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 | + | <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, | + | <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 | + | <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: | + | <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 | + | <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). |
− | + | 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 | + | 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
Contents
Problem
Triangle lies in the first quadrant. Points , , and are reflected across the line to points , , and , respectively. Assume that none of the vertices of the triangle lie on the line . Which of the following statements is not always true?
Triangle lies in the first quadrant.
Triangles and have the same area.
The slope of line is .
The slopes of lines and are the same.
Lines and are perpendicular to each other.
Solution
Let's analyze all of the options separately.
: Clearly is true, because a point in the first quadrant will have non-negative - and -coordinates, and so its reflection, with the coordinates swapped, will also have non-negative - and -coordinates.
: The triangles have the same area, since and are the same triangle (congruent). More formally, we can say that area is invariant under reflection.
: If point has coordinates , then will have coordinates . The gradient is thus , so this is true. (We know since the question states that none of the points , , or lies on the line , so there is no risk of division by zero).
: Repeating the argument for , we see that both lines have slope , so this is also true.
: By process of elimination, this must now be the answer. Indeed, if point has coordinates and point has coordinates , then and will, respectively, have coordinates and . The product of the gradients of and is , so in fact these lines are never perpendicular to each other (using the "negative reciprocal" condition for perpendicularity).
Thus the answer is .
Counterexamples
If and , then the slope of , , is , while the slope of , , is . is the reciprocal of , but it is not the negative reciprocal of . To generalize, let denote the coordinates of point , let denote the coordinates of point , let denote the slope of segment , and let denote the slope of segment . Then, the coordinates of are , and of are . Then, , and . If and , , 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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