Area Of The Medial Triangle

Relevant wiki: Area of Triangles - Problem Solving - Medium

I will offer a couple of different methods to solve this problem. For notation purposes let $[\triangle ABC]$ denote the area of $\triangle ABC$ .

Method 1: Areal/Barycentric Coordinates

$A,B,C$ have areal coordinates $(1,0,0),(0,1,0),(0,0,1)$

$D,E,F$ will have the areal coordinates $\left(0,\dfrac{1}{2},\dfrac{1}{2} \right ),\left(\dfrac{1}{2},0,\dfrac{1}{2} \right ),\left(\dfrac{1}{2},\dfrac{1}{2},0 \right )$ .

The area of a triangle from its areal coordinates can be calculated from the determinate of the matrix of its coordinates so:

$\dfrac{X}{Y}=\dfrac{\begin{vmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{vmatrix}}{\begin{vmatrix} 0 & 0.5 & 0.5\\ 0.5 & 0 & 0.5\\ 0.5 & 0.5 & 0 \end{vmatrix}}=\dfrac{1}{\frac{1}{4}}=\color{#20A900}{\boxed{\boxed{4}}}$

Method 2: Similar Triangles.

Let $D,E,F$ be the midpoints of $AB,AC,BC$ respectively.

An enlargement about $A$ of scale factor $2$ carries $D$ to $B$ and $E$ to $C$ . It therefore follows that $\triangle ADE \cong \triangle ABC$ .

As the sides are in ratio $1:2$ it follows $\dfrac{[\triangle ADE]}{[\triangle ABC]}=\dfrac{1}{4} \implies [\triangle ADE]=\dfrac{[\triangle ABC]}{4}$ .

We have similar results for $\triangle BDF$ and $\triangle CEF$ .

$[\triangle BDF]+[\triangle CEF]+[\triangle ADE]=\dfrac{3 [\triangle ABC]}{4}$

$[\triangle DEF]=[\triangle ABC]-([\triangle BDF]+[\triangle CEF]+[\triangle ADE])=\dfrac{[\triangle ABC]}{4}$

We therefore have $\dfrac{X}{Y}=\dfrac{1}{\frac{1}{4}}=\color{#20A900}{\boxed{\boxed{4}}}$

Let $AC=a$ and $CB=b$ so the area of $\triangle ABC=X=\frac {ab}{2}$ and $\triangle DEF=Y=\frac {ab}{8}$

Then $\frac{X}{Y}=\frac{\frac {ab} {2}}{\frac {ab}{8}}=4$