2023 USAMO Problems/Problem 1: Difference between revisions
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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 10:32, 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
