Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2021 Fall AMC 10A Problems/Problem 14

Revision as of 20:54, 2 November 2024 by Goofytaipan91 (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

From the first equation, we can express $y$ in terms of $x$:

$3y = 9 - x^2$ which is $y = (9 - x^2)/3$

The second equation can be rewritten as:

$|x| + |y| - 4 = \pm 1$.

This gives us two scenarios to examine:

1. $|x| + |y| = 5$

2. $|x| + |y| = 3$

Case 1:

Substituting $y$,

$|x| + |(9 - x^2)/3| = 5$.

First, consider the case when $9 - x^2 \geq 0$. Then,

$|y| = (9 - x^2)/3$ which is $|x| + (9 - x^2)/3 = 5$.

Multiplying by 3, we get

$3|x| + 9 - x^2 = 15$ which is $3|x| - x^2 = 6$ which is $x^2 - 3|x| + 6 = 0$.

However, the discriminant of this quadratic is $(-3)^2 - 4 * 1 * 6 = 9 - 24 = -15$, which indicates there are no real solutions in this scenario.

Now, we can consider when $9 - x^2 < 0$

$|y| = -(9 - x^2)/3 = (x^2 - 9)/3$ which is $|x| + (x^2 - 9)/3 = 5$.

Multiplying by 3, we get

$3|x| + x^2 - 9 = 15$ which is $x^2 + 3|x| - 24 = 0$.

Now, let $u = |x|$, which gives $u^2 + 3u - 24 = 0$. When we calculate the discriminant, we get $3^2 - 4 * 1 * (-24) = 9 + 96 = 105$. So, the roots are $u = (-3 \pm \sqrt(105))/2$.

Both roots give positive values for $u$, resulting in two values of $x$ for each root (one positive and one negative).

Case 2: \( |x| + |y| = 3 \)

Substituting $y$:

$|x| + |(9 - x^2)/3| = 3$.

If $9 - x^2 \geq 0$, then $|y| = (9 - x^2)/3$ which is |$x| + (9 - x^2)/3 = 3$.

Multiplying by 3, we get

$3|x| + 9 - x^2 = 9$ which is $3|x| = x^2$ which is $x^2 - 3|x| = 0$.

Thus, $|x|(|x| - 3) = 0$, leading to $x = 0$ or $x = 3$ (both giving corresponding $y$ values).

If $9 - x^2 < 0$, then $|y| = (x^2 - 9)/3$ which is $|x| + (x^2 - 9)/3 = 3$.

When we multiply through, we get $3|x| + x^2 - 9 = 9$ which is $x^2 + 3|x| - 18 = 0$.

The discriminant here is $3^2 - 4 * 1 * (-18) = 9 + 72 = 81$.

This gives two more real roots for $u = |x|$.

Now,

- Case 1 contributes 2 solutions

- Case 2 contributes 1 solution from $x = 0$ and $x = 3$, and 2 solutions from the second sub-case

Thus, counting all solutions gives us a total of 5 unique ordered pairs, and the answer is $\boxed{\textbf{(D) } 5}$.

~goofytaipan

Video Solution

https://youtu.be/yASY-XL9vtI

~Education, the Study of Everything

Video Solution

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

Video Solution by WhyMath

https://youtu.be/5SVmxNrZUbY

~savannahsolver

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_14&oldid=231282"