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

Revision as of 11:36, 6 November 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

Video Solution by TheBeautyofMath

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

~IceMatrix

Video Solution by WhyMath

https://youtu.be/5SVmxNrZUbY

~savannahsolver

@@ Line 87: / Line 87: @@
 ~KingRavi
-==Solution 2 (Unrigorous but Feasible)==
-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.
-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.
-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.
-Adding these two cases together, we get our final answer of <math>\boxed{\textbf{(D) } 5}</math>.
 ==Video Solution by TheBeautyofMath==

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