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| == Problem ==
| | #redirect [[2012 USAMO Problems/Problem 5]] |
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| Let <math>P</math> be a point in the plane of triangle <math>ABC</math>, and <math>\gamma</math> a line passing through <math>P</math>. Let <math>A'</math>, <math>B'</math>, <math>C'</math> be the points where the reflections of lines <math>PA</math>, <math>PB</math>, <math>PC</math> with respect to <math>\gamma</math> intersect lines <math>BC</math>, <math>AC</math>, <math>AB</math>, respectively. Prove that <math>A'</math>, <math>B'</math>, <math>C'</math> are collinear.
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| ==Solution==
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| By the [[Law_of_Sines|law of sines]] on triangle <math>AB'P</math>,
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| <cmath>\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},</cmath>
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| so
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| <cmath>AB' = AP \cdot \frac{\sin \angle APB'}{\sin \angle AB'P}.</cmath>
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| <asy>
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| import graph;
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| import geometry;
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| unitsize(0.5 cm);
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| pair[] A, B, C;
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| pair P, R;
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| A[0] = (2,12);
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| B[0] = (0,0);
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| C[0] = (14,0);
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| P = (4,5);
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| R = 5*dir(70);
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| A[1] = extension(B[0],C[0],P,reflect(P + R,P - R)*(A[0]));
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| B[1] = extension(C[0],A[0],P,reflect(P + R,P - R)*(B[0]));
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| C[1] = extension(A[0],B[0],P,reflect(P + R,P - R)*(C[0]));
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| draw((P - R)--(P + R),red);
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| draw(A[1]--B[1]--C[1]--cycle,blue);
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| draw(A[0]--B[0]--C[0]--cycle);
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| draw(A[0]--P);
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| draw(B[0]--P);
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| draw(C[0]--P);
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| draw(P--A[1]);
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| draw(P--B[1]);
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| draw(P--C[1]);
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| draw(A[1]--B[0]);
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| draw(A[1]--B[0]);
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| label("$A$", A[0], N);
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| label("$B$", B[0], S);
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| label("$C$", C[0], SE);
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| dot("$A'$", A[1], SW);
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| dot("$B'$", B[1], NE);
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| dot("$C'$", C[1], W);
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| dot("$P$", P, SE);
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| label("$\gamma$", P + R, N);
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| </asy>
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| Similarly,
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| <cmath>
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| \begin{align*}
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| B'C &= CP \cdot \frac{\sin \angle CPB'}{\sin \angle CB'P}, \\
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| CA' &= CP \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P}, \\
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| A'B &= BP \cdot \frac{\sin \angle BPA'}{\sin \angle BA'P}, \\
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| BC' &= BP \cdot \frac{\sin \angle BPC'}{\sin \angle BC'P}, \\
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| C'A &= AP \cdot \frac{\sin \angle APC'}{\sin \angle AC'P}.
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| \end{align*}
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| </cmath>
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| Hence,
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| <cmath>
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| \begin{align*}
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| &\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} \\
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| &= \frac{\sin \angle APB'}{\sin \angle AB'P} \cdot \frac{\sin \angle CB'P}{\sin \angle CPB'} \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P} \cdot \frac{\sin \angle BA'P}{\sin \angle BPA'} \cdot \frac{\sin \angle BPC'}{\sin \angle BC'P} \cdot \frac{\sin \angle AC'P}{\sin \angle APC'}.
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| \end{align*}
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| </cmath>
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| Since angles <math>\angle AB'P</math> and <math>\angle CB'P</math> are supplementary or equal, depending on the position of <math>B'</math> on <math>AC</math>,
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| <cmath>\sin \angle AB'P = \sin \angle CB'P.</cmath>
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| Similarly,
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| <cmath>
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| \begin{align*}
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| \sin \angle CA'P &= \sin \angle BA'P, \\
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| \sin \angle BC'P &= \sin \angle AC'P.
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| \end{align*}
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| </cmath>
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| By the reflective property, <math>\angle APB'</math> and <math>\angle BPA'</math> are supplementary or equal, so
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| <cmath>\sin \angle APB' = \sin \angle BPA'.</cmath>
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| Similarly,
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| <cmath>
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| \begin{align*}
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| \sin \angle CPA' &= \sin \angle APC', \\
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| \sin \angle BPC' &= \sin \angle CPB'.
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| \end{align*}
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| </cmath>
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| Therefore,
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| <cmath>\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} = 1,</cmath>
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| so by [[Menelaus'_Theorem|Menelaus's theorem]], <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear.
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| {{alternate solutions}}
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| ==See also==
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| *[[USAJMO Problems and Solutions]]
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| {{USAJMO newbox|year=2012|num-b=5|aftertext=|after=Last Problem}}
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| {{MAA Notice}}
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