2019 AIME I Problems/Problem 12

Problem 12

Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ , $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ .

Solution 1

Notice that we must have $\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in i\mathbb R .$ However, $f(t)-t=t(t-20)$ , so $\begin{align*} \frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\ &=\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\ &=(z-19)(z+1)\\ &=(z-9)^2-100. \end{align*}$ Then, the real part of $(z-9)^2$ is $100$ . Since $\text{Im}(z-9)=\text{Im}(z)=11$ , let $z-9=a+11i$ . Then, $100=\text{Re}((a+11i)^2)=a^2-121\implies a=\pm\sqrt{221}.$ It follows that $z=9+\sqrt{221}+11i$ , and the requested sum is $9+221=\boxed{230}$ .

(Solution by TheUltimate123)

Solution 2

We will use the fact that segments $AB$ and $BC$ are perpendicular in the complex plane if and only if $\frac{a-b}{b-c}\in i\mathbb{R}$ . To prove this, when dividing two complex numbers you subtract the angle of one from the other, and if the two are perpendicular, subtracting these angles will yield an imaginary number with no real part.

Now to apply this: $\frac{f(z)-z}{f(f(z))-f(z)}\in i\mathbb{R}$ $\frac{z^2-19z-z}{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}$ $\frac{z^2-20z}{z^4-38z^3+341z^2+380z}$ $\frac{z(z-20)}{z(z+1)(z-19)(z-20)}$ $\frac{1}{(z+1)(z-19)}\in i\mathbb{R}$

The factorization of the nasty denominator above is made easier with the intuition that $(z-20)$ must be a divisor for the problem to lead anywhere. Now we know $(z+1)(z-19)\in i\mathbb{R}$ so using the fact that the imaginary part of $z$ is $11i$ and calling the real part r,

$(r+1+11i)(r-19+11i)\in i\mathbb{R}$ $r^2-18r-140=0$

solving the above quadratic yields $r=9+\sqrt{221}$ so our answer is $9+221=\boxed{230}$

Solution 3

I would like to use a famous method namely coni method.

Statement .If we consider there complex number  in argand plane then .

According to the question given, we can assume , $A= f(f(z)),B=f(z),C= x$ respectively.

WLOG, $Z_1= \frac{f(f(z))-f(z)}{z-f(z)}$ . According to the question $\arg{Z_1}=\frac{\pi}{2}$ .

So, $\Re (Z_1)=0$ .

Now, $Z_1=\frac{z(z-19)(z+1)(z-20)}{-z(z-20)}$ .

$\implies Z_1= -(z^2-18z-19)$ . WLOG, $z=a+11i$ .where $a=m+\sqrt{n}$ .

So, $\Re (Z_1)= -(a^2-18a-140)$ . Solving, $a^2-18a-140=0$ .get ,

$a=9$ ± $\sqrt{221}$ . So, possible value of $a=9+\sqrt{221}$ .

$\boxed{m+n=230}$ . ~ftheftics.

Solution 4

It is well known that $AB$ is perpendicular to $CD$ iff $\frac{d-c}{b-a}$ is a pure imaginary number. Here, we have that $A=z$ , $B,C=f(z)$ , and $D=f(f(z))$ . This means that this is equivalent to $\frac{f(f(z))-f(z)}{f(z)-z}$ being a pure imaginary number. Plugging in $f(z)=z^2-19z$ , we have that $\frac{(z^2-19z)-19(z^2-19z)-(z^2-19z)}{z^2-19z-z}$ being pure imaginary. Factoring and simplifying, we find that this is simply equivalent to $(z-19)(z+1)$ being pure imaginary. We let $z=a+bi$ , so this is equivalent to $(a+bi-19)(a+bi+1)$ being pure imaginary. Expanding the product, this is equivalent to $a^2+abi+a+abi-b^2+bi-19a-19bi-19$ being pure imaginary. Taking the real part of this, and setting this equal to $0$ , we have that $a^2-18a-b^2-19=0$ . Since $b=11$ , we have that $a^2-18a-140=0$ . By the quadratic formula, $a=9 \pm \sqrt{221}$ , and taking the positive root gives that $a=9+ \sqrt{221}$ , so the answer is $9+221=230$

~smartninja2000

2019 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search