Comparing the inradii

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Atonu Roy Chowdhury
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Comparing the inradii

Unread post by Atonu Roy Chowdhury » Thu Apr 13, 2017 8:24 am

Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $A_1,B_1$ and $C_1$ be respectively the midpoints of the arcs $BAC,CBA$ and $ACB$ of $\Gamma$. Show that the inradius of triangle $A_1B_1C_1$ is not less than the inradius of triangle $ABC$.

User avatar
Atonu Roy Chowdhury
Posts: 40
Joined: Fri Aug 05, 2016 7:57 pm
Location: Chittagong, Bangladesh

Re: Comparing the inradii

Unread post by Atonu Roy Chowdhury » Thu Apr 13, 2017 8:43 am

My solution
Trivial angle chasing yields that the angles of $\triangle A_1B_1C_1$ are $\frac{\angle A + \angle B}{2}$ , $\frac{\angle B + \angle C}{2}$ and $\frac {\angle C + \angle A}{2}$ . We know that $r = 4R \sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) $. So, it remains to show that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le \sin(\frac{A+B}{4}) \sin(\frac{B+C}{4}) \sin(\frac{C+A}{4})$
By AM-GM, $\sin(\frac{A}{4})\cos(\frac{B}{4}) + \sin(\frac{B}{4})\cos(\frac{A}{4}) \ge 2 \sqrt{ \sin(\frac{A}{4})\cos(\frac{B}{4}) \sin(\frac{B}{4})\cos(\frac{A}{4})}$. Similarly we get two other ineqs. By multiplying them, we get the result.

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