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Corners are sliced off a unit cube so that the six faces each become

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Math Expert
Joined: 02 Sep 2009
Posts: 55149
Corners are sliced off a unit cube so that the six faces each become  [#permalink]

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14 Mar 2019, 04:31
10
00:00

Difficulty:

85% (hard)

Question Stats:

17% (02:22) correct 83% (02:31) wrong based on 18 sessions

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TOUGH QUESTION:

Corners are sliced off a unit cube so that the six faces each become regular octagons. What is the total volume of the removed tetrahedra?

(A) (5√2 − 7)/3

(B) (10 −7√2)/3

(C) (3 − 2√2)/3

(D) (8√2 − 11)/3

(E) (6 − 4√2)/3

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Math Expert
Joined: 02 Sep 2009
Posts: 55149
Re: Corners are sliced off a unit cube so that the six faces each become  [#permalink]

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17 May 2019, 00:57
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BUMPING FOR DISCUSSION.
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Joined: 19 Oct 2018
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Corners are sliced off a unit cube so that the six faces each become  [#permalink]

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17 May 2019, 04:04
total number of tetrahedron=8 from each corner
let s be the side of the regular octagon, that left
edge of cube =$$s/(2)^{1/2} + s + s/(2)^{1/2} = 1$$
$${(2)^{1/2} +1}$$s=1
$$s=1/{(2)^{1/2} +1}=(2)^{1/2} -1$$

Area of base which is equilateral triangle= (3^{1/2}/4)*s^2

height of tetrahedron= $$[(slant edge of tetrahedron)^2 - (circumcenter of base triangle)^2]^{1/2}$$
height of tetrahedron= $$[(s^2/2)^-(s^2/3)]^{1/2}$$= $$s/(6)^{1/2}$$

Volume of tetrahedron= 1/3 * base area * height

Volume of tetrahedron =$$1/3*(3^{1/2}/4)*s^2*s/(6)^{1/2}= s^3/12*(2)^{1/2}$$

$$s^3= {(2)^{1/2} -1}^3$$= $${5*2^{1/2} -7}$$

Volume of tetrahedron= $${5*2^{1/2} -7}/12*(2)^{1/2}$$
Total cutoff area= $$8*{5*2^{1/2} -7}/12*(2)^{1/2}$$
= $$(10-7*2^{1/2})/3$$

Enough Mathematics for today..... Took me more than 30 minutes to solve it
Bunuel wrote:
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Corners are sliced off a unit cube so that the six faces each become   [#permalink] 17 May 2019, 04:04
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