Difference between revisions of "2013 AMC 12B Problems/Problem 17"
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The difference between these two values is <math>\boxed{\textbf{(D)} \ \frac{16}{3}}</math>. | The difference between these two values is <math>\boxed{\textbf{(D)} \ \frac{16}{3}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | (no Cauchy-Schwarz) | ||
+ | |||
+ | From the first equation, we know that <math>c=2-a-b</math>. We substitute this into the second equation to find that | ||
+ | <cmath>a^2+b^2+(2-a-b)^2=12.</cmath> | ||
+ | This simplifies to <math>2a^2+2b^2-4a-4b+2ab=8</math>, which we can write as the quadratic <math>a^2+(b-2)a+(b^2-2b-4)=0</math>. We wish to find real values for <math>a</math> and <math>b</math> that satisfy this equation. Therefore, the discriminant is nonnegative. Hence, | ||
+ | <cmath>(b-2)^2-4(b^2-2b-4)\ge0,</cmath> | ||
+ | or <math>-3b^2+4b+20\ge 0</math>. This factors as <math>-(3b-10)(b+2)\ge 0</math>. Therefore, <math>-2\le b\le \frac{10}{3}</math>, and by symmetry this must be true for <math>a</math> and <math>c</math> as well. | ||
+ | |||
+ | 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>. | ||
== See also == | == See also == |
Revision as of 04:37, 8 November 2016
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 .
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.