KVPY 2016 SA Question 7

Geometry Level 2

Consider all the cuboids all of whose edges are integers and whose base is a square. Suppose the sum of all of its edges is numerically equal to the sum of the areas of all its six faces.

Find the sum of all of its edges.

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16 24 18 15

2 solutions

Fidel Simanjuntak
Jun 13, 2017

Let the length of the edges is $s$ . Then, the sum of all of its edges is $12s$ , since it's a cube. Then, the sum of the areas of all its six faces is $6s^2$ .

Given, $12s = 6s^2$ .

Solving, we have $s = 2$ . Then, the answer is $12 \times 2 = 24$ .

But in this question, it is given that the geometry is cuboid, how all the sides can be equal, 8 sides (edges of the square at bottom and top will be equal), please clarify the answer

Naman Gupta - 3 years, 8 months ago

You can show that the cuboid must be a cube. See my solution above.

Ron Gallagher - 7 months, 1 week ago

Exactly it's a cuboid not a cube

A Former Brilliant Member - 3 years, 7 months ago

Ron Gallagher
Nov 3, 2020

Let x be the length and width of the cuboid, and let z be the height. The sum of the lengths of the edges is then 8 x + 4 z, and the sum of the areas of the faces is 2 x^2 + 4 x*z. Setting these equal and solving for z gives:

z = x (x-4)/(2 (1-x))

(Note that x =1 is not compatible with the equations, so that division by 1-x is allowable).

Hence, the sum of the edges is:

4 x + 2 z = 4 x + 2 x (x-4)/(2 (1-x)) = 3 (x^2)/(x-1) = 3 x + 3 + 3/(x-1)

Since x and z are both integers, this last quantity must be an integer. Hence, 3 must be divisible by x-1. This can only happen if x = 2. The equation for z then implies z = 2 (-2)/(2 (-1)) = 2. Hence,

sum of the lengths of the edges = 4 x + 8 z = 8 + 16 = 24

KVPY 2016 SA Question 7

Try my set KVPY 2016 SA Questions

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