2012 USAMO Problems/Problem 5
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(Modified by Evan Chen)
\paragraph{Third solution (barycentric), by Catherine Xu}
We will perform barycentric coordinates on the triangle ,
with
,
, and
.
Set
,
,
as usual.
Since
,
,
are collinear,
we will define
and
.
\begin{claim*}
Lineis the angle bisector of
,
, and
.
\end{claim*} \begin{proof}
Sinceis the reflection of
across
, etc.
\end{proof}
Thus is the intersection of the isogonal of
with respect to
with the line
; that is,
\[ B' = \left( \frac pk \frac{b^2}{\ell}
: \frac{b^2}{\ell} : \frac{c^2}{q} \right). \]
Analogously, is the intersection of the isogonal of
with respect to
with the line
; that is,
\[ A' = \left( \frac{p}{\ell} \frac{b^2}{k}
: \frac{b^2}{k} : \frac{c^2}{q} \right). \]
The ratio of the first to third coordinate in these two points
is both , so it follows
,
, and
are collinear.
~peppapig_
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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