What is the ratio of the area of a triangle $ABC$ to the area of the triangle whose sides are equal in length to the medians of triangle $ABC$?
I see an obvious method of brute-force wherein I can impose a coordinate system onto the figure. But is there a better solution?
Best Answer
Let $X$, $Y$ and $Z$ be the side midpoints. Construct the parallelogram PYBZ and connect PC.
By construction, PY = BZ = AZ, PY || AZ and then PYZA is also a parallelogram, which leads to AP || ZY || XC and AP = XC, and in turn the parallelogram APCX. Thus, PC = AX and the sides of the triangle PCZ are the medians of ABC.
The parallelogram PYZA also yields AQ = QY = $\frac14$AC and then
$$\frac{Area_{PCZ}}{Area_{ABC}}=\frac{Area_{QCZ}}{Area_{AZC}}=\frac{QC}{AC} = \frac34 $$