1972 AHSME Problems/Problem 34

Problem 34

Three times Dick's age plus Tom's age equals twice Harry's age. Double the cube of Harry's age is equal to three times the cube of Dick's age added to the cube of Tom's age. Their respective ages are relatively prime to each other. The sum of the squares of their ages is

$\textbf{(A) }42\qquad \textbf{(B) }46\qquad \textbf{(C) }122\qquad \textbf{(D) }290\qquad \textbf{(E) }326$

Solution

$t=2h-3d$ $3d^3+t^3=2h^3$

First, substitute in t into the second equation and get $3d^3+8h^3-12h^2d+18hd^2-27d^3=2h^3$

Page

Toolbox

Search

1972 AHSME Problems/Problem 34

Problem 34

Solution