Difference between revisions of "2021 AMC 10A Problems/Problem 19"

Latest revision as of 20:05, 29 October 2023

Problem

The area of the region bounded by the graph of $x^2+y^2 = 3|x-y| + 3|x+y|$ is $m+n\pi$ , where $m$ and $n$ are integers. What is $m + n$ ?

$\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54$

Solution 1

In order to attack this problem, we can use casework on the sign of $|x-y|$ and $|x+y|$ .

Case 1: $|x-y|=x-y, |x+y|=x+y$

Substituting and simplifying, we have $x^2-6x+y^2=0$ , i.e. $(x-3)^2+y^2=3^2$ , which gives us a circle of radius $3$ centered at $(3,0)$ .

Case 2: $|x-y|=y-x, |x+y|=x+y$

Substituting and simplifying again, we have $x^2+y^2-6y=0$ , i.e. $x^2+(y-3)^2=3^2$ . This gives us a circle of radius $3$ centered at $(0,3)$ .

Case 3: $|x-y|=x-y, |x+y|=-x-y$

Doing the same process as before, we have $x^2+y^2+6y=0$ , i.e. $x^2+(y+3)^2=3^2$ . This gives us a circle of radius $3$ centered at $(0,-3)$ .

Case 4: $|x-y|=y-x, |x+y|=-x-y$

One last time: we have $x^2+y^2+6x=0$ , i.e. $(x+3)^2+y^2=3^2$ . This gives us a circle of radius $3$ centered at $(-3,0)$ .

After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like:

$[asy] size(10cm); Label f; f.p=fontsize(7); xaxis(-8,8,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 1.0)); draw(arc((-3,0),3,90,270) -- cycle, gray); draw(arc((0,3),3,0,180) -- cycle, gray); draw(arc((3,0),3,-90,90) -- cycle, gray); draw(arc((0,-3),3,-180,0) -- cycle, gray); draw((-3,3)--(3,3)--(3,-3)--(-3,-3)--cycle, grey); [/asy]$ Now, the area of the shaded region is just a square with side length $6$ with four semicircles of radius $3$ . The area is $6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi$ . The answer is $36+18$ which is $\boxed{\textbf{(E) }54}$

~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 $90^{\circ}$ about the origin. This allows us to quickly sketch the region after solving Case 1.

Upon simplifying Case 1, we obtain $(x-3)^2 + y^2 = 3^2$ which is a circle of radius 3 centered at $(3,0)$ . We remark that only the points on the semicircle where $x \ge 3$ work here, since Case 1 assumes $x-y \ge 0$ and $x+y \ge 0$ . Next, we observe that an ordered pair is a solution to the given equation if and only if any of its $90^{\circ}$ rotations about the origin is a solution. This follows as the value of $x^2+y^2-3(|x-y|+|x+y|)$ is invariant to $90^{\circ}$ rotations, since $x^2+y^2$ simply represents the square of the distance to the origin (which is unchanged upon rotation), and $|x-y|+|x+y|$ is the sum of the distances to the lines $y=x$ and $y=-x$ , multiplied by $\sqrt{2}$ (also unchanged upon $90^{\circ}$ rotation).

By the above observation, we can quickly sketch the remainder of the region, and the area is $\boxed{\textbf{(E) }54}$ as above.

~scrabbler94

Solution 3 (Guessing)

Assume $y$ = $0$ . We get that $x$ = $6$ . That means that this figure must contain the points $(0,6), (6,0), (0, -6), (-6, 0)$ . Now, assume that $x$ = $y$ . We get that $x$ = $3 \sqrt 3$ . We get the points $(3,3), (3,-3), (-3, 3), (-3, -3)$ .

Since this contains $x^2 + y^2$ , assume that there are circles. Therefore, we can guess that there is a center square with area $6 \cdot 6$ = $36$ and $4$ semicircles with radius $3$ . We get $4$ semicircles with area $4.5 \pi$ , and therefore the answer is $36+18$ = $\boxed {(E)54}$

~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 $a^2 + b^2 = 3\sqrt2|a| + 3\sqrt2|b|$ . 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

~ 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

@@ Line 5: / Line 5: @@
 == Solution 1 ==
-In order to attack this problem, we need to consider casework:
+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>
@@ Line 40: / Line 40: @@
 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>
-Solution by Bryguy
+~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).
+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 ==
@@ Line 46: / Line 62: @@
 [[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 (Using Absolute Value Properties to Graph) ==
+== Video Solution by OmegaLearn (Using Absolute Value Properties to Graph) ==
 https://youtu.be/EHHpB6GIGPc

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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