefactor = 0.5;  
−
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
+
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);  
−
pen dotstyle = black; /* point style */
+
pen dotstyle = black;  
−
real xmin = -1.278031073276777, xmax = 26.735740814463117, ymin = -9.456108920092317, ymax = 4.7809371214468275;  /* image dimensions */
+
real xmin = -1.278031073276777, xmax = 26.735740814463117, ymin = -9.456108920092317, ymax = 4.7809371214468275;   
 
pen qqwuqq = rgb(0.,0.39215686274509803,0.);  
 
pen qqwuqq = rgb(0.,0.39215686274509803,0.);  
 
draw((13.41604889400778,-6.856056531808553)--(13.294446243957523,-6.529198735935188)--(12.967588448084157,-6.650801385985445)--(13.089191098134414,-6.97765918185881)--cycle, linewidth(2.) + qqwuqq);  
 
draw((13.41604889400778,-6.856056531808553)--(13.294446243957523,-6.529198735935188)--(12.967588448084157,-6.650801385985445)--(13.089191098134414,-6.97765918185881)--cycle, linewidth(2.) + qqwuqq);  
Line 37: Line 37:
 
label("<math>\alpha = 90^\\circ</math>", (13.156295051978871,-6.792827789919868), NE * labelscalefactor,qqwuqq);  
 
label("<math>\alpha = 90^\\circ</math>", (13.156295051978871,-6.792827789919868), NE * labelscalefactor,qqwuqq);  
 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  
 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  
−
/* end of picture */
+
 
 
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>.  
 
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>.  
  

Revision as of 12:31, 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 labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -1.278031073276777, xmax = 26.735740814463117, ymin = -9.456108920092317, ymax = 4.7809371214468275; pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((13.41604889400778,-6.856056531808553)--(13.294446243957523,-6.529198735935188)--(12.967588448084157,-6.650801385985445)--(13.089191098134414,-6.97765918185881)--cycle, linewidth(2.) + qqwuqq); draw((9.638133559035012,2.2984960680144826)--(5.734005553668605,-4.81861693653532), linewidth(2.)); draw((5.734005553668605,-4.81861693653532)--(18.84453746580399,-4.836466959731463), linewidth(2.)); draw((18.84453746580399,-4.836466959731463)--(9.638133559035012,2.2984960680144826), linewidth(2.)); draw((9.638133559035012,2.2984960680144826)--(13.089191098134414,-6.97765918185881), linewidth(2.)); draw(circle((10.344923836854495,-2.718588274420166), 5.066624891969315), linewidth(2.)); draw((9.638133559035012,2.2984960680144826)--(14.950106647313914,-4.831164682073189), linewidth(2.)); draw((9.638133559035012,2.2984960680144826)--(9.628436372158689,-4.823919214193597), linewidth(2.)); draw((18.84453746580399,-4.836466959731463)--(13.089191098134414,-6.97765918185881), linewidth(2.)); draw((5.734005553668605,-4.81861693653532)--(13.089191098134414,-6.97765918185881), linewidth(2.)); draw((12.294120103174464,-1.2663343070293533)--(12.289271509736299,-4.827541948133391), linewidth(2.)); dot((9.638133559035012,2.2984960680144826),dotstyle); label("$A$", (9.703893586940504,2.462896137778213), NE * labelscalefactor); dot((5.734005553668605,-4.81861693653532),dotstyle); label("$B$", (5.807611933540062,-4.655626882991359), NE * labelscalefactor); dot((18.84453746580399,-4.836466959731463),dotstyle); label("$C$", (18.910297493709486,-4.6720668899677325), NE * labelscalefactor); dot((12.289271509736299,-4.827541948133391),linewidth(4.pt) + dotstyle); label("$M$", (12.350734710136587,-4.688506896944105), NE * labelscalefactor); dot((13.089191098134414,-6.97765918185881),linewidth(4.pt) + dotstyle); label("$P$", (13.156295051978871,-6.842147810848988), NE * labelscalefactor); dot((14.950106647313914,-4.831164682073189),linewidth(4.pt) + dotstyle); label("$Q$", (15.01401584030904,-4.7049469039204785), NE * labelscalefactor); dot((12.294120103174464,-1.2663343070293533),linewidth(4.pt) + dotstyle); label("$N$", (12.36717471711296,-1.1374653900475058), NE * labelscalefactor); dot((9.628436372158689,-4.823919214193597),linewidth(4.pt) + dotstyle); label("$X$", (9.687453579964131,-4.688506896944105), NE * labelscalefactor); label("$\alpha = 90^\\circ$ (Error compiling LaTeX. Unknown error_msg)", (13.156295051978871,-6.792827789919868), NE * labelscalefactor,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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