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DIAMOND PUZZLE or Double Pyramid Puzzle
https://www.puzzlemaster.ca/browse/inventors/vaclav/5813-double-pyramid
A diamond game piece is composed of two regular square
pyramids which have the same base. The side of the base is
6.5 cm and the heights of the diamond puzzle is 8.5 cm.
a) Which is the surface of this piece ?
b) Knowing that the object is made of pinewood (density:
410g/𝑑𝑚3
) prove that the object weighs less then
0.5g.
Solution:
a) The lateral area of the SABCD pyramid is A =
𝑃 𝑏 ∙ 𝑎 𝑝
2
.
The area of the wood piece is 2A.
SO=
𝑆𝑉
2
= 4.25 𝑐𝑚., where SV = the height of the wood piece = 8.5cm, {O} = BD ∩AC
AB=6.5 ⟹OM=
𝐴𝐵
2
⟹OM=3.5
𝑆𝑂 ⊥ (𝐴𝐵𝐶)
𝑂𝑀 ⊂ (𝐴𝐵𝐶)𝑤ℎ𝑒𝑟𝑒 𝑂𝑀 𝑖𝑠 𝑎𝑝𝑜𝑡ℎ𝑒𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
} ⟹ SO ⊥ OM
𝑇.𝑃𝑖𝑡𝑎𝑔𝑜𝑟𝑎
⇒
𝑆𝑀2
= 𝑆𝑂2
+ 𝑂𝑀2
⟹ 𝑆𝑀2
= 4.252
+ 3.252
= 18.0625 + 10.5625 = 28.625 ⟹
𝑆𝑀 ≈ 5.35
A = 4∙6.5 ∙ 5.35 = 139.1 . A ≈ 139.1
b) The volume of the piece = the volume of SABCD pyramid + the volume of VABCD pyramid=V
V =
𝐴𝑏∙(ℎ1+ℎ2)
3
=
𝐴 𝑏∙2∙𝑆𝑂
3
=
𝐴 𝑏∙𝑆𝑉
3
, h1=SO ,h2=VO
V =
6.52∙8.5
3
=
42.25 ∙ 8.5
3
=
359.125
3
= 119.7 𝑐𝑚3
V ≈ 120 ∙10−3
=0.12 𝑑𝑚3
𝑚 = ρ ∙ 𝑉 = 410 ∙ 0.12 = 49.2 𝑔 < 50𝑔
𝒎 = 𝟒𝟗. 𝟐 𝒈 < 𝟓𝟎𝒈

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Diamond puzzle or double pyramid puzzle

  • 1. DIAMOND PUZZLE or Double Pyramid Puzzle https://www.puzzlemaster.ca/browse/inventors/vaclav/5813-double-pyramid A diamond game piece is composed of two regular square pyramids which have the same base. The side of the base is 6.5 cm and the heights of the diamond puzzle is 8.5 cm. a) Which is the surface of this piece ? b) Knowing that the object is made of pinewood (density: 410g/𝑑𝑚3 ) prove that the object weighs less then 0.5g. Solution: a) The lateral area of the SABCD pyramid is A = 𝑃 𝑏 ∙ 𝑎 𝑝 2 . The area of the wood piece is 2A. SO= 𝑆𝑉 2 = 4.25 𝑐𝑚., where SV = the height of the wood piece = 8.5cm, {O} = BD ∩AC AB=6.5 ⟹OM= 𝐴𝐵 2 ⟹OM=3.5 𝑆𝑂 ⊥ (𝐴𝐵𝐶) 𝑂𝑀 ⊂ (𝐴𝐵𝐶)𝑤ℎ𝑒𝑟𝑒 𝑂𝑀 𝑖𝑠 𝑎𝑝𝑜𝑡ℎ𝑒𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 } ⟹ SO ⊥ OM 𝑇.𝑃𝑖𝑡𝑎𝑔𝑜𝑟𝑎 ⇒ 𝑆𝑀2 = 𝑆𝑂2 + 𝑂𝑀2 ⟹ 𝑆𝑀2 = 4.252 + 3.252 = 18.0625 + 10.5625 = 28.625 ⟹ 𝑆𝑀 ≈ 5.35
  • 2. A = 4∙6.5 ∙ 5.35 = 139.1 . A ≈ 139.1 b) The volume of the piece = the volume of SABCD pyramid + the volume of VABCD pyramid=V V = 𝐴𝑏∙(ℎ1+ℎ2) 3 = 𝐴 𝑏∙2∙𝑆𝑂 3 = 𝐴 𝑏∙𝑆𝑉 3 , h1=SO ,h2=VO V = 6.52∙8.5 3 = 42.25 ∙ 8.5 3 = 359.125 3 = 119.7 𝑐𝑚3 V ≈ 120 ∙10−3 =0.12 𝑑𝑚3 𝑚 = ρ ∙ 𝑉 = 410 ∙ 0.12 = 49.2 𝑔 < 50𝑔 𝒎 = 𝟒𝟗. 𝟐 𝒈 < 𝟓𝟎𝒈