Difference between revisions of "2012 USAMO Problems/Problem 5"
Knowingant (talk | contribs) (deleting my solution: i found the points a', b', and c' i claimed the coordinates of in the solution are not actually collinear. i messed up some algebra somewhere and i'm not going through this thing again.) |
Knowingant (talk | contribs) (Add back cartesian bash) |
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~peppapig_ | ~peppapig_ | ||
+ | |||
+ | ==Solution 3, Cartesian== | ||
+ | |||
+ | We will use coordinates. Without loss of generality, let <math>P = (0,0)</math> and let <math>\gamma</math> be the line <math>x = 0</math>. Let <math>A = (x_1,y_1)</math>, <math>B = (x_2,y_2)</math>, and <math>C = (x_3,y_3)</math>. Then <math>A'</math> 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 | ||
+ | <cmath>\left(\frac{x_1(x_2y_3 - y_2x_3)}{x_1y_2 + y_1x_2 - x_1y_3 - x_3y_1},\frac{y_1(y_2x_3 - x_2y_3)}{x_1y_2 + y_1x_2 - x_1y_3 - x_3y_1}\right).</cmath> | ||
+ | To check collinearity, we need the determinant of the matrix with rows | ||
+ | \[\begin{bmatrix} | ||
+ | x_1(x_2y_3 - y_2x_3) & y_1(y_2x_3 - x_2y_3) & x_1y_2 + y_1x_2 - x_1y_3 - x_3y_1 | ||
+ | \end{bmatrix}\] | ||
+ | to be zero, which is true because the columns sum to zero. | ||
==See also== | ==See also== |
Revision as of 12:19, 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 with rows
\[\begin{bmatrix}
x_1(x_2y_3 - y_2x_3) & y_1(y_2x_3 - x_2y_3) & x_1y_2 + y_1x_2 - x_1y_3 - x_3y_1
\end{bmatrix}\] to be zero, which is true because the columns sum to zero.
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.