Difference between revisions of "2023 USAMO Problems/Problem 1"
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− | + | == Solution 1 == | |
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 11:32, 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
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