A Carpenter's Tetradecahedron

Geometry Level 3

The artistic carpenter forms a tetradecahedron from a cube with side length 3". For each of the cube's vertices, he measures an inch out on the edges, and cuts out the resulting right tetrahedron from each corner, leaving 6 octagonal faces, and 8 triangular faces. The volume is a b \frac{a}{b} where a a and b b are relatively prime positive integers and the surface area is c + d e c+d\sqrt{e} , and e e is prime. What is the sum a + b + c + d + e a+b+c+d+e ?


The answer is 129.

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Rob Matuschek by Rob Matuschek , 54, USA

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1 solution

Rob Matuschek
Sep 10, 2015

The volume of the cube is 27 − 8 × ( 1 6 ) = 77 3 27-8\times(\frac{1}{6}) = \frac{77}{3} The surface area is 6 × 3 × 3 − 24 × 1 2 ( 1 ) ( 1 ) + 8 × 3 4 × 2 2 = 42 + 4 3 . 6\times3\times3 - 24\times\frac{1}{2}(1)(1)+8\times\frac{\sqrt{3}}{4}\times\sqrt{2}^{2} =42+4\sqrt{3}. Thus a = 77 , b = 3 , c = 42 , d = 4 , e = 3 a = 77, b = 3, c = 42, d = 4, e = 3 and a + b + c + d + e = 129. a+b+c+d+e = 129.

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Where does the square root of 2 squared term come from?

Geoff Pilling - 5 years, 2 months ago

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