# 2012 AMC 12B Problems/Problem 19

## Problem

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?

$[asy] import three; size(7.5cm); triple eye = (-4, -8, 3); currentprojection = perspective(eye); triple[] P = {(1, -1, -1), (-1, -1, -1), (-1, 1, -1), (-1, -1, 1), (1, -1, -1)}; // P[0] = P[4] for convenience triple[] Pp = {-P[0], -P[1], -P[2], -P[3], -P[4]}; // draw octahedron triple pt(int k){ return (3*P[k] + P[1])/4; } triple ptp(int k){ return (3*Pp[k] + Pp[1])/4; } draw(pt(2)--pt(3)--pt(4)--cycle, gray(0.6)); draw(ptp(2)--pt(3)--ptp(4)--cycle, gray(0.6)); draw(ptp(2)--pt(4), gray(0.6)); draw(pt(2)--ptp(4), gray(0.6)); draw(pt(4)--ptp(3)--pt(2), gray(0.6) + linetype("4 4")); draw(ptp(4)--ptp(3)--ptp(2), gray(0.6) + linetype("4 4")); // draw cube for(int i = 0; i < 4; ++i){ draw(P[1]--P[i]); draw(Pp[1]--Pp[i]); for(int j = 0; j < 4; ++j){ if(i == 1 || j == 1 || i == j) continue; draw(P[i]--Pp[j]); draw(Pp[i]--P[j]); } dot(P[i]); dot(Pp[i]); dot(pt(i)); dot(ptp(i)); } label("P_1", P[1], dir(P[1])); label("P_2", P[2], dir(P[2])); label("P_3", P[3], dir(-45)); label("P_4", P[4], dir(P[4])); label("P'_1", Pp[1], dir(Pp[1])); label("P'_2", Pp[2], dir(Pp[2])); label("P'_3", Pp[3], dir(-100)); label("P'_4", Pp[4], dir(Pp[4])); [/asy]$

$\textbf{(A)}\ \frac{3\sqrt{2}}{4}\qquad\textbf{(B)}\ \frac{7\sqrt{6}}{16}\qquad\textbf{(C)}\ \frac{\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{2\sqrt{3}}{3}\qquad\textbf{(E)}\ \frac{\sqrt{6}}{2}$

## Solution 1

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 14:55, 26 February 2012 (EST)

## Solution 2

Standard 3D geometry, no coordinates.

Let the tip of the octahedron on side $P_1P_3$ be $K_1$ and the opposite vertex be $K_2$. Our key is to examine the trapezoid $P_1K_1K_2P_3$.

Let the side length of the octahedron be $s$. Then $P_1K_1 = \frac{s}{\sqrt{2}}$ and $P_3K_2 = 1 - \frac{2}{\sqrt{2}}$. Then, we have $P_1P_3 = \sqrt{2}$. Finally, we want to find $K_1K_2$, which is just double the height of half the octahedron. We can use Pythagorean Theorem to find that height as $\sqrt{2}s$. Now, we use the Pythagorean Theorem on the trapezoid. We get

$$(\sqrt{2})^2 + (2\sqrt{2}-1)^2 = (s\sqrt{2})^2$$ $$s = \frac{3\sqrt{2}}{4}.$$

~superagh

## Solution 3

Let the length of $P_1A = a$, $P_1B = b$

$AB = a^2 + b^2$, $AB' = (1-b)^2 + (1-a)^2 + 1$, $AB = AB'$

$a^2 + b^2 = (1-b)^2 + (1-a)^2 + 1$, $a^2 + b^2 = 1 - 2b + b^2 + 1 - 2a + a^2 + 1$, $a+ b = \frac32$

As $AC = BC$, $a^2 + P_1C^2 = b^2 + P_1C^2$, $a = b$, $a = \frac34$

$AB = \boxed{\textbf{(A) } \frac{3\sqrt{2}}{4}}$