Difference between revisions of "2012 USAMO Problems/Problem 5"
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==Solution 3, Coordinate Bash== | ==Solution 3, Coordinate Bash== | ||
− | Fix <math>P</math> to be at <math>(0,0)</math> and <math>\gamma</math> to be the line <math>x = 0</math>. Let the coordinates of point <math>Z \in \{A,B,C,A',B',C'\}</math> be <math>(x_Z,y_Z)</math>. | + | Fix <math>P</math> to be at <math>(0,0)</math> and <math>\gamma</math> to be the line <math>x = 0</math>. Let the coordinates of point <math>Z \in \{A,B,C,A',B',C'\}</math> be <math>(x_Z,y_Z)</math>. Let |
+ | |||
+ | <cmath>\sum_{ABC}f(x_A,x_B,x_C,y_A,y_B,y_C) = f(x_A,x_B,x_C,y_A,y_B,y_C) + f(x_B,x_C,x_A,y_B,y_C,y_A) + f(x_C,x_A,x_B,y_C,y_A,y_B)\text{.}</cmath> | ||
The reflection of line <math>PA</math> with respect to <math>\gamma</math> has equation <math>y = -\frac{x_A}{y_A}x</math>. Line <math>BC</math> has equation <math>y - y_B = \frac{y_2-y_3}{x_2-x_3}(x - x_B)</math>. <math>A'</math> is the intersection of these two points. We now find <math>x_{A'}</math>. | The reflection of line <math>PA</math> with respect to <math>\gamma</math> has equation <math>y = -\frac{x_A}{y_A}x</math>. Line <math>BC</math> has equation <math>y - y_B = \frac{y_2-y_3}{x_2-x_3}(x - x_B)</math>. <math>A'</math> is the intersection of these two points. We now find <math>x_{A'}</math>. | ||
− | + | ||
− | + | <cmath>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}</cmath> | |
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− | |||
− | |||
Then, <cmath>y_{A'} = -\frac{x_A}{y_A}x_{A'} = \frac{x_A(x_By_C - x_Cy_B)}{y_Ay_B - y_Ay_C + x_Ax_B - x_Ax_C} \text{.}</cmath> | Then, <cmath>y_{A'} = -\frac{x_A}{y_A}x_{A'} = \frac{x_A(x_By_C - x_Cy_B)}{y_Ay_B - y_Ay_C + x_Ax_B - x_Ax_C} \text{.}</cmath> | ||
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Similarly, we can find the coordinates of <math>B'</math> and <math>C'</math>, which are | Similarly, we can find the coordinates of <math>B'</math> and <math>C'</math>, which are | ||
− | + | <cmath>x_{B'} = \frac{y_B(x_Ay_C - x_Cy_A)}{y_By_C - y_By_A + x_Bx_C - x_Bx_A}</cmath> | |
− | + | <cmath>y_{B'} = \frac{x_B(x_Cy_A - x_Ay_C)}{y_By_C - y_By_A + x_Bx_C - x_Bx_A}</cmath> | |
− | + | <cmath>x_{C'} = \frac{y_A(x_By_A - x_Ay_B)}{y_Cy_A - y_Cy_B + x_Cx_A - x_Cx_B}</cmath> | |
− | + | <cmath>y_{C'} = \frac{x_C(x_Ay_B - x_By_A)}{y_Cy_A - y_Cy_B + x_Cx_A - x_Cx_B}\text{.}</cmath> | |
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We can now find the slope of line <math>A'B'</math>. | We can now find the slope of line <math>A'B'</math>. | ||
− | \begin{ | + | <cmath>\begin{align*} |
− | |||
m_{A'B'} &= \frac{y_{A'}-y_{B'}}{x_{A'}-x_{B'}} \ | 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{\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{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)} | &= \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{ | + | \end{align*}</cmath> |
− | |||
Similarly, <cmath>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)}\text{.}</cmath> | Similarly, <cmath>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)}\text{.}</cmath> | ||
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Then, | Then, | ||
− | \begin{ | + | <cmath>\begin{align*} |
− | |||
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} \ | 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} \ | &= \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{.} | &= m_{A'B'}\text{.} | ||
− | \end{ | + | \end{align*}</cmath> |
− | |||
Since the slope of line <math>B'C'</math> is equal to the slope of line <math>A'B'</math>, points <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear. <math>\blacksquare</math> | Since the slope of line <math>B'C'</math> is equal to the slope of line <math>A'B'</math>, points <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear. <math>\blacksquare</math> |
Revision as of 15:31, 26 April 2023
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
. Let
The reflection of line with respect to
has equation
. Line
has equation
.
is the intersection of these two points. We now find
.
Then,
Similarly, we can find the coordinates of and
, which are
We can now find the slope of line .
Similarly,
Then,
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.