Problem 4087 from Crux Mathematicorum

Problem

Problem 4087 from Crux Mathematicorum

Solution

Let $A'$ be the reflection of $A\,$ in $M,\,$ the midpoint of $BC.\,$ Then $AA'=2m_a,\,$ $A'B=AC=b,\,$ $\angle ABA'=\pi-A,\,$ $[\Delta ABA']=[\Delta ABC]=S,\,$ $\sin\angle ABA'=\sin A.$

Problem 4087 from Crux Mathematicorum, solution

Denote the inradius and the circumradius of $\Delta ABA'\,$ as $r'\,$ and $R',\,$ respectively, and let $m=2m_a.\,$ We need to prove that $m(b+c)+m^2\ge 8S\sin \angle ABA',\,$ which is the same as

$\displaystyle\begin{align} &m(m+b+c)\ge 8S\sin\angle ABA'\,\Longleftrightarrow\\ &\frac{2mS}{r'}\ge 8S\sin\angle ABA'\,\Longleftrightarrow\\ &\frac{m}{\sin\angle ABA'}\ge 4r'\,\Longleftrightarrow\\ &R'\ge 2r', \end{align}$

which is Euler's inequality. Equality is achieved when $\Delta ABA'\,$ is equilateral, i.e., when $b=c\,$ and $\angle A=120^{\circ}.$

Acknowledgment

This is Problem 4087 from Crux Mathematicorum (Solution, v 42,n 9, November 2016) which was kindly sent to me by Leo Giugiuc, along wit a solution of his. The problem is by Lorian Saceanu. The editors indicate that they received seven correct submissions of which they chose to present Leo Giugiuc's.

 

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