2017 AIME I Problems/Problem 10

Problem 10

Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$.

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 $\frac{z_3 - z_1}{z_2 - z_1}$ as some complex number with form $r_1 (\cos \theta_1 + i \sin \theta_1).$ Similarly, we can write $\frac{z-z_2}{z-z_3}$ as some $r_2 (\cos \theta_2 + i \sin \theta_2).$

The product must be real, so we have that $r_1 r_2 (\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2)$ is real. $r_1 r_2$ is real by definition, so dividing the real number above by $r_1 r_2$ will still yield a real number. (Note that we can see that $r_1 r_2 \not= 0$ from the definitions of $z_1,$ $z_2,$ and $z_3$). Thus we have

\[(\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2) = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i(\cos \theta_1 \sin \theta_2 + \cos \theta_2 \sin \theta_1)\]

is real. The imaginary part of this is $(\cos \theta_1 \sin \theta_2 + \cos \theta_2 \sin \theta_1),$ which we recognize as $\sin(\theta_1 + \theta_2).$ This is only $0$ when $\theta_1 + \theta_2$ is some multiple of $\pi.$ In this problem, this implies $z_1, z_2, z_3$ and $z$ must form a cyclic quadrilateral, so the possibilities of $z$ lie on the circumcircle of $z_1, z_2$ and $z_3.$

To maximize the imaginary part of $z,$ it must lie at the top of the circumcircle, which means the real part of $z$ 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 $56,$ so the real part of $z$ is $56,$ and thus our answer is $\boxed{056}.$

Bashy Solution :)

We know that\[z_3-z_1 = (78+99i)-(16+83i) = 60 + 16i.\]\[z_2-z_1=(18+39i)-(18+83i) = -44i.\]Hence,\[\frac{z_3-z_1}{z_2-z_1} = \frac{60 + 16i}{-44i} = \frac{15i-4}{11} = \frac{c}{15i+4}\]where $c \in R$. Let $z = ai+b$. Then, \[\frac{z-z_2}{z-z_3} = \frac{(a+bi)-(18+39i)}{(a+bi)-(78+99i)} =\]\[\frac{z-z_2}{z-z_3} = \frac{(a-18)+i(b-39)}{(a-78)+i(b-99)} =\]\[\frac{z-z_2}{z-z_3} = \frac{((a-18)+i(b-39)((a-78)+i(99-b))}{((a-78)+i(b-99))((a-78)+i(99-b)} =\]\[\frac{z-z_2}{z-z_3} = \frac{((a-18)+i(b-39)((a-78)+i(99-b))}{k}\]The numerator is: \[(a-18)(a-78)+(b-39)(99-b)+i((b-39)(a-78) + (99-b)(a-18))=\]\[a^2+b^2-96a-138b+18 \cdot 78 - 39 \cdot 99 + i(60a - 60b + 39 \cdot 78 + 18 \cdot 99)\]The ratio of the imaginary part to the real part must be $\frac{15}{4}$ because $\frac{z_3-z_1}{z_2-z_1} = \frac{c}{15i+4}.$ Hence, \[\frac{60a - 60b + 39 \cdot 78 - 18 \cdot 99}{a^2+b^2-96a-138b+18 \cdot 78 - 39 \cdot 99} = \frac{15}{4} \implies\]\[\frac{4a-4b+84}{a^2+b^2-96a-138b+9(2 \cdot 78 - 39 \cdot 11)} = \frac{1}{4} \implies\]\[16a - 16b + 33b = a^2 + b^2 -96a - 138b - 2457 \implies\]\[0 = a^2 + b^2 -112a-122b-2793.\]Evidentlty, $b$ is maximized when $112a-a^2$ is minimized or when $a = \boxed{56}.$

~AopsUser101

Solution 2

Algebra Bash

First we calculate $\frac{z_3 - z_1}{z_2 - z_1}$ , which becomes $\frac{15i-4}{11}$.

