2017 AIME I Problems/Problem 10
Contents
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 1 (Coordinates, Geometry)
This problem is pretty obvious how to bash, and indeed many of the solutions below explain how to do that. But there’s no fun in that, and let’s see if we can come up with a slicker solution that will be more enjoyable.
Instead of thinking of complex numbers as purely a real plus a constant times , let’s graph them and hope that the geometric visualization adds insight to the problem.
Note that when we subtract two vectors, the geometric result is the line segment between the two endpoints of the vectors. Thus we can fill in and as so;
looks similar to , so let’s try to prove that they are congruent. We can show this in two ways;
Solution 1.1
Let’s look back at the information given to us. The problem states that is a real number. Let the real number be some constant, . Rearranging yields . But how do we relate this expression to our angles? Well, let’s take a look at the divisions themselves.
The subtraction of two vectors yields a vector, and we can write any vector division as where is a complex number, as the division of two vectors also yields a vector. We can rewrite this as . We can think of this expression as transforming directly on to , and is the transformation function. However, this transformation must be some kind of rotation, which means that the degree measure of is equal to the angle between the two vectors since we need to rotate by that angle to lay flat on .
Thus we can rewrite our previous equation as , where the angle of equals the angle between and and likewise for . More precisely, we can write as and , respectively by Euler’s formula. Then is the claim we wish to prove.
We can now do some simple algebra to prove this;
is obviously real, so must be real as well. But the only way that can happen is if .
Solution 1.2
Let us write as some complex number with form Similarly, we can write as some
The product must be real, so we have that is real. is real by definition, so dividing the real number above by will still yield a real number. (Note that we can see that from the definitions of and ). Thus we have
is real. The imaginary part of this is which we recognize as This is only when for some integer . Here represents the major angle, and the angle we want is the minor angle, so we can rewrite the equation as . We can see from the diagram that both and are obtuse, so therefore .
Solution 1 Rejoined
Now that we’ve proven this fact, we know that all four points lie on a circle (intuitively one can also observe this because and satisfy the property in the question, and techincally gives no imaginary part), so let’s draw that in;
While are fixed, can be anywhere on the circle because those are the only values of that satisfy the problem requirements. However, we want to find the real part of the with the maximum imaginary part. This Z would lie directly above the center of the circle, and thus the real part would be the same as the x-value of the center of the circle. So all we have to do is find this value and we’re done.
Consider the perpendicular bisectors of and . Since any chord can be perpendicularly bisected by a radius of a circle, these two lines both intersect at the center. Since is vertical, the perpendicular bisector will be horizontal and pass through the midpoint of this line, which is (18, 61). Therefore the equation for this line is . is nice because it turns out the differences in the x and y values are both equal (60) which means that the slope of the line is 1. The slope of the perpendicular bisector is therefore -1 and it passes through the midpoint, (48,69), so the equation of this line is . Finally, equating the two yields
~KingRavi
~Anonymous(Solution 1.2)
Bashy Solution :)
We know thatHence,where . Let . Then, The numerator is: The ratio of the imaginary part to the real part must be because Hence, Evidently, is maximized when is maximized or when
~AopsUser101
Solution 3
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 4 (algebra but much cleaner)
We see that .
Now, let , in which case and .
We now have that is real, meaning that: is also real.
We see that: so therefore, is real.
This means that , so we now have that:
This can be rewritten as: In order to maximize we want to maximize , and in order to maximize we want and , so . (Note: is the imaginary part of , and is the real part of ) ~Stormersyle
Solution 5
We will just bash. Let where . We see that after doing some calculations. We also see that We note that is a multiple of because the numerator has to be real. Thus, expanding it out, we see that Hence, To maximize the imaginary part, must equal so hence, .
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.