2012 AMC 12B Problems/Problem 19
Problem
A unit cube has vertices and . Vertices , , and are adjacent to , and for vertices and are opposite to each other. A regular octahedron has one vertex in each of the segments , , , , , and . What is the octahedron's side length?
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 corner of the unit cube. The other three dots have been placed exactly x units from the 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 are each ) from each other. The same is true for the three dots that are near There is a unique for which the rectangle drawn in red becomes a square. This will occur when the distance from to is ).
Using the distance formula we find the distance between the two points to be: = . Equating this to ) and squaring both sides, we have the equation:
=
= .
Since the length of each side is ), we have a final result of . Thus, Answer choice 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 be and the opposite vertex be . Our key is to examine the trapezoid .
Let the side length of the octahedron be . Then and . Then, we have . Finally, we want to find , which is just double the height of half the octahedron. We can use Pythagorean Theorem to find that height as . Now, we use the Pythagorean Theorem on the trapezoid. We get
~superagh
Solution 3
Let the length of ,
, ,
, ,
As , , ,
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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