2012 USAMO Problems/Problem 5
Contents
[hide]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, Coordinate Bash
Fix to be at
and
to be the line
. Let the coordinates of point
be
.
The reflection of line with respect to
has equation
. Line
has equation
.
is the intersection of these two points. We now find
.
\begin{equation*} \begin{split}
x_{A'} & = \frac{\frac{y_B-y_C}{x_B-x_C}x_B - y_B}{\frac{y_B-y_C}{x_B-x_C} + \frac{x_A}{y_A}} = \frac{x_B(y_B - y_C) - y_B(x_B - x_C)}{y_B - y_C + \frac{x_A}{y_A}(x_B - x_C)} = \frac{y_A(x_Cy_B - x_By_C)}{y_Ay_B - y_Ay_C + x_Ax_B - x_Ax_C}
\end{split} \end{equation*}
Then,
Similarly, we can find the coordinates of and
, which are
\begin{align*}
x_{B'} &= \frac{y_B(x_Ay_C - x_Cy_A)}{y_By_C - y_By_A + x_Bx_C - x_Bx_A} \\ y_{B'} &= \frac{x_B(x_Cy_A - x_Ay_C)}{y_By_C - y_By_A + x_Bx_C - x_Bx_A} \\ x_{C'} &= \frac{y_A(x_By_A - x_Ay_B)}{y_Cy_A - y_Cy_B + x_Cx_A - x_Cx_B} \\ y_{C'} &= \frac{x_C(x_Ay_B - x_By_A)}{y_Cy_A - y_Cy_B + x_Cx_A - x_Cx_B}\text{.}
\end{align*}
Define
We can now find the slope of line .
\begin{equation*} \begin{split}
m_{A'B'} &= \frac{y_{A'}-y_{B'}}{x_{A'}-x_{B'}} \\ &= \frac{\frac{x_A(x_By_C - x_Cy_B)}{y_Ay_B - y_Ay_C + x_Ax_B - x_Ax_C} - \frac{x_B(x_Cy_A - x_Ay_C)}{y_By_C - y_By_A + x_Bx_C - x_Bx_A}}{\frac{y_A(x_Cy_B - x_By_C)}{y_Ay_B - y_Ay_C + x_Ax_B - x_Ax_C} - \frac{y_B(x_Ay_C - x_Cy_A)}{y_By_C - y_By_A + x_Bx_C - x_Bx_A}} \\ &= \frac{x_A(x_By_C - x_Cy_B)(y_By_C - y_By_A + x_Bx_C - x_Bx_A) - x_B(x_Cy_A - x_Ay_C)(y_Ay_B - y_Ay_C + x_Ax_B - x_Ax_C)}{y_A(x_Cy_B - x_By_C)(y_By_C - y_By_A + x_Bx_C - x_Bx_A) - y_B(x_Ay_C - x_Cy_A)(y_Ay_B - y_Ay_C + x_Ax_B - x_Ax_C)} \\ &= \frac{\sum_{ABC}(x_A^2x_Bx_Cy_B + x_Ax_Cy_Ay_B^2 - x_A^2x_Bx_Cy_C - x_Ax_Cy_B^2y_C)}{\sum_{ABC}(x_Ax_Cy_By_C + x_Ay_Ay_By_C^2 - x_A^2x_By_By_C - x_Ay_Ay_B^2y_C)}
\end{split} \end{equation*}
Similarly,
Then,
\begin{equation*} \begin{split}
m_{B'C'} &= \frac{\sum_{ABC}x_B^2x_Cx_Ay_C + x_Bx_Ay_By_C^2 - x_B^2x_Cx_Ay_A - x_Bx_Ay_C^2y_A}{\sum_{ABC}x_Bx_Ay_Cy_A + x_By_By_Cy_A^2 - x_B^2x_Cy_Cy_A - x_By_By_C^2y_A} \\ &= \frac{\sum_{ABC}x_A^2x_Bx_Cy_B + x_Ax_Cy_Ay_B^2 - x_A^2x_Bx_Cy_C - x_Ax_Cy_B^2y_C}{\sum_{ABC}x_Ax_Cy_By_C + x_Ay_Ay_By_C^2 - x_A^2x_By_By_C - x_Ay_Ay_B^2y_C} \\ &= m_{A'B'}\text{.}
\end{split} \end{equation*}
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 |
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