Difference between revisions of "2012 USAMO Problems/Problem 5"
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<cmath>\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} = 1,</cmath> | <cmath>\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} = 1,</cmath> | ||
so by [[Menelaus'_Theorem|Menelaus's theorem]], <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear. | so by [[Menelaus'_Theorem|Menelaus's theorem]], <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear. | ||
+ | |||
+ | ==Solution 2(Modified by Evan Chen)== | ||
+ | |||
+ | \paragraph{Third solution (barycentric), by Catherine Xu} | ||
+ | We will perform barycentric coordinates on the triangle <math>PCC'</math>, | ||
+ | with <math>P=(1,0,0)</math>, <math>C'=(0,1,0)</math>, and <math>C=(0,0,1)</math>. | ||
+ | Set <math>a = CC'</math>, <math>b = CP</math>, <math>c = C'P</math> as usual. | ||
+ | Since <math>A</math>, <math>B</math>, <math>C'</math> are collinear, | ||
+ | we will define <math>A = (p : k : q)</math> and <math>B = (p : \ell : q)</math>. | ||
+ | |||
+ | \begin{claim*} | ||
+ | Line <math>\gamma</math> is the angle bisector of | ||
+ | <math>\angle APA' </math>, <math>\angle BPB'</math>, and <math>\angle CPC'</math>. | ||
+ | \end{claim*} | ||
+ | \begin{proof} | ||
+ | Since <math>A'P</math> is the reflection of <math>AP</math> across <math>\gamma</math>, etc. | ||
+ | \end{proof} | ||
+ | |||
+ | Thus <math>B'</math> is the intersection of the isogonal of <math>B</math> with respect to <math>\angle P</math> | ||
+ | with the line <math>CA</math>; that is, | ||
+ | \[ B' = \left( \frac pk \frac{b^2}{\ell} | ||
+ | : \frac{b^2}{\ell} : \frac{c^2}{q} \right). \] | ||
+ | Analogously, <math>A'</math> is the intersection of the isogonal of <math>A</math> with respect to <math>\angle P</math> | ||
+ | with the line <math>CB</math>; 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 <math>b^2pq : c^2k\ell</math>, so it follows <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear. | ||
+ | |||
+ | ~peppapig_ | ||
==See also== | ==See also== |
Revision as of 10:39, 13 March 2023
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.