Difference between revisions of "2023 USAMO Problems/Problem 1"
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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>. | 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 == | == Solution 1 == | ||
− | [asy | + | [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.); | 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.); | 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.); |
Revision as of 13:42, 13 April 2023
In an acute triangle , let
be the midpoint of
. Let
be the foot of the perpendicular from
to
. Suppose that the circumcircle of triangle
intersects line
at two distinct points
and
. Let
be the midpoint of
. Prove that
.
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("",(9.062733314861951,3.1698164377622584),NE*lsf); dot(B,ds); label("
",(6.070652045162033,-4.836466959731464),NE*lsf); dot(C,ds); label("
",(19.058257556496844,-4.836466959731464),NE*lsf); dot(M,linewidth(4.pt)+ds); label("
",(12.564454800829438,-4.86934697368421),NE*lsf); dot(P,linewidth(4.pt)+ds); label("
",(13.616615247317322,-7.253147985258316),NE*lsf); dot(Q,linewidth(4.pt)+ds); label("
",(16.066176286796924,-4.86934697368421),NE*lsf); dot((12.5,-1.),linewidth(4.pt)+ds); label("
",(12.564454800829438,-0.8744252784255355),NE*lsf); dot(X,linewidth(4.pt)+ds); label("
",(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
be the foot from
to
. By definition,
. Thus,
, and
.
From this, we have , as
. Thus,
is also the midpoint of
.
Now, iff
lies on the perpendicular bisector of
. As
lies on the perpendicular bisector of
, which is also the perpendicular bisector of
(as
is also the midpoint of
), we are done.
~ Martin2001, ApraTrip