Trisecting Hypotenuse

By Phthagores Theorem $AB^{ 2 }+AC^{ 2 }=BC^{ 2 }=6$ and, $AP^{ 2 }=(2/3\quad \times \quad AB)^{ 2 }+(1/3\quad \times \quad AC)^{ 2 }$ $AQ^{ 2 }=(1/3\quad \times \quad AB)^{ 2 }+(2/3\quad \times \quad AC)^{ 2 }$ so, $AB^{ 2 }+AP^{ 2 }+AQ^{ 2 }+AC^{ 2 }=14/9\quad \times \quad (AB^{ 2 }+AC^{ 2 })$ $= 14/9 \times 6 = 28/3$

First of all we get CQ = QP = PB with each of them equal to $\frac { \sqrt { 6 } }{ 3 } =\quad \frac { \sqrt { 2 } }{ \sqrt { 3 } }$ Now ${ AC }^{ 2 }\quad +\quad { AB }^{ 2 }\quad =\quad { BC }^{ 2 }\quad (By\quad Pythagores\quad Theorem)$ So ${ AC }^{ 2 }\quad +\quad { AB }^{ 2 }\quad =\quad 6$ Now we need some constructions. let X be the midpoint of BC. Join AX which is the median. Length of AX = BX = CX. So AX = $\frac { \sqrt { 6 } }{ 2 }$ ${ AQ }^{ 2 }\quad =\quad { AX }^{ 2\quad }+\quad { XQ }^{ 2 }\quad -\quad 2AX.XQ.Cos\theta \quad \quad (Where\quad \theta \quad =\quad \angle AXQ)$ and ${ AP }^{ 2 }\quad =\quad { AX }^{ 2\quad }+\quad XP^{ 2 }\quad -\quad 2AX.XP.Cos(\pi -\theta )\\ { AP }^{ 2 }\quad =\quad { AX }^{ 2\quad }+\quad XP^{ 2 }\quad +\quad 2AX.XP.Cos\theta \quad \quad \quad \quad \quad (Because\quad Cos(\pi -\theta )\quad =\quad -Cos\theta )$ Adding the two equations ${ AP }^{ 2 }\quad +\quad { AQ }^{ 2 }\quad =\quad 2({ AX }^{ 2\quad }+\quad XP^{ 2 })\\ { AP }^{ 2 }\quad +\quad { AQ }^{ 2 }\quad =\quad 2[\frac { 3 }{ 2 } \quad +\quad { (\frac { \sqrt { 6 } }{ 2 } -\quad \frac { \sqrt { 6 } }{ 3 } ) }^{ 2 }]\\ { AP }^{ 2 }\quad +\quad { AQ }^{ 2 }\quad =\quad \frac { 10 }{ 3 }$ ${ AP }^{ 2 }\quad +\quad { AQ }^{ 2 }\quad +\quad { AC }^{ 2 }\quad +\quad { AB }^{ 2 }\quad =\quad 6\quad +\quad \frac { 10 }{ 3 } \\ { AP }^{ 2 }\quad +\quad { AQ }^{ 2 }\quad +\quad { AC }^{ 2 }\quad +\quad { AB }^{ 2 }\quad =\quad \frac { 28 }{ 3 }$

I took AC=0 . So AB=BC and AP=2/3 AB,AQ=1/3 AB and got result easily.

By Apollonius Theorem, $AB^{2} + AP^{2} = 2AQ^{2} +\frac{ PC^{2}}{2}$ and $AC^{2} + AQ^{2} = 2AP^{2} +\frac{ PC^{2}}{2}$

Solving we get, $AP^{2} + AQ^{2}=\frac{10}{3}$ and thus answer= $\frac{10}{3}+6=\boxed{\frac{28}{3}}$

We know that $BP=PQ=QC=\frac{\sqrt{6}}{3}$

From the median line theorem (forgot the name)...

$AP^{2} + AC^{2} = 2(AQ)^2 + 2(\frac{\sqrt{6}}{3})^{2}$

$AB^{2} + AQ^{2} = 2(AP)^2 + 2(\frac{\sqrt{6}}{3})^{2}$

Sum these 2 together we get

$AB^{2} + AP^{2} + AQ^{2} + AC^{2} = 2(AP)^2 + 2(AQ)^2 + \frac{8}{3}$

$AB^{2} + AC^{2} = AP^{2} + AQ^{2} + \frac{8}{3}$

Pythagoras;

$6 = AP^{2} + AQ^{2} + \frac{8}{3}$

$AP^{2} + AQ^{2} = \frac{10}{3}$

Therefore, $AB^{2} + AP^{2} + AQ^{2} + AC^{2} = 6 + \frac{10}{3} = \boxed{\frac{28}{3}}$ ~~~

CB is the diameter of circumcircle of triangle ABC. D the centre of the circle and C-Q-D-P-B. It is obvious that: $\\CQ = \frac {1}{3} \sqrt 6\\CD=\frac{1}{3}\sqrt 6\\CP=\frac{2}{3}\sqrt 6\\CB=\sqrt 6\\DA=\frac{1}{2}\sqrt 6\\$ A can be chosen to lie anywhere on the circumference since the only insistance is that CB is fixed; Let DA be perpendicular to DB $\\AP^2=AQ^2=\frac{6}{4}+(\frac{1}{2}\sqrt 6-\frac{1}{3}\sqrt 6)^2=\frac{10}{3}\\AB^2+AC^2+AQ^2+AP^2=6+\frac{10}{3}=\frac{28}{3}$

Kumar Gupta, this problem was marvellous!!! Got the answer 28/3

By Stewart's Theorem, $AB^{2}(PC)+AC^{2}(PB)=BC(AP^{2}+PB * PC)$ $AB^{2}\left(\frac{2\sqrt{6}}{3} \right)+AC^{2}\left(\frac{\sqrt{6}}{3} \right)=\sqrt{6}\left(AP^{2}+\frac{12}{9}\right)$ Which simplifies to $2AB^{2}+AC^{2}=3AP^{2}+4$ Again, using Stewart's Theorem, we can get $AB^{2}+2AC^{2}=3AQ^{2}+4$ Adding both the equation gives $3(AB^{2}+AC^{2})=3(AP^{2}+AQ^{2})+8$ But since $AB^{2}+AC^{2}=6$ , we get $3(AP^{2}+AQ^{2})+8=18$ , $AP^{2}+AQ^{2}=\frac{10}{3}$ Therefore, $AB^{2}+AP^{2}+AQ^{2}+AC^{2}=6+\frac{10}{3}= \boxed{\frac{28}{3}}$