Difference between revisions of "2013 AMC 12B Problems/Problem 17"
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Now <math>a=b=2</math> and <math>c=-2</math> satisfy both equations, so we see that <math>c=-2</math> must be the minimum possible value of <math>c</math>. Also, <math>c=\frac{10}{3}</math> and <math>a=b=-\frac{2}{3}</math> satisfy both equations, so we see that <math>c=\frac{10}{3}</math> is the maximum possible value of <math>c</math>. The difference between these is <math>\frac{10}{3}-(-2)=\frac{16}{3}</math>, or <math>\boxed{\textbf{(D)}}</math>. | Now <math>a=b=2</math> and <math>c=-2</math> satisfy both equations, so we see that <math>c=-2</math> must be the minimum possible value of <math>c</math>. Also, <math>c=\frac{10}{3}</math> and <math>a=b=-\frac{2}{3}</math> satisfy both equations, so we see that <math>c=\frac{10}{3}</math> is the maximum possible value of <math>c</math>. The difference between these is <math>\frac{10}{3}-(-2)=\frac{16}{3}</math>, or <math>\boxed{\textbf{(D)}}</math>. | ||
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+ | ==Solution 4== | ||
+ | We take a geometrical approach. | ||
+ | |||
+ | From the given, we have <math> a + b = 2 - c</math> and <math>a^2 + b^2 = 12 - c^2 </math>. The first equation is a line with x and y intercepts of <math>2-c</math> and the second equation is a circle centered at the origin with radius <math>\sqrt{12-c^2}</math>. Intuitively, if we want to find the minimum / maximum <math>c</math> such that there still exist real solutions, the two graphs of the equations should be tangent. | ||
+ | |||
+ | Thus, we have that <math>\sqrt{2} \cdot \sqrt{12-c^2} = 2-c</math>, which simplifies to <math>3c^2-4c-20=0</math>. Solving the quadratic, we get that the values of <math>c</math> for which the two graphs are tangent are <math>c=-2</math> and <math>c=\frac{10}{3}</math>. Thus, our answer is <math>\boxed{\frac{16}{3}}</math>. | ||
+ | |||
== See also == | == See also == |
Revision as of 02:52, 14 January 2018
Problem
Let and be real numbers such that
What is the difference between the maximum and minimum possible values of ?
Solution 1
. Now, by Cauchy-Schwarz, we have that . Therefore, we have that . We then find the roots of that satisfy equality and find the difference of the roots. This gives the answer, .
Solution 2
This is similar to the first solution but is far more intuitive. From the given, we have This immediately suggests use of the Cauchy-Schwarz inequality. By Cauchy, we have Substitution of the above results and some algebra yields This quadratic inequality is easily solved, and it is seen that equality holds for and .
The difference between these two values is .
Solution 3
(no Cauchy-Schwarz)
From the first equation, we know that . We substitute this into the second equation to find that This simplifies to , which we can write as the quadratic . We wish to find real values for and that satisfy this equation. Therefore, the discriminant is nonnegative. Hence, or . This factors as . Therefore, , and by symmetry this must be true for and as well.
Now and satisfy both equations, so we see that must be the minimum possible value of . Also, and satisfy both equations, so we see that is the maximum possible value of . The difference between these is , or .
Solution 4
We take a geometrical approach.
From the given, we have and . The first equation is a line with x and y intercepts of and the second equation is a circle centered at the origin with radius . Intuitively, if we want to find the minimum / maximum such that there still exist real solutions, the two graphs of the equations should be tangent.
Thus, we have that , which simplifies to . Solving the quadratic, we get that the values of for which the two graphs are tangent are and . Thus, our answer is .
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.