Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(double slash)
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>.
 +
 +
 +
 +
 +
 +
 +
 +
 +
 
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 10:32, 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 $NB=NC$.





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. -ApraTrip, Martin2001

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_USAMO_Problems/Problem_1&oldid=192055"