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Difference between revisions of "2021 AMC 10A Problems/Problem 24"
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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.
 
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
 
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)==
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For the equation <math>(x+ay)^2 = 4a^2,</math> the cases are
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(1) <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>
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(2) <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>
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For the equation <math>(ax-y)^2 = a^2,</math> the cases are
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(1)* <math>ax-y=a.</math> This is a line with <math>x</math>-intercept <math>1</math>, <math>y</math>-intercept <math>-a</math>, and slope <math>a.</math>
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(2)* <math>ax-y=-a.</math> This is a line with <math>x</math>-intercept <math>-1</math>, <math>y</math>-intercept <math>a</math>, and slops <math>a.</math>
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Since the slopes of the intersecting lines (from the 4 above equations) are negative reciprocals, the quadrilateral is a rectangle.
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Plugging <math>a=2</math> into the 4 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 <cmath>(x,y)=\left(\frac 85, \frac 65\right), (0,2), (0,-2), \left(-\frac 85, -\frac 65\right).</cmath>
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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>
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Plugging <math>a=2</math> into the choices gives
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<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>
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The answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math>
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~MRENTHUSIASM
  
 
== Video Solution by OmegaLearn (System of Equations and Shoelace Formula) ==
 
== Video Solution by OmegaLearn (System of Equations and Shoelace Formula) ==

Revision as of 22:59, 11 February 2021

Problem 24

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 $a > 0$?

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

Solution

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 slops $a.$

Since the slopes of the intersecting lines (from the 4 above equations) are negative reciprocals, the quadrilateral is a rectangle.

Plugging $a=2$ into the 4 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 \[(x,y)=\left(\frac 85, \frac 65\right), (0,2), (0,-2), \left(-\frac 85, -\frac 65\right).\] 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 \[\left(\frac{8\sqrt5}{5}\right)\left(\frac{4\sqrt5}{5}\right)=\frac{32}{5}.\] Plugging $a=2$ into the choices gives

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

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

~MRENTHUSIASM

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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