Difference between revisions of "2021 Fall AMC 10A Problems/Problem 14"
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<math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7</math> | <math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7</math> | ||
| − | ==Solution (Graphing)== | + | ==Solution 1 (Graphing)== |
The second equation is <math>(|x|+|y| - 4)^2 = 1</math>. We know that the graph of <math>|x| + |y|</math> is a very simple diamond shape, so let's see if we can reduce this equation to that form: <cmath>(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5\}.</cmath> | The second equation is <math>(|x|+|y| - 4)^2 = 1</math>. We know that the graph of <math>|x| + |y|</math> is a very simple diamond shape, so let's see if we can reduce this equation to that form: <cmath>(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5\}.</cmath> | ||
Revision as of 19:59, 24 November 2021
Problem
How many ordered pairs 
of real numbers satisfy the following system of equations?
Solution 1 (Graphing)
The second equation is 
. We know that the graph of 
is a very simple diamond shape, so let's see if we can reduce this equation to that form: ![\[(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5\}.\]](http://latex.artofproblemsolving.com/4/e/a/4eae12e93fc1db2f8b8038aebdd206a7a6b5c323.png)
We now have two separate graphs for this equation and one graph for the first equation, so let's put it on the coordinate plane:
![[asy] Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 1.0,0.5)); yaxis(-8,8,Ticks(f, 1.0,0.5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return -x-3; } draw(graph(f,0,-3)); real f(real x) { return 5-x; } draw(graph(f,0,5)); real f(real x) { return 5+x; } draw(graph(f,0,-5)); real f(real x) { return x-5; } draw(graph(f,0,5)); real f(real x) { return -x-5; } draw(graph(f,0,-5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return (-x^2)/3+3; } draw(graph(f,-5,5)); [/asy]](http://latex.artofproblemsolving.com/5/1/b/51bfd5ebea99685462e153dbecd9a825da8b2056.png)
We see from the graph that there are 5 intersections, so the answer is 
.
~KingRavi
Solution 2 (Unrigorous but pretty standard)
We can manipulate the first equation to get 
. From the second equation, we have that 
or 
. We will consider each case separately.
If , then 
. The graph of this is a square with vertices 
, 
, 
and 
. The parabola from the first equation is downwards facing, and its vertex is inside this square; the parabola will clearly intersect the square twice. Therefore, this case gives us 
solutions.
If , then 
. The graph of this is a square with vertices 
, 
, 
and 
. The vertex of the parabola from the first equation is on one of the corners of this square (in particular, ). Also, at 
, the parabola has 
intercepts of 
; the square passes through both of those points. If we continue to move down, the square narrows in, while the parabola continues to expand. Therefore, these are our only 
intersection points in this case: , and . This case gives us 
solutions.
Adding these two cases together, we get our final answer of .
See Also
| 2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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. 