Next, we define $z$ to be $a+bi$ for some real numbers $a$ and $b$. Then, $\frac {z-z_2}{z-z_3}$ can be written as $\frac{(a-18)+(b-39)i}{(a-78)+(b-99)i}.$ Multiplying both the numerator and denominator by the conjugate of the denominator, we get:

$\frac {[(a-18)(a-78)+(b-39)(b-99)]+[(a-78)(b-39)-(a-18)(b-99)]i}{(a-78)^2+(b-99)^2}$

In order for the product to be a real number, since both denominators are real numbers, we must have the numerator of $\frac {z-z_2}{z-z_3}$ be a multiple of the conjugate of $15i-4$, namely $-15i-4$ So, we have $(a-18)(a-78)+(b-39)(b-99) = -4k$ and $(a-78)(b-39)-(a-18)(b-99) = -15k$ for some real number $k$.

Then, we get:$(a-18)(a-78)+(b-39)(b-99) = \frac{4}{15}[(a-78)(b-39)-(a-18)(b-99)]$

Expanding both sides and combining like terms, we get:

$a^2 - 112a +b^2 - 122b + \frac{1989}{5} = 0$

which can be rewritten as:

$(a-56)^2 + (b-61)^2 = \frac{32296}{5}$

Now, common sense tells us that to maximize $b$, we would need to maximize $(b-61)^2$. Therefore, we must set $(a-56)^2$ to its lowest value, namely 0. Therefore, $a$ must be $\boxed{056}.$

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, but same as solution 1)

The $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ just means $z$ is on the circumcircle of $\triangle{z_1 z_2 z_3}$ 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 $x$-coordinate of the circumcenter of $(18, 83), (18, 39), (78, 99)$ (just get the intersection of the perpendicular bisectors $y=61$ and $y=-x+117$) and we get the $x$-coordinate of the circumcenter is $\boxed{056}$.

~First

Solution 4 (algebra but much cleaner)

We see that $\frac{z_3-z_1}{z_2-z_1}=\frac{60+16i}{-44i}=\frac{15i-4}{11}$. Now, let $z-z_3=a+bi$, in which case $z=(a+78)+(b+99)i$ and $z-z_2=(a+60)+(b+60)i$. We now have that $(\frac{(a+60)+(b+60)i}{a+bi})(\frac{15i-4}{11})$ is real, meaning that $((a+60)+(b+60)i)(\frac{15i-4}{a+bi})$ is real.

We see that $\frac{15i-4}{a+bi}=\frac{(15i-4)(a-bi)}{a^2+b^2}=\frac{(15b-4a)+(15a+4b)i}{a^2+b^2}$, so therefore $x=((a+60)+(b+60)i)((15b-4a)+(15a+4b)i)$ is real.

This means that $\Im(x)=0$, so we now have that $(a+60)(15a+4b)+(b+60)(15b-4a)=15a^2+15b^2+660a+1140b=0$, so $a^2+b^2+44a+76b=0$, which can be rewritten as$(a+22)^2+(b+38)^2=22^2+38^2$. In order to maximize $\Im(z)$ we want to maximize $b$, and in order to maximize $b$ we want $a+22=0$ and $a=-22$, so $\Re(z)=a+78=-22+78=\boxed{056}$. (Note: $\Im(\omega)$ is the imaginary part of $\omega$, and $\Re(\omega)$ is the real part of $\omega$) ~Stormersyle

Solution 5

We will just bash. Let $z=a+bi$ where $a,b\in\mathbb{R}$. We see that $\frac{z_3-z_1}{z_2-z_1}=\frac{-4+15i}{11}$ after doing some calculations. We also see that $\frac{[(a-18)+(b-39)i][(a-78)-(b-99)i]}{\text{some real stuff}}.$ We note that $[(a-18)+(b-39)i][(a-78)-(b-99)i]$ is a multiple of $-4-15i$ because the numerator has to be real. Thus, expanding it out, we see that $(a-18)(a-78)+(b-39)(b-99)=-4k \\ (a-78)(b-39)-(a-18)(b-99)=-15k.$ Hence, $(a-18)(a-78)+(b-39)(b-99)=\frac{4}{15}[(a-78)(b-39)-(a-18)(b-99)] \implies a^2-96a+b^2-138b+5625=16a-16b+336 \\ (a-56)^2+(b-61)^2=1568.$ To maximize the imaginary part, $(a-56)^2$ must equal $0$ so hence, $a=\boxed{56}$.

(Solution by pleaseletmetwin, but not added to the wiki by pleaseletmetwin)

See Also

2017 AIME I (ProblemsAnswer KeyResources)
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

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