A unit cube has vertices $P_1,P_2,P_3,P_4,P_1',P_2',P_3',$ and $P_4'$. Vertices $P_2$, $P_3$, and $P_4$ are adjacent to $P_1$, and for $1\le i\le 4,$ vertices $P_i$ and $P_i'$ are opposite to each other. A regular octahedron has one vertex in each of the segments $P_1P_2$, $P_1P_3$, $P_1P_4$, $P_1'P_2'$, $P_1'P_3'$, and $P_1'P_4'$. What is the octahedron's side length?
[asy] import three;  size(7.5cm); triple eye = (-4, -8, 3); currentprojection = perspective(eye);  triple[] P = {(1, -1, -1), (-1, -1, -1), (-1, 1, -1), (-1, -1, 1), (1, -1, -1)}; // P[0] = P[4] for convenience triple[] Pp = {-P[0], -P[1], -P[2], -P[3], -P[4]};  // draw octahedron triple pt(int k){ return (3*P[k] + P[1])/4; } triple ptp(int k){ return (3*Pp[k] + Pp[1])/4; } draw(pt(2)--pt(3)--pt(4)--cycle, gray(0.6)); draw(ptp(2)--pt(3)--ptp(4)--cycle, gray(0.6)); draw(ptp(2)--pt(4), gray(0.6)); draw(pt(2)--ptp(4), gray(0.6)); draw(pt(4)--ptp(3)--pt(2), gray(0.6) + linetype("4 4")); draw(ptp(4)--ptp(3)--ptp(2), gray(0.6) + linetype("4 4"));  // draw cube for(int i = 0; i < 4; ++i){ 	draw(P[1]--P[i]); draw(Pp[1]--Pp[i]); 	for(int j = 0; j < 4; ++j){ 		if(i == 1 || j == 1 || i == j) continue; 		draw(P[i]--Pp[j]); draw(Pp[i]--P[j]); 	} 	dot(P[i]); dot(Pp[i]); 	dot(pt(i)); dot(ptp(i)); }  label("$P_1$", P[1], dir(P[1])); label("$P_2$", P[2], dir(P[2])); label("$P_3$", P[3], dir(-45)); label("$P_4$", P[4], dir(P[4])); label("$P'_1$", Pp[1], dir(Pp[1])); label("$P'_2$", Pp[2], dir(Pp[2])); label("$P'_3$", Pp[3], dir(-100)); label("$P'_4$", Pp[4], dir(Pp[4])); [/asy]
$\textbf{(A)}\ \frac{3\sqrt{2}}{4}\qquad\textbf{(B)}\ \frac{7\sqrt{6}}{16}\qquad\textbf{(C)}\ \frac{\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{2\sqrt{3}}{3}\qquad\textbf{(E)}\ \frac{\sqrt{6}}{2}$
[2012 AMC-12B-19.jpg](https://artofproblemsolving.com/wiki/index.php/File:2012_AMC-12B-19.jpg)
Observe the diagram above. Each dot represents one of the six vertices of the regular octahedron. Three dots have been placed exactly x units from the $(0,0,0)$ corner of the unit cube. The other three dots have been placed exactly x units from the $(1,1,1)$ corner of the unit cube. A red square has been drawn connecting four of the dots to provide perspective regarding the shape of the octahedron. Observe that the three dots that are near $(0,0,0)$ are each $(x)(\sqrt{2}$) from each other. The same is true for the three dots that are near $(1,1,1).$ There is a unique $x$ for which the rectangle drawn in red becomes a square. This will occur when the distance from $(x,0,0)$ to $(1,1-x, 1)$ is $(x)(\sqrt{2}$).

Using the distance formula we find the distance between the two points to be: $\sqrt{{(1-x)^2} + {(1-x)^2} + 1}$ = $\sqrt{2x^2 - 4x +3}$. Equating this to $(x)(\sqrt{2}$) and squaring both sides, we have the equation:
$2{x^2} - 4x + 3$ = $2{x^2}$
$-4x + 3 = 0$
$x$ = $\frac{3} {4}$.

Since the length of each side is $(x)(\sqrt{2}$), we have a final result of $\frac{3 \sqrt{2}}{4}$. Thus, Answer choice $\boxed{\text{A}}$ is correct.

(If someone can draw a better diagram with the points labeled P1,P2, etc., I would appreciate it).
--[Jm314](https://artofproblemsolving.com/wiki/index.php?title=User:Jm314&action=edit&redlink=1) 14:55, 26 February 2012 (EST)