Difference between revisions of "2021 AMC 10A Problems/Problem 19"
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− | ==Problem | + | ==Problem== |
The area of the region bounded by the graph of<cmath>x^2+y^2 = 3|x-y| + 3|x+y|</cmath>is <math>m+n\pi</math>, where <math>m</math> and <math>n</math> are integers. What is <math>m + n</math>? | The area of the region bounded by the graph of<cmath>x^2+y^2 = 3|x-y| + 3|x+y|</cmath>is <math>m+n\pi</math>, where <math>m</math> and <math>n</math> are integers. What is <math>m + n</math>? | ||
Line 5: | Line 5: | ||
== Solution 1 == | == Solution 1 == | ||
− | This is what the diagram looks like: | + | In order to attack this problem, we can use casework on the sign of <math>|x-y|</math> and <math>|x+y|</math>. |
+ | |||
+ | Case 1: <math>|x-y|=x-y, |x+y|=x+y</math> | ||
+ | |||
+ | Substituting and simplifying, we have <math>x^2-6x+y^2=0</math>, i.e. <math>(x-3)^2+y^2=3^2</math>, which gives us a circle of radius <math>3</math> centered at <math>(3,0)</math>. | ||
+ | |||
+ | Case 2: <math>|x-y|=y-x, |x+y|=x+y</math> | ||
+ | |||
+ | Substituting and simplifying again, we have <math>x^2+y^2-6y=0</math>, i.e. <math>x^2+(y-3)^2=3^2</math>. This gives us a circle of radius <math>3</math> centered at <math>(0,3)</math>. | ||
+ | |||
+ | Case 3: <math>|x-y|=x-y, |x+y|=-x-y</math> | ||
+ | |||
+ | Doing the same process as before, we have <math>x^2+y^2+6y=0</math>, i.e. <math>x^2+(y+3)^2=3^2</math>. This gives us a circle of radius <math>3</math> centered at <math>(0,-3)</math>. | ||
+ | |||
+ | Case 4: <math>|x-y|=y-x, |x+y|=-x-y</math> | ||
+ | |||
+ | One last time: we have <math>x^2+y^2+6x=0</math>, i.e. <math>(x+3)^2+y^2=3^2</math>. This gives us a circle of radius <math>3</math> centered at <math>(-3,0)</math>. | ||
+ | |||
+ | After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like: | ||
+ | |||
<asy> | <asy> | ||
size(10cm); | size(10cm); | ||
Line 20: | Line 39: | ||
Now, the area of the shaded region is just a square with side length <math>6</math> with four semicircles of radius <math>3</math>. | Now, the area of the shaded region is just a square with side length <math>6</math> with four semicircles of radius <math>3</math>. | ||
The area is <math>6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi</math>. The answer is <math>36+18</math> which is <math>\boxed{\textbf{(E) }54}</math> | The area is <math>6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi</math>. The answer is <math>36+18</math> which is <math>\boxed{\textbf{(E) }54}</math> | ||
− | |||
+ | ~Bryguy | ||
+ | |||
+ | ==Solution 2 == | ||
+ | A somewhat faster variant of solution 1 is to use a bit of symmetry in order to show that the remaining three cases are identical to Case 1 in the above solution, up to rotations by <math>90^{\circ}</math> about the origin. This allows us to quickly sketch the region after solving Case 1. | ||
− | + | Upon simplifying Case 1, we obtain <math>(x-3)^2 + y^2 = 3^2</math> which is a circle of radius 3 centered at <math>(3,0)</math>. We remark that only the points on the semicircle where <math>x \ge 3</math> work here, since Case 1 assumes <math>x-y \ge 0</math> and <math>x+y \ge 0</math>. Next, we observe that an ordered pair is a solution to the given equation if and only if any of its <math>90^{\circ}</math> rotations about the origin is a solution. This follows as the value of <math>x^2+y^2-3(|x-y|+|x+y|)</math> is invariant to <math>90^{\circ}</math> rotations, since <math>x^2+y^2</math> simply represents the square of the distance to the origin (which is unchanged upon rotation), and <math>|x-y|+|x+y|</math> is the sum of the distances to the lines <math>y=x</math> and <math>y=-x</math>, multiplied by <math>\sqrt{2}</math> (also unchanged upon <math>90^{\circ}</math> rotation). | |
− | == Video Solution (Using | + | By the above observation, we can quickly sketch the remainder of the region, and the area is <math>\boxed{\textbf{(E) }54}</math> as above. |
+ | |||
+ | ~scrabbler94 | ||
+ | |||
+ | ==Solution 3 (Guessing)== | ||
+ | Assume <math>y</math> = <math>0</math>. We get that <math>x</math> = <math>6</math>. That means that this figure must contain the points <math>(0,6), (6,0), (0, -6), (-6, 0)</math>. Now, assume that <math>x</math> = <math>y</math>. We get that <math>x</math> = <math>3 \sqrt 3</math>. We get the points <math>(3,3), (3,-3), (-3, 3), (-3, -3)</math>. | ||
