# When Is Triangle Equilateral: Marian Dinca's Criterion

### Proof 1

Let $BC=a=x+y,\,$ $AC=b=x+z,\,$ and $AB=c=y+z.\,$ $x,y,z\gt 0;\,$ $\displaystyle s=\frac{a+b+c}{2}=x+y+z.$

In the barycentric coordinates,

$\displaystyle D=\left(0,\frac{s-b}{a},\frac{s-c}{a}\right),\,E=\left(\frac{s-a}{b},0,\frac{s-c}{b}\right),\,F=\left(\frac{s-a}{c},\frac{s-b}{c},0\right).\,$ The barycenters of triangles $ABC\,$ and $DEF\,$ are, respectively,

\displaystyle\begin{align} G &= \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)\\ &= \left(\frac{1}{3}\left(\frac{s-a}{b}+\frac{s-a}{c}\right),\frac{1}{3}\left(\frac{s-b}{a}+\frac{s-b}{c}\right),\frac{1}{3}\left(\frac{s-c}{a}+\frac{s-c}{b}\right)\right), \end{align}

yielding three equations:

$\displaystyle\frac{s-a}{b}+\frac{s-a}{c}=1,\,\frac{s-b}{a}+\frac{s-b}{c}=1,\,\frac{s-c}{a}+\frac{s-c}{b}=1,$

or,

$\displaystyle\frac{x}{x+z}+\frac{x}{y+z}=1,\,\frac{y}{x+y}+\frac{y}{y+z}=1,\,\frac{z}{x+y}+\frac{z}{x+z}=1,$

$\displaystyle\frac{x+z}{x+z}+\frac{x+y}{y+z}+\frac{y+z}{x+y}=3,$

or,

$\displaystyle\frac{x+y}{y+z}+\frac{y+z}{x+y}=2.$

By the AM-GM inequality, $\displaystyle\frac{x+y}{y+z}=\frac{y+z}{x+y}=1$ so that $x+y=y+z\,$ and $x=z.\,$ Further, $\displaystyle\frac{x}{x+z}+\frac{x}{y+z}=1\,$ reduces to $\displaystyle\frac{x}{2x}+\frac{x}{y+x}=1,\,$ i.e., $\displaystyle\frac{x}{y+x}=1,\,$ from which $x=y.\,$ Thus $x=y=z\,$ and so $a=b=c.$

### Proof 2

Let $s-a=x,\,s-b=y,\,$ and $s-c=z.\,$ It's well known that $\displaystyle\frac{\overrightarrow{DB}}{\overrightarrow{DC}}=-\frac{y}{z},\,$ so that, for any point $X,\,$ $\displaystyle\overrightarrow{XD}=\frac{z\overrightarrow{XB}+y\overrightarrow{XC}}{y+z}.\,$ Similarly, $\displaystyle\overrightarrow{XE}=\frac{z\overrightarrow{XA}+x\overrightarrow{XC}}{x+z}\,$ and $\displaystyle\overrightarrow{XF}=\frac{y\overrightarrow{XA}+x\overrightarrow{XB}}{x+y}.\,$

Since $ABC\,$ and $DEF\,$ share the same centroid, $\overrightarrow{XA}+\overrightarrow{XB}+\overrightarrow{XC}=\overrightarrow{XD}+\overrightarrow{XE}+\overrightarrow{XF}.$ Thus, for $X=A,\,$ we get

$\displaystyle\overrightarrow{AB}+\overrightarrow{AC}=\left(\frac{z}{y+z}+\frac{x}{x+y}\right)\overrightarrow{AB}+\left(\frac{y}{y+z}+\frac{x}{x+z}\right)\overrightarrow{AC},\,$

and from here

$\displaystyle\frac{z}{y+z}+\frac{x}{x+y}=1,\;\frac{y}{y+z}+\frac{x}{x+z}=1.$

One reduces to $y^2=xz,\,$ the other to $z^2=xy\,$ so that $\displaystyle\left(\frac{y}{z}\right)^3=1,\,$ or $y=z.\,$ Similarly, $x=y\,$ so that $x=y=z\,$ and, subsequently, $a=b=c.$

### Acknowledgment

The problem and Proof 1 are due to Marian Dinca. Proof 2 is by Leo Giugiuc.

A third proof is a consequence of a more general statement.

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• Equilateral Triangles On Sides of a Parallelogram
• Pompeiu's Theorem
• Pairs of Areas in Equilateral Triangle
• The Eutrigon Theorem
• Equilateral Triangle in Equilateral Triangle
• Seven Problems in Equilateral Triangle
• Spiral Similarity Leads to Equilateral Triangle
• Parallelogram and Four Equilateral Triangles
• A Pedal Property in Equilateral Triangle
• Miguel Ochoa's van Schooten Like Theorem
• Two Conditions for a Triangle to Be Equilateral
• Incircle in Equilateral Triangle
• Barycenter of Cevian Triangle
• Excircle in Equilateral Triangle
• Converse Construction in Pompeiu's Theorem
• Wonderful Trigonometry In Equilateral Triangle
• 60o Angle And Importance of Being The Other End of a Diameter
• One More Property of Equilateral Triangles
• Van Khea's Quickie
• Equilateral Triangle from Three Centroids