Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Counterexamples)
(→‎Solution)
 
(7 intermediate revisions by 4 users not shown)
Line 24: Line 24:
 
<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.
  
−
<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).
+
<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>.
  
 
==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 '''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.
+
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.
 +
 
 +
==Video Solution==
 +
https://youtu.be/XKSZ9o54dg8
 +
 
 +
~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 21:24, 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)
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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_5&oldid=199187"