The incenter of triangle ABC is I and inradius is 2. What is the smallest possible
value of AI+BI+CI?
Incenter of Triangle
- asif e elahi
- Posts:185
- Joined:Mon Aug 05, 2013 12:36 pm
- Location:Sylhet,Bangladesh
Re: Incenter of Triangle
Take the triange $ABC$ for which it attains the minimum value.Consider the case when $CA\neq CB$. Now let $CI$ meets the incirle at $D$ so that $I$ lies between $C$ and $D$. Let the tangent at $D$ meet $AI,BI$ at $A_0,B_0$ resp. Then prove that $AI+BI \geq A_0+B_0$(You will need to go through some messy trigonometric calculation). The rest is easy.
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: Incenter of Triangle
Let $D, E, F$ be the perpendiculars from $I$ to $BC, CA, AB$ respectively. Each of them are equal to the inradius and thus has length $2$. Now, $AI=2\csc{\dfrac{A}{2}}, BI=2\csc{\dfrac{B}{2}}, CI=2\csc{\dfrac{C}{2}}$. Now,
\(AI+BI+CI=2\left(\csc{\dfrac{A}{2}}+\csc{\dfrac{B}{2}}+\csc{\dfrac{C}{2}}\right)\)
Using Jensen's inequality as $\csc \theta$ is convex in $0<\theta<\dfrac{\pi}{2}$ and since $0< A,B,C<\pi$
\(AI+BI+CI\ge 6 \left(\csc\left(\frac{A/2+B/2+C/2}{3}\right)\right)=6\csc\dfrac{\pi}{6}=12\)
The equality occurs when $\Delta ABC$ is equilateral.
\(AI+BI+CI=2\left(\csc{\dfrac{A}{2}}+\csc{\dfrac{B}{2}}+\csc{\dfrac{C}{2}}\right)\)
Using Jensen's inequality as $\csc \theta$ is convex in $0<\theta<\dfrac{\pi}{2}$ and since $0< A,B,C<\pi$
\(AI+BI+CI\ge 6 \left(\csc\left(\frac{A/2+B/2+C/2}{3}\right)\right)=6\csc\dfrac{\pi}{6}=12\)
The equality occurs when $\Delta ABC$ is equilateral.
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.