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

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so by [[Menelaus'_Theorem|Menelaus's theorem]], <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear.
 
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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==Solution 2, Barycentric (Modified by Evan Chen)==
 +
 
 +
We will perform barycentric coordinates on the triangle <math>PCC'</math>, with <math>P=(1,0,0)</math>, <math>C'=(0,1,0)</math>, and <math>C=(0,0,1)</math>. Set <math>a = CC'</math>, <math>b = CP</math>, <math>c = C'P</math> as usual. Since <math>A</math>, <math>B</math>, <math>C'</math> are collinear, we will define <math>A = (p : k : q)</math> and <math>B = (p : \ell : q)</math>.
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Claim: Line <math>\gamma</math> is the angle bisector of <math>\angle APA' </math>, <math>\angle BPB'</math>, and <math>\angle CPC'</math>.
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This is proved by observing that since <math>A'P</math> is the reflection of <math>AP</math> across <math>\gamma</math>, etc.
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Thus <math>B'</math> is the intersection of the isogonal of <math>B</math> with respect to <math>\angle P</math>
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with the line <math>CA</math>; that is,
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<cmath> B' = \left( \frac pk \frac{b^2}{\ell}: \frac{b^2}{\ell} : \frac{c^2}{q} \right). </cmath>
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Analogously, <math>A'</math> is the intersection of the isogonal of <math>A</math> with respect to <math>\angle P</math>
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with the line <math>CB</math>; that is,
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<cmath> A' = \left( \frac{p}{\ell} \frac{b^2}{k} : \frac{b^2}{k} : \frac{c^2}{q} \right). </cmath>
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The ratio of the first to third coordinate in these two points
 +
is both <math>b^2pq : c^2k\ell</math>, so it follows <math>A'</math>, <math>B'</math>, and <math>C'</math> are collinear.
 +
 
 +
~peppapig_
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 +
==Solution 3, Cartesian==
 +
 
 +
We will use coordinates. Without loss of generality, let <math>P = (0,0)</math> and let <math>\gamma</math> be the line <math>x = 0</math>. Let <math>A = (x_1,y_1)</math>, <math>B = (x_2,y_2)</math>, and <math>C = (x_3,y_3)</math>. Then <math>A'</math> is the intersection of the lines
 +
\begin{align*}
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y &= -\frac{y_1}{x_1}x \
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y - y_2 &= \frac{y_2 - y_3}{x_2 - x_3}(x - x_2).
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\end{align*}
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Solving the system of equations, we see it is
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<cmath>\left(\frac{x_1(x_2y_3 - y_2x_3)}{x_1y_2 + y_1x_2 - x_1y_3 - x_3y_1},\frac{y_1(y_2x_3 - x_2y_3)}{x_1y_2 + y_1x_2 - x_1y_3 - x_3y_1}\right).</cmath>
 +
To check collinearity, we need the determinant of the matrix
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<cmath>\begin{bmatrix}
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x_1(x_2y_3 - y_2x_3) & y_1(y_2x_3 - x_2y_3) & x_1y_2 + y_1x_2 - x_1y_3 - x_3y_1 \
 +
x_2(x_3y_1 - y_3x_1) & y_2(y_3x_1 - x_3y_1) & x_2y_3 + y_2x_3 - x_2y_1 - x_1y_2 \
 +
x_3(x_1y_2 - y_1x_2) & y_3(y_1x_2 - x_1y_2) & x_3y_1 + y_3x_1 - x_3y_2 - x_2y_3
 +
\end{bmatrix}</cmath>
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to be zero, which is true because the columns sum to zero.
 +
 
 +
~knowingant
  
 
==See also==
 
==See also==

Latest revision as of 12:23, 26 April 2025

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_

Solution 3, Cartesian

We will use coordinates. Without loss of generality, let $P = (0,0)$ and let $\gamma$ be the line $x = 0$. Let $A = (x_1,y_1)$, $B = (x_2,y_2)$, and $C = (x_3,y_3)$. Then $A'$ is the intersection of the lines y=−y1x1xy−y2=y2−y3x2−x3(x−x2). Solving the system of equations, we see it is \[\left(\frac{x_1(x_2y_3 - y_2x_3)}{x_1y_2 + y_1x_2 - x_1y_3 - x_3y_1},\frac{y_1(y_2x_3 - x_2y_3)}{x_1y_2 + y_1x_2 - x_1y_3 - x_3y_1}\right).\] To check collinearity, we need the determinant of the matrix \[\begin{bmatrix} x_1(x_2y_3 - y_2x_3) & y_1(y_2x_3 - x_2y_3) & x_1y_2 + y_1x_2 - x_1y_3 - x_3y_1 \\ x_2(x_3y_1 - y_3x_1) & y_2(y_3x_1 - x_3y_1) & x_2y_3 + y_2x_3 - x_2y_1 - x_1y_2 \\ x_3(x_1y_2 - y_1x_2) & y_3(y_1x_2 - x_1y_2) & x_3y_1 + y_3x_1 - x_3y_2 - x_2y_3 \end{bmatrix}\] to be zero, which is true because the columns sum to zero.

~knowingant

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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