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

Latest revision as of 19:27, 5 November 2023

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 1

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 $90^{\circ}$ . This 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 ~ShawnX (diagram)

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

Solution 5 (Rotation Matrix)

First note that the given line goes through $(20,20)$ with a slope of $\frac{3}{5}$ . This means that $(25,23)$ is on the line. Now consider translating the graph so that $(20,20)$ goes to the origin, then $(25,23)$ becomes $(5,3)$ . We now rotate the line $45^\circ$ about the origin using a rotation matrix. This maps $(5,3)$ to $\begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix} 5 \\ 3\end{bmatrix}=\begin{bmatrix}\sqrt{2} \\ 4\sqrt{2}\end{bmatrix}$ The line through the origin and $(\sqrt{2}, 4\sqrt{2})$ has slope $4$ . Translating this line so that the origin is mapped to $(20,20)$ , we find that the equation for the new line is $4x-60$ , meaning that the $x$ -intercept is $x=\frac{60}{4}=\boxed{\textbf{(B) }15}$ .

Solution 6 (Angle Bisector)

Note $(20,20)$ is on the line. Construct the perpendicular line $5x+3y-160=0$ . This creates a right triangle that intersects the x-axis at $(\frac{-40}{3},0)$ and $(32,0)$ a distance of $136/3$ apart. The $45^\circ$ transformation will bisect the right angle. The angle bisector theorem tells us the $136$ will split in ratio to the lengths of the sides. These are $\sqrt{12^2+20^2}$ and $\sqrt{\frac{100}{3}^2 + 20^2} = 4\sqrt{34}$ and $\frac{20}{3}\sqrt{34}$ . Thus the x intercept will split the line from $\frac{-40}{3}$ to $32$ into a ratio of $5:3$ making the x-intercept $15$ .

Solution 7 (Complex Numbers)

Converting to the complex plane, we can see that two numbers on the line are $8i$ and $5+11i$ . Translating $20+20i$ to the origin, we get $8i-20-20i = -20-12i$ and $5+11i-20-20i = -15-9i$ . Multiplying each of them by $e^{\pi i/4}$ , we get $-4\sqrt 2 - 16 \sqrt2 i$ and $-3\sqrt 2 - 12 \sqrt 2 i$ . This line has a slope of $4$ . Now, back to the cartesian plane. We have a line passing through $(20, 20)$ with slope $4$ which gives the equation as $y = 4x-60$ which implies the $x$ coordinate of the $x$ intercept is $60/4 = 15$ .

~rocketsri (minor error corrected by kn07)

Solution 8 (quick)

A quick check tells us that $(20,20)$ falls on the given line. Common sense tells us that if the slope of the original line is $1$ , or 45 degrees from the horizontal, a 45 counter clockwise rotation will result in a vertical line, and the x-intercept will be $(20,0)$ . Thinking of a 45 degree counter clockwise rotation as a 90 degree counter clockwise rotation that is bisected will helps in visualizing this line of reasoning. Therefore, it follows that if the original line is made steeper, then the x-intercept will move away from $(20,0)$ to the right. If the original line is made lower, then the opposite will happen. Our given line has slope $3/5$ , so the answer must be $A$ or $B$ . $A$ can be eliminated because an x-intercept of $(10,0)$ can only occur when the original line is horizontal. In conclusion, the answer must be $\boxed{\textbf{(B) }15}$ .

