Find the volume of a cuboid whose shortest side has a length of 34, whose longest side is the sum of the other two side lengths, and whose inner triangle formed by its three different diagonals has an area of 2418.
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I took somewhat of a brute force approach, solving the equation ∣ ∣ ( L , 0 , H ) × ( 0 , W , H ) ∣ ∣ = ∣ ∣ ( 3 4 + W , 0 , 3 4 ) × ( 0 , W , 3 4 ) ∣ ∣ = 2 × 2 4 1 8 . After a bit of computation we find W ( W + 3 4 ) + 3 4 2 = 4 8 3 6 so W ( W + 3 4 ) = 3 6 8 0 and the volume of the cuboid is 3 4 W ( W + 3 4 ) = 1 2 5 1 2 0 .
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The dimensions of this cuboid is 3 4 × W × ( 3 4 + W ) , where W ≥ 3 4 .
By Pythagorean theorem , the side lengths of the triangle in question are 3 4 2 + W 2 , 3 4 2 + ( 3 4 + W ) 2 , W 2 + ( 3 4 + W ) 2 . Let a , b , c denote the these side lengths, respectively. Then, using one of the formula of Heron's formula , we have 2 4 1 8 = 4 1 ( a 2 + b 2 + c 2 ) 2 − 2 ( a 4 + b 4 + c 4 ) . With ⎩ ⎪ ⎨ ⎪ ⎧ a 2 = 3 4 2 + W 2 b 2 = 3 4 2 + ( 3 4 + W ) 2 c 2 = W 2 + ( 3 4 + W ) 2 , the equation simplifies to ( 2 4 1 8 ⋅ 4 ) 2 = 4 ( W 2 + 3 4 W + 1 1 5 6 ) 2 ⟹ W = 4 6 . The volume of the cuboid is simply 3 4 W ( 3 4 + W ) = 1 2 5 1 2 0 .