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Difference between revisions of "2019 AMC 10B Problems/Problem 5"

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Let's analyze all of the options separately.  
 
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 (+,+)
+
<math>\textbf{(A)}</math>: Clearly <math>\textbf{(A)}</math> is true, because a point in the first quadrant will have non-negative <math>x</math>- and <math>y</math>-coordinates, and so its reflection, with the coordinates swapped, will also have non-negative <math>x</math>- and <math>y</math>-coordinates.
  
−
B: The triangles have the same area, it's the same triangle.
+
<math>\textbf{(B)}</math>: The triangles have the same area, since <math>\triangle ABC</math> and <math>\triangle A'B'C'</math> are the same triangle (congruent). More formally, we can say that area is ''invariant'' under reflection.
  
−
C: If coordinate A has (x,y), then its inverse will have (y,x). (x-y)/(y-x)=-1, so this is true.
+
<math>\textbf{(C)}</math>: If point <math>A</math> has coordinates <math>(p,q)</math>, then <math>A'</math> will have coordinates <math>(q,p)</math>. The gradient is thus <math>\frac{p-q}{q-p} = -1</math>, so this is true. (We know <math>p \neq q</math> since the question states that none of the points <math>A</math>, <math>B</math>, or <math>C</math> lies on the line <math>y=x</math>, so there is no risk of division by zero).
  
−
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.  
+
<math>\textbf{(D)}</math>: Repeating the argument for <math>\textbf{(C)}</math>, we see that both lines have slope <math>-1</math>, so this is also 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).  
+
<math>\textbf{(E)}</math>: By process of elimination, this must now be the answer. Indeed, if point <math>A</math> has coordinates <math>(p,q)</math> and point <math>B</math> has coordinates <math>(r,s)</math>, then <math>A'</math> and <math>B'</math> will, respectively, have coordinates <math>(q,p)</math> and <math>(s,r)</math>. The product of the gradients of <math>AB</math> and <math>A'B'</math> is <math>\frac{s-q}{r-p} \cdot \frac{r-p}{s-q} = 1 \neq -1</math>, so in fact these lines are '''never''' perpendicular to each other (using the "negative reciprocal" condition for perpendicularity).
  
−
Clearly, the answer is <math>\boxed{\textbf{(E)}}</math>.
+
Thus the answer is <math>\boxed{\textbf{(E)}}</math>.
  
 
==Counterexamples==
 
==Counterexamples==
−
If <math>(x_1,y_1) = (2,3)</math> and <math>(x_2,y_2) = (7,1)</math>, then the slope of <math>AB</math>, <math>m_{AB}</math>, is <math>\frac{1 - 3}{7 - 2} = -\frac{2}{5}</math>, while the slope of <math>A'B'</math>, <math>m_{A'B'}</math>, is <math>\frac{7 - 2}{1 - 3} = -\frac{5}{2}</math>. <math>m_{A'B'}</math> is the <math>\textbf{reciprocal}</math> of <math>m_{AB}</math>, but it is not the negative reciprocal of <math>m_{AB}</math>. To generalize, let <math>(x_1,y_1)</math> denote the coordinates of point A, let <math>(x_2, y_2)</math> denote the coordinates of point B, let <math>m_{AB}</math> denote the slope of segment <math>\overline{AB}</math>, and let <math>m_{A'B'}</math> denote the slope of segment <math>\overline{A'B'}</math>. Then, the coordinate pair for <math>A'</math> is <math>(y_1, x_1)</math>, and the pair for <math>B'</math> is <math>(y_2, x_2)</math>. Then, <math>m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}</math>, and <math>m_{A'B'} = \frac{x_2 - x_1}{y_2 - y_1} = \frac{1}{m_{ab}}</math>. If <math>y_1 \neq y_2</math> and <math>x_1 \neq x_2</math>, <math>\frac{1}{m_{AB}} \neq \frac{1}{m_{A'B'}} \Rightarrow m_{AB} \neq m_{A'B'}</math>, and in these cases, the condition is false.
+
If <math>(x_1,y_1) = (2,3)</math> and <math>(x_2,y_2) = (7,1)</math>, then the slope of <math>AB</math>, <math>m_{AB}</math>, is <math>\frac{1 - 3}{7 - 2} = -\frac{2}{5}</math>, while the slope of <math>A'B'</math>, <math>m_{A'B'}</math>, is <math>\frac{7 - 2}{1 - 3} = -\frac{5}{2}</math>. <math>m_{A'B'}</math> is the '''reciprocal''' of <math>m_{AB}</math>, but it is not the negative reciprocal of <math>m_{AB}</math>. To generalize, let <math>(x_1,y_1)</math> denote the coordinates of point <math>A</math>, let <math>(x_2, y_2)</math> denote the coordinates of point <math>B</math>, let <math>m_{AB}</math> denote the slope of segment <math>\overline{AB}</math>, and let <math>m_{A'B'}</math> denote the slope of segment <math>\overline{A'B'}</math>. Then, the coordinates of <math>A'</math> are <math>(y_1, x_1)</math>, and of <math>B'</math> are <math>(y_2, x_2)</math>. Then, <math>m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}</math>, and <math>m_{A'B'} = \frac{x_2 - x_1}{y_2 - y_1} = \frac{1}{m_{ab}}</math>. If <math>y_1 \neq y_2</math> and <math>x_1 \neq x_2</math>, <math>\frac{1}{m_{AB}} \neq \frac{1}{m_{A'B'}} \Rightarrow m_{AB} \neq m_{A'B'}</math>, and in these cases, the condition is false.
  
 
==See Also==
 
==See Also==

Revision as of 19:19, 17 February 2019

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.

$\textbf{(A)}$: Clearly $\textbf{(A)}$ 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.

$\textbf{(B)}$: The triangles have the same area, since $\triangle ABC$ and $\triangle A'B'C'$ are the same triangle (congruent). More formally, we can say that area is invariant under reflection.

$\textbf{(C)}$: If point $A$ has coordinates $(p,q)$, then $A'$ will have coordinates $(q,p)$. The gradient is thus $\frac{p-q}{q-p} = -1$, so this is true. (We know $p \neq q$ 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).

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

$\textbf{(E)}$: By process of elimination, this must now be the answer. Indeed, if point $A$ has coordinates $(p,q)$ and point $B$ has coordinates $(r,s)$, then $A'$ and $B'$ will, respectively, have coordinates $(q,p)$ and $(s,r)$. The product of the gradients of $AB$ and $A'B'$ is $\frac{s-q}{r-p} \cdot \frac{r-p}{s-q} = 1 \neq -1$, so in fact these lines are never perpendicular to each other (using the "negative reciprocal" condition for perpendicularity).

Thus 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 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 coordinates of $A'$ are $(y_1, x_1)$, and of $B'$ are $(y_2, x_2)$. 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 by
Problem 4 		Followed by
Problem 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

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