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

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<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.
 
<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.
  
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<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).
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<math>\textbf{(E)}</math>: This is the only one left, presumably the answer. To prove: 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).
  
 
Thus the answer is <math>\boxed{\textbf{(E)}}</math>.
 
Thus the answer is <math>\boxed{\textbf{(E)}}</math>.
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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>, so slopes arent negative reciprocals of each other.
 
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>, so slopes arent negative reciprocals of each other.
  
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==Video Solution (CREATIVE PROBLEM SOLVING!!!)==
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==Video Solution==
 
https://youtu.be/XKSZ9o54dg8
 
https://youtu.be/XKSZ9o54dg8
  
 
~Education, the Study of Everything
 
~Education, the Study of Everything
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==Video Solution==
 
==Video Solution==

Latest revision as of 20:24, 5 October 2023

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.

$\textbf{(E)}$: This is the only one left, presumably the answer. To prove: 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 $m_{A'B'} = \frac{x_2 - x_1}{y_2 - y_1} = \frac{1}{m_{AB}}$, so slopes arent negative reciprocals of each other.

Video Solution

https://youtu.be/XKSZ9o54dg8

~Education, the Study of Everything

Video Solution

https://youtu.be/dYn6jYRgEv4

~savannahsolver

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
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Problem 4 		Followed by
Problem 6
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All AMC 10 Problems and Solutions

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