2012 USAMO Problems/Problem 5: Difference between revisions

Revision as of 11:19, 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 \begin{align*}

 y &= -\frac{y_1}{x_1}x \\
 y - y_2 &= \frac{y_2 - y_3}{x_2 - x_3}(x - x_2).

\end{align*} 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 with rows \[\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

\end{bmatrix}\] to be zero, which is true because the columns sum to zero.

2012 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6
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