MHB Minimal Possible Values of Triangle Point Distances - POTW #266 Jun 7th, 2017

[anemone]

Here is this week's POTW:

-----

Let $PQR$ be a triangle such that $PQ=3,\,QR=4$ and $PR=5$. Let $X$ be a point in the triangle. Find the minimal possible values of $PX^2+QX^2+RX^2$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to kaliprasad for his correct solution, and you can find the suggested solution as follows::)

Spoiler

Let the perpendicular distance from $X$ to $QR$ and $QP$ be $x$ and $y$ respectively. Then we have

$\begin{align*}PX^2+QX^2+RX^2&=x^2+y^2+(3-x)^2+(4-y)^2+x^2+y^2\\&=3\left(y-\dfrac{4}{3}\right)^2+3(x-1)^2+\dfrac{50}{3}\end{align*}$

This suggests that the minimal possible value of $PX^2+QX^2+RX^2$ is $\dfrac{50}{3}$.