Difference between revisions of "2017 AIME I Problems/Problem 10"
Stormersyle (talk | contribs) (→Solution 4 (algebra but much cleaner)) |
Stormersyle (talk | contribs) (→Solution 4 (algebra but much cleaner)) |
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==Solution 4 (algebra but much cleaner)== | ==Solution 4 (algebra but much cleaner)== | ||
− | We see that <math>\frac{z_3-z_1}{z_2-z_1}=\frac{60+16i}{-44i}=\frac{15i-4}{11}</math>. Now, let <math>z-z_3=a+bi</math>, in which case <math>z=(a+78)+(b+99)i</math> and <math>z-z_2=(a+60)+(b+60)i</math>. We now have that <math>(\frac{(a+60)+(b+60)i}{a+bi})(\frac{15i-4}{11})</math> is real, meaning that <math>((a+60)+(b+60)i)(\frac{15i-4}{a+bi}</math> is real. | + | We see that <math>\frac{z_3-z_1}{z_2-z_1}=\frac{60+16i}{-44i}=\frac{15i-4}{11}</math>. Now, let <math>z-z_3=a+bi</math>, in which case <math>z=(a+78)+(b+99)i</math> and <math>z-z_2=(a+60)+(b+60)i</math>. We now have that <math>(\frac{(a+60)+(b+60)i}{a+bi})(\frac{15i-4}{11})</math> is real, meaning that <math>((a+60)+(b+60)i)(\frac{15i-4}{a+bi})</math> is real. |
We see that <math>\frac{15i-4}{a+bi}=\frac{(15i-4)(a-bi)}{a^2+b^2}=\frac{(15b-4a)+(15a+4b)i}{a^2+b^2}</math>, so therefore <math>x=((a+60)+(b+60)i)((15b-4a)+(15a+4b)i)</math> is real. | We see that <math>\frac{15i-4}{a+bi}=\frac{(15i-4)(a-bi)}{a^2+b^2}=\frac{(15b-4a)+(15a+4b)i}{a^2+b^2}</math>, so therefore <math>x=((a+60)+(b+60)i)((15b-4a)+(15a+4b)i)</math> is real. | ||
− | This means that <math>\Im(x)=0</math>, so we now have that <math>(a+60)(15a+4b)+(b+60)(15b-4a)=15a^2+15b^2+660a+1140b=0</math>, so <math>a^2+b^2+44a+76b=0</math>, which can be rewritten as<math>(a+22)^2+(b+38)^2=22^2+38^2</math>. In order to maximize <math>\Im(z)</math> we want to maximize <math>b</math>, and in order to maximize <math>b</math> we want <math>a+22=0</math> and <math>a=-22</math>, so <math>\Re(z)=a+78=-22+78=\boxed{056}</math>. | + | This means that <math>\Im(x)=0</math>, so we now have that <math>(a+60)(15a+4b)+(b+60)(15b-4a)=15a^2+15b^2+660a+1140b=0</math>, so <math>a^2+b^2+44a+76b=0</math>, which can be rewritten as<math>(a+22)^2+(b+38)^2=22^2+38^2</math>. In order to maximize <math>\Im(z)</math> we want to maximize <math>b</math>, and in order to maximize <math>b</math> we want <math>a+22=0</math> and <math>a=-22</math>, so <math>\Re(z)=a+78=-22+78=\boxed{056}</math>. |
+ | (Note: <math>\Im(\omega)</math> is the imaginary part of <math>\omega</math>, and <math>\Re(\omega)</math> is the real part of <math>\omega</math>) | ||
~Stormersyle | ~Stormersyle |
Revision as of 12:11, 9 March 2019
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
(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 complex 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 (Not well explained)
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 circumcenter of (just get the intersection of the perpendicular bisectors) and we get the -coordinate of the circumcenter is .
~First
Solution 4 (algebra but much cleaner)
We see that . Now, let , in which case and . We now have that is real, meaning that is real.
We see that , so therefore is real.
This means that , so we now have that , so , which 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
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.