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

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:

$[asy] Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 1.0,0.5)); yaxis(-8,8,Ticks(f, 1.0,0.5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return -x-3; } draw(graph(f,0,-3)); real f(real x) { return 5-x; } draw(graph(f,0,5)); real f(real x) { return 5+x; } draw(graph(f,0,-5)); real f(real x) { return x-5; } draw(graph(f,0,5)); real f(real x) { return -x-5; } draw(graph(f,0,-5)); [/asy]$

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

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

Revision as of 21:18, 23 November 2021

Contents

Problem

Solution (Graphing)

Solution 2 (Unrigorous but pretty standard)

See Also

@@ Line 58: / Line 58: @@
 real f(real x)
 {
-return 5-3;
+return x-5;
 }
 draw(graph(f,0,5));
 real f(real x)
 {
-return -5-3;
+return -x-5;
 }
 draw(graph(f,0,-5));