Difference between revisions of "2019 AIME I Problems/Problem 12"
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~smartninja2000 | ~smartninja2000 | ||
− | + | ==Solution 5 (Official MAA)== | |
+ | The arguments of the two complex numbers differ by <math>90^\circ</math> if the ratio of the numbers is a pure imaginary number. Thus three distinct complex numbers <math>A,\,B,</math> and <math>C</math> form a right triangle at <math>B</math> if and only if <math>\tfrac{C-B}{B-A}</math> has real part equal to <math>0.</math> Hence <cmath>\begin{align*} | ||
+ | \frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}{(z^2-19z)-z}\ | ||
+ | &=\frac{(z^2-19z)(z^2-19z-19-1)}{z^2-20z}\ | ||
+ | &=\frac{z(z-19)(z+1)(z-20)}{z(z-20)} \ | ||
+ | &=z^2-18z-19 | ||
+ | \end{align*}</cmath> must have real part equal to <math>0.</math> If <math>z=x+11i,</math> the real part of <math>z^2-18z-19</math> is <math>x^2-11^2-18x-19,</math> which is <math>0</math> when <math>x=9\pm\sqrt{221}.</math> The requested sum is <math>9+221=230.</math> | ||
==See also== | ==See also== | ||
{{AIME box|year=2019|n=I|num-b=11|num-a=13}} | {{AIME box|year=2019|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:31, 25 February 2021
Contents
[hide]Problem
Given , there are complex numbers
with the property that
,
, and
are the vertices of a right triangle in the complex plane with a right angle at
. There are positive integers
and
such that one such value of
is
. Find
.
Solution 1
Notice that we must have However,
, so
Then, the real part of
is
. Since
, let
. Then,
It follows that
, and the requested sum is
.
(Solution by TheUltimate123)
Solution 2
We will use the fact that segments and
are perpendicular in the complex plane if and only if
. 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:
The factorization of the nasty denominator above is made easier with the intuition that must be a divisor for the problem to lead anywhere. Now we know
so using the fact that the imaginary part of
is
and calling the real part r,
solving the above quadratic yields so our answer is
Solution 3
I would like to use a famous method, namely the coni method.
Statement .If we consider there complex numberin argand plane then
.
According to the question given, we can assume , respectively.
WLOG,.
According to the question
.
So,.
Now, .
.
WLOG,
.where
.
So,.
Solving,
.get ,
±
.
So, possible value of
.
.
~ftheftics.
Proof of this method
Note that if we translate a triangle, the measures of all of its angles stay the same. So we can translate on the complex plane so that
. Let the images of
be
respectively. Then, we can use the formula:
. (This is known as Euler's Theorem.)
Using Euler's theorem (represent each complex number in polar form, then use exponent identities), we can show that . So this method is valid.
~Math4Life2020
Solution 4
It is well known that is perpendicular to
iff
is a pure imaginary number. Here, we have that
,
, and
. This means that this is equivalent to
being a pure imaginary number. Plugging in
, we have that
being pure imaginary. Factoring and simplifying, we find that this is simply equivalent to
being pure imaginary. We let
, so this is equivalent to
being pure imaginary. Expanding the product, this is equivalent to
being pure imaginary. Taking the real part of this, and setting this equal to
, we have that
. Since
, we have that
. By the quadratic formula,
, and taking the positive root gives that
, so the answer is
~smartninja2000
Solution 5 (Official MAA)
The arguments of the two complex numbers differ by if the ratio of the numbers is a pure imaginary number. Thus three distinct complex numbers
and
form a right triangle at
if and only if
has real part equal to
Hence
must have real part equal to
If
the real part of
is
which is
when
The requested sum is
See also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.