Cube Puzzle

Geometry Level 4

A cube with side length a a has its eight vertices specified in the Cartesian space as follows:

A ( 0 , 0 , 0 ) , B ( a , 0 , 0 ) , C ( a , a , 0 ) , D ( 0 , a , 0 ) , E ( 0 , 0 , a ) , F ( a , 0 , a ) , G ( a , a , a ) , H ( 0 , a , a ) A(0, 0, 0) , B(a, 0, 0), C(a, a, 0), D(0, a, 0), E(0, 0, a), F(a, 0, a), G(a, a, a), H(0, a, a)

There is a point P = ( x , y , z ) P = (x, y, z) inside the cube such that

P A = 70 , P B = 97 , P C = 88 , P E = 43 PA = \sqrt{70}, PB = \sqrt{97}, PC = \sqrt{88}, PE = \sqrt{43}

Find the sum a + x + y + z a + x + y + z .


The answer is 23.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

Chew-Seong Cheong
Jun 10, 2020

By Pythagorean theorem , we have:

{ x 2 + y 2 + z 2 = 70 . . . ( 1 ) ( x − a ) 2 + y 2 + z 2 = 97 . . . ( 2 ) ( x − a ) 2 + ( y − a ) 2 + z 2 = 88 . . . ( 3 ) x 2 + y 2 + ( z − a ) 2 = 43 . . . ( 4 ) \begin{cases} x^2+y^2+z^2 = 70 & ...(1) \\ (x-a)^2+y^2+z^2 = 97 & ...(2) \\ (x-a)^2+(y-a)^2+z^2 = 88 & ...(3) \\ x^2+y^2+(z-a)^2 = 43 & ...(4) \end{cases}

( 1 ) − ( 2 ) : 2 a x − a 2 = − 27 ⟹ x = a 2 − 27 2 a ( 2 ) − ( 3 ) : 2 a y − a 2 = 9 ⟹ y = a 2 + 9 2 a ( 1 ) − ( 4 ) : 2 a z − a 2 = 27 ⟹ z = a 2 + 27 2 a \begin{array} {lll} (1)-(2): & 2ax - a^2 = -27 & \implies x = \dfrac {a^2 - 27}{2a} \\ (2)-(3): & 2ay - a^2 = 9 & \implies y = \dfrac {a^2 +9}{2a} \\ (1)-(4): & 2az - a^2 = 27 & \implies z = \dfrac {a^2 +27}{2a} \end{array}

Substitute x x , y y , and z z in ( 1 ) (1) :

( a 2 − 27 2 a ) 2 + ( a 2 + 9 2 a ) 2 + ( a 2 + 27 2 a ) 2 = 70 3 a 4 − 262 a 2 + 1539 = 0 ( 3 a 2 − 19 ) ( a 2 − 81 ) = 0 \begin{aligned} \left(\frac {a^2 - 27}{2a} \right)^2 + \left(\frac {a^2 +9}{2a} \right)^2 + \left(\frac {a^2 + 27}{2a} \right)^2 & = 70 \\ 3a^4-262a^2 + 1539 & = 0 \\ (3a^2-19)(a^2-81) & = 0 \end{aligned}

⟹ { a 2 = 19 3 ⟹ a = 19 3 a 2 = 81 ⟹ a = 9 \implies \begin{cases} a^2 = \dfrac {19}3 & \implies a = \sqrt{\dfrac {19}3} \\ a^2 = 81 & \implies a = 9 \end{cases}

For positive x x , a = 9 a = 9 and { x = a 2 − 27 2 a = 3 y = a 2 + 9 2 a = 5 z = a 2 + 27 2 a = 6 \begin{cases} x = \dfrac {a^2 - 27}{2a} & = 3 \\ y = \dfrac {a^2 +9}{2a} & = 5 \\ z = \dfrac {a^2 +27}{2a} & = 6 \end{cases}

Therefore a + x + y + z = 9 + 3 + 5 + 6 = 23 a+x+y+z = 9+3+5+6 = \boxed {23} .

3 Helpful 0 Interesting 2 Brilliant 0 Confused

@Chew-Seong Cheong , a diagram here would be helpful to see. Is it possible that you can add one? I would have definitely made one but I don't know how to make one in 3D. Thanks!

Mahdi Raza - 1 year ago

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Oops! I'm not able too.

Chew-Seong Cheong - 1 year ago

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