Difference between revisions of "2021 AMC 10A Problems/Problem 24"
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The conditions <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math> give <math>|x+ay| = |2a|</math> and <math>|ax-y| = |a|</math> or <math>x+ay = \pm 2a</math> and <math>ax-y = \pm a</math>. The slopes here are perpendicular, so the quadrilateral is a rectangle. | The conditions <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math> give <math>|x+ay| = |2a|</math> and <math>|ax-y| = |a|</math> or <math>x+ay = \pm 2a</math> and <math>ax-y = \pm a</math>. The slopes here are perpendicular, so the quadrilateral is a rectangle. | ||
Plug in <math>a=1</math> and graph it. We quickly see that the area is <math>2\sqrt{2} \cdot \sqrt{2} = 4</math>, so the answer can't be <math>A</math> or <math>B</math> by testing the values they give (test it!). Now plug in <math>a=2</math>. We see using a ruler that the sides of the rectangle are about <math>\frac74</math> and <math>\frac72</math>. So the area is about <math>\frac{49}8 = 6.125</math>. Testing <math>C</math> we get <math>\frac{16}3</math> which is clearly less than <math>6</math>, so it is out. Testing <math>D</math> we get <math>\frac{32}5</math> which is near our answer, so we leave it. Testing <math>E</math> we get <math>\frac{16}5</math>, way less than <math>6</math>, so it is out. So, the only plausible answer is <math>\boxed{D}</math> ~firebolt360 | Plug in <math>a=1</math> and graph it. We quickly see that the area is <math>2\sqrt{2} \cdot \sqrt{2} = 4</math>, so the answer can't be <math>A</math> or <math>B</math> by testing the values they give (test it!). Now plug in <math>a=2</math>. We see using a ruler that the sides of the rectangle are about <math>\frac74</math> and <math>\frac72</math>. So the area is about <math>\frac{49}8 = 6.125</math>. Testing <math>C</math> we get <math>\frac{16}3</math> which is clearly less than <math>6</math>, so it is out. Testing <math>D</math> we get <math>\frac{32}5</math> which is near our answer, so we leave it. Testing <math>E</math> we get <math>\frac{16}5</math>, way less than <math>6</math>, so it is out. So, the only plausible answer is <math>\boxed{D}</math> ~firebolt360 | ||
+ | |||
+ | ==Solution 2 (Casework)== | ||
+ | For the equation <math>(x+ay)^2 = 4a^2,</math> the cases are | ||
+ | |||
+ | (1) <math>x+ay=2a.</math> This is a line with <math>x</math>-intercept <math>2a</math>, <math>y</math>-intercept <math>2</math>, and slope <math>-\frac 1a.</math> | ||
+ | |||
+ | (2) <math>x+ay=-2a.</math> This is a line with <math>x</math>-intercept <math>-2a</math>, <math>y</math>-intercept <math>-2</math>, and slope <math>-\frac 1a.</math> | ||
+ | |||
+ | For the equation <math>(ax-y)^2 = a^2,</math> the cases are | ||
+ | |||
+ | (1)* <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> | ||
+ | |||
+ | (2)* <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 slops <math>a.</math> | ||
+ | |||
+ | Since the slopes of the intersecting lines (from the 4 above equations) are negative reciprocals, the quadrilateral is a rectangle. | ||
+ | |||
+ | Plugging <math>a=2</math> 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 <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> The area we seek is <cmath>\left(\frac{8\sqrt5}{5}\right)\left(\frac{4\sqrt5}{5}\right)=\frac{32}{5}.</cmath> | ||
+ | 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> | ||
+ | |||
+ | The answer is <math>\boxed{\textbf{(D)} ~\frac{8a^2}{a^2+1}}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
== Video Solution by OmegaLearn (System of Equations and Shoelace Formula) == | == Video Solution by OmegaLearn (System of Equations and Shoelace Formula) == |
Revision as of 23:59, 11 February 2021
Contents
Problem 24
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
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 slops
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
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.