2019 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 5 Easy

Triangle ABC lies in the first quadrant. Points A, B, and C are reflected across the line y=x to points A^\prime, B^\prime, and C^\prime, 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? (2019 AMC 10B Problem, Question#5)

  • A.

    Tiangle A^\prime B^\prime C^\prime lies in the first quadrant.

  • B.

    Tiangles ABC and A^\prime B^\prime C^\prime have the same area.

  • C.

    The slope of line AA^\prime is -1.

  • D.

    The slopes of lines AA^\prime and CC^\prime are the same.

  • E.

    Lines AB and A^\prime B^\primeare perpendicular to each other.

Answer:E

Let's analyze all of the options separately.

\rm (A): Clearly \rm (A) is true, because a point in the first quadrant will have non-negative x- and y-. coordinates, and so its reflection, with the coordinates swapped, will also have non-negative x- and y- coordinates.

\rm (B): The triangles have the same area, since \triangle ABC and \triangle A^\prime B^\prime C^\prime are the same tiangle (congruent). More formally, we can say that area is invariant under reflection.

\rm (C): If point A has coordinates (p, q), then A^\prime will have coordinates (q.p). The gradient is thus \frac{p-q}{q-p}=-1, so this is true. (We know p\ne q since the question states that none of the points A, B,or C lies on the line y=x, so there is no risk of division by zero).

\rm (D): Repeating the argument for \rm (C), we see that both lines have slope-1, so this is also true.

\rm (E): By process of eliminaton,this must now be the answer, Indeed, it point A has

coordinates (p,q) and point B has coordinates (r,s), then A^\prime and B^\prime will, respectively, have coordinates (q,p) and (s,r). The product of the gradients of AB and A^{ \prime }B^{ \prime } is \frac{s-q}{r-p} \cdot \frac{r-p}{s-q}=1 \neq -1, so in fact these lines are never perpendicuar to each other (using the "negative reciprocal" condition for perpendicularity). Thus the anewer is \rm (E).

~Counterexamples

If (x_1,1)=(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^\prime B^\prime, m_{A^\prime B^\prime}, is \frac{7-2}{1-3}=- \frac{5}{2}. m_{A^\prime B^\prime} is the 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 mA^\prime B^\prime denote the slope of segment \overline{A^\prime B^\prime},. Then, the coordinates of A^\prime are (y_1,x_1), and of B^\prime are (y_2,x_2).

Then, m_{AB}=\frac{y_2-y_1}{x_2-x_1}, and m_{A^{ \prime }B^{ \prime }}= \frac{x_{2}-x_{1}}{y_{2}-y_{1}}= \frac{1}{m_{ab}}.

If y_1\ne y_2 and x_1\ne x_2, \frac{1}{m_{AB}} \ne \frac{1}{m_{A^{ \prime }B^{ \prime }}} \Rightarrow m_{AB} \ne m_{A^\prime B^\prime}, and in these cases, the condition is false.

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