Maths Olympiad Prep

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Algebra Difficulty 3.0 Junior Find the answer

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

Pick one

Solution

t=2h−3dt=2h-3d
3d3+t3=2h33d^3+t^3=2h^3
First, substitute in t into the second equation and get 3d3+8h3−36h2d+54hd2−27d3=2h33d^3+8h^3-36h^2d+54hd^2-27d^3=2h^3. That turns into h3−6h2d+9hd2−4d3=0h^3-6h^2d+9hd^2-4d^3=0 which is factored into (h−4d)(h−d)2=0.(h-4d)(h-d)^2 =0. WLOG, d=1d=1 and consequently h=4h=4. Then t=8−3=5t=8-3=5. Everything appears to be relatively prime already. The answer is thus 1+16+25=(A) 42.1+16+25=\boxed{\textbf{(A) }42}. ~lopkiloinm

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.