Difference between revisions of "2020 AMC 12B Problems/Problem 7"
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~JHawk0224 | ~JHawk0224 | ||
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+ | ==Solution 2== | ||
+ | Place on coordinate plane. | ||
+ | Lines are <math>y=mx, y=6mx.</math> | ||
+ | Goes through <math>(0,0),(1,m),(1,6m),(1,0).</math> | ||
+ | So by law of sines, <math>\frac{5m}{\sin{45^{\circ}}} = \frac{\sqrt{1+m^2}}{1/(\sqrt{1+36m^2})},</math> lettin <math>a=m^2</math> we want <math>6a.</math> | ||
+ | Simplifying gives <math>50a = (1+a)(1+36a),</math> so <math>36a^2-13a+1=0 \implies 36(a-1/4)(a-1/9)=0,</math> so max <math>a=1/4,</math> and <math>6a=3/2 \quad \boxed{(C)}.</math> | ||
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+ | Law of sines on the green triangle with the red angle (45 deg) and blue angle, where sine blue angle is <math>1/(\sqrt{1+36m^2})</math> from right triangle w vertices <math>(0,0),(1,0),(1,6m).</math> | ||
+ | |||
+ | ~ccx09 | ||
==Video Solution== | ==Video Solution== |
Revision as of 20:27, 8 February 2020
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
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 2
Place on coordinate plane.
Lines are
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
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.