Difference between revisions of "2021 AMC 10A Problems/Problem 19"
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Scrabbler94 (talk | contribs) (add alternate solution without casework) |
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== Solution 1 == | == Solution 1 == | ||
− | In order to attack this problem, we | + | 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> | Case 1: <math>|x-y|=x-y, |x+y|=x+y</math> | ||
Line 42: | Line 42: | ||
~Bryguy | ~Bryguy | ||
− | ==Solution 2 (Guessing)== | + | ==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, since Case 1 assumes <math>x-y \ge 0</math> and <math>x+y \ge 0</math>. Let <math>(x_0,y_0)</math> be an arbitrary point in the Cartesian plane, and let <math>(x_1,y_1)</math> be a point obtained by rotations of <math>90^{\circ}</math> about the origin such that <math>(x_1,y_1)</math> satisfies the conditions of Case 1. | ||
+ | |||
+ | We claim that <math>(x_0,y_0)</math> is a solution to the given equation if and only if <math>(x_1,y_1)</math> is also a solution. The proof is to note that rotation by <math>90^{\circ}</math> about the origin preserves both the value of <math>x^2+y^2</math> (as the distance to the origin is the same), as well as the value of <math>|x-y|+|x+y|</math> (as this represents the sum of distances from a point to the lines <math>y=x</math> and <math>y=-x</math>, multiplied by <math>\sqrt{2}</math>). Therefore the value of <math>x^2+y^2 - 3(|x-y|+|x+y|)</math> is invariant to rotations by <math>90^{\circ}</math>, establishing the claim. | ||
+ | We obtain <math>\boxed{(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>. | 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>. | ||
Revision as of 06:13, 28 October 2022
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, since Case 1 assumes
and
. Let
be an arbitrary point in the Cartesian plane, and let
be a point obtained by rotations of
about the origin such that
satisfies the conditions of Case 1.
We claim that is a solution to the given equation if and only if
is also a solution. The proof is to note that rotation by
about the origin preserves both the value of
(as the distance to the origin is the same), as well as the value of
(as this represents the sum of distances from a point to the lines
and
, multiplied by
). Therefore the value of
is invariant to rotations by
, establishing the claim.
We obtain
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:
Video Solution (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.