Difference between revisions of "2021 AMC 10A Problems/Problem 24"
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==Solution== | ==Solution== | ||
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> | + | 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 |
== Video Solution by OmegaLearn (System of Equations and Shoelace Formula) == | == Video Solution by OmegaLearn (System of Equations and Shoelace Formula) == |
Revision as of 23:22, 11 February 2021
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
Video Solution by OmegaLearn (System of Equations and Shoelace Formula)
~ pi_is_3.14