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.) |
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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, 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>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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, | ||
+ | <cmath> B' = \left( \frac pk \frac{b^2}{\ell}: \frac{b^2}{\ell} : \frac{c^2}{q} \right). </cmath> | ||
+ | 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, | ||
+ | <cmath> A' = \left( \frac{p}{\ell} \frac{b^2}{k} : \frac{b^2}{k} : \frac{c^2}{q} \right). </cmath> | ||
+ | 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== |
Latest revision as of 22:32, 2 May 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,
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_
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.