Difference between revisions of "2021 AMC 10A Problems/Problem 24"
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− | ==Problem | + | ==Problem== |
The interior of a quadrilateral is bounded by the graphs of <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math>, where <math>a</math> a positive real number. What is the area of this region in terms of <math>a</math>, valid for all <math>a > 0</math>? | The interior of a quadrilateral is bounded by the graphs of <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math>, where <math>a</math> a positive real number. What is the area of this region in terms of <math>a</math>, valid for all <math>a > 0</math>? | ||
<math>\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}</math> | <math>\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}</math> | ||
− | ==Solution== | + | ==Diagram== |
+ | Graph in Desmos: https://www.desmos.com/calculator/satawguqsc | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 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. | 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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For the equation <math>(x+ay)^2 = 4a^2,</math> the cases are | For the equation <math>(x+ay)^2 = 4a^2,</math> the cases are | ||
− | (1) | + | <math>(1) \ x+ay=2a.</math> This is a line with <math>x</math>-intercept <math>2a,y</math>-intercept <math>2,</math> and slope <math>-\frac 1a.</math> |
− | (2) | + | <math>(2) \ x+ay=-2a.</math> This is a line with <math>x</math>-intercept <math>-2a,y</math>-intercept <math>-2,</math> and slope <math>-\frac 1a.</math> |
For the equation <math>(ax-y)^2 = a^2,</math> the cases are | For the equation <math>(ax-y)^2 = a^2,</math> the cases are | ||
− | (1) | + | <math>(1') \ ax-y=a.</math> This is a line with <math>x</math>-intercept <math>1,y</math>-intercept <math>-a,</math> and slope <math>a.</math> |
− | (2) | + | <math>(2') \ ax-y=-a.</math> This is a line with <math>x</math>-intercept <math>-1,y</math>-intercept <math>a,</math> and slope <math>a.</math> |
− | |||
− | |||
− | |||
− | |||
Plugging <math>a=2</math> into the choices gives | Plugging <math>a=2</math> into the choices gives | ||
<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> | ||
+ | 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 <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> | ||
The answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> | The answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> | ||
~MRENTHUSIASM | ~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 <b><i>consecutive</i></b> vertices | ||
+ | <cmath>\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*}</cmath> | ||
+ | |||
+ | The area formula is | ||
+ | <cmath>\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*}</cmath> | ||
+ | Therefore, the answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> | ||
+ | |||
+ | 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) <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> | ||
+ | |||
+ | (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> | ||
+ | |||
+ | (3)* <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> | ||
+ | |||
+ | (4)* <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> | ||
+ | |||
+ | The area of the rectangle created by the four equations can be written as <math>2a\cdot \cos A\cdot4\sin A</math> | ||
+ | |||
+ | = <math>8a\cos A \cdot \sin A</math> | ||
+ | |||
+ | = <math>8a\cdot~\frac{1}{\sqrt{a^2+1}}\cdot~\frac{a}{\sqrt{a^2+1}}</math> | ||
+ | |||
+ | = <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> | ||
+ | |||
+ | (Note: <math>\tan A=</math> slope <math>a</math>) | ||
+ | |||
+ | -fnothing4994 | ||
+ | |||
+ | ==Solution 4 (bruh moment solution)== | ||
+ | |||
+ | 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 10:04, 25 March 2021
Contents
Problem
The interior of a quadrilateral is bounded by the graphs of and , where a positive real number. What is the area of this region in terms of , valid for all ?
Diagram
Graph in Desmos: https://www.desmos.com/calculator/satawguqsc
~MRENTHUSIASM
Solution 1
The conditions and give and or and . The slopes here are perpendicular, so the quadrilateral is a rectangle. Plug in and graph it. We quickly see that the area is , so the answer can't be or by testing the values they give (test it!). Now plug in . We see using a ruler that the sides of the rectangle are about and . So the area is about . Testing we get which is clearly less than , so it is out. Testing we get which is near our answer, so we leave it. Testing we get , way less than , so it is out. So, the only plausible answer is ~firebolt360
Solution 2 (Casework)
For the equation the cases are
This is a line with -intercept -intercept and slope
This is a line with -intercept -intercept and slope
For the equation the cases are
This is a line with -intercept -intercept and slope
This is a line with -intercept -intercept and slope
Plugging into the choices gives
Plugging into the four above equations and solving systems of equations for intersecting lines [ and and and and ], we get the respective solutions 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 and The area we seek is
The answer is
~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
The area formula is Therefore, the answer is
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) This is a line with -intercept , -intercept , and slope
(2) This is a line with -intercept , -intercept , and slope
(3)* This is a line with -intercept , -intercept , and slope
(4)* This is a line with -intercept , -intercept , and slope
The area of the rectangle created by the four equations can be written as
=
=
=
(Note: slope )
-fnothing4994
Solution 4 (bruh moment solution)
Trying narrows down the choices to options , and . Trying and eliminates and , to obtain as our answer. -¢
Video Solution by OmegaLearn (System of Equations and Shoelace Formula)
~ 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.