Volume and surface area (2)

Geometry Level 3

The square base of a right pyramid has sides of integral length. The height of the pyramid is also an integer. The volume of the pyramid, which is also an integer, is equal to it's total surface area. What is the height of the pyramid?


The answer is 8.

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2 solutions

Chris Lewis
May 20, 2019

A square pyramid with height h h and whose base has side-length a a has volume 1 3 a 2 h \frac{1}{3} a^2 h and surface area a 2 + a a 2 + 4 h 2 a^2+a\sqrt{a^2+4h^2} .

Equating these and doing a bit of tidying, we find a 2 = 36 h h 6 a^2 = \frac{36h}{h-6}

So we require 36 h h 6 \frac{36h}{h-6} , where h h is a positive integer, to be a perfect square. Clearly we need h > 6 h>6 .

For h > 6 h>6 , 36 h h 6 \frac{36h}{h-6} is a decreasing function in h h . As h h \to \infty , 36 h h 6 3 6 + \frac{36h}{h-6} \to 36^+ . So as soon as 36 h h 6 < 49 \frac{36h}{h-6}<49 , we can stop searching, as there are no perfect squares in the interval ( 36 , 49 ) (36,49) .

Solving the inequality, we find this means that h < 23 h<23 . We can just check the values of 36 h h 6 \frac{36h}{h-6} for each h h from h = 7 h=7 to h = 22 h=22 . Several of these give integers, but the only one that gives a square is h = 8 h=\boxed8 . (The corresponding side-length is a = 12 a=12 ).

Kyle T
May 20, 2019
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<?php
for($l=1;$l<=100;$l++){
    for($h=1;$h<=100;$h++){
        $v = volume($l,$h);
        if($v == floor($v)){
            $sa = surface_area($l,$h);
            if($sa == $v){
                echo 'base:'.$l.',  height:'.$h.',  volume:'.$v.',  surface area:'.$sa.'<br>';
                echo 'answer: '.$h.'<br>';
                exit;
            }
        }
    }
}

function volume($l,$h){
    return (pow($l,2)*$h)/3;
}

function surface_area($l,$h){
    return pow($l,2) + (2*($l * sqrt(pow($l/2,2)+pow($h,2))));
}

?>
// prints the following: 
// base:12, height:8, volume:384, surface area:384
// answer: 8

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