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 "2021 AMC 10A Problems/Problem 24"

m (→‎Solution 2.2 (Shoelace Formula): Underlined -> italicized.)
m (→‎Solution 2 (Casework))
Line 30: Line 30:
 
<math>\textbf{(A)} ~\frac{32}{9}\qquad\textbf{(B)} ~\frac{8}{3}\qquad\textbf{(C)} ~\frac{16}{3}\qquad\textbf{(D)} ~\frac{32}{5}\qquad\textbf{(E)} ~\frac{16}{5}</math>
 
<math>\textbf{(A)} ~\frac{32}{9}\qquad\textbf{(B)} ~\frac{8}{3}\qquad\textbf{(C)} ~\frac{16}{3}\qquad\textbf{(D)} ~\frac{32}{5}\qquad\textbf{(E)} ~\frac{16}{5}</math>
  
−
Plugging <math>a=2</math> into the four above equations and solving systems of equations for intersecting lines [<math>(1)</math> and <math>(1'), (1)</math> and <math>(2'), (2)</math> and <math>(1'), (2)</math> and <math>(2')</math>], we get the respective solutions <cmath>(x,y)=\left(\frac 85, \frac 65\right), (0,2), \left(-\frac 85, -\frac 65\right), (0,-2).</cmath>  
+
Plugging <math>a=2</math> into the four above equations and solving systems of equations for intersecting lines [<math>(1)</math> and <math>(1'), (1)</math> and <math>(2'), (2)</math> and <math>(1'), (2)</math> and <math>(2')</math>], we get the respective solutions <cmath>(x,y)=\left(\frac 85, \frac 65\right), (0,2), \left(-\frac 85, -\frac 65\right), (0,-2).</cmath>
 +
Two solutions follow from here:
 +
 
===Solution 2.1 (Rectangle)===
 
===Solution 2.1 (Rectangle)===
 
Since the slopes of the intersecting lines (from the four above equations) are negative reciprocals, the quadrilateral is a rectangle. Finally, by the Distance Formula, the length and width of the rectangle are <math>\frac{8\sqrt5}{5}</math> and <math>\frac{4\sqrt5}{5}.</math> The area we seek is <cmath>\left(\frac{8\sqrt5}{5}\right)\left(\frac{4\sqrt5}{5}\right)=\frac{32}{5}.</cmath>
 
Since the slopes of the intersecting lines (from the four above equations) are negative reciprocals, the quadrilateral is a rectangle. Finally, by the Distance Formula, the length and width of the rectangle are <math>\frac{8\sqrt5}{5}</math> and <math>\frac{4\sqrt5}{5}.</math> The area we seek is <cmath>\left(\frac{8\sqrt5}{5}\right)\left(\frac{4\sqrt5}{5}\right)=\frac{32}{5}.</cmath>

Revision as of 09:04, 25 March 2021

Problem

The interior of a quadrilateral is bounded by the graphs of $(x+ay)^2 = 4a^2$ and $(ax-y)^2 = a^2$, where $a$ a positive real number. What is the area of this region in terms of $a$, valid for all $a > 0$?

$\textbf{(A)} ~\frac{8a^2}{(a+1)^2}\qquad\textbf{(B)} ~\frac{4a}{a+1}\qquad\textbf{(C)} ~\frac{8a}{a+1}\qquad\textbf{(D)} ~\frac{8a^2}{a^2+1}\qquad\textbf{(E)} ~\frac{8a}{a^2+1}$

Diagram

Graph in Desmos: https://www.desmos.com/calculator/satawguqsc

~MRENTHUSIASM

Solution 1

The conditions $(x+ay)^2 = 4a^2$ and $(ax-y)^2 = a^2$ give $|x+ay| = |2a|$ and $|ax-y| = |a|$ or $x+ay = \pm 2a$ and $ax-y = \pm a$. The slopes here are perpendicular, so the quadrilateral is a rectangle. Plug in $a=1$ and graph it. We quickly see that the area is $2\sqrt{2} \cdot \sqrt{2} = 4$, so the answer can't be $A$ or $B$ by testing the values they give (test it!). Now plug in $a=2$. We see using a ruler that the sides of the rectangle are about $\frac74$ and $\frac72$. So the area is about $\frac{49}8 = 6.125$. Testing $C$ we get $\frac{16}3$ which is clearly less than $6$, so it is out. Testing $D$ we get $\frac{32}5$ which is near our answer, so we leave it. Testing $E$ we get $\frac{16}5$, way less than $6$, so it is out. So, the only plausible answer is $\boxed{D}$ ~firebolt360

Solution 2 (Casework)

For the equation $(x+ay)^2 = 4a^2,$ the cases are

$(1) \ x+ay=2a.$ This is a line with $x$-intercept $2a,y$-intercept $2,$ and slope $-\frac 1a.$

$(2) \ x+ay=-2a.$ This is a line with $x$-intercept $-2a,y$-intercept $-2,$ and slope $-\frac 1a.$

For the equation $(ax-y)^2 = a^2,$ the cases are

$(1') \ ax-y=a.$ This is a line with $x$-intercept $1,y$-intercept $-a,$ and slope $a.$

$(2') \ ax-y=-a.$ This is a line with $x$-intercept $-1,y$-intercept $a,$ and slope $a.$

Plugging $a=2$ into the choices gives

$\textbf{(A)} ~\frac{32}{9}\qquad\textbf{(B)} ~\frac{8}{3}\qquad\textbf{(C)} ~\frac{16}{3}\qquad\textbf{(D)} ~\frac{32}{5}\qquad\textbf{(E)} ~\frac{16}{5}$

Plugging $a=2$ into the four above equations and solving systems of equations for intersecting lines [$(1)$ and $(1'), (1)$ and $(2'), (2)$ and $(1'), (2)$ and $(2')$], we get the respective solutions \[(x,y)=\left(\frac 85, \frac 65\right), (0,2), \left(-\frac 85, -\frac 65\right), (0,-2).\] Two solutions follow from here:

Solution 2.1 (Rectangle)

Since the slopes of the intersecting lines (from the four above equations) are negative reciprocals, the quadrilateral is a rectangle. Finally, by the Distance Formula, the length and width of the rectangle are $\frac{8\sqrt5}{5}$ and $\frac{4\sqrt5}{5}.$ The area we seek is \[\left(\frac{8\sqrt5}{5}\right)\left(\frac{4\sqrt5}{5}\right)=\frac{32}{5}.\]

The answer is $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$

~MRENTHUSIASM

Solution 2.2 (Shoelace Formula)

Even if we do not recognize that the solutions form the vertices of a rectangle, we can apply the Shoelace Formula on consecutive vertices \begin{align*} (x_1,y_1) &= \left(\frac 85, \frac 65\right), \\ (x_2,y_2) &= (0,2), \\ (x_3,y_3) &= \left(-\frac 85, -\frac 65\right), \\ (x_4,y_4) &= (0,-2). \end{align*}

The area formula is \begin{align*} A &= \frac{1}{2} \left|(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)\right| \\ &= \frac{1}{2} \left|\left[\frac85\cdot2+0\cdot\left(-\frac65\right)+\left(-\frac{8}{5}\right)\cdot(-2)+0\cdot\frac65\right] - \left[\frac65\cdot0+2\cdot\left(-\frac85\right)+\left(-\frac65\right)\cdot0+(-2)\cdot\frac85\right]\right| \\ &= \frac{1}{2} \left|\left[\frac{16}{5}+\frac{16}{5}\right]-\left[-\frac{16}{5}-\frac{16}{5}\right]\right| \\ &= \frac{1}{2} \left|\frac{64}{5}\right| \\ &= \frac{32}{5}. \end{align*} Therefore, the answer is $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$

Suggested Reading for the Shoelace Formula: https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem

~MRENTHUSIASM

Solution 3 (Geometry)

Similar to Solution 2, we will use the equations of the four cases:

(1) $x+ay=2a.$ This is a line with $x$-intercept $2a$, $y$-intercept $2$, and slope $-\frac 1a.$

(2) $x+ay=-2a.$ This is a line with $x$-intercept $-2a$, $y$-intercept $-2$, and slope $-\frac 1a.$

(3)* $ax-y=a.$ This is a line with $x$-intercept $1$, $y$-intercept $-a$, and slope $a.$

(4)* $ax-y=-a.$ This is a line with $x$-intercept $-1$, $y$-intercept $a$, and slope $a.$

The area of the rectangle created by the four equations can be written as $2a\cdot \cos A\cdot4\sin A$

= $8a\cos A \cdot \sin A$

= $8a\cdot~\frac{1}{\sqrt{a^2+1}}\cdot~\frac{a}{\sqrt{a^2+1}}$

= $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$

(Note: $\tan A=$ slope $a$)

-fnothing4994

Solution 4 (bruh moment solution)

Trying $a = 1$ narrows down the choices to options $\textbf{(C)}$, $\textbf{(D)}$ and $\textbf{(E)}$. Trying $a = 2$ and $a = 3$ eliminates $\textbf{(C)}$ and $\textbf{(E)}$, to obtain $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$ as our answer. -¢

Video Solution by OmegaLearn (System of Equations and Shoelace Formula)

https://youtu.be/2iohPYkZpkQ

~ pi_is_3.14

See also

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 10 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=2021_AMC_10A_Problems/Problem_24&oldid=150254"