6 Through the midpoints of two sides of a triangle $A B C$ we draw the medians and connect the points that trisect the third side with the opposite vertex. Thus, inside, a kite (two triangles joined by a vertex) is obtained. We are asked to calculate the fraction of the total area of the triangle that the kite represents.
Solution. Suppose, without loss of generality, that the area of triangle $ABC$ is one. That is, $[ABC]=1$. We have
![](https://cdn.mathpix.com/cropped/2024_04_26_95c41fcefd175860f118g-2.jpg?height=426&width=1258&top_left_y=1469&top_left_x=362)

The medians divide the triangle into six equal areas, so: $[AMF]=[FNC]=\frac{1}{6}$ and $[AFC]=2 \cdot \frac{1}{6}=\frac{1}{3}$. We draw, from $B, A, P$ and $Q$, perpendiculars to the median $CM$ and let $X, Y, Z$ and $V$ be their respective feet, as shown in the figure on the right. In this figure, working with the median $CM$:

- $AMY \equiv BMX$. Since $AM=MB$, then $AY=XB=h$.
- $AYC \cong PZC \cong QVC$. Since $AP=PQ=QC$, then $PZ=\frac{2}{3}h$ and $QV=\frac{1}{3}h$.
- $PEZ \cong XBE$. Since $PZ=\frac{2}{3}XB \rightarrow PE=\frac{2}{3}EB \rightarrow PE=\frac{2}{5}PB$ and $EB=\frac{3}{5}PB$.
- $QHV \cong XBH$. Since $QV=\frac{1}{3}QB \rightarrow QH=\frac{1}{3}HB \rightarrow QH=\frac{1}{4}QB$.

Working analogously ${ }^{1}$ with the other median $AN$, we obtain

$$
QG=\frac{2}{5}QB \quad GB=\frac{3}{5}QB \quad \text{and} \quad PD=\frac{1}{4}PB
$$

We focus, for example, on the cevian $PB$ and can now determine the proportion in which the points $\boldsymbol{D}$ and $\boldsymbol{E}$ divide it. We have

$$
DE=PE-PD=\left(\frac{2}{5}-\frac{1}{4}\right)PB=\frac{3}{20}PB
$$

Therefore,

$$
PD=\frac{5}{20}PB \quad DE=\frac{3}{20}PB \quad \text{and} \quad EB=\frac{12}{20}PB
$$

Now we draw the auxiliary line $AE$. On the cevian $PB$, $ABP$ of known area is divided into three triangles, $ABE, AED$ and $ADP$, of identical height. Therefore,

$$
\frac{[ADP]}{[ABP]}=\frac{PD}{PB}=\frac{5}{20}, \quad \frac{[AED]}{[ABP]}=\frac{DE}{PB}=\frac{3}{20}, \quad \frac{[ABE]}{[ABP]}=\frac{EB}{PB}=\frac{12}{20}
$$

Since

$$
[ABP]=\frac{1}{3} \rightarrow [ADP]=\frac{5}{60}, \quad [AED]=\frac{3}{60} \quad \text{and} \quad [ABE]=\frac{12}{60}
$$

On the other hand, since $\boldsymbol{EM}$ is the median of triangle $\boldsymbol{ABE}$, we have $[AME]=\frac{1}{2}[ABE]=\frac{6}{60}$. Finally,

$$
\frac{1}{6}=[AMF]=[AME]+[AED]+[DEF]=\frac{6}{60}+\frac{3}{60}+[DEF]
$$

from which the area of the half-butterfly is obtained. That is, $[DEF]=\frac{1}{60}$. Similarly, $[\boldsymbol{FGH}]=\frac{1}{60}$, and the required area is

$$
[DEF]+[FGH]=\frac{1}{30}=\frac{1}{30}[ABC]
$$[^0]