+ | |||
+ | Since this contains <math>x^2 + y^2</math>, assume that there are circles. Therefore, we can guess that there is a center square with area <math>6 \cdot 6</math> = <math>36</math> and <math>4</math> semicircles with radius <math>3</math>. We get <math>4</math> semicircles with area <math>4.5 \pi</math>, and therefore the answer is <math>36+18</math> = <math>\boxed {(E)54}</math> | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | == Remark == | ||
+ | This problem asks for the area of <b>the union of these four circles</b>: | ||
+ | |||
+ | [[File:Image 2021-02-11 111327.png|center|600px]] | ||
+ | [[File:Screenshot 2023-06-14 194749]] | ||
+ | |||
+ | == Rotation by 45 degrees == | ||
+ | |||
+ | with the help of rotation https://www.wolframalpha.com/input/?i=rotate+45+degrees, we can simplify the equation to <math>a^2 + b^2 = 3\sqrt2|a| + 3\sqrt2|b|</math>. Then follow the previous question | ||
+ | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_21 | ||
+ | |||
+ | ~aliciawu | ||
+ | |||
+ | == Video Solution by OmegaLearn (Using Absolute Value Properties to Graph) == | ||
https://youtu.be/EHHpB6GIGPc | https://youtu.be/EHHpB6GIGPc | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution by The Power Of Logic (Graphing) == | ||
+ | https://youtu.be/-pa72wBA85Y | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/U6obY_kio0g | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2021|ab=A|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:05, 29 October 2023
Contents
Problem
The area of the region bounded by the graph ofis
, where
and
are integers. What is
?
Solution 1
In order to attack this problem, we can use casework on the sign of and
.
Case 1:
Substituting and simplifying, we have , i.e.
, which gives us a circle of radius
centered at
.
Case 2:
Substituting and simplifying again, we have , i.e.
. This gives us a circle of radius
centered at
.
Case 3:
Doing the same process as before, we have , i.e.
. This gives us a circle of radius
centered at
.
Case 4:
One last time: we have , i.e.
. This gives us a circle of radius
centered at
.
After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like:
Now, the area of the shaded region is just a square with side length
with four semicircles of radius
.
The area is
. The answer is
which is
~Bryguy
Solution 2
A somewhat faster variant of solution 1 is to use a bit of symmetry in order to show that the remaining three cases are identical to Case 1 in the above solution, up to rotations by about the origin. This allows us to quickly sketch the region after solving Case 1.
Upon simplifying Case 1, we obtain which is a circle of radius 3 centered at
. We remark that only the points on the semicircle where
work here, since Case 1 assumes
and
. Next, we observe that an ordered pair is a solution to the given equation if and only if any of its
rotations about the origin is a solution. This follows as the value of
is invariant to
rotations, since
simply represents the square of the distance to the origin (which is unchanged upon rotation), and
is the sum of the distances to the lines
and
, multiplied by
(also unchanged upon
rotation).
By the above observation, we can quickly sketch the remainder of the region, and the area is as above.
~scrabbler94
Solution 3 (Guessing)
Assume =
. We get that
=
. That means that this figure must contain the points
. Now, assume that
=
. We get that
=
. We get the points
.
Since this contains , assume that there are circles. Therefore, we can guess that there is a center square with area
=
and
semicircles with radius
. We get
semicircles with area
, and therefore the answer is
=
~Arcticturn
Remark
This problem asks for the area of the union of these four circles:
File:Screenshot 2023-06-14 194749
Rotation by 45 degrees
with the help of rotation https://www.wolframalpha.com/input/?i=rotate+45+degrees, we can simplify the equation to . Then follow the previous question
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_21
~aliciawu
Video Solution by OmegaLearn (Using Absolute Value Properties to Graph)
~ pi_is_3.14
Video Solution by The Power Of Logic (Graphing)
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.