Difference between revisions of "2014 AMC 12B Problems/Problem 13"
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In order for us the find the lowest possible value for <math>b</math>, we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. | In order for us the find the lowest possible value for <math>b</math>, we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. | ||
Thus we get <cmath>a=b-1</cmath> and <cmath>\frac{1}{a} + \frac{1}{b}=1</cmath> | Thus we get <cmath>a=b-1</cmath> and <cmath>\frac{1}{a} + \frac{1}{b}=1</cmath> | ||
| − | Substituting, we get <cmath>\frac{2b-1}{b(b-1)} = 1</cmath> | + | Substituting, we get |
| + | <cmath>\frac{1}{b-1}+\frac{1}{b}=1</cmath> | ||
| + | <cmath>\frac{b+b-1}{b(b-1)}</cmath> | ||
| + | <cmath>\frac{2b-1}{b(b-1)} = 1</cmath> | ||
| + | <cmath>2b-1=b^2-b</cmath> | ||
Solving for <math>b</math> using the quadratic equation, we get | Solving for <math>b</math> using the quadratic equation, we get | ||
<cmath>b^2-3b+1=0 \implies b = \boxed{\textbf{(C)} \ \frac{3+\sqrt{5}}{2}}</cmath> | <cmath>b^2-3b+1=0 \implies b = \boxed{\textbf{(C)} \ \frac{3+\sqrt{5}}{2}}</cmath> | ||
Revision as of 20:31, 13 February 2016
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
![\[\frac{1}{b-1}+\frac{1}{b}=1\]](http://latex.artofproblemsolving.com/0/4/6/046c0b838337a95ac6bad02d2d397af8cc840484.png)
![\[\frac{b+b-1}{b(b-1)}\]](http://latex.artofproblemsolving.com/5/b/7/5b72be3d81863b51f939d2ef1a8bd85f9f1b920c.png)
![\[\frac{2b-1}{b(b-1)} = 1\]](http://latex.artofproblemsolving.com/e/1/a/e1a91ccb76bef11a34069e903d584155406f248d.png)
![\[2b-1=b^2-b\]](http://latex.artofproblemsolving.com/0/c/8/0c8a2355ef9f70d0133b08b9af3537dbb0e5c01f.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. 
