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

(→‎Solution 3 (Trigonometry): Reformatted the list and equation blocks so they appear neat.)
(Rearranged the solutions based on education values. I will push the PURE guess-and-check solutions to the end. If anyone disagrees with this, please PM me, and we will figure something out.)
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 1==
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==Solution 1 (Casework: Rectangle)==
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The conditions <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math> give <math>|x+ay| = |2a|</math> and <math>|ax-y| = |a|</math> or <math>x+ay = \pm 2a</math> and <math>ax-y = \pm a</math>. The slopes here are perpendicular, so the quadrilateral is a rectangle.
 
−
Plug in <math>a=1</math> and graph it. We quickly see that the area is <math>2\sqrt{2} \cdot \sqrt{2} = 4</math>, so the answer can't be <math>A</math> or <math>B</math> by testing the values they give (test it!). Now plug in <math>a=2</math>. We see using a ruler that the sides of the rectangle are about <math>\frac74</math> and <math>\frac72</math>. So the area is about <math>\frac{49}8 = 6.125</math>. Testing <math>C</math> we get <math>\frac{16}3</math> which is clearly less than <math>6</math>, so it is out. Testing <math>D</math> we get <math>\frac{32}5</math> which is near our answer, so we leave it. Testing <math>E</math> we get <math>\frac{16}5</math>, way less than <math>6</math>, so it is out. So, the only plausible answer is <math>\boxed{D}</math> ~firebolt360
 
−
 
 
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==Solution 2 (Casework: Rectangle)==
 
 
The cases for <math>(x+ay)^2 = 4a^2</math> are <math>x+ay = \pm2a,</math> or two parallel lines. We rearrange each case and construct the table below:
 
The cases for <math>(x+ay)^2 = 4a^2</math> are <math>x+ay = \pm2a,</math> or two parallel lines. We rearrange each case and construct the table below:
 
<cmath>\begin{array}{c||c|c|c|c}
 
<cmath>\begin{array}{c||c|c|c|c}
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Two solutions follow from here:
 
Two solutions follow from here:
  
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===Solution 2.1 (Distance Between Parallel Lines)===
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===Solution 1.1 (Distance Between Parallel Lines)===
 
Recall that for constants <math>A,B,C_1</math> and <math>C_2,</math> the distance <math>d</math> between parallel lines
 
Recall that for constants <math>A,B,C_1</math> and <math>C_2,</math> the distance <math>d</math> between parallel lines
 
<math>\begin{cases}
 
<math>\begin{cases}
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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===Solution 2.2 (Answer Choices)===
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===Solution 1.2 (Answer Choices)===
 
Plugging <math>a=2</math> into the answer choices gives  
 
Plugging <math>a=2</math> into the answer choices gives  
  
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 3 (Trigonometry)==
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==Solution 2 (Trigonometry)==
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Similar to Solution 2, we will use the equations from the four cases:
+
Similar to Solution 1, we will use the equations from the four cases:
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
 
   <li><math>x+ay=2a.</math> This is a line with <math>x</math>-intercept <math>2a,</math> <math>y</math>-intercept <math>2,</math> and slope <math>-\frac 1a.</math></li><p>
 
   <li><math>x+ay=2a.</math> This is a line with <math>x</math>-intercept <math>2a,</math> <math>y</math>-intercept <math>2,</math> and slope <math>-\frac 1a.</math></li><p>
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~MRENTHUSIASM (Code Adjustments)
 
~MRENTHUSIASM (Code Adjustments)
  
−
==Solution 4 (bruh moment solution)==
+
==Solution 3 (Observations Version 1)==
 +
The conditions <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math> give <math>|x+ay| = |2a|</math> and <math>|ax-y| = |a|</math> or <math>x+ay = \pm 2a</math> and <math>ax-y = \pm a</math>. The slopes here are perpendicular, so the quadrilateral is a rectangle.
 +
Plug in <math>a=1</math> and graph it. We quickly see that the area is <math>2\sqrt{2} \cdot \sqrt{2} = 4</math>, so the answer can't be <math>\textbf{(A)}</math> or <math>\textbf{(B)}</math> by testing the values they give (test it!). Now plug in <math>a=2</math>. We see using a ruler that the sides of the rectangle are about <math>\frac74</math> and <math>\frac72</math>. So the area is about <math>\frac{49}8 = 6.125</math>. Testing <math>\textbf{(C)}</math> we get <math>\frac{16}3</math> which is clearly less than <math>6</math>, so it is out. Testing <math>\textbf{(D)}</math> we get <math>\frac{32}5</math> which is near our answer, so we leave it. Testing <math>\textbf{(E)}</math> we get <math>\frac{16}5</math>, way less than <math>6</math>, so it is out. So, the only plausible answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}</math>.
 +
 
 +
~firebolt360
 +
 
 +
==Solution 4 (Observations Version 2)==
 +
Trying <math>a = 1</math> narrows down the choices to options <math>\textbf{(C)}</math>, <math>\textbf{(D)}</math> and <math>\textbf{(E)}</math>. Trying <math>a = 2</math> and <math>a = 3</math> eliminates <math>\textbf{(C)}</math> and <math>\textbf{(E)}</math>, to obtain <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}</math> as our answer.
  
−
Trying <math>a = 1</math> narrows down the choices to options <math>\textbf{(C)}</math>, <math>\textbf{(D)}</math> and <math>\textbf{(E)}</math>. Trying <math>a = 2</math> and <math>a = 3</math> eliminates <math>\textbf{(C)}</math> and <math>\textbf{(E)}</math>, to obtain <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> as our answer. -¢
+
~¢
  
 
== Video Solution by OmegaLearn (System of Equations and Shoelace Formula) ==
 
== Video Solution by OmegaLearn (System of Equations and Shoelace Formula) ==

Revision as of 14:23, 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 (Casework: Rectangle)

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 this 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 (Answer Choices)

Plugging $a=2$ 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 respective solutions to systems $(1)\cap(1*), (1)\cap(2*), (2)\cap(1*), (2)\cap(2*)$ are \[(x,y)=\left(\frac 85, \frac 65\right), (0,2), (0,-2), \left(-\frac 85, -\frac 65\right).\]

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

Finally, 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 (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 3 (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 4 (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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