Difference between revisions of "2021 Fall AMC 10A Problems/Problem 14"
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How many ordered pairs <math>(x,y)</math> of real numbers satisfy the following system of equations? | How many ordered pairs <math>(x,y)</math> of real numbers satisfy the following system of equations? | ||
− | <cmath>x^2+3y=9 | + | <cmath>\begin{align*} |
− | + | x^2+3y&=9 \\ | |
+ | (|x|+|y|-4)^2 &= 1 | ||
+ | \end{align*}</cmath> | ||
+ | <math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7</math> | ||
− | + | ==Solution 1 (Graphing)== | |
− | |||
− | |||
− | ==Solution (Graphing)== | ||
− | |||
− | |||
− | |||
− | |||
+ | The second equation is <math>(|x|+|y| - 4)^2 = 1</math>. We know that the graph of <math>|x| + |y|</math> is a very simple diamond shape, so let's see if we can reduce this equation to that form: <cmath>(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5\}.</cmath> | ||
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: | 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; | Label f; | ||
f.p=fontsize(6); | f.p=fontsize(6); | ||
− | xaxis(-8,8,Ticks(f, | + | xaxis(-8,8,Ticks(f, 1.0,0.5)); |
− | yaxis(-8,8,Ticks(f, | + | 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) | real f(real x) | ||
{ | { | ||
− | return | + | 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, | + | draw(graph(f,0,5)); |
real f(real x) | real f(real x) | ||
{ | { | ||
− | return | + | 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,- | + | draw(graph(f,0,-5)); |
real f(real x) | real f(real x) | ||
{ | { | ||
− | return x | + | return 3-x; |
} | } | ||
− | draw(graph(f, | + | 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> | </asy> | ||
+ | We see from the graph that there are <math>5</math> intersections, so the answer is <math>\boxed{\textbf{(D) } 5}</math>. | ||
+ | |||
+ | ~KingRavi | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | From the first equation, we can express <math>y</math> in terms of <math>x</math>: | ||
+ | |||
+ | <math>3y = 9 - x^2</math> which is <math> y = (9 - x^2)/3</math> | ||
+ | |||
+ | The second equation can be rewritten as: | ||
+ | |||
+ | <math>|x| + |y| - 4 = \pm 1</math>. | ||
+ | |||
+ | This gives us two scenarios to examine: | ||
+ | |||
+ | 1. <math>|x| + |y| = 5</math> | ||
+ | |||
+ | 2. <math>|x| + |y| = 3</math> | ||
+ | |||
+ | Case 1: | ||
+ | |||
+ | Substituting <math>y</math>, | ||
+ | |||
+ | <math>|x| + |(9 - x^2)/3| = 5</math>. | ||
+ | |||
+ | First, consider the case when <math>9 - x^2 \geq 0</math>. Then, | ||
+ | |||
+ | <math>|y| = (9 - x^2)/3</math> which is <math>|x| + (9 - x^2)/3 = 5</math>. | ||
+ | |||
+ | Multiplying by 3, we get | ||
+ | |||
+ | <math>3|x| + 9 - x^2 = 15</math> which is <math>3|x| - x^2 = 6</math> which is <math>x^2 - 3|x| + 6 = 0</math>. | ||
+ | |||
+ | However, the discriminant of this quadratic is <math>(-3)^2 - 4 * 1 * 6 = 9 - 24 = -15</math>, which indicates there are no real solutions in this scenario. | ||
+ | |||
+ | Now, we can consider when <math>9 - x^2 < 0</math> | ||
+ | |||
+ | <math>|y| = -(9 - x^2)/3 = (x^2 - 9)/3</math> which is <math>|x| + (x^2 - 9)/3 = 5</math>. | ||
+ | |||
+ | Multiplying by 3, we get | ||
+ | |||
+ | <math>3|x| + x^2 - 9 = 15</math> which is <math>x^2 + 3|x| - 24 = 0</math>. | ||
+ | |||
