2018 AIME I Problems/Problem 5
Contents
[hide]Problem 5
For each ordered pair of real numbers satisfying
there is a real number
such that
Find the product of all possible values of
.
Solution 1
Using the logarithmic property , we note that
That gives
upon simplification and division by
. Factoring
by Simon's Favorite Factoring Trick gives
Then,
From the second equation,
If we take
, we see that
. If we take
, we see that
. The product is
.
-expiLnCalc
Solution 2
Do as done in Solution 1 to get
Do as done in Solution 1 to get
$$ (Error compiling LaTeX. Unknown error_msg)\implies \frac{x}{y}=$$ (Error compiling LaTeX. Unknown error_msg)\frac{-2\pm \sqrt{4-24(1-K)}}{12}
\implies \frac{x}{y}=\frac{-2\pm 2\sqrt{6K-5}}{12}=\frac{-1\pm \sqrt{6K-5}}{6}
\frac{x}{y}=1
1=\frac{-1\pm \sqrt{6K-5}}{6}
\implies 6=-1\pm \sqrt{6K-5}
\implies 7=\pm \sqrt{6K-5}
\implies 49=6K-5
\implies K=9
\frac{x}{y}=-2
-2=\frac{-1\pm \sqrt{6K-5}}{6}
\implies -12=-1\pm \sqrt{6K-5}
\implies -11=\sqrt{6K-5}
\implies 121=6K-5
\implies 126=6K
\implies K=21$$ (Error compiling LaTeX. Unknown error_msg). Hence our final answer is
-vsamc
-minor edit:einsteinstudent
-style edit: yeaboi
Solution 3 (Official MAA)
Because the right side of the first equation is real. It follows that the left side of the equation is also real, so
and
Thus
which implies that
Therefore either
or
and because
must be positive and
Similarly,
If
then
when
If
then
when
The requested product is
Video Solution
https://www.youtube.com/watch?v=iE8paW_ICxw
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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