Difference between revisions of "2021 AMC 10A Problems/Problem 24"
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(2)* <math>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>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 | |
+ | |||
+ | <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 [(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> | Plugging <math>a=2</math> 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 <cmath>(x,y)=\left(\frac 85, \frac 65\right), (0,2), (0,-2), \left(-\frac 85, -\frac 65\right).</cmath> | ||
− | |||
− | |||
− | <math>\ | + | 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> |
Revision as of 05:18, 16 February 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 ?
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
(1) This is a line with -intercept -intercept and slope
(2) This is a line with -intercept -intercept and slope
For the equation the cases are
(1)* This is a line with -intercept -intercept and slope
(2)* 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 [(1) and (1)*, (1) and (2)*, (2) and (1)*, (2) and (2)*], we get the respective solutions
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 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.