Difference between revisions of "2023 USAMO Problems/Problem 1"
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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. | ||
~ Martin2001, ApraTrip | ~ Martin2001, ApraTrip | ||
+ | ==See also== | ||
+ | {{Usamo box|year=2023|num-b=8|num-a=10|n=I}} |
Revision as of 16:06, 16 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
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