# 2019 AMC 10B Problems/Problem 5

## Problem

Triangle $ABC$ lies in the first quadrant. Points $A$, $B$, and $C$ are reflected across the line $y=x$ to points $A'$, $B'$, and $C'$, respectively. Assume that none of the vertices of the triangle lie on the line $y=x$. Which of the following statements is not always true?

$\textbf{(A) }$ Triangle $A'B'C'$ lies in the first quadrant.

$\textbf{(B) }$ Triangles $ABC$ and $A'B'C'$ have the same area.

$\textbf{(C) }$ The slope of line $AA'$ is $-1$.

$\textbf{(D) }$ The slopes of lines $AA'$ and $CC'$ are the same.

$\textbf{(E) }$ Lines $AB$ and $A'B'$ are perpendicular to each other.

## Solution

Let's analyze all of the options separately.

A: Clearly A is true, because a coordinate in the first quadrant will have (+,+), and its inverse would also have (+,+)

B: The triangles have the same area, it's the same triangle.

C: If coordinate A has (x,y), then its inverse will have (y,x). (x-y)/(y-x)=-1, so this is true.

D: Likewise, if coordinate A has (x1,y1), and AA' has a slope of -1, then coordinate B, with (x2,y2), will also have a slope of -1. This is true.

E: By process of elimination, this is the answer, but if coordinate A has (x1,y1) and coordinate B has (x2,y2), then their inverses will be (y1,x1), (y2,x2), and it is not necessarily true that (y2-y1)/(x2-x1)=-(y2-y1)/(x2-x1). (Negative inverses of each other).

Clearly, the answer is $\boxed{\textbf{(E)}}$.

## Counterexamples

If $(x_1,y_1) = (2,3)$ and $(x_2,y_2) = (7,1)$, then the slope of $AB$, $m_{AB}$, is $\frac{1 - 3}{7 - 2} = -\frac{2}{5}$, while the slope of $A'B'$, $m_{A'B'}$, is $\frac{7 - 2}{1 - 3} = -\frac{5}{2}$. $m_{A'B'}$ is the $\textbf{reciprocal}$ of $m_{AB}$, but it is not the negative reciprocal of $m_{AB}$. To generalize, let $(x_1,y_1)$ denote the coordinates of point A, let $(x_2, y_2)$ denote the coordinates of point B, let $m_{AB}$ denote the slope of segment $\overline{AB}$, and let $m_{A'B'}$ denote the slope of segment $\overline{A'B'}$. Then, the coordinate pair for $A'$ is $(y_1, x_1)$, and the pair for $B'$ is $(y_2, x_2)$. Then, $m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}$, and $m_{A'B'} = \frac{x_2 - x_1}{y_2 - y_1} = \frac{1}{m_{ab}}$. If $y_1 \neq y_2$ and $x_1 \neq x_2$, $\frac{1}{m_{AB}} \neq \frac{1}{m_{A'B'}} \Rightarrow m_{AB} \neq m_{A'B'}$, and in these cases, the condition is false.

## See Also

 2019 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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