Triangle Inequality Problem - AMC 12B, 2014 - Problem 13

Try this beautiful problem from American Mathematics Competition - 12B ,2014, Problem Number - 13 based on Triangle inequality

Problem - Triangle Inequality

• $\frac{3+\sqrt{3}}{2}$
• $\frac{5}{2}$
• $\frac{3+\sqrt{5}}{2}$
• $\frac{3+\sqrt{6}}{2}$

Key Concepts

Triangle Inequality

Inequality

Geometry

Check the Answer

Answer: \(\frac {3+\sqrt 5}{2}\)

Try with Hints

Here is the first hint where you can start this sum:

It is given $1>\frac{1}{a}>\frac{1}{b}$. Use Triangle Inequality here :

$$
\begin{aligned}
& a+1>b \
& a>b-1 \
& \frac{1}{a}+\frac{1}{b}>1
\end{aligned}
$$

If we want to find the lowest possible value of $b$, we create we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. Thus we get : $a=b-1$

Now try to do the rest of the sum……………………

Here is the rest of steps to check your problem :

$$
2 b-1=b^2-b
$$

Now Solving for $\mathbf{b}$ using the quadratic equation, we get

$$
\begin{aligned}
& b^2-3 b+1=0 \
& b=\frac{3+\sqrt{5}}{2} \text { (Answer) }
\end{aligned}
$$

Other useful links

• https://www.cheenta.in/triangle-inequality-mathematical-circles-problem-no-5/
• https://www.youtube.com/watch?v=1O-BxsyPw6E&list=PLTDTcDkWcXuyidnUii1hUUmu6y8E8M3lu&index=4

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