Difference between revisions of "2014 AMC 12B Problems/Problem 13"
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Real numbers <math>a</math> and <math>b</math> are chosen with <math>1<a<b</math> such that no triangle with positive area has side lengths <math>1</math>, <math>a</math>, and <math>b</math> or <math>\frac{1}{b}</math>, <math>\frac{1}{a}</math>, and <math>1</math>. What is the smallest possible value of <math>b</math>? | Real numbers <math>a</math> and <math>b</math> are chosen with <math>1<a<b</math> such that no triangle with positive area has side lengths <math>1</math>, <math>a</math>, and <math>b</math> or <math>\frac{1}{b}</math>, <math>\frac{1}{a}</math>, and <math>1</math>. What is the smallest possible value of <math>b</math>? | ||
| − | <math> \textbf{(A)}\ \frac{3+\sqrt{3}}{2}\qquad\textbf{(B)}\ \frac{5}{2}\qquad\textbf{(C)}\ \frac{3+\sqrt{5}}{2}\qquad\textbf{(D) | + | <math> \textbf{(A)}\ \frac{3+\sqrt{3}}{2}\qquad\textbf{(B)}\ \frac{5}{2}\qquad\textbf{(C)}\ \frac{3+\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{3+\sqrt{6}}{2}\qquad\textbf{(E)}\ 3 </math> |
==Solution== | ==Solution== | ||
Revision as of 10:14, 3 March 2015
Problem
Real numbers 
and 
are chosen with 
such that no triangle with positive area has side lengths 
, , and or 
, 
, and . What is the smallest possible value of ?
Solution
Notice that 
. Using the triangle inequality, we find
![\[a+1 > b \implies a>b-1\]](http://latex.artofproblemsolving.com/9/0/5/9059181d4c2754b73bf1c41a41965c53a1cc8447.png)
![\[\frac{1}{a}+\frac{1}{b} > 1\]](http://latex.artofproblemsolving.com/d/c/3/dc32c35ab9185374efdc82a7a5b45f683f7cefad.png)
In order for us the find the lowest possible value for , we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side.
Thus we get ![\[a=b-1\]](http://latex.artofproblemsolving.com/6/f/3/6f3041f077b976ffc37055a870542c7d0f26d83c.png)
and ![\[\frac{1}{a} + \frac{1}{b}=1\]](http://latex.artofproblemsolving.com/0/9/7/0974cd3d7f89d6bde4fb19e184bcde625bdc9697.png)
Substituting, we get ![\[\frac{2b-1}{b(b-1)} = 1\]](http://latex.artofproblemsolving.com/e/1/a/e1a91ccb76bef11a34069e903d584155406f248d.png)
Solving for using the quadratic equation, we get
![\[b^2-3b+1=0 \implies b = \boxed{\textbf{(C)} \ \frac{3+\sqrt{5}}{2}}\]](http://latex.artofproblemsolving.com/a/4/5/a45924a7146c220f53694631b0bcce80d43c8439.png)
See also
| 2014 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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. 
