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

Revision as of 02:23, 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 (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

@@ Line 27: / Line 27: @@
 ==Solution 3 (Cheap)==
-Using the protractor you brought, carefully graph the equation and rotate the given line <math>45^{\circ}</math> counter-clockwise about the line. 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.
 It should then become fairly obvious that the x intercept is <math>x=\boxed{15}</math> (only use this as a last resort).

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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