Problem 1

Let \(m_a, m_b, m_c\) be the medians of the triangle \(ABC\) and \(r_a, r_b, r_c\) be its exradii. Then prove that \(m_a+m_b+m_c\le r_a+r_b+r_c\).

AoPS forum

Tags: triangle calculation inequality geometric inequality

Solution

First we'll prove that \(r_a+r_b+r_c=4R+r\) where \(R\) and \(r\) are the circumradius and the inradius of \(ABC\). Let \(I\) and \(I_a\) be the incenter and the \(A\)-excenter of \(ABC\). Let \(W\) be the midpoint of \(II_a\). \(BI\) and \(BI_a\) are the internal and external bisectors of angle \(B\) so \(\angle IBI_a=90^\circ\). Similarly \(\angle ICI_a=90^\circ\) and \(B, I, C, I_a\) lie on a circle with center \(W\). Hence \(\angle BWC=2\angle BI_aC=180^\circ-\angle A\) and \(W\) lies on th circle \(ABC\). Thus \[\frac{r_a}{\sin{\frac{\angle A}{2}}}-\frac{r}{\sin{\frac{\angle A}{2}}}=AI_a-AI=II_a=2WI=2WB= 4R\sin{\frac{\angle A}{2}}\] and \(r_a=r+2R(1-\cos{\angle A})\). Summing up the other two similar equalities we get by Carnot's theorem \[r_a+r_b+r_c=3r+6R-2R(\cos{\angle A} + \cos{\angle B} + \cos{\angle C})=3r+6R-2(R+r)=4R+r.\] Now denote by \(A', B', C'\) the midpoints of respective sides of \(ABC\) and by \(O\) its circumcenter. According to the triangle inequality \(R+OA'=AO+OA'\ge AA'=m_a\). Summing up the other two similar inequalities we get \[m_a+m_b+m_c\le 3R+OA'+OB'+OC'=4R+r=r_a+r_b+r_c.\]