Difference between revisions of "2019 AMC 10B Problems/Problem 5"
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==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 <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. | ||
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==See Also== | ==See Also== |
Revision as of 11:14, 15 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.
A: Clearly A is true, because a coordinate in the first quadrant will have (+,+), and its inverse would also have (+,+)
B: The triangles have the same area, it's the same triangle.
C: If coordinate A has (x,y), then its inverse will have (y,x). (x-y)/(y-x)=-1, so this is true.
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.
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).
Clearly, the answer is .
Counterexamples
If and , then the slope of , , is , while the slope of , , is . is the of , but it is not the negative reciprocal of . To generalize, let denote the coordinates of point A, let denote the coordinates of point B, let denote the slope of segment , and let denote the slope of segment . Then, the coordinate pair for is , and the pair for is . 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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