Difference between revisions of "1972 AHSME Problems/Problem 34"
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<cmath>3d^3+t^3=2h^3</cmath> | <cmath>3d^3+t^3=2h^3</cmath> | ||
| − | First, substitute in t into the second equation and get <math>3d^3+8h^3-36h^2d+54hd^2-27d^3=2h^3</math>. That turns into <math>h^3-6h^2d+9hd^2-4d^3=0</math> which is factored into <math>(h-4d)(h-d)^2 =0.</math> WLOG, <math>d=1</math> and consequently <math>h=4</math>. Then <math>t=5</math>. The answer is | + | First, substitute in t into the second equation and get <math>3d^3+8h^3-36h^2d+54hd^2-27d^3=2h^3</math>. That turns into <math>h^3-6h^2d+9hd^2-4d^3=0</math> which is factored into <math>(h-4d)(h-d)^2 =0.</math> WLOG, <math>d=1</math> and consequently <math>h=4</math>. Then <math>t=8-3=5</math>. Everything appears to be relatively prime already. The answer is thus <math>1+16+25=\boxed{\textbf{(A) }42}.</math> ~lopkiloinm |
Latest revision as of 04:17, 7 November 2020
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
Solution
![\[t=2h-3d\]](http://latex.artofproblemsolving.com/d/9/1/d9140c0d6861ceb1425ac8f55dd342704c5e6a86.png)
![\[3d^3+t^3=2h^3\]](http://latex.artofproblemsolving.com/0/5/6/0569e6c22a5c1b20af5ae982f1845dff5b883690.png)
First, substitute in t into the second equation and get 
. That turns into 
which is factored into 
WLOG, 
and consequently 
. Then 
. Everything appears to be relatively prime already. The answer is thus 
~lopkiloinm
