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 09:14, 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
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
and
Substituting, we get
Solving for
using the quadratic equation, we get
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.