Difference between revisions of "2021 AMC 10A Problems/Problem 24"
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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> | <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> | ||
− | Next, we plug <math>a=2</math> into the four lines' equations. The respective solutions for <math>(1) | + | Next, we plug <math>a=2</math> into the four lines' equations. The respective solutions for <math>(1)\cap(1*), (1)\cap(2*), (2)\cap(1*), (2)\cap(2*)</math> are <cmath>(x,y)=\left(\frac 85, \frac 65\right), (0,2), (0,-2), \left(-\frac 85, -\frac 65\right).</cmath> |
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> respectively. The area we seek is <cmath>\frac{8\sqrt5}{5}\cdot\frac{4\sqrt5}{5}=\frac{32}{5}.</cmath> | 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> respectively. The area we seek is <cmath>\frac{8\sqrt5}{5}\cdot\frac{4\sqrt5}{5}=\frac{32}{5}.</cmath> |
Revision as of 08:14, 24 April 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: Rectangle)
The cases for are
We rearrange each case and construct the table below:
The cases for
are
We rearrange each case and construct the table below:
Since the slopes of all intersecting lines are negative reciprocals, the quadrilateral is a rectangle.
Two solutions follow from here:
Solution 2.1 (Formulas)
For constants and
recall that the distance between the parallel lines
is
From this formula:
- The distance between
and
is
- The distance between
and
is
The area we seek is
~MRENTHUSIASM
Solution 2.2 (Answer Choices)
Plugging into the answer choices gives
Next, we plug into the four lines' equations. The respective solutions for
are
Finally, by the Distance Formula, the length and width of the rectangle are and
respectively. The area we seek is
Therefore, 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.