Difference between revisions of "2012 USAMO Problems/Problem 5"

 
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==Solution==
 
==Solution==
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 +
By the [[Law_of_Sines|sine law]] 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.
  
 
==See also==
 
==See also==
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{{USAMO newbox|year=2012|num-b=4|num-a=6}}
 
{{USAMO newbox|year=2012|num-b=4|num-a=6}}
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{{MAA Notice}}
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[[Category:Olympiad Geometry Problems]]

Latest revision as of 13:01, 18 June 2020

Problem

Let $P$ be a point in the plane of triangle $ABC$, and $\gamma$ a line passing through $P$. Let $A'$, $B'$, $C'$ be the points where the reflections of lines $PA$, $PB$, $PC$ with respect to $\gamma$ intersect lines $BC$, $AC$, $AB$, respectively. Prove that $A'$, $B'$, $C'$ are collinear.

Solution

By the sine law on triangle $AB'P$, \[\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},\] so \[AB' = AP \cdot \frac{\sin \angle APB'}{\sin \angle AB'P}.\]

[asy] import graph; import geometry;  unitsize(0.5 cm);  pair[] A, B, C; pair P, R;  A[0] = (2,12); B[0] = (0,0); C[0] = (14,0); P = (4,5); R = 5*dir(70); A[1] = extension(B[0],C[0],P,reflect(P + R,P - R)*(A[0])); B[1] = extension(C[0],A[0],P,reflect(P + R,P - R)*(B[0])); C[1] = extension(A[0],B[0],P,reflect(P + R,P - R)*(C[0]));  draw((P - R)--(P + R),red); draw(A[1]--B[1]--C[1]--cycle,blue); draw(A[0]--B[0]--C[0]--cycle); draw(A[0]--P); draw(B[0]--P); draw(C[0]--P); draw(P--A[1]); draw(P--B[1]); draw(P--C[1]); draw(A[1]--B[0]); draw(A[1]--B[0]);  label("$A$", A[0], N); label("$B$", B[0], S); label("$C$", C[0], SE); dot("$A'$", A[1], SW); dot("$B'$", B[1], NE); dot("$C'$", C[1], W); dot("$P$", P, SE); label("$\gamma$", P + R, N); [/asy]

Similarly, \begin{align*} B'C &= CP \cdot \frac{\sin \angle CPB'}{\sin \angle CB'P}, \\ CA' &= CP \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P}, \\ A'B &= BP \cdot \frac{\sin \angle BPA'}{\sin \angle BA'P}, \\ BC' &= BP \cdot \frac{\sin \angle BPC'}{\sin \angle BC'P}, \\ C'A &= AP \cdot \frac{\sin \angle APC'}{\sin \angle AC'P}. \end{align*} Hence, \begin{align*} &\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} \\ &= \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'}. \end{align*}

Since angles $\angle AB'P$ and $\angle CB'P$ are supplementary or equal, depending on the position of $B'$ on $AC$, \[\sin \angle AB'P = \sin \angle CB'P.\] Similarly, \begin{align*} \sin \angle CA'P &= \sin \angle BA'P, \\ \sin \angle BC'P &= \sin \angle AC'P. \end{align*}

By the reflective property, $\angle APB'$ and $\angle BPA'$ are supplementary or equal, so \[\sin \angle APB' = \sin \angle BPA'.\] Similarly, \begin{align*} \sin \angle CPA' &= \sin \angle APC', \\ \sin \angle BPC' &= \sin \angle CPB'. \end{align*} Therefore, \[\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} = 1,\] so by Menelaus's theorem, $A'$, $B'$, and $C'$ are collinear.

See also

2012 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All USAMO Problems and Solutions

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