Difference between revisions of "2021 AMC 10A Problems/Problem 24"
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2* & ax-y+a=0 & -1 & a & a \\ [0.75ex] | 2* & ax-y+a=0 & -1 & a & a \\ [0.75ex] | ||
\end{array}</cmath> | \end{array}</cmath> | ||
− | Since the slopes of intersecting lines <math>(1)\cap(1*), (1)\cap(2*), (2)\cap(1*),</math> and <math>(2)\cap(2*)</math> are negative reciprocals, we get four right angles, from which | + | Since the slopes of intersecting lines <math>(1)\cap(1*), (1)\cap(2*), (2)\cap(1*),</math> and <math>(2)\cap(2*)</math> are negative reciprocals, we get four right angles, from which this quadrilateral is a rectangle. |
Two solutions follow from here: | Two solutions follow from here: | ||
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From this formula: | From this formula: | ||
− | * The distance between lines <math>(1)</math> and <math>(2)</math> is <math>\frac{4a}{\sqrt{1+a^2}},</math> the length of | + | * The distance between lines <math>(1)</math> and <math>(2)</math> is <math>\frac{4a}{\sqrt{1+a^2}},</math> the length of this rectangle. |
− | * The distance between lines <math>(1*)</math> and <math>(2*)</math> is <math>\frac{2a}{\sqrt{a^2+1}},</math> the width of | + | * The distance between lines <math>(1*)</math> and <math>(2*)</math> is <math>\frac{2a}{\sqrt{a^2+1}},</math> the width of this rectangle. |
The area we seek is <cmath>\frac{4a}{\sqrt{1+a^2}}\cdot\frac{2a}{\sqrt{a^2+1}}=\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</cmath> | The area we seek is <cmath>\frac{4a}{\sqrt{1+a^2}}\cdot\frac{2a}{\sqrt{a^2+1}}=\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</cmath> | ||
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At <math>a=2,</math> the respective solutions to systems <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> | At <math>a=2,</math> the respective solutions to systems <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> | ||
− | By the Distance Formula, the length and width of | + | 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. |
Finally, 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> | Finally, 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> |
Revision as of 02:24, 31 May 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 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.
Two solutions follow from here:
Solution 2.1 (Distance Between Parallel Lines)
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 2.2 (Answer Choices)
Plugging into the answer choices gives
At the respective solutions to systems are
By the Distance Formula, the length and width of this rectangle are and respectively.
Finally, the area we seek is from which the answer is
~MRENTHUSIASM
Solution 3 (Trigonometry)
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.