# An Angle Inequality in Triangle with Perpendicular Medians

### Source

### Problem

In $\Delta ABC,$ the medians m_b and m_c are perpendicular. Prove that

$\displaystyle \cos A\ge \frac{4}{5}.$

### Solution 1

We shall use a well known result that holds due to the two medians being perpendicular, viz., $b^2+c^2=5a^2.$ From this, $b^2+c^2-a^2=4a^2,$ or $\displaystyle \frac{b^2+c^2-a^2}{2bc}=\frac{4a^2}{2bc},$ which, by the Law of Cosines, means that $\displaystyle \displaystyle \cos A=\frac{4a^2}{2bc}.$

Now, using the AM-GM inequality, and again $b^2+c^2=5a^2,$

$\displaystyle \frac{4a^2}{2bc}\ge\frac{4a^2}{b^2+c^2}=\frac{4a^2}{5a^2}=\frac{4}{5}.$

Equality is attained for $b=c$ so that $5a^2=2b^2,$ i.e., for the triples $(t\sqrt{2},t\sqrt{5},t\sqrt{5}),$ $t\gt 0.$

### Solution 2

Let $\angle A=\phi$, $AB=2p$, and $AC=2q$. Choosing the $+X$ along $AB$,

$\begin{align} \vec{AB}&=2p\hat{x},~\vec{AC}=2q\cos\phi\hat{x}+2q\sin\phi\hat{y}, \\ \vec{AM_c}&=p\hat{x},~\vec{AM_b}=q\cos\phi\hat{x}+q\sin\phi\hat{y}. \end{align}$

$\begin{align} \vec{CM_c}&=\vec{AM_c}-\vec{AC}=(p-2q\cos\phi)\hat{x}-2q\sin\phi\hat{y}, \\ \vec{BM_b}&=\vec{AM_b}-\vec{AB}=(q\cos\phi-2p)\hat{x}+q\sin\phi\hat{y}. \end{align}$

From perpendicularity,

$\begin{align} &\vec{CM_c}\cdot\vec{BM_b}=0 \\ \Rightarrow~&(p-2q\cos\phi)(q\cos\phi-2p)-2q^2\sin^2\phi=0 \\ \Rightarrow~&\cos\phi=\frac{(2p^2+2q^2)}{5pq}\geq\frac{4}{5}~\text{(AM-GM)}. \end{align}$

### Acknowledgment

The problem by Miguel Ochoa Sanchez that was posted at the Peru Geometrico facebook group was kindly communicated to me by Leo Giugiuc, along with a solution of his. More solutions can be found at the site of the original posting.

Solution 2 is by Amit Itagi.

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