# Segments AB, AC, and AD are edges of a cube, and AG is a diagonal through the center of the cube. Point P satisfies PB = 60`sqrt(10)` , PC = 60`sqrt(5)` , PD = 120`sqrt(2)` , and PG =36`sqrt(7)` . What is PA?

PA is 192.

Let's consider the coordinate system linked with the cube: A is the origin; B, C, and D are along axes. Denote the length of the cube side as `a ` and the coordinates of P as `x , y , z .`

Then we have four equations that represent length data:

`x^2 + y^2 + ( z - a )^2 = 10 * 60^2`
`x^2 + ( y - a )^2 + z^2 = 5 * 60^2`
`( x - a )^2 + y^2 + z^2 = 2 * 120^2 = 8 * 60^2`
`( x - a )^2 + ( y - a )^2 + ( z - a )^2 = 7 * 36^2 .`

And the question is, what is `sqrt ( x^2 + y^2 + z^2 ) ? ` To find `x^2 + y^2 + z^2 , ` sum up the first three equations and subtract the fourth:

`2 ( x^2 + y^2 + z^2 ) = ( 10 + 5 + 8 ) * 60^2 - 7 * 36^2 ,`

so

`x^2 + y^2 + z^2 = 1 / 2 ( 23 * 60^2 - 7 * 36^2 ) = 18 * ( 23 * 100 - 7 * 36 ) = 18 * 2048 = 3^2 * 2^12 .`

Now we know that `PA = sqrt ( 3^2 * 2^12 ) = 3 * 2^6 = 3 * 64 = 192 .`

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