2000 AIME I Problems/Problem 7
Suppose that and are three positive numbers that satisfy the equations and Then where and are relatively prime positive integers. Find .
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques
We can rewrite as .
Substituting into one of the given equations, we have
We can substitute back into to obtain
We can then substitute once again to get Thus, , so .
Thus . So .
Since , so . Also, by the second equation. Substitution gives , , and , so the answer is 4+1 which is equal to .
(Hybrid between 1/2)
Because and . Substituting and factoring, we get , , and . Multiplying them all together, we get, , but is , and by the Identity property of multiplication, we can take it out. So, in the end, we get . And, we can expand this to get , and if we make a substitution for , and rearrange the terms, we get This will be important.
Now, lets add the 3 equations , and . We use the expand the Left hand sides, then, we add the equations to get Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus We move all constant terms to the right, and all linear terms to the left, to get , so which gives an answer of
Get rid of the denominators in the second and third equations to get and . Then, since , we have and . Then, since we know that , we can subtract these two equations to get that . The result follows that and , so , and the requested answer is
Rewrite the equations in terms of x.
Now express in terms of x.
This evaluates to , giving us . We can now plug x into the other equations to get and .
, and we are done. ~MC413551
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