document.write( "Question 843221: planets A, B, C orbit their star every 2,7, and 12 months respectivly. if three plants are now in the same straight line what is the smallest number of months that must pass before they line up again? explain how you arrived at your answer. \n" ); document.write( "

Algebra.Com's Answer #508003 by richard1234(7193) $\"\"$ $\"About$

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The \"obvious\" answer would be lcm(2,7,12) = 84 months but this is actually incorrect, because that assumes they must do full orbits before they are in a line.\r
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\n" ); document.write( "\n" ); document.write( "After t months, the angle that each planet makes with respect to its starting location and the star is

, and

respectively. For the planets to be collinear, these three angles, when taken pairwise, must differ by an integer multiple of pi. In particular

and

. The smallest t for which these three differences are each integer multiples of pi is t = 42, so the answer is 42 months.\r
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\n" ); document.write( "\n" ); document.write( "This problem is really similar to 2007 AIME I #4, if you want to check that out. \n" ); document.write( "

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