Difference between revisions of "2000 AIME I Problems/Problem 7"
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+ | ===Solution 5=== | ||
+ | Get rid of the denominators in the second and third equations to get <math>xz-5z=-1</math> and <math>xy-29x=-1</math>. Then, since <math>xyz=1</math>, we have \(\frac 1y-5z=-1\) and \(\frac 1z-29x=-1\). Then, since we know that \(\frac 1z+x=5\), we can subtract these two equations to get that <math>30x=6\implies x=5</math>. The result follows that \(z=\frac 5{24}\) and <math>y=24</math>, so \(\frac 14+\frac 5{24}=\frac 14\), and the requested answer is <math>1+4=\boxed{005}.</math> | ||
== See also == | == See also == |
Revision as of 15:55, 16 June 2020
Contents
Problem
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
Solution 1
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
.
Solution 2
Let .
Thus . So
.
Solution 3
Since , so
. Also,
by the second equation. Substitution gives
,
, and
, so the answer is 4+1 which is equal to
.
Solution 4
(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
-AlexLikeMath
Solution 5
Get rid of the denominators in the second and third equations to get and
. Then, since
, we have \(\frac 1y-5z=-1\) and \(\frac 1z-29x=-1\). Then, since we know that \(\frac 1z+x=5\), we can subtract these two equations to get that
. The result follows that \(z=\frac 5{24}\) and
, so \(\frac 14+\frac 5{24}=\frac 14\), and the requested answer is
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.