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 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. ~AlcBoy1729

2019 AMC 10B (Problems • Answer Key • Resources)
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2019 AMC 10B Problems/Problem 5

Contents

Problem

Solution

Counterexamples

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