2012 AMC 12B Problems/Problem 19

Revision as of 08:52, 7 April 2013 by JWK750 (talk | contribs) (See Also)

Solution

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 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)

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
Invalid username
Login to AoPS