Difference between revisions of "2020 AMC 12B Problems/Problem 7"
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Using this same method for the other answer choices, we eventually find that the answer is <math>\boxed{\textbf{(C)}\ \frac32}</math> since our slopes are <math>\frac12</math> and <math>3</math> which forms a perfect 45 degree angle. | Using this same method for the other answer choices, we eventually find that the answer is <math>\boxed{\textbf{(C)}\ \frac32}</math> since our slopes are <math>\frac12</math> and <math>3</math> which forms a perfect 45 degree angle. | ||
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+ | ==Solution 7== | ||
+ | If you have this formula memorized then it will be very easy to do: If <math>\theta</math> is the angle between two lines with slopes <math>m_1</math> and <math>m_2</math>, then <math>\tan(\theta)=\frac{m_1-m_2}{1+m_1m_2}</math>. | ||
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+ | Now let the smaller slope be <math>m</math>, thus the other slope is <math>6m</math>. Using our formula above: <cmath>\tan(45^\circ)=\frac{6m-m}{1+6m^2} \implies 6m^2-5m+1=0 \implies (2m-1)(3m-1)=0.</cmath> Therefore the two possible values for <math>m</math> are <math>\tfrac{1}{2}</math> and <math>\tfrac{1}{3}</math>. We choose the larger one and thus our answer is <cmath>6m^2=6 \cdot \frac{1}{4} = \frac{3}{2} \implies \boxed{\mathbf{(C)}}.</cmath> | ||
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+ | ~AngelaLZ | ||
==Video Solution== | ==Video Solution== | ||
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Revision as of 00:04, 29 January 2021
Contents
[hide]Problem
Two nonhorizontal, non vertical lines in the -coordinate plane intersect to form a angle. One line has slope equal to times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
Solution 1 (complex)
Let the intersection point is the origin. Let be a point on the line of lesser slope. The mutliplication of by cis 45. and therefore lies on the line of greater slope.
Then, the rotation of by 45 degrees gives a line of slope .
We get the equation and this gives our answer to be .
~jeffisepic
Solution 2 (vector products)
Intersect at the origin and select a point on each line to define vectors . Note that gives equal magnitudes of the vector products
Substituting coordinate expressions for vector products, we find Divide this equation by to obtain where is the slope of line . Taking , we obtain The latter solution gives the largest product of slopes
~fairpark
Solution 3 (bash)
Place on coordinate plane. Lines are The intersection point at the origin. Goes through So by law of sines, lettin we want Simplifying gives so so max and
Law of sines on the green triangle with the red angle (45 deg) and blue angle, where sine blue angle is from right triangle w vertices
~ccx09
Solution 4 (tan)
Let one of the lines have equation . Let be the angle that line makes with the x-axis, so . The other line will have a slope of . Since the slope of one line is times the other, and is the smaller slope, . If , the other line will have slope . If , the other line will have slope . The first case gives the bigger product of , so our answer is .
~JHawk0224
Solution 5 (matrix transformation)
Multiply by the rotation transformation matrix where
Solution 6 (Cheating)
Let the smaller slope be , then the larger slope is . Since we want the greatest product we begin checking each answer choice, starting with (E).
.
.
This gives and . Checking with a protractor we see that this does not form a 45 degree angle.
Using this same method for the other answer choices, we eventually find that the answer is since our slopes are and which forms a perfect 45 degree angle.
Solution 7
If you have this formula memorized then it will be very easy to do: If is the angle between two lines with slopes and , then .
Now let the smaller slope be , thus the other slope is . Using our formula above: Therefore the two possible values for are and . We choose the larger one and thus our answer is
~AngelaLZ
Video Solution
Two solutions https://youtu.be/6ujfjGLzVoE
~IceMatrix
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.