~jackshi2006

@@ Line 5: / Line 5: @@
 <math>\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30</math>
-== Solution ==
+== Solution 1==
+[[File:AMC 12A 2020 Question-12 Solution-1.png|500px|right]]
 The slope of the line is <math>\frac{3}{5}</math>. We must transform it by <math>45^{\circ}</math>.
-<math>45^{\circ}</math> creates an isosceles right triangle since the sum of the angles of the triangle must be <math>180^{\circ}</math> and one angle is <math>90^{\circ}</math> which means the last leg angle must also be <math>45^{\circ}</math>.
-In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of <math>\frac{3}{5}</math> slope on graph paper. That line with <math>\frac{3}{5}</math> slope starts at <math>(0,0)</math> and will go to <math>(5,3)</math>, the vector <math><5,3></math>. Construct another line from <math>(0,0)</math> to <math>(3,-5)</math>, the vector <math><3,-5></math>. This is <math>\perp</math> and equal to the original line segment. The difference between the two vectors is <math><2,8></math>, which is the slope <math>4</math>, and that is the slope of line <math>k</math>.
+<math>45^{\circ}</math> creates an isosceles right triangle, since the sum of the angles of the triangle must be <math>180^{\circ}</math> and one angle is <math>90^{\circ}</math>. This means the last leg angle must also be <math>45^{\circ}</math>.
-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
+In the isosceles right triangle, the two legs are congruent. We can therefore construct an isosceles right triangle with a line of <math>\frac{3}{5}</math> slope on graph paper. That line with <math>\frac{3}{5}</math> slope starts at <math>(0,0)</math> and will go to <math>(5,3)</math>, the vector <math><5, 3 ></math>.
+Construct another line from <math>(0,0)</math> to <math>(3,-5)</math>, the vector <math><3,-5></math>. This is <math>\perp</math> and equal to the original line segment. The difference between the two vectors is <math><2,8></math>, which is the slope <math>4</math>, and that is the slope of line <math>k</math>.
+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 ~ShawnX (diagram)
+==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
+==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.
+It should then become fairly obvious that the x intercept is <math>x=\boxed{15}</math> (only use this as a last resort).
+~Silverdragon
+==Solution 5 (Rotation Matrix)==
+First note that the given line goes through <math>(20,20)</math> with a slope of <math>\frac{3}{5}</math>. This means that <math>(25,23)</math> is on the line. Now consider translating the graph so that <math>(20,20)</math> goes to the origin, then <math>(25,23)</math> becomes <math>(5,3)</math>. We now rotate the line <math>45^\circ</math> about the origin using a rotation matrix. This maps <math>(5,3)</math> to
+<cmath>\begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix} 5 \\ 3\end{bmatrix}=\begin{bmatrix}\sqrt{2} \\ 4\sqrt{2}\end{bmatrix}</cmath>
+The line through the origin and <math>(\sqrt{2}, 4\sqrt{2})</math> has slope <math>4</math>. Translating this line so that the origin is mapped to <math>(20,20)</math>, we find that the equation for the new line is <math>4x-60</math>, meaning that the <math>x</math>-intercept is <math>x=\frac{60}{4}=\boxed{\textbf{(B) }15}</math>.
+==Solution 6 (Angle Bisector)==
+Note <math>(20,20)</math> is on the line.  Construct the perpendicular line <math>5x+3y-160=0</math>.  This creates a right triangle that intersects the x-axis at <math>(\frac{-40}{3},0)</math> and <math>(32,0)</math> a distance of <math>136/3</math> apart.  The <math>45^\circ</math> transformation will bisect the right angle.
+The angle bisector theorem tells us the <math>136</math> will split in ratio to the lengths of the sides.
+These are <math>\sqrt{12^2+20^2}</math> and <math>\sqrt{\frac{100}{3}^2 + 20^2} = 4\sqrt{34}</math> and <math>\frac{20}{3}\sqrt{34}</math>.  Thus the x intercept will split the line from <math>\frac{-40}{3}</math> to <math>32</math> into a ratio of <math>5:3</math> making the x-intercept <math>15</math>.
+==Solution 7 (Complex Numbers)==
+Converting to the complex plane, we can see that two numbers on the line are <math>8i</math> and <math>5+11i</math>. Translating <math>20+20i</math> to the origin, we get <math>8i-20-20i = -20-12i</math> and <math>5+11i-20-20i = -15-9i</math>. Multiplying each of them by <math>e^{\pi i/4}</math>, we get <math>-4\sqrt 2 - 16 \sqrt2 i</math> and <math>-3\sqrt 2 - 12 \sqrt 2 i</math>. This line has a slope of <math>4</math>. Now, back to the cartesian plane. We have a line passing through <math>(20, 20)</math> with slope <math>4</math> which gives the equation as <math>y = 4x-60</math> which implies the <math>x</math> coordinate of the <math>x</math> intercept is <math>60/4 = 15</math>.
+~rocketsri (minor error corrected by kn07)
+==Solution 8 (quick)==
+A quick check tells us that <math>(20,20)</math> falls on the given line. Common sense tells us that if the slope of the original line is <math>1</math>, or 45 degrees from the horizontal, a 45 counter clockwise rotation will result in a vertical line, and the x-intercept will be <math>(20,0)</math>.
+Thinking of a 45 degree counter clockwise rotation as a 90 degree counter clockwise rotation that is bisected will helps in visualizing this line of reasoning.
+Therefore, it follows that if the original line is made steeper, then the x-intercept will move away from <math>(20,0)</math> to the right. If the original line is made lower, then the opposite will happen. Our given line has slope <math>3/5</math>, so the answer must be <math>A</math> or <math>B</math>. <math>A</math> can be eliminated because an x-intercept of <math>(10,0)</math> can only occur when the original line is horizontal. In conclusion, the answer must be <math>\boxed{\textbf{(B) }15}</math>.
+~jackshi2006
+==See Also==
+{{AMC12 box|year=2020|ab=A|num-b=11|num-a=13}}
+[[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

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