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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 14"

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==Solution 2 (Unrigorous but pretty standard)==
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We can manipulate the first equation to get <math>y = -\frac{x^{2}}{3} + 3</math>. From the second equation, we have that <math>|x|+|y|-4 = 1</math> or <math>|x|+|y|-4 = -1</math>. We will consider each case separately.
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If <math>|x|+|y|-4 = 1</math>, then <math>|x|+|y| = 5</math>. The graph of this is a square with vertices <math>(5,0)</math>, <math>(-5,0)</math>, <math>(0,5)</math> and <math>(0,-5)</math>. 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 <math>\underline{2}</math> solutions.
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If <math>|x|+|y|-4 = -1</math>, then <math>|x|+|y| = 3</math>. The graph of this is a square with vertices <math>(3,0)</math>, <math>(-3,0)</math>, <math>(0,3)</math> and <math>(0,-3)</math>. The vertex of the parabola from the first equation is on one of the corners of this square (in particular, <math>(0,3)</math>). Also, at <math>y = 0</math>, the parabola has <math>x</math> intercepts of <math>\pm 3</math>; 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 <math>3</math> intersection points in this case: <math>(0,3)</math>, <math>(3,0)</math> and <math>(-3,0)</math>. This case gives us <math>\underline{3}</math> solutions.
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Adding these two cases together, we get our final answer of <math>\boxed{\textbf{(D) } 5}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=13|num-a=15}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:18, 23 November 2021

Problem

How many ordered pairs $(x,y)$ of real numbers satisfy the following system of equations? \[x^2+3y=9\] \[(|x|+|y|-4)^2 = 1\]

$\textbf{(A )} 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7$


Solution (Graphing)

The second equation is $(|x|+|y| - 4)^2 = 1$. We know that the graph of $|x| + |y|$ is a very simple diamond shape, so let's see if we can reduce this equation to that form.

$(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5$}.

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,-2,2)); [/asy]

Solution 2 (Unrigorous but pretty standard)

We can manipulate the first equation to get $y = -\frac{x^{2}}{3} + 3$. From the second equation, we have that $|x|+|y|-4 = 1$ or $|x|+|y|-4 = -1$. We will consider each case separately.

If $|x|+|y|-4 = 1$, then $|x|+|y| = 5$. The graph of this is a square with vertices $(5,0)$, $(-5,0)$, $(0,5)$ and $(0,-5)$. 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 $\underline{2}$ solutions.

If $|x|+|y|-4 = -1$, then $|x|+|y| = 3$. The graph of this is a square with vertices $(3,0)$, $(-3,0)$, $(0,3)$ and $(0,-3)$. The vertex of the parabola from the first equation is on one of the corners of this square (in particular, $(0,3)$). Also, at $y = 0$, the parabola has $x$ intercepts of $\pm 3$; 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 $3$ intersection points in this case: $(0,3)$, $(3,0)$ and $(-3,0)$. This case gives us $\underline{3}$ solutions.

Adding these two cases together, we get our final answer of $\boxed{\textbf{(D) } 5}$.

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

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