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

Revision as of 01:50, 7 August 2022

Problem

How many ordered pairs $(x,y)$ of real numbers satisfy the following system of equations? $\begin{align*} x^2+3y&=9 \\ (|x|+|y|-4)^2 &= 1 \end{align*}$ $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7$

Solution 1 (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: $[asy] Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 1.0,0.5)); yaxis(-8,8,Ticks(f, 1.0,0.5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return -x-3; } draw(graph(f,0,-3)); real f(real x) { return 5-x; } draw(graph(f,0,5)); real f(real x) { return 5+x; } draw(graph(f,0,-5)); real f(real x) { return x-5; } draw(graph(f,0,5)); real f(real x) { return -x-5; } draw(graph(f,0,-5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return (-x^2)/3+3; } draw(graph(f,-5,5)); [/asy]$ We see from the graph that there are $5$ intersections, so the answer is $\boxed{\textbf{(D) } 5}$ .

~KingRavi

Solution 2 (Unrigorous but Feasible)

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 $\boxed{\textbf{(D) } 5}$ .

Video Solution by TheBeautyofMath

https://youtu.be/zq3UPu4nwsE?t=974

~IceMatrix

Video Solution by WhyMath

https://youtu.be/5SVmxNrZUbY

~savannahsolver

@@ Line 103: / Line 103: @@
 ==Video Solution by WhyMath==
-https://youtu.be/5SVmxNrZUbY ~savannahsolver
+https://youtu.be/5SVmxNrZUbY
+~savannahsolver
 ==See Also==
 {{AMC10 box|year=2021 Fall|ab=A|num-b=13|num-a=15}}
 {{MAA Notice}}

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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