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AMC 12 2012b Problem 19

  • Paper

A unit cube has vertices P1,P2,P3,P4,P1′,P2′,P3′,P_1,P_2,P_3,P_4,P_1',P_2',P_3', and P4′P_4'. Vertices P2P_2, P3P_3, and P4P_4 are adjacent to P1P_1, and for 1≤i≤4,1\le i\le 4, vertices PiP_i and Pi′P_i' are opposite to each other. A regular octahedron has one vertex in each of the segments P1P2P_1P_2, P1P3P_1P_3, P1P4P_1P_4, P1′P2′P_1'P_2', P1′P3′P_1'P_3', and P1′P4′P_1'P_4'. What is the octahedron's side length?

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("P1P_1", P[1], dir(P[1]));
label("P2P_2", P[2], dir(P[2]));
label("P3P_3", P[3], dir(-45));
label("P4P_4", P[4], dir(P[4]));
label("P1′P'_1", Pp[1], dir(Pp[1]));
label("P2′P'_2", Pp[2], dir(Pp[2]));
label("P3′P'_3", Pp[3], dir(-100));
label("P4′P'_4", Pp[4], dir(Pp[4]));

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

Dataset text AMC 12 2000 to 2025. Quoted with source.

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A

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