Difference between revisions of "2017 AIME I Problems/Problem 10"
(→Solution) |
m (→See Also) |
||
| Line 17: | Line 17: | ||
==See Also== | ==See Also== | ||
| − | {{AIME box|year=2017|n=I|num-b= | + | {{AIME box|year=2017|n=I|num-b=9|num-a=11}} |
| + | {{MAA Notice}} | ||
Revision as of 20:23, 8 March 2017
Problem 10
Let 
and 
where 
Let 
be the unique complex number with the properties that 
is a real number and the imaginary part of is the greatest possible. Find the real part of .
Solution
(This solution's quality may be very poor. If one feels that the solution is inadequate, one may choose to improve it.)
Let us write 
be some imaginary number with form 
Similarly, we can write 
as some 
The product must be real, so we have that 
is real. Of this, 
must be real, so the imaginary parts only arise from the second part of the product. Thus we have
![\[(\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2) = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i(\cos \theta_1 \sin \theta_2 + \cos \theta_2 \sin \theta_1)\]](http://latex.artofproblemsolving.com/0/5/3/0534ff4b47ba01bdd1b475562ba36c690d377f61.png)
is real. The imaginary part of this is 
which we recognize as 
This is only 
when 
is some multiple of 
In this problem, this implies 
and must form a cyclic quadrilateral, so the possibilities of lie on the circumcircle of 
and 
To maximize the imaginary part of 
it must lie at the top of the circumcircle, which means the real part of is the same as the real part of the circumcircle. The center of the circumcircle can be found in various ways, (such as computing the intersection of the perpendicular bisectors of the sides) and when computed gives us that the real part of the circumcircle is 
so the real part of is and thus our answer is 
See Also
| 2017 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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. 
