Finding the Distance from Point P to AC on an ABCD.EFGH Cube

[Monoxdifly]

Given an ABCD.EFGH cube whose its side length is 8 cm. The point P is within AB so that AP = 3PH. The distance between P to AC is ...
A. $$2\sqrt3$$ cm
B. $$3\sqrt3$$ cm
C. $$2\sqrt6$$ cm
D. $$3\sqrt6$$ cm
E. $$4\sqrt6$$ cm

So AH is $$8\sqrt2$$ cm and PH = $$6\sqrt2$$. What do I do now? Is the triangle HPC a right triangle?

 

[Opalg]

Given an ABCD.EFGH cube whose its side length is 8 cm. The point P is within AB so that AP = 3PH. The distance between P to AC is ...
A. $$2\sqrt3$$ cm
B. $$3\sqrt3$$ cm
C. $$2\sqrt6$$ cm
D. $$3\sqrt6$$ cm
E. $$4\sqrt6$$ cm

So AH is $$8\sqrt2$$ cm and PH = $$6\sqrt2$$. What do I do now? Is the triangle HPC a right triangle?

This sounds wrong. If P lies on AB then it must be closer to A than to H. So the condition AP = 3PH is impossible.

 

[Monoxdifly]

Sorry, I meant P is within AH. Sigh, why did I make a lot of typos last night?

 

[Opalg]

My method is to use coordinates, taking the vertices of the cube as
$A = (8,0,0),$
$B = (0,0,0),$
$C = (0,8,0),$
$D = (8,8,0),$
$E = (8,0,8),$
$F = (0,0,8),$
$G = (0,8,8),$
$H = (8,8,8).$
(Of course, that is not the only way to assign the vertices. I was working from a sketch in which it was convenient to take $B$ as the origin.)

The point $P$ is then $(8,6,6)$. A point on $AC$ is given by $(t,8-t,0)$. The distance $d$ from that point to $P$ satisfies $d^2 = (t-8)^2 + (2-t)^2 + 6^2$. If you minimise that by calculus, you find that $t=5$, and $d = 3\sqrt6$.

Edit. If you don't want to use calculus, you can do it algebraically by completing the square: $$d^2 = (t-8)^2 + (2-t)^2 + 6^2 = 2t^2 - 20t + 104 = 2(t-5)^2 + 54.$$ That clearly has minimum value 54, when $t=5$. So $d = \sqrt{54} = 3\sqrt6$.

 

[Monoxdifly]

I ended up solving the problem with trigonometry instead of algebra and calculus. Thanks, anyway. It's always nice to see different approaches.