Dan-Stefan Marinescu Inequality

Leo Giugiuc has posted the following problem due to Dan-Stefan Marinescu at the CutTheKnotMath facebook page

Points $D,E,F$ are either inside of on the border of $\Delta ABC$ and have the property that the two triangles $ABC$ and $DEF$ have the same centroid. Then

$MA + MB + MC \ge MD+ME+MF,$

for any point $M$ in the plane.

Proof

For simplicity, let's rewrite the problem as follows:

Points $B_i,\;i=1,2,3$ are either inside or on the border of $\Delta A_{1}A_{2}A_{3}$ and have the property that the two triangles $A_{1}A_{2}A_{3}$ and $B_{1}B_{2}B_{3}$ have the same centroid. Then

$\displaystyle\sum_{i=1}^{3}MA_{i}\ge\sum_{i=1}^{3}MB_{i},$

for any point $M$ in the plane.

Thinking of the barycentric coordinate, the condition that points $B$ are within $\Delta A_{1}A_{2}A_{3}$ means that

$\displaystyle B_{i}=\sum_{j=1}^{3}\alpha_{ij}A_{j},\;\sum_{j=1}^{3}\alpha_{ij}=1,\;i=1,2,3,$

with $\alpha_{ij}\ge 0,$ $i,j=1,2,3.$

Centroid of three points is their center of gravity, implying that, since, $\displaystyle\frac{1}{3}\sum_{i=1}^{3}A_{i}=\frac{1}{3}\sum_{i=1}^{3}B_{i},$ so too $\displaystyle\sum_{i=1}^{3}A_{i}=\sum_{i=1}^{3}B_{i}.$

To simplify, I shall be writing say $\displaystyle\sum_{k},$ omitting the end points of counting in $\displaystyle\sum_{k=1}^{3},$ for all the sums that we encounter are from $1$ to $3.$ So, we continue

$\begin{align}\displaystyle \sum_{i}A_{i} &=\sum_{i}B_{i}\\ &=\sum_{i}\sum_{j}\alpha_{ij}A_{j}\\ &=\sum_{j}\sum_{i}\alpha_{ij}A_{j}\\ &=\sum_{j}A_{j}\sum_{i}\alpha_{ij}, \end{align}$

from which $\displaystyle\sum_{i}\alpha_{ij}=1,\;j=1,2,3$ because of the uniqueness of the representation in barycentric coordinates.

Now, for any point $M,$

$\begin{align}\displaystyle \sum_{i}|M-B_{i}| &=\sum_{i}\bigg|\sum_{j}\alpha_{ij}A_{j}-M\bigg|\\ &=\sum_{i}\bigg|\sum_{j}\alpha_{ij}A_{j}-\sum_{j}\alpha_{ij}M\bigg|\\ &=\sum_{i}\bigg|\sum_{j}\alpha_{ij}(A_{j}-M)\bigg|\\ &\le\sum_{i}\sum_{j}\alpha_{ij}|A_{j}-M|\\ &=\sum_{j}\sum_{i}\alpha_{ij}|A_{j}-M|\\ &=\sum_{j}|A_{j}-M|\sum_{i}\alpha_{ij}\\ &=\sum_{j}|A_{j}-M|.\\ \end{align}$

Exactly as required.

Note that the condition that point $M$ was supposed to lie in the plane has been never used.


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