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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>
  
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==Solution==
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==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  
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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>
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==Solution 2 (similar to Solution 1) ==
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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.
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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>.
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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).
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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>.
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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>.
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Note: this solution assumes that <math>a</math> is not less than <math>1.618</math>, the points at which the last inequality.
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~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 $a$ and $b$ are chosen with $1<a<b$ such that no triangle with positive area has side lengths $1$, $a$, and $b$ or $\frac{1}{b}$, $\frac{1}{a}$, and $1$. What is the smallest possible value of $b$?

$\textbf{(A)}\ \frac{3+\sqrt{3}}{2}\qquad\textbf{(B)}\ \frac{5}{2}\qquad\textbf{(C)}\ \frac{3+\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{3+\sqrt{6}}{2}\qquad\textbf{(E)}\ 3$

Solution 1

Notice that $1>\frac{1}{a}>\frac{1}{b}$. Using the triangle inequality, we find \[a+1 > b \implies a>b-1\] \[\frac{1}{a}+\frac{1}{b} > 1\] In order for us the find the lowest possible value for $b$, we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. Thus we get \[a=b-1\] and \[\frac{1}{a} + \frac{1}{b}=1\] Substituting, we get \[\frac{1}{b-1}+\frac{1}{b}=\frac{b+b-1}{b(b-1)}=1\] \[\frac{2b-1}{b(b-1)} = 1\] \[2b-1=b^2-b\] Solving for $b$ using the quadratic equation, we get \[b^2-3b+1=0 \implies b = \boxed{\textbf{(C)} \ \frac{3+\sqrt{5}}{2}}\]

Solution 2 (similar to Solution 1)

We can form degenerate triangles from the given information. Since $b>a>1$, we know that $a,b,1$ must satisfy $a + 1 \le b$ to be degenerate in the first scenario.


Similarly, since $\frac{1}{n} + 1 > 1$ for all positive integers $n$, the only condition that is guaranteed to form a degenerate triangle from the second instance is $\frac{1}{a} + \frac{1}{b} \le 1$.


Solving the second inequality for $a$ in terms of $b$ (for more-easily apparent inequality manipulation) yields $b\ge \frac{a}{a-1}$. Notice that $b> a+1 \ge \frac{a}{a-1}$ for almost all positive real numbers $a>1$ (see note at the end).


Thus, it suffices to solve the inequality $a+1 \ge \frac{a}{a-1}$ for $a$ whose only positive solution is $a \ge \frac{1 + \sqrt{5}}{2}$.


Substituting back into the first inequality yields $\boxed{b \ge \frac{3 + \sqrt{5}}{2}}$ and thus the answer is $\text{C}$.


Note: this solution assumes that $a$ is not less than $1.618$, 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

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