Difference between revisions of "2017 AIME I Problems/Problem 10"
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==Solution== | ==Solution== | ||
− | ( | + | Let us write <math>\frac{z_3 - z_1}{z_2 - z_1}</math> as some complex number with form <math>r_1 (\cos \theta_1 + i \sin \theta_1).</math> Similarly, we can write <math>\frac{z-z_2}{z-z_3}</math> as some <math>r_2 (\cos \theta_2 + i \sin \theta_2).</math> |
− | + | The product must be real, so we have that <math>r_1 r_2 (\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2)</math> is real. <math>r_1 r_2</math> is real by definition, so dividing the real number above by <math>r_1 r_2</math> will still yield a real number. (Note that we can see that <math>r_1 r_2 \not= 0</math> from the definitions of <math>z_1,</math> <math>z_2,</math> and <math>z_3</math>). Thus we have | |
− | |||
− | The product must be real, so we have that <math>r_1 r_2 (\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2)</math> is real. | ||
<cmath>(\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)</cmath> | <cmath>(\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)</cmath> | ||
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To maximize the imaginary part of <math>z,</math> it must lie at the top of the circumcircle, which means the real part of <math>z</math> 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 <math>56,</math> so the real part of <math>z</math> is <math>56,</math> and thus our answer is <math>\boxed{056}.</math> | To maximize the imaginary part of <math>z,</math> it must lie at the top of the circumcircle, which means the real part of <math>z</math> 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 <math>56,</math> so the real part of <math>z</math> is <math>56,</math> and thus our answer is <math>\boxed{056}.</math> | ||
− | ==Solution 2== | + | ==Bashy Solution :)== |
+ | We know that<cmath>z_3-z_1 = (78+99i)-(16+83i) = 60 + 16i.</cmath><cmath>z_2-z_1=(18+39i)-(18+83i) = -44i.</cmath>Hence,<cmath>\frac{z_3-z_1}{z_2-z_1} = \frac{60 + 16i}{-44i} = \frac{15i-4}{11} = \frac{c}{15i+4}</cmath>where <math>c \in R</math>. | ||
+ | Let <math>z = a+bi</math>. Then, | ||
+ | <cmath>\frac{z-z_2}{z-z_3} = \frac{(a+bi)-(18+39i)}{(a+bi)-(78+99i)} =</cmath><cmath>\frac{z-z_2}{z-z_3} = \frac{(a-18)+i(b-39)}{(a-78)+i(b-99)} =</cmath><cmath>\frac{z-z_2}{z-z_3} = \frac{((a-18)+i(b-39)((a-78)+i(99-b))}{((a-78)+i(b-99))((a-78)+i(99-b)} =</cmath><cmath>\frac{z-z_2}{z-z_3} = \frac{((a-18)+i(b-39)((a-78)+i(99-b))}{k}</cmath>The numerator is: | ||
+ | <cmath>(a-18)(a-78)+(b-39)(99-b)+i((b-39)(a-78) + (99-b)(a-18))=</cmath><cmath>a^2+b^2-96a-138b+18 \cdot 78 - 39 \cdot 99 + i(60a - 60b + 39 \cdot 78 + 18 \cdot 99)</cmath>The ratio of the imaginary part to the real part must be <math>\frac{15}{4}</math> because <math>\frac{z_3-z_1}{z_2-z_1} = \frac{c}{15i+4}.</math> Hence, | ||
+ | <cmath>\frac{60a - 60b + 39 \cdot 78 - 18 \cdot 99}{a^2+b^2-96a-138b+18 \cdot 78 - 39 \cdot 99} = \frac{15}{4} \implies</cmath><cmath>\frac{4a-4b+84}{a^2+b^2-96a-138b+9(2 \cdot 78 - 39 \cdot 11)} = \frac{1}{4} \implies</cmath><cmath>16a - 16b + 33b = a^2 + b^2 -96a - 138b - 2457 \implies</cmath><cmath>0 = a^2 + b^2 -112a-122b-2793.</cmath>Evidently, <math>b</math> is maximized when <math>112a-a^2</math> is minimized or when <math>a = \boxed{056}.</math> | ||
+ | |||
+ | ~AopsUser101 | ||
+ | |||
+ | ==Solution 3== | ||
Algebra Bash | Algebra Bash | ||
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First we calculate <math>\frac{z_3 - z_1}{z_2 - z_1}</math> , which becomes <math>\frac{15i-4}{11}</math>. | First we calculate <math>\frac{z_3 - z_1}{z_2 - z_1}</math> , which becomes <math>\frac{15i-4}{11}</math>. | ||
− | Next, we define <math>z</math> to be <math>a | + | Next, we define <math>z</math> to be <math>a+bi</math> for some real numbers <math>a</math> and <math>b</math>. Then, <math>\frac {z-z_2}{z-z_3}</math> can be written as <math>\frac{(a-18)+(b-39)i}{(a-78)+(b-99)i}.</math> Multiplying both the numerator and denominator by the conjugate of the denominator, we get: |
<math>\frac {[(a-18)(a-78)+(b-39)(b-99)]+[(a-78)(b-39)-(a-18)(b-99)]i}{(a-78)^2+(b-99)^2}</math> | <math>\frac {[(a-18)(a-78)+(b-39)(b-99)]+[(a-78)(b-39)-(a-18)(b-99)]i}{(a-78)^2+(b-99)^2}</math> | ||
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~stronto | ~stronto | ||
− | ==Solution | + | ==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{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>. | ||
+ | (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 | ||
− | + | ==Solution 5== | |
+ | We will just bash. Let <math>z=a+bi</math> where <math>a,b\in\mathbb{R}</math>. We see that <math>\frac{z_3-z_1}{z_2-z_1}=\frac{-4+15i}{11}</math> after doing some calculations. We also see that <math>\frac{[(a-18)+(b-39)i][(a-78)-(b-99)i]}{\text{some real stuff}}.</math> We note that <math>[(a-18)+(b-39)i][(a-78)-(b-99)i]</math> is a multiple of <math>-4-15i</math> because the numerator has to be real. Thus, expanding it out, we see that <math>(a-18)(a-78)+(b-39)(b-99)=-4k \\ (a-78)(b-39)-(a-18)(b-99)=-15k.</math> Hence, <math>(a-18)(a-78)+(b-39)(b-99)=\frac{4}{15}[(a-78)(b-39)-(a-18)(b-99)] \implies a^2-96a+b^2-138b+5625=16a-16b+336 \\ (a-56)^2+(b-61)^2=1568.</math> To maximize the imaginary part, <math>(a-56)^2</math> must equal <math>0</math> so hence, <math>a=\boxed{56}</math>. | ||
==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}} |
Latest revision as of 17:52, 29 March 2021
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
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 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
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 minimized 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 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
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.