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

m (cartesian bash to distinguish it from bary bash)
(deleting my solution: i found the points a', b', and c' i claimed the coordinates of in the solution are not actually collinear. i messed up some algebra somewhere and i'm not going through this thing again.)
 
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~peppapig_
 
~peppapig_
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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
 
  
 
==See also==
 
==See also==

Latest revision as of 21:32, 2 May 2023

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.

Solution 2, Barycentric (Modified by Evan Chen)

We will perform barycentric coordinates on the triangle $PCC'$, with $P=(1,0,0)$, $C'=(0,1,0)$, and $C=(0,0,1)$. Set $a = CC'$, $b = CP$, $c = C'P$ as usual. Since $A$, $B$, $C'$ are collinear, we will define $A = (p : k : q)$ and $B = (p : \ell : q)$.

Claim: Line $\gamma$ is the angle bisector of $\angle APA'$, $\angle BPB'$, and $\angle CPC'$. This is proved by observing that since $A'P$ is the reflection of $AP$ across $\gamma$, etc.

Thus $B'$ is the intersection of the isogonal of $B$ with respect to $\angle P$ with the line $CA$; that is, \[B' = \left( \frac pk \frac{b^2}{\ell}: \frac{b^2}{\ell} : \frac{c^2}{q} \right).\] Analogously, $A'$ is the intersection of the isogonal of $A$ with respect to $\angle P$ with the line $CB$; that is, \[A' = \left( \frac{p}{\ell} \frac{b^2}{k} : \frac{b^2}{k} : \frac{c^2}{q} \right).\] The ratio of the first to third coordinate in these two points is both $b^2pq : c^2k\ell$, so it follows $A'$, $B'$, and $C'$ are collinear.

~peppapig_

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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