Difference between revisions of "2012 USAMO Problems/Problem 5"
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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)== | + | ==Solution 2, Barycentric (Modified by Evan Chen)== |
− | + | 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>. | |
− | 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>. | ||
− | + | Claim: Line <math>\gamma</math> is the angle bisector of <math>\angle APA' </math>, <math>\angle BPB'</math>, and <math>\angle CPC'</math>. | |
− | + | This is proved by observing that since <math>A'P</math> is the reflection of <math>AP</math> across <math>\gamma</math>, etc. | |
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Thus <math>B'</math> is the intersection of the isogonal of <math>B</math> with respect to <math>\angle P</math> | Thus <math>B'</math> is the intersection of the isogonal of <math>B</math> with respect to <math>\angle P</math> |
Revision as of 10:41, 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, 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,
\[ B' = \left( \frac pk \frac{b^2}{\ell}
: \frac{b^2}{\ell} : \frac{c^2}{q} \right). \]
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.