In $\triangle PQR$ with $QR^2+PR^2=5PQ^2$, medians $PM$ and $QN$ meet at $G$. Find $\angle QGM$.

Question

This problem becomes much simpler if we solve it using coordinate geometry but i am seeking a solution which involves pure geometry.

Consider a $\triangle PQR$ in which the relation $$QR^2+PR^2=5PQ^2$$ holds. Let $G$ be the point of intersection of medians $PM$ and $QN$. Then find the value of $\angle QGM$.

Answer 1

$QR^2+PR^2=5PQ^2$

median lengths

$$PM=\frac{1}{2}\sqrt{2(PQ^2+PR^2)-QR^2}$$ $$PM=\frac{1}{2}\sqrt{2\left(\frac{x^2+y^2}{5}+y^2\right)-x^2}=\frac{1}{2} \sqrt{\frac{12 y^2}{5}-\frac{3 x^2}{5}}$$ $$QN=\frac{1}{2}\sqrt{2\left(\frac{x^2+y^2}{5}+x^2\right)-y^2}=\frac{1}{2} \sqrt{\frac{12 x^2}{5}-\frac{3 y^2}{5}}$$

$$GM=\frac{1}{3}PM=\frac{1}{6} \sqrt{\frac{12 y^2}{5}-\frac{3 x^2}{5}}$$ $$GQ=\frac{2}{3}QN=\frac{1}{3} \sqrt{\frac{12 x^2}{5}-\frac{3 y^2}{5}}$$ $$GM^2+GQ^2=\frac{x^2}{4}=QM^2\to \widehat{QGM}=90°$$

Answer 2

From the triangle median theorem $$NQ^2= \frac12( RQ^2+ PQ^2)-\frac14RP^2, \>\>\> MP^2= \frac12( RP^2+ PQ^2)-\frac14 RQ^2 $$ Combine the two equalities $$NQ^2 +MP^2 = PQ^2 +\frac14(RQ^2+RP^2)= \frac94 PQ^2$$

Then, substitute it into the cosine rule below \begin{align} \cos\angle MGN= \frac{NG^2+MG^2- NM^2}{2NG\cdot MG}= \frac{\frac19 (NQ^2+MP^2)- \frac14 PQ^2}{2NG\cdot MG}=0 \end{align}Thus, $\angle QGM = \angle MGN =90^\circ$.