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Difference between revisions of "2021 AMC 10A Problems/Problem 24"

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2* & ax-y+a=0 & -1 & a & a \\ [0.75ex]
 
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Since the slopes of intersecting lines <math>(1)\cap(1*), (1)\cap(2*), (2)\cap(1*),</math> and <math>(2)\cap(2*)</math> are negative reciprocals, we get four right angles, from which this quadrilateral is a rectangle.
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Since the slopes of intersecting lines <math>(1)\cap(1*), (1)\cap(2*), (2)\cap(1*),</math> and <math>(2)\cap(2*)</math> are negative reciprocals, we get four right angles, from which the quadrilateral is a rectangle.
  
 
Two solutions follow from here:
 
Two solutions follow from here:

Revision as of 22:59, 28 June 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 (Generalized Value of a)

The cases for $(x+ay)^2 = 4a^2$ are $x+ay = \pm2a,$ or two parallel lines. We rearrange each case and construct the table below: \[\begin{array}{c||c|c|c|c} & & & & \\ [-2.5ex] \textbf{Case} & \textbf{Line's Equation} & \boldsymbol{x}\textbf{-intercept} & \boldsymbol{y}\textbf{-intercept} & \textbf{Slope} \\ [0.5ex] \hline & & & & \\ [-1.5ex] 1 & x+ay-2a=0 & 2a & 2 & -\frac1a \\ [2ex] 2 & x+ay+2a=0 & -2a & -2 & -\frac1a \\ [0.75ex] \end{array}\] The cases for $(ax-y)^2 = a^2$ are $ax-y=\pm a,$ or two parallel lines. We rearrange each case and construct the table below: \[\begin{array}{c||c|c|c|c} & & & & \\ [-2.5ex] \textbf{Case} & \textbf{Line's Equation} & \boldsymbol{x}\textbf{-intercept} & \boldsymbol{y}\textbf{-intercept} & \textbf{Slope} \\ [0.5ex] \hline & & & & \\ [-1.5ex] 1* & ax-y-a=0 & 1 & -a & a \\ [2ex] 2* & ax-y+a=0 & -1 & a & a \\ [0.75ex] \end{array}\] Since the slopes of intersecting lines $(1)\cap(1*), (1)\cap(2*), (2)\cap(1*),$ and $(2)\cap(2*)$ are negative reciprocals, we get four right angles, from which the quadrilateral is a rectangle.

Two solutions follow from here:

Solution 1.1 (Distance Between Parallel Lines)

Recall that for constants $A,B,C_1$ and $C_2,$ the distance $d$ between parallel lines $\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$ is \[d=\frac{\left|C_2-C_1\right|}{\sqrt{A^2+B^2}}.\] From this formula:

  • The distance between lines $(1)$ and $(2)$ is $\frac{4a}{\sqrt{1+a^2}},$ the length of this rectangle.
  • The distance between lines $(1*)$ and $(2*)$ is $\frac{2a}{\sqrt{a^2+1}},$ the width of this rectangle.

The area we seek is \[\frac{4a}{\sqrt{1+a^2}}\cdot\frac{2a}{\sqrt{a^2+1}}=\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.\] ~MRENTHUSIASM

Solution 1.2 (Distance Between Points)

The solutions to systems $(1)\cap(1*), (1)\cap(2*), (2)\cap(2*), (2)\cap(1*)$ are \[(x,y)=\left(\frac{a(a+2)}{a^2+1},\frac{a(2a-1)}{a^2+1}\right), \left(-\frac{a(a-2)}{a^2+1},\frac{a(2a+1)}{a^2+1}\right), \left(-\frac{a(a+2)}{a^2+1},-\frac{a(2a-1)}{a^2+1}\right), \left(\frac{a(a-2)}{a^2+1},-\frac{a(2a+1)}{a^2+1}\right),\] respectively, which are the consecutive vertices of this rectangle.

By the Distance Formula, the length and width of this rectangle are $\frac{4a\sqrt{a^2+1}}{a^2+1}$ and $\frac{2a\sqrt{a^2+1}}{a^2+1},$ respectively.

The area we seek is \[\frac{4a\sqrt{a^2+1}}{a^2+1}\cdot\frac{2a\sqrt{a^2+1}}{a^2+1}=\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.\] ~MRENTHUSIASM

Solution 2 (Specified Value of a)

In this solution, we will refer to equations $(1),(2),(1*),$ and $(2*)$ from Solution 1.

Substituting $a=2$ into the answer choices gives

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

At $a=2,$ the solutions to systems $(1)\cap(1*), (1)\cap(2*), (2)\cap(2*), (2)\cap(1*)$ are \[(x,y)=\left(\frac 85, \frac 65\right), (0,2), \left(-\frac 85, -\frac 65\right), (0,-2),\] respectively, which are the consecutive vertices of the quadrilateral.

Two solutions follow from here:

Solution 2.1 (Rectangle)

From the tables in Solution 1, we conclude that the quadrilateral is a rectangle.

By the Distance Formula, the length and width of this rectangle are $\frac{8\sqrt5}{5}$ and $\frac{4\sqrt5}{5},$ respectively.

The area we seek is \[\frac{8\sqrt5}{5}\cdot\frac{4\sqrt5}{5}=\frac{32}{5},\] from which the answer is $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$

~MRENTHUSIASM

Solution 2.2 (Shoelace Theorem)

Even if we do not recognize that the quadrilateral is a rectangle, we can apply the Shoelace Theorem to its 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 we seek is \[A = \frac{1}{2} \left|(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)\right| = \frac{32}{5}.\] from which the answer is $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.$

~MRENTHUSIASM

Solution 3 (Trigonometry)

Similar to Solution 1, we will use the equations from 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 $a.$

Let $\tan A=a.$ The area of the rectangle created by the four equations can be written as \begin{align*} 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}}. \end{align*} ~fnothing4994 (Solution)

~MRENTHUSIASM (Code Adjustments)

Solution 4 (Observations Version 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 $\textbf{(A)}$ or $\textbf{(B)}$ 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 $\textbf{(C)}$, we get $\frac{16}3$ which is clearly less than $6$, so it is out. Testing $\textbf{(D)}$, we get $\frac{32}5$ which is near our answer, so we leave it. Testing $\textbf{(E)}$, we get $\frac{16}5$, way less than $6$, so it is out. So, the only plausible answer is $\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}$.

~firebolt360

Solution 5 (Observations Version 2)

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

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