Difference between revisions of "2017 AIME I Problems/Problem 10"
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==Solution 3== | ==Solution 3== | ||
If you don't already(and can consistently solve at least <math>5</math> questions on the AIME or more), you should learn how to complex bash for Olympiads. The <math>\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}</math> just means <math>z</math> is on the circumcircle of <math>\triangle{z_1 z_2 z_3}</math> and we just want the highest point on the circle in terms of imaginary part. Convert to Cartesian coordinates and we just need to compute the <math>x</math>-coordinate of the circumcircle of <math>(18, 83), (18, 39), (78, 99)</math>(just get the intersection of the perpendicular bisectors) and we get the <math>x</math>-coordinate of the circumcircle is <math>\boxed{056}</math>. | If you don't already(and can consistently solve at least <math>5</math> questions on the AIME or more), you should learn how to complex bash for Olympiads. The <math>\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}</math> just means <math>z</math> is on the circumcircle of <math>\triangle{z_1 z_2 z_3}</math> and we just want the highest point on the circle in terms of imaginary part. Convert to Cartesian coordinates and we just need to compute the <math>x</math>-coordinate of the circumcircle of <math>(18, 83), (18, 39), (78, 99)</math>(just get the intersection of the perpendicular bisectors) and we get the <math>x</math>-coordinate of the circumcircle is <math>\boxed{056}</math>. | ||
+ | |||
+ | ~First | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2017|n=I|num-b=9|num-a=11}} | {{AIME box|year=2017|n=I|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:44, 27 June 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
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 circumcenter. 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 circumcenter is
so the real part of
is
and thus our answer is
Solution 2
Algebra Bash
First we calculate , which becomes
.
Next, we define to be
for some real numbers
and
. Then,
can be written as
Multiplying both the numerator and denominator by the conjugate of the denominator, we get:
In order for the product to be a real number, since both denominators are real numbers, we must have the numerator of be a multiple of the conjugate of
, namely
So, we have
and
for some real number
.
Then, we get:
Expanding both sides and combining like terms, we get:
which can be rewritten as:
Now, common sense tells us that to maximize , we would need to maximize
. Therefore, we must set
to its lowest value, namely 0. Therefore,
must be
You can also notice that the ab terms cancel out so all you need is the x-coordinate of the center and only expand the a parts of the equation.
~stronto
Solution 3
If you don't already(and can consistently solve at least questions on the AIME or more), you should learn how to complex bash for Olympiads. The
just means
is on the circumcircle of
and we just want the highest point on the circle in terms of imaginary part. Convert to Cartesian coordinates and we just need to compute the
-coordinate of the circumcircle of
(just get the intersection of the perpendicular bisectors) and we get the
-coordinate of the circumcircle is
.
~First
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.