Tetrahedral Water Tank - Part 1

Geometry Level 3

A water tank in the shape of a tetrahedron with altitude 100 cm 100\text{ cm} is resting flat on the ground on one of its faces. It contains water to a height of 50 cm . 50\text{ cm}. Now, the tank is turned upside down such that it rests on one of its vertices, with the opposing base lying horizontally above it, parallel to the ground.

What will be the new height of water level above the bottom vertex in centimeters?
(The answer will surprise you.)

Inspiration


The answer is 95.647.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

Hosam Hajjir
Dec 29, 2017

Let the volume of water = W W , original level of water = y = 50 = y = 50 cm, altitude of tetrahedron = h = 100 = h = 100 cm.

The volume of the whole tetrahedron is related to its altitude by

V = a h 3 V = a h^3 , for some constant a.

Therefore, the water volume is related to y y by:

W = V − a ( h − y ) 3 = a h 3 − a ( h − y ) 3 W = V - a (h - y)^3 = a h^3 - a (h- y)^3

Let y ′ y' be the new water level in the flipped tetrahedron, then

W = a y ′ 3 W = a y'^3

Hence,

a y ′ 3 = a ( h 3 − ( h − y ) 3 ) a y'^3 = a ( h^3 - (h - y)^3 )

dividing through by a h 3 a h^3 ,

( y ′ h ) 3 = 1 − ( 1 − y h ) 3 ) (\dfrac{y'}{h} )^3 = 1 - (1 - \dfrac{y}{h})^3 )

Substituting the given values,

( y ′ 100 ) 3 = 1 − ( 1 − 50 100 ) 3 = 1 − 1 8 = 7 8 (\dfrac{y'}{100})^3 = 1 - ( 1 - \dfrac{50}{100})^3 = 1 - \dfrac{1}{8} = \dfrac{7}{8}

Therefore,

y ′ = 100 7 8 3 = 95.647 y' = 100 \sqrt[3]{ \dfrac{7}{8} }= 95.647 cm

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