Problem: A B ‾ , A C ‾ A B , A C A D ‾ A D A G ‾ A G P P B P = 60 10 , C P = 60 5 , D P = 120 2 B P = 6 0 1 0 ​ , C P = 6 0 5 ​ , D P = 1 2 0 2 ​ G P = 36 7 G P = 3 6 7 ​ A P A P
Solution:
Let the cube have side length s s 3 3 A ( 0 , 0 , 0 ) , B ( s , 0 , 0 ) , C ( 0 , s , 0 ) A ( 0 , 0 , 0 ) , B ( s , 0 , 0 ) , C ( 0 , s , 0 ) D ( 0 , 0 , s ) D ( 0 , 0 , s ) G ( s , s , s ) G ( s , s , s ) P P ( x , y , z ) ( x , y , z )
B P 2 = ( x − s ) 2 + y 2 + z 2 = ( 60 10 ) 2 C P 2 = x 2 + ( y − s ) 2 + z 2 = ( 60 5 ) 2 D P 2 = x 2 + y 2 + ( z − s ) 2 = ( 120 2 ) 2 G P 2 = ( x − s ) 2 + ( y − s ) 2 + ( z − s ) 2 = ( 36 7 ) 2 . ​ B P 2 = ( x − s ) 2 + y 2 + z 2 = ( 6 0 1 0 ​ ) 2 C P 2 = x 2 + ( y − s ) 2 + z 2 = ( 6 0 5 ​ ) 2 D P 2 = x 2 + y 2 + ( z − s ) 2 = ( 1 2 0 2 ​ ) 2 G P 2 = ( x − s ) 2 + ( y − s ) 2 + ( z − s ) 2 = ( 3 6 7 ​ ) 2 . ​
Adding the first three equations and subtracting the fourth equation yields
2 ( x 2 + y 2 + z 2 ) = 1 2 2 ⋅ [ ( 5 10 ) 2 + ( 5 5 ) 2 + ( 10 2 ) 2 − ( 3 7 ) 2 ] = 1 2 2 ⋅ ( 250 + 125 + 200 − 63 ) = 2 ⋅ 1 2 2 ⋅ 1 6 2 = 2 ⋅ 19 2 2 2 ( x 2 + y 2 + z 2 ) = 1 2 2 ⋅ [ ( 5 1 0 ​ ) 2 + ( 5 5 ​ ) 2 + ( 1 0 2 ​ ) 2 − ( 3 7 ​ ) 2 ] = 1 2 2 ⋅ ( 2 5 0 + 1 2 5 + 2 0 0 − 6 3 ) = 2 ⋅ 1 2 2 ⋅ 1 6 2 = 2 ⋅ 1 9 2 2 ​
Therefore A P = x 2 + y 2 + z 2 = 192 A P = x 2 + y 2 + z 2 ​ = 1 9 2 ​