Difference between revisions of "2020 AMC 12A Problems/Problem 12"
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Furthermore, the equation <math>3x-5y+40=0</math> passes straight through <math>(20,20)</math> since <math>3(20)-5(20)+40=60-100+40=0</math>, which means that any rotations about <math>(20,20)</math> would contain <math>(20,20)</math>. We can create a line of slope <math>4</math> through <math>(20,20)</math>. The <math>x</math>-intercept is therefore <math>20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}</math>~lopkiloinm | Furthermore, the equation <math>3x-5y+40=0</math> passes straight through <math>(20,20)</math> since <math>3(20)-5(20)+40=60-100+40=0</math>, which means that any rotations about <math>(20,20)</math> would contain <math>(20,20)</math>. We can create a line of slope <math>4</math> through <math>(20,20)</math>. The <math>x</math>-intercept is therefore <math>20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}</math>~lopkiloinm | ||
+ | ==Solution 2== | ||
+ | Since the slope of the line is <math>\frac{3}{5}</math>, and the angle we are rotating around is x, then <math>\tan x = \frac{3}{5}</math> | ||
+ | <math>\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4</math> | ||
+ | |||
+ | Hence, the slope of the rotated line is <math>4</math>. Since we know the line intersects the point <math>(20,20)</math>, then we know the line is <math>y=4x-60</math>. Set <math>y=0</math> to find the x-intercept, and so <math>x=\boxed{15}</math> | ||
+ | |||
+ | ~Solution by IronicNinja | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2020|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2020|ab=A|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:09, 2 February 2020
Contents
[hide]Problem
Line in the coordinate plane has equation . This line is rotated counterclockwise about the point to obtain line . What is the -coordinate of the -intercept of line
Solution
The slope of the line is . We must transform it by .
creates an isosceles right triangle since the sum of the angles of the triangle must be and one angle is which means the last leg angle must also be .
In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of slope on graph paper. That line with slope starts at and will go to , the vector .
Construct another line from to , the vector . This is and equal to the original line segment. The difference between the two vectors is , which is the slope , and that is the slope of line .
Furthermore, the equation passes straight through since , which means that any rotations about would contain . We can create a line of slope through . The -intercept is therefore ~lopkiloinm
Solution 2
Since the slope of the line is , and the angle we are rotating around is x, then
Hence, the slope of the rotated line is . Since we know the line intersects the point , then we know the line is . Set to find the x-intercept, and so
~Solution by IronicNinja
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.