Difference between revisions of "2021 AMC 10A Problems/Problem 24"
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By the Distance Formula, the length and width of this rectangle are <math>\frac{8\sqrt5}{5}</math> and <math>\frac{4\sqrt5}{5},</math> respectively. | By the Distance Formula, the length and width of this rectangle are <math>\frac{8\sqrt5}{5}</math> and <math>\frac{4\sqrt5}{5},</math> respectively. | ||
− | + | The area we seek is <cmath>\frac{8\sqrt5}{5}\cdot\frac{4\sqrt5}{5}=\frac{32}{5},</cmath> from which the answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> | |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 18:18, 28 June 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 (Generalized Value of a)
The cases for are or two parallel lines. We rearrange each case and construct the table below: The cases for are or two parallel lines. We rearrange each case and construct the table below: Since the slopes of intersecting lines and are negative reciprocals, we get four right angles, from which this quadrilateral is a rectangle.
Recall that for constants and the distance between parallel lines is From this formula:
- The distance between lines and is the length of this rectangle.
- The distance between lines and is the width of this rectangle.
The area we seek is ~MRENTHUSIASM
Solution 2 (Specified Value of a)
In this solution, we will refer to equations and from Solution 1.
Substituting into the answer choices gives
At the respective solutions to systems are Two solutions follow from here:
Solution 2.1 (Rectangle)
From the tables in Solution 1, we conclude that the quadrilateral is a rectangle.
By the Distance Formula, the length and width of this rectangle are and respectively.
The area we seek is from which the answer is
~MRENTHUSIASM
Solution 2.2 (Shoelace Theorem)
Even if we do not recognize that the quadrilateral is a rectangle, we can apply the Shoelace Theorem to its consecutive vertices from which Therefore, the answer is
~MRENTHUSIASM
Solution 3 (Trigonometry)
Similar to Solution 1, we will use the equations from the four cases:
- This is a line with -intercept -intercept and slope
- This is a line with -intercept -intercept and slope
- This is a line with -intercept -intercept and slope
- This is a line with -intercept -intercept and slope
Let The area of the rectangle created by the four equations can be written as ~fnothing4994 (Solution)
~MRENTHUSIASM (Code Adjustments)
Solution 4 (Observations Version 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 5 (Observations Version 2)
Trying narrows down the choices to options , and . Trying and eliminates and , to obtain as our answer.
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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.