Difference between revisions of "2012 USAMO Problems/Problem 5"
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to be zero, which is true because the columns sum to zero. | to be zero, which is true because the columns sum to zero. | ||
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+ | ~knowingant | ||
==See also== | ==See also== |
Revision as of 12:23, 26 April 2025
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
We will use coordinates. Without loss of generality, let and let
be the line
. Let
,
, and
. Then
is the intersection of the lines
\begin{align*}
y &= -\frac{y_1}{x_1}x \\
y - y_2 &= \frac{y_2 - y_3}{x_2 - x_3}(x - x_2).
\end{align*}
Solving the system of equations, we see it is
To check collinearity, we need the determinant of the matrix
to be zero, which is true because the columns sum to zero.
~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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.