Difference between revisions of "2012 USAMO Problems/Problem 5"
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− | ==Solution 3, | + | ==Solution 3, Cartesian bash== |
Fix <math>P</math> to be at <math>(0,0)</math> and <math>\gamma</math> to be the line <math>x = 0</math>. Let the coordinates of point <math>Z \in \{A,B,C,A',B',C'\}</math> be <math>(x_Z,y_Z)</math>. Let | Fix <math>P</math> to be at <math>(0,0)</math> and <math>\gamma</math> to be the line <math>x = 0</math>. Let the coordinates of point <math>Z \in \{A,B,C,A',B',C'\}</math> be <math>(x_Z,y_Z)</math>. Let | ||
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==See also== | ==See also== | ||
*[[USAMO Problems and Solutions]] | *[[USAMO Problems and Solutions]] |
Revision as of 16:35, 26 April 2023
Contents
Problem
Let be a point in the plane of triangle
, and
a line passing through
. Let
,
,
be the points where the reflections of lines
,
,
with respect to
intersect lines
,
,
, respectively. Prove that
,
,
are collinear.
Solution
By the sine law on triangle ,
so
Similarly,
Hence,
Since angles and
are supplementary or equal, depending on the position of
on
,
Similarly,
By the reflective property, and
are supplementary or equal, so
Similarly,
Therefore,
so by Menelaus's theorem,
,
, and
are collinear.
Solution 2, Barycentric (Modified by Evan Chen)
We will perform barycentric coordinates on the triangle , with
,
, and
. Set
,
,
as usual. Since
,
,
are collinear, we will define
and
.
Claim: Line is the angle bisector of
,
, and
.
This is proved by observing that since
is the reflection of
across
, etc.
Thus is the intersection of the isogonal of
with respect to
with the line
; that is,
Analogously,
is the intersection of the isogonal of
with respect to
with the line
; that is,
The ratio of the first to third coordinate in these two points
is both
, so it follows
,
, and
are collinear.
~peppapig_
Solution 3, Cartesian bash
Fix to be at
and
to be the line
. Let the coordinates of point
be
. Let
The reflection of line with respect to
has equation
. Line
has equation
.
is the intersection of these two points. We now find
.
Then,
Similarly, we can find the coordinates of and
, which are
We can now find the slope of line .
Similarly,
Then,
Since the slope of line is equal to the slope of line
, points
,
, and
are collinear.
~KnowingAnt
See also
2012 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.