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− | ==Solution 3, Cartesian bash==
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− | 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
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− | <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>
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− | 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>.
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− | <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>
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− | Similarly, we can find the coordinates of <math>B'</math> and <math>C'</math>, which are
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− | <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>
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− | <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>
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− | <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>
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− | <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>.
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− | <cmath>\begin{align*}
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− | m_{A'B'} &= \frac{y_{A'}-y_{B'}}{x_{A'}-x_{B'}} \
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− | &= \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}} \
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− | &= \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)} \
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− | &= \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)}
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− | \end{align*}</cmath>
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− | 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,
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− | <cmath>\begin{align*}
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− | 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} \
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− | &= \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} \
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− | &= m_{A'B'}\text{.}
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− | \end{align*}</cmath>
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− | 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>
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− | ~KnowingAnt
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| ==See also== | | ==See also== |
Latest revision as of 21:32, 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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.