Difference between revisions of "2021 AMC 10A Problems/Problem 24"
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(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> | (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 sin A\ | + | The area of the rectangle created by the four equations can be written as <math>2a\cdot \sin A\cdot4\sin A</math> |
− | = <math>8a(sin A)^2</math> | + | = <math>8a(\sin A)^2</math> |
= <math>8a(~\frac{1}{\sqrt(a^2+1)})^2</math> | = <math>8a(~\frac{1}{\sqrt(a^2+1)})^2</math> | ||
Line 52: | Line 52: | ||
= <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> | = <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> | ||
− | (Note: <math>tan A=</math> slope <math>a</math>) | + | (Note: <math>\tan A=</math> slope <math>a</math>) |
-fnothing4994 | -fnothing4994 |
Revision as of 14:56, 13 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
Since the slopes of the intersecting lines (from the 4 above equations) are negative reciprocals, the quadrilateral is a rectangle.
Plugging 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 Finally, by the Distance Formula, the length and width of the rectangle are and The area we seek is Plugging into the choices gives
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
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.