Difference between revisions of "2014 AMC 12B Problems/Problem 13"
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<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> | <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 1== |
Notice that <math>1>\frac{1}{a}>\frac{1}{b}</math>. Using the triangle inequality, we find | Notice that <math>1>\frac{1}{a}>\frac{1}{b}</math>. Using the triangle inequality, we find | ||
Line 18: | Line 18: | ||
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> | ||
+ | |||
+ | ==Solution 2 (similar to Solution 1) == | ||
+ | |||
+ | |||
+ | We can form degenerate triangles from the given information. Since <math>b>a>1</math>, we know that <math>a,b,1</math> must satisfy <math>a + 1 \le b</math> to be degenerate in the first scenario. | ||
+ | |||
+ | |||
+ | Similarly, since <math>\frac{1}{n} + 1 > 1</math> for all positive integers <math>n</math>, the only condition that is guaranteed to form a degenerate triangle from the second instance is <math>\frac{1}{a} + \frac{1}{b} \le 1</math>. | ||
+ | |||
+ | |||
+ | Solving the second inequality for <math>a</math> in terms of <math>b</math> (for more-easily apparent inequality manipulation) yields <math>b\ge \frac{a}{a-1}</math>. Notice that <math>b> a+1 \ge \frac{a}{a-1}</math> for almost all positive real numbers <math>a>1</math> (see note at the end). | ||
+ | |||
+ | |||
+ | Thus, it suffices to solve the inequality <math>a+1 \ge \frac{a}{a-1}</math> for <math>a</math> whose only positive solution is <math>a \ge \frac{1 + \sqrt{5}}{2}</math>. | ||
+ | |||
+ | |||
+ | Substituting back into the first inequality yields <math>\boxed{b \ge \frac{3 + \sqrt{5}}{2}}</math> and thus the answer is <math>\text{C}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | Note: this solution assumes that <math>a</math> is not less than <math>1.618</math>, the points at which the last inequality. | ||
+ | |||
+ | |||
+ | |||
+ | ~baldeagle123 | ||
== See also == | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=12|num-a=14}} | {{AMC12 box|year=2014|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:24, 5 December 2020
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
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.