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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>?
  
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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>
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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>
  
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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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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>
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Substituting, we get <cmath>\frac{2b-1}{b(b-1)} = 1</cmath>
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Substituting, we get
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<cmath>\frac{1}{b-1}+\frac{1}{b}=\frac{b+b-1}{b(b-1)}=1</cmath>
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<cmath>\frac{2b-1}{b(b-1)} = 1</cmath>
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<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>
  
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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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 +
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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
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==Solution 3==
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By the triangle inequality we have <math>a+b \ge 1</math>, <math>a + 1 \ge b</math>, <math>b+1 \ge a</math> and <math>b(1+a) \ge a</math>, <math>a(1+b) \ge b</math>, <math>a+b \ge ab</math>. Clearly a triangle can't have negative area so, our triangle must have <math>0</math> area which means equality must be obtained for all the inequalities above. Since <math>a = 1+b</math> we know that <math>(1+b)^2 = b</math> which simplifies to <math>b^2 - 3b + 1 = 0</math>. Solving for <math>b</math> we get <math>\frac{3 + \sqrt{5}}{2}</math>. So the answer is C.
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~coolmath_2018
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==Solution 4==
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<math>1+a \leq b, \quad \frac{1}{b} + \frac{1}{a} \leq 1</math>
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<math>\implies a \leq b-1, \quad \frac{a+b}{ab} \leq 1</math>
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<math>\implies a+b \leq ab \implies b \leq ab-a \implies b \leq(b-1)a \implies a \geq \frac{b}{b-1}</math>
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<math>\therefore \frac{b}{b-1} \leq a \leq b-1</math>
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<math>\frac{b}{b-1} \leq b-1 \implies b \leq b^2-2b+1 \implies b^2-3b+1 \geq 0</math>
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<math>b= \frac{3 \pm \sqrt{9-4}}{2}= \frac{3 \pm \sqrt{5}}{2}</math>
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<math>b=\frac{3 + \sqrt{5}}{2}</math>
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select C
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~ Ji Yang
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== 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}}

Latest revision as of 23:18, 29 October 2024

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

Solution 3

By the triangle inequality we have $a+b \ge 1$, $a + 1 \ge b$, $b+1 \ge a$ and $b(1+a) \ge a$, $a(1+b) \ge b$, $a+b \ge ab$. Clearly a triangle can't have negative area so, our triangle must have $0$ area which means equality must be obtained for all the inequalities above. Since $a = 1+b$ we know that $(1+b)^2 = b$ which simplifies to $b^2 - 3b + 1 = 0$. Solving for $b$ we get $\frac{3 + \sqrt{5}}{2}$. So the answer is C. ~coolmath_2018

Solution 4

$1+a \leq b, \quad \frac{1}{b} + \frac{1}{a} \leq 1$

$\implies a \leq b-1, \quad \frac{a+b}{ab} \leq 1$

$\implies a+b \leq ab \implies b \leq ab-a \implies b \leq(b-1)a \implies a \geq \frac{b}{b-1}$

$\therefore \frac{b}{b-1} \leq a \leq b-1$

$\frac{b}{b-1} \leq b-1 \implies b \leq b^2-2b+1 \implies b^2-3b+1 \geq 0$

$b= \frac{3 \pm \sqrt{9-4}}{2}= \frac{3 \pm \sqrt{5}}{2}$

$b=\frac{3 + \sqrt{5}}{2}$

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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