# Triangle by HM segments

### What Might This Be About?

11 September 2015, Created with GeoGebra

### Problem

The following has been posted by Oai Thanh Dào at the CutTheKnotMath facebook page:

The assumption is of course that $\Delta ABC$ is scalene.

### Preliminaries

Let $P\in BC$ such that $\overrightarrow{BP}=-k\overrightarrow{CP}.$

Then

$\displaystyle \overrightarrow{AP}=\frac{\overrightarrow{AB}+k\overrightarrow{AC}}{1+k}.$

Indeed, $\overrightarrow{AP}-\overrightarrow{AC}=\overrightarrow{CP}.$ Therefore,

\begin{align} \overrightarrow{AP}-\overrightarrow{AB}&=\overrightarrow{BP}\\ &=-kCP\\ &=-k(\overrightarrow{AP}-\overrightarrow{AC}) \end{align}

which is he same as $(1+k)\overrightarrow{AP}=\overrightarrow{AB}+k\overrightarrow{AC},$ as required.

In $\Delta ABC,$ $\overrightarrow{BH_a}=-k\overrightarrow{CH_a}$ where $\displaystyle k=\frac{\cot B}{\cot C}.$

As a consequence,

$\displaystyle\overrightarrow{AH_a}=\frac{\overrightarrow{AB}\cdot\cot C+\overrightarrow{AC}\cdot\cot B}{\cot B+\cot C}.$

### Solution

Without loss of generality, we'll assume that $A\gt B\gt C.$ As we have seen

$\displaystyle\overrightarrow{AH_a}=\frac{\overrightarrow{AB}\cdot\cot C+\overrightarrow{AC}\cdot\cot B}{\cot B+\cot C}$

and

$\displaystyle\overrightarrow{AM_a}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}.$

It follows that

\displaystyle\begin{align} \overrightarrow{H_aM_a}&=\frac{(\cot C-\cot B)(\overrightarrow{AC}-\overrightarrow{AB})}{2(\cot B+\cot C)} &=\frac{(\cot C-\cot B)(\overrightarrow{BC})}{2(\cot B+\cot C)}. \end{align}

Next

\displaystyle\begin{align} H_aM_a&=|\overrightarrow{H_aM_a}|\\ &=\left|\frac{(\cot C-\cot B)(\overrightarrow{BC})}{2(\cot B+\cot C)}\right|\\ &=\frac{(\cot C-\cot B)\cdot a}{2(\cot B+\cot C)}\\ &=\frac{a\cdot\sin(B-C)}{2\sin(B+C)}\\ &=\frac{a\cdot\sin(B-C)}{2\sin(A)}\\ &=R\cdot\sin (B-C), \end{align}

where $R$ is the circumradius of $\Delta ABC.$ Similarly, $H_bM_b=\sin (A-C)$ and $H_cM_c=R\cdot\sin (A-B).$

Thus the problem is reduced to showing that the three positive numbers $\sin (B-C),$ $\sin (A-C),$ and $\sin (A-B)$ are a triangle side lengths, i.e., satisfy three triangle inequalities.

$1.\;\sin (B-C)+\sin (A-B)\gt\sin (A-C)$

This is equivalent to $\sin (B-C)\gt\sin (A-C)-\sin (A-B),$ or,

$\displaystyle2\cdot\sin\left(\frac{B-C}{2}\right)\cos\left(\frac{B-C}{2}\right)\gt 2\sin\left(\frac{B-C}{2}\right)\cos\left(A-\frac{B+C}{2}\right)$

which simplifies to

$\displaystyle\cos\left(\frac{B-C}{2}\right)\gt\cos\left(A-\frac{B+C}{2}\right),$

or, $\displaystyle A-\frac{B+C}{2}\gt\frac{B-C}{2}$ which is just $A\gt B,$ as was assumed.

$2.\;\sin (B-C)+\sin (A-C)\gt\sin (A-B)$

This is equivalent to $\sin (B-C)\gt\sin (A-B)-\sin (A-C),$ or,

$\displaystyle2\cdot\sin\left(\frac{B-C}{2}\right)\cos\left(\frac{B-C}{2}\right)\gt -2\sin\left(\frac{B-C}{2}\right)\cos\left(A-\frac{B+C}{2}\right)$

or, $\displaystyle\cos\left(\frac{B-C}{2}\right)+\cos\left(A-\frac{B+C}{2}\right)\gt 0,$ i.e., $\displaystyle 2\cos\left(\frac{A-B}{2}\right)\cos\left(\frac{A-C}{2}\right)\gt 0,$ which is obviously true.

$3.\;\sin (A-C)+\sin (A-B)\gt\sin (B-C)$

This reduces to $\displaystyle 2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A-C}{2}\right)\gt 0,$ which is also true.

### Acknowledgment

The solution by Claudia Nanuti, Diana Trailescu, Dan Sitaru and Leo Giugiuc was supplied by Leo Giugiuc. I am grateful to Leo for helping me get through the details of the solution.