2014 AMC 12B Problems/Problem 13
Contents
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 1
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
Solution 2 (similar to Solution 1)
We can form degenerate triangles from the given information. Since , we know that must satisfy to be degenerate in the first scenario.
Similarly, since for all positive integers , the only condition that is guaranteed to form a degenerate triangle from the second instance is .
Solving the second inequality for in terms of (for more-easily apparent inequality manipulation) yields . Notice that for almost all positive real numbers (see note at the end).
Thus, it suffices to solve the inequality for whose only positive solution is .
Substituting back into the first inequality yields and thus the answer is .
Note: this solution assumes that is not less than , the points at which the last inequality.
~baldeagle123
Solution 3
By the triangle inequality we have , , and , , . Clearly a triangle can't have negative area so, our triangle must have area which means equality must be obtained for all the inequalities above. Since we know that which simplifies to . Solving for we get . So the answer is C. ~coolmath_2018
Solution 4
select C
~ Ji Yang
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 |
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