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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: 58464
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:

75% (hard)

Question Stats:

20% (01:55) correct 80% (02:35) wrong based on 20 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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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 =$$s/\sqrt{(2)} + s + s/\sqrt{(2)} = 1$$
[$${\sqrt{(2)} +1}$$]s=1
$$s=1/{\sqrt{(2)} +1}=\sqrt{(2)} -1$$

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

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

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

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

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

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

Enough Mathematics for today..... Took me more than 30 minutes to solve it
Bunuel wrote:
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BUMPING FOR DISCUSSION.

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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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