Find the sum of the squares of the medians of triangle APC below

Question

For reference: The diameter of a circle measures 8m. On this diameter are located the points A and B equidistant 1 m from the center; through B draw any chord PC , determine the sum of the squares of the medians of triangle APC.(Answer: $73,5$)

My progress: sum of the squares of the median

$4(PE^2+CD^2+AB^2) = 3(AC^2+AP^2+CP^2)\therefore\\ PE^2+CD^2+AB^2=\frac{3}{4}(AC^2+AP^2+CP^2)\\ AC = AP \implies PE^2+CD^2+AB^2=\frac{3}{4}(2AC^2+CP^2)\\ CB=\frac{CP}{2} \\ \triangle ABC: 4+(\frac{CP}{2})^2=AC^2\implies \frac{16+CP^2}{4} =AC^2\\ \therefore PE^2+CD^2+AB^2 = \frac{3}{8}(16+3CP^2) $

...??

Answer 1

If $O$ is the center of the circle, $OP$ is median of $\triangle APB$. Radius of the circle is $4$.

So, $4 OP^2 = 64 = 2 AP^2 + 2 PB^2 - AB^2$
$ \implies AP^2 + PB^2 = 34$
Similarly, $AC^2 + BC^2 = 34$

Now by intersecting chords theorem, $PB \cdot BC = 5 \cdot 3 = 15$

So, $AP^2 + AC^2 + PB^2 + BC^2 + 2 PB \cdot BC = 98$
$\implies AP^2 + AC^2 + CP^2 = 98$

Given you know the sum of squares of the sides of the triangle, you can find the sum of squares of the medians.

Answer 2

Here's a proof for the general case where $\small PC$ is not necessarily perpendicular to $\small AB$:

Given that we have already proven $$\mathsf{4.(sum\ of\ squares\ of\ medians)=3.(sum\ of\ squares\ of\ sides)}$$ using Appollonius theorem, continue from there.

So let's calculate $\small AP^2+AC^2+PC^2\tag{*}$

Drop the perp from $\small A$ to $\small PC$ and the foot $\small I$ must lie on the small circle (angle of half circle=$\small 90^\circ$).

Now using power of point,

$\small \begin{align}PI\cdot PB&=3\cdot 5\\ PI\cdot(PI+IB)&=15\\PI^2+PI\cdot IB&=15\end{align}$

See $\small PI=BC$.

Also note that $\small PI\cdot IC=PI\cdot(PI+IB)$.

From Pythagoras' theorem $\small AP^2=AI^2+PI^2$, $\small AC^2=AI^2+IC^2$ and $\small AB^2=AI^2+IB^2$.

Now substitute these in (*).

$\small \begin{align}&=AP^2+AC^2+PI^2+IC^2+2\cdot PI\cdot IC\\ &=2AI^2+2PI^2+2IC^2+30\\&=2(AB^2-IB^2+PI^2+IC^2+15)\\&=2(2^2+PI^2+(IC^2-IB^2)+15)\\&=2(4+PI^2+(IC-IB)(IC+IB)+15)\\&=2(4+PI^2+PI(PI+2IB)+15)\\&=2(4+2(PI^2+PI\cdot IB)+15)\\&=2(4+2\cdot15+15)\\&=98\end{align}$

Now using the known identity, sum of squares of medians

$=98\cdot\dfrac34=73.5$