Difference between revisions of "2020 AMC 12B Problems/Problem 7"

Revision as of 19:29, 8 February 2020

Problem

Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?

$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\ \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$

Solution

Let one of the lines have equation $y=ax$ . Let $\theta$ be the angle that line makes with the x-axis, so $\tan(\theta)=a$ . The other line will have a slope of $\tan(45^{\circ}+\theta)=\frac{\tan(45^{\circ})+\tan(\theta)}{1-\tan(45^{\circ})\tan(\theta)} = \frac{1+a}{1-a}$ . Since the slope of one line is $6$ times the other, and $a$ is the smaller slope, $6a = \frac{1+a}{1-a} \implies 6a-6a^2=1+a \implies 6a^2-5a+1=0 \implies a=\frac{1}{2},\frac{1}{3}$ . If $a = \frac{1}{2}$ , the other line will have slope $\frac{1+\frac{1}{2}}{1-\frac{1}{2}} = 3$ . If $a = \frac{1}{3}$ , the other line will have slope $\frac{1+\frac{1}{3}}{1-\frac{1}{3}} = 2$ . The first case gives the bigger product of $\frac{3}{2}$ , so our answer is $\boxed{\textbf{(C)}\ \frac32}$ .

~JHawk0224

Solution 2 (bash)

Place on coordinate plane. Lines are $y=mx, y=6mx.$ The intersection point at the origin. Goes through $(0,0),(1,m),(1,6m),(1,0).$ So by law of sines, $\frac{5m}{\sin{45^{\circ}}} = \frac{\sqrt{1+m^2}}{1/(\sqrt{1+36m^2})},$ lettin $a=m^2$ we want $6a.$ Simplifying gives $50a = (1+a)(1+36a),$ so $36a^2-13a+1=0 \implies 36(a-1/4)(a-1/9)=0,$ so max $a=1/4,$ and $6a=3/2 \quad \boxed{(C)}.$

Law of sines on the green triangle with the red angle (45 deg) and blue angle, where sine blue angle is $1/(\sqrt{1+36m^2})$ from right triangle w vertices $(0,0),(1,0),(1,6m).$

~ccx09

Solution 3 (complex)

Let the intersection point is the origin. Let $(a,b)$ be a point on the line of lesser slope. The mutliplication of $a+bi$ by cis 45. $(a+bi)(\frac{1}{\sqrt 2 }+i*\frac{1}{\sqrt 2 })=\frac{1}{\sqrt 2 }((a-b)+(a+b)*i)$ and therefore $(a-b, a+b)$ lies on the line of greater slope.

Then, the rotation of $(a,b)$ by 45 degrees gives a line of slope $\frac{a+b}{a-b}$ .

We get the equation $\frac{6b}{a}=\frac{a+b}{a-b}\implies a^2-5ab+6b^2=(a-3b)(a-2b)=0$ and this gives our answer.

Video Solution

Two solutions https://youtu.be/6ujfjGLzVoE

~IceMatrix

@@ Line 9: / Line 9: @@
 ~JHawk0224
-==Solution 2==
+==Solution 2 (bash)==
 Place on coordinate plane.
-Lines are <math>y=mx, y=6mx.</math>
+Lines are <math>y=mx, y=6mx.</math> The intersection point at the origin.
 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>
@@ Line 19: / Line 19: @@
 ~ccx09
+==Solution 3 (complex)==
+Let the intersection point is the origin. Let <math>(a,b)</math> be a point on the line of lesser slope. The mutliplication of <math>a+bi</math> by cis 45.
+<math>(a+bi)(\frac{1}{\sqrt 2 }+i*\frac{1}{\sqrt 2 })=\frac{1}{\sqrt 2 }((a-b)+(a+b)*i)</math> and therefore <math>(a-b, a+b)</math> lies on the line of greater slope.
+Then, the rotation of <math>(a,b)</math> by 45 degrees gives a line of slope <math>\frac{a+b}{a-b}</math>.
+We get the equation <math>\frac{6b}{a}=\frac{a+b}{a-b}\implies a^2-5ab+6b^2=(a-3b)(a-2b)=0</math> and this gives our answer.
 ==Video Solution==

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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