Difference between revisions of "2023 USAMO Problems/Problem 1"
Martin2001 (talk | contribs) (Created page with "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 \...") |
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Now, <math>NB = NC</math> iff <math>N</math> lies on the perpendicular bisector of <math>\overline{BC}</math>. As <math>N</math> lies on the perpendicular bisector of <math>\overline{XQ}</math>, which is also the perpendicular bisector of <math>\overline{BC}</math> (as <math>M</math> is also the midpoint of <math>XQ</math>), we are done. | Now, <math>NB = NC</math> iff <math>N</math> lies on the perpendicular bisector of <math>\overline{BC}</math>. As <math>N</math> lies on the perpendicular bisector of <math>\overline{XQ}</math>, which is also the perpendicular bisector of <math>\overline{BC}</math> (as <math>M</math> is also the midpoint of <math>XQ</math>), we are done. | ||
− | + | -ApraTrip, Martin2001 |
Revision as of 10:31, 13 April 2023
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. -ApraTrip, Martin2001