Diagonal Triangle of a Cuboid

Geometry Level 3

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.


The answer is 125120.

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2 solutions

Pi Han Goh
Dec 28, 2018

The dimensions of this cuboid is 34 × W × ( 34 + W ) 34 \times W \times (34 + W) , where W 34 W\geq 34 .

By Pythagorean theorem , the side lengths of the triangle in question are 3 4 2 + W 2 , 3 4 2 + ( 34 + W ) 2 , W 2 + ( 34 + W ) 2 . \sqrt{34^2 + W^2}, \sqrt{34^2 + (34 + W)^2} , \sqrt{W^2 + (34 + W)^2} . Let a , b , c a,b,c denote the these side lengths, respectively. Then, using one of the formula of Heron's formula , we have 2418 = 1 4 ( a 2 + b 2 + c 2 ) 2 2 ( a 4 + b 4 + c 4 ) . 2418 = \dfrac14 \sqrt{(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 + ( 34 + W ) 2 c 2 = W 2 + ( 34 + W ) 2 , \begin{cases} a^2= 34^2 + W^2 \\ b^2 = 34^2 + (34 + W)^2\\ c^2 = W^2 + (34 + W)^2 , \end{cases} the equation simplifies to ( 2418 4 ) 2 = 4 ( W 2 + 34 W + 1156 ) 2 W = 46. (2418\cdot 4)^2 = 4(W^2 + 34W+ 1156)^2 \implies W = 46. The volume of the cuboid is simply 34 W ( 34 + W ) = 125120 34W(34 + W) = \boxed{125120} .

Otto Bretscher
Dec 26, 2018

I took somewhat of a brute force approach, solving the equation ( L , 0 , H ) × ( 0 , W , H ) = ( 34 + W , 0 , 34 ) × ( 0 , W , 34 ) = 2 × 2418 ||(L,0,H) \times (0,W,H)||=||(34+W,0,34) \times (0,W,34)||=2\times 2418 . After a bit of computation we find W ( W + 34 ) + 3 4 2 = 4836 W(W+34)+34^2=4836 so W ( W + 34 ) = 3680 W(W+34)=3680 and the volume of the cuboid is 34 W ( W + 34 ) = 125120 34W(W+34)=\boxed{125120} .

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