Difference between revisions of "2023 USAMO Problems/Problem 1"
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Now, <math>NB = NC</math> if <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> if <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 | ~ Martin2001 | ||
+ | ~ SomebodyST (minor edits) | ||
==Solution 2 == | ==Solution 2 == |
Revision as of 02:53, 31 August 2024
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, if
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
~ SomebodyST (minor edits)
Solution 2
We are going to use barycentric coordinates on . Let
,
,
, and
,
,
. We have
and
so
and
. Since
, it follows that
Solving this gives
so
The equation for
is
Plugging in
and
gives
. Plugging in
gives
so
Now let
where
so
. It follows that
. It suffices to prove that
. Setting
, we get
. Furthermore we have
so it suffices to prove that
which is valid.
~KevinYang2.71
See also
2023 USAMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |