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Difference between revisions of "2012 USAMO Problems/Problem 5"
(→‎Solution)
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==Solution==
 
==Solution==
 +
 +
By the [[Law_of_Sines|sine law]] on triangle <math>AB'P</math>,
 +
<cmath>\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},</cmath>
 +
so
 +
<cmath>AB' = AP \cdot \frac{\sin \angle APB'}{\sin \angle AB'P}.</cmath>
 +
 +
<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,
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
Hence,
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
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>,
 +
<cmath>\sin \angle AB'P = \sin \angle CB'P.</cmath>
 +
Similarly,
 +
<cmath>
 +
\begin{align*}
 +
\sin \angle CA'P &= \sin \angle BA'P, \\
 +
\sin \angle BC'P &= \sin \angle AC'P.
 +
\end{align*}
 +
</cmath>
 +
 +
By the reflective property, <math>\angle APB'</math> and <math>\angle BPA'</math> are supplementary or equal, so
 +
<cmath>\sin \angle APB' = \sin \angle BPA'.</cmath>
 +
Similarly,
 +
<cmath>
 +
\begin{align*}
 +
\sin \angle CPA' &= \sin \angle APC', \\
 +
\sin \angle BPC' &= \sin \angle CPB'.
 +
\end{align*}
 +
</cmath>
 +
Therefore,
 +
<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.
  
 
==See also==
 
==See also==

Revision as of 09:13, 26 April 2012

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