+ | Now, let <math>u = |x|</math>, which gives <math>u^2 + 3u - 24 = 0</math>. When we calculate the discriminant, we get <math>3^2 - 4 * 1 * (-24) = 9 + 96 = 105</math>. So, the roots are <math>u = (-3 \pm \sqrt(105))/2</math>. | ||
+ | |||
+ | Both roots give positive values for <math>u</math>, resulting in two values of <math>x</math> for each root (one positive and one negative). | ||
+ | |||
+ | Case 2: \( |x| + |y| = 3 \) | ||
+ | |||
+ | Substituting <math>y</math>: | ||
+ | |||
+ | <math>|x| + |(9 - x^2)/3| = 3</math>. | ||
+ | |||
+ | If <math>9 - x^2 \geq 0</math>, then <math>|y| = (9 - x^2)/3</math> which is |<math>x| + (9 - x^2)/3 = 3</math>. | ||
+ | |||
+ | Multiplying by 3, we get | ||
+ | |||
+ | <math>3|x| + 9 - x^2 = 9</math> which is <math>3|x| = x^2</math> which is <math>x^2 - 3|x| = 0</math>. | ||
+ | |||
+ | Thus, <math>|x|(|x| - 3) = 0</math>, leading to <math>x = 0</math> or <math>x = 3</math> (both giving corresponding <math>y</math> values). | ||
+ | |||
+ | If <math>9 - x^2 < 0</math>, then <math>|y| = (x^2 - 9)/3</math> which is <math>|x| + (x^2 - 9)/3 = 3</math>. | ||
+ | |||
+ | When we multiply through, we get <math>3|x| + x^2 - 9 = 9</math> which is <math>x^2 + 3|x| - 18 = 0</math>. | ||
+ | |||
+ | The discriminant here is <math>3^2 - 4 * 1 * (-18) = 9 + 72 = 81</math>. | ||
+ | |||
+ | This gives two more real roots for <math>u = |x|</math>. | ||
+ | |||
+ | Now, | ||
+ | |||
+ | - Case 1 contributes 2 solutions | ||
+ | |||
+ | - Case 2 contributes 1 solution from <math>x = 0</math> and <math>x = 3</math>, 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 <math>\boxed{\textbf{(D) } 5}</math>. | ||
+ | |||
+ | ~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== | ==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}} |
Latest revision as of 20:54, 2 November 2024
Contents
Problem
How many ordered pairs of real numbers satisfy the following system of equations?
Solution 1 (Graphing)
The second equation is . We know that the graph of is a very simple diamond shape, so let's see if we can reduce this equation to that form: 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: We see from the graph that there are intersections, so the answer is .
~KingRavi
Solution 2
From the first equation, we can express in terms of :
which is
The second equation can be rewritten as:
.
This gives us two scenarios to examine:
1.
2.
Case 1:
Substituting ,
.
First, consider the case when . Then,
which is .
Multiplying by 3, we get
which is which is .
However, the discriminant of this quadratic is , which indicates there are no real solutions in this scenario.
Now, we can consider when
which is .
Multiplying by 3, we get
which is .
Now, let , which gives . When we calculate the discriminant, we get . So, the roots are .
Both roots give positive values for , resulting in two values of for each root (one positive and one negative).
Case 2: \( |x| + |y| = 3 \)
Substituting :
.
If , then which is |.
Multiplying by 3, we get
which is which is .
Thus, , leading to or (both giving corresponding values).
If , then which is .
When we multiply through, we get which is .
The discriminant here is .
This gives two more real roots for .
Now,
- Case 1 contributes 2 solutions
- Case 2 contributes 1 solution from and , 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 .
~goofytaipan
Video Solution
~Education, the Study of Everything
Video Solution
https://youtu.be/zq3UPu4nwsE?t=974
Video Solution by WhyMath
~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.