Calculate $PA^2 + PB^2 + PC^2$ in the triangle below;

Question

On the arc, $AB$, of the circumscribed circle, to an equilateral triangle, $ABC$, of height, $h$, units, is taken to point P. Calculate: $PA^2 + PB^2 + PC^2$ (Answer:$\frac{8h^2}{3}$)

My progress: $PA^2+PB^2+PC^2 = ?\\ \boxed{}APBC\rightarrow PB = PA+PC\\ AB^2 = PB.BD\\ AB.BC = BH.2R\\ HO = \frac{h}{3}\\ \triangle AOH: tg30^o = \frac{HO}{R}\implies R \sqrt3 =3HO \therefore R = \sqrt3HO=\frac{h\sqrt3}{3} \\ PA^2+PC^2 = 2PH^2+\frac{AC^2}{2}\\ \triangle ABH: tg60^o = \frac{h}{\frac{AC}{2}}\implies AC =\frac{2h\sqrt3}{3}=AB=BC\\ \therefore PA^2+PC^2 = 2PH^2 + \frac{2h^2}{3}$

I need to find line $PH$ as a function of $h$.

Answer 1

If you are looking for a solution without trigonometry, see the below diagram as one of the approaches.

$BH = h \implies AC = \frac{2h}{\sqrt3}$. Note $AOCG$ is a rhombus and $AG = CG = \frac{2h}{3}$

Applying Ptolemy's theorem in $APGC$,

$AG \cdot PC = AC \cdot PG + PA \cdot CG$

and you can obtain, $PG \sqrt3 = (PC - PA)$

Squaring, $3 (BG^2 - PB^2) = PC^2 + PA^2 - 2 PA \cdot PC \tag 1$

Now applying Ptolemy's theorem in $ABCP$, you already obtained that $PB = PA + PC$

i.e $ ~ PB^2 = PA^2 + PC^2 + 2 PA \cdot PC \tag 2$

Adding $(1)$ and $(2)$ and simplifying, you would obtain the answer.

Answer 2

You're given the height $h$, then the circumradius is $R = \dfrac{2}{3} h $.

Construct a circle (the circumcircle), then we can the vertices to be

$A = R(0, 1) $

$ B = R (\sin \alpha, \cos \alpha) $

$C = R (-\sin \alpha, \cos \alpha)$

where $\alpha = \dfrac{2\pi}{3}$ and

$P = R (\sin \theta, \cos \theta ) $

$\begin{equation} \begin{split} PA^2 + PB^2 + PC^2 &= R^2 ( \sin^2 \theta + (\sin \theta - \sin \alpha)^2 + (\sin \theta + \sin \alpha)^2 \\ &+ (\cos \theta - 1)^2 + 2(\cos \theta - \cos \alpha)^2 ) \end{split} \end{equation}$

Using $\sin^2 \theta + \cos^2 \theta = \sin^2 \alpha + \cos^2 alpha = 1 $ , the above equation simplifies to

$PA^2 + PB^2 + PC^2 = R^2 (6 - 2 \cos \theta - 4 \cos \theta \cos \alpha )$

But $\cos \alpha = -\dfrac{1}{2} $

So the terms containing $\cos \theta $ cancel out and the expression reduces to

$PA^2 + PB^2 + PC^2 = 6 R^2 = 6 \left( \dfrac{4}{9} h^2 \right) = \left(\dfrac{8}{3}\right) h^2 $

Answer 3

Assume $\angle AOP = 2\theta$, then

$PA = 2 R \sin\theta$

$\angle COP = 120^\circ - 2\theta, PC = 2R \sin (60^\circ - \theta)$

$\angle BOP = 120^\circ + 2 \theta, PB = 2R \sin(60^\circ + \theta)$

Now can you find $PA^2 + PB^2 + PC^2$ in terms of $R$?

Finally given equilateral triangle, use the relation between $R$ and $h$, where $h$ is altitude of $\triangle ABC$ from any of the vertices.