Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 12A Problems/Problem 12"

m (→‎Solution 3 (Cheap))
(Square)
Line 25: Line 25:
 
~Solution by IronicNinja
 
~Solution by IronicNinja
  
−
==Solution 3 (Cheap)==
+
==Solution 3==
 +
 
 +
<asy>
 +
draw((0,0)--(20, 0)--(20, 20)--(0, 20)--cycle);
 +
draw((20, 20)--(0, 8));
 +
draw((15, 0)--(20, 20));
 +
 
 +
dot("$P$", (20, 20));
 +
dot("$A$", (0, 8), dir(75));
 +
dot("$B$", (15, 0), dir(45));
 +
dot("$X$", (20, 0));
 +
dot("$Y$", (0, 20), dir(50));
 +
</asy>
 +
 
 +
Let <math>P</math> be <math>(20, 20)</math> and <math>X, Y</math> be <math>(20, 0)</math> and <math>(0, 20)</math> respectively. Since the slope of the line is <math>3/5</math> we know that <math>\tan{\angle{YPA}} = 3/5.</math> Segments <math>\overline{PA}</math> and <math>\overline{PB}</math> represent the before and after of rotating <math>l</math> by 45 counterclockwise. Thus, <math>\angle{XPB} = 45 - \angle{YPA}</math> and <cmath>BX = 20 \tan{\angle{XPB}} = 20 \cdot \frac{1 - 3/5}{1 + 3/5} = 5</cmath> by tangent addition formula. Since <math>BX</math> is 5 and the sidelength of the square is 20 the answer is <math>20 - 5 \implies \boxed{\textbf{B}}.</math>
 +
 
 +
==Solution 4 (Cheap)==
  
 
Using the protractor you brought, carefully graph the equation and rotate the given line <math>45^{\circ}</math> counter-clockwise about the point <math>(20,20)</math>. Scaling everything down by a factor of 5 makes this process easier.
 
Using the protractor you brought, carefully graph the equation and rotate the given line <math>45^{\circ}</math> counter-clockwise about the point <math>(20,20)</math>. Scaling everything down by a factor of 5 makes this process easier.

Revision as of 14:13, 2 February 2020

Problem

Line $l$ in the coordinate plane has equation $3x-5y+40=0$. This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$. What is the $x$-coordinate of the $x$-intercept of line $k?$

$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$

Solution

The slope of the line is $\frac{3}{5}$. We must transform it by $45^{\circ}$.

$45^{\circ}$ creates an isosceles right triangle since the sum of the angles of the triangle must be $180^{\circ}$ and one angle is $90^{\circ}$ which means the last leg angle must also be $45^{\circ}$.

In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of $\frac{3}{5}$ slope on graph paper. That line with $\frac{3}{5}$ slope starts at $(0,0)$ and will go to $(5,3)$, the vector $<5,3>$.

Construct another line from $(0,0)$ to $(3,-5)$, the vector $<3,-5>$. This is $\perp$ and equal to the original line segment. The difference between the two vectors is $<2,8>$, which is the slope $4$, and that is the slope of line $k$.

Furthermore, the equation $3x-5y+40=0$ passes straight through $(20,20)$ since $3(20)-5(20)+40=60-100+40=0$, which means that any rotations about $(20,20)$ would contain $(20,20)$. We can create a line of slope $4$ through $(20,20)$. The $x$-intercept is therefore $20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}$~lopkiloinm

Solution 2

Since the slope of the line is $\frac{3}{5}$, and the angle we are rotating around is x, then $\tan x = \frac{3}{5}$ $\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4$

Hence, the slope of the rotated line is $4$. Since we know the line intersects the point $(20,20)$, then we know the line is $y=4x-60$. Set $y=0$ to find the x-intercept, and so $x=\boxed{15}$

~Solution by IronicNinja

Solution 3

[asy] draw((0,0)--(20, 0)--(20, 20)--(0, 20)--cycle); draw((20, 20)--(0, 8)); draw((15, 0)--(20, 20)); dot("$P$", (20, 20)); dot("$A$", (0, 8), dir(75)); dot("$B$", (15, 0), dir(45)); dot("$X$", (20, 0)); dot("$Y$", (0, 20), dir(50)); [/asy]

Let $P$ be $(20, 20)$ and $X, Y$ be $(20, 0)$ and $(0, 20)$ respectively. Since the slope of the line is $3/5$ we know that $\tan{\angle{YPA}} = 3/5.$ Segments $\overline{PA}$ and $\overline{PB}$ represent the before and after of rotating $l$ by 45 counterclockwise. Thus, $\angle{XPB} = 45 - \angle{YPA}$ and \[BX = 20 \tan{\angle{XPB}} = 20 \cdot \frac{1 - 3/5}{1 + 3/5} = 5\] by tangent addition formula. Since $BX$ is 5 and the sidelength of the square is 20 the answer is $20 - 5 \implies \boxed{\textbf{B}}.$

Solution 4 (Cheap)

Using the protractor you brought, carefully graph the equation and rotate the given line $45^{\circ}$ counter-clockwise about the point $(20,20)$. Scaling everything down by a factor of 5 makes this process easier.

It should then become fairly obvious that the x intercept is $x=\boxed{15}$ (only use this as a last resort).

~Silverdragon

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_12&oldid=